Download SQA CfE Higher Human Biology Unit 1: Human Cells

SCHOLAR Study Guide
SQA CfE Higher Human Biology
Unit 1: Human Cells
Authored by:
Eoin McIntyre
Reviewed by:
Sheena Haddow
Previously authored by:
Mike Cheung
Eileen Humphrey
Eoin McIntyre
Jim McIntyre
Heriot-Watt University
Edinburgh EH14 4AS, United Kingdom.
First published 2014 by Heriot-Watt University.
This edition published in 2014 by Heriot-Watt University SCHOLAR.
Copyright © 2014 Heriot-Watt University.
Members of the SCHOLAR Forum may reproduce this publication in whole or in part for
educational purposes within their establishment providing that no profit accrues at any stage,
Any other use of the materials is governed by the general copyright statement that follows.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system
or transmitted in any form or by any means, without written permission from the publisher.
Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the
information contained in this study guide.
Distributed by Heriot-Watt University.
SCHOLAR Study Guide Unit 1: SQA CfE Higher Human Biology
1. SQA CfE Higher Human Biology
ISBN 978-1-909633-16-2
Printed and bound in Great Britain by Graphic and Printing Services, Heriot-Watt University,
Edinburgh.
Acknowledgements
Thanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and
created these materials, and to the many colleagues who reviewed the content.
We would like to acknowledge the assistance of the education authorities, colleges, teachers
and students who contributed to the SCHOLAR programme and who evaluated these materials.
Grateful acknowledgement is made for permission to use the following material in the
SCHOLAR programme:
The Scottish Qualifications Authority for permission to use Past Papers assessments.
The Scottish Government for financial support.
All brand names, product names, logos and related devices are used for identification purposes
only and are trademarks, registered trademarks or service marks of their respective holders.
i
Contents
1 Stem cells
1.1 Introduction . . . . . . . . . .
1.2 What are stem cells? . . . . .
1.3 Embryonic stem cells . . . .
1.4 Tissue (adult) stem cells . . .
1.5 Learning points . . . . . . . .
1.6 Extended response question
1.7 End of topic test . . . . . . .
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2 Differentiation in cells
2.1 Mitosis and meiosis . . . . . . . . . . .
2.2 Differentiation in somatic cells . . . . . .
2.3 Human body tissues . . . . . . . . . . .
2.4 Formation of the body organs . . . . . .
2.5 Differentiation of germline cells . . . . .
2.6 Mutations in germline and somatic cells
2.7 Learning points . . . . . . . . . . . . . .
2.8 Extended response question . . . . . .
2.9 End of topic test . . . . . . . . . . . . .
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3 Research: Stem cells and cancer cells
3.1 Introduction . . . . . . . . . . . . . . . . . . . . .
3.2 Stem cell research . . . . . . . . . . . . . . . . .
3.3 Therapeutic uses of stem cells . . . . . . . . . .
3.4 Ethical issues and the regulation of stem cell use
3.5 The biology of cancer cells . . . . . . . . . . . .
3.6 Learning points . . . . . . . . . . . . . . . . . . .
3.7 Extended response question . . . . . . . . . . .
3.8 End of topic test . . . . . . . . . . . . . . . . . .
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4 DNA structure and replication
4.1 Introduction . . . . . . . . . . . . . . .
4.2 DNA structure . . . . . . . . . . . . . .
4.3 Arrangement of DNA in chromosomes
4.4 DNA replication . . . . . . . . . . . . .
4.5 Learning points . . . . . . . . . . . . .
4.6 Extended response question . . . . .
4.7 End of topic test . . . . . . . . . . . .
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5 Gene expression in human cells
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77
ii
CONTENTS
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
Introduction . . . . . . . . . . . . . . . . . .
Gene expression through protein synthesis
Structure and functions of RNA . . . . . . .
Transcription of DNA into mRNA . . . . . .
Translation of mRNA into a polypeptide . .
Single gene, several proteins . . . . . . . .
Learning points . . . . . . . . . . . . . . . .
Extended response question . . . . . . . .
End of topic test . . . . . . . . . . . . . . .
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6 Genes and proteins in health and disease
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Different levels of protein structure . . . . . . . . . . . . . . . .
6.4 Functions of proteins . . . . . . . . . . . . . . . . . . . . . . . .
6.5 Mutations and genetic disorders . . . . . . . . . . . . . . . . .
6.6 Single gene mutations . . . . . . . . . . . . . . . . . . . . . . .
6.7 The effect of mutations on the structure and function of proteins
6.8 Chromosome structure mutations . . . . . . . . . . . . . . . . .
6.9 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.10 Extended response question . . . . . . . . . . . . . . . . . . .
6.11 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . .
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7 Human genomics
7.1 Introduction . . . . . . . . . . . . . . . . . . . .
7.2 Sequencing DNA . . . . . . . . . . . . . . . . .
7.3 Bioinformatics . . . . . . . . . . . . . . . . . . .
7.4 Systematics . . . . . . . . . . . . . . . . . . . .
7.5 Personalised medicine . . . . . . . . . . . . . .
7.6 Amplification and detection of DNA sequences
7.7 DNA probes . . . . . . . . . . . . . . . . . . . .
7.8 Medical and forensic applications . . . . . . . .
7.9 Learning points . . . . . . . . . . . . . . . . . .
7.10 Extended response question . . . . . . . . . .
7.11 End of topic test . . . . . . . . . . . . . . . . .
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9 Cellular respiration I: Glycolysis
9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Summary of glucose breakdown . . . . . . . . . . . . . . . . . . . . . .
9.3 Roles of ATP in the cell . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8 Cell metabolism
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Types of metabolic pathway . . . . . . . . . . . . . . . . .
8.3 Control of metabolic pathways - the action of enzymes . .
8.4 The role of the active site . . . . . . . . . . . . . . . . . .
8.5 Control of metabolic pathways through enzyme inhibition
8.6 Learning points . . . . . . . . . . . . . . . . . . . . . . . .
8.7 Extended response question . . . . . . . . . . . . . . . .
8.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . .
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© H ERIOT-WATT U NIVERSITY
CONTENTS
9.4
9.5
9.6
9.7
9.8
iii
Glycolysis . . . . . . . . . . . . . . . .
Regulation of the respiratory pathway
Learning points . . . . . . . . . . . . .
Extended response question . . . . .
End of topic test . . . . . . . . . . . .
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11 Cellular Respiration III: Electron transport system
11.1 Introduction . . . . . . . . . . . . . . . . . . . . .
11.2 The electron transport chain . . . . . . . . . . . .
11.3 ATP synthesis . . . . . . . . . . . . . . . . . . . .
11.4 Learning points . . . . . . . . . . . . . . . . . . .
11.5 Extended response question . . . . . . . . . . .
11.6 End of topic test . . . . . . . . . . . . . . . . . .
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10 Cellular Respiration II: Citric acid cycle
10.1 Introduction . . . . . . . . . . . . . .
10.2 The mitochondrion . . . . . . . . . .
10.3 The citric acid cycle . . . . . . . . .
10.4 Alternative substrates for respiration
10.5 Learning points . . . . . . . . . . . .
10.6 Extended response question . . . .
10.7 End of topic test . . . . . . . . . . .
12 Energy systems in muscle cells
12.1 Introduction . . . . . . . . . . .
12.2 Creatine phosphate . . . . . .
12.3 Lactic acid metabolism . . . . .
12.4 Types of skeletal muscle fibres
12.5 Learning points . . . . . . . . .
12.6 Extended response question .
12.7 End of topic test . . . . . . . .
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13 End of unit test
255
Glossary
263
Answers to questions and activities
1
Stem cells . . . . . . . . . . . . . . . . . . . . . .
2
Differentiation in cells . . . . . . . . . . . . . . . .
3
Research: Stem cells and cancer cells . . . . . .
4
DNA structure and replication . . . . . . . . . . .
5
Gene expression in human cells . . . . . . . . . .
6
Genes and proteins in health and disease . . . . .
7
Human genomics . . . . . . . . . . . . . . . . . .
8
Cell metabolism . . . . . . . . . . . . . . . . . . .
9
Cellular respiration I: Glycolysis . . . . . . . . . .
10 Cellular Respiration II: Citric acid cycle . . . . . .
11 Cellular Respiration III: Electron transport system
12 Energy systems in muscle cells . . . . . . . . . .
13 End of unit test . . . . . . . . . . . . . . . . . . . .
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© H ERIOT-WATT U NIVERSITY
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1
Topic 1
Stem cells
Contents
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 What are stem cells? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
3
1.3 Embryonic stem cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Tissue (adult) stem cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
7
1.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.6 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
10
Learning Objectives
By the end of this topic, you should be able to:
• explain the term 'cellular differentiation';
• describe the properties of stem cells;
• describe the differences between embryonic and tissue stem cells;
• list the types of cell into which bone marrow cells can develop.
2
TOPIC 1. STEM CELLS
1.1
Introduction
Learning Objective
By the end of this section, you should be able to:
• briefly describe the process of embryonic development after fertilisation.
All living things are characterised by levels of organisation that are hierarchical. The cell
is the lowest level of organisation that can exist independently. Multicellular organisms,
like humans, have cells organised into groups of cells called tissues, the next level
of organisation. Tissues are formed from specialised cells that carry out a particular
function. The columnar cells in the lining of the intestine, for example, are specialised
for absorption (of nutrients), the stomach is made up of mucus membrane tissue, muscle
tissue and a layer of tissue lining the abdomen. Tissues can themselves become
grouped together to form an organ.
Most organs, such as the heart, lung and liver, are also specialised for a certain function.
The final level of organisation is the organ system, where a group of organs work
together at a particular function. The systems in our bodies include integumentary (skin,
hair, nails), cardiovascular (circulatory), digestive, muscular and the nervous system.
Organisation: cells to tissues to organs to systems
After fertilisation, the zygote undergoes repeated mitotic divisions to form first a solid
ball of cells and then a hollow ball with a fluid-filled interior. Early in this process, the
cells are unspecialised and are capable of developing into any of the body's cell types.
Later, the cells lose this general ability and become increasingly limited in terms of their
potential functions. However, it is important to remember that all the cells of the body
carry the same genetic information in the nucleus.
Cellular differentiation is the process by which an unspecialised cell develops specific
functions. It does this by expressing only those genes which are characteristic of that
type of cell. At the same time, genes which would give it other capabilities are switched
off.
© H ERIOT-WATT U NIVERSITY
TOPIC 1. STEM CELLS
3
Thus, once a cell becomes differentiated, it only expresses the genes that produce the
proteins characteristic of that type of cell. In this way, a developing muscle cell activates
the genes which control the production of the contractile proteins, but switches off those
which control the production of a digestive enzyme such as amylase.
Introduction: Questions
Q1: What is meant by the term 'body tissue'?
..........................................
Q2: "The heart is an organ". Explain this statement.
..........................................
Q3: What is a human body system? Give two examples.
..........................................
1.2
What are stem cells?
Learning Objective
By the end of this section, you should be able to:
• state that stem cells are found only in multicellular organisms;
• explain how stem cells are different from other body cells;
• state that stem cells can self-renew and differentiate.
In the early embryo all the cells are able to divide, whereas in the adult body only
relatively few have this ability. Such cells are called stem cells.
As the embryo grows, different groups of cells start to become specialised to carry
out particular functions. At the same time, their potential to do other things is lost
as the genes associated with these activities are turned off. This is the process of
differentiation.
Stem cells are different from other body cells because they have the following
characteristics:
• undifferentiated (unspecialised cell type), allowing them to divide and maintain a
supply of stem cells for the body;
• found in all multicellular organisms;
• self-renewing and can differentiate: in some organs, like the gut, stem cells
regularly divide to repair and replace worn out or damaged tissues.
© H ERIOT-WATT U NIVERSITY
4
TOPIC 1. STEM CELLS
The two types of stem cells found in humans are:
1. embryonic stem cells;
2. tissue (adult) stem cells.
A stem cell line is a group of constantly dividing cells from a single parent group of
stem cells. They can be obtained from human or animal tissues, and can replicate
for long periods of time in vitro (within glass; or, commonly, in the lab, in an artificial
environment). These cell cultures may be used for a wide range of research purposes
and may even (occasionally) be used to clone entire organisms.
Stem cells: Questions
Q4:
What do you understand by the term 'differentiation' in regard to stem cells?
..........................................
Q5: What is the unique property of stem cells which makes them different from
specialised cells?
..........................................
Q6:
Give examples of differentiated cells and their functions.
..........................................
1.3
Embryonic stem cells
Learning Objective
By the end of this section, you should be able to:
• explain what a blastocyst is;
• explain why embryonic stem cells are pluripotent;
• list some diseases that might be treated by transplanting cells generated from
human embryonic stem cells.
The first successful isolation and culturing of embryonic stem cells took place in the
1980s. This work involved mouse embryos, but by the late 1990s a method to extract
stem cells from human embryos had been successfully developed. Scientists were able
to use embryos initially created for reproductive purposes through in vitro fertilisation.
Many more eggs are fertilised in this process than are eventually implanted into the
mother's womb. With the informed consent of the donor, the extra embryos are made
available for research. At this stage, the embryos are balls of 70-100 undifferentiated
cells. However, there are considerable ethical issues concerning the use of human
embryos in this way.
Most embryonic stem cells are derived from embryos that are ready for implantation.
The diagram shows stages in the development of the embryo up to the point of
implantation.
© H ERIOT-WATT U NIVERSITY
TOPIC 1. STEM CELLS
Stages in the early development of the human embryo
After fertilisation, the zygote undergoes rapid cell division (stage A) and produces a
multicellular hollow ball of cells which contains a fluid-filled cavity, the blastocoel (stage
B). Once this ball comprises 70-100 cells, there is an inner cell mass at one end
which will eventually develop into the embryo. At this point, the ball of cells is called
a blastocyst and it is ready to implant into the endometrium of the uterus. Once
implanted, the inner cell mass starts to differentiate and is known as the epiblast (stage
C).
At the pre-implantation stage, the inner cell mass consists of undifferentiated cells which
are pluripotent, i.e. they have the potential to form any cell in the body (apart from the
extra-embryonic tissues such as the amnion). It is these cells which are used for stem
cell research.
Human embryonic stem cells can be formed by transferring cells from a preimplantationstage embryo into a culture dish that contains a nutrient broth, known as culture medium.
As well as having the ability to undergo cell division, embryonic stem cells are able to
undergo cell differentiation to generate new functional cells. In more recent research into
embryonic stem cells, scientists have reliably directed the differentiation of embryonic
stem cells into specific cell types. They are able to use the resulting differentiated cells
to treat certain diseases.
Diseases that might be treated by transplanting cells generated from human embryonic
stem cells include Parkinson's disease, diabetes, traumatic spinal cord injury, vision and
hearing loss, Duchenne's muscular dystrophy and heart disease. The techniques used
for this research work will be studied in a later topic.
© H ERIOT-WATT U NIVERSITY
5
6
TOPIC 1. STEM CELLS
Embryonic stem cells: Questions
Q7: Put the following stages into the correct order to explain how to use human
embryonic stem cells to form specialised cells:
• Formation of mass of undifferentiated stem cells
• Stem cell cultured in the laboratory
• Embryo stem cell removed
• Undifferentiated stem cells cultured in different culture conditions
• Formation of specialised cells: nerve cell, muscle cell, gut cells
• Early human embryo Blastocyst
..........................................
Q8:
What is a blastocyst?
..........................................
Q9:
Where do we find embryonic stem cells?
..........................................
Q10: For what reason are the cells of the inner cell mass useful for stem cell research?
..........................................
Q11: List three diseases that might be treated using transplanted cells originated from
embryonic stem cells.
..........................................
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. STEM CELLS
1.4
7
Tissue (adult) stem cells
Learning Objective
By the end of this section, you should be able to:
• state that bone marrow contains two types of tissue stem cell;
• explain why tissue stem cells are multipotent;
• explain that tissue stem cells can only make certain types of specialised cells.
Tissue stem cells are undifferentiated cells which can be found among differentiated
cells in a tissue or organ. The primary roles of tissue stem cells in a living organism are
to maintain and repair the tissue in which they are found.
Work on tissue stem cells began in the 1950s when researchers discovered that bone
marrow contains stem cells which develop into red blood cells, platelets, and the various
forms of phagocytes and lymphocytes.
Tissue stem cells have limited capabilities compared with embryonic stem cells. Tissue
stem cells can only make the type of cells they belong to: they are called multipotent.
They can only make certain types of specialised cells, not all of the kinds of cells in the
body, so they are limited in their ability to differentiate.
For example, bone tissue contains two types of tissue stem cells: haematopoietic stem
cells, which form all the types of blood cells in the body; bone marrow stem cells,
which generate bone, cartilage, fat, cells that support the formation of blood, and fibrous
connective tissue.
Tissue stem cells have been identified in many organs and tissues, including brain, bone
marrow, peripheral blood, blood vessels, skeletal muscle and skin. Although they are
found in so many types of tissues, only a very small number of stem cells actually occur
in these tissues. This makes the study and the generation of tissue stem cells outside
the body very challenging.
Tissue stem cells are most commonly obtained from the outside part of the pelvis called
the iliac crest. A needle is inserted into the iliac bone and bone marrow is withdrawn.
It is likely that several samples may be obtained from this area. These stem cells are
then separated from other cells in the marrow. Under ideal laboratory conditions they
are grown, or expanded, in a process that can take between 7 and 21 days.
After the formation of these 'new' stem cells, they are placed in a specific tissue
environment such as the bone. The tissue stem cells can become activated and they
start to divide. A set of new stem cells and second generation (progenitor) cells is
formed. It is these progenitor cells that differentiate into newer cells with the same
phenotype as the host tissue.
In 2006, it was shown that tissue stem cells isolated from adult mice could be induced to
become pluripotent, i.e. capable of dividing and differentiating to form any type of body
cell. When, one year later, it was found that human cells could be similarly induced, this
opened the door to much wider research using stem cells without the controversy which
surrounds the use of embryos.
© H ERIOT-WATT U NIVERSITY
8
TOPIC 1. STEM CELLS
Tissue stem cells: Questions
Q12: For each term or statement listed below, decide if it refers to embryonic or tissue
stem cells:
1. Pluripotent
2. Develop into all cell types
3. Ability to differentiate into some cells in the body
4. Give rise to a limited range of cell types
5. Develop into cell types that are closely related to the tissue in which they are found
6. Found in developing embryo - blastocyst
7. Have the capacity to become all cell types
8. Multipotent
9. Found in body tissues
10. Ability to differentiate into all of the cell types
..........................................
Q13: What are the two types of tissue stem cells found in bone marrow? What types of
tissues can they generate?
..........................................
Q14: What do you understand when tissue stem cells are described as multipotent?
..........................................
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. STEM CELLS
1.5
9
Learning points
Summary
• Cellular differentiation is the process by which a cell develops more
specialised functions.
• Cellular differentiation occurs by expressing only the genes characteristic of
that type of cell.
• Once a cell becomes differentiated, it only expresses the genes that
produce the proteins characteristic for that type of cell.
• Stem cells are relatively unspecialised cells.
• Stem cells are able to continue to divide to produce more stem cells or cells
which will differentiate into specialised cells of one or more types.
• During embryological development, the unspecialised cells of the early
embryo differentiate into cells with specialised functions.
• Embryonic stem cells are found in the inner cell mass (epiblast) of the
blastocyst.
• Embryonic stem cells are capable of developing into any of the body's cell
types (pluripotent).
• Tissue (adult) stem cells replenish differentiated cells that need to be
replaced.
• Tissue (adult) stem cells are found in very small numbers throughout the
body tissues.
• Tissue (adult) stem cells give rise to a limited range of cell types typical of
the organ of which they are a part (multipotent).
1.6
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of embryonic and tissue stem cells before
attempting the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended response question: Embryonic and tissue stem cells
Compare the location and functions of embryonic and tissue (adult) stem cells. (7 marks)
..........................................
© H ERIOT-WATT U NIVERSITY
15 min
10
TOPIC 1. STEM CELLS
1.7
End of topic test
End of Topic 1 test
5 min
Q15: Complete the statements by matching the parts on the left with the parts on the
right. (12 marks)
Cellular differentiation:
adult stem cells.
Muscle cells only express
stem cells.
Produced from bone marrow stem
cells:
inner cell mass of the blastocyst.
Cells capable of repeated division:
the genes characteristic of that type
of cell.
Stem cells:
replenish differentiated cells.
Dividing stem cells produce
dividing stem cells.
Specialised cells are produced from
embryonic stem cells.
Embryonic stem cells are located in
the
red blood cells, platelets,
phagocytes and lymphocytes.
Through the body:
cell develops more specialised
functions.
Tissue stem cells
a limited range of cell types.
Can develop into any cell type:
more stem cells.
Tissue stem cells can produce
relatively unspecialised.
© H ERIOT-WATT U NIVERSITY
TOPIC 1. STEM CELLS
..........................................
Q16: What is meant by the term 'cellular differentiation'? (1 mark)
..........................................
Q17: Why does a muscle cell only produce the proteins typical of its cell type? (1 mark)
..........................................
Q18: List the types of blood cell into which bone marrow stem cells can differentiate. (1
mark)
..........................................
Q19: List two general features of stem cells. (2 marks)
..........................................
Q20: When they divide, what are the two types of cell that a stem cell can produce? (2
marks)
..........................................
Q21: Where are embryonic stem cells found? (1 mark)
..........................................
Q22: Where are tissue (adult) stem cells found? (1 mark)
..........................................
Q23: State the function of tissue stem cells. (1 mark)
..........................................
Q24: Compare the potential of the cells produced by embryonic and tissue stem cells.
(1 mark)
..........................................
© H ERIOT-WATT U NIVERSITY
11
12
TOPIC 1. STEM CELLS
© H ERIOT-WATT U NIVERSITY
13
Topic 2
Differentiation in cells
Contents
2.1 Mitosis and meiosis . . . . . . . . . . .
2.1.1 Mitosis . . . . . . . . . . . . . . .
2.1.2 Meiosis . . . . . . . . . . . . . .
2.2 Differentiation in somatic cells . . . . . .
2.3 Human body tissues . . . . . . . . . . .
2.3.1 Epithelial tissue . . . . . . . . . .
2.3.2 Connective tissue . . . . . . . .
2.3.3 Muscle tissue . . . . . . . . . . .
2.3.4 Nervous tissue . . . . . . . . . .
2.4 Formation of the body organs . . . . . .
2.5 Differentiation of germline cells . . . . .
2.6 Mutations in germline and somatic cells
2.7 Learning points . . . . . . . . . . . . . .
2.8 Extended response question . . . . . .
2.9 End of topic test . . . . . . . . . . . . .
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14
15
16
20
21
21
22
23
25
28
28
29
31
32
32
Prerequisite knowledge
In the previous topic, the process by which the relatively unspecialised stem cells in the
embryo and adult tissues are able to differentiate into all the cell types of the body was
outlined. These cells belong to two fundamental types: germline cells, which give rise
to gametes by dividing by mitosis and meiosis, and somatic cells, which produce all the
other cells of the body by mitosis. In what follows in this topic, we will look at some of the
ways in which the various types of cell in the body become specialised to their different
functions.
Learning Objectives
By the end of this topic, you should be able to:
• describe the production and differentiation of somatic cells;
• describe the specialisations and functions of the four types of body tissue;
• describe the types of cell division in germline cells which produce more germline
cells and gametes;
• explain the significance of mutations in somatic and germline cells.
14
TOPIC 2. DIFFERENTIATION IN CELLS
2.1
Mitosis and meiosis
Learning Objective
By the end of this section, you should be able to:
• outline the process of mitosis and explain its significance in the production of
diploid somatic and germline cells;
• outline the process of meiosis and explain its significance in the production of
haploid gametes.
In this topic you will be learning about somatic cells and germline cells, and how these
cells are involved in the process of differentiation.
From previous study, you should be familiar with the two types of cell division: mitosis
and meiosis. The details of these processes are not required for this course, but it is
helpful to be familiar with them to fully understand their significance in relation to somatic
and germline cells.
Every normal human body cell has 46 chromosomes arranged in 23 pairs. These
cells are known as diploid cells (because they contain a double set of chromosomes).
Every normal human gamete (sex cell) contains only 23 unpaired chromosomes. Such
cells are called haploid (because they contain a single set of chromosomes). After
fertilisation, the newly formed zygote now contains two sets of chromosomes. In other
words, it is diploid. The zygote continues to divide by mitosis and eventually develops
into an adult. The cycle of sexual reproduction can now begin all over again. These
events are summarised in the following diagram.
The sexual life cycle in animals
Haploid cells are produced by a special type of cell division known as meiosis.
Both mitosis and meiosis are types of nuclear division. At the end of each process,
however, the cytoplasm divides and daughter cells are formed.
It is important to understand that mitosis and meiosis serve fundamentally different
purposes.
© H ERIOT-WATT U NIVERSITY
TOPIC 2. DIFFERENTIATION IN CELLS
Mitosis produces two cells which are genetically identical to the parent cell. The process
of copying and allocating the DNA to the daughter cells is virtually foolproof; when
it rarely does fail, there are cellular checking systems which correct any errors. This
ensures that all the cells of the body contain exactly the same inherited information and
so can be relied upon to react to stimuli like hormones in a consistent and predictable
way.
Meiosis produces gametes (sex cells) which are intended to fuse with another gamete
to produce a zygote, and hence a new individual with a complete genetic complement.
To achieve this, every gamete must contain an example of each gene and each
chromosome. This is brought about in Meiosis I where the diploid number of
chromosomes in the gamete mother cells is reduced to the haploid number in the
gametes by the pairing and separation of the homologous chromosomes.
The crossing over of portions of chromatid in Meiosis I, which is an important source of
variation, also causes the two chromatids of each chromosome to no longer be identical.
In Meiosis II, a division that is very similar to mitosis, the two dissimilar chromatids of
each chromosome are separated into daughter cells, ensuring that every gamete only
carries one allele of each gene.
The products of mitosis are two identical diploid cells; in contrast, meiosis produces four
non-identical haploid cells.
In order to help you understand the differences between mitosis and meiosis, we will
very quickly review the stages involved in mitosis.
2.1.1
Mitosis
Mitosis produces two daughter cells which are genetically identical to the parent cell.
Stages of mitosis
The stages can be described as follows:
1. Shows the cell just before it begins mitosis. The chromosomes as single
chromatids have replicated so that each chromosome now consists of two
chromatids joined at the centromere.
2. Shows the chromosomes have coiled up to become shorter and thicker and can
be seen to be double stranded. Each double stranded chromosome consists of a
pair of chromatids joined together at the centromere. The nuclear membrane is
beginning to break down.
3. Shows the spindle forming and the double-stranded chromosomes migrating to,
and line up along, the equator of the cell. Each chromosome becomes attached
to a separate spindle fibre.
© H ERIOT-WATT U NIVERSITY
15
16
TOPIC 2. DIFFERENTIATION IN CELLS
4. Shows the chromatids being pulled apart to opposite poles of the cell by the
spindle fibres.
5. Shows a new nuclear membrane forming around each set of single-stranded
chromosomes. Division of the cytoplasm follows.
Two identical daughter cells have been produced from a diploid parent cell.
Mitosis: Question
Q1:
Decide on the correct order of the illustrations to show the stages of mitosis:
5 min
..........................................
2.1.2
Meiosis
All of our body cells are diploid, that is they contain 23 pairs of chromosomes (46 in
all). Gametes, however, are haploid, containing 23 unpaired chromosomes. This is
important so that a diploid zygote (fertilised egg) is produced at fertilisation. Meiosis is
the type of nuclear division which reduces the number of chromosomes, thus producing
haploid gametes. Meiosis consists of two separate divisions, the first and second meiotic
divisions, and therefore results in four haploid daughter cells. We will look at each
division in turn.
Stages of the first meiotic division during meiosis
The stages can be described as follows:
1. Shows the cell just before meiosis begins. It is at this point that the chromosomes
are replicated.
2. Shows that the chromosomes have shortened and thickened. They can be seen
to consist of pairs of chromatids joined at the centromere.
© H ERIOT-WATT U NIVERSITY
TOPIC 2. DIFFERENTIATION IN CELLS
3. Shows the chromosomes finding their homologous partners and moving very close
to each other. The chromatids become intertwined at points called chiasmata. At
these points the chromatids may break and rejoin so that some genes may cross
over from one chromatid to the other. This does not happen in mitosis.
4. Shows the pairs of homologous chromosomes migrating to the equator of the
cell as a unit and lining up independently of any other pair. The spindle forms and
each homologous pair of chromosomes becomes attached to the same spindle
fibre.
5. Shows the homologous chromosomes (still double-stranded) being pulled apart
towards opposite ends of the cell by the motor proteins of the spindle fibres.
6. Shows the formation of new nuclear membranes around each set of
chromosomes. Then the cell divides in two. Note that each new cell is haploid,
that is they possess only a single set of (double-stranded) chromosomes.
Stages of the second meiotic division during meiosis
Each cell now undergoes what is essentially a mitotic division to separate the doublestranded chromosomes:
1. Shows the cells just before they begin the second meiotic division. Note that there
is no replication of the chromosomes since they are already double-stranded.
2. Shows the nuclear membranes of the cells breaking down, the spindle forming, the
chromosomes migrating to the equator, and each chromosome becoming attached
to a separate spindle fibre.
3. Shows the chromosomes (now single-stranded) being pulled to opposite poles of
the cells.
4. Shows a new nuclear membrane forming around each set of chromosomes. The
cytoplasm divides again, producing four haploid gametes.
© H ERIOT-WATT U NIVERSITY
17
18
TOPIC 2. DIFFERENTIATION IN CELLS
Meiosis: Questions
10 min
Q2: Decide on the correct order of the illustrations to show the stages of the first
meiotic division during meiosis:
..........................................
Q3: Decide on the correct order of the illustrations to show the stages of the second
meiotic division during meiosis:
..........................................
Q4: There are 46 chromosomes (23 pairs of chromosomes) in a human cell. How
many chromosomes are present in each daughter cell after mitosis?
..........................................
Q5: How many chromosomes are present in each daughter cell after the first meiotic
division?
..........................................
Q6: How many chromosomes are present in each daughter cell after the second
meiotic division?
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 2. DIFFERENTIATION IN CELLS
A cell entering meiosis
Q7: Which label in the illustration indicates a pair of homologous chromosomes?
..........................................
Q8: Which label in the illustration indicates a single chromosome?
..........................................
Q9: Which label in the illustration indicates a chromatid?
..........................................
Q10: Which label in the illustration indicates a centromere?
..........................................
Q11: Which label in the illustration indicates a chiasma?
..........................................
..........................................
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2.2
Differentiation in somatic cells
Learning Objective
By the end of this section, you should be able to:
• define a somatic cell;
• give examples of somatic cells.
A somatic cell is a non-sex cell; they make up all the cells in the human body except
the reproductive cells (gametes). Somatic cells are the differentiated cells that form the
different types of human body tissues. Somatic cells make up all the organs, skin, bones
and blood. Some examples of somatic cells are smooth muscle cells, red blood cells, B
lymphocytes, and epithelial cells.
Somatic cells in the body contain the diploid number of chromosomes and some can
undergo mitosis, giving rise to daughter cells. These daughter cells then grow, and may
themselves divide to form more cells. The result of these two processes is an increase
in the size of the body and its organs.
A micrograph of typical stained somatic cells
Differentiation in somatic cells: Questions
Q12: How many chromosomes are found in somatic cells?
..........................................
Q13: What are differentiated cells?
..........................................
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2.3
21
Human body tissues
Learning Objective
By the end of this section, you should be able to:
• list the four types of body tissues as epithelial, connective, muscle and nervous;
• give examples of each type of tissue;
• state the functions of each type of tissue.
The somatic cells of the body of all multicellular animals form four types of tissue (each
of which can be subdivided). These tissues are:
1. connective - gives shape to organs and supports them, including blood and bone;
2. epithelial - covers the organ surfaces, including the skin and the lining of the gut;
3. muscle - causes locomotion or movement within organs by the action of contractile
cells;
4. nervous - transmits messages within the central nervous system and between the
central nervous system and the rest of the body.
2.3.1
Epithelial tissue
Epithelial tissues cover the whole surface of the body. They are made up of cells closely
packed together and arranged in one or more layers. These tissues are specialised to
form the covering or lining of all internal and external body surfaces, e.g. the skin, the
lining of the airways, the gut lining and the reproductive tracts.
Depending on its location and function, the epithelium may have various structures.
Epithelial tissue has two general functions:
1. protection, which it does by having very tight junctions between its cells, e.g. the
skin epidermis, the epidermis lining of the mouth, and the surface layer of the
endometrium;
2. secretion and absorption, e.g. secretion of digestive enzymes by the intestinal
glands in the small intestine and absorption into the villi of the small intestine or
absorption into the alveoli of the lung.
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Different types of epithelium
Epithelial tissue: Question
Q14: List the main functions of epithelial tissues.
..........................................
2.3.2
Connective tissue
The main function of connective tissue is to support the human body and to connect
tissues together. It is strong enough to give a mechanical framework for the skeleton
and so it also plays an important role in body movement.
Connective tissue has three main components, all of which are embedded in the body
fluids:
1. cells;
2. fibres;
3. extracellular matrix (the extracellular part of the animal tissue that provides
structural support).
Cells, called fibroblasts, are responsible for the production of many forms of connective
tissue.
Examples of the different functions of connective tissues are:
• connecting body organs - blood;
• linking connective tissue to muscle tissue - cartilage (in tendons);
• protection - skull bones;
• structural framework - ribs;
• storage of energy - adipose tissue.
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TOPIC 2. DIFFERENTIATION IN CELLS
Connective tissue of different types may be grouped together, e.g. in an artery wall.
Structure of an artery
Cartilage is a good example of connective tissue. As well as forming the basic structure
of all bones, it is found in the joints between bones, the rib cage, the ear, the nose,
the elbow, the knee and the ankle. Over 150 disorders have been identified related to
connective tissue, e.g. cellulitis (a result of an infection), scars (injuries to connective
tissue) and genetic disorders such as Marfan syndrome.
Connective tissue: Questions
Q15: Where do you find cartilage connective tissues?
..........................................
Q16: List the main functions of connective tissue.
..........................................
2.3.3
Muscle tissue
The function of muscle cells is to produce force, and, as a result, motion. This is
achieved by protein filaments in the cells sliding past each other, changing both the
length and the shape of the cell. Muscle cells can only contract and relax, they cannot
push; hence the need for antagonistic pairs of muscles at joints (e.g. the biceps and
triceps at the elbow).
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TOPIC 2. DIFFERENTIATION IN CELLS
There are three types of muscle tissue:
1. Skeletal (voluntary) muscle is attached by cartilage to different bones at two or
more points, which allows it to bring about locomotion and to maintain posture.
This muscle appears striped under the microscope because of the arrangement of
the contractile proteins. The contraction of this muscle is under voluntary control;
although posture is maintained by unconscious reflexes, the muscles can also
be contracted voluntarily. Examples are the quadriceps of the upper leg, and the
biceps and triceps of the upper arm. The '-ceps' refers to the number of attachment
points (origins) at the inner end of the muscle.
Skeletal muscle structure
2. Smooth (involuntary) muscle is found widely throughout the body in the walls of
structures, e.g. arteries (contributing to blood flow in major arteries and controlling
access to capillary beds in arterioles), the stomach and all of the alimentary canal
(causing peristalsis), and the bladder (causing elimination of urine). The erector
pili, which cause 'goose flesh' when we are cold, are also smooth muscle, as are
the muscles which control the diameter of the pupil of the eye by means of the iris.
Smooth muscle is not under conscious control.
Smooth muscle structure
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3. Cardiac muscle is a form of involuntary muscle which is unique to the heart. Like
voluntary muscle, it has a striped appearance under the microscope which is
caused by the arrangement of the contractile proteins within the cells. It has a
very high resistance to fatigue.
Cardiac muscle structure
Muscle tissue: Question
Q17: Use the word list to complete the gaps in the sentences below. Words may be
used more than once.
,
and
The body contains three types of muscle tissue:
.
is unique in that it is only found in one organ, the
.
Two of the muscle types are not under conscious control and are called
muscle.
, is under conscious control and is called
The other type of muscle,
muscle.
and
.
The two muscle types which are striped are
.
The muscle type which forces food through the small intestine is
, but its control is both
The muscle type which maintains posture is
and
.
Word list: cardiac, heart, involuntary, skeletal, smooth, voluntary.
..........................................
2.3.4
Nervous tissue
The human nervous system consists of the central nervous system (CNS), comprising
the brain and spinal cord, and the peripheral nervous system (PNS), comprising the
cells which connect the CNS to the rest of the body.
At the most primitive level, the tissue which makes up the nervous system is specialised
to react to stimuli and to conduct impulses to various organs in the body, bringing about
a response to each stimulus. It comprises two groups of cell: neurons and glial cells.
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TOPIC 2. DIFFERENTIATION IN CELLS
Neurons
Neurons (often referred to loosely as 'nerve cells'), which conduct impulses, have three
types:
1. Sensory - carry information as impulses from sense receptors into the central
nervous system (CNS);
2. Motor - carry impulses from the CNS out to muscles and glands (effectors);
3. Interneurons - form connections between neurons in the CNS.
Sensory neurons are adapted to take messages to the brain from sensory receptors in
the skin, the specialised sense organs, the nose, tongue and ears.
Note that the cell body sits part way along the axon and there are no dendrites on the
cell body.
A motor neuron has many dendrites protruding from the cell body, which is end-on to
the axon.
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TOPIC 2. DIFFERENTIATION IN CELLS
An interneuron has many dendrites and axon terminals to connect with other neurons.
Glial cells
Glial cells (neuroglia) also have several forms and functions:
• to maintain a constant environment for the neurons by supplying nutrients and
removing waste;
• make the myelin which ensheaths some neurons and provides insulation;
• provide support and protection in the nervous tissue.
Nervous tissue: Question
Q18: Complete the following sentences using terms from the list below.
consists of neurons and glial cells.
conduct impulses into the CNS.
conduct impulses out from the CNS.
connect other neurons.
maintain a constant environment for neurons.
Word list: Glial cells, Interneurons, Motor neurons, Nervous tissue, Sensory neurons.
..........................................
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2.4
Formation of the body organs
Learning Objective
By the end of this section, you should be able to:
• explain that body organs can be formed from different body tissues.
There are almost 70 organs in a human body which vary according to their sizes,
functions or actions. Each organ has its own recognisable structure like the heart, lungs,
liver, eyes, ears and stomach, and each performs its own specific functions.
Tissues are grouped together so as to perform a single or several functions in the human
body. For example, the gall bladder contains different types of tissues such as the
epithelial tissues, forming a lining to protect the organ from the effects of bile, and
muscle tissue that contracts and expels bile.
Formation of the body organs: Question
Q19: Why is the gall bladder a good example of an organ?
..........................................
2.5
Differentiation of germline cells
Learning Objective
By the end of this section, you should be able to:
• define a germline cell;
• state the functions of a germline cell.
The germline of an individual is the sequence, or line, of cells that is capable of passing
its genetic material on to the next generation. A germ cell is one which can give rise to
gametes.
The germline cells segregate from the other body cells very early in the formation
of embryo. As the body grows and develops, the germline cells release hormones
into the blood circulation; in turn, they are controlled by hormones released by the
pituitary gland.
Thus, the body develops the characteristics of the male or the female and the germline
cells are stimulated to start meiosis as well as mitosis so that, in addition to producing
more diploid germline cells, haploid gametes are formed.
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TOPIC 2. DIFFERENTIATION IN CELLS
The sexual life cycle in animals
In contrast to most cells, which can only divide 30-50 times in the growing embryo,
germline cells (like stem cells) can divide indefinitely.
2.6
Mutations in germline and somatic cells
Mutations are changes to the genetic material, affecting either chromosomes or
individual genes. They may have no effect, alter the product of a gene or metabolic
process, or prevent a gene from functioning properly. The details of these various
mutations are dealt with in a later topic.
Mutations have a bad press. This is undeserved; without mutations, and the consequent
production of new combinations of DNA within genes and new arrangements of genes
on chromosomes, natural selection would have had no material to work on and evolution
would not have been possible.
Some mutations arise without the action of outside influences. They may be caused by
errors in the DNA replication or repair processes, or by bonds breaking by hydrolysis
in a natural process of chemical decay.
Alternatively, mutations can arise as a result of exposure to outside agents and the list
of such mutagenic agents, or mutagens, is long. It includes:
• various forms of radiation, e.g. X-rays, gamma rays, ultraviolet light, radon;
• a wide range of chemicals, e.g. nitrosamines (tobacco), mustard gas, bromine,
benzene;
• several metals and their compounds, e.g. arsenic, cadmium, chromium;
• some bacteria and viruses, e.g. Helicobacter pylori, hepatitis B virus.
The effects and significance of mutations which take place in germline and somatic cells
are very different.
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Germline cell mutation
After the egg and sperm fuse at fertilisation, the zygote divides by mitosis to give rise
to all the other cells of the body. If a mutation has occurred in the germline cells that
produced one of these gametes, this mutation will be transmitted to every single cell of
the resulting organism. Germline mutations can be found in every cell which originates
from the zygote to which the gamete containing the mutation contributed.
An oft-quoted example of such a mutation involves Queen Victoria. There is no record
of haemophilia in her family history, yet her youngest son was a haemophiliac and two
of her five daughters were carriers. The mutation must have occurred in the gamete
mother cells of Victoria’s father and he then passed the mutant allele on to his only child,
thus making her a carrier of the condition.
(The expected outcome of such a situation, in which the mother is a carrier and the
father is unaffected, would be that, on average, half of her sons would be haemophiliacs
and half of her daughters would be carriers. Victoria had four sons and five daughters
so one out of four and two out of five is well within the expected range.)
Somatic cell mutation
In somatic cell mutation, only cells within the body of the person affected will contain
the mutation. If the cell containing the mutation has already undergone differentiation,
then only it will be affected; if the cell is a stem cell, then all of its progeny will also carry
the mutation. Compared to germline mutations, this type of mutation is localised in its
presence within the body and is non-inheritable.
Cancer tumours are the result of one type of somatic cell mutation. In this case,
the mutation affects a gene involved in cell division (a proto-oncogene), causing
uncontrolled cell division and leading to the collection of undifferentiated cells that is
a tumour. Although the mutation is restricted to this tissue, its effects may of course
extend to the whole body.
Mutations in germline and somatic cells: Question
Q20: State one similarity and two differences between mutations to germline and
somatic cells.
..........................................
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TOPIC 2. DIFFERENTIATION IN CELLS
2.7
Learning points
Summary
• All of the cells of the body are somatic cells apart from gametes and the
cells which produce them.
• Examples of somatic cells are skin epidermis, B lymphocytes and motor
neurons.
• Somatic cells are produced by mitosis.
• Some somatic cells are able to divide by mitosis to form more somatic cells.
• Dividing by mitosis ensures that somatic cells maintain the diploid
chromosome number and the genetic complement.
• Somatic cells can differentiate to form different types of tissue.
• The four types of tissue found in the body are epithelial, connective, muscle
and nervous.
• The functions of epithelial tissue are protection (e.g. skin epidermis),
secretion (e.g. intestinal glands), and absorption (e.g. villi of small
intestine).
• The functions of connective tissue are protection (e.g. skull bones),
structural framework (e.g. ribs), storage of energy (e.g. adipose tissue),
connecting body organs (e.g. blood), and linking connective tissue to
muscle tissue (cartilage of tendons).
• The function of muscle tissue is to produce force and motion, e.g. skeletal
muscle (movement), smooth muscle (arteriole contraction) and cardiac
muscle (heart contractions).
• The function of nervous tissue is to react to stimuli and conduct impulses
to various parts of the body, e.g. neurons (carry impulses) and glial cells
(maintain a constant environment for neurons).
• Body organs are all formed from a variety of these tissues.
• A germline cell is one which is capable of passing on its genetic information
to the next generation by giving rise to gametes.
• Some germline cells divide by mitosis to produce more diploid germline
cells.
• Some germline cellsdivide by meiosis to produce haploid gametes.
• Mutations that occur in germline cells will be passed to offspring.
• Mutations in somatic cells will not be passed to offspring.
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TOPIC 2. DIFFERENTIATION IN CELLS
2.8
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of the four types of human body tissue before
attempting the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended answer question: Four types of human body tissue
15 min
Give an account, with examples, of the different body tissue types and their functions.
(8 marks)
..........................................
2.9
End of topic test
End of Topic 2 test
20 min
Q21: Complete the sentences by matching the parts on the left with the parts on the
right. (8 marks)
Somatic cells are found
differentiation.
Germline cells are
B-lympbocytes and motor neurons.
Examples of somatic cells are
more somatic cells.
Dividing somatic cells produce
cell or tissue involved.
Mitosis
germline.
Somatic cells form different types of
tissue by
throughout the body.
Cells capable of producing gametes
are
maintains the chromosome number.
Mutations in somatic cells affect the
gamete mother cells.
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TOPIC 2. DIFFERENTIATION IN CELLS
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..........................................
Q22: Complete the statements by matching the definitions on the left with the terms
and values on the right. (7 marks)
Somatic cells divide:
23
Number of types of tissue in
stomach:
mitosis
Produces more germline cells:
four
Germline cells produce gametes:
affect all cells of offspring
Germline chromosome complement:
mitosis
Gamete chromosome complement:
diploid
Mutations in germline cells:
meiosis
..........................................
Q23: Complete the paragraph using the words from the list. (11 marks)
tissue, the function of which is
.
Skin epidermis is an example of
tissue is used to store
and is an example of
tissue.
muscle has the function of producing
and is found in
.
cells are found in
tissue where they make the
of
some neurons.
Word list: adipose, artery walls, connective, energy, epithelial, force and motion, glial,
myelin sheath, nervous, protection, smooth.
..........................................
Q24: Where are somatic cells found? (1 mark)
..........................................
Q25: Name two examples of somatic cells. (1 mark)
..........................................
Q26: When they divide, what do some somatic cells produce? (1 mark)
..........................................
Q27: State the type of cell division found in somatic cells. (1 mark)
..........................................
Q28: Explain the significance of somatic cells dividing in this way. (1 mark)
..........................................
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TOPIC 2. DIFFERENTIATION IN CELLS
Q29: Name the process by which somatic cells form different types of tissue. (1 mark)
..........................................
Q30: Complete the table by listing the four types of body tissue, a function for each, and
an example of each. (12 marks)
Tissue
Example
Function
..........................................
Q31: Which tissues form an organ such as the stomach? (1 mark)
..........................................
Q32: State the meaning of the term 'germline cell'. (1 mark)
..........................................
Q33: Name the process by which germline cells produce more germline cells. (1 mark)
..........................................
Q34: Name the process by which germline cells produce gametes. (1 mark)
..........................................
Q35: List the terms used to describe the chromosome complement of germline cells
and gametes, and explain their meaning. (2 marks)
..........................................
Q36: Compare the significance of mutations which take place in the somatic cells and
germline cells. (1 mark)
..........................................
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Topic 3
Research: Stem cells and cancer
cells
Contents
3.1 Introduction . . . . . . . . . . . . . . .
3.2 Stem cell research . . . . . . . . . . .
3.2.1 Cell growth and differentiation .
3.2.2 Gene regulation . . . . . . . .
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3.2.3 Drug testing . . . . . . . . . . . . . . . . .
3.2.4 Development of disease . . . . . . . . . .
3.3 Therapeutic uses of stem cells . . . . . . . . . .
3.4 Ethical issues and the regulation of stem cell use
3.5 The biology of cancer cells . . . . . . . . . . . .
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3.6 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
52
53
Prerequisite knowledge
You should remember from Topic 1 that stem cells in multicellular organisms have the
ability to develop into different types of body cells during early embryonic development
and growth. In certain types of body tissues, tissue stem cells can become an internal
repair system; they can divide without limit to replenish other cells.
Learning Objectives
By the end of this topic, you should be able to:
Stem cells
• describe the contribution of stem cell research to our understanding of cell
processes such as growth, differentiation and gene regulation;
• explain why stem cells have therapeutic uses;
• describe the use of stem cells in the repair of damaged or diseased organs and
tissues;
• explain the use of stem cells as models to study disease development;
36
TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS
• describe the use of stem cells in drug testing;
• outline the principal ethical issues associated with stem cell use in research;
• describe the legal regulation of stem cell use;
Cancer cells
• describe the differences between cancer and normal cells;
• explain the formation of a tumour;
• explain the differences between malignant and benign tumours;
• state the meaning of the terms primary tumour and secondary tumour.
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TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS
3.1
37
Introduction
In an earlier topic, you learned about embryonic and tissue stem cells. In this topic, you
will study the use of stem cell research and therapy. You will also consider some of the
moral and ethical issues related to the wider use of stem cells, especially the use of
embryonic stem cells.
With cancer being one of the major causes of death in the UK, in this topic you will also
study the biology of cancer cells and the development of the different types of tumour.
3.2
Stem cell research
Learning Objective
By the end of this section, you should be able to:
• give examples of research work done on stem cells;
• define stem cell lines;
• describe the use of stem cells in skin grafts;
• briefly describe how stem cells can be used in bone marrow transplants;
• briefly describe how stem cells can be used in cornea transplants;
• describe how induced pluripotent stem cells (iPSC) are produced;
• give examples of the use of stem cells for drug testing.
Scientists have been interested in the study of cell biology since the discovery of
microscopes in the 1800s. In the early 1900s, researchers discovered that blood cells,
i.e. white blood cells, red blood cells and platelets, all originated from a particular stem
cell type. In the 1950s, bone marrow transplants involving adult stem cells were first
used in patients receiving radiation and chemotherapy.
The world's first human embryonic stem cell line was developed by James Thomson in
1998 at the University of Wisconsin in Madison. He successfully removed cells from
spare embryos at fertility clinics and cultured them in the laboratory.
Several thousand research papers on embryonic and adult stem cells are published
annually in scientific journals. Although clinical trials using embryonic stem cells are
still in their infancy, tissue stem cells are already being used in the treatment of many
conditions including leukaemia and heart disease.
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TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS
Since James Thomson's work in 1998, there has been large amount of research work
done on stem cells. Here are just a few of the significant findings in stem cell research
over the last decade:
• 2000s: Several reports of adult stem cell plasticity are published.
• 2005: Researchers at Kingston University in UK discovered a third category of
stem cell called cord-blood-derived embryonic-like stem cells (CBEs). These cells
came from the blood of the umbilical cord. The researchers claimed that the stem
cells were capable of differentiating into more types of tissue than adult stem cells.
• October 2006: Scientists at Newcastle University in UK used umbilical cord blood
stem cells to create the first ever artificial liver cells.
• October 2007: Capecchi, Evans and Smithies won the Nobel Prize for Physiology
or Medicine. They were rewarded for their work on mouse embryonic stem cells in
which they used gene targeting strategies to produce genetically engineered mice
for gene research. They called the mice "Knockout mice".
• October 2008: German scientist Conrad and colleagues generated pluripotent
stem cells from sperm mother cells in the adult human testis.
• 2011: Israeli scientist Inbar Friedrich Ben-Nun and his team produced the first
stem cells from endangered species; this could be a breakthrough that could save
animals in danger of extinction.
• 2013: A team from the University of Pittsburgh, Pennsylvania, grew human heart
stem cells in culture dishes and created conditions in which they formed a very
primitive heart.
• 2013: Mark Post of Maastricht University produced meat from muscle stem cells,
which was eaten as a burger.
The research work on stem cell differentiation and development can provide us with a
lot of information about the human body. If we can understand which genes turn 'on' and
'off', and the effect this has on our development, we will have a better understanding of
diseases like cancer and other growth abnormalities of the body's tissues and organs.
The reasons why stem cells are desirable for research are that they are pluripotent
and can be cultured indefinitely in the laboratory. This means that stem cells with
particular properties can be maintained in repositories and appropriate cells can be
supplied for particular research projects. Such lines include cells with different genetic
abnormalities, including Duchenne muscular dystrophy and thalassaemia. Studying
these cells enables recognition of mechanisms at the molecular level that might possibly
be blocked and so impede the disease progression. Thus embryonic stem cell lines
originating from embryos with genetic and chromosomal abnormalities enhance our
understanding of the pathways of genetic defects.
© H ERIOT-WATT U NIVERSITY
TOPIC 3. RESEARCH: STEM CELLS AND CANCER CELLS
What are stem cell lines?
You should remember that embryonic stem cells are derived from blastocysts. These
human blastocysts, used to produce stem cell lines, have been donated for research
from in vitro fertilisation clinics.
In order to obtain an embryonic stem cell line, scientists have to remove the cells from
the inner cell mass region of a blastocyst. This is because these cells are pluripotent.
Once these cells are removed, they are placed in culture plates with the correct nutrients
and growth factors. An embryonic cell line is established when these cells multiply and
divide. Under the right conditions, these cell lines can be maintained indefinitely and
can be used in all types of stem cell research. The next stage involves the addition of
different growth factors to induce these cells to differentiate into different cell types. The
newly formed cells can then be used in therapies to replace damaged cells or organs.
Stages of stem cell line production
1. Stages 1 and 2 involve the isolation of the embryonic stem cells.
2. In stages 3 and 4, the cells were tested or characterised. This should provide
enough information to let the researchers know if these cells were stem cells and
if they have the potential to differentiate.
3. Finally, stage 5 involves scaling up so as to provide enough cells for further clinical
studies.
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The production of induced pluripotent stem cells (iPSC)
An alternative method of producing pluripotent stems cells is by using a process called
induced pluripotent stem cells (iPSC).
Normally, the embryos are destroyed if we are to harvest stem cells from donated
embryos. This provokes wide discussions about the ethical and moral issues of using
embryos in stem cell work.
Research work has shown that by using differentiated adult cells, like skin cells, it is
possible to 'reprogramme' these cells by switching off the genes that make another adult
differentiated cell. Through the use of a virus as carrier (vector) to transfer genes, the
reprogramming will change the adult skin cell into one that has the properties of an
embryonic stem cell. The reprogrammed pluripotent stem cells behave like embryo
stem cells, but are actually from an adult. Although these cells meet the defining criteria
for pluripotent stem cells, it is not known if iPSCs and embryonic stem cells are clinically
different.
The iPSC process is illustrated in the following diagram:
Induced pluripotent stem cell (iPSC)
A major problem with the use of viruses to alter the genome of the somatic cell is that
it can trigger the expression of oncogenes, which cause tumours to develop. Recently,
it has been found that it is possible to induce somatic cells to become pluripotent by
treating them with certain proteins, thus opening the door to many more applications of
iPSCs.
Researchers are hoping to use iPSC in transplantation medicine, and also for drug
development and the modelling of diseases. At present, viruses are used for the
reprogramming of embryonic stem cells into adult cells; the process must be carefully
controlled and tested before this technique can be used for the treatment of human
diseases. The production of iPSC creates pluripotent stem cells which, with the studies
of other types of pluripotent stem cells, will help researchers learn how to reprogramme
cells to repair damaged tissues in the human body.
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Stem cell research: Questions
Q1: What are stem cell lines?
..........................................
Q2: Where do embryonic stems cells come from in the production of a stem cell line?
..........................................
Q3: Where do the cells used for iPSC originate from?
..........................................
Q4: Why do you think people who are against embryonic stem cell research would like
to see further studies in iPSC?
..........................................
3.2.1
Cell growth and differentiation
In 2013, researchers at the Austrian Academy of Sciences successfully grew complex
human brain tissue which formed 'mini-brains' with a cerebral cortex and retina. Their
research began with human embryonic stem cell lines and induced pluripotent stem
cells from mouse embryos. By carefully adjusting the nutrient medium in which they
were cultured, the scientists caused the cells to divide and differentiate into nervous
tissue. These fragments were attached to a 3D scaffold, and, within 3-4 weeks, defined
brain regions had formed. These mini-brains were maintained for several months.
The value of this work is its potential to give new insights into developmental brain
disorders. Already, it has been used to analyse the onset of microcephaly, a genetic
condition in which the brain is severely reduced in size.
3.2.2
Gene regulation
At several universities, the molecular causes of Rett syndrome are being studied. Rett
syndrome is an autism spectrum disorder which is the second most frequent form of
intellectual disability in girls after Down's syndrome. In most cases, this is caused by
the mutation of an X-linked gene, the protein product of which functions as a regulator
of gene expression.
The researchers are using induced pluripotent stem cells from Rett syndrome patients
to study the effect of key hormones and their inhibitors on neurone function, in particular
the growth of dendrites.
3.2.3
Drug testing
One of the important potential outcomes of embryonic stem cell research is that these
cells might be used to create very specific cell cultures for the initial stages of drug
testing. For example, by maintaining them in an appropriate nutrient solution, embryonic
stem cells could be stimulated to generate brain cells showing the effects of Alzheimer's
disease. Researchers could then screen millions of potential drugs against this more
realistic cellular model and identify the most effective compounds before going forward
to animal testing and clinical trials.
The speed and accuracy of drug screening would be improved by the use of real human
cells that mimic diseases.
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3.2.4
Development of disease
A potentially exciting use for iPS cells is the development of cell models of Parkinson's
disease. In theory, scientists could use cells from people living with Parkinson's disease
to create iPS cell models of the disease that have the same intrinsic cellular machinery
as a Parkinson's patient. Researchers could use these cell models to evaluate genetic
and environmental factors implicated in Parkinson's disease.
Many scientists use animals to model human diseases (mice can be obese or display
symptoms of Parkinson's disease, while rats get Alzheimer's and diabetes), but animal
models are seldom perfect and so scientists are looking at the use of induced pluripotent
stem cells (iPS cells).
People with Parkinson's disease don't have enough dopamine, a chemical that allows
messages to be sent to the parts of the brain that control movement. The disease kills
dopamine-producing nerve cells, or neurons, in part of the brain called the substantia
nigra. Parkinson's is also linked to formation of clumps of a protein called alphasynuclein in the brain.
Although the underlying cause of Parkinson's disease is unknown, scientists do know
which cells and areas of the brain are involved. Researchers are already using stem
cells to grow dopamine-producing nerve cells in the lab so that they can study the
disease. Because a single, well-defined type of cell is affected, it may also be possible
to treat Parkinson's by replacing the lost nerve cells with healthy new ones.
In the study of motor neuron disease, iPS cells can create motor neurons that grow in a
Petri dish and show the same characteristic blobs and tangles in the long fibres of the
nerve cells as are found in patients with the condition.
3.3
Therapeutic uses of stem cells
Learning Objective
By the end of this section, you should be able to:
• briefly describe the therapeutic use of stem cells in skin grafts;
• list other therapeutic uses of stem cells - bone marrow and cornea grafts.
'Therapeutic uses' simply refers to use in medical treatments. The generation of
cells and tissues used for cell-based therapies is one of the most important potential
applications of human stem cells. The availability of donated organs and tissues is
limited compared to the demand for transplant operations. The possibility of using stem
cells which have been stimulated to differentiate into specific cell types can be a valuable
source for replacing ailing or damaged cells and tissues. Stem cells are already available
for therapeutic use in diabetes, osteoarthritis, Alzheimer's disease, stroke, burns, and
rheumatoid arthritis.
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Here are three examples of the therapeutic use of stem cells:
1. Skin grafts
At present, patients with a full thickness burn to the skin, where the epidermis and the
delicate tissue of the dermis are damaged, are often treated using skin grafts of their
own skin (autologous grafts). During the period of 3 weeks that it takes for the graft to
be prepared, the patient is at risk of dehydration and infection.
The diagram shows a section through the human skin and the position of the major
tissues, the epidermis and the dermis.
A section of human skin
To prepare a graft, a section of healthy skin is removed from an undamaged part of
the patient's body. This is then grown in special nutrient solutions, which encourage
the epidermal stem cells to divide, and a graft for the burn site is produced from this
culture. It takes up to three weeks between the harvesting of the skin and growth of
sufficient new epidermis for grafting. An alternative procedure is to use skin donated
from deceased donors, but this risks problems of rejection.
In order to improve the outcome of the skin grafting treatment, studies have been
conducted into the possibility of using embryonic stem cells to generate a ready source
of skin cells for temporary grafts while patients are waiting for their autologous grafts.
The process involves culturing embryonic stem cells on a framework made of collagen
and the fibroblast cells which form the underlying structure of tissues and are involved
in healing. The stem cells are induced to develop into epidermal cells by maintaining
them in a specialised medium that encourages cell differentiation. After several rounds
of subculturing and replication, a layer of tissue develops which can be used to quickly
provide a temporary graft. It is hoped that, in the future, these cells can be stimulated to
generate a graft containing the dermis layer, where the nerve endings, hair follicles, oil
glands and sweat glands are located.
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2. Bone marrow transplants
A bone marrow transplant involves taking healthy stem cells from the bone marrow of
one person and transferring them to the bone marrow of another person, or back into
the donor at a later date. The cells removed are the haematopoietic stem cells from the
red bone marrow, most often from the upper edge of the pelvis at the iliac crest. These
are multipotent stem cells which can divide and differentiate into all of the various types
of blood cell.
The patient's own bone marrow cells must be killed, either by radiation or drugs, then
the donor stem cells are infused. If they are compatible with the patient, they will travel
to the bone marrow and start to produce new blood cells.
Bone marrow transplants are often used to treat conditions that damage bone marrow
so that it is no longer able to produce blood cells. In cases where a bone marrow
transplant is necessary to treat congenital defects or autoimmune diseases, the stem
cells are transplanted from a donor. Examples of conditions treated in this way are:
• severe aplastic anaemia (bone marrow failure);
• certain genetic blood and immune system disorders, such as sickle cell anaemia,
thalassaemia and some severe immune system diseases.
In the case of patients with cancer who require a bone marrow transplant, the
patient's own marrow is harvested and reintroduced into their body after radiation or
chemotherapy has controlled the disease. Examples of conditions treated in this way
are:
• some forms of leukaemia - cancer of the white blood cells;
• some non-Hodgkin's lymphomas - cancer of the lymphatic system.
3. Corneal repair
The cornea is the transparent front part of the eye that covers the iris and pupil. It
consists of a layer of epithelial tissue called the corneal epithelium. If a person is to
have excellent vision, the corneal epithelium must be transparent and not damaged in
any way. The conjunctiva covers the white of the eye and is connected to the corneal
epithelium by a group of tissues called the limbus. It is in the limbus that stem cells are
found.
If the limbus is damaged, then the person is unable to make corneal epithelial stem
cells to replace any damaged area of the cornea. The cornea of the eye will be covered
with blood vessels and eventually the person can lose his or her sight. Traditionally,
the person would have to wait for donated corneal tissues to become available. In
this treatment, the limbus of the person remains damaged and unable to produce new
corneal epithelium.
With the therapeutic use of stem cells, it is possible to isolate stem cells from the patient's
healthy eye and culture them in the laboratory. The new tissue is then transplanted back
into the damaged eye where the new cells will be able to repair the damaged area of the
corneal epithelium.
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Therapeutic uses of stem cells: Questions
Q5: Explain why embryonic stem cells are potentially more useful as sources of tissue
for skin grafts than epidermal stem cells.
..........................................
Q6: Why would the production of tissue for temporary skin grafts improve the outcome
for burns patients awaiting a graft of their own tissue?
..........................................
Q7: Why is it that blood stem cells are not suitable for other types of bone marrow
transplant?
..........................................
Q8: Why is it important that bone marrow used in transplants is of the same genetic
tissue type as the patient?
..........................................
Q9: Why is corneal repair using stem cells better than the traditional organ donor
scheme for damaged corneas?
..........................................
3.4
Ethical issues and the regulation of stem cell use
Learning Objective
By the end of this section, you should be able to:
• explain that the ethical objections to the use of embryonic stem cells are
principally that a zygote is a human being with a right to life, and that human
beings in the form of embryos should not be artificially created;
• state that in the UK, stem cell research is regulated by The Human Tissue
Authority and The Human Fertilisation and Embryology Authority.
Ethics
Tissue stem cells have been used in medicine for many years. It is the use of embryonic
stem cells in research and their possible application in medicine that has raised ethical
concerns.
One of the main issues is the sourcing of embryonic stem cells. Until recently, these
came from embryos donated for research because they were the unused embryos from
in vitro fertilisation (IVF) programmes. As up to thirty eggs may be fertilised in the
procedure, couples using IVF have four choices for the left-over embryos: have them
destroyed, give them to other couples, store them for an annual fee, or donate them to
research.
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Rather than have the surplus embryos simply destroyed, many parents prefer them to
be used for research purposes. Storage allows for future use by the couple, but also
puts off taking the decision about the unused embryos.
The key ethical objections to the use of embryonic stem cells that have been raised are
as follows:
• Human life should not be artificially created, an objection which also applies to in
vitro fertilisation. This is the very strictest stance; others find it acceptable to use
IVF as long as the gametes used are from the parents to be, but the use of donor
sperm would not be permissible.
• Life begins at fertilisation, so a fertilised egg (and, of course, blastocyst) is a human
being with the same right to life as anyone else. Others argue that life begins at
implantation, and here the debate shifts to the early termination of pregnancy by
abortion.
Within the law, it is up to individuals to decide whether the rights of an embryo or foetus
outweigh the potential benefits from stem cell research and therapeutic work.
The range of types of stem cell available for research today is wide, and, in some
situations, tissue stem cells and induced pluripotent stem cells provide a preferable,
less problematic, alternative to embryonic stem cells.
Legal
In the UK, there are very strict laws controlling the use of embryonic stem cells. All
stem cell research work must be licensed by The Human Tissue Authority and The
Human Fertilisation and Embryology Authority. The embryo must be less than the
development stage of 14 days, and researchers must justify the creation of such an
embryo for research work. Most recently, the law has changed to allow the creation of
human embryos specifically for research work. These embryos must only be used for
research and never for implantation into a womb.
All work on therapeutic treatments with stem cells is restricted to tissue stem cells in
the UK; the use of embryonic stem cells for therapeutic use remains a subject of great
debate.
Is using human admixed embryos by means of the nuclear transfer technique a
possible solution?
This is a relatively new process involving the creation of human admixed embryos. The
process involves mixing human tissue stem cells with the embryonic stem cell of an
animal. These could provide a source of embryonic stem cells without the need for
human eggs.
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Admixed embryo production
The following provides a summary of admixed embryo production.
Human admixed embryos
..........................................
Ethical issues and the regulation of stem cell use: Question
Q10: List two arguments for the use of embryonic stem cells in research and two
arguments against their use.
..........................................
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3.5
The biology of cancer cells
Learning Objective
By the end of this section, you should be able to:
• state that, compared to normal cells, cancer cells divide uncontrollably to
produce a mass of abnormal cells called a tumour;
• state that, compared to normal cells, cancer cells do not respond to normal
regulatory signals which would instruct them to stop dividing, when necessary;
• explain that a tumour develops when mutations take place in the genes
controlling cell division, programmed cell death, and differentiation;
• state that a benign tumour is one which does not invade neighbouring tissues
or spread throughout the body;
• explain that a malignant tumour is one in which the cells lose the molecules
on their surface that normally hold them in place so they can therefore become
detached from their neighbours and spread through the body in the blood or
lymphatic circulation;
• state that the original tumour is called the primary tumour and those forming
from the detached cells are called secondary tumours.
The term 'cancer' refers to a broad group of over 200 diseases which are characterised
by unregulated cell growth. The cancer cells divide excessively to produce a mass of
cells called a tumour. They do not respond to regulatory signals and may fail to attach
to each other.
If these cells invade neighbouring tissues and spread to other parts of the body in the
blood or lymphatic systems, they are classed as malignant tumours. The tumours which
develop as a result of this spread are known as secondary tumours.
If the tumour cells do not invade other tissues or spread, they are classed as benign,
e.g. moles. So-called benign tumours, though not actually cancers, can of course cause
very serious or fatal problems if they develop in areas where they put pressure on other
tissues, nerves or blood vessels, e.g. in the brain.
Causes of cancer
The causes of cancer are only partially understood because they are very varied and
complicated. An increased risk of cancer is associated with:
• ionising radiation (X-rays, gamma-rays, radioactive substances);
• UV radiation (sunlight, sunbeds);
• chemicals (tobacco smoke contains over 50 known carcinogens, asbestos,
benzene);
• viruses (human papilloma virus, hepatitis B, hepatitis C);
• poor diet, physical inactivity and obesity;
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• inheritance (breast cancer, ovarian cancer);
• hormones (oestrogen and progesterone with breast and ovarian cancer,
testosterone with prostate cancer).
Development of a tumour
Cancers develop when the regulation of tissue growth fails as a result of the mutation of
the genes which control cell division and differentiation. The genes involved are:
• Tumour suppressor genes, which produce proteins that slow down the cell division
cycle and stimulate programmed cell death;
• Proto-oncogenes, which promote cell growth and differentiation, and on mutation
can give rise to tumour forming oncogenes.
The diagram shows the steps in the development of a tumour.
Steps in the development of a tumour
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Under the microscope, tumour cells show several differences from normal cells in that
they have little cytoplasm, multiple nuclei, large nucleoli, and coarse chromosomal
material.
The most significant differences between them lie in their behaviour:
• the cells divide uncontrollably to produce a mass of abnormal cells;
• they do not respond to normal regulatory signals which would instruct them to stop
dividing when necessary;
• in a malignant tumour, they lose the molecules on their surface that normally hold
them in place; they can therefore become detached from their neighbours and
spread through the body to form secondary tumours.
The biology of cancer cells: Questions
Q11: State two characteristics of cancer cells.
..........................................
Q12: Explain the steps by which a malignant tumour produces secondary tumours.
..........................................
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3.6
Learning points
Summary
Stem cells:
• Stem cells are useful for research because they are capable of giving rise
to a range of cell types and can be cultured indefinitely in the laboratory.
• Stem cell research has contributed to our understanding of cell processes
such as growth, differentiation and gene regulation.
• Stem cells have therapeutic uses because they are pluripotent and can be
stimulated in culture to form a wide variety of tissues.
• Stem cells can be used in the repair of damaged or diseased organs and
tissues.
• In corneal repair, the use of corneal stem cells from the patient's own tissue
speeds up the process, creates new healthy tissue, and reduces the chance
of rejection.
• In skin grafting, the use of epidermal stem cells allows the much more rapid
production of a temporary graft, which reduces the risk of dehydration or
infection.
• Stem cells cultured from patients suffering from particular conditions can be
used as models to study disease development, e.g. Parkinson's disease,
motor neuron disease.
• Stem cells cultured from patients suffering from particular conditions can
be used to produce very specific cultures for use in testing a wide range of
drugs, e.g. Alzheimer's disease.
• The principal ethical issues associated with stem cell use in research apply
to embryonic stem cell use.
• The ethical objections to the use of embryonic stem cells are principally that
a zygote is a human being with a right to life and that human beings in the
form of embryos should not be artificially created.
• In the UK, stem cell research is regulated by The Human Tissue Authority
and The Human Fertilisation and Embryology Authority.
Cancer cells:
• Compared to normal cells, cancer cells:
– divide uncontrollably to produce a mass of abnormal cells called a
tumour;
– do not respond to normal regulatory signals which would instruct them
to stop dividing, when necessary.
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Summary Continued
• A tumour develops when mutations take place in the genes controlling cell
division, programmed cell death and differentiation.
• A benign tumour is one which does not invade neighbouring tissues or
spread throughout the body.
• A malignant tumour is one in which the cells lose the molecules on their
surface that normally hold them in place; they can therefore become
detached from their neighbours and spread through the body in the blood
or lymphatic circulation.
• The original tumour is called the primary tumour and those forming from the
detached cells are called secondary tumours.
3.7
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of stem cell research work before attempting the
question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended answer question: Stem cell research work
Describe the way in which the use of stem cells has contributed to:
20 min
A) corneal transplants; (3 marks)
B) skin grafts. (3 marks)
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..........................................
3.8
End of topic test
End of Topic 3 test
Q13: Complete the sentences by matching the parts on the left with the parts on the
right. (10 marks)
One reason stem cells are useful
for research is that
motor neuron disease.
Cells which can differentiate into all
cell types are
reducing the chance of rejection.
Stem cell research has contributed
to understanding
reducing the chance of dehydration.
The use of stem cells has improved
corneal repair by
Alzheimer's disease.
The use of stem cells has improved
skin graft procedures by
they can be cultured indefinitely.
The development of this can be
studied using stem cells
The Human Tissue Authority.
The use of stem cell cultures to test
a wide range of drugs helps
treatment of
embryonic stem cells.
Ethical objections have been raised
against the use of
a zygote is a human being.
An ethical objection against the use
of stem cells is that
pluripotent.
This regulates cell research in the
cell growth and differentiation.
UK
..........................................
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Q14: Complete the sentences by matching the parts on the left with the parts on the
right. (8 marks)
Unlike normal cells, cancer cells
mutations.
Cancer cells do not respond to
a mass of abnormal cells.
A tumour is
divide uncontrollably.
Tumours develop as a result of
its cells lose the molecules that
normally hold them in place.
The genes involved in tumour
development control cause
normal regulatory signals.
A tumour is benign if
slower cell division and programmed
cell death.
A tumour becomes malignant if
cells detach from a primary tumour
and spread through the body.
it does not invade neighbouring
tissues.
..........................................
Secondary tumours develop when
Q15: List two reasons that stem cells are useful for research. (2 marks)
..........................................
Q16: To which cell processes has stem cell research contributed most? (2 marks)
..........................................
Q17: Describe the improvement that the use of stem cells has made to the therapeutic
procedures of:
i
corneal repair;
ii
skin grafts.
(2 marks)
..........................................
Q18: Name two diseases, the development of which can be studied using stem cell
lines. (1 mark)
..........................................
Q19: How can stem cell cultures contribute to the treatment of Alzheimer's disease? (1
mark)
..........................................
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Q20: The scientific use of which stem cells has given rise to ethical objections? (1
mark)
..........................................
Q21: State two of the ethical objections to the use of stem cells. (2 marks)
..........................................
Q22: Name the authorities which regulate stem cell research in the UK. (2 marks)
..........................................
Q23: State the two key differences in behaviour between cancer cells and normal cells.
(2 marks)
..........................................
Q24: What causes a tumour to develop? (1 mark)
..........................................
Q25: What would cause a tumour to be classed as benign? (1 mark)
..........................................
Q26: How does a tumour become malignant? (1 mark)
..........................................
Q27: How do secondary tumours develop? (2 marks)
..........................................
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Topic 4
DNA structure and replication
Contents
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 DNA structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.3 Arrangement of DNA in chromosomes . . . . . . . . . . . . . . . . . . . . . . .
4.4 DNA replication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.6 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Learning Objectives
By the end of this topic, you should be able to:
• explain the role that DNA plays in the inheritance of characteristics;
• state that DNA directs its own replication;
• describe the structure of a DNA molecule;
• describe the arrangement of DNA molecules in chromosomes;
• explain the process of DNA replication.
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TOPIC 4. DNA STRUCTURE AND REPLICATION
4.1
Introduction
Learning Objective
By the end of this section, you should be able to:
• state that DNA is the molecule of inheritance and can direct its own replication;
• explain that all cells store their genetic information in the base sequence of DNA;
• state that the genotype is determined by the sequence of DNA bases;
• state that DNA is found on chromosomes.
Proteins are essential to the functioning of all biological systems. They perform a vast
array of functions. As enzymes and hormones, proteins are involved in controlling cell
activities, e.g. catalase and growth hormone. Some are structural proteins, e.g. collagen
in cartilage and bone, keratin in hair, feathers and scales, and elastin in artery walls. Yet
others are involved in cell signalling and transport.
Each protein is composed of amino acids joined together by peptide bonds into a unique
sequence. The type and order of the amino acids give the protein its characteristics.
They are determined by the deoxyribonucleic acid (DNA) of the gene which codes for
that protein. Specifically, it is the sequence of the bases in the DNA which stores the
genetic information, the genotype and the genome, of the organism.
DNA, or its close relative RNA (ribonucleic acid), is found in all life-forms on Earth, and
wherever it occurs, its structure and function are fundamentally the same. Not only
is DNA the molecule of inheritance, but, by the fact that it controls the production of
proteins, it also directs its own replication.
Introduction: Questions
Q1:
Of what are proteins made?
..........................................
Q2:
List three examples of proteins and state their general functions.
..........................................
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4.2
DNA structure
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Learning Objective
By the end of this section, you should be able to:
• state that the basic unit of DNA is the nucleotide;
• explain that a DNA nucleotide is composed of deoxyribose sugar, phosphate
and an organic nitrogenous base;
• state that the 5 carbon atoms of the deoxyribose molecule are numbered (1 =
one prime) according to their position in the molecule;
• state that nucleotides are joined by strong covalent bonds to form a sugarphosphate backbone;
• state that there are four bases, adenine (A), thymine (T), guanine (G) and
cytosine (C);
• state that these bases join to the equivalent base on the opposite strand by
weak hydrogen bonds;
• explain that each base can only bond to one other of the four to form a
complementary pair;
• state that the complementary pairs are A-T and G-C;
• explain that the complementary base pairs hold the two strands by hydrogen
bonds forming a double helix;
• state that DNA is a double-stranded, antiparallel structure, with the strands
running in opposite directions;
• explain that the direction of the strand is identified by the carbon atoms of the
deoxyribose which are found at the end of the strand;
• state that one strand goes from 3 to 5 the other from 5 to 3 ;
• explain that deoxyribose is found at the 3 end of the DNA strand and phosphate
is found at the 5 end of the DNA strand.
Chromosomes are thread-like structures found in the nucleus of the cell; they contain
DNA, which is tightly wound around alkaline proteins called histones.
On the following page is the famous 'Photograph 51', an X-ray crystallography image
which was taken in 1952 by Raymond Gosling, under the supervision of Rosalind
Franklin. It was she who concluded, from measurements of the structures in the
photograph, that the DNA molecule was a helix. Based on this, James Watson and
Francis Crick developed their double helix model in 1953, which was further refined by
Maurice Wilkins. This work was all conducted in labs in Cambridge, and Crick, Watson
and Wilkins were awarded the Nobel Prize in 1962 for it; Franklin died in 1958 and so
was not eligible.
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Photograph 51
Watson and Crick's model of the DNA molecule comprises two strands of repeating
units, called nucleotides, hydrogen-bonded together to form a double helical structure.
These two strands run in opposite, or antiparallel, directions.
Interactive DNA structure
An activity that shows the double helix structure of DNA as an interactive 3D model is
available in the online materials at this point. The following illustration gives an idea of
what to expect.
The double helix structure of DNA
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TOPIC 4. DNA STRUCTURE AND REPLICATION
Nucleotide structure
Each nucleotide consists of the following three parts.
• A central 5-carbon sugar; deoxyribose in the case of DNA and ribose in the case
of RNA. These carbons are numbered 1 to 5, with four of them being arranged in
a ring that is completed by an oxygen atom, and the fifth standing off to the side.
In the diagram below, the carbon atoms are represented by the numbers 1 (one
prime) to 5 , and the other elements by their symbols.
A central 5-carbon sugar
If there was an OH attached at carbon 2 , rather than one of the hydrogens, then
the sugar would be ribose.
• A phosphate group attached to carbon 5 .
• An organic nitrogenous base attached to carbon 1 , of which there are four:
adenine, thymine, guanine and cytosine.
The components of a nucleotide
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Nucleotide bases
A strong chemical (covalent) bond forms between the phosphate group of one nucleotide
and carbon 3 of the deoxyribose sugar of another. These bonds are not easily broken
and join neighbouring nucleotides into a permanent strand with a sugar-phosphate
backbone.
The DNA molecule comprises two of these strands, joined together by the weak
hydrogen bonds that form between their bases. These hydrogen bonds can easily be
broken so that the strands can be separated when necessary. Each type of base can
only bond with one other base: adenine (A) always bonds with thymine (T), by two
hydrogen bonds; guanine (G) always bonds with cytosine (C), by three hydrogen bonds.
A-T and G-C are called complementary base pairs, and the bases A and T, and G and
C, are complementary to each other.
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TOPIC 4. DNA STRUCTURE AND REPLICATION
Part of a double-stranded DNA molecule
The resultant double-stranded molecule is DNA, and its two strands are arranged in a
twisted coil called a double helix. It is like a spiral ladder in which the sugar-phosphate
molecules form the upright parts and the base pairs form the rungs. If the DNA in a
single human diploid cell was completely unravelled, it could extend to about 1.8 metres.
The following diagram provides a more detailed view of the chemical structure of DNA.
Notice that each set of base pairs have a total of three 'rings' so are a constant width,
but that the numbers of hydrogen bonds between the base pairs are different.
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The chemical structure of DNA
Notice also that the two strands are running in opposite, or antiparallel, directions. The
ends of the sugar-phosphate backbones are different. On the left, the strand ends with
the phosphate group on the 5 carbon of the deoxyribose, whereas on the right, the
strand ends with the 3 carbon of the sugar. These are referred to as the 5 (five-prime)
and 3 (three-prime) ends of the strands.
Here are two fascinating facts.
1. DNA is not restricted to the nucleus. It is also found in the mitochondrion (and the
chloroplast in plants), reflecting the ancient incorporation of these organelles as
bacteria into the eukaryote cell. The DNA which controls the mitochondria is split
between the nucleus and the mitochondria, and it is thought that the genes now
found in the nucleus have 'migrated' there from the mitochondria. We all inherit
our mitochondria from our mothers in the cytoplasm of the egg.
2. ATP (adenosine tri-phosphate) is very similar to a nucleotide. This ubiquitous
co-enzyme energy carrier comprises a ribose sugar with an adenine base
attached to carbon 1 , and a chain of three phosphate groups attached to carbon
5 .
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DNA structure: Questions
Q3: What shape is the DNA molecule?
..........................................
Q4: What type of bonding holds the two strands together in DNA?
..........................................
Q5: What feature of these bonds allows the two strands of DNA to be easily
separated?
..........................................
Q6: Complete the table by matching each DNA base with its complementary base.
DNA Base
Complementary base
Adenine
Cytosine
Guanine
Thymine
..........................................
Q7: Which base pair is more easily separated and why?
..........................................
Q8: State the components of an RNA nucleotide.
..........................................
Q9: What term describes the opposite directions of the two strands of DNA?
..........................................
Q10: What are found at the 3 and 5 ends of the DNA molecule?
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4.3
Arrangement of DNA in chromosomes
Learning Objective
By the end of this section, you should be able to:
• explain that chromosomes are made up of DNA;
• explain that chromosomes consist of tightly coiled DNA, packaged with
associated proteins.
Chromosomes tend to be illustrated as condensed and elongated structures during the
process of cell division. However, when cells are not undergoing division, chromosomes
do not look like this at all. For example, in nerve cells, which do not divide, the
chromosomes never assume a condensed shape.
The main function of DNA is to carry the genetic information which determines the
proteins that make up the human body. Each human cell contains nearly 2 metres of
DNA, and this has to be packaged into the nucleus which is about 6 μm (micrometre)
in diameter. This is like packaging about 50 km of fine threads into a tennis ball. The
only way that this can be achieved is for specialised proteins to bind and fold the DNA,
forming a series of loops and coils, but at the same time preventing the DNA from
becoming all tangled together.
Although DNA is compact in the cell, it still has to allow for cell replication, repair, and
for the functioning of genes. The DNA in the human body is split between different
chromosomes. The human genome contains about 3.2 × 10 7 nucleotides; these are
distributed between 24 different chromosomes. Each chromosome has a single linear
DNA molecule with the proteins that fold and pack the strands of DNA into a compact
structure. The materials from which chromosomes are made are chromatin, histones,
and non-histone proteins.
The packaging of DNA into a chromosome can be divided into four stages:
1. Nucleosomes, or 'beads on a string', are the basic unit of DNA packaging,
consisting of a segment of DNA wound around histone proteins. This stage is
often compared to a thread wrapped around a spool.
2. Formation of chromatin fibre. The nucleosomes are coiled and stacked.
3. Looped fibres. Chromatin fibres are looped along the protein scaffold.
4. Compacted chromosome. More folds and coiling of coils result in the most
compacted chromosome. The most condensed form of DNA is packaged during
metaphase in cell division.
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Arrangement of DNA from double helix to chromosome
Arrangement of DNA in chromosomes: Questions
Q11: By what means is so much DNA packed into the cell nucleus?
..........................................
Q12: What forms the scaffold around which DNA is arranged?
..........................................
Q13: A human somatic cell contains 46 chromosomes, but the genome is distributed
between 24 different chromosomes. Explain this apparent contradiction.
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4.4
DNA replication
Learning Objective
By the end of this section, you should be able to:
• state that DNA is unwound and unzipped by an enzyme (helicase);
• explain that this forms two template strands;
• state that DNA polymerase is the enzyme which adds nucleotides to the new
DNA strand;
• explain that DNA polymerase needs a primer to start replication;
• state that DNA polymerase can only add complementary nucleotides to the
deoxyribose / 3 end of the DNA strand;
• explain that this results in 3 strand / the leading strand being continuously
replicated;
• state that the 5 strand / the lagging strand is replicated in fragments;
• state that these fragments are joined together by the enzyme ligase.
The two processes that require new cells to be formed in the body are growth and repair.
The replication of DNA allows cells to pass this information to the new cells. In order for
an organism to function in a controlled and organised way, the information in the nucleus
of all cells must be the same. For this to be possible, the DNA replication process (and
the cell division process) must take place in a highly structured manner which virtually
eliminates mistakes. Occasional errors do occur in the process of DNA replication, but
there are systems in place to check for and correct them. As each daughter molecule
contains one of the strands from the original DNA molecule, it is referred to as being
'semi-conservative'.
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TOPIC 4. DNA STRUCTURE AND REPLICATION
General diagram of DNA replication
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Summary of DNA replication
The stages of DNA replication are summarised below. This account includes rather
more detail than is strictly required for the syllabus, but it is given to provide a more
complete, and hopefully comprehensible, picture of the process.
1. The double helix of the DNA molecule is unwound by the enzyme topoisomerase.
2. The two strands are separated by the enzyme helicase, and binding proteins
stabilise the single strands of DNA.
3. Two replicating forks form, so named because of the forked shape of the unzipped
DNA strands. The exposed bases of these strands form the templates for the
construction of new complementary DNA strands.
4. Because of the antiparallel arrangement of the strands, one of the separated
strands ends with an exposed deoxyribose (3 ) end; this is designated the leading
strand. The other ends with a 5 end; this is designated the lagging strand.
Detailed model of DNA replication
5. As DNA polymerase can only attach complementary bases to the deoxyribose (3 )
end of a DNA strand, primers must be made. These are RNA (ribonucleic acid)
nucleotides, which are attached by the enzyme DNA primase to exposed bases
at the end of the leading DNA strand and at various points along the lagging DNA
strand. They provide the necessary 3 starting points for the enzyme.
6. DNA polymerase replaces DNA primase, begins to attach nucleotides to the
primers, and so synthesises a new strand starting from the 3 end. This is
continuous on the leading strand, but discontinuous on the lagging strand as it
has to be read in reverse order.
7. The gaps in the sugar phosphate backbone of the new strands are then closed by
the enzyme ligase. The result is two identical double strands of DNA.
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DNA replication: Question
Q14: Complete the sentences by matching the parts on the left with the parts on the
right.
DNA is unwound and
unzipped to form
the enzyme helicase.
DNA is unzipped by
ligase.
At the end of the 3 strand is
two template strands.
At the end of the 5 strand is
leading 3 strand.
To start attaching
nucleotides, DNA
polymerase requires
deoxyribose.
DNA polymerase
attaches nucleotides to
the
lagging 5 strand.
The strand that is
continuously replicated is
the
ATP and free nucleotides.
The strand replicated in
fragments is the
3 end of the DNA strand.
The enzyme which joins
DNA fragment is
a primer.
In addition to a DNA
molecule and enzymes,
phosphate.
DNA replication needs
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4.5
Learning points
Summary
DNA
• DNA is the molecule of inheritance and can direct its own replication.
• All cells store their genetic information in the base sequence of DNA.
• The genotype is determined by the sequence of DNA bases.
• DNA is a component of chromosomes.
• Chromosomes consist of tightly coiled DNA, packaged with associated
proteins.
Structure of DNA
• The basic unit of DNA is the nucleotide.
• A DNA nucleotide is composed of deoxyribose sugar, phosphate and an
organic nitrogenous base.
• The 5 carbon atoms of the deoxyribose molecule are numbered (e.g. 1 =
one prime) according to their position in the molecule.
• Nucleotides are joined by strong covalent bonds to form a sugar-phosphate
backbone.
• There are four bases: adenine (A), thymine (T), guanine (G) and cytosine
(C).
• These bases join to the equivalent base on the opposite strand by weak
hydrogen bonds.
• Each base can only bond to one other of the four to form a complementary
pair.
• The complementary pairs are A-T and G-C.
• The complementary base pairs hold the two strands of the DNA molecule
by hydrogen bonds, forming a double helix.
• DNA is a double-stranded, antiparallel structure with the strands running in
opposite directions.
• The direction of the strand is identified by the carbon atoms of the
deoxyribose which are found at the end of the strand.
• One strand goes from 3 to 5 the other from 5 to 3 .
• Deoxyribose is found at the 3 end of the DNA strand and phosphate is
found at the 5 end of the DNA strand.
DNA replication
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TOPIC 4. DNA STRUCTURE AND REPLICATION
Summary Continued
• DNA is unwound and unzipped by an enzyme (helicase).
• This forms two template strands.
• DNA polymerase is the enzyme which adds nucleotides to the new DNA
strand.
• DNA polymerase needs a primer to start replication.
• DNA polymerase can only add complementary nucleotides to the
deoxyribose / 3 end of the DNA strand.
• This results in the 3 strand / the leading strand being continuously
replicated.
• The 5 strand / the lagging strand is replicated in fragments.
• These fragments are joined together (by the enzyme ligase).
4.6
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of the location and structure of DNA, and DNA
replication, before attempting the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended response question: The location and structure of DNA, and
DNA replication
A) Describe the location and structure of DNA. (8 marks)
B) Give an account of the replication of DNA. (7 marks)
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4.7
End of topic test
End of Topic 4 test
Q15: Complete the sentences by matching the parts on the left with the parts on the
right. (10 marks)
DNA is the molecule of inheritance
and is
held together by hydrogen bonds.
The base sequence of DNA is
able to direct its own replication.
DNA is stored on chromosomes and
is
adenine, cytosine, guanine and
thymine.
The basic unit of DNA is
used by cells to store genetic
information.
A nucleotide is
identified by their position in the
molecule.
The carbons in deoxyribose are
tightly coiled with associated proteins.
The sugar-phosphate backbone is
the nucleotide.
The bases of DNA are
adenine and thymine, guanine and
cytosine.
The base pairings of DNA are
composed of deoxyribose, phosphate
and a base.
The two strands of DNA are
held together by strong covalent
bonds.
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TOPIC 4. DNA STRUCTURE AND REPLICATION
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Q16: Complete the table using the words and phrases from the list. (8 marks)
The shape of the DNA molecule is a
The two strands of DNA are
At the 3 and 5 ends of the DNA strands are
The enzyme that adds nucleotides to the new DNA strand is
Before DNA replication can start, what is required is a
Complementary nucleotides are added to the 3 end of the
The DNA strand which is continuously replicated is the
The enzyme that joins fragments of DNA together is
Word/phrase list: antiparallel, deoxyribose and phosphate, DNA polymerase, double
helix, ligase, new DNA strand, primer, the 3 or leading strand.
..........................................
Q17: Explain the significance of DNA. (1 mark)
..........................................
Q18: How do cells store their genetic information? (1 mark)
..........................................
Q19: How is DNA stored in chromosomes? (1 mark)
..........................................
Q20: Name the basic unit of DNA and describe its composition. (2 marks)
..........................................
Q21: How are the carbon atoms on deoxyribose identified? (1 mark)
..........................................
Q22: How is the sugar-phosphate backbone of DNA held together? (1 mark)
..........................................
Q23: Name the nitrogenous bases found in DNA and list the complementary pairs they
form. (2 marks)
..........................................
Q24: How are the two strands of DNA held together? (1 mark)
..........................................
Q25: Why are the two strands of DNA relatively easy to separate? (1 mark)
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TOPIC 4. DNA STRUCTURE AND REPLICATION
Q26: What shape is the DNA molecule? (1 mark)
..........................................
Q27: What term refers to the fact that the strands of the DNA molecule run in opposite
directions? (1 mark)
..........................................
Q28: What is found at the 3 and 5 ends of the DNA strand? (1 mark)
..........................................
Q29: How is the DNA molecule formed into two template strands? (1 mark)
..........................................
Q30: Name the enzyme which adds nucleotides to the new DNA strand. (1 mark)
..........................................
Q31: What is the function of a primer? (1 mark)
..........................................
Q32: To where on a DNA strand can complementary nucleotides be added? (1 mark)
..........................................
Q33: Which DNA strand is continuously replicated? (1 mark)
..........................................
Q34: How are the fragments of DNA replicated from the other strand joined together?
(1 mark)
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Topic 5
Gene expression in human cells
Contents
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Gene expression through protein synthesis . . . . . . . . . . . . . . . . . . . .
78
79
5.3 Structure and functions of RNA . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Transcription of DNA into mRNA . . . . . . . . . . . . . . . . . . . . . . . . . .
81
84
5.5 Translation of mRNA into a polypeptide . . . . . . . . . . . . . . . . . . . . . .
91
5.6 Single gene, several proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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101
5.8 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.9 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103
104
Prerequisite knowledge
You will remember from previous studies that DNA codes for proteins in the cell. The
phenotype of a person is the characteristics, or traits, of that person, and is determined
by their genome, which is the entire genetic structure of the individual. This is contained
in the nucleus of each cell. The replication of DNA allows cells to pass this information
to the new cells.
Learning Objectives
By the end of this topic, you should be able to:
• explain the link between the genome and the phenotype of an individual;
• describe the structure and the functions of RNA;
• describe the transcription of DNA into messenger RNA (mRNA);
• describe the translation of mRNA into polypeptides and proteins;
• explain the fact that a single gene can give rise to several different proteins.
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TOPIC 5. GENE EXPRESSION IN HUMAN CELLS
5.1
Introduction
The many thousands of proteins that our cells use are synthesised inside the cells by
a complex process involving the nucleic acids DNA and RNA, and structures called
ribosomes which are mainly found on the rough endoplasmic reticulum. In addition,
free nucleotides, amino acids, many enzymes and ATP are required.
You will remember from previous studies that a protein is composed of amino acids
joined together in a specific sequence. The information that determines this sequence
is contained in the DNA in the nucleus.
In this topic, we will study how the instructions for making a protein are transferred
into the cytoplasm using a form of RNA, and how a polypeptide is first constructed on
the ribosomes and then modified to form the final protein. We will also see that the
old maxim of 'one gene - one enzyme' has had to be modified to 'one gene - many
enzymes' as we have discovered more about the mechanism of converting the DNA
code into protein.
Gene expression involves two major stages. The first process is transcription, in which
the DNA is copied to produce an RNA molecule called messenger RNA (mRNA).
Although there are some differences, this mRNA carries the same code in its base
sequence as the DNA of the original gene. This takes place in the nucleus.
The second stage is known as translation because the code of the base sequence
of the nucleic acid is used to produce a sequence of amino acids in the form of a
polypeptide chain. This takes place on the ribosomes, which are mainly located on
the rough endoplasmic reticulum. The polypeptide is then modified in various ways, e.g.
folding or adding phosphate, to produce the final protein.
An analogy with writing might help you to remember these terms. If you write down a
sentence from a computer screen, you have taken information in one form and put the
same information into a different form: you have written it across, or trans-scribed it. If
you then turn the sentence from English into Gaelic, you have taken the meaning of the
information and constructed something new: you have carried the meaning across, or
trans-lated it.
DNA
Process:
Occurs in:
makes
makes
mRNA
transcription
translation
nucleus
ribosomes in
cytoplasm
protein
Protein synthesis summary
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5.2
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Gene expression through protein synthesis
Learning Objective
At the end of this section, you should be able to:
• state that genes are expressed by using their DNA code to produce a protein;
• state that phenotype is determined by the proteins produced by gene
expression;
• state that in any one cell, only a fraction of the genome is expressed;
• explain how gene expression is influenced by intracellular factors, e.g.
phosphofructokinase activated by high levels of ADP;
• explain how gene expression is influenced by extracellular factors, e.g. low
blood sugar levels activating the alpha cells of the islets of Langerhans in the
pancreas to produce glucagon;
• state that gene expression has two major stages, transcription and translation;
• explain that transcription takes place in the nucleus and is the transfer of the
genetic code from the DNA of the gene to mRNA;
• state that translation takes place at the ribosome in the cytoplasm and is the
formation of a polypeptide from the code carried on the mRNA;
• describe how gene expression may be controlled at any of the many steps of
both transcription and translation;
• state that the code of the gene is a series of triplets of bases in DNA;
• explain that there are 64 possible combinations of three bases, hence 64
different triplets.
Chromosomes carry the genetic information of the cell in the form of genes. Gene
expression is the process by which the information stored in a gene is used to synthesise
a product, which is almost always a protein. Each gene has a specific base sequence
which is hundreds or thousands of nucleotides long. This sequence codes for the linear
order of amino acids in a polypeptide chain. Proteins are made up of one or more
polypeptide chains.
The nucleus of every cell carries the entire genome of the individual, but only a
proportion of these genes will be expressed in any one cell. All cells will have the
genes controlling glycolysis, but only a few will activate the gene to produce amylase.
Inactivation at this level takes place during differentiation.
Gene expression may also be switched on and off temporarily, by both extra- and
intracellular influences. There are many ways in which this is achieved, including
regulating the production or activity of an enzyme.
The regulation of blood sugar levels is an example of extracellular regulation. The
detection of low blood glucose levels by the alpha cells of the islets of Langerhans, in the
pancreas, causes the gene for the production of the peptide hormone glucagon to be
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'switched on' so that the hormone is manufactured and released into the blood stream.
The glucagon is detected when it attaches to glucagon receptors, present on the cell
membrane of liver and muscle cells, where it triggers the activation of the enzymes that
break down glycogen to glucose.
At the intracellular level, the enzyme phosphofructokinase, which controls a key step
in the sequence of reactions that is glycolysis, is activated by increasing levels of ADP.
As ADP is formed when ATP is broken down, its presence at high levels indicates a low
level of available energy within the cell in the form of ATP.
Although it is convenient to refer to genes being switched 'on' or 'off', as later sections
in this topic will show, the process of gene expression is much more complex than
this simple picture suggests. Transcription and translation involve several steps, each
depending on the action of several enzymes, all of which may be used to regulate the
process of turning the triplet code of the DNA into the gene-product as a protein.
The genetic code and protein synthesis
Inherited characteristics are the result of many biochemical processes controlled by
enzymes. In humans, for example, certain enzymes control the biochemical pathways
which cause a particular eye colour to develop. Enzymes are one form of protein. Their
exact molecular structure, shape and ability to carry out functions all depend on the
sequence of their amino acids. This sequence, in turn, is determined by the sequence
of bases in the organism's DNA. In this way, DNA controls the structure of enzymes and,
in so doing, determines the organism's inherited characteristics.
Remember that there are about 20 different amino acids. DNA contains only four
different nucleotides which are defined by their bases. If each were to specify a single
amino acid, only four amino acids could be coded for. A two base code would give
16 (42 ) possible arrangements - still not enough to code for all 20 amino acids. The
shortest unit that can code for all amino acids is a triplet code, which produces 64 (4 3 )
possibilities - more than enough!
Gene expression through protein synthesis: Questions
Q1:
A gene encodes the information for a specific protein.
a) True
b) False
..........................................
Q2: The regulation of blood sugar levels by hormones is an example of which type of
regulation of gene expression?
..........................................
Q3:
High levels of which chemical in the cell activates phosphofructokinase?
..........................................
Q4:
At what stages of gene expression may regulation of gene expression take place?
..........................................
Q5: How many more possible combinations of bases does a triplet code provide than
a code based on only two bases?
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5.3
Structure and functions of RNA
Learning Objective
At the end of this section, you should be able to:
• state that ribonucleic acid (RNA) is a single strand of nucleic acid;
• describe the structure of RNA as comprising ribose sugar, a phosphate group
and one of four bases;
• explain that RNA contains three of the bases in DNA (namely adenine, cytosine,
and guanine), plus uracil, which replaces the thymine of DNA;
• state that RNA bases pair A-U and G-C, both with DNA and other RNA
molecules;
• state that there are three forms of RNA;
• state that messenger RNA (mRNA) carries the code of the gene from the DNA
of the nucleus out to a ribosome in the cytoplasm;
• state that transfer (tRNA) carries specific amino acids to the mRNA, attached to
a ribosome;
• describe how base pairing between different parts of the mRNA strand folds the
molecule, exposing three bases at one end (the anticodon) and an attachment
point for an amino acid at the other end;
• state that ribosomal RNA (rRNA) reads the mRNA code in the ribosome;
Ribonucleic acid (RNA) provides a bridge between DNA and protein synthesis.
RNA consists of nucleotides with the generalised structure shown below.
An RNA nucleotide
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Important points to remember about RNA ribonucleotide structure:
• RNA nucleotides contain the sugar ribose as opposed to the deoxyribose of DNA,
a small difference that significantly alters the shape of the molecule;
• the RNA strand has A (adenine), U (uracil), G (guanine) and C (cytosine) as its
bases;
• RNA has the nitrogenous base (U) uracil rather than its close relative (T) thymine
(as found in DNA);
• RNA molecules are usually single-stranded.
There are three main types of RNA involved in protein synthesis:
• mRNA (messenger RNA), which transcribes the code from the DNA molecule and
carries it out of the nucleus to the ribosomes in the cytoplasm where the proteins
are synthesised;
• tRNA (transfer RNA), molecules of which are found in the cytoplasm, becoming
attached to specific amino acids and bringing them to the ribosomes where they
are joined to form a polypeptide;
• rRNA (ribosomal RNA), which catalyses the combination of mRNA and tRNA
during translation.
Messenger RNA (mRNA)
For the synthesis of a particular protein, the particular
sequence of bases on the DNA is first transcribed into
the complementary sequence of mRNA. This messenger
RNA can then carry the information for a particular
protein through the nuclear envelope to the sites of
protein synthesis (the ribosomes).
Transfer RNA (tRNA)
This type of RNA is responsible for the transport and
transfer of individual amino acids during protein
synthesis. Amino acids are transported by specific tRNA
molecules, which recognise the genetic code presented
by the mRNA. The 3 bases exposed at the bottom form
the anticodon. This is the complementary base
sequence to the base sequence on mRNA coding for a
particular amino acid.
Ribosomal RNA (rRNA)
This type of RNA is bound to structural proteins to form a
ribosome. The ribosome is used in the synthesis of
proteins.
The three forms of RNA
The triplet codes which define the proteins are carried in the DNA of the nucleus; this
code is then carried in the matching codons of the mRNA to the ribosomes in the
cytoplasm. Just as living cells carry their inherited information in the form of nucleic acid
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(usually DNA, occasionally RNA, e.g. polio virus), ribosomes are also found in all living
cells. They are complex molecular machines which build proteins, and they comprise
two major parts: a small subunit, which reads the mRNA, and a larger subunit, which
constructs the amino acid chain of the polypeptide. Each of these units consists of rRNA
and a variety of proteins.
Structure and functions of RNA: Questions
Q6: Complete the diagram of the RNA nucleotide structure using the labels and
structural components shown below.
..........................................
Q7:
Complete the table using the properties from the list.
RNA
DNA
Structure
Preferred form
Number of types
Present in
Nitrogenous bases
Properties: >1; 1; adenine, thymine, guanine, cytosine; adenine, uracil, guanine,
cytosine; double helix; not a double helix; double-stranded; single-stranded; the nucleus;
the cytoplasm and the nucleus.
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Q8: Name the three types of RNA found in the cell.
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Q9:
What are main functional differences between mRNA and tRNA?
..........................................
Q10: Nucleotides are the building blocks of:
a)
b)
c)
d)
DNA only
RNA only
both DNA and RNA
neither DNA nor RNA
..........................................
Q11: Name the two components of a ribosome.
..........................................
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5.4
Transcription of DNA into mRNA
Learning Objective
At the end of this section, you should be able to:
• define transcription as the transfer of the gene information from the triplet code
on the DNA template to mRNA;
• state that the DNA double helix is unwound and unzipped by the enzyme RNA
polymerase;
• state that RNA nucleotides pair with the appropriate exposed bases of the DNA
template;
• state that the base on each RNA nucleotide pairs with its complementary DNA
base;
• state that RNA polymerase causes the RNA nucleotides to bind into a single
strand;
• state that this strand of mRNA is the primary transcript;
• state that the primary transcript contains both protein-coding regions (exons)
and non-coding regions (introns);
• explain that, after transcription, the introns are removed from the primary
transcript and the exons are joined together;
• state that the joining of the exons is called RNA splicing;
• state that the final form of the mRNA is the mature transcript.
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The nucleus of every cell contains one or more structures called nucleoli. Each
nucleolus is composed of RNA, DNA and protein, but is not bounded by a membrane.
Nucleoli are the site of the synthesis of RNA and other components of ribosomes. These
components pass out of the nucleus through pores and are assembled in the cytoplasm.
Polypeptides are made on the ribosomes in the cytoplasm. The instructions (or code)
telling the ribosomes what proteins to make are contained in the DNA, which never
leaves the nucleus. This code is carried from the nucleus to the ribosomes by a form of
RNA called messenger RNA (mRNA). The DNA code is transcribed from one of the DNA
strands using free RNA nucleotides present in the cell's nucleus. Transcription is the first
step in protein synthesis. During DNA replication, the DNA double helix is unwound by
the enzyme helicase as DNA polymerase in unable to do this. In mRNA synthesis, the
DNA is unwound to expose the DNA template strand by the RNA polymerase enzyme
itself.
Free RNA nucleotides in the nucleus bind to complementary DNA nucleotides using the
base pair rules shown in the following table.
DNA nucleotide
RNA nucleotide
A
pairs with
U
T
pairs with
A
G
pairs with
C
C
pairs with
G
Base pairing rules
The enzyme RNA polymerase then joins the nucleotides to each other by attaching the
phosphate group of one nucleotide to the ribose of the next, and the strand separates
from the DNA template. The RNA molecule produced during transcription of DNA is
messenger RNA (mRNA) and is called the primary transcript. It contains both protein
coding sections, called exons, and non-coding sections, called introns.
As the synthesis of a new strand of mRNA is in the 5 to 3 direction, only the exposed
DNA strand which runs in the 3 to 5 direction is used as a template to build the mRNA.
This apparent oddity results from the fact that, in order to be recognised when it reaches
the ribosome, the mRNA must start with a special cap which only attaches to the 5 end
of the molecule (5 cap). This and other changes to the original mRNA strand, such as
the removal of introns, are called post-transcriptional regulation, and they produce the
mature transcript which leaves the nucleus for the ribosomes. The removal of introns
and the joining-up of the exons are achieved by the process of RNA splicing.
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Transcription of DNA into mRNA: Steps
10 min
The following provides a summary of nucleotide structure and the steps involved in the
process of RNA transcription.
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Most cells contain thousands of ribosomes in their cytoplasm. Ribosomes are important
as the site of protein synthesis in the cell and contain enzymes essential for protein
synthesis. Cells that have a lot of ribosomes synthesise more protein than those that
have fewer ribosomes.
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Transcription of DNA into mRNA: Questions
Q12: What process takes place on ribosomes?
..........................................
Q13: What are the functions of the enzymes helicase and RNA polymerase?
..........................................
Q14: Which of the following best describes transcription?
a)
b)
c)
d)
DNA → DNA
DNA → RNA
DNA → protein
RNA → DNA
..........................................
Q15: The complementary messenger RNA strand that would be synthesized from the
DNA base sequence of CTGAC would be:
a)
b)
c)
d)
GACTG
UGACU
AGTUC
GACUG
..........................................
Q16: Why is only one strand of DNA transcribed into mRNA?
..........................................
Q17: State two changes which take place to the primary transcript to produce the
mature transcript.
..........................................
Q18: Name the process by which the mature transcript is produced.
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91
Translation of mRNA into a polypeptide
Learning Objective
At the end of this section, you should be able to:
• define translation as the conversion of the gene information in the form of mRNA
into a polypeptide at a ribosome;
• state that ribosomes comprise rRNA and proteins;
• state that mRNA carries the genetic code in the form of sets of three bases,
called codons;
• state that each tRNA has three exposed bases, called an anticodon;
• state that each amino acid is carried by a different type of tRNA;
• state that, at the ribosome, rRNA reads the incoming mRNA;
• explain that one codon acts as start signal and others act as stop signals,
indicating the start and finish of the codons to be translated;
• explain that, when the start codon is reached, the ribosome starts attaching
tRNA bases to the mRNA strand;
• state that the anticodons of the tRNA molecules are matched to complementary
codons on the mRNA;
• state that the enzymes in a ribosome join the amino acids together with peptide
bonds;
• state that the tRNA then disengages from the amino acid and the ribosome, and
goes off to collect another molecule of its specific amino acid;
• explain that, when the stop codon is reached, the ribosome disassociates and
the chain of amino acids is released as a polypeptide.
Once the DNA of a gene has been transcribed into mRNA, the mRNA molecules
pass through the nuclear pores into the cytoplasm. Here, translation of mRNA into
polypeptides takes place on ribosomes. This involves a third type of RNA, transfer RNA
(tRNA).
Each molecule of tRNA is a single-stranded sequence of between 70 and 90
nucleotides, which is folded into a clover-leaf shape as a result of base pairing, using
hydrogen bonds, between different regions of the strand. At the 3 end of the strand is
a site to which an amino acid attaches by a covalent bond, and at the opposite end of
the molecule are three nucleotides with exposed bases.
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tRNA
These exposed bases match up against the codons of the mRNA and are called
anticodons. As it is possible for any of the four RNA bases (A, U, G and C) to be
present on each of these nucleotides, there are 64 possible combinations of bases in
the anticodon.
tRNA
15 min
The following provides structural information about a tRNA molecule, in this case with
the amino acid serine attached to it.
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However, in the translation of the mRNA there must be codons which signal where
translation should start (AUG, also methionine) and stop (UAG, UAA and UGA). It
is changes to the DNA which produce such 'stop' combinations that lead to the
nonsense mutation mentioned in the next topic.
This leaves far more combinations than are minimally necessary to code for the 20
common amino acids. As a result, several mRNA codons signify the same amino acid
and several tRNA molecules with the corresponding anticodons will attach to it.
The genetic code table of amino acids and codon letters
The process of translation begins when the small subunit of a ribosome binds to the
5 end of the mRNA strand. The small subunit then moves along the mRNA until it
encounters the AUG start codon, at which point it is joined by the large subunit of the
ribosome. A tRNA molecule with anticodon bases complementary to the first codon of
the mRNA transcript is then attached to the exposed bases by the large subunit.
The subunits of the ribosome then move along the mRNA strand, attaching the
appropriate tRNA molecules to the exposed codons. As they do so, enzymes in the
large subunit form peptide bonds between the amino acids, creating the polypeptide
molecule. The tRNA then releases its amino acid, disengages from the mRNA and
goes off to recharge with another amino acid.
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When the stop codon is reached, rather than another tRNA attaching, release factors are
produced which cause the ribosome subunits to disassociate, releasing the polypeptide.
Translation: Steps
15 min
The following provides a summary of the steps involved in the translation of mRNA into
a protein.
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Translation of mRNA into a polypeptide: Questions
Q19: Select the correct answers from the options listed after each question.
1. The triplet code in the nucleus is stored in: mRNA, tRNA, DNA or rRNA?
2. The nucleic acid which carries amino acids is: mRNA, tRNA, DNA or rRNA?
3. The cloverleaf shape of tRNA is caused by base-pairing using: covalent bonds,
peptide bonds or hydrogen bonds?
4. At the 3 end of a charged tRNA is found the: codon, anticodon or amino acid?
5. The codon AUG indicates: start, stop or nonsense?
6. The mRNA is read by the: ribosome, small subunit or large subunit?
7. The type of nucleic acid found in the small subunit of the ribosome is: mRNA, tRNA,
DNA or rRNA?
8. The bonds formed between the amino acids are: covalent bonds, peptide bonds or
hydrogen bonds?
9. The stop codon causes: tRNA attachment, tRNA detachment, polypeptide release
or ribosome subunits disassociation?
..........................................
Q20: Try using the genetic code table of amino acids and codon letters to complete the
sequence of each of the proteins by typing the appropriate amino acid name into each
of the answer boxes
The genetic code table of amino acids and codon letters
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5.6
Single gene, several proteins
Learning Objective
At the end of this section, you should be able to:
• state that the human body contains many more types of protein and mRNA than
there are genes in the DNA;
• state that one gene must be able to code for several proteins;
• state that different mRNA molecules can be produced from the same primary
transcript;
• explain that this is achieved by different mRNA segments being treated as exons
and introns;
• state that this is called alternative RNA splicing;
• state that proteins may be modified by the cutting and combining of polypeptide
chains;
• state that proteins may be modified by adding phosphate or carbohydrate
groups;
• state that these are examples of post-translational modification of protein
structure.
Recent studies about the human genome have revealed that there are about 21,000
protein-coding genes. Yet, in the human body, the number of proteins are in excess
of 25,000, and earlier estimates were that the human genome would comprise 100,000
genes. In fact, only about 1.5% of the genome represents protein-coding genes, the rest
being associated with non-coding RNA (e.g. tRNA), the regulation of gene expression,
and introns. There are also many sequences for which no function has yet been found.
Therefore, within the human body, one gene must be able to encode many proteins,
i.e. a variety of proteins can be expressed by the same gene. This is achieved by two
processes: alternative RNA splicing and post-translational modification (PTM).
Alternative RNA splicing
This process allows a variety of different proteins to be expressed from one gene.
The human genome consists of exons and introns. These terms might seem a little
counterintuitive, but exon is short for 'expressed region' and intron for 'intragenic region'.
The coding sections of a gene (the exons) can be split into several sections by noncoding sections (the introns). The name 'exon' is short for 'expressed region'; 'intron' is
perhaps best remembered as the 'in-between' sections.
Splicing occurs after transcription when the introns are removed from the primary
transcript. For many introns, splicing is done in a series of reactions which are catalysed
by a complex of small, nuclear ribonucleoproteins (snRNPs). NB: snRNPs are not in the
syllabus, but are included here because of their name, 'snurps'.
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Particular exons can either be included or excluded from the mature transcript. It is also
possible that two splice sites are produced at one end of an exon, making it possible to
produce multiple transcripts by 'alternative splicing'. The joining of different splice sites
allows individual genes to express multiple mRNAs that encode proteins with diverse
and even antagonistic functions.
The process allows a single gene to express a variety of proteins. The advantage
of the splicing process is that a single primary transcript can be spliced into different
mRNA by the inclusion of different sets of exons. The process of alternative splicing
explains how the 21,000 genes in the human genome are able to encode the 120,000
different translated mRNA molecules reported to exist in the human cells. In humans, it
is estimated that alternative splicing occurs in more than 60% of genes.
Alternative RNA splicing
In the example in the diagram above, when the introns are removed, exons 1 and 2 may
either be spliced with exon 3 or with exon 4.
Post-translational modification
The chemical modification of a polypeptide after its translation is called post-translational
modification (PTM). The process is a late stage of protein synthesis which also allows
a gene to produce many different proteins. PTM of the polypeptide alters the function
of the protein. This is achieved by either adding other biochemical groups, such as
phosphate, lipids and carbohydrates, or cutting the polypeptide chain and reassembling
it into a modified version. Also, the amino acid coded by the start codon, methionine, is
usually removed.
Effectively PTM will change the structural or functional nature of the protein. Two
examples of PTM are as follows.
• A protease enzyme cuts the polypeptide chain in two places. The middle section
of the protein is then removed and the structure re-formed. An example of this
is pro-insulin→insulin formation. The succession of changes leading to the final
mature version of insulin begins in the membrane of the endoplasmic reticulum,
continues in the Golgi apparatus, and is completed in the secretory vesicle, which
will excrete the insulin from the cell.
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• Proteins are able to be phosphorylated with the addition of a phosphate group
by a kinase enzyme. This is a way to control the activity of many enzymes and
receptors. The addition of a phosphate group causes a change in the structure of
the protein, allowing enzymes and receptors to be switched 'on' or 'off'.
Another post-translational modification is 'protein folding'. The correct three-dimensional
structure is determined by interactions between the amino acids and is essential to the
correct functioning of the protein.
Single gene, several proteins: Question
Q21: Complete the paragraph using the words from the list.
as a result of RNA
and
One gene can be expressed as many
. Different
molecules are produced from the
post-translational
transcript depending on which RNA segments are treated as exons
same
. Protein structure can be altered post-translation by:
and
chains, e.g. the formation of insulin from
;
cutting and combining
or
groups to the protein.
adding
Word list: carbohydrate, introns, modification, mRNA, phosphate, polypeptide, primary,
pro-insulin, proteins, splicing.
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5.7
101
Learning points
Summary
Gene expression
• Genes are expressed by using their DNA code to produce a protein.
• The phenotype is determined by the proteins produced by gene expression.
• In any one cell, only a fraction of the genome is expressed.
• Gene expression is influenced by intracellular
phosphofructokinase activated by high levels of ADP.
factors,
e.g.
• Gene expression is influenced by extracellular factors, e.g. low blood sugar
levels activating the alpha cells of the islets of Langerhans in the pancreas
to produce glucagon.
• Gene expression has two major stages: transcription and translation.
• Transcription takes place in the nucleus and is the transfer of the genetic
code from the DNA of the gene to mRNA.
• Translation takes place at the ribosome in the cytoplasm and is the formation
of a polypeptide from the code carried on the mRNA.
• Gene expression may be controlled at any of the many steps of both
transcription and translation.
• The code of the gene is a series of triplets of bases in DNA.
• There are 64 possible combination of three bases, so 64 different triplets.
Structure and functions of RNA
• Ribonucleic acid (RNA) is a single strand of nucleic acid.
• RNA comprises ribose sugar, a phosphate group and one of four bases.
• RNA contains 3 of the bases in DNA, namely adenine, guanine and
cytosine, plus uracil which replaces the thymine of DNA.
• RNA bases pair A-U and G-C, both with DNA and other RNA molecules.
• There are three forms of RNA.
• Messenger RNA (mRNA) carries the code of the gene from the DNA of the
nucleus out to a ribosome in the cytoplasm.
• Transfer (tRNA) carries specific amino acids to the mRNA attached to a
ribosome.
• Base pairing between different parts of the mRNA strand fold the molecule,
exposing three bases at one end (the anticodon) and an attachment point
for an amino acid at the other end.
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Summary Continued
• Ribosomal RNA (rRNA) reads the mRNA code in the ribosome.
Transcription of DNA into mRNA
• Transcription is the transfer of the gene information from the triplet code on
the DNA template to mRNA.
• The DNA double helix is unwound and unzipped by the enzyme RNA
polymerase.
• RNA nucleotides pair with the appropriate exposed bases of the DNA
template.
• The base on each RNA nucleotide pairs with its complementary DNA base.
• RNA polymerase causes the RNA nucleotides to bind into a single strand.
• This strand of mRNA is the primary transcript.
• The primary transcript contains both protein-coding regions (exons) and
non-coding regions (introns).
• After transcription, the introns are removed from the primary transcript and
the exons are joined together.
• The joining of the exons is called RNA splicing.
• The final form of the mRNA is the mature transcript.
Translation of mRNA into polypeptides and proteins
• Translation is the conversion of the gene information, in the form of mRNA,
into a polypeptide at a ribosome.
• Ribosomes comprise rRNA and proteins.
• mRNA carries the genetic code in the form of sets of 3 bases called codons.
• Each tRNA has three exposed bases called an anticodon.
• Each amino acid is carried by a different type of tRNA.
• At the ribosome, rRNA reads the incoming mRNA.
• One codon acts as start signal and others act as stop signals, indicating the
start and finish of the codons to be translated.
• When the start codon is reached, the ribosome starts attaching tRNA bases
to the mRNA strand.
• The anticodons of the tRNA molecules are matched to complementary
codons on the mRNA.
• Enzymes in ribosome join the amino acids by peptide bonds.
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Summary Continued
• The tRNA then disengages from the amino acid and the ribosome, and goes
off to collect another molecule of its specific amino acid.
• When the stop codon is reached, the ribosome disassociates and the chain
of amino acids is released as a polypeptide.
One gene, many proteins
• The human body contains many more types of protein and mRNA than there
are genes in the DNA.
• One gene must be able to code for several proteins.
• Different mRNA molecules can be produced from the same primary
transcript.
• This is achieved by different mRNA segments being treated as exons and
introns.
• This is called alternative RNA splicing.
• Proteins may be modified by the cutting and combining of polypeptide
chains.
• Proteins may be modified by adding phosphate or carbohydrate groups.
• These are examples of post-translational modification (PTM) of protein
structure.
5.8
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of transcription and post-translational
modification (PTM) before attempting the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended response question: Transcription and post-translational
modification (PTM)
Give an account of gene expression under the following headings.
A) Transcription (6 marks)
B) Post-translational modification (PTM) (4 marks)
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5.9
End of topic test
End of Topic 5 test
Q22: Complete the paragraph using the words from the list. (10 marks)
.
The individual's phenotype is determined by the proteins produced by gene
in the nucleus of each cell, but in any
These proteins are generated from the
is expressed. Intracellular factors can influence
one cell only a fraction of the
is activated by high levels of ADP.
factors
gene expression, e.g.
may also have an effect, e.g. low blood blood sugar levels activate the production of
by the alpha cells of the pancreas. Gene expression has two stages,
in the nucleus and translation at the
, both of which may be used to control the
of bases in the DNA. A
process. The code of a gene consists of a series of
triplet of bases, of which there are
types, can represent 64 different instructions.
Word list: DNA, expression, extracellular, four, genome, phosphofructokinase, glucagon,
ribosome, transcription, triplets.
..........................................
Q23: Complete the table using the words and phrases from the list. (9 marks)
RNA:
Phosphate, base, ...
Adenine, cytosine, guanine, ...
G-C, ...
Messenger, transfer, ...
mRNA:
tRNA:
rRNA:
Anticodon:
Word/phrase list: 3 exposed bases on tRNA, A-U, amino acid to ribosome, anticodon,
nucleus to cytoplasm, reads mRNA code, ribose sugar, ribosomal, single strand of
nucleic acid, uracil.
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Q24: Complete the table by matching the parts on the left with the parts on the right. (9
marks)
Transcription:
protein-coding regions of mRNA.
RNA base uracil:
joining up of exons.
RNA polymerase:
DNA triplet code to mRNA.
mRNA codon:
final form of mRNA.
Primary transcript:
DNA unwound and unzipped/RNA
nucleotides join.
Exons:
DNA base adenine pairs with it.
Introns:
DNA triplet of bases matches with it.
RNA splicing:
first RNA strand formed from DNA.
Mature transcript:
non-coding regions of mRNA.
..........................................
Q25: Rearrange the following steps to describe the order in which translation takes
place. (9 marks)
• Enzymes in ribosome join the amino acids by peptide bonds.
• Translation takes place at the ribosome.
• When a start codon is reached, the ribosome starts attaching tRNA bases to the
mRNA.
• tRNA disengages and goes off to find another of its amino acids.
• Ribosomes consist of rRNA and proteins.
• Each tRNA molecule carries a specific amino acid.
• At the ribosome, rRNA reads the incoming mRNA.
• When a stop codon is reached, the ribosome diassociates and the polypeptide is
released.
• mRNA has sets of three bases called codons and tRNA has matching sets called
anticodons.
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Q26: Complete the sentences by matching the parts on the left with the parts on the
right. (6 marks)
The body contains many more
proteins than there are
exons and introns.
One gene must be able to code for
several different mRNA molecules.
One primary transcript can produce
post-translational modification.
Alternative RNA splicing treats
different RNA segments as
carbohydrate groups.
Post-translational modification may
involve adding
genes.
Cutting and combining of
several proteins.
polypeptides is another form of
..........................................
Q27: By what means is phenotype linked to genotype? (1 mark)
..........................................
Q28: Give an example of an intracellular factor influencing gene expression. (2 marks)
..........................................
Q29: Describe the way in which low blood sugar levels act as an extracellular factor. (2
marks)
..........................................
Q30: Explain why only 4 bases can give rise to 64 different codes for amino acids. (1
mark)
..........................................
Q31: When RNA is being formed using DNA as a template, what are the three base
pairings which can occur (letters are acceptable)? (1 mark)
..........................................
Q32: Name the three forms of RNA involved in gene expression (letters are not
acceptable except for RNA). (2 marks)
..........................................
Q33: State the meaning of the term 'transcription'. (1 mark)
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Q34: How is the mature transcript produced from the primary transcript? (2 marks)
..........................................
Q35: What is RNA splicing? (1 mark)
..........................................
Q36: Describe the structure of tRNA. (2 marks)
..........................................
Q37: Explain the fact that 20 amino acids are coded for by 64 possible combinations of
bases in the anticodons of tRNA. (1 mark)
..........................................
Q38: Explain what happens when the end of the mRNA section to be translated is
reached? (1 mark)
..........................................
Q39: Explain how different mRNA molecules can be produced from the same primary
transcript. (1 mark)
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Topic 6
Genes and proteins in health and
disease
Contents
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Different levels of protein structure . . . . . . . . . . . . . . . . .
Functions of proteins . . . . . . . . . . . . . . . . . . . . . . . . .
Mutations and genetic disorders . . . . . . . . . . . . . . . . . .
Single gene mutations . . . . . . . . . . . . . . . . . . . . . . . .
The effect of mutations on the structure and function of proteins .
Chromosome structure mutations . . . . . . . . . . . . . . . . . .
Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Extended response question . . . . . . . . . . . . . . . . . . . .
End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . .
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112
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135
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Prerequisite knowledge
You will have come across proteins before and covered some of the many roles they
undertake in the human body. Proteins are essential in all biological systems because
they control most of the work going on in the cell. They are also important components
of cellular structures and, as enzymes and hormones, are involved in controlling cell
activities.
As described in the previous topic, proteins are specified by the DNA code of a gene;
anything which changes that code will almost always cause changes to the protein it
defines, with possibly fatal results.
Learning Objectives
By the end of this topic, you should be able to:
• state that proteins show a large variety of structures and shapes, resulting in a
wide range of functions;
• state that amino acids combine to form polypeptide chains;
• explain that these chains can fold into three-dimensional shapes;
• state that genetic disorders are caused by changes to genes or chromosomes;
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• state that single gene mutations involve the alteration of a DNA nucleotide
sequence;
• explain that the alteration of a DNA nucleotide sequence is a result of the
substitution, insertion or deletion of nucleotides;
• describe the altered protein structure and function in genetic disorders;
• understand the link between altered chromosomal structure and genetic disorders.
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6.1
Introduction
Proteins are essential in all biological systems. They control most of the work going on
in the cell and are an important component of cellular structures.
Protein groups
In this topic we will explore how the great diversity in the structure and function of
proteins arises.
Any changes to the sequence of amino acids in the protein have the potential to
cause very significant changes to its structure and function. Such changes arise from
mutations to DNA code which determines the amino acid sequence. In the second half
of the topic, we will examine the range of mutations which can affect individual genes
or whole chromosomes, and consider some examples of the medical conditions which
result.
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6.2
Proteins
Learning Objective
By the end of this section, you should be able to:
• describe the structure of proteins.
Proteins are organic compounds which are made up of the elements carbon, hydrogen,
oxygen, nitrogen, and sometimes other elements such as sulfur. They are complex
molecules with three dimensional structures which are unique for each protein.
Proteins are composed of building blocks called amino acids, of which twenty kinds
are commonly found in natural proteins. From these twenty, the cell can synthesise
thousands of different proteins. The amino acids are linked together by strong bonds,
called peptide bonds, to form polypeptide chains. The polypeptide chains can be
any size, several can be linked together, and they may have other groups such as
carbohydrates, phosphates and metal ions bound into them.
Proteins: Questions
Q1:
List the elements found in all proteins.
..........................................
Q2: Name the basic building blocks of proteins and state how many common forms of
them exist.
..........................................
Q3:
Name the type of strong bonds which link amino acids into chains.
..........................................
Q4:
What name is given to the chains of amino acids which make up proteins?
..........................................
Q5:
Explain the three sources of variation in structure and function of proteins.
..........................................
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Different levels of protein structure
Learning Objective
By the end of this section, you should be able to:
• state that function of an enzyme depends on its three-dimensional structure;
• explain that an area of the enzyme molecule is folded to form an 'active site';
• state that the active site is exposed so that it readily combines with its substrate;
• state that proteins are organised at four different structural levels;
• explain that:
1. primary structure consists of the sequence of amino acids;
2. secondary structure involves the folding of the polypeptide chains, which
are then held in place by weak hydrogen bonds;
3. tertiary structure involves further folding held in place by strong (disulphide)
bonds between amino acids;
4. quaternary structure is formed when two or more polypeptide chains
associate together to form the final 3D protein.
The function of a protein depends on its three-dimensional (3D) structure. Proteins are
organised at four different structural levels.
Different levels of protein structure: Visualisations
Primary and secondary level protein structure
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Tertiary protein structure
Quaternary protein structure
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Mature proteins fall into two main groups as shown below.
Mature protein types
..........................................
Structures of hexokinase and haemoglobin
The tertiary structure for the protein hexokinase is shown. Hexokinase is an enzyme
that phosphorylates (adds phosphate to) glucose molecules in the start of its oxidation.
You can see the substrate (glucose) at the active site of the upper enzyme. The function
of an enzyme depends on its three-dimensional structure. One part of the molecule has
been folded to form an 'active site', to which the substrate molecules temporarily become
attached. The active site is exposed so that it readily combines with its substrate(s).
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Structure of hexokinase
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An interactive molecular model of amylase, an enzyme that causes hydrolysis of
polysaccharides to glucose molecules (presented in 3D), is available in the online
materials at this point.
Haemoglobin is an example of a protein with a quaternary structure. It is found in red
blood cells and is responsible for carrying oxygen around the body. It is composed of
four chains of the protein globin, with two each of alpha- and beta-globin.
An interactive molecular model showing the structure of haemoglobin (presented in 3D)
is available in the online materials at this point. A diagram of the structure is shown
below.
The structure of haemoglobin
..........................................
Different levels of protein structure: Question
Q6: Complete the table using the phrases from the list.
Level of structure
Description
Primary:
Secondary:
Tertiary:
Quaternary:
Phrase list: held by strong (disulphide) bonds between amino acids, held by weak
hydrogen bonds between amino acids, the sequence of amino acids, two or more
polypeptide chains associated together.
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6.4
Functions of proteins
Learning Objective
By the end of this section, you should be able to:
• list the functions of proteins as enzymes, some hormones, muscular contraction,
transport, antibodies and structural support;
• give at least one example of each.
Proteins are fundamental to life on earth; they are the other side of the coin to nucleic
acids. Consequently, they are to be found involved in all the functions of the body, a
few of which are mentioned below. The most ubiquitous of functions, that of enzyme, is
covered in other topics. The table shows the variety of proteins found in human cells.
Protein group
Example
Function
structural
collagen, elastin
form part of cellular
structures
contractile
actin and myosin
allow muscle cells to
shorten
hormones
insulin
involved in controlling
blood sugar levels
receptors
insulin receptor
recognises insulin on the
cell membrane
transport proteins
glucose transporter
allows glucose to enter the
cell
defence proteins
antibodies
recognise molecules of
invading organisms
enzymes
lipase, amylase
cut, assemble, digest
molecules
Protein diversity
Hormones
Hormones are chemical messengers produced by special glands. They are secreted
into the bloodstream and are carried to their site of action where they play an important
role in regulating growth and metabolism. Many hormones, for example insulin and
ADH (anti-diuretic hormone), are proteins. Glucagon, oxytocin and prolactin are also
peptide hormones.
Muscles
Muscles are unique; their fibres are made up from a number of different proteins, for
example actin and myosin, arranged in a very specific manner. The appearance of
skeletal and cardiac muscle is characterised by the presence of a number of relatively
light and dark bands across the filaments.
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Transport
Many proteins transport substances around the body. Myoglobin and haemoglobin, for
example, are proteins involved in the transport of oxygen. Other proteins found in cell
membranes transport chemicals across the membrane.
Antibodies
Antibodies are proteins produced by certain white blood cells, called lymphocytes, and
are very important in preventing and fighting disease. This will be discussed in more
detail in a later topic.
Structural
Bones, tendons, ligaments, skin and hair are all mainly composed of structural proteins
such as collagen, elastin and keratin. Collagen is the most common protein in the
body, contributing up to a third of the total protein content. It is found in all connective
tissue, as well as being a major component of tendons, ligaments and cartilage. As its
name implies, elastin forms fibres which will contract to their original length after being
stretched. It is found in artery walls, the lungs, the bladder, skin and the elastic cartilage
of the nose and ears.
Keratin is a key structural protein, making up the much of the outer layers of the skin
as well as hair and nails. In the animal kingdom, it also is found in horns, feathers and
scales.
Interactive antibody structure
An interactivity showing the molecular model of an antibody (presented in 3D) is
available in the online materials at this point.
..........................................
Functions of proteins: Questions
Q7: Complete the table using the types of protein from the list.
20 min
Type of protein
Example
collagen, elastin
actin and myosin in muscle cells
insulin
insulin receptor in liver cells (forming part
of the structure of the plasma membrane)
transporter of glucose into the cell
immunoglobulins
lipase, pepsin, maltase
Type of proteins: contractile, defence proteins, enzymes, hormones, receptors,
structural, transport proteins.
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6.5
Mutations and genetic disorders
Learning Objective
By the end of this section, you should be able to:
• define the term mutagenic agent (mutagens);
• list examples of mutagens.
The occurrence of mutations
A mutation is a change in genetic material. An individual gene expressing a mutation is
referred to as a mutant. The tissue in which the mutation occurs is of great importance.
A mutation in a somatic cell or a tissue stem cell will only affect that cell or its daughter
cells; only if it affects the proto-oncogenes and leads to the development of a tumour
is this likely to be serious. However, a mutation that occurs in the germline cells of an
organism can be inherited from one generation to the next; if it is, it will be present in all
the cells of the offspring.
Mutation provides a source of variation in a species. Whereas the shuffling of genes
during meiosis provides new combinations of existing alleles, mutation is the only source
of new alleles and genes. Without mutation, evolution would not be possible.
Given that only a very small percentage of the total DNA in the nucleus is protein-coding,
most changes to the order of bases are unlikely to change a gene. However, changes
to start and stop codons can have profound effects, as can alterations to the promoters
which provide an attachment for RNA polymerase near a gene.
Types of mutation
There are different types of mutation: they might cause a change in the structure of a
chromosome or alter a single base in a gene. The effect of such changes can range
from none to being fatal. In humans, a huge range of inherited disorders are the result
of single gene mutations. Despite this, the fact is that most mutations have no effect on
survival.
The majority of mutations in single genes are recessive. A recessive mutation in one
allele of a gene cannot normally be observed in an individual because the mutated
allele is masked by the original allele that is still present. Only if two individuals who are
heterozygous for the mutant allele have offspring will there be a possibility (1 in 4) that
an affected individual will be produced.
Mutagenic agents
A mutagen is an agent that increases the rate of mutation. In any individual, there is a
low rate of spontaneous mutation caused, for example, by decay of the bonds in DNA or
errors in replication but, as cells have very efficient checking systems for errors in DNA
replication, any mistakes are normally very quickly rectified.
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A wide variety of chemicals (e.g. benzene, arsenic) and some types of radiation, such as
ultra-violet (UV) light and X-rays, are mutagenic. The major source of human exposure
to UV light is the sun. The two-fold increase in the incidence of skin cancer (which is
caused by mutations in skin cells) in Scotland over the last twenty years is thought to be
due, at least in part, to the rising number of holidays taken in sunny destinations such
as Spain. Another significant factor is the increased use of sun or tanning beds.
Treatment of bacteria with UV light: Question
A freshly grown culture of bacteria was diluted, and samples were spread out onto
agar plates to give approximately 300 bacterial cells per plate. Each plate was then
exposed for up to 3 minutes to a UV light source. Following UV exposure, the plates
were incubated for two days before counting the bacterial colonies that grew on them.
UV exposure
Q8: Describe the relationship between the number of bacteria that are killed and the
duration of exposure to UV light.
..........................................
Mutations and genetic disorders: Questions
Q9: What is a mutagen?
..........................................
Q10: Give two examples of mutagens.
..........................................
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6.6
Single gene mutations
Learning Objective
By the end of this section, you should be able to:
• state that gene mutations are due to the alteration of a DNA nucleotide
sequence;
• describe the effects of the different types of gene mutation: substitution, deletion
and insertion;
• state that a mutation that affects a single nucleotide is called a point mutation;
• explain that the substitution of one nucleotide will alter a maximum of one amino
acid (and may alter none);
• explain that insertions and deletions may alter from no amino acids to the whole
DNA sequence;
• state that nucleotide insertions or deletions can also result in an expansion of a
nucleotide sequence repeat;
• state that Huntingdon's disease is a condition associated with nucleotide repeat
expansion.
Single gene mutations are those which take place within the DNA of a single gene or
regions of the DNA which affect its expression. Such mutations in the sequence of bases
in DNA can be described according to the effect that they have on the base sequence.
Substitution - in this type of mutation, one nucleotide is replaced with another.
A substitution mutation
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Single-nucleotide substitutions include:
• missense, in which one nucleotide is changed in the DNA of the gene, and hence
the complementary mRNA codon is altered and may code for a different amino
acid;
• nonsense, when the change to the DNA triplet code causes an amino acid mRNA
codon to be replaced by a stop codon so that during translation no amino acid is
attached and the process stops;
• splice-site mutations, which either create or destroy codons for exon-intron
splicing.
A substitution mutation within a protein-coding gene may not always lead to a change in
the amino acid sequence of the encoded protein. In this case, the mutation is described
as silent. As there are 64 possible combinations of bases in a DNA base triplet, many
of the 20 amino acids are coded for by more than one combination. Consequently,
a change to a single nucleotide may leave the amino acid sequence, and hence the
protein, unaltered.
Insertion - this describes the addition of one or more nucleotides into the DNA.
An insertion mutation
If an insertion mutation occurs in a protein-coding gene, it usually has a dramatic effect
on the amino acid sequence of the encoded protein. The diagram below illustrates this
point.
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A frameshift mutation
In part (a), the DNA is represented as a repeat sequence of the base triplet GAT. The
DNA is transcribed into mRNA to produce a sequence of codons (in this case CUA,
encoding the amino acid leucine).
An insertion mutation - the addition of a C between the first and second base triplets in
the DNA - is shown in part (b). This mutation disturbs the pattern of the DNA triplets and
alters the way in which they are transcribed into mRNA (see part (c)).
All of the codons in the mRNA that appear after the mutation have changed, and, as a
result, the amino acids in the protein are also different. A type of mutation which has
this effect is called a frameshift mutation.
Deletion - this refers to the removal of one or more nucleotides from the DNA.
A deletion mutation
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As with an insertion mutation, a deletion mutation alters the pattern of base triplets in
the DNA and is also an example of a frameshift mutation.
Effects of gene mutations on amino acid sequences
A gene is a region of DNA which consists of a specific sequence of nucleotide bases
arranged in triplets. Every amino acid is coded for by one or more of these triplets.
Therefore, the sequence of bases determines the sequence in which amino acids are
joined together to form a polypeptide or protein. Thus, a gene codes for a particular
protein or polypeptide.
A gene mutation is a change in the sequence or type of nucleotide bases in a strand of
DNA. This can lead to a change in the sequence of amino acids, and thus to a change
in the protein which is synthesised in the ribosomes.
Sometimes a mutation causes only a minor change, perhaps affecting only one amino
acid. If this is the result of a substitution, then a maximum of one amino acid is altered
and the protein produced may be little altered.
Other forms of point mutation, i.e. insertions or deletions, can cause a major change
affecting the coding for many amino acids. Such a mutation is known as a frameshift
mutation and leads to a completely different protein being produced, which cannot carry
out the required function.
Each of the three types of gene mutation are described in the following diagrams, with
particular respect to changes in the amino acid sequences, which show the codons of
part of a DNA strand and the amino acids which are coded for by them. Remember that
the same amino acid can be coded for by more than one DNA triplet.
Q11:
This is a deletion mutation. The first 'A' nucleotide in the original DNA strand has been
removed (or deleted). Answer the following questions:
How many amino acids in this sequence have been changed?
Is a deletion mutation a frameshift mutation?
..........................................
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Q12:
This is an insertion mutation. The 'T' nucleotide has been inserted between the 'C' and
'A' nucleotides in the original DNA strand. Answer the following questions:
How many amino acids in this sequence have been changed?
Is a deletion mutation a frameshift mutation?
..........................................
Q13:
This is a substitution mutation. The 'T' nucleotide in the middle of the original DNA strand
has been replaced (or substituted) by a 'G' nucleotide. Answer the following questions:
How many amino acids in this sequence have been changed?
Is a substitution mutation a frameshift mutation?
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Point and frameshift mutations
Any mutation involving a single nucleotide is a point mutation. Substitutions, and
insertions or deletions involving a single nucleotide, are all point mutations. However,
their impact on the make-up of the ensuing polypeptide can be very different.
If the number of nucleotides inserted or deleted is not a multiple of three, the impact
of the mutation is likely to be seen in all the subsequent triplets and hence the amino
acids for which they code. The pattern of base triplets in the DNA that is transcribed into
mRNA is called the reading frame; an insertion that changes the reading frame is called
a frameshift mutation.
A frameshift mutation will also alter the first triplet that signals a stop codon, potentially
producing a much longer polypeptide than intended; equally, if it generates a new stop
codon, a much shorter polypeptide will result. Neither will function correctly.
Expansion of a nucleotide sequence repeat
Repeat sequences of DNA nucleotide triplets (trinucleotides) are common in genes,
where they code for a sequence of the same amino acid being repeated several times
in the polypeptide. Because there is a repeating set of bases, during DNA replication it
is possible for the copying process to 'lose the place', causing sections of the new DNA
strand to loop out and additional new trinucleotides to be added against the original
template strand. The result is known as a trinucleotide repeat expansion.
The occurrence of deletion and insertion point mutations can have the effect of triggering
this slippage in the DNA replication process, increasing the number of repeated amino
acids in the final polypeptide. The greater the number of repeats added, the greater
the effect of the mutation. A small expansion usually has no noticeable effect, but a
characteristic of this type of mutation is that the expansion tends to increase with each
generation. Eventually, the protein produced fails to function normally and symptoms
become apparent.
Two conditions resulting from trinucleotide repeat expansions are Huntingdon's disease
and Fragile X syndrome.
Huntingdon's disease is a trinucleotide repeat disorder of the HTT gene on chromosome
4. The repeated base sequence is CAG, which codes for glutamic acid. Individuals
with 26 or fewer CAG repeats are not affected, whereas 40 and over show progressive
degeneration of the nervous system, affecting muscle coordination and cognitive ability.
It typically only becomes noticeable in mid-adult life.
Fragile X syndrome is associated with the expansion of a sequence of repeated CGG
trinucleotides (coding for arginine) in one of the genes on the X-chromosome. It is the
most common single-gene cause of autism and inherited intellectual disability, although
there is a wide spectrum of associated effects. As with any condition controlled by
genes on the X-chromosome, it affects males at a higher rate than females (although
the difference is much less than might be expected).
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Gene mutations: Examples
The following provides examples of the three types of gene mutations.
30 min
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Gene mutations: Questions
Q14: A substitution mutation involves:
20 min
a) one nucleotide
b) two nucleotides
c) three nucleotides
..........................................
Q15: A mutation which results in a new stop codon:
a) missense
b) nonsense
c) splice-site
..........................................
Q16: A mutation which affects exons and introns:
a) missense
b) nonsense
c) splice-site
..........................................
Q17: A mutation which causes a change to one mRNA codon:
a) missense
b) nonsense
c) splice-site
..........................................
Q18: A substitution mutation will lead to a different amino acid:
a) always
b) sometimes
c) never
..........................................
Q19: An insertion mutation affects the triplets which come:
a) before it
b) after it
c) at any point
..........................................
Q20: A deletion mutation affects the triplets which come:
a) before it
b) after it
c) at any point
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Q21: The minimum number of amino acids affected by a point mutation:
a) 0
b) 1
c) 2
..........................................
Q22: A frameshift mutation can alter codons for:
a)
b)
c)
d)
amino acids
start
stop
all of these
..........................................
Complete the sequences of amino acids after mutation in each of the examples by
referring to the genetic code table.
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Q23:
..........................................
Q24:
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Q25:
..........................................
Q26: Why does a frameshift mutation arise only if the number of nucleotides inserted
or deleted is not a multiple of three?
..........................................
Q27: Describe a nucleotide sequence repeat (also known as a trinucleotide repeat
sequence)?
..........................................
Q28: What may trigger the expansion of a trinucleotide repeat sequence?
..........................................
Q29: Name a trinucleotide repeat disorder.
..........................................
..........................................
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6.7
133
The effect of mutations on the structure and function of
proteins
Learning Objective
By the end of this section, you should be able to:
• give examples of inherited diseases caused by different single gene mutations.
As mentioned in the previous section, single gene mutations can have effects ranging
from none to being fatal, depending on which proteins are affected and the impact of
that effect on their functioning.
The following is a list of inherited diseases caused by single gene mutations. The
syllabus does not specify any that should be known about in particular, but you should
be able to quote details about a couple of examples with different causes.
• Sickle-cell disease (mis-sense, autosomal codominant) - causes the production
of abnormal haemoglobin which makes the red cells take on a rigid, sickle shape,
causing obstructions in small vessels as well as a wide range of other serious
effects. The condition survives because people who are heterozygous for it
suffer much reduced symptoms if infected with malaria, compared to people
homozygous for the normal haemoglobin gene.
• Phenylketonuria (PKU) (mis-sense, autosomal recessive) - causes the production
of an inactive form of the enzyme (phenylalanine hydroxylase, PAH) which breaks
down the amino acid phenylalanine, resulting in a range of symptoms including
impaired brain function.
• Beta (β) thalassemia (splice-site, autosomal recessive) - causes reduced, or no,
production of one of the polypeptide chains (the beta chain) which make up
haemoglobin. The symptoms range from none to severe anaemia, depending
on where the mutation occurs on the gene.
• Duchenne muscular dystrophy (DMD) (nonsense, X-linked recessive) - caused
by mutation to the largest gene on the X chromosome, resulting in failure to
produce dystrophin, an important structural protein in muscle. It causes muscle
degeneration and early death.
• Tay-Sachs syndrome (frameshift insertion, autosomal recessive) - caused by
mutation to the hexoaminidase A, or HEXA, gene on chromosome 15, resulting
in progressive destruction of the central nervous system and early death.
• Cystic fibrosis (frameshift deletion, autosomal recessive) - caused by a mutation to
the gene CFTR which regulates the movement of chloride and sodium ions across
epithelial membranes. It results in the secretory glands of the lungs producing
very sticky mucus, poor growth and frequent chest infections. Untreated, it leads
to death in infancy.
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The effect of mutations on the structure and function of proteins:
Question
Q30: Complete the table using the causes from the list.
Inherited disease
Beta thalassemia
Cystic fibrosis
Cause
Duchenne muscular dystrophy (DMD)
Phenylketonuria (PKU)
Sickle-cell disease
Tay-Sachs syndrome
Cause list: frameshift deletion, autosomal recessive; frameshift insertion, autosomal
recessive; missense, codominant; missense, autosomal recessive; nonsense, X-linked
recessive; splice-site, autosomal recessive.
..........................................
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6.8
Chromosome structure mutations
135
Learning Objective
By the end of this section, you should be able to:
• state that chromosome structural mutations are due to the addition, removal or
reversal of segments of chromosome;
• explain that any mutation which alters gene expression is likely to have a serious
effect, e.g. by altering stop/start codons, exons, introns and promoters;
• describe the different forms of chromosome mutation, i.e. deletion, duplication,
translocation and inversion;
• give examples of conditions which result from each type of chromosome
mutation.
Chromosome mutations occur during cell division, both in mitosis or meiosis. They
can cause a change in either the number or structure of the chromosomes, leading to
numerical or structural abnormalities. Numerical abnormalities arise during meiosis I
or II and lead to conditions such as Down's syndrome (3 of chromosome 21), Edwards
syndrome (3 of chromosome 18), Turner syndrome (only one sex chromosome, X), and
Klinefelter syndrome (male with an extra X, so XXY).
Here we consider only (some of) the mutations which produce structural abnormalities.
• Deletion: a segment of a chromosome, and its associated genes, is lost during
crossover in meiosis I. The effect depends on exactly which sections of DNA are
lost and varies from no effect to being fatal. An example is Cri-du-chat syndrome,
which arises from deletion of 10-20% from the end of the short arm of chromosome
5. It is so named as the cry of affected children sounds like the meowing of
a cat because of problems with the nervous system and the larynx. There
are many other symptoms, including poor growth, severe speech and cognitive
delays, behavioural problems, and heart defects. A second example is Williams
syndrome, which is caused by a deletion of about 26 genes from chromosome
7. It results in a neurodevelopmental disorder, generally linked with an unusually
cheerful character and ease with strangers, delayed mental development coupled
with strong language skills, and cardiovascular problems.
• Duplication: a chromosome gains a segment that has been lost by another
chromosome, so extra genetic material is present as genes are repeated on the
same chromosome. It may arise in various ways, e.g. during crossing-over in
meiosis I or during DNA replication. As the extra genes are non-functional, they
can mutate without any deleterious effect on the individual, and thus provide a
valuable source of genetic variation for natural selection. Duplication of oncogenes
is a cause of many cancers. Being restricted to the somatic cells, such mutations
are not inherited, but have potentially fatal effects for the individual concerned.
Common cancers caused are breast, cervical, colorectal, cervical, ovarian and
gastric.
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• Inversion: a segment of a chromosome becomes reversed because it has broken
off, turned round, and become reattached during meiosis I. This may well go
unnoticed unless a gene is broken, in which case it may well be fatal. The most
common effect is a reduction in fertility.
• Translocation: segments of two (or more) chromosomes are exchanged or an
entire chromosome is attached to another. An example is chronic myeloid
leukaemia, which results from a gene from chromosome 22 fusing with a gene
on chromosome 9. It causes a cancer of stem cells in the bone marrow, which
leads to overproduction of certain white blood cells.
These mutations are illustrated in the following diagram, where the colours represent
sections of chromosome rather than individual genes.
Chromosome mutations
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Chromosome structure mutations: Question
Q31: Complete the sentences by matching the parts on the left with the parts on the
right.
Deletion and duplication are examples of
deletion.
Chromosome mutations occur during
reduced fertility.
Deletion is loss of a segment of chromosome
during
structural abnormality.
The condition Cri-du-Chat is caused by a
myeloid leukaemia.
Extra chromosomal material will be present
because of
oncogenes.
Duplication of these genes is the cause of many
cancers
meiosis I.
The most common effect of an inversion
mutation is
translocation.
A gene from chromosome 22 fusing a gene on
chromosome 9
duplication.
cell division.
A cancer of stem cells in the bone marrow
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6.9
Learning points
Summary
Proteins
• Proteins are composed of building blocks called amino acids.
• Amino acids are linked together by strong bonds, called peptide bonds, to
form polypeptide chains.
• The diversity of structure and function among proteins is due to:
– the arrangement of amino acids in polypeptide chains;
– the association, in many cases, of more than one polypeptide chain in
the mature protein;
– the inclusion of non-protein structures such as phosphates, metal ions
and carbohydrates.
• The 3D structure of the enzyme is vital to its function.
• An area of the enzyme is folded so that an 'active site' is exposed, which
readily combines with its substrate.
• Proteins are organised at four different structural levels:
– primary structure, consisting of the sequence of amino acids;
– secondary structure, involving the folding of the polypeptide chains,
which are then held in place by weak hydrogen bonds;
– tertiary structure, involving further folding held in place by strong bonds
between amino acids;
– quaternary structure, formed when two or more polypeptide chains
associate together to form the final three dimensional protein structure.
• Proteins have many functions, including: enzymes, some hormones,
muscular contraction, transport, antibodies and structural support.
• Many hormones, for example insulin and ADH, are proteins.
• Many proteins transport substances around the body.
• Myoglobin and haemoglobin, for example, are involved in the transport of
oxygen.
• Bones, tendons, ligaments, skin and hair are all mainly composed of
structural proteins such as collagen, elastin and keratin.
Gene mutations
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Summary Continued
• A gene mutation is a random change in the number or sequence of bases.
• A gene mutation results in no protein, or a faulty protein, being produced.
• Mutations that involve a single nucleotide are called point mutations.
• A substitution mutation occurs when one base is replaced by another.
• Single-nucleotide substitutions include:
– missense, when one amino acid codon replaces another, e.g. sickle
cell disease;
– nonsense, when an amino acid codon is replaced with a stop codon,
e.g. Duchenne muscular dystrophy (DMD);
– splice-site mutations, resulting in the creation or destruction of the
codons for exon-intron splicing, e.g. Beta (β) thalassemia.
• Substitution mutations do not usually have a large effect on the amino acid
sequence coded unless they involve stop or start codons, introns or exons.
• A deletion mutation involves the removal of a base.
• An insertion mutation involves the addition of a base.
• Insertion and deletion mutations are known as frameshift mutations
because they alter the entire sequence of amino acids coded after them.
• The effect of a frameshift mutation is greater the earlier in the gene that it
occurs as more codons, introns, exons and stop codons will be affected.
• Frameshift mutations are likely to result in no protein being produced, and
so may often be fatal.
• An example of a condition caused by a frameshift insertion mutation is TaySachs syndrome.
• An example of a condition caused by a frameshift deletion mutation is Cystic
Fibrosis;
• Nucleotide insertions or deletions can also result in an expansion of a
nucleotide sequence repeat - a condition associated with nucleotide repeat
expansion is Huntingdon's disease.
Chromosome mutations
• Chromosome structural mutations are due to the addition, removal or
reversal of segments of chromosome.
• Any mutation which alters gene expression is likely to have a serious effect,
e.g. by altering start and stop codons, exons, introns and promoters.
• Chromosome mutations can take the form of:
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Summary Continued
– a deletion, involving the loss of a segment of a chromosome;
– duplication, when a segment of chromosome is repeated;
– translocation, the rearrangement of chromosomal material involving
two or more chromosomes;
– inversion, the reversal of a segment of a chromosome.
• The substantial changes in chromosome mutations often make them lethal.
• Cri-du-chat syndrome is an example of deletion: a part of the short arm of
chromosome 5 is missing.
• Common cancers, e.g. breast, colorectal and gastric, are often associated
with duplication.
• Chronic myeloid leukaemia (CML) is an example of reciprocal translocation:
a gene from chromosome 22 is fused with a gene on chromosome 9.
• Inversion mutations usually have no noticeable effects, although they may
reduce fertility.
6.10
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of proteins and gene mutation before attempting
the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended response question: Proteins and gene mutation
A) Give an account of the structure and function of proteins. (8 marks)
B) Give an account of gene mutation. (9 marks)
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6.11
End of topic test
End of Topic 6 test
Q32: Complete the paragraph using the words from the list. Some of the words may be
used more than once each. (12 marks)
joined together by
to form
Proteins are composed of chains of
. They are held in complex
shapes by
bonds and
a
. In addition, several
chains may
interactions between individual
and metal ions may be
associate together and non-protein groups such as
bound into the final protein structure. Proteins have many functions including acting as
),
(e.g insulin), transport (e.g.
), and
enzymes (e.g.
).
structural components (e.g.
Word list: amino acids, carbohydrates, collagen, haemoglobin, hormones, hydrogen,
peptide bonds, phosphofructokinase, polypeptide, three-dimensional.
..........................................
Q33: Complete the table about gene mutations using the phrases from the list. (8
marks)
Term
Point mutation
Substitution
Missense
Nonsense
Splice-site
Definition
Deletion
Insertion
Frameshift
Phrase list: addition of a base, base sequence changed, new exon-intron codons, new
stop codon, one base replaced, one codon replaced, removal of a base, single base
involved.
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Q34: Complete the table about chromosome mutations by selecting the correct answers
from the options listed. (8 marks)
addition / removal /
reversal / all of these
Which type of mutation results from the loss of a segment of deletion / duplication /
chromosome?
translocation
exons / introns / stop
Which codons are potentially altered by structural mutations?
codons / all of these
deletion / duplication /
Which type of mutation causes Cri-du-Chat?
translocation /
inversion
deletion / duplication /
Which type of mutation is associated with breast cancer?
translocation /
inversion
deletion / duplication /
Which type of mutation causes chronic myeloid leukaemia?
translocation /
inversion
deletion / duplication /
Which type of mutation usually has not noticeable effect?
translocation /
inversion
What is the likely effect of substantial changes to the
no effect / moderate
effect / lethal effect
chromosome?
..........................................
Of what are chromosome structural mutations the result?
Q35: What is the name of the bonds which hold amino acids together? (1 mark)
..........................................
Q36: What is a polypeptide? (1 mark)
..........................................
Q37: Explain the function of the protein structure of hydrogen bonds. (1 mark)
..........................................
Q38: Name two types of non-protein groups that may be bound into the protein
structure. (2 marks)
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Topic 7
Human genomics
Contents
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Sequencing DNA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
144
144
7.3 Bioinformatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4 Systematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
146
148
7.5 Personalised medicine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
149
7.6 Amplification and detection of DNA sequences . . . . . . . . . . . . . . . . . .
7.7 DNA probes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
151
155
7.8 Medical and forensic applications . . . . . . . . . . . . . . . . . . . . . . . . . .
7.9 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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161
7.10 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.11 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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164
Learning Objectives
By the end of this topic, you should be able to:
• understand DNA sequencing and how the sequence of bases can be determined
for individual genes and entire genomes;
• describe the use of bioinformatics;
• explain the role of systematics;
• describe how the analysis of an individual's genome may lead to personalised
medicine through understanding the genetic component of risk of disease;
• describe the amplification and detection of DNA sequences;
• explain the Polymerase Chain Reaction (PCR);
• explain the use of DNA probes;
• give examples of medical and forensic applications through amplification and
detection of DNA sequences.
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7.1
Introduction
DNA can now be sequenced, synthesised and recombined in the laboratory. These
important advances in technology have led to the new sciences of bioinformatics
and genomics with their implications for understanding evolutionary relationships and
developing personalised medicine. The ability to replicate DNA in vitro has implications
for fundamental research and for medical and forensic applications, and provides
opportunities for informed discussion of the associated social, moral and ethical issues
facing society.
There are undoubtedly huge benefits to be gained by using DNA technology, but it also
raises many moral and ethical issues. Two techniques that are very important to the
Human Genome Project are described in detail here, namely the polymerase chain
reaction (PCR) and DNA probes.
7.2
Sequencing DNA
Learning Objective
By the end of this section, you should be able to:
• explain that sequencing DNA and the sequence of bases can be used to
determine individual genes and entire genomes.
The Human Genome Project began in October 1990 and was completed in 2003. The
project involved the discovery of all the estimated 20,000 human genes, making them
available for further studies. The project also led to the discovery of the complete
sequence of the 3 billion DNA sub-units (the bases in the human genome). In April
2003, the completion of the human DNA sequence coincided with the 50th anniversary
of Crick and Watson's description of the DNA structure in 1953.
Only about 3 percent of the human genome is actually used as the set of instructions
to make proteins. These regions are called coding regions. At present, little is known
functionally for most of the remaining 97 percent of the genome. These regions are
called non-coding regions, but that should not be interpreted as indicating that they
have no function, e.g. some act as introns, which can have a wide variety of functions,
some are gene promoters, which initiate DNA transcription, and the function of the rest
is the subject of much current research.
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Human proteins categorized by function, given both as number of encoding genes and
percentage of all genes
Single nucleotide polymorphisms, or SNPs (pronounced "snips"), are DNA sequence
variations that occur when a single nucleotide (A, C, G, or T) in the genome sequence
is altered. For example, a SNP might change the DNA sequence ACGGCTCA to
ATGGCTCA. Such substitution point mutations are the simplest form of mutation,
affecting at most a single amino acid in the protein for which a gene codes. Generally
they have little or no effect on the phenotype, but they are more frequent in non-coding
regions of the genome as these are not subjected to the pressures of natural selection.
A variation is considered a SNP when it occurs in at least 1% of the population. SNPs
make up about 90% of all human genetic variation and occur every 100 to 300 bases
along the 3-billion-base human genome. About 66% of SNPs involve the replacement
of cytosine (C) with thymine (T), which can occur in coding and non-coding regions of
the genome.
The genome of Craig Venter (American biologist and entrepreneur, and founder of the
Celera Genomics) was published in 2007. His genome contains 4.1 million variations,
3.2 million of which were SNPs. Today, the emphasis has shifted from sequencing the
entire genome to focussing solely on the exons, the so-called exome, as it is in these
coding regions that 85% of human inherited disorders are found.
Although SNPs have no effect on cell function, SNP maps help to identify the multiple
genes associated with complex ailments such as cancer, diabetes, vascular disease,
and some forms of mental illness. They are also very important in the process of DNA
fingerprinting.
Although SNPs do not usually cause diseases, they can help determine the likelihood
of someone developing a particular illness, e.g. apolipoprotein E, or ApoE, a gene
associated with Alzheimer's disease. It is also believed that SNPs may help in the
discovery and cataloguing of the unique sets of changes involved in different types of
cancers, and so offer new methods of combatting such diseases.
Scientists are trying to identify all the different SNPs in the human genome. They are
sequencing the genomes of a large number of people and then comparing the base
sequences to discover SNPs. The sequence data is being stored in computers that can
generate a single map of the human genome containing all possible SNPs.
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Sequencing DNA: Questions
Q1:
In terms of base pairs, how big is the human genome (in billions)?
..........................................
Q2:
What are SNPs?
..........................................
Q3: What do we call the sections of the human genome which are used as a set of
instructions to make proteins?
..........................................
7.3
Bioinformatics
Learning Objective
By the end of this section, you should be able to:
• explain that the enormous amounts of data produced by DNA and protein
sequencing can be managed and analysed using computer technology, an area
known as bioinformatics.
Bioinformatics is a field of study which draws on the techniques of computer science,
statistics, mathematics and engineering to process biological data. It uses software
tools to develop and improve methods for storing, retrieving, organising and analysing
information. In the context of this topic, computer technology has been used to identify
DNA sequences.
DNA and protein sequencing generate vast quantities of data which can only be
managed by the use of computer technology. This has been used to identify gene
sequences by searching for DNA sequences which are similar to known genes, start
codons, or sequences lacking stop codons. In the same way, base sequences can be
identified which correspond to the amino acid sequence of a particular protein.
A good example of bioinformatics in action is the '1000 Genome Project' which was
launched in 2008. Its aim was to catalogue human genetic variation by sequencing the
genomes of at least 1000 anonymous volunteers from a variety of ethnic groups from
around the world.
By 2011, over 1150 prokaryote and 310 eukaryote genomes had been published. At
least two-thirds of genes known to be involved with cancer have also been found in fruit
flies, Drosophila melanogaster. Finding out how these genes work in a much simpler
organism should provide researchers a better understanding of how they operate in
humans. Ultimately, this should give us a better understanding of how to control or
prevent the onset of certain types of cancer.
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The table shows a comparison of genome sizes between human and some other
organisms.
Estimated size
(base pairs)
Chromosome
number
Estimated gene
number
3 billion
46
≈20,000
Fruit fly
(Drosophila
melanogaster)
165 million
8
13,000
Mouse
(Mus musculus)
2.9 billion
40
≈25,000
Gorilla
2.9 billion
48
≈25,000
Roundworm
(Caenorhabditis
elegans)
97 million
12
19,000
Bacteria
(Escherichia coli)
4.6 million
1
3,200
Yeast
(Saccharomyces
cerevisiae)
12 million
32
6,000
Organism
Human
(Homo sapiens)
Comparison of genome sizes between different organisms
In the future, bioinformatics will allow us to predict:
• how genes are regulated;
• how the proteins will be synthesised;
• how they will fold and function in the cell;
• how they will interact with other molecules, be they natural or synthetic.
Bioinformatics: Questions
Q4: What techniques does bioinformatics draw on to make sense of biological data?
..........................................
Q5: What do computer programs use to identify gene sequences?
..........................................
Q6: How do computer programs identify protein sequences?
..........................................
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7.4
Systematics
Learning Objective
By the end of this section, you should be able to:
• explain that systematics, that is the comparison of human genome sequence
data (and genomes of other species), provides information on evolutionary
relationships and origins.
Systematics is the comparison of human genome sequence data. When the Human
Genome Project was being proposed, some argued that only the 3% of the genome
which encodes proteins should be sequenced. This would involve collecting mRNAs to
make cDNA libraries. Since some genes are only expressed at certain times during
development and in certain cell types, this approach could mean some genes would be
missed. Other sequences important for gene expression could also be missed, which
is why it was decided to sequence the entire genome in the Human Genome Project.
Another argument for sequencing the entire genome is that it provides the opportunity
to study chromosome structure and evolution.
In addition to mapping the human genome, the genomes of other species are also
being mapped. These include species important to biological research and agriculture,
such as the mouse, chicken, pig, cow, rice, wheat, Caenorhabditis elegans (nematode),
Drosophila melanogaster (fruit fly), Saccharomyces cerevisiae (yeast), Escherichia coli
(E. coli), and other prokaryotes. The genomes of some of these organisms, such as
E. coli, yeast, the nematode and the fruit fly, have now been completely mapped and
sequenced. These maps can be used to locate homologous genes in the human
genome and to help in determining gene function.
Comparative genome analysis is being used to find out more about evolution. In the
early 1950s, insulin became the first protein to have its amino acid sequence worked
out. The amino acid sequence for a protein in one species can be compared with that
from another species. The number of differences in an amino acid sequence can be
used to calculate the time since two species diverged from a common ancestor. If there
are lots of differences between the maps, it can be deduced that the species diverged a
longer time ago than if there are only a few differences. This type of information is used
alongside other methods for measuring the rate of evolution.
Gene maps can be used to predict gene order. If gene X is found next to gene Y and Z
in one species, the likelihood is that it will be found next to the same two genes in
another closely related species. Comparative maps will be used to find candidate
genes for phenotypes mapped in species as diverse as chicken and human.
Systematics: Questions
Q7: Comparative genome analysis has shown that some genes have been conserved
in evolution. Genes that have evolved from a common ancestor are described as:
a)
b)
c)
d)
Complementary
Linked
Homologous
Identical
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Q8: Using comparative genome analysis, how is it possible to work out if two species
are from the common ancestor?
..........................................
7.5
Personalised medicine
Learning Objective
By the end of this section, you should be able to:
• explain that the analysis of an individual's genome may lead to personalised
medicine as a result of understanding the genetic component of risk of disease.
The completion of the Human Genome Project in 2003 has led to the further
understanding of genetic contributions to human health; the field of genomic medicine
was born as a result. Genetics can play an increasingly important role in the diagnosis,
monitoring, and treatment of diseases. It has given us a better understanding of the
importance of distinguishing between neutral mutations and harmful mutations, and
also the complex nature of many diseases. Pharmacogenetics is the use of genome
information in the choice of effective drugs.
Pharmaceutical companies maintain that pharmacogenetics has several advantages in
the treatment of human diseases:
• it allows the production of more 'powerful' medicines through the discovery of
drugs based on the enzymes, proteins and RNA molecules associated with genes
and diseases;
• it allows the production of better and safer drugs more quickly than the standard
trial and error method;
• it gives more accurate methods of determining drug dosages;
• it allows for advanced screening for a particular disease;
• knowledge of their genetic code will allow individuals to make adequate lifestyle
and environmental changes at an early age, thus avoiding or lessening the severity
of a genetic disorder;
• pharmaceutical companies will be able to discover potential therapies more easily
using genome targets.
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However, there are challenges for pharmacogenetics:
• it is a complex process to find gene variations that affect drug response because
millions of SNPs must be identified and analysed to determine their involvement
(if any) in drug response;
• there are limited drug alternatives because only one or two approved drugs are
available for treatment of a particular condition;
• drug companies are reluctant to make multiple pharmacogenetics products;
• the introduction of multiple pharmacogenetic products to treat the same condition
will complicate the process of prescribing and dispensing drugs, and it will not be
cost effective to educate healthcare providers.
Personalised medicine: Question
Q9: List four advantages of using pharmacogenetics in the treatment of human
diseases. (4 marks)
..........................................
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7.6
151
Amplification and detection of DNA sequences
Learning Objective
By the end of this section, you should be able to:
• state that the polymerase chain reaction (PCR) is a technique for amplifying
(increasing in quantity) a sample of DNA;
• explain that PCR is useful in any area requiring larger samples of DNA, e.g.
forensics and archaeology;
• state that PCR is a four step process, involving:
1. Mixing together the components of the PCR (template DNA, primers,
nucleotides and the Taq DNA polymerase).
2. Heating (90 - 95 ◦ C) separates the DNA strands.
3. Annealing (54 ◦ C) is the binding of the primers, which mark the start and
end of the sequence to be amplified as well as preventing reattachment of
the separated DNA strands.
4. Extension (72◦ C) of the primers to complete the complementary strand is
carried out by heat stable DNA polymerase.
• describe how repeated cycles of heating and cooling amplify this region of DNA;
• explain that PCR uses the same mechanism for copying the DNA as our cells
use for DNA replication;
• state that PCR uses the enzyme DNA polymerase to replicate, or 'amplify', the
DNA strand.
For a range of purposes, it is essential to copy samples of DNA many times to provide
sufficient material for analysis. The main areas involved are:
• medical applications, e.g. genetic testing, tissue typing;
• the study of infectious disease, e.g.
tuberculosis (TB);
human immumodeficiency virus (HIV),
• forensic applications, e.g. genetic fingerprinting;
• research applications, e.g. analysis of DNA from extinct species, study of gene
expression.
The technique most frequently used is the Polymerase Chain Reaction (PCR), although
there are now a number of alternative approaches.
PCR can be used to amplify a desired DNA sequence of any origin (virus, bacteria, plant
or animal) millions of times in a matter of hours. It is especially useful because:
• it is highly specific;
• it is easily automated;
• it is capable of amplifying minute amounts of a sample.
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Before considering PCR, you should know the following:
• double-stranded DNA can be separated into single strands by heating to 90 - 95 ◦ C:
the DNA is then said to be 'denatured';
• under the correct conditions, complementary base sequences will bond or
'hybridise' with each other;
• 'primers' are short sequences of single-stranded DNA;
• DNA polymerases are enzymes required for the synthesis of DNA;
• Taq polymerase is a special DNA polymerase - it is an enzyme that is not
denatured by heating to 90 - 95 ◦ C;
• DNA polymerases can only synthesise new DNA in one direction, starting from the
3 end of a primer.
PCR depends on special heat-stable DNA polymerase enzymes from thermophilic
bacteria, which can withstand temperatures of 95 ◦ C and have an optimal temperature
for activity of 72◦ C.
The PCR process
The whole process can be divided into four stages.
1. The components of the PCR (template DNA, primers, nucleotides and the Taq
DNA polymerase) are mixed together.
2. Heating (90 - 95 ◦ C) - the first step in the PCR is to separate (denature) the doublestranded DNA containing the target DNA sequence for amplification.
3. Annealing (54 ◦ C) - the temperature is then reduced to allow the primers to bind
(by forming hydrogen bonds) to their complementary sequences on the separated
strands. Primers are short DNA sequences or oligonucleotides (each about 20
bases long) that bind at either side of the target sequence, one on each of the
complementary strands of the target DNA.
4. Extension (72◦ C) - Taq polymerase now adds nucleotides to the 3 ends of the
primers and extends them into new complementary strands.
The result is that the amount of DNA has now doubled.
Repeated heating and cooling cycles multiply the target DNA exponentially since each
new double strand separates to become two templates for further synthesis.
In about one hour, 20 PCR cycles can amplify the target DNA a million-fold.
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The Polymerase Chain Reaction (PCR) process: Steps
The following summarises the steps involved in the PCR process.
Stage 1: Mixing of the DNA template, primers, nucleotides and Taq polymerase.
Stage 2: Heating (90-94 ◦ C) - the double-stranded template DNA is split into single
strands.
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Stage 3: Annealing (54 ◦ C) - the primers anneal to complementary sites on the single
stranded template DNA.
Stage 4: Extension (72 ◦ C) - the DNA polymerase binds to the annealed primers and
extends them in a 5 to 3 direction.
Result: the amount of DNA has now doubled.
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Amplification and detection of DNA sequences: Question
Q10: Complete the paragraph using the words from the list.
of the components in the PCR process, three further stages follow.
After the
Heating to
which denatures the DNA by separating its two strands. The
, which allows
to bind to their
temperature is then dropped to
complementary sequences on the separated strands. Primers are short
sequences of about 20 bases that bind on either side of the
sequence. This
stage is known as
. The temperature is then raised again to
in the
stage. Taq
adds
to the 3 ends of the primers and extends
them into new complementary strands. Continued cycles of heating and cooling cause
cycles will amplify the
the DNA to be doubled repeatedly, so that in one hour,
target DNA more than 1 million times.
Word list: 20, 54◦ C, 72◦ C, 95◦ C, annealing, DNA, extension, mixing, nucleotides,
polymerase, primers, target.
..........................................
7.7
DNA probes
Learning Objective
By the end of this section, you should be able to:
• state that a DNA probe is a short length of DNA with a known nucleotide (base)
sequence, which has the ability to pair (bind) with the complementary sequence
on a DNA strand;
• state that DNA probes contain radioactive phosphorus, which can be detected
by autoradiography on X-ray film;
• state that DNA probes are used for locating specific pieces of DNA;
• explain that DNA probes can also be tagged with fluorescent markers which
allow detection of specific DNA sequences under the microscope.
To identify a gene (specific section of DNA), we can use a gene probe. A gene probe
is a single strand of DNA with a base sequence complementary to that of the gene
which needs to be located. It will therefore bind to it. The phosphate groups in the DNA
backbone of the manufactured gene probe are made with radioactive phosphorus, 32 P,
so the probe can be detected when it has bound to the gene, i.e. the radioactivity is
easy to detect using X-ray film. This process is called autoradiography.
A fluorescent label can be added instead of a radioactive label. The probe would then
be detected using laser scanning rather than autoradiography.
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DNA probes: Steps
The following summarises the steps involved in DNA probes.
..........................................
DNA probes: Questions
Q11: A DNA probe can be used to:
a)
b)
c)
d)
insert a gene into the bacterium E. coli.
construct a new gene.
identify a particular sequence of DNA.
produce DNA.
..........................................
Q12: Which of the following statements concerning gene probes is false?
a) A gene probe is a single strand of DNA.
b) The pentose sugar ring of a gene probe is labelled with radioactive 14 C.
c) A gene probe has a complementary base sequence to that of the gene which needs
to be located.
d) Gene probes can be located by autoradiography.
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7.8
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Medical and forensic applications
Learning Objective
By the end of this section, you should be able to:
• explain that the diagnosis of disease status or risk of disease onset can be
made by screening a cell sample from a patient for the presence or absence of
a particular sequence;
• state that this information can be used to diagnose a condition or describe the
likelihood of a condition developing;
• state that women with a family history of cancer can be tested for mutations in
the BRCA 1 and BRCA 2 genes;
• explain that DNA profiling allows the identification of individuals by means of
comparison of regions of the genome with highly variable numbers of repetitive
sequences of DNA;
• explain that if a tiny sample of blood is left at a crime scene, the DNA it contains
can be amplified using PCR;
• explain that a probe can then be created, which binds to one of these highly
variable sequences;
• explain that if a suspect is apprehended, their DNA can be tested with the same
probe to look for the sequence in question.
Medical application
Some human disorders are inherited and are frequently associated with congenital
abnormalities. Thousands of inherited disorders have been known for many years
because they run in families. Many of these are severe and are due to single gene
defects. Fortunately, most of these are rare disorders. Modern genetics provides a way
of accurately diagnosing who is at risk of inheriting some of these conditions.
It is possible to make a diagnosis of disease status or risk of disease onset. One
example is that of using a cell sample from a patient to screen for the presence or
absence of a particular sequence, e.g. a mutation in a gene. The information can be
used to diagnose a condition or describe the likelihood of a condition developing. For
example, women with a family history of cancer can be tested for mutations in the BRCA
1 (BReast CAncer gene one) and BRCA 2 (BReast CAncer gene two) genes. Mutations
in these genes can increase the risk of developing breast or ovarian cancer.
Forensic application
The DNA of every individual is unique (except for identical twins). We will discuss how
this variation is used by forensic scientists to identify individuals in criminal cases and
paternity disputes. Using PCR, the forensic scientist can amplify DNA extracted from
tiny fragments of skin, bloodstains or semen samples left by a criminal at the scene of
a crime. DNA can even be extracted from the saliva left on a stamp! This DNA can be
matched against that of any suspect.
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Dr. Alec Jeffries, a scientist at Leicester University, developed the original technique of
DNA profiling in 1984. By treating DNA in a certain way, Dr. Jeffries was able to produce
patterns similar to the bar codes we see on goods from the supermarket. The likelihood
of DNA from two individuals producing the same pattern is so low that it can be used to
identify individuals.
DNA profiling depends on the use of special DNA sequences called minisatellites.
If the number of repeat units within the minisatellite shows variation (polymorphism)
between individuals, it is called a variable number tandem repeat, or VNTR.
VNTRs can be detected by the changes they cause in the length of DNA fragments
following digestion by restriction enzymes. The patterns produced are called restriction
fragment length polymorphisms, or RFLPs. The steps involved in creating a DNA
fingerprint using a VNTR are summarised below.
1. Extraction of DNA.
2. Digestion of DNA using a restriction enzyme.
3. Separation of the DNA fragments by size using gel electrophoresis.
4. Blotting, or transfer, of the DNA fragments onto a filter.
5. Hybridisation, or binding, of a labelled copy of a chosen VNTR to the DNA on the
filter.
The accuracy of the technique is increased by using VNTRs that are multi-locus, or
present at more than one site in the genome. The DNA profiling patterns produced using
multi-locus VNTRs are much more complex than those produced using single-locus
VNTRs. The technique can also be made more reliable by doing several fingerprints
using different VNTRs.
The technique relies on amplification by PCR of small segments of DNA which carry a
particular microsatellite, or short tandem repeat (STR).
The steps involved in creating a DNA profile using STRs are summarised below.
1. Isolation of DNA sample.
2. Amplification of DNA by PCR using primers that bind to the unique DNA sequence
at either side of the STR.
3. Repetition of the process to amplify three to four different loci.
4. Labelling of the amplified DNA product with fluorescent dyes.
5. Analysis of the DNA product using an automated DNA sequencer (the chart
produced by the sequencer gives peaks indicating the length of each amplified
fragment).
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Medical and forensic applications: Questions
Q13: Complete the flow chart by using each of the six phrases to fill in the empty boxes.
30 min
..........................................
Read the passage below and then answer the questions that follow.
A terrible crime has taken place. A middle-aged man was murdered during a burglary.
Forensic scientists collected a blood sample from the victim and a sample from the
scene of the crime. Blood samples were also taken from four suspects identified by the
police. Forensic scientists have prepared the autoradiograph shown from DNA extracted
from the blood samples. The prosecution wishes to use it as evidence in a court case.
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DNA profiles
Q14: In the correct order, briefly list the six steps used to produce an autoradiogram.
The first and last steps have been done for you. The technique was developed by Ed
Southern at Edinburgh University.
1.
2.
3.
4.
5.
6.
DNA extracted from blood sample.
Autoradiography.
..........................................
Q15: Who should be charged with the crime?
a)
b)
c)
d)
Suspect 1
Suspect 2
Suspect 3
Suspect 4
..........................................
Q16: The prosecution wishes to use the forensic evidence in court, but is concerned
that the autoradiograph is not conclusive proof. Why?
..........................................
Q17: How could the accuracy of the analysis be improved?
..........................................
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7.9
Learning points
Summary
Sequencing DNA
• Sequencing DNA and the sequence of bases can be used to determine
individual genes and entire genomes.
• Bioinformatics - enormous amounts of data produced by DNA and protein
sequencing can be managed and analysed using computer technology.
• Systematics - comparison of human genome sequence data (and genomes
of other species) provides information on evolutionary relationships and
origins.
• The analysis of an individual's genome may lead to personalised medicine
through understanding the genetic component of risk of disease.
Polymerase chain reaction (PCR)
• This is a technique for amplifying (increasing in quantity) a sample of DNA.
• It is therefore useful in any area requiring larger samples of DNA, e.g.
forensics and archaeology.
• It is a four step process, involving mixing, heating, annealing and extension:
– Mixing together the components of the PCR (template DNA, primers,
nucleotides and the Taq DNA polymerase).
– Heating (90 - 95 ◦ C) separates the DNA strands.
– Annealing (54 ◦ C) is the binding of the primers, which mark the start
and end of the sequence to be amplified, as well as preventing reattachment of the separated DNA strands.
– Extension (72◦ C) (of the primers to complete the complementary
strand) is carried out by heat stable DNA polymerase.
• Repeated cycles of heating and cooling amplify this region of DNA.
• PCR uses the same mechanism for copying the DNA as our cells use for
DNA replication.
• PCR uses the enzyme DNA polymerase to replicate, or 'amplify', the DNA
strand.
DNA probes
• A DNA probe is a short length of DNA with a known nucleotide (base)
sequence, which has the ability to pair (bind) with the complementary
sequence on a DNA strand.
• It has radioactive phosphorus, which can be detected by autoradiography
on X-ray film.
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Summary Continued
• They are therefore useful for locating specific pieces of DNA.
• DNA probes can also be tagged with fluorescent markers which allow
detection of specific DNA sequences under the microscope.
Medical and forensic applications
• The diagnosis of disease status or risk of disease onset can be made by
screening a cell sample from a patient for the presence or absence of a
particular sequence.
• This information can be used to diagnose a condition or describe the
likelihood of a condition developing.
• Women with a family history of cancer can be tested for mutations in the
BRCA 1 and BRCA 2 genes.
• DNA profiling allows the identification of individuals through comparison of
regions of the genome with highly variable numbers of repetitive sequences
of DNA.
• If a tiny sample of blood is left at a crime scene, the DNA it contains can be
amplified using PCR.
• A probe can then be created, which binds to one of these highly variable
sequences.
• If a suspect is apprehended, their DNA can be tested with the same probe
to look for the sequence in question.
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7.10
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of the PCR process before attempting the
question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended response question: The PCR process
Give an account of the amplification of DNA sequences by the Polymerase Chain
Reaction (PCR). (8 marks)
..........................................
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7.11
End of topic test
End of Topic 7 test
Q18: Complete the sentences by matching the descriptions on the left with the terms
on the right. (8 marks)
The sequence of bases can be
determined for individual genes by
using
bioinformatics.
The use of computer technology to
identify DNA sequences is known as
polymerase chain reaction (PCR).
Comparison of the human genome
with genomes of other species is
known as
DNA probes.
An understanding of the genetic
component of the risk of disease is
required for
DNA profiling.
A technique for the amplification of
DNA in vitro is known as
systematics.
Short sequences of single-stranded
DNA that are complementary to
specific target sequences are known
as
DNA sequencing.
Short single stranded fragments of
DNA are known as
personalised medicine.
Comparison of highly variable
primers.
repetitive sequences of DNA is
known as
..........................................
Q19: Rearrange the following steps to describe the order in which the polymerase chain
reaction (PCR) takes place. (6 marks)
• Temperature of the reaction adjusted to 72 ◦ C.
• Temperature of the reaction adjusted to 55 ◦ C.
• Penetration of the double-stranded DNA.
• Temperature of the reaction adjusted to 95 ◦ C.
• Synthesis of DNA by the enzyme DNA polymerase.
• Annealing of the primers to the single stranded DNA.
..........................................
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Q20: The genome of an organism is: (1 mark)
a)
b)
c)
d)
the full complement of DNA possessed by the organism.
a description of the genes possessed by the organism.
a physical map of the chromosomes possessed by an organism.
a genetic map of the organism.
..........................................
Q21: The DNA polymerase used in the polymerase chain reaction (PCR) possesses a
particular characteristic that makes it ideally suited to the purpose. What is it? (1 mark)
a)
b)
c)
d)
It contains DNA ligase.
The enzyme is relatively stable at high temperatures.
The enzyme is very accurate at copying DNA from a template.
The enzyme synthesises DNA at a very rapid rate.
..........................................
occurs at 72 ◦ C. (pick from 'annealing',
Q22: During the process of PCR,
'exponential', 'extension' and 'heating') (1mark)
..........................................
Q23: During the process of PCR, DNA primers bind to DNA single strands during
. (pick from 'annealing', 'exponential', 'extension' and 'heating') (1mark)
..........................................
separates the DNA into single strands.
Q24: During the process of PCR,
(pick from 'annealing', 'exponential', 'extension' and 'heating') (1mark)
..........................................
Q25: Starting with a single molecule of DNA, the polymerase chain reaction was
allowed to go through three complete cycles. How many molecules of DNA would be
produced? (1 mark)
a)
b)
c)
d)
4
8
16
32
..........................................
Q26: Which of the following statements with regard to the primers involved in the
polymerase chain reaction is incorrect? (1 mark)
a)
b)
c)
d)
Primers allow DNA probes to attach to single stranded DNA.
Primers allow DNA polymerase to attach.
Primers prevent strands re-joining.
Primers start addition of nucleotides.
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Topic 8
Cell metabolism
Contents
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Types of metabolic pathway . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169
170
8.2.1 Anabolic and catabolic pathways . . . . . . . . . . . . . . . . . . . . . .
8.2.2 Reversible and irreversible pathways . . . . . . . . . . . . . . . . . . . .
170
171
8.3 Control of metabolic pathways - the action of enzymes . . . . . . . . . . . . . .
175
8.3.1 Control by regulation of enzyme production . . . . . . . . . . . . . . . .
8.3.2 Regulation of the activity of continuously present enzymes . . . . . . .
175
177
8.4 The role of the active site . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4.1 The induced fit model . . . . . . . . . . . . . . . . . . . . . . . . . . . .
178
178
8.4.2 Reversible reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4.3 Multi-enzyme complexes . . . . . . . . . . . . . . . . . . . . . . . . . .
182
182
8.5 Control of metabolic pathways through enzyme inhibition . . . . . . . . . . . .
8.5.1 Competitive and non-competitive enzyme inhibition . . . . . . . . . . . .
183
183
8.5.2 Feedback inhibition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.6 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
185
187
8.7 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
189
189
Prerequisite knowledge
You should already know that:
• enzymes are biological catalysts made by all living cells;
• enzymes are made of protein;
• enzymes lower the energy input required for chemical reactions;
• enzymes can speed up chemical reactions but remain unchanged;
• enzymes are involved in degradation and in synthesis reactions (you should be
able to give examples);
• the characteristic shape of an enzyme molecule is complementary to its substrate.
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You should also remember that chemical reactions in the human body involve the
breakdown or synthesis of molecules to provide energy and building blocks to be used
in other reactions.
Learning Objectives
By the end of this topic, you should be able to:
• describe the role of metabolic pathways in the cell;
• explain the role of enzymes in metabolic pathways;
• describe the control of metabolic pathways.
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Introduction
Metabolism is the term used to describe all the enzyme-catalysed reactions that take
place within the cells of the body. A metabolic pathway is a sequence of reactions
controlled by enzymes which change one metabolite into another. The metabolite
produced at each stage, the product, is used as the substrate for the next stage of
the pathway. For the cell, and the body of which it is part, to function coherently, these
pathways must be integrated and subject to control processes.
Metabolism: Question
Q1: Match each phrase on the left with the relevant term on the right. (4 marks)
All the chemical reactions of the body:
substrate
The chemical on which an enzyme acts:
metabolites
Chemical produced by an enzyme's action:
metabolism
product
Chemicals involved in cells reactions:
..........................................
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8.2
Types of metabolic pathway
Learning Objective
By the end of this section, you should be able to:
• state that anabolic pathways are biosynthetic, building up more complex
molecules from simpler ones and requiring the input of energy;
• state that catabolic pathways break down molecules, providing building blocks
for other reactions and usually releasing energy;
• explain that metabolic pathways may have reversible and irreversible steps;
• explain that metabolic pathways may contain alternative routes that by-pass
steps in the pathway.
8.2.1
Anabolic and catabolic pathways
There are two types of enzyme-catalysed reaction:
1. biosynthetic reactions (also known as synthesis reactions), which are also referred
to as anabolism;
2. breakdown reactions, which are also referred to as catabolism.
Anabolism
Anabolic reactions require the input of energy to proceed, which is usually obtained from
the breakdown of ATP to ADP and the transfer of the P i group (inorganic phosphate) to
the substrate molecule.
Example 1: glycogen synthetase is an enzyme that adds glucose residues to the storage
polysaccharide, glycogen. It adds a high-energy form of glucose, uridine-diphosphateglucose (UDP-glucose), to the existing chain of glucose residues that make up the
glycogen molecule.
glycogen
synthetase
UDPglucose
+
glycogen
(n
residues)
→
UDP
+
glycogen
(n+1
residues)
Example 2: in the replication of DNA, the enzyme DNA polymerase adds free
nucleotides to the 3 end of the newly-forming DNA strand. This results in elongation of
the strand in the 5 → 3 direction.
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Catabolism
Catabolic reactions involve the breakdown of complex molecules into simpler ones. This
may also involve the release of energy, which is used to add an additional inorganic
phosphate group to ADP to form ATP.
Example: glycogen phosphorylase catalyses the removal of glucose-1-phosphate from
glycogen in the liver and muscles, which can then be passed into the glycolysis pathway.
Anabolic and catabolic pathways: Question
Q2: Complete the table using the words from the list. (4 marks)
Anabolic
Catabolic
Biosynthesis
Requires energy input
ATP → ADP + Pi
DNA polymerase
Word list: ADP → Pi → ATP, Breakdown, Glycogen phosphorylase, May release energy.
..........................................
8.2.2
Reversible and irreversible pathways
Metabolic pathways are chains of enzyme catalysed reactions in which the product of
one reaction is the substrate for the next. In some cases these steps can be reversed,
whereas others are irreversible, so that once that reaction has taken place the metabolite
must pass onto a new stage in the pathway.
The following activity illustrates the start of the respiration process in all cells. It includes
the first five stages in the glycolysis pathway, known as the (energy) investment, or
preparatory, phase. This subject will be revisited in the next topic, but as it illustrates so
many of the key points of this section, it is useful to consider it here.
There are two important points to be aware of: firstly, although initially it looks very
complicated, the logic of the activity and diagram should become clear with explanation;
secondly, you will not asked to remember it.
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Glycolysis: Steps
The following provides a summary of the steps involved in the energy investment phase
of glycolysis.
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General points
• The molecular diagrams show the arrangement of the carbon, hydrogen and
oxygen atoms in each molecule, and the position of the phosphate groups which
become attached as the metabolites pass along the chain.
• The names of the enzymes are certainly more complicated that the pepsin and
amylase of earlier courses, but they are also much more informative once the code
is understood. An important example in this pathway are the kinases, a family of
enzymes which catalyse the addition to molecules of phosphate derived from the
breakdown of ATP to ADP, in a process called phosphorylation.
• Isomerase enzymes alter the chemical structure of a molecule without changing
the chemical formula.
• The first part of the enzyme name identifies the substrate, so phospho-fructokinase is a kinase enzyme which acts on fructose that has been phosphorylated.
• The product of the action of phosphofructokinase is fructose 1,6-biphosphate, i.e.
a fructose sugar molecule which has phosphate groups attached to carbons 1 and
6.
• The second part of glycolysis pathway, the pay-off phase, which releases energy
at some steps to make ATP, has as its final product two molecules of pyruvate for
every one of glucose used. This is covered in detail in the next topic.
Reversible and irreversible steps
• The steps in the chain which involve the addition of phosphate (phosphorylation)
also use up energy. These steps are irreversible.
• The step from glucose 6-phosphate to fructose 6-phosphate is reversible, with
the phosphoglucose isomerase enzyme re-arranging the atoms of the glucose to
form fructose, its isomer, without the addition of other atoms or groups. Under
normal conditions, the fructose 6-phosphate is used up in the next step, so its
concentration is low and presence of glucose 6-phosphate causes the reaction
to proceed from left to right. However, if the fructose 6-phospoate builds up, the
reaction will run the other way, changing it back to glucose 6-phosphate.
• The final step in this part of the chain sees the action of an aldolase enzyme
which forms, or in this case breaks, carbon-carbon bonds to split fructose 1,6biphosphate into two 3-carbon molecules which are isomers of each other.
• Of the isomers so-formed, only glyceraldehyde 3-phosphate is the substrate for the
next step so the dihydroxyacetone phosphate must be converted into it by another
isomerase enzyme.
• In times of excess lipid or protein energy sources, certain reactions in the
glycolysis pathway may run in reverse in order to produce glucose 6-phosphate
which is then used for storage as glycogen.
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Alternative routes
• An alternative route exists, called the pentose phosphate pathway, which takes
glucose 6-phosphate and eventually produces NADPH, important in reductive
processes such as the synthesis of fatty acids, and ribose, the 5-carbon sugar
in nucleic acids.
• For those metabolic pathways that have stages that cannot be reversed, there are
often alternative pathways that can overcome any blockage.
..........................................
Reversible and irreversible pathways: Questions
Q3: State two key differences between reversible and irreversible reactions in the
investment phase of glycolysis.
..........................................
Q4: State the conditions which will determine the direction in which a reversible
reaction will run.
..........................................
Q5:
Describe an alternative pathway to the investment phase of glycolysis.
..........................................
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8.3
175
Control of metabolic pathways - the action of enzymes
Learning Objective
By the end of this section, you should be able to:
• state that metabolic pathways may be controlled by the production, activation or
inhibition of particular enzymes;
• state that metabolic pathways may be controlled by the regulation of the activity
of key enzymes in the pathway;
• state that the rate of action of particular enzymes in a pathway may be regulated
by both intra- and extracellular signal molecules;
• state that the genes for some enzymes are continuously expressed, and so
these enzymes are always present in the cell;
• explain that control of such continuously present enzymes is achieved through
regulation of their rate of reaction;
• state that most metabolic reactions are reversible;
• explain that the direction of reversible pathways is determined by the presence
of the substrate and the relative concentration of the product of the reaction.
In all cells, there is the potential for a huge range of enzyme-assisted reactions to take
place. Control of the metabolic pathways of the cell is achieved by adjusting the activity
of the enzymes which catalyse the reactions in the pathway.
During the process of differentiation, many of the genes which code for enzymes will
have been inactivated. In addition, in any particular cell, only a subset of the total
possible enzymes for that cell will be required to be functioning at any one time. Of the
rest, some will not be produced at all as their genes have been 'switched off' temporarily;
others will be present in the cytoplasm, but either not activated or inhibited from acting.
8.3.1
Control by regulation of enzyme production
Inborn errors of metabolism
The ultimate examples of the control of metabolic pathways by the production of
enzymes are the inherited diseases caused by single gene mutations mentioned in Topic
6.6 - the so-called inborn errors of metabolism.
One such example is phenylketonuria (PKU), which results from a mutation to the gene
that codes for the conversion of the amino acid phenylalanine into tyrosine. The entire
metabolic pathway is summarised as:
phenylalanine → tyrosine → LDOPA → dopamine → noradrenaline → adrenaline
Phenylalanine is an 'essential' amino acid, i.e. one that cannot be synthesised in
the body and must be obtained from the diet. Tyrosine, on the other hand, can be
synthesised by the above pathway and so is 'non-essential'. The last three compounds
in the pathway are all important neurotransmitters.
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People with PKU lack the enzyme phenylalanine dehydroxylase and cannot carry out
the conversion of phenylalanine to tyrosine. They can obtain tyrosine from their diet in
the form of high protein foods and so can produce the neurotransmitters. However, a
build-up of phenylalanine in the body will lead to intellectual disabilities and seizures
unless its intake in proteins is controlled.
Regulation of gene expression
Some enzymes are only required occasionally or at specific times, and so are not
continuously present in the cytoplasm. Their genes are expressed in response to
particular circumstances, which are usually the presence of certain signal molecules
in the tissue fluid.
One example involves the hormone estrogen. It is most often associated with its
production in the ovaries, triggering the development of secondary sexual characteristics
in girls, and with the homeostatic control of the menstrual cycle. However, estrogen is
also produced in smaller quantities in the liver, adrenal glands and the breasts, and is
involved in the induction of a wide range of changes in the body. It is also produced in
men, where it contributes to the maturation of sperm.
The production of estrogen varies widely and it is the different levels of estrogen reaching
cells that influence their activity. Because of its hydrophobic nature, estrogen is able to
cross the phospholipid bilayer of the cell membrane and so enter all cells, but it will only
become effective in cells which contain estrogen receptor molecules in the cytoplasm.
The estrogen molecules attach to these receptors and then they move as a complex to
the nucleus. This complex binds to its specific target gene and activates its transcription
to form the enzymes which initiate the desired reactions.
Regulation by intra- and extracellular signal molecules
In cellular terms, signal molecules transmit information either between or within cells. In
the example given in the text above, estrogen carries information to the cells of the body,
causing them to initiate various cellular activities. Although the receptors for estrogen
are located within the cell, other receptors for signal molecules are found in the cell
membrane.
Such receptors span the membrane and set off a sequence of protein-protein
interactions within the cell when a signal molecule attaches to them. These proteins
act as intracellular signal molecules which ultimately trigger the expression of the target
gene in the nucleus. Such complex transduction pathways offer opportunities for:
• feedback, through the level of products affecting earlier steps on the pathway;
• signal amplification, where one signal molecule leads to many responses in the
cell; and
• interactions between pathways, whereby products from reactions in one pathway
are passed to others.
Examples are the cytokines, which are involved in the inflammatory response. Here,
the extracellar signal molecule is the cytokine molecule and the intracellular signal
molecules are the cascade of interacting proteins within the cell which influence gene
transcription.
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Control by regulation of enzyme production: Questions
Q6: What event causes the condition phenylketonuria?
..........................................
Q7: How does PKU affect the metabolic pathway leading to adrenaline?
..........................................
Q8: Explain the difference between extra- and intracellular signal molecules.
..........................................
Q9: Give one example of: (i) an extracellular signal molecule; (ii) an intracellular signal
molecule.
..........................................
8.3.2
Regulation of the activity of continuously present enzymes
Enzymes that are continuously present in the cytoplasm can be regulated in a number
of ways.
• As described in a later section, the concentration of the end-product of a reaction
may act as a negative feedback mechanism to control a reversible reaction.
• Post-translational modification may also regulate enzyme activity.
The
phosphorylation of glycogen synthase in response to the hormone insulin
determines whether glycogen is built up or broken down.
• The local environment may also affect the action of enzymes, e.g. changing pH.
• Localisation within the cell. Enzymes which carry out related functions are held
in groups in particular parts of the cell, e.g. in the Golgi apparatus or the
mitochondrion.
Enzyme regulation: Question
Q10: Match each phrase on the left with the relevant phrase on the right. (4 marks)
Concentration of end-product:
glycogen synthase
Post-translational modification:
negative feedback
Local environment:
Golgi apparatus
changing pH
Localisation within the cell:
..........................................
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8.4
The role of the active site
Learning Objective
By the end of this section, you should be able to:
• explain the induced fit concept of enzyme action;
• explain that the active site orientates reactants, lowers the activation energy of
the transition state, and releases products with a low affinity for the active site;
• explain that most metabolic reactions are reversible;
• explain that the presence of a substrate or the relative concentration of a product
will determine the direction of reversible reactions;
• explain that enzymes often act in groups or as multi-enzyme complexes.
8.4.1
The induced fit model
Molecules need a certain minimum energy before a reaction will take place. The energy
needed to allow a reaction to occur is called the activation energy. The presence of a
catalyst ensures that the initial energy requirement for a reaction is lowered, therefore
more molecules have this energy and the reaction takes place faster.
It is like rolling a boulder down a hill, but having to push it up a small hump first - this
initial push takes energy, but after that the boulder rolls on. Enzymes make the small
hump even smaller!
Activation energy hill analogy
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Activation energy
The action of an enzyme can be represented by the following equation:
E
enzyme
+
S
→
substrate
→
ES
EP
enzymeenzymesubstrate
product
complex
complex
→
E
enzyme
+
P
product
This emphasises that, for a reaction to take place, the substrate must collide with the
enzyme and that the reaction takes place on the surface of the enzyme. The activity of
enzymes depends on their flexible and dynamic shape. There is an affinity of substrate
molecules for the active site of an enzyme; once in position there are two proposed
theories of how enzymes work.
1. The lock and key hypothesis: the substrate and enzyme fit exactly.
2. The induced fit hypothesis: the substrate alters the enzyme's active site, causing
a reaction and allowing the products to leave.
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The lock and key hypothesis: Steps
5 min
The following provides a summary of the lock and key hypothesis - the substrate is the
exact fit of the enzyme's active site in the same way as a key fits exactly into a lock.
Lock and key hypothesis
..........................................
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The induced fit hypothesis: Steps
The following provides a summary of the induced fit hypothesis - the substrate molecule
does not fit exactly but instead induces a slight change in the shape of the active site to
allow the substrate molecule to fit perfectly. This conformational change in shape of the
active site facilitates the reaction.
Induced fit hypothesis
..........................................
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8.4.2
Reversible reactions
As mentioned above, in the glycolysis pathway there are some steps which can be
reversed and others which cannot. Those which are reversible do not involve the
addition of phosphate from the breakdown of ATP. For example, the step from glucose
6-phosphate to fructose 6-phosphate is reversible, the phosphoglucose isomerase
enzyme re-arranging the atoms of the glucose to form fructose, its isomer, without the
addition of other atoms or groups.
The direction of such reversible reactions is determined by two factors. In the example
mentioned in the text above, under normal conditions the fructose 6-phosphate product
is used up as the substrate in the next step so its concentration is low. The presence of
glucose 6-phosphate then causes the reaction to proceed from left to right. However, if
the concentration of fructose 6-phosphate builds up, the reaction will run the other way,
changing it back to glucose 6-phosphate.
Reversible reactions: Question
Q11: State the two factors which can determine the direction of a reversible reaction.
..........................................
8.4.3
Multi-enzyme complexes
Enzymes often act in groups, or as multi-enzyme complexes, which are a form of
quaternary protein structure. In these complexes, the proteins are linked by weak bonds,
and are held together by membrane proteins and the proteins of the cell cytoskeleton.
Where the enzymes in the complex form a sequence which is a metabolic pathway; they
are referred to as a metabolon.
These complexes are a vital part of most cellular processes, e.g. synthesis of DNA and
RNA, glycolysis, and the Krebs/Citric Acid/Tricarboxylic Acid (TCA) cycle. The presence
of the enzymes in close proximity to each other means that not only is the product of the
action of one enzyme immediately available as the substrate for the next enzyme in the
chain, but it is also presented so that it fits immediately into the active site.
Multi-enzyme complexes: Questions
Q12: Give an example of a multi-enzyme complex (metabolon).
..........................................
Q13: State two ways in which the metabolon makes such complex biochemical
pathways possible.
..........................................
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8.5
Control of metabolic pathways through enzyme
inhibition
183
Learning Objective
By the end of this section, you should be able to:
• describe non-competitive inhibition or stimulation of enzyme activity by the
binding of molecules that change the shape of the active site;
• describe how competitive inhibitors compete for the active site and resemble the
substrate;
• state that competitive inhibition is reversible;
• explain how competitive inhibition is reduced by increasing the substrate
concentration;
• explain how end products can inhibit an enzyme which catalyses a reaction
early in a pathway, and therefore control the pathway.
Genes for some enzymes are continuously expressed, leading to the permanent
presence of these enzymes within cells. These enzymes are always present in the cell
and their control involves regulation of their rate of reaction. There are many different
ways in which gene expression is controlled.
Enzymes may have their activity reduced or stopped, either temporarily or permanently,
by different forms of inhibition. Alternatively, they may be 'switched on' by enzyme
activators.
8.5.1
Competitive and non-competitive enzyme inhibition
An inhibitor is a substance which slows down, or stops, the activity of an enzyme. Two
different types of inhibition are competitive and non-competitive.
Competitive inhibition
In competitive inhibition, a molecule that is very close in shape to the true substrate
competes for the active site of the enzyme and effectively reduces the concentration
of the available enzyme. The inhibitor attaches temporarily to the active site, blocking
the substrate from attaching. This type of inhibition can be reversed by increasing the
concentration of the substrate. At high concentration, there is a greater likelihood of a
substrate molecule attaching to an active site than an inhibitor molecule.
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Competitive inhibition
The following provides a summary of the steps involved in the process of competitive
inhibition.
..........................................
Non-competitive inhibition
An inhibitor molecule binds to the enzyme at a different area (away from the active site),
changing the conformation (the shape) of the enzyme and altering the shape of the
active site. As a result, the catalytic efficiency of the enzyme is reduced; indeed, the
shape of the active site may be altered so much that the reaction cannot occur at all.
Increasing substrate concentration will not increase the rate of reaction.
The binding site of a non-competitive inhibitor molecule is called an allosteric site. On the
enzyme molecule, there will actually be allosteric sites for both inhibitors and activators.
This allows the enzyme to have its activity switched 'on' and 'off' by the presence of the
relevant inhibitor and activator molecules. In the same way that an inhibitor molecule
binding to an allosteric site alters the shape of the enzyme molecule so that the active
site no longer fits the substrate molecule, an activator molecule binding to an activator
site will cause the active site to alter so that it does fit the substrate.
Non-competitive inhibition: Steps
The following provides a summary of the steps involved in the process of noncompetitive inhibition.
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..........................................
In non-competitive inhibition the active site is not available to the substrate, thus the
rate of reaction is not generally influenced by the substrate concentration. Increasing
the substrate concentration will have little or no effect on enzyme activity because the
enzyme molecules have been denatured by the non-competitive inhibitor.
Competitive and non-competitive inhibition of enzymes: Question
Q14: Complete the table using the labels provided.
..........................................
8.5.2
Feedback inhibition
Metabolic pathways can be controlled by feedback inhibition. When the product of
a series of enzymatic reactions, for example an amino acid, starts to accumulate in
the cell, it can inhibit the action of the first enzyme involved in its synthesis. Further
action of the enzyme is halted. This is called feedback inhibition. As the level of the
end product drops, the inhibition is removed and the pathway will proceed. Feedback
inhibition prevents a wasteful build-up of product occurring.
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Feedback inhibition: Summary
The following provides a summary of the process of feedback inhibition.
..........................................
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An example of feedback inhibition in a synthetic enzyme pathway is the transamination
(change in the R group) of the amino acid threonine to isoleucine. Isoleucine, the endproduct in the sequence, can inhibit threonine deaminase, the initial enzyme in the
sequence. The following diagram illustrates this.
Feedback inhibition example
8.6
Learning points
Summary
Types of metabolic pathway
• Anabolic pathways are biosynthetic, building up more complex molecules
from simpler ones and requiring the input of energy.
• Catabolic pathways break down molecules, providing building blocks for
other reactions and usually releasing energy.
• Metabolic pathways may have reversible and irreversible steps.
• Metabolic pathways may contain alternative routes that by-pass steps in the
pathway.
Control of metabolic pathways - the action of enzymes
• Metabolic pathways may be controlled by the production, activation or
inhibition of particular enzymes.
• Metabolic pathways may be controlled by the regulation of the activity of key
enzymes in the pathway.
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Summary Continued
• The rate of action of particular enzymes in a pathway may be regulated by
both intra- and extracellular signal molecules.
• The genes for some enzymes are continuously expressed and so these
enzymes are always present in the cell.
• Control of such continuously present enzymes is achieved through
regulation of their rate of reaction.
• Most metabolic reactions are reversible.
• The direction of reversible pathways is determined by the presence of the
substrate and the relative concentration of the product of the reaction.
The role of the active site
• By the induced fit concept of enzyme action, the substrate entering the
active site makes the site change and fit more closely, causing a reaction
before allowing the products to leave.
• The active site orientates reactants, lowers the activation energy of the
transition state, and releases products with a low affinity for the active site.
• Most metabolic reactions are reversible.
• The presence of a substrate, or relative concentration of a product, will
determine the direction of reversible reactions.
• Enzymes often act in groups or as multi-enzyme complexes.
Control of metabolic pathways through enzyme inhibition
• The binding of molecules that change the shape of the active site may cause
non-competitive inhibition or stimulation of enzyme activity.
• Competitive inhibitors compete for the active site and resemble the
substrate.
• Competitive inhibition is reduced by increasing the substrate concentration.
• Competitive inhibition is reversible.
• Feedback control acts through end products, inhibiting an enzyme which
catalyses a reaction early in a metabolic pathway.
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8.7
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of the control of the action of enzymes before
attempting the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended response question: The control of the action of enzymes
Describe the control of the action of enzymes which are continuously present in the cell.
(10 marks)
..........................................
8.8
End of topic test
End of Topic 8 test
Q15: Complete the table using the words and phrases from the list. (6 marks)
Biosynthetic means
Anabolic pathways
Catabolic pathways
Steps which require energy are
Reactions which can run both ways are
Alternative routes
Word and phrase list: building up complex molecules, by-pass steps in a pathway,
irreversible, require energy input, reversible, usually release energy.
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..........................................
Q16: Complete the sentences by matching the parts on the left with the parts on the
right. (7 marks)
Regulation of the activity of key enzymes
enables control of
substrate.
Extracellular signal molecules regulate
continuously
expressed genes.
Protein-protein interactions are set off by
metabolic pathways.
Some enzymes are always present in cells as a
result of
product.
A reaction may be initiated by the presence of
hormones.
A reaction may be reversed by the build-up of
intracellular signal
molecules.
..........................................
Q17: Complete the paragraph using the words from the list. (10 marks)
site is the point on the enzyme molecule to which the
molecule(s)
The
temporarily binds. In many cases, the active site does not exactly fit the substrate and it
of the
is the substrate molecule itself which causes the necessary change in the
fit.
enzyme molecule. This is called
reactants, lowers
energy, and releases products with low
The active site
for the active site.
and the presence of a substrate or relative
Many metabolic reactions are
of a product will determine their direction.
.
Glycolysis is an example of metabolic pathway controlled by a metabolon, or
Word list: activation, active, affinity, concentration, conformation, induced, orientates,
multi-enzyme complex, reversible, substrate.
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..........................................
Q18: Complete the definitions by matching the statements on the left with the
descriptions and terms on the right. (6 marks)
Sites where activators may bind:
competitive.
Inhibitors bind to sites away from the active site:
allosteric.
Inhibitors which resemble substrate:
reversible.
Increase reduces competitive inhibition:
feedback control.
Competitive inhibition:
substrate
concentration.
End-products inhibit earlier enzymes:
non-competitive.
..........................................
Q19: Explain the difference between anabolic and catabolic pathways. (4 marks)
..........................................
Q20: What controls the direction of a reversible pathway? (2 marks)
..........................................
Q21: Explain the role of the active site. (3 marks)
..........................................
Q22: Explain feedback control of a metabolic pathway. (2 marks)
..........................................
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Topic 9
Cellular respiration I: Glycolysis
Contents
9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Summary of glucose breakdown . . . . . . . . . . . . . . . . . . . . . . . . . .
194
195
9.3 Roles of ATP in the cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3.1 What is ATP? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
197
197
9.3.2 ATP and energy transport . . . . . . . . . . . . . . . . . . . . . . . . . .
198
9.3.3 ATP and phosphorylation . . . . . . . . . . . . . . . . . . . . . . . . . .
9.4 Glycolysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
199
200
9.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.4.2 The energy investment stage . . . . . . . . . . . . . . . . . . . . . . . .
200
201
9.4.3 The energy pay-off stage . . . . . . . . . . . . . . . . . . . . . . . . . .
9.5 Regulation of the respiratory pathway . . . . . . . . . . . . . . . . . . . . . . .
203
206
9.5.1 Feedback inhibition and regulation of the glycolytic pathway . . . . . . .
9.5.2 Synchronisation of the rates of glycolysis and the citric acid cycle . . . .
206
206
9.6 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.7 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
208
210
9.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
210
Learning Objectives
By the end of this topic, you should be able to:
• give a general description of cellular respiration;
• describe the structure, formation and functions of ATP;
• describe the process of glycolysis;
• explain how the respiratory pathway is regulated.
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In this topic, some of the material included is not in the syllabus. The purpose of this
is to set the required content into its proper context so that it may be learnt through an
understanding of the processes, rather than memorised by means of repetition. The
questions in each section largely focus on the syllabus material. The learning points
and end-of-topic test address only the knowledge required by the syllabus.
9.1
Introduction
Learning Objective
By the end of this section, you should be able to:
• explain the central role of respiration in cell metabolism.
One of the laws which govern the behaviour of energy in the universe is the Second Law
of Thermodynamics, a simple interpretation of which is that complex structures will tend
to break down and release their energy to other simpler structures which are part of the
same system. Living things constantly oppose this tendency by building more complex
molecules from simpler ones. The energy for these anabolic reactions and for other
purposes, such as movement or keeping warm, is provided by respiration, which is a
catabolic process.
Respiration provides the fuel for all of the metabolic processes that require an energy
input in the cell. Furthermore, being a long chain of reactions, the metabolic pathways
of respiration are connected to many other pathways because intermediate metabolites
can be passed down alternative routes.
Introduction: Questions
Q1:
Why is respiration central to all metabolic processes?
..........................................
Q2: What name describes reactions which build up complex molecules from simpler
ones?
..........................................
Q3: What name describes reactions which break down complex molecules to simpler
ones?
..........................................
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9.2
195
Summary of glucose breakdown
Learning Objective
By the end of this section, you should be able to:
• describe in outline the process of glucose breakdown in respiration.
Respiration can be divided into several linked stages. The initial one is the breakdown
of glucose to pyruvate in the cytoplasm, which is called glycolysis (from 'glycose', an
earlier name for glucose, and 'lysis', meaning 'splitting'). It releases a relatively small
quantity of energy in the form of two molecules of ATP per glucose molecule and two
carrier molecules with hydrogen attached.
Subsequently, different pathways are available depending on whether oxygen is
available or not. In anaerobic conditions, in the absence of oxygen, pyruvate enters
an anaerobic pathway which has lactic acid as its final product. In aerobic conditions,
when oxygen is present, pyruvate enters the mitochondria where it reacts to form acetyl
Co-enzyme A and passes into a cycle of reactions known variously as:
• the Krebs Cycle, named after Hans Adolf Krebs, the Nobel Prize-winning
biochemist who first worked out the steps in the cycle;
• the Citric Acid Cycle, named after the molecule which is first formed after acetyl
Co-enzyme A reaches the cycle, the name being derived from the citrus fruit from
which it was first identified;
• the Tricarboxylic Acid (TCA) Cycle, using the modern name for citric acid.
In the course of the cycle, dehydrogenase enzymes remove hydrogen ions and electrons
to carriers which transport them to the final stage of aerobic respiration. Additionally,
carbon dioxide is released, ATP (or GTP) is generated, and the initial acceptor molecule
is re-established.
In aerobic conditions, the last stage of respiration is the cytochrome/hydrogen
transfer/electron transfer system, which is also located in the mitochondrion. The
alternative names stem from:
• cytochrome, a respiratory pigment involved as a hydrogen receptor in the system
(a related pigment is active in the chloroplast);
• hydrogen and (high energy) electrons which are passed along a chain of receptors,
giving out energy as they go. The energy released during aerobic respiration is
used to generate a total of 38 molecules of ATP from each glucose molecule;
• finally, the hydrogen is combined with oxygen and forms water, the other 'waste'
product of the pathway.
Glycolysis and the citric acid cycle are considered to be very ancient, probably having
evolved along with the earliest organisms. Interestingly, some direct descendents of
these ancient organisms, the bacteria-like Archaea which live in extreme conditions such
as hot springs in volcanic regions, are able to run the TCA cycle in reverse, generating
carbon compounds from carbon dioxide and water. They use hydrogen and sulfides to
supply the electrons to drive the process.
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Summary of glucose breakdown: Question
Q4:
Choose words from the list to complete the labelling of the respiration diagram.
Word list: 2, 36, acetyl CoA, citric acid cycle, electron transport system, glycolysis,
hydrogen, lactic acid, oxygen present, pyruvate.
..........................................
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9.3
Roles of ATP in the cell
197
Learning Objective
By the end of this section, you should be able to:
• describe the role of ATP in the transfer of energy within the cell;
• state that ATP is used to transfer energy to synthetic pathways and other cellular
processes where energy is required;
• explain the role of ATP in phosphorylation.
9.3.1
What is ATP?
ATP is adenosine triphosphate and, in common with nucleic acids, glycolysis and the
citric acid cycle, it is found in virtually all living cells. Typically, although only 250g of ATP
are present in the adult human body at any one time, in the course of a day the body will
make its own weight of the chemical as it is being constantly created and broken down.
ATP's principal function is very short-term and local: it transports energy from the site
of its release in the cell to areas where anabolic reactions are occurring in that cell. If
energy is to be stored in the body, it is put into the form of carbohydrate, as glycogen, or
fat; for transport between cells, it is in the form of glucose, lipids (triglycerides) or fatty
acids.
Chemically, ATP consists of three parts: the 5-carbon sugar ribose; the base adenine,
as found in DNA and RNA, attached to the 1 carbon atom of the ribose; and three
phosphate groups, attached in a chain to the 5 carbon of the ribose. Adenosine is the
term for the adenine base attached to the ribose sugar. In simple form, the molecular
structure of ATP is:
adenine - ribose - phosphate - phosphate - phosphate
A more accurate depiction of this structure is as follows:
The structure of ATP
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A little background information
Adenosine monophosphate (AMP) is also present in the cell, both as the adenine
nucleotide in nucleic acids and, in a cyclic form (cAMP), as an important messenger
in intracellular signalling.
Of the other nitrogenous bases found in nucleic acids, guanine has the same basic tworing structure as adenine; together they are called purines. Cytosine and thymine/uracil
are single-ring molecules and called pyrimidines. Guanine is also found acting as part
of an energy transport molecule as guanosine triphosphate (GTP) but, unlike ATP, it
is only involved in protein synthesis and in gluconeogenesis, the formation of glucose
from non-carbohydrate substrates (vital to maintaining constant blood sugar levels in
humans, but also found in all organisms). GTP is also essential to signal transduction,
in which extracellular signal molecules attach to receptors on the cell membrane and
secondary messengers amplify the signal into the cell.
What is ATP: Question
Q5:
In simple terms, describe the structure of the ATP molecule.
..........................................
9.3.2
ATP and energy transport
Because the third phosphate group in ATP is furthest from the ribose sugar, the
formation and breaking of its bond with the second phosphate involves more energy
than either of the bonds of the other two phosphates. In the watery environment of the
cell, ATP is relatively unstable, and decomposes to ADP (adenosine diphosphate) and
Pi (inorganic phosphate).
It is the ready interconversion of ATP and ADP (adenosine diphosphate) + P i (inorganic
phosphate) which allows ATP to act as the cell's principal energy transporter. ATP is
reconstituted by phosphorylation, i.e. the enzyme-assisted addition of a phosphate
group, which reforms the third phosphate bond.
The interconversion of ATP and ADP
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TOPIC 9. CELLULAR RESPIRATION I: GLYCOLYSIS
ATP is generated both by glycolysis and from the electron transport/cytochrome system.
It then travels to those areas of the cell which require energy to power their activity; this
might be anabolic processes such as protein synthesis or movement in muscle cells, or
signal transduction through the cell membrane.
ATP and energy transport: Question
Q6: Explain why ATP is used as an energy transporter in the cell.
..........................................
9.3.3
ATP and phosphorylation
Phosphorylation is the addition of a phosphate group to a molecule. The formation
of ATP from ADP, in glycolysis and the electron transport chain, is therefore a
phosphorylation. However, when ATP is hydrolysed to ADP and P i , the energy released
can be used to add the phosphate group to another molecule.
The first step in the glycolysis pathway involves such a phosphorylation. The enzyme
hexokinase transfers the phosphate group from ATP to the 6 carbon of the glucose
molecule.
The first step in the glycolysis pathway
The effect of this is twofold. Firstly, it prepares the glucose for the next step in the chain
of reactions. Secondly, it reduces the concentration of glucose in the cell and, at the
same time, prevents its loss from the cell as glucose 6-phosphate is much less able to
cross the cell membrane.
Phosphorylation is a common means of switching enzymes 'on' and 'off' as the addition
of the phosphate group alters their structure and activity. This is a form of posttranslational modification of the proteins.
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ATP and phosphorylation: Question
Q7:
What is meant by the term phosphorylation?
..........................................
Q8:
When glucose is phosphorylated, what is formed?
..........................................
Q9: What effect does post-translational modification by phosphorylation have on
enzymes?
..........................................
9.4
Glycolysis
Learning Objective
By the end of this section, you should be able to:
• summarise the main points of the glycolysis pathway;
• explain the key aspects of the energy investment and pay-off stages;
• explain the role of phosphorylation in glycolysis;
• describe the alternative pathways connected to the glycolytic pathway;
• describe the role of phosphofructokinase;
• describe the progression pathways in the presence and absence of oxygen;
• describe the ways in which the glycolytic pathway is regulated.
9.4.1
Introduction
Glycolysis takes place in the cytoplasm. It is a metabolic pathway involving ten steps,
each of which is enzyme-assisted. Such a sequence would be impossible if these
enzymes were randomly distributed in the cytoplasm; in fact, they are held in multienzyme complexes that are attached to the structural proteins of the cytoskeleton so
that the product of each step is readily available as the substrate for the next enzyme in
the chain.
In summary, during glycolysis a 6-carbon molecule of glucose is split into two molecules
of 3-carbon pyruvate. In the course of the pathway, two molecules of ATP are used up
in the initial 'energy investment' stage and four are released in the later 'energy pay-off'
stage. Two hydrogen ions are also released and, in the presence of oxygen, they are
transported by carrier molecules (NADH) to the cytochrome system. In the absence of
oxygen, they are used to turn pyruvate into lactic acid.
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Glycolysis: Questions
Q10: Where in the cell does glycolysis take place?
..........................................
Q11: How does the arrangement of the enzymes of glycolysis in a multi-enzyme
complex improve their efficiency?
..........................................
Q12: What are the final products when one molecule of glucose is passed along the
glycolytic pathway?
..........................................
9.4.2
The energy investment stage
The first part of glycolysis, the energy investment stage, is so-called because it uses
up two molecules of ATP. It starts with a glucose molecule as the first substrate and
finishes with two molecules of glyceraldehyde 3-phosphate as its final product. The
6-carbon glucose is split into two 3-carbon molecules.
During this stage, two phosphorylations take place. In each of these, a molecule of ATP
is converted to ADP; the phosphate group which is released is added to the substrate
molecule. The first occurs when glucose is converted to glucose 6-phosphate by the
enzyme hexokinase. This step is irreversible, but the glucose 6-phosphate can be used
in a variety of metabolic pathways:
• it can progress to the next step in the glycolysis pathway;
• it can enter the pentose phosphate pathway, generating the reducing agent
NADPH, which supplies the hydrogen necessary for several reactions, such as
fatty acid synthesis, and the precursor of nucleotides for DNA synthesis in the
form of ribose phosphate;
• it can be converted to glycogen for storage.
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The energy investment stage
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The second phosphorylation is the conversion of fructose 6-phosphate to fructose 1,6biphosphate, which adds a second phosphate at the 1’carbon position. This step is
catalysed by the enzyme phosphofructokinase, is irreversible, and can only lead on to
the glycolytic pathway.
The energy investment stage: Questions
Q13: Why is the energy investment stage so-called?
..........................................
Q14: Apart from energy, how does ATP contribute to the phosphorylation reactions in
this stage during glycolysis?
..........................................
Q15: Comparing the phosphorylation reactions to the other reactions in the pathway,
suggest what makes the phosphorylations irreversible.
..........................................
9.4.3
The energy pay-off stage
During this stage, a total of 4 ATPs are generated, two from the processing of each
molecule of glyceraldehyde 6-phosphate. This gives a net gain of two molecules of ATP
for each glucose molecule, as this stage runs twice for each glucose molecule entering
the pathway.
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The energy pay-off stage
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Like the energy investment stage, the pay-off stage consists of several enzyme-assisted
reactions, some of which are reversible and others which are not. Two of the reactions
are phosphorylations, catalysed by kinase enzymes, but, in this case, the phosphate
group is being transferred to ADP, forming ATP, rather than from ATP to a substrate
molecule. This is known as substrate-level phosphorylation because the phosphate
group comes directly from the substrate (as opposed to the oxidative phosphorylation
that takes place in the cytochrome system).
The first step in the pay-off stage is catalysed by a dehydrogenase enzyme, and the
hydrogen released is attached to the coenzyme carrier molecule NAD + to form NADH.
In the presence of oxygen, this enters the electron transport chain of the mitochondrion
and generates ATP. In the absence of oxygen, it is used to convert pyruvate to lactic acid
in the cytoplasm.
The final product of this stage is pyruvate, which has links to several metabolic pathways:
• in the presence of oxygen, pyruvate can enter a mitochondriom and be converted
to an acetyl group attached to Co-enzyme A (acetyl-coA) by the loss of carbon
dioxide and hydrogen. Acetyl-coA then enters the citric acid cycle;
• in the absence of oxygen, it can be converted by fermentation to lactic acid with
the addition of the hydrogen released in glycolysis. No further ATP is generated;
• it can be converted to carbohydrates, by the process of gluconeogenesis, fatty
acids, and the amino acid alanine.
The energy investment stage: Questions
Q16: How many ATP molecules are generated from each pyruvate molecule?
..........................................
Q17: What is the net yield of ATP molecules from each glucose molecule?
..........................................
Q18: What type of reaction produces ATP?
..........................................
Q19: Why is this type of ATP formation called substrate-level phosphorylation?
..........................................
Q20: In the presence of oxygen, how may further energy be released from pyruvate?
..........................................
Q21: In the absence of oxygen, how may pyruvate be removed from the cell?
..........................................
Q22: Apart from glucose and lactic acid, name two other substances into which
pyruvate may be converted.
..........................................
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9.5
Regulation of the respiratory pathway
Learning Objective
By the end of this section, you should be able to:
• explain what is meant by 'feedback inhibition';
• describe the regulation of the glycolysis pathway;
• explain how the rates of glycolysis and the citric acid cycle are synchronised.
9.5.1
Feedback inhibition and regulation of the glycolytic pathway
Feedback inhibition in a metabolic pathway occurs when an enzyme's activity is reduced
by an increase in the concentration of a product from a later step in the pathway.
This ensures that the potentially damaging build-up of metabolites is avoided, and that
resources, e.g. energy in the form of ATP, are not expended unnecessarily.
In the glycolytic pathway, there is a good example of feedback inhibition. In the energy
investment stage, the enzyme phosphofructokinase catalyses the phosphorylation of
fructose 6-phosphate to fructose 1,6-biphosphate. This step is irreversible and can
only lead on to the energy pay-off stage, serving as a key point to control the whole
respiration process.
Phosphofructokinase is inhibited by ATP, which binds to a part of the enzyme other
than the active site (an allosteric site) and reduces its affinity for its substrate,
fructose 6-phosphate. Consequently, if the ATP concentration rises in the cell,
phosphofructokinase will become less active and the production of ATP from both
glycolysis and the cytochrome system will be reduced. As the concentration of ATP
falls, the concentration of AMP (adenosine monophosphate) rises; as this reverses the
effect of ATP on the enzyme, the pathway will be activated again.
Feedback inhibition and regulation of the glycolytic pathway: Questions
Q23: What is meant by the term 'feedback inhibition' in a metabolic pathway?
..........................................
Q24: Of what advantage to the cell is feedback inhibition?
..........................................
Q25: Explain the inhibition of phosphofructokinase by ATP.
..........................................
9.5.2
Synchronisation of the rates of glycolysis and the citric acid cycle
The energy pay-off stage also includes steps which may be regulated. Both of the
phosphorylation reactions which generate ATP are sensitive to the concentration of ADP
in the cytoplasm. Low levels of ADP, which would indicate that ATP is not being used
up and is accumulating, inevitably mean that these steps will be held up, controlling the
production of ATP from glycolysis and the supply of pyruvate to the mitochondrion.
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This, together with the feedback inhibition of phosphofructokinase by ATP, ensures that
glycolysis is only supplying pyruvate to the mitochondria at a rate which is suited to the
ATP availability within the cell, and hence the cell's rate of use of energy. When the
concentration of ATP is high, resources will be conserved; when the ATP concentration
is low, indicating high activity in the cell, pyruvate will be made available at the maximum
rate.
It has been suggested that phosphofructokinase is also inhibited by high citrate levels
(citrate being a salt of citric acid). Such conditions would indicate that the citric acid
cycle was absorbing the Acetyl CoA produced from pyruvate but not using it up, and
so it was diffusing out into the cytoplasm. This would then inhibit phosphofructokinase,
slowing the glycolysis pathway and the production of pyruvate.
However, this is open to debate. While it can be demonstrated in the lab (in vitro) that
high levels of citrate will indeed inhibit phosphofructokinase, such high levels have not
been found to occur in living cells (in vivo). In the same way, in vivo ATP concentrations
do not seem to vary sufficiently for ATP to act as an effective inhibitor.
These are good examples of the need to evaluate findings from experimental situations
in the light of the conditions which exist within living organisms, and the difficulty of
accurately simulating the conditions in the cell during experiments.
Synchronisation of the rates of glycolysis and the citric acid cycle:
Questions
Q26: How does the cell's use of energy influence the pay-off stage of glycolysis?
..........................................
Q27: Why is it important that the rates of glycolysis and the citric acid cycle are
synchronised?
..........................................
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9.6
Learning points
Summary
Respiration and metabolism
• Respiration provides the energy for all of the cell's anabolic reactions,
movement and heat release.
• The respiration pathways provide intermediate compounds which may be
used in several other metabolic pathways.
• Carbon dioxide and water are released as waste products.
Summary of glucose breakdown
• During the respiration process, glucose is broken down.
• Hydrogen ions and electrons are removed by dehydrogenase enzymes.
• ATP is formed and made available to the cell.
Roles of ATP
• ATP is adenosine triphosphate.
• ATP is readily hydrolysed to ADP and P i (adenosine diphosphate and
inorganic phosphate), releasing energy.
• ADP can be phosphorylated to ATP by the addition of P i and the necessary
input of energy.
• ATP is generated through the glycolysis pathway and the cytochrome
system.
• ATP travels from where it is formed to those areas of the cell which need
energy.
• ATP provides energy for anabolic processes (e.g. protein synthesis),
movement, and signal transduction across the cell membrane.
• Phosphorylation is the addition of a phosphate group to a molecule.
• The phosphorylation of glucose to glucose 1-phosphate at the first step of
glycolysis uses the phosphate released from the hydrolysis of ATP to ADP
and Pi .
Glycolysis
• Glycolysis takes place in the cytoplasm.
• In summary, one molecule of glucose yields two molecules of pyruvate, two
molecules of ATP and two hydrogen ions as 2NADH.
• Glycolysis consists of two stages: an energy investment stage followed by
an energy pay-off stage.
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Summary Continued
• The energy investment stage includes two phosphorylations, both of which
are irreversible.
• The first phosphorylation adds phosphate to glucose to form glucose 6phosphate.
• Glucose 6-phosphate can continue into several different metabolic
pathways.
• The second phosphorylation involves the enzyme phosphofructokinase; its
product can only progress further down the glycolytic pathway.
• The energy pay-off stage generates two molecules of ATP per pyruvate
molecule.
• These ATP molecules are formed by the transfer of phosphate directly
from the substrate to ADP, hence they are called substrate-level
phosphorylations.
• The pyruvate formed from the energy pay-off stage can be used as a
metabolite in several metabolic pathways.
• In the presence of oxygen, pyruvate may enter the mitochondria and the
citric acid cycle.
• In the absence of oxygen, pyruvate may be converted to lactic acid.
Regulation of the respiratory pathway
• Feedback inhibition occurs when an enzyme's activity is reduced by an
increase in the concentration of a product from a later step in the pathway.
• Phosphofructokinase, in the energy investment stage of glycolysis, is
inhibited by high levels of ATP.
• This will reduce the production of pyruvate, and therefore of ATP, when the
cell's need for energy is low.
• Phosphofructokinase may also be inhibited by high citrate levels in the cell.
• Glycolysis and the citric acid cycle are synchronised by the feedback
inhibition of phosphofructokinase in the energy investment stage, and the
supply of ADP to the phosphorylation reactions in the energy pay-off stage.
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9.7
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of feedback inhibition before attempting the
question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended answer question: Feedback inhibition
10 min
State what is meant by the term 'feedback inhibition', give an example of it, and explain
its advantage to the cell. (7 marks)
..........................................
9.8
End of topic test
End of Topic 9 test
Q28: Complete the paragraph using the words from the list. (14 marks)
for all the cell's
reactions,
, and
Respiration provides
release. In addition,
compounds may be used in many other
pathways.
respiration,
is broken down,
is used up, and carbon
During
are released.
ions and
are removed by
dioxide and
enzymes. As a result,
is formed and made available to the cell.
Word list: ATP, aerobic, anabolic, dehydrogenase, electrons, energy, glucose, heat,
hydrogen, intermediate, metabolic, movement, oxygen, water.
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..........................................
Q29: Complete the table using the phrases from the list. (5 marks)
ATP is hydrolysed to
ATP is formed by
Glycolysis and the cytochrome system
Energy for anabolic processes is
Glucose to glucose 1-phosphate occurs
by
Phrase list: ADP and Pi , generate ATP, phosphorylation, provided by ATP, the addition
of Pi to ADP.
..........................................
Q30: Complete the sentences by matching the parts on the left with the parts on the
right. (10 marks)
Glycolysis takes place
forms two molecules of ATP per
pyruvate.
Glycolysis of one glucose molecule
uses up two molecules of ATP.
The energy investment stage
pyruvate can be converted to lactic
acid.
The energy pay-off stage
can join several metabolic pathways.
The energy investment stage
contains
yields two molecules of pyruvate.
Glucose 6-phosphate is formed
in the cytoplasm.
Glucose 6-phosphate and pyruvate
two irreversible phosphorylations.
The product of phosphofructokinase
by phosphorylation of glucose.
Transfer of phosphate directly to ADP
can only proceed to glycolysis.
In the absence of oxygen
substrate-level phosphorylation.
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..........................................
Q31: Complete the paragraph using the words from the list. (13 marks)
Glycolysis and the
cycle are synchronised by the
inhibition of
in the energy investment stage and the supply of
to the phosphorylation
stage. Feedback
occurs when
's
reactions in the energy
in the concentration of a product from
step in
activity is reduced by
stage of glycolysis is inhibited
the pathway. Phosphofructokinase in the energy
levels of
. This will reduce the production of
, and therefore
by
of ATP, when the cell's need for energy is low.
Word list: a later, an enzyme, an increase, ADP, ATP, citric acid, feedback, high,
inhibition, investment, pay-off, phosphofructokinase, pyruvate.
..........................................
Q32: Apart from providing energy, explain the other important role of the respiratory
pathways. (1 mark)
..........................................
Q33: List the products of glycolysis. (3 marks)
..........................................
Q34: Explain why the formation of ATP in glycolysis is a substrate-level phosphorylation.
(1 mark)
..........................................
Q35: Explain how glycolysis and the citric acid cycle are synchronised. (2 marks)
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Topic 10
Cellular Respiration II: Citric acid
cycle
Contents
10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 The mitochondrion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
214
215
10.2.1 Structure of the mitochondrion . . . . . . . . . . . . . . . . . . . . . . .
215
10.2.2 Fascinating facts about mitochondria . . . . . . . . . . . . . . . . . . . .
10.3 The citric acid cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
216
217
10.3.1 The steps of the citric acid cycle . . . . . . . . . . . . . . . . . . . . . .
10.3.2 The enzymes and carriers of the citric acid cycle . . . . . . . . . . . . .
217
219
10.4 Alternative substrates for respiration . . . . . . . . . . . . . . . . . . . . . . . .
10.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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224
10.6 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Learning Objectives
By the end of this topic, you should be able to:
• describe the structure of the mitochondrion and the location of processes within it;
• describe the citric acid cycle, its generation of ATP, and its release of CO 2 and
hydrogen;
• list the different substrates available for respiration;
• explain how these different substrates are incorporated into the respiratory
pathway.
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10.1
Introduction
In active cells in aerobic conditions, most of the pyruvate produced by glycolysis is used
in the generation of ATP to provide the energy for the cell's other functions.
This is achieved by passing the pyruvate from the cytoplasm into the mitochondrion
(plural: mitochondria), where it is first converted to acetyl coenzyme A and then passed
round the citric acid cycle, from which carbon dioxide and hydrogen ions are released.
The carbon dioxide passes out of the cell as a waste product, but the hydrogen ions
(along with high energy electrons) are transferred by carriers to the electron transport
system where they are passed along a series of acceptor molecules, generating the
energy to form ATP as they go. The hydrogen is finally accepted by oxygen, so forming
water which is the other waste product of aerobic respiration. Each of these many
steps is catalysed by a different enzyme. Although we shall mainly consider aerobic
respiration, which uses glucose as its substrate, there are several alternative respiratory
substrates; these are dealt with after the basic process has been described.
In this topic, we will consider the detail of the citric acid cycle, leaving the electron
transport system/chain to the next topic. Despite the fact that both pathways are found in
the mitochondrion, this has been done to reduce the quantity of material to be mastered
at any one time. There is also the fact that, as with glycolysis, the processes are also
fairly discrete.
There is a point of possible confusion which should be clarified. Although the pathway
is called the citric acid cycle, it is in fact the negatively charged ion (anion) citrate which
is formed and passed into the cycle. In the same way, it is pyruvate (rather than pyruvic
acid) which enters the mitochondrion. These anions are in the watery solution of the cell,
and not linked to the positive hydrogen ions (cations) which would turn them into their
acid equivalents. The names of the acids and the anions are often used interchangeably,
and the distinction is not in the syllabus.
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10.2
The mitochondrion
215
Learning Objective
By the end of this section, you should be able to:
• describe the structure of the mitochondrion as consisting of an outer boundary
membrane, an inner membrane, which is folded to form cristae, and the central
matrix;
• explain that pyruvate enters the matrix of the mitochondrion and is broken down
to an acetyl group which is attached to co-enzyme A, forming acetyl co-enzyme
A (acetyl-coA);
• explain that acetyl-coA passes the acetate into the citric acid cycle in the matrix
of the mitochondrion;
• explain that hydrogen and high energy electrons are released from the citric acid
cycle and are taken by carrier molecules to the electron transport chain on the
cristae.
10.2.1
Structure of the mitochondrion
The mitochondrion is bounded by a double layer of membranes, each of which is
itself composed of a phospholipid bilayer. The outer membrane surrounds the
mitochondrion and acts as a boundary; although it is highly permeable to small
molecules, the entry of larger molecules is strictly regulated. The inner membrane is
folded, forming structures called cristae.
Structure of the mitochondrion
Pyruvate crosses the double membrane and enters the inner space of the
mitochondrion, which is called the matrix. Here it is oxidised and, in combination with
co-enzyme A as acetyl co-enzyme A (or acetyl-coA), it enters the citric acid cycle. The
co-enzymes which accept the hydrogen ions that are released from the citric acid cycle
then move to the cristae of the inner membrane, where the hydrogen/electron transport
chain and its enzymes is located. It is here that nearly all of the ATP is generated.
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10.2.2
Fascinating facts about mitochondria
The most likely theory of the formation of mitochondria is that, about 2 billion years ago,
one particular type of bacterium survived being engulfed into another. The evidence
for this is the high degree of similarity between the mitochondria of different species of
living organism, and between these mitochondria and certain bacteria which still exist
today, namely the rickettsia.
These similarities lie not only in their structure, but also in their DNA. Although many
of the genes controlling the activity of the mitochondrion are now located in the
nuclear DNA, key enzymes are still controlled by the circular DNA molecules in the
mitochondria. Also, mitochondria control their own replication and respond to increased
energy demands in the cell by dividing more frequently.
The DNA of the mitochondrion is interesting from a more recent evolutionary
perspective. At the point of fertilisation, the egg destroys the mitochondria which enter
it from the sperm - we all carry only the mitochondria received from our mothers. As
mitochondrial DNA mutates relatively slowly and passes down the generations without
any of the genetic mixing associated with the formation of gametes, it can be used
to find relationships between individuals widely separated in time. It is also useful in
considering the evolutionary history of populations.
Mitochondria range in size from 0.5 to 1.0 μm, and in numbers from over a thousand in
active cells, such as those in the liver, to relatively few in cells such as skin epidermis,
and only one in some single-celled organisms. In addition, active cells have a much
more intensively folded inner membrane, creating large numbers of cristae and hence
more sites for ATP synthesis.
Mitochondria: Question
Q1: Complete the pairs of items concerning mitochondria by matching the parts on
the left with the parts on the right.
Glycolysis
citric acid cycle
Pyruvate to acetyl-coA
cristae of mitochondrion
Mitochondrial matrix
multi-enzyme complex in cytoplasm
Electron transport chain
mitochondrial matrix
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10.3
217
The citric acid cycle
The citric acid cycle is located in the matrix of the mitochondrion. As will be shown
below, it is classically described as a cycle with a starting point at citrate.
However, this is really a false impression of what actually goes on. The acetyl-coA,
which enters the cycle to form citrate, can be derived from several sources other than
pyruvate. Amino acids, and certain fatty acids, can be synthesised into intermediate
compounds in the cycle. Finally, several of the intermediates can be transferred to other
synthetic metabolic pathways. Some of these points are explored in a later section.
10.3.1
The steps of the citric acid cycle
Learning Objective
By the end of this section, you should be able to:
• state that pyruvate, produced by glycolysis, enters the mitochondrion and is
broken down to form acetate with the release of carbon dioxide;
• explain that this acetate is combined with coenzyme A to form acetyl coenzyme
A, or acetyl-coA;
• explain that coenzyme A passes the acetyl group on to oxaloacetate, forming
citrate, the first stage of the citric acid cycle;
• state that the citric acid cycle results in the direct generation of ATP, the release
of carbon dioxide, the transfer of hydrogen ions to carriers, and the regeneration
of oxaloacetate.
For every glucose molecule, glycolysis takes place once, but the succeeding stages of
aerobic respiration take place twice.
Once inside the mitochondrial matrix, the pyruvate is converted to an acetyl group by
the removal of hydrogen and carbon dioxide, a step known as pyruvate decarboxylation.
The acetyl group is then attached to a carrier molecule, coenzyme A, to form acetyl
coenzyme A (acetyl-coA) which then joins the citric acid cycle.
The steps of the cycle are shown in the following diagram.
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The citric acid cycle
The steps of the citric acid cycle are as follows:
1. the 3-C pyruvate loses one carbon to CO 2 and one hydrogen to NAD + , forming
NADH - the remaining 2-C acetyl group is attached to coenzyme A, forming acetylcoA;
2. coenzyme A passes the 2-C acetyl group to oxaloacetate (4-C), forming citrate
(6-C) and releasing coenzyme A to collect another acetyl group;
3. at successive steps in the cycle, two further carbon molecules are lost to CO2 and
three hydrogen ions are transferred to NAD +, forming NADH;
4. sufficient energy is released at one step in the cycle to form one molecule of ATP
(or GTP);
5. a second hydrogen carrier, FAD, removes two hydrogen ions as FADH 2;
6. finally, the 4-C oxaloacetate is formed and is ready to accept another 2-C acetyl
group from acetyl-coA.
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219
Although a knowledge of the numbers of carbon atoms in the various compounds in the
cycle is not required, they are included here to show how the cycle finally regenerates
the 4-carbon (4-C) molecule of oxaloacetate, which combines with the 2-C acetyl group
from acetyl-coA to form the 6-C citrate.
The steps of the citric acid cycle: Questions
Q2: For each molecule of glucose, how many times does glycolysis take place?
..........................................
Q3: For each molecule of glucose, how many times does the citric acid cycle take
place?
..........................................
Q4: Name the molecule to which pyruvate is converted before entering the citric acid
cycle.
..........................................
Q5: To which molecule is the 2-C compound, formed from pyruvate, joined when it
enters the citric acid cycle?
..........................................
Q6: List the three products of the citric acid cycle.
..........................................
Q7: Why is the citric acid cycle actually called a cycle?
..........................................
10.3.2
The enzymes and carriers of the citric acid cycle
Learning Objective
By the end of this section, you should be able to:
• state that each step in the citric acid cycle is mediated by a different enzyme;
• state that dehydrogenase enzymes remove hydrogen ions and electrons;
• state that coenzymes NAD+ and FAD accept these hydrogen ions and electrons
to form NADH and FADH 2;
• state that NADH and FADH 2 release high-energy electrons to the electron
transport chain on the cristae, resulting in the generation of ATP.
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The citric acid cycle consists of many reactions.
The citric acid cycle
Enzymes
Each step in the cycle is catalysed by a different enzyme which is specific to the
substrate involved. In particular, hydrogen is removed at four points in the cycle;
although the substrates are different in each case, the enzymes involved all belong
to a class of enzyme called dehydrogenases. In simple terms, these enzymes transfer
hydrogen ions and high energy electrons to carrier molecules which then convey them
to the electron transport system on the cristae.
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Hydrogen Carriers
The hydrogen ions and high energy electrons released by the various dehydrogenases
are picked by coenzyme carrier molecules of two types.
Most of the hydrogen attaches to NAD + , often simply referred to as NAD; the plus sign
('+ ') indicates what is called the formal charge on the NAD molecule, and shows that it
requires a negatively charged electron to balance its charge. In the citric acid cycle, it
picks up one hydrogen ion and one electron to form NADH.
NAD stands for nicotinamide adenine dinucleotide. When this complex name is broken
down it reveals some 'old friends': nicotinamide is a combination of nicotinic acid, or
vitamin B3 , and an amino acid; adenine is the base that is also found in nucleic acids;
and the dinucleotide indicates that there are two nucleotides in the molecule.
FAD stands for Flavin Adenine Dinucleotide, which also contains adenine and two
nucleotides; in this case, the vitamin is riboflavin, also known as vitamin B2 . FAD accepts
two hydrogen ions and two electrons from the citric acid cycle to form FADH 2 .
NADH and FADH 2 carry the hydrogen and high energy electrons from the citric acid
cycle to the electron transport system on the cristae, where the electrons pass along a
chain of receptors to generate the energy to form ATP.
The enzymes and carriers of the citric acid cycle: Questions
Q8: By what is each step in the citric acid cycle mediated?
..........................................
Q9: Name the enzymes which remove hydrogen ions and high energy electrons from
the citric acid cycle.
..........................................
Q10: Name the two coenzymes which accept the hydrogen and high energy electrons
from the citric acid cycle.
..........................................
Q11: When they accept hydrogen ions, what do these coenzymes form?
..........................................
Q12: Where do these coenzymes take the hydrogen that they carry?
..........................................
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10.4
Alternative substrates for respiration
Learning Objective
By the end of this section, you should be able to:
• state that the storage carbohydrate glycogen can be broken down into glucose
for use as a respiratory substrate;
• state that, in order to serve as respiratory substrates, other sugars must be
converted into glucose or intermediate compounds in the glycolysis pathway;
• state that proteins can be broken down into amino acids, which can be converted
into intermediate compounds in glycolysis and the citric acid cycle;
• state that fats can be broken down into glycerol and fatty acids;
• state that glycerol can be converted into glucose in the liver;
• state that fatty acids can be converted into acetyl-coA.
The average diet in economically more developed countries contains carbohydrates,
proteins and fats, all of which can be used to provide respiratory substrates. Under what
might be called 'normal' conditions, by which it is meant that there is no stress on the
body in terms of food supply, temperature or exercise, glucose is the substrate used for
respiration by all cells. Any excess of glucose beyond the body's immediate respiratory
needs is stored either as fat or glycogen. Excess fats are stored in the adipose tissue,
and any amino acids which are in excess of the immediate demand for protein building
are broken down in the liver. This forms the waste product urea, for excretion, and
glucose for storage.
The human body, like that of other animals, is adapted to store energy when there is
more available in the diet than is needed to meet everyday needs. We are programmed
to store fat because only a few generations ago people had to rely on that energy store
to carry us through the times in the year when food was in short supply. As anyone who
tries to reduce their weight by switching to a low calorie diet will discover, the body has
clever mechanisms for adapting to a reduced energy intake. Without increasing physical
activity alongside the diet, the kilograms will not melt away.
In times of real famine, the different energy sources in the body are used in turn. First,
the glycogen from the liver and the muscles is used up, then the fat from the various fat
depots, and finally the protein of the muscles is used to generate respiratory substrates.
The detailed picture is inevitably much more complex, but this is the general pattern.
Glycogen
Glucose is readily interconvertible with glycogen. After a meal, as digestion proceeds,
glucose is absorbed into the blood and blood sugar levels rise. This is detected by the
islets of Langerhans in the pancreas, and they increase the secretion of the hormone
insulin into the blood. This, in turn, stimulates the liver and muscle cells to absorb
glucose and convert it to glycogen. Once the blood glucose level has stabilised, the
liver then reconverts the stored glycogen to glucose, under the influence of glucagon
(also from the pancreas), so that the concentration of glucose in the blood, and hence
the tissue fluid bathing the body's cells, remains relatively constant. Glycogen is the
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immediately available source of glucose which is used flexibly throughout the day to
maintain blood sugar levels.
In endurance sporting events, such as marathon running, distance cycling or crosscountry skiing, the body draws on the glycogen of the liver and muscles. As the muscles
lack an essential enzyme which allows them to export glucose, their glycogen stores
cannot contribute to the maintenance of the blood sugar level, but they do provide
glucose to glycolysis in the muscle cells themselves. If these stores become depleted
before the end of an event, the athlete suddenly experiences extreme fatigue, a condition
known as 'hitting the wall'.
Other sugars
The diet of many in economically more developed countries contains relatively high
concentrations of sucrose sugar (the sort which is put into coffee, tea, and a surprising
range of other foods). In the small intestine, sucrose is digested by the enzyme sucrase
into two simpler sugars: glucose and fructose. Like glucose, fructose is absorbed into
the blood and is converted into an intermediate during glycolysis in the liver. This may
either be used to form glucose (and then glycogen) or pyruvate.
Proteins
The basic units of proteins are amino acids, of which twenty are common. Protein is
digested in the gut by various protease and peptidase enzymes to form these amino
acids. These are then absorbed into the blood and pass round the body, with cells
taking in what they require for protein synthesis. Excess amino acids are taken into the
liver and are broken down, forming the waste product urea and compounds that are fed
into the citric acid cycle and its associated pathways.
Depending on their composition, these amino acids enter the citric acid cycle by several
routes: many are converted to pyruvate, or acetyl-coA, others to various intermediates,
and some even to oxaloacetate. As well as fuelling the citric acid cycle, some of these
intermediates may alternatively be converted into glucose, lipids, or both.
Amino acids are only used as a major respiratory substrate when the other fuels,
glycogen and fat, are exhausted, i.e. in starvation conditions.
Fats/lipids (more strictly, triglycerides)
Fats are digested by the enzyme lipase and are converted into fatty acids and glycerol.
Following absorption within the small intestine, these substances are reformed into fat
which is carried as small particles, called chylomicrons, first in the lymphatic system and
then in the blood, to fat depots in the adipose tissue, to muscles, and to the liver. The
processing of fat in the liver is a complex story which is covered in Unit 2 of the course.
Glycerol and fatty acids also circulate freely in the blood and, along with chylomicrons,
they supply cells with fats/lipids and their components. In the cytoplasm of cells, the
lipids are broken down by hydrolysis to glycerol and fatty acids. Glycerol is converted
to an intermediate in glycolysis, ready to enter the energy pay-off stage, whereas the
fatty acids are converted into acetyl-coA. Glycerol can also be converted into glucose
in the liver. In all cells, fatty acids are synthesised from citrate which is exported from
the mitochondria back into the cytoplasm before entering the metabolic pathway which
forms fatty acids.
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Alternative substrates for respiration: Questions
Q13: Name the storage carbohydrate in humans.
..........................................
Q14: Why is this useful in the control of blood sugar levels?
..........................................
Q15: State two possible fates in the cell for sugars other than glucose.
..........................................
Q16: How is energy released from amino acids in the cell?
..........................................
Q17: Where is glycerol converted to glucose?
..........................................
Q18: How is glycerol respired in cells?
..........................................
Q19: How do fatty acids become available for respiration in the cell?
..........................................
10.5
Learning points
Summary
The mitochondrion
• The mitochondrion consists of an outer boundary membrane, an inner
membrane, which is folded to form cristae, and the central matrix.
• Pyruvate enters the matrix of the mitochondrion and is broken down to an
acetyl group which is attached to co-enzyme A, forming acetyl co-enzyme
A (acetyl-coA).
• Acetyl-coA passes the acetate into the citric acid cycle in the matrix of the
mitochondrion.
• Hydrogen and high energy electrons are released from the citric acid cycle
and are taken by carrier molecules to the electron transport chain on the
cristae.
The steps of the citric acid cycle
• Pyruvate, produced by glycolysis, enters the mitochondrion and is broken
down to form acetate with the release of carbon dioxide.
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Summary Continued
• This acetate is combined with coenzyme A to form acetyl coenzyme A, or
acetyl-coA.
• Coenzyme A passes the acetyl group on to oxaloacetate, forming citrate,
the first stage of the citric acid cycle.
• The citric acid cycle results in the direct generation of ATP, the release
of carbon dioxide, the transfer of hydrogen ions to carriers, and the
regeneration of oxaloacetate.
The enzymes and carriers of the citric acid cycle
• Each step in the citric acid cycle is mediated by a different enzyme.
• Dehydrogenase enzymes remove hydrogen ions and electrons.
• Coenzymes NAD+ and FAD accept these hydrogen ions and electrons to
form NADH and FADH 2 .
• NADH and FADH 2 release high-energy electrons to the electron transport
chain on the cristae, resulting in the generation of ATP.
Alternative substrates for respiration
• The storage carbohydrate glycogen can be broken down into glucose for
use as a respiratory substrate.
• In order to serve as respiratory substrates, other sugars must be converted
into glucose or intermediate compounds in the glycolysis pathway.
• Proteins can be broken down into amino acids, which can be converted to
intermediate compounds in glycolysis and the citric acid cycle.
• Fats can be broken down into glycerol and fatty acids.
• Glycerol can be converted into glucose in the liver.
• Fatty acids can be converted into acetyl-coA.
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10.6
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of enzymes and coenzymes before attempting
the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended answer question: Enzymes and coenzymes
10 min
Give an account of the enzymes and coenzymes involved in the citric acid cycle. (6
marks)
..........................................
10.7
End of topic test
End of Topic 10 test
Q20: Complete the paragraph using the words from the list. (10 marks)
membrane, an inner membrane which
The mitochondrion consists of an outer
, and the central
.
enters the matrix of the
is folded to form
, and attached to
to form acetyl comitochondrion, is converted to
). This passes the acetate into the Citric Acid cycle in the matrix of
enzyme A (
and high energy
are released from the citric acid
the mitochondrion.
chain on the cristae
cycle and are taken by carrier molecules to the electron
of the inner membrane of the mitochondrion.
Word list: acetate, acetyl-coA, boundary, co-enzyme A, cristae, electrons, hydrogen,
matrix, pyruvate, transport.
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..........................................
Q21: Complete the sentences by matching the parts on the left with the parts on the
right. (10 marks)
Pyruvate is produced by
oxaloacetate.
Pyruvate is broken down to form
NAD+ and FAD.
Acetyl-coA passes the acetyl group to
dehydrogenases.
Oxaloacetate combines with the
acetyl group to form
ATP.
The citric acid cycle directly
generates
glycolysis.
Hydrogen is removed from the citric
acid cycle by
cristae.
Hydrogen is transferred from the citric
acid cycle to
citrate.
High energy electrons are carried by
an acetyl group.
NADH and FADH 2 carry hydrogen to
the
coenzymes.
The electron transport chain is
located on the
electron transport chain.
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..........................................
Q22: Complete the table by matching the descriptions on the left with the terms on the
right. (6 marks)
A storage carbohydrate:
fructose
A sugar which can be converted into
a glycolysis intermediate:
amino acids
Some can be converted to citric acid
cycle intermediates:
glycogen
Formed from glycerol and fatty acids:
fatty acids
Converted into glucose in the liver:
lipids
glycerol
Converted into acetyl-coA:
..........................................
Q23: Summarise the end results of the citric acid cycle. (4 marks)
..........................................
Q24: Explain the role of coenzymes in the citric acid cycle. (4 marks)
..........................................
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Topic 11
Cellular Respiration III: Electron
transport system
Contents
11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 The electron transport chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
230
231
11.3 ATP synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
234
11.4 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.5 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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237
11.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
237
Learning Objectives
By the end of this topic, you should be able to:
• describe the electron transport chain;
• explain the generation of ATP by the electron transport chain.
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TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM
11.1
Introduction
The electron transport chain is the point in the process of aerobic respiration at which
the bulk of the energy in the respiratory substrate is finally made available as ATP. The
pathway is located on the cristae of the inner membrane of the mitochondrion. Although
it has a variety of names, including the cytochrome system and the hydrogen carrier
system, we shall use the term electron transport system because it is high energy
electrons which drive it.
There are two parts to the system which are said to be coupled, meaning that one
cannot occur without the other:
• the electron transport chain, which comprises a series of enzyme-controlled
electron donors and electron acceptors - each transfer of an electron provides
the energy to pass a hydrogen ion (or proton) across the inner membrane into the
space between the inner and outer membranes of the mitochondrion;
• the ATP synthesis stage, which involves the enzyme ATP synthase - the
protons are passed back from the intermembrane space into the matrix of the
mitochondrion, releasing energy which is used to form ATP.
In this, and the preceding topics, reference has been made to hydrogen ions and
electrons. Hydrogen is the simplest of all atoms and is found throughout the universe,
being by far the commonest element. It consists of a nucleus comprising a single particle
which carries a positive electrical charge, a proton, and an orbiting negatively charged
particle (an electron). The atom is therefore electrically neutral. If the proton is separated
from its electron, it becomes a hydrogen ion with a positive charge, H + .
Electrons are always negatively charged, but they can exist at different energy levels.
They can absorb energy and be raised to a higher energy level, or give out energy and
drop to a lower energy level.
If a compound combines with hydrogen ions or gains electrons, it is said to be reduced.
Conversely, and perhaps confusingly, the loss of hydrogen or electrons is known as
oxidation, as is the combination with oxygen. In the summary equation of aerobic
respiration, glucose is oxidised to carbon dioxide and oxygen is reduced to water:
C6 H12 O6 + 6O2 → 6CO2 + 6H2 O
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11.2
231
The electron transport chain
Learning Objective
By the end of this section, you should be able to:
• state that coenzymes NAD+ and FAD collect hydrogen ions and high energy
electrons from glycolysis and the citric acid cycle;
• state that, as NADH and FADH 2, the reduced coenzymes travel to the inner
mitochondrial membrane;
• state that, at the inner mitochondrial membrane, NADH and FADH 2 pass the
hydrogen ions and high energy electrons to carriers in the electron transport
chain;
• state that the electron transport chain is a collection of proteins attached to the
inner mitochondrial membrane;
• state that as the high energy electrons released from NADH and FADH 2 pass
along the chain, they release their energy in stages;
• state that the energy so released is used by enzymes to pump hydrogen ions
(H+ ) across the inner mitochondrial membrane into the intermembrane space;
• state that the final electron acceptor is oxygen, which combines with hydrogen
ions to form water.
The reduced coenzymes NADH and FADH 2 bring the hydrogen ions (protons) and high
energy electrons to the inner membrane of the mitochondrion. Here the protons are
released and the electrons are passed to a series of protein acceptors and donors which
are arranged in three closely linked complexes.
When the electrons are passed from donor to acceptor, energy is released. This energy
is used to pump the hydrogen ions across the inner membrane into the intermembrane
space, building up a concentration there that can be up to ten times that of the matrix.
As the electrons pass along the chain, their energy level declines until eventually they
reach their lowest energy level. At that point they are released to combine first with
oxygen, and then with two hydrogen ions to from a water molecule (H 2 O).
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The following illustration shows the stages in the electron transport chain. The electrons
are shown passing from their coenzymes to carriers in the transport chain. The changing
height of the carriers across the diagram indicates the relative energy level of the
electrons.
The electron transport chain
1. NADH carries high energy electrons and hydrogen to the inner mitochondrial
membrane, where the electrons are passed to an electron carrier in the protein
complex on the membrane. Each transfer is catalysed by a different enzyme.
2. FADH2 passes electrons to a carrier slightly further along the chain. For those
interested in counting molecules of ATP generated, this means that FADH 2
generates less ATP than NADH.
3. As the electrons pass from carrier to carrier, the energy released is used
by enzymes to pump hydrogen ions across the inner membrane into the
intermembrane space. This leads to a large difference in the hydrogen ion
concentration between the matrix and the intermembrane space, which can be
sufficient to cause a difference of 1pH unit.
4. Having released their energy, the electrons are transferred to an oxygen atom
which then carries two negative charges. It combines with two hydrogen ions from
the matrix to form water, the second waste product from respiration (the other
being carbon dioxide).
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The electron transport chain: Questions
Q1: What do NADH and FADH 2 carry?
..........................................
Q2: To where do NADH and FADH 2 go?
..........................................
Q3: Of what type of compound is the electron transport chain composed?
..........................................
Q4: What is the function of these compounds in relation to electrons?
..........................................
Q5: What is the function of these compounds in relation to hydrogen ions?
..........................................
Q6: What is the final electron acceptor in the electron transport chain?
..........................................
Q7: Name the waste product from the electron transport chain.
..........................................
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11.3
ATP synthesis
Learning Objective
By the end of this section, you should be able to:
• state that as electrons pass along the electron transport chain, energy is
released;
• explain that this energy helps enzymes pump hydrogen ions across the inner
membrane of the mitochondrion into the intermembrane space;
• state that the return flow of these H + ions rotates part of the enzyme ATP
synthase;
• explain that the rotation of ATP synthase catalyses the formation of ATP;
• state that this process generates most of the ATP in aerobic respiration.
As the electrons pass along the electron transport chain, energy is released as the
electrons pass from donor to acceptor molecule. Eventually, the electrons reach
their lowest energy level and are passed to the final acceptor, which is oxygen. The
energy released is used to enzymatically pump H + ions across the inner mitochondrial
membrane into the intermembrane space. As the membrane is impermeable to protons,
this results in a build-up in their concentration in the intermembrane space compared to
the matrix, leading to as much as a tenfold difference (equivalent to 1pH unit).
This considerable difference in the concentration of positively charged hydrogen ions
causes a large electrochemical gradient between the intermembrane space and the
matrix. The only route back for the H+ ions is through molecules of the enzyme ATP
synthase, which span the inner membrane. The ATP synthase molecule is roughly rodshaped, with a central rotating subunit, or rotor, and three active sites in its circular
base.
As the H+ ions pass through the molecule, they cause the central subunit to rotate inside
the base. As it does so, it alters the shape of the active sites, which have three states.
The first state is 'open', in which ADP and P i can enter the site; the second is 'loose',
in which the site closes loosely around the two substrates; and the final state is 'tight',
during which the site forces the ADP and P i together to form ATP. As the site returns to
its 'open' state, the newly-formed ATP molecule is ejected.
It is a beautiful example of a biochemical machine!
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TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM
Synthesis of ATP by ATP synthase
i
Hydrogen ions accumulate in the intermembrane space, resulting in an
electrochemical gradient across the inner membrane between the intermembrane
space and the matrix.
ii
This, in turn, provides the energy necessary to make ATP synthase function. It
causes the rotor and rod of the ATP synthase to rotate inside the circular base
unit.
iii The mechanical energy from this rotation is converted into chemical energy as
phosphate is added to ADP to form ATP.
Unlike the formation of ATP in the energy pay-off stage of glycolysis, in which
phosphate is directly transferred from a substrate molecule to ADP (hence substrate
level phosphorylation), this phosphorylation involves the removal of hydrogen and
electrons from substrate molecules. Hence, it is known as oxidative phosphorylation.
ATP synthesis: Questions
Q8: What is released as electrons pass along the electron transport chain?
..........................................
Q9: What happens to the H + ions released from the NADH and FADH 2?
..........................................
Q10: Name the enzyme which forms ATP from ADP and P i ?
..........................................
Q11: What causes part of this molecule to rotate?
..........................................
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Q12: What is the outcome of this rotation?
..........................................
Q13: From where does most of the ATP generated in aerobic respiration come?
..........................................
Once you have completed this topic, it would be a good idea to go back to Topic 9 Section
2, just to remind yourself of the overall picture of aerobic respiration. Although the total
yield of energy from one molecule of glucose is theoretically sufficient to generate 38
molecules of ATP (2 from direct phosphorylation during glycolysis and 36 from oxidative
phosphorylation), in practice the yield is closer to 30 molecules as there are, just as in
industrial power systems, losses of energy at various stages in the process.
11.4
Learning points
Summary
The electron transport chain
• Coenzymes NAD+ and FAD collect hydrogen ions and high energy electrons
from glycolysis and the citric acid cycle.
• As NADH and FADH 2, they travel to the inner mitochondrial membrane.
• At the inner mitochondrial membrane, NADH and FADH 2 pass the hydrogen
ions and high energy electrons to carriers in the electron transport chain.
• The electron transport chain is a collection of proteins attached to the inner
mitochondrial membrane.
• As the high energy electrons released from NADH and FADH 2 pass along
the chain, they release their energy in stages.
• The energy so released is used by enzymes to pump hydrogen ions (H + )
across the inner mitochondrial membrane into the intermembrane space.
• The final electron acceptor is oxygen, which combines with hydrogen ions
to form water.
ATP synthesis
• As electrons pass along the electron transport chain, energy is released.
• This energy helps enzymes pump hydrogen ions across the inner
membrane of the mitochondrion into the intermembrane space.
• The return flow of these H + ions rotates part of the enzyme ATP synthase.
• The rotation of ATP synthase catalyses the formation of ATP.
• This process generates most of the ATP in aerobic respiration.
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11.5
237
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of the electron transport chain before attempting
the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended answer question: The electron transport chain
Give an account of the electron transport chain. (8 marks)
..........................................
11.6
End of topic test
End of Topic 11 test
Q14: Complete the paragraph using the words from the list. (14 marks)
attached to the
The electron transport chain is a collection of
membrane.
NAD + and
collect hydrogen ions and
electrons
and the citric acid cycle, forming NADH and
. They travel to
from
and high energy electrons to
the inner mitochondrial membrane and pass the
in the electron transport chain. As the high energy electrons released from
and FADH 2 pass along the chain of electron acceptors and
, they
which is used by
to pump hydrogen ions (H + ) across the inner
release
space. The final electron acceptor is
mitochondrial membrane into the
which combines with hydrogen ions to form water.
Word list: carriers, coenzymes, donors, energy, enzymes, FAD, FADH2, glycolysis, high
energy, hydrogen ions, inner mitochondrial, intermembrane, NADH, oxygen, proteins.
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TOPIC 11. CELLULAR RESPIRATION III: ELECTRON TRANSPORT SYSTEM
..........................................
Q15: Complete the sentences by matching the parts on the left with the parts on the
right. (5 marks)
Energy is released as electrons pass
along
part of ATP synthase.
Enzymes pump H+ ions into
the formation of ATP.
The return flow of H + ions into the
matrix rotates
the ATP from aerobic respiration.
ATP synthase catalyses
the electron transport chain.
The electron transport chain
the intermembrane space.
produces most of
..........................................
Q16: Explain the role of electrons in the electron transport chain. (6 marks)
..........................................
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Topic 12
Energy systems in muscle cells
Contents
12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2 Creatine phosphate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
240
242
12.3 Lactic acid metabolism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.4 Types of skeletal muscle fibres . . . . . . . . . . . . . . . . . . . . . . . . . . .
244
246
12.4.1 Slow twitch (Type 1) fibres . . . . . . . . . . . . . . . . . . . . . . . . . .
247
12.4.2 Fast twitch (Type 2) fibres . . . . . . . . . . . . . . . . . . . . . . . . . .
12.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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250
12.6 Extended response question . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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252
Learning Objectives
By the end of this topic, you should be able to:
• describe the unique demands made on the respiratory pathway by muscle cells;
• explain the roles of aerobic and anaerobic respiration in muscle cells;
• describe creatine phosphate metabolism;
• describe lactic acid metabolism;
• describe the different types of muscle fibres and explain their role in different
sports.
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12.1
Introduction
Muscle tissue has as its function the exertion of force by contraction. This may be to
reduce the diameter of arteries or the gut (involuntary smooth muscle), to contract the
heart chambers (cardiac muscle), or to move parts of the skeleton (voluntary striped
muscle). Muscle tissue is never totally at rest, being constantly stimulated to contract at
a low level of intensity to maintain muscle tone. Whereas smooth muscle is not generally
required to contract suddenly or violently, cardiac and skeletal muscle do both, and
skeletal muscle may go from its relaxed state to strong contraction very rapidly. In this
topic, we shall limit our consideration to skeletal muscle and its use in different types of
sport.
Skeletal muscle is required to work in a range of different ways. When we are doing
anything other than lying flat out, many sets of muscles must work antagonistically
to maintain our posture by supporting our body weight and keeping us upright. Any
movement, from picking up a pencil to sprinting for a bus, similarly requires co-ordinated
activity by different muscles, all of which is controlled by the cerebellum in the brain.
The way in which muscle tissue gains the energy for its activity varies with the duration
of the exercise in which the body is engaged. There are four potential energy sources.
• ATP, which is by far the most important supplier of energy for cell activities, is
synthesised in response to demand, and is only present in small quantities in the
cell. The adult human body only contains about 250g of ATP at any one time,
although during the course of a day the body will use up the equivalent of its own
weight of ATP. There is no store of ATP in muscle cells so the immediately available
ATP is very rapidly depleted when sudden activity takes place.
• Creatine phosphate (or phosphocreatine) is found in tissues with a high and
variable energy demand, such as muscle and brain. For a brief period (2-10s),
it can supply phosphate to convert ADP to ATP without the need for oxygen
(anaerobically).
• Lactic acid metabolism occurs when the oxygen supply is insufficient to support
the electron transport chain. Pyruvate can be converted to lactic acid thus allowing
glycolysis to proceed and continue to generate ATP, albeit in relatively small
quantities.
• The citric acid cycle and the electron transport chain, in the presence of an
adequate oxygen supply, yield at least 15 times more ATP per glucose molecule
than lactic acid metabolism. However, creatine phosphate and lactic acid
metabolism can provide ATP 50 times faster, and so are vital for short bursts of
strenuous activity. The glucose fuel must come either from stored glycogen or
from the blood.
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The following diagram shows the relative importance of these different energy sources
as the duration of exercise changes.
The four potential energy sources during exercise
In terms of sport, we can see that events which involve an explosive burst of activity,
such as weight-lifting or discus throwing, are going to depend on ATP and creatine
phosphate. On the other hand, endurance events, such as triathlon or marathon, rely
on aerobic respiration of glucose derived from glycogen stored in the muscles.
Not only are there different energy sources for these different types of event, but also
the muscle best suited to each differs as well. Explosive sports require muscle which
will contract fast, but for a short time only; these are known as 'fast twitch' muscle fibres.
Endurance events demand muscles which will contract repeatedly over a long period of
time, even though they may contract more slowly; these are 'slow twitch' fibres.
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12.2
Creatine phosphate
Learning Objective
By the end of this section, you should be able to:
• state that muscle cells do not contain any reserve of ATP which will fuel more
than 2s of activity;
• state that creatine phosphate is an energy reserve that is found in cells with high
and variable energy demands, such as muscle and brain;
• state that creatine phosphate breaks down anaerobically to release energy and
phosphate, which is used to convert ADP and P i to ATP at a fast rate;
• state that there is only sufficient creatine phosphate to fuel strenuous muscle
activity for a maximum of 10s;
• explain that when energy demands are low, creatine phosphate reserves are
restored using ATP provided by aerobic respiration.
Apart from the protein from which they are made, muscles contain three types of energy
store.
• ATP. Although it is not a storage compound, a small quantity of ATP is held within
the muscle fibres. It is sufficient to fuel about 2s of intense activity.
• Glycogen is the storage carbohydrate in the human body, with about 100g in the
liver (used in the control of blood sugar levels) and 300g in the muscles (used as
a fuel reserve). It is made from glucose that is supplied to the cells by the blood.
• Creatine phosphate, which is found in cells which intermittently have a high energy
demand (e.g. muscle and brain), sufficient quantities of which are stored in the
muscle fibres to support up to 10s of vigorous exercise. It can be obtained from
two sources: ingestion of meat and internal production by the liver and kidneys. It
is also taken as a supplement by some athletes, although it is only of relevance
to sports requiring short bursts of energy and, more importantly, no studies have
been carried out on long-term effects on the body of taking such supplements.
Creatine phosphate is reversibly converted to creatine and phosphate by another kinase
enzyme, creatine kinase, which carries out a direct phosphorylation by transferring
the phosphate group to ADP, so reforming ATP. This is another example of coupled
reactions. They are anaerobic, meaning that this source of ATP is not dependent on the
delivery of oxygen to the tissues by the blood.
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Creatine phosphate energy store
When muscle energy demand is reduced, ATP from aerobic respiration allows creatine
kinase to run the reaction in the opposite direction, transferring phosphate from ATP
to creatine to reform creatine phosphate and ADP. The dephosphorylation of ATP is
catalysed by the enzyme ATPase. Notice that, like many enzymes, creatine kinase
is able to catalyse the interconversion of creatine and creatine phosphate in both
directions, depending on conditions such as the concentrations of reactants. Again,
these are coupled reactions but, being dependent on the supply of ATP from the electron
transport chain, they only occur when the cell has an oxygen supply.
Creatine phosphate: Questions
Q1: For how many seconds will stored ATP sustain muscle contraction?
..........................................
Q2: Name two tissues which have creatine phosphate as an energy store.
..........................................
Q3: Why do these tissues have creatine phosphate as an energy store?
..........................................
Q4: What are formed when creatine phosphate is anaerobically broken down?
..........................................
Q5: What is the maximum time in seconds for which creatine phosphate can sustain
strenuous muscle activity?
..........................................
Q6: How are creatine phosphate reserves restored?
..........................................
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12.3
Lactic acid metabolism
Learning Objective
By the end of this section, you should be able to:
• explain that during vigorous exercise, muscle cells do not get enough oxygen to
support the electron transport chain;
• state that under these conditions, pyruvate is converted to lactic acid;
• state that this conversion involves the transfer of hydrogen from the NADH
formed during glycolysis to pyruvate, forming lactate;
• state that the use of NADH to form lactate regenerates the NAD + that is required
to maintain ATP production by glycolysis;
• state that the accumulation of lactate causes muscle fatigue;
• state that lactate is reconverted to pyruvate in the muscles and to glucose in the
liver;
• state that the reconversion of lactate to pyruvate or glucose requires ATP that is
formed by aerobic respiration;
• explain that this additional respiratory activity creates an oxygen debt.
The demands of vigorous exercise on a muscle's energy supply are such that aerobic
respiration cannot keep up the necessary flow of ATP. The supply of oxygen from the
blood limits the rate at which the electron transport chain can generate it. Whereas
ATP and creatine phosphate reserves can provide fuel for some 10s, it is anaerobic
respiration that must meet the need for additional supplies of ATP for up to 90s thereafter.
Anaerobic respiration is also known as (lactate) fermentation.
Lactate formation
In the absence of oxygen as a final hydrogen acceptor, the electron transport chain
cannot function so that pyruvate and NADH accumulate in the cytoplasm. Pyruvate has
been shown to inhibit the enzyme phosphofructokinase, which catalyses the second
irreversible reaction in the energy investment stage of glycolysis. Likewise, a lack of
NAD+ limits the first reaction in the energy pay-off stage. For both of these reasons, it is
vital to the continued activity of the muscle cell that pyruvate is removed and that NADH
is oxidised back to NAD+ by removing hydrogen. This is achieved by converting it to
lactate.
Lactate is continuously formed by muscle cells whether oxygen is present or not. It is
also the only source of energy for red blood cells, which lack mitochondria. However,
when the energy demand in muscle cells exceeds the ability of aerobic respiration to
supply all of the ATP needed and anaerobic respiration increases, lactate production
exceeds the cell's ability to use it or export it, and so it accumulates. This leads to a
lowering of the cell's pH and glycolysis is inhibited, resulting in muscle fatigue. The
associated burning pain is a signal to allow the body to clear the muscles of lactate
and the other metabolites which have built up in the cells. Persisting with the exercise
can result in nausea and vomiting. However, build-up of lactate is not the cause of the
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soreness and stiffness that may be felt in the succeeding days; that is a result of minor
muscle damage.
Lactate removal
Lactate is excreted from muscle (and other) cells and is carried in the blood to the liver
cells. Here it is first converted to pyruvate and then passed into the mitochondria, where
several of the steps in glycolysis are reversed to form glucose. Muscle cells are also
capable of using up lactate and converting it to pyruvate but, instead of reconverting it
to glucose, they pass it into the citric acid cycle in the mitochondria.
Both of these processes require an input of ATP so can only take place to any significant
extent when the vigorous exercise has ceased, in which case aerobic respiration can
provide a surplus of ATP beyond the metabolic needs of the body. The effect of this is
noticed as a need to keep breathing deeply after the exercise is over in order to supply
the oxygen to the cells so that they can carry out this additional respiratory activity. This
gives rise to the term 'oxygen debt' and the idea that it is being repaid by deep breathing.
Lactic acid metabolism: Questions
Q7: Why do muscle cells increase lactic acid formation during vigorous activity?
..........................................
Q8: What is converted to form lactate/lactic acid?
..........................................
Q9: What does this conversion regenerate for glycolysis?
..........................................
Q10: What does the accumulation of lactic acid cause in the muscle?
..........................................
Q11: To what is lactate reconverted in the (i) liver, (ii) muscles?
..........................................
Q12: What must aerobic respiration supply for lactate to be reconverted?
..........................................
Q13: Why does this create an oxygen debt?
..........................................
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12.4
Types of skeletal muscle fibres
Learning Objective
By the end of this section, you should be able to:
• state that there are two types of muscle fibre, namely slow and fast twitch;
• state that all skeletal muscles contain a mixture of the two types of fibre;
• explain that athletes show patterns of muscle fibres that reflect their events.
There are two basic types of muscle fibre found in skeletal muscle, namely slow and
fast twitch fibres (or Type 1 and Type 2 fibres). These are so named because fast
twitch fibres contract 2-3 times faster that slow twitch. Muscles always contain a mixture
of the two fibres, but the proportions differ considerably between different muscles.
The actual proportions in any individual person appear to be genetically determined,
although training can force one type of fibre to develop more of the characteristic of
the other. Slow twitch fibres are especially suited to endurance activities, whereas fast
twitch fibres are required for power sports which demand an explosive burst of activity.
Types of skeletal muscle fibres: Questions
Q14: Name the two types of muscle fibre.
..........................................
Q15: How many of these types of fibre are found in any one muscle?
..........................................
Q16: What type of fibre is relatively common in endurance athletes?
..........................................
Q17: What type of fibre is relatively common in power athletes?
..........................................
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12.4.1
Slow twitch (Type 1) fibres
247
Learning Objective
By the end of this section, you should be able to:
• state that slow twitch (Type 1) fibres contract relatively slowly, but can sustain
contractions longer;
• explain that slow twitch fibres are good for endurance sporting activities, such
as long distance cycling or running;
• state that slow twitch fibres rely on aerobic respiration to generate ATP;
• explain that slow twitch fibres have many mitochondria, a large blood supply,
and a high concentration of the oxygen-storing protein myoglobin, which makes
them dark red in colour;
• explain that slow twitch fibres use both glycogen and fat as fuel depending on
the intensity of the exercise.
As we have seen in the earlier sections of this topic, a muscle cell has four sources of
ATP to fuel its contraction:
1. ATP stored in small quantities in the muscle fibres, able to support up to 2s of
contraction;
2. creatine phosphate, anaerobically fuelling up to 10s of activity;
3. lactic acid metabolism, able to sustain up to 90s of activity anaerobically;
4. aerobic respiration, able to sustain strenuous activity until the stores of glycogen
run out, which is generally after about 4hr.
From this, it is obvious that any endurance event will require muscle to be operating
aerobically. The type of events which are in this category are running any distance over
800m (but especially longer races such as 10 000m and marathon), longer distance
cycling, kayak, swimming or ski races, and mixed discipline distance events like triathlon.
The type of muscle fibre which is suited to the demands of these events is the slow twitch
(Type 1) fibre. Although they contract more slowly than fast twitch fibres, they are able
to contract for much longer before fatiguing. To do this, they rely on aerobic respiration
which demands a high oxygen supply. This is provided by a good blood supply through
an extensive capillary network, and a high concentration of the respiratory protein
myoglobin. Myoglobin is related to the haemoglobin found transporting oxygen in red
blood cells; it too stores oxygen and is deep red in colour.
Aerobic respiration is usually fuelled by either glucose or, less frequently, fatty acids
and glycerol. Glucose may be supplied in the blood (it can be quickly absorbed from
energy drinks), but essentially during endurance events it must come from stored fat
and especially glycogen.
Interestingly, the balance of supply between fats and glycogen varies with the intensity
of the exercise. If the athlete is operating at 75% or more of their maximum capability,
then glycogen is the sole source of fuel in the form of glucose.
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Below this level, at roughly the intensity of exercise which puts one out of breath, but
not stressed, with a heart rate of roughly 130bpm (this depends on age), the balance
shifts much more in favour of fat as the fuel source. At this point, fat consumption is at
its maximum, making up 40-60% of the energy supply. For those trying to lose some
excess fat, this is the level at which to exercise. At still less demanding levels of exercise,
the balance shifts back to glucose, either derived from glycogen in the muscles or liver,
or from food in the gut.
Slow twitch (Type 1) fibres: Questions
Q18: Describe the contraction of slow twitch muscle fibres.
..........................................
Q19: To which sporting events are Type 1 muscle fibres best suited?
..........................................
Q20: On which source of ATP do slow twitch fibres rely?
..........................................
Q21: Complete the table to show the sources of fuel for slow twitch fibres at different
levels of exercise.
Level of sustained exercise
Source of respiratory substrate
Light
Moderate
High
..........................................
Q22: Explain the concentrations of myoglobin, mitochondria and capillaries in muscles
with a high proportion of slow twitch fibres.
..........................................
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TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS
12.4.2
Fast twitch (Type 2) fibres
249
Learning Objective
By the end of this section, you should be able to:
• state that fast twitch (Type 2) fibres contract relatively quickly, but for short
periods;
• explain that fast twitch fibres are good for power sporting activities such as
sprinting or weight-lifting;
• explain that fast twitch fibres generate ATP through glycolysis and the
breakdown of creatine phosphate;
• state that fast twitch fibres have few mitochondria and a relatively low blood
supply;
• state that fast twitch fibres have a low level of myoglobin, making them light pink
in colour;
• explain that fast twitch fibres use both glycogen and creatine phosphate as fuel.
In man's evolutionary past, there must have been many occasions when it was
necessary to contract muscles faster than could be achieved if the necessary ATP was
supplied by aerobic respiration alone. One need only think of avoiding a predator or
one's enemies, running after prey, or lifting rocks and tree trunks, to imagine the sort
of sudden movement required. As mentioned in a previous section, using the small
reserves of ATP and creatine phosphate, followed by anaerobic respiration, can provide
the sudden burst of muscular contraction needed to cope with such situations for up to
90s.
These types of activity, modelled by sports activities such as sprinting and weight-lifting,
demand a different type of muscle fibre to the slow twitch (Type 1) fibre required by
endurance activities, namely the fast twitch (Type 2) fibre. In adapting to contracting
much faster than slow twitch fibres, these Type 2 fibres have lost the ability to sustain
their contraction. Also, relying largely on anaerobic sources of ATP, i.e. creatine
phosphate and glycolysis, they have little myoglobin, so are pink or even white, and
have a relatively poor capillary blood supply because they do not need to store or be
supplied with large quantities of oxygen.
Of course, fast twitch fibres do carry out aerobic respiration, both to maintain their
general metabolism and, after exertion, to provide the ATP necessary to reconstitute
creatine phosphate from creatine, and pyruvate from lactate. However, they do not use
aerobic respiration to the same extent that slow twitch fibres do, and so they have far
fewer mitochondria. Because of their greater reliance on anaerobic respiration, they
have a relatively higher concentration of glycolytic enzymes.
The energy stores in fast twitch fibres are creatine phosphate and glycogen, which can
be rapidly broken down to glucose 6-phosphate and fed into the glycolysis pathway.
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Fast twitch (Type 2) fibres: Questions
Q23: Describe the contraction of fast twitch muscle fibres.
..........................................
Q24: To which sporting events are Type 2 muscle fibres best suited?
..........................................
Q25: On which sources of ATP do slow twitch fibres rely?
..........................................
Q26: Which compounds are used as energy stores in fast twitch fibres?
..........................................
Q27: Compared to slow twitch fibres, explain the relative concentrations of myoglobin,
mitochondria and capillaries in fast twitch fibres.
..........................................
12.5
Learning points
Summary
Creatine phosphate
• Muscle cells do not contain any reserve of ATP which will fuel more than 2s
of contraction.
• Creatine phosphate is an energy reserve found in cells with high and
variable energy demands, such as muscle and brain.
• Creatine phosphate breaks down anaerobically to release energy and
phosphate, which is used to convert ADP and P i to ATP at a fast rate.
• There is only sufficient creatine phosphate to fuel strenuous muscle activity
for a maximum of 10s.
• When energy demands are low, creatine phosphate reserves are restored
using ATP provided by aerobic respiration.
Lactic acid metabolism
• During vigorous exercise, muscle cells do not get enough oxygen to supply
all their energy needs through the electron transport chain.
• Under these conditions, pyruvate is converted to lactic acid.
• This conversion involves the transfer of hydrogen from the NADH formed
during glycolysis to pyruvate, forming lactate.
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TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS
Summary Continued
• The use of NADH to form lactate regenerates the NAD + required to maintain
ATP production by glycolysis.
• The accumulation of lactate causes muscle fatigue.
• Lactate is reconverted to pyruvate in the muscles, and to glucose in the liver.
• The reconversion of lactate to pyruvate or glucose requires ATP formed by
aerobic respiration.
• This additional respiratory activity creates an oxygen debt.
Types of skeletal muscle fibres
• There are two types of muscle fibre, namely slow and fast twitch.
• All skeletal muscles contain a mix of the two types of fibre.
• Athletes show patterns of muscle fibres that reflect their events.
• Slow twitch (Type 1) fibres:
– contract relatively slowly but can sustain contractions longer;
– are good for endurance sporting activities, such as long distance
cycling or running;
– rely on aerobic respiration to generate ATP;
– have many mitochondria, a large blood supply, and a high
concentration of the oxygen-storing protein myoglobin, which makes
them dark red in colour;
– use both glycogen and fat as fuel depending on the intensity of the
exercise.
• Fast twitch (Type 2) fibres:
– contract relatively quickly, but for short periods;
– are good for power sporting activities, such as sprinting or weightlifting;
– generate ATP through glycolysis and the breakdown of creatine
phosphate;
– have few mitochondria and a relatively low blood supply;
– have a low level of myoglobin, making them light pink in colour;
– use both glycogen and creatine phosphate as fuel.
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TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS
12.6
Extended response question
The activity which follows presents an extended response question similar to the style
that you will encounter in the examination.
You should have a good understanding of creatine phosphate in muscles before
attempting the question.
You should give your completed answer to your teacher or tutor for marking, or try to
mark it yourself using the suggested marking scheme.
Extended answer question: Creatine phosphate in muscles
Give an account of the role of creatine phosphate in muscles. (8 marks)
10 min
..........................................
12.7
End of topic test
End of Topic 12 test
Q28: Complete the paragraph using the words from the list. (10 marks)
which will fuel more than
of
Muscle cells do not contain any reserve of
activity. Creatine phosphate is an energy reserve found in cells with high and
energy demands, such as the muscles and the
. There is only sufficient creatine
.
phosphate to fuel strenuous muscle activity for a maximum of
to release energy and
which is used
Creatine phosphate breaks down
to ATP at a fast rate. When energy demands are
,
to convert ADP and
respiration.
creatine phosphate reserves are restored using ATP provided by
Word list: 2s, 10s, aerobic, aerobically, anaerobic, anaerobically, ATP, brain, low, P i ,
phosphate, variable.
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..........................................
Q29: Complete the sentences by matching the descriptions on the left with the terms
on the right. (10 marks)
During vigorous exercise, muscle
cells cannot support
muscles.
Without oxygen, pyruvate is
converted to
muscle fatigue.
Conversion of pyruvate involves
transferring hydrogen from
the electron transport chain.
The use of NADH to form lactate
regenerates
lactate.
NAD+ is required to maintain ATP
production by
NADH.
Accumulation of lactate causes
oxygen debt.
Lactate is reconverted to pyruvate in
the
AD+ .
Lactate is reconverted to glucose in
the
glycolysis.
Reconversion of lactate requires ATP
from
liver.
Reconversion of lactate causes
aerobic respiration.
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TOPIC 12. ENERGY SYSTEMS IN MUSCLE CELLS
..........................................
Q30: Complete the statements by matching the parts on the left with the parts on the
right. (12 marks)
There are two types of muscle fibre,
namely
endurance events.
All skeletal muscles contain
short periods only.
Athletes show patterns of muscle
fibres that
slow and fast twitch.
Slow twitch fibres contract slowly but
aerobic respiration.
Slow twitch fibres are good for
a high concentration of myoglobin.
Slow twitch fibres rely on
glycogen and fat.
A large blood supply and many
mitochondria are linked to
power events.
Slow twitch fibres use fuel stores of
little myoglobin.
Fast twitch fibres contract quickly but
for
reflect their events.
Fast twitch fibres are good for
creatine phosphate and glycolysis.
Fast twitch fibres generate ATP by
both types of muscle fibre.
Few mitochondria and a low blood
sustain contraction longer.
supply are linked to
..........................................
Q31: Explain the occurrence of creatine phosphate in the body. (2 marks)
..........................................
Q32: Describe the supply of energy to muscle cells as they undertake a minute of
intense activity. (3 marks)
..........................................
Q33: Explain the difference between the colour of slow and fast twitch muscle fibres. (6
marks)
..........................................
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Topic 13
End of unit test
Contents
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TOPIC 13. END OF UNIT TEST
End of Unit 1 test
Q1: Complete the statements by matching the parts on the left with the parts on the
right. (12 marks)
Cellular differentiation means that
cells
regulate cell research in the UK.
Muscle cells can only
maintain the chromosome number.
Embryonic stem cells can
express genes characteristic of their
cell type.
Tissue stem cells can
develop more specialised functions.
Germline cells are able to
exemplify epithelial tissue.
Mitosis provides the ability to
develop into any cell type.
Mutations in germline cells
produce a limited range of cell types.
Skin epidermis can be used to
invade neighbouring tissues.
Stem cell research lets scientists
produce gametes.
The Human Tissue Authority has a
remit to
respond to normal regulatory signals.
Cancer cells do not
affect all cells of offspring.
A tumour is benign if it does not
understand cell growth and
differentiation.
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TOPIC 13. END OF UNIT TEST
257
..........................................
Q2: Complete the statements by matching the parts on the left with the parts on the
right. (12 marks)
A nucleotide is composed of
deoxyribose,
strong covalent bonds.
The sugar-phosphate backbone is
held together by
deoxyribose and phosphate.
The base pairings of DNA are
adenine and thymine,
ligase.
At the 3 and 5 ends of the DNA
strands are
uracil.
The enzyme which joins fragments of
DNA together is
the primary transcript.
The formation of mRNA from a DNA
template is
guanine and cytosine.
The base in RNA which replaces
thymine is
carbohydrate groups.
The first strand of RNA formed from
DNA is
a stop codon.
The protein-coding regions of mRNA
are
phosphate and a nitrogenous base.
The polypeptide is released when
reaching
several different mRNA molecules.
One primary transcript can produce
exons.
Post-translational modification may
involve adding
transcription.
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TOPIC 13. END OF UNIT TEST
..........................................
Q3: Complete the statements by matching the parts on the left with the parts on the
right. (12 marks)
Pathways that need an energy input
are called
pyruvate.
The active site lowers
oxaloacetate.
Glycolysis is catalysed by
activation energy.
The product of glycolysis is
fructose.
The product of phosphofructokinase
only enters
a multi-enzyme complex.
Acetyl-coA passes the acetyl group to
the intermembrane space.
A substrate that is converted into a
glycolysis intermediate is
NAD+ and FAD.
Enzymes pump H+ ions into
anabolic.
High energy electrons are carried by
glycolysis.
..........................................
Q4:
Complete the paragraph using the words from the list. (7 marks)
is converted to
.
Without oxygen,
.
The use of NADH to form lactate regenerates
.
Lactate is reconverted to glucose in the
reserves are restored by aerobic respiration.
events.
Slow twitch fibres are good for
Fast twitch fibres generate ATP by creatine phosphate and
.
Word list: creatine phosphate, endurance, glycolysis, lactate, liver, NAD+ , pyruvate.
..........................................
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TOPIC 13. END OF UNIT TEST
The following questions relates to different types of body cell.
The diagrams show two types of somatic cell.
Q5: Name the type of cell division by which somatic cells are formed. (1 mark)
..........................................
Q6: As somatic cells have developed, they have differentiated. What is meant by
'differentiated'? (1 mark)
..........................................
Q7: Name a type of somatic tissue other than those to which the cells above belong
and state its function. (1 mark)
..........................................
Q8: State two differences between cancer cells and normal tissue cells. (2 marks)
..........................................
Q9: Compare the significance of mutations to somatic and germline cells. (2 marks)
..........................................
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TOPIC 13. END OF UNIT TEST
The following questions relate to nucleic acids.
DNA is made up of nucleotides.
Q10: List the components of a nucleotide. (1 mark)
..........................................
Q11: Name the type of bond which holds DNA strands together. (1 mark)
..........................................
Q12: Explain what is meant by the antiparallel structure of DNA. (3 marks)
..........................................
Q13: State one feature of the diagram above which indicates that DNA replication rather
than RNA synthesis is taking place. (1 mark)
..........................................
Q14: Name the enzyme principally involved with DNA replication. (1 mark)
..........................................
Q15: What does this enzyme need before it can start DNA replication? (1 mark)
..........................................
Q16: The enzyme can only add nucleotides to one end of the DNA strand. How does
this affect the replication of the DNA strands? (1 mark)
..........................................
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TOPIC 13. END OF UNIT TEST
The following questions concern mutations.
Q17: What are mutations? (1 mark)
..........................................
Q18: What effect do mutations have? (1 mark)
..........................................
Q19: Which type of chromosome mutation involves the loss of a segment of
chromosome? (1 mark)
a) Deletion
b) Duplication
c) Translocation
..........................................
Q20: Which type of chromosome mutation involves the repetition of a segment or
section of chromosome? (1 mark)
a) Deletion
b) Duplication
c) Translocation
..........................................
Q21: Which type of chromosome mutation involves the rearrangement of chromosomal
material involving two or more chromosomes? (1 mark)
a) Deletion
b) Duplication
c) Translocation
..........................................
Q22: State the most likely outcome of chromosome mutations and suggest an
explanation. (2 marks)
..........................................
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TOPIC 13. END OF UNIT TEST
The following questions relate to metabolic pathways and their control.
Q23: Metabolic pathways are either anabolic or catabolic. Choose one of these terms
and explain what it means. (2 marks)
..........................................
Q24: Control of metabolic pathways may involve the active site of enzymes. Explain the
role of the active site. (2 marks)
..........................................
Q25: Give an example of a pathway using a multi-enzyme complex and state the
advantage of such an arrangement. (3 marks)
..........................................
Q26: Feedback inhibition is very important in regulating the rate of production of
metabolites. State the meaning of the term 'feedback inhibition' and give an example. (2
marks)
..........................................
The following questions relate to the processes of respiration.
Q27: In aerobic respiration, the product of glycolysis enters the mitochondrion. Name
the product. (1 mark)
..........................................
Q28: State what happens to this product first in the mitochondrion. (2 marks)
..........................................
Q29: Lactic acid metabolism is very important to muscle tissue under certain
circumstances. Describe these circumstances. (1 mark)
..........................................
Q30: When their event is over, athletes may be left gasping for breath. Explain this odd
situation, as they are no longer exerting themselves. (3 marks)
..........................................
..........................................
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GLOSSARY
Glossary
Activation energy
the minimum energy required by reactants to allow a reaction to occur
ADH
(anti-diuretic hormone) a hormone produced by the pituitary gland which promotes
water retention by the kidney
Adipose
connective tissue composed mainly of cells that store fat, found beneath the skin,
around internal organs, in bone marrow, and in breast tissue
ADP
adenosine diphosphate
Anabolic
a reaction which builds up complex molecules from simpler ones
ATP
adenosine triphosphate
Bile
a liquid produced by the liver which helps the digestion of fats/lipids by emulsifying
these molecules
Blastocyst
an embryo of about 150 cells (preimplantation) produced by cell division following
fertilisation
Carcinogens
any agent directly involved in causing cancer
Carrier
an individual who is heterozygous for a particular recessive characteristic and
shows a dominant phenotype
Catabolic
a reaction which breaks down complex molecules to simpler ones
cDNA
stands for complementary DNA, or copy DNA, and can be single-stranded or
double-stranded; it is synthesized in vitro from an mRNA template
Centromere
region of the chromosome where two chromosomes are closed together
Chiasma
the point where two homologous chromosomes are touching
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GLOSSARY
Chromatid
daughter chromosomes
Chromatin
a complex of DNA, histones, and non-histone proteins found in the nucleus of a
human cell. This is the material from which chromosomes are made.
Codominant
of alleles, when they are both fully expressed in the phenotype of the heterozygote,
e.g. in the AB blood group.
Codon
set of three adjacent bases on messenger RNA which correspond to a DNA triplet
of bases and an anticodon on transfer RNA
Co-enzyme
organic molecule that acts as a temporary carrier of atoms being removed from or
added to a substrate during a reaction. They aid the action of the enzyme to which
they are loosely bound. Co-enzymes belong to a larger group of molecules called
cofactors which enzymes require for their proper functioning.
Collagen
a tough, fibrous protein that is a major component of connective tissue.
Covalent bond
strong chemical bond involving the sharing of electrons between atoms
Cytokines
substances secreted by specific cells of the immune system which will stimulate
specific immune responses
Cytoskeleton
system of microfilaments and microtubules made of protein that give the cell its
structure and shape
Differentiation
a process, which takes place after cell division, where cells become specialised
cells
Endometrium
inner lining of the uterus
Eukaryote
an organism in which the DNA of the cells is contained within a nucleus and
enclosed in a membrane
FAD
flavin adenine dinucleotide
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GLOSSARY
Gall bladder
the organ in the body which stores bile
Gene
a unit in the hereditary materials of an organism
Glucagon
a hormone produced by the pancreas which raises blood glucose levels, acting in
opposition to insulin
Glycolysis
cell process which converts glucose to pyruvate and occurs in the cytoplasm of
the cells of nearly all organisms
GTP
guanosine triphosphate, which performs a similar role to ATP, but is more specific
Haploid cells
contain only a single set of 23 chromosomes
Histones
the basic proteins around which DNA is coiled and supercoiled. Histones and
non-histone proteins are the chief protein components of chromatin.
Homologous chromosome
a pair of chromosomes which share the same gene sequences, each derived from
one parent
Homologous genes
genes of the same type
Hydrolysis
the breaking of chemical bonds by reaction with water
Hydrophobic
of a molecule which is repelled by water, e.g. lipids (oils and fats)
Insulin
a hormone produced by the pancreas which lowers blood glucose levels
In vitro fertilisation
a technique that unites the egg and sperm in a laboratory instead of inside the
female body
Isomer
molecule with the same chemical formula but different molecular structure
Metabolite
these are intermediates and products of metabolism
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GLOSSARY
Micrometre
(μm), 10-6 m
Minisatellites
highly repetitive DNA consisting of sequences between 10 and 100 base pairs
long, repeated in tandem arrays which vary in size from 0.5 to 40 kbp
Multipotent stem cells
these are true stem cells that can only differentiate into a limited number of types
Mutation
occurs when a DNA gene is damaged or changed in such a way as to alter the
genetic message carried by that gene
Myoglobin
a conjugated protein, resembling haemoglobin, involved in oxygen transport in
muscles
Negative feedback mechanism
self-regulating process in which the build-up of a product inhibits earlier steps in
the pathway that generates it
Neurotransmitters
a range of chemicals which convey messages between neurons, or neurons and
effectors
Nonsense mutation
a change to a single nucleotide in DNA which results in a non-functional protein
being produced
Nucleolus
a small, generally spherical, body found within the nucleus of eukaryotic cells
Oxytocin
a hormone produced by the pituitary gland which stimulates the uterus to contract
during childbirth
Phosphofructokinase
enzyme which catalyses a key regulatory step in glycolysis
Phospholipid bilayer
thin membrane made up of two layers of phospholipid molecules, orientated
such that their hydrophilic ('water-loving') heads are on the outside and their
hydrophobic ('water-hating') tails are on the inside
Pituitary gland
small, pea-sized gland attached to the base of the brain which secretes many
hormones and is central to several homeostatic processes in the body
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GLOSSARY
Pluripotent stem cells
these are stem cells with the potential to make any differentiated cell in the body
Prokaryote
organisms which do not have any organelles contained within membranes in their
cytoplasm, e.g. bacteria
Prolactin
a hormone produced by the pituitary gland which stimulates milk production and
secretion
Ribosomes
structures found in the cytoplasm and attached to ER where protein synthesis
occurs
Rough endoplasmic reticulum
extensive double-membrane sheets in the cytoplasm which are studded with
ribosomes and continuous with the nuclear membrane
Scaffold
organ scaffolds are organs from which all the cells have been removed. What
remains is the extracellular matrix - the part of the organ that supports its shape.
Stem cells
undifferentiated cells that can differentiate into specialised cells or divide by mitosis
to produce more stem cells
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ANSWERS: TOPIC 1
Answers to questions and activities
1 Stem cells
Introduction: Questions (page 3)
Q1:
Groups of the same kind of cells with a common structure and function.
Q2: It consists of several different tissues and it carries out a specific function within a
body system.
Q3: A group of organs working together to perform a particular function, e.g. any two
from:
• circulatory;
• digestive;
• endocrine;
• excretory;
• immune;
• integumentary;
• lymphatic;
• musculoskeletal;
• nervous;
• reproductive;
• urinary;
• vascular;
• vestibular.
Stem cells: Questions (page 4)
Q4:
The process by which an unspecialised cell develops specific functions.
Q5:
They are capable of repeated cell division.
Q6:
Any from:
• Muscle cell: movement;
• nerve cell: carries impulses;
• red blood cell: transport of oxygen.
If you have different answers, check them with your teacher as they may still be correct.
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 1
Embryonic stem cells: Questions (page 6)
Q7: How to use human embryonic stem cells to form specialised cells:
1. Early human embryo Blastocyst
2. Embryo stem cell removed
3. Stem cell cultured in the laboratory
4. Formation of mass of undifferentiated stem cells
5. Undifferentiated stem cells cultured in different culture conditions
6. Formation of specialised cells: nerve cell, muscle cell, gut cells
Q8: The hollow ball of (70-100) cells which implants into the endometrium.
Q9: In the epiblast (inner cell mass) of the blastocyst.
Q10: They are pluripotent, which means that they are able to differentiate into any type
of body cell.
Q11: Any three from:
• diabetes;
• Duchenne's;
• muscular dystrophy;
• heart disease;
• Parkinson's disease;
• traumatic spinal cord injury;
• vision and hearing loss.
If you have different answers, check them with your teacher as they may still be correct.
Tissue stem cells: Questions (page 8)
Q12: Each term or statement listed below refers to the type of stem cells shown in bold:
1. Pluripotent - embryonic
2. Develop into all cell types - embryonic
3. Ability to differentiate into some cells in the body - tissue
4. Give rise to a limited range of cell types - tissue
5. Develop into cell types that are closely related to the tissue in which they are found
- tissue
6. Found in developing embryo - blastocyst - embryonic
7. Have the capacity to become all cell types - embryonic
8. Multipotent - tissue
9. Found in body tissues - tissue
10. Ability to differentiate into all of the cell types - embryonic
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ANSWERS: TOPIC 1
Q13: The two types of tissue stem cells found in bone marrow are:
1. bone marrow stem cells, which can generate bone, cartilage, fat, cells that support
the formation of blood, and fibrous connective tissue.
2. haematopoietic stem cells, which can form all the types of blood cells in the body.
Q14: They are able to give rise to other types of cells, but are limited in their ability to
differentiate.
Extended response question: Embryonic and tissue stem cells (page 9)
Suggested marking scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of seven marks can be gained.
1. Stem cells are undifferentiated cells capable of repeated division to both more
stem cells and cells that will later differentiate to form specialised cells.
2. Embryonic stem cells (ESCs) are found in the inner cell mass of the blastocyst.
3. ESCs are capable of repeated division. . .
4. . . .to form more ESCs and other cell types.
5. ESCs are: capable of forming all the other cell types of the body / pluripotent
6. Tissue stem cells (TSCs) are found throughout the juvenile and adult body.
7. TSCs are capable of repeated division to form more TSCs and other cell types.
8. TSCs: can only form cells of the organ to which they belong / are multipotent.
9. e.g. bone marrow tissue cells can only give rise to bone marrow cells, red blood
cells, platelets, phagocytes and lymphocytes.
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ANSWERS: TOPIC 1
271
End of Topic 1 test (page 10)
Q15:
Cellular differentiation:
cell develops more specialised functions.
Muscle cells only express
the genes characteristic of that type of
cell.
Produced from bone marrow stem cells:
red blood cells, platelets, phagocytes
and lymphocytes.
Cells capable of repeated division:
stem cells.
Stem cells:
relatively unspecialised.
Dividing stem cells produce
more stem cells.
Specialised cells are produced from
dividing stem cells.
Embryonic stem cells are located in the
inner cell mass of the blastocyst.
Through the body:
adult stem cells.
Tissue stem cells
replenish differentiated cells.
Can develop into any cell type:
embryonic stem cells.
Tissue stem cells can produce
a limited range of cell types.
Q16: The process by which a cell develops more specialised functions.
Q17: By expressing only the genes characteristic of that type of cell.
Q18: Red blood cells, platelets, and the various forms of phagocytes and lymphocytes.
Q19: Capable of repeated division, relatively unspecialised.
Q20: More stem cells, specialised cells.
Q21: The inner cell mass of the blastocyst.
Q22: Throughout the body.
Q23: Replenish differentiated cells that need to be replaced.
Q24: Embryonic stem cells are capable of developing into any of the body's cell types
whereas tissue (adult) stem cells give rise to a limited range of cell types typical of the
organ of which they are a part.
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ANSWERS: TOPIC 2
2 Differentiation in cells
Mitosis: Question (page 16)
Q1:
Meiosis: Questions (page 18)
Q2:
Q3:
Q4:
46
Q5:
23
Q6:
23
Q7:
E
Q8:
A
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 2
Q9: D
Q10: C
Q11: B
Differentiation in somatic cells: Questions (page 20)
Q12: 46
Q13: Cells that have become specialised to carry out their various specific functions.
Epithelial tissue: Question (page 22)
Q14: Protection, secretion and absorption.
Connective tissue: Questions (page 23)
Q15: Joints between bones, the rib cage, the ear, the nose, the elbow, the knee and
the ankle.
Q16: Storage of energy, protection of organs, providing a structural framework for the
body, connection of body tissues and connecting epithelial to muscle tissue.
Muscle tissue: Question (page 25)
Q17: The body contains three types of muscle tissue: skeletal, smooth and cardiac.
Cardiac is unique in that it is only found in one organ, the heart.
Two of the muscle types are not under conscious control and are called involuntary
muscle.
The other type of muscle, skeletal, is under conscious control and is called voluntary
muscle.
The two muscle types which are striped are skeletal and cardiac.
The muscle type which forces food through the small intestine is smooth.
The muscle type which maintains posture is skeletal, but its control is both voluntary
and involuntary.
Nervous tissue: Question (page 27)
Q18: Nervous tissue consists of neurons and glial cells.
Sensory neurons conduct impulses into the CNS.
Motor neurons conduct impulses out from the CNS.
Interneurons connect other neurons.
Glial cells maintain a constant environment for neurons.
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ANSWERS: TOPIC 2
Formation of the body organs: Question (page 28)
Q19: The gall bladder is formed from epithelial and muscle tissues, each type of tissue
performing a different function in the organ.
Mutations in germline and somatic cells: Question (page 30)
Q20: Similarity:
• both involve changes to the genetic material.
Differences:
• germline mutations affect all cells in the body whereas somatic mutations only
affect one cell or its descendents;
• germline mutations are inherited whereas somatic mutations are not.
Extended answer question: Four types of human body tissue (page 32)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of eight marks can be gained.
1. Body tissue cells derive from somatic stem cells . . .
2. . . . by repeated mitosis.
3. Epithelial tissue - covers the organ surfaces.
4. Protection - skin / secretion - intestinal glands / absorption - villi.
5. Connective tissue - gives shape to organs and supports them.
6. Protection - skull bones / structural framework - ribs / storage of energy - adipose
tissue / connecting body organs - blood / connecting epithelial to muscle tissue cartilage (in tendons)
7. Muscle tissue - which causes locomotion or movement within organs.
8. Skeletal muscle - locomotion / smooth muscle - in arterioles control of access to
capillary bed / cardiac muscle - contraction of the heart.
9. Nervous tissue - which transmits messages between the central nervous system
and the rest of the body (and within the central nervous system).
10. Neurons - conduct impulses / glial cells - maintain a constant environment for
neurons.
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ANSWERS: TOPIC 2
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End of Topic 2 test (page 32)
Q21:
Somatic cells are found
throughout the body.
Germline cells are
gamete mother cells.
Examples of somatic cells are
B-lympbocytes and motor neurons.
Dividing somatic cells produce
more somatic cells.
Mitosis
maintains the chromosome number.
Somatic cells form different types of
tissue by
differentiation.
Cells capable of producing gametes are
germline.
Mutations in somatic cells affect the
cell or tissue involved.
Q22:
Somatic cells divide:
mitosis
Number of types of tissue in stomach:
four
Produces more germline cells:
mitosis
Germline cells produce gametes:
meiosis
Germline chromosome complement:
diploid
Gamete chromosome complement:
23
Mutations in germline cells:
affect all cells of offspring
Q23: Skin epidermis is an example of epithelial tissue, the function of which is
protection.
Adipose tissue is used to store energy and is an example of connective tissue.
Smooth muscle has the function of producing force and motion and is found in artery
walls.
Glial cells are found in nervous tissue where they make the myelin sheath of some
neurons.
Q24: Throughout the body, apart from the gamete mother cells.
Q25: Any two from:
• skin epidermis;
• B lymphocytes;
• motor neurons;
• any other named body cells.
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ANSWERS: TOPIC 2
Q26: More somatic cells.
Q27: Mitosis
Q28: It maintains the diploid chromosome number and the genetic complement.
Q29: Differentiation
Q30:
Tissue
Function
Example
Epithelial
absorption /
protection /
secretion
intestinal glands /
skin epidermis /
villi
Connective
connecting body organs /
connecting epithelial to
muscle tissue /
protection /
storage of energy /
structural framework
adipose tissue /
blood /
cartilage of tendons /
ribs /
skull bones
Muscle
produce force and motion
cardiac - heart /
skeletal - biceps /
smooth - artery walls
Nervous
react to stimuli, carrying
impulses to various parts
of the body
glial cells /
inter-neurons /
motor /
sensory
Q31: All four types.
Q32: A cell that is capable of passing on its genetic information to the next generation
by giving rise to gametes.
Q33: Mitosis
Q34: Meiosis
Q35: Germline cells: diploid, which means two of each type of chromosome.
Gametes: haploid, which means one of each type of chromosome.
Q36: Mutations in somatic cells only affect the cell or the cell line involved. In germline
cells, mutations would be passed on to all cells of the offspring.
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 3
3 Research: Stem cells and cancer cells
Stem cell research: Questions (page 41)
Q1: They are produced from a single group of embryonic stem cells and can be used
for research work in stem cells.
Q2: They are formed from donated spare embryos which are at the blastocyst stage.
Q3: Tissue stem cells
Q4: It does not involve the use of embryonic stem cells, which is a contentious issue
when using stem cells in research work.
Therapeutic uses of stem cells: Questions (page 45)
Q5: Embryonic stem cells are pluripotent / capable of differentiating into any body cell.
Epidermal stem cells are multipotent / only capable of forming a limited range of cell
types.
Q6: Reduced possibility of dehydration, infection or rejection.
Q7: They are unable to differentiate into other types of cells for transplant.
Q8: The matching of genetic types is to make sure that the patient will not reject the
transplantation.
Q9: Advantages of corneal repair using stem cells:
• the cornea is repaired and the limbus is able to produce new epithelial tissues to
replace further damage to the corneal epithelium.
• matching is required for the donor scheme, and the limbus of the person is still
damaged and unable to produce corneal epithelium.
Ethical issues and the regulation of stem cell use: Question (page 47)
Q10:
For:
• surplus embryos from IVF would otherwise be destroyed without any value being
put on them;
• as they are pluripotent, they can be used to study a wide range of conditions
affecting many different cell types.
Against:
• human life should not be artificially created, especially for research;
• a zygote or blastocyst is a human being with the same right to life as anyone else.
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ANSWERS: TOPIC 3
The biology of cancer cells: Questions (page 50)
Q11: Cancer cells divide excessively and do not respond to regulatory signals.
Q12: The tumour cells fail to attach to each other and can spread throughout the body
in the blood and lymphatic systems.
Extended answer question: Stem cell research work (page 52)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of six marks can be gained.
A) Corneal transplants (maximum of 3 marks):
1. Traditionally, the person would have to wait for donated corneal tissues to
become available.
2. With the use of stem cells isolated from the patient's healthy eye, new tissue
can be cultured. . .
3. . . .which can be transplanted back into the damaged eye,
4. and will repair the damaged corneal epithelium.
B) Skin grafts (maximum of 3 marks):
i
Traditionally, a section of the patient's healthy skin is cultured over 3 weeks.
ii
During this time, dehydration and infection are risks for the patient.
iii Skin stem cells from the patient can be cultured to produce new epidermal
cells much more quickly.
iv This reduces the time during which the patient is at risk.
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ANSWERS: TOPIC 3
279
End of Topic 3 test (page 53)
Q13:
One reason stem cells are useful
for research is that
they can be cultured indefinitely.
Cells which can differentiate into all
cell types are
pluripotent.
Stem cell research has contributed
to understanding
cell growth and differentiation.
The use of stem cells has improved
corneal repair by
reducing the chance of rejection.
The use of stem cells has improved
skin graft procedures by
reducing the chance of dehydration.
The development of this can be
studied using stem cells
motor neuron disease.
The use of stem cell cultures to test
a wide range of drugs helps
treatment of
Alzheimer's disease.
Ethical objections have been raised
against the use of
embryonic stem cells.
An ethical objection against the use
of stem cells is that
a zygote is a human being.
This regulates cell research in the
UK
The Human Tissue Authority.
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ANSWERS: TOPIC 3
Q14:
Unlike normal cells, cancer cells
divide uncontrollably.
Cancer cells do not respond to
normal regulatory signals.
A tumour is
a mass of abnormal cells.
Tumours develop as a result of
mutations.
The genes involved in tumour
development control cause
slower cell division and programmed
cell death.
A tumour is benign if
it does not invade neighbouring
tissues.
A tumour becomes malignant if
its cells lose the molecules that
normally hold them in place.
Secondary tumours develop when
cells detach from a primary tumour
and spread through the body.
Q15: They are pluripotent (capable of differentiating into all cell types).
They can be cultured indefinitely in the laboratory.
Q16: Cell growth, differentiation and gene regulation.
Q17: Stem cells have improved the therapeutic procedures of:
i
corneal repair by speeding up the process, creating new healthy tissue and
reducing the chance of rejection;
ii
skin grafts by the much more rapid production of a temporary graft, reducing the
risk of dehydration or infection.
Q18: Parkinson's disease, motor neuron disease.
Q19: They can be used to produce very specific cultures for use in testing a wide range
of drugs.
Q20: Embryonic stem cells
Q21: A zygote is a human being with a right to life.
Human beings, in the form of embryos, should not be artificially created.
Q22: The Human Tissue Authority.
The Human Fertilisation and Embryology Authority.
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 3
Q23: Cancer cells divide uncontrollably to produce a mass of abnormal cells called a
tumour.
Cancer cells do not respond to normal regulatory signals which would instruct them to
stop dividing when necessary.
Q24: Mutations in the genes controlling cell division, programmed cell death, and
differentiation.
Q25: It does not invade neighbouring tissues or spread throughout the body.
Q26: The cells lose the molecules on their surface that normally hold them in place.
Q27: Cells become detached from a primary tumour, spreading through the body in the
blood or lymphatic circulation.
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ANSWERS: TOPIC 4
4 DNA structure and replication
Introduction: Questions (page 58)
Q1:
Amino acids joined by peptide bonds.
Q2: Catalase - enzyme; collagen/elastin/keratin - structural; insulin - cell signalling.
If you have different answers, check them with your teacher as they may still be correct.
DNA structure: Questions (page 65)
Q3:
Double helix
Q4:
Hydrogen
Q5:
Weakness
Q6:
DNA Base
Complementary base
Adenine
Thymine
Cytosine
Guanine
Guanine
Cytosine
Thymine
Adenine
Q7:
A-T, because it only has two hydrogen bonds whereas C-G has three.
Q8:
Ribose sugar, phosphate, base.
Q9:
Antiparallel
Q10: 3 end - deoxyribose; 5 end - phosphate.
Arrangement of DNA in chromosomes: Questions (page 67)
Q11: It is tightly looped and coiled.
Q12: Histones and non-histone proteins.
Q13: There are 22 homologous pairs of chromosomes, with each member of each pair
carrying the same genes, and the two sex chromosomes, which are X and X in a female,
and X and Y in a male. Therefore, the total number of different chromosomes is 22 + 2
= 24.
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 4
283
DNA replication: Question (page 71)
Q14:
DNA is unwound and
unzipped to form
two template strands.
DNA is unzipped by
the enzyme helicase.
At the end of the 3 strand is
deoxyribose.
At the end of the 5 strand is
phosphate.
To start attaching
nucleotides, DNA
polymerase requires
a primer.
DNA polymerase
attaches nucleotides to
the
3 end of the DNA strand.
The strand that is
continuously replicated is
the
leading 3 strand
The strand replicated in
fragments is the
lagging 5 strand
The enzyme which joins
DNA fragment is
ligase
In addition to a DNA
molecule and enzymes,
DNA replication needs
ATP and free nucleotides
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ANSWERS: TOPIC 4
Extended response question: The location and structure of DNA, and DNA
replication (page 73)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of fifteen marks can be gained.
A) Describe the location and structure of DNA. (maximum of 8 marks):
1. DNA is located on chromosomes in the nucleus.
2. A DNA molecule consists of two strands wound in a double helix.
3. Each strand consists of subunits called nucleotides.
4. A nucleotide consists of a deoxyribose sugar molecule, a phosphate group
and an organic / nitrogenous base.
5. The deoxyribose and phosphate are linked to their neighbours to form a
sugar-phosphate backbone.
6. There are four organic bases: adenine (A), thymine (T), guanine (G) and
cytosine (C).
7. Bases are linked in complementary pairs A-T and G-C.
8. Bases link the two DNA strands by hydrogen bonds.
9. The two DNA strands have an antiparallel structure / explanation.
10. Deoxyribose is found at the 3 end of each strand and phosphate is found at
the 5 end.
B) Give an account of the replication of DNA. (maximum of 7 marks):
i
DNA is unwound and unzipped. . .
ii
. . .by the enzyme helicase. . .
iii . . .to form two template strands.
iv DNA polymerase is the enzyme which adds nucleotides to the new DNA
strand.
v
DNA polymerase needs a primer to start replication.
vi DNA polymerase can only add complementary nucleotides to the
deoxyribose / 3 end of the DNA strand.
vii This results in one strand / the leading strand being continuously replicated,
viii and the other strand / the lagging strand being replicated in fragments,
ix which are joined together (by the enzyme ligase).
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End of Topic 4 test (page 74)
Q15:
DNA is the molecule of inheritance
and is
able to direct its own replication.
The base sequence of DNA is
used by cells to store genetic
information.
DNA is stored on chromosomes and
is
tightly coiled with associated proteins.
The basic unit of DNA is
the nucleotide.
A nucleotide is
composed of deoxyribose, phosphate
and a base.
The carbons in deoxyribose are
identified by their position in the
molecule.
The sugar-phosphate backbone is
held together by strong covalent
bonds.
The bases of DNA are
adenine, cytosine, guanine and
thymine.
The base pairings of DNA are
adenine and thymine, guanine and
cytosine.
The two strands of DNA are
held together by hydrogen bonds.
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Q16:
The shape of the DNA molecule is a
The two strands of DNA are
double helix.
antiparallel.
At the 3 and 5 ends of the DNA strands are
deoxyribose and
phosphate.
The enzyme that adds nucleotides to the new DNA strand is
DNA polymerase.
Before DNA replication can start, what is required is a
primer.
Complementary nucleotides are added to the
3
end of the
The DNA strand which is continuously replicated is the
new DNA strand.
3 or leading strand.
The enzyme that joins fragments of DNA together is
ligase.
Q17: It is the molecule of inheritance and can direct its own replication.
Q18: In the base sequence of DNA.
Q19: It is tightly coiled and packaged with associated proteins.
Q20: Name: nucleotide
Composition: deoxyribose sugar, phosphate, and organic / nitrogenous base.
Q21: They are numbered according to their position on the molecule, e.g. 1 .
Q22: By strong (covalent) bonds between deoxyribose and phosphate groups.
Q23: Bases: adenine (A), thymine (T), guanine (G) and cytosine (C). (NB: The question
asks for the names so the letters alone would not be correct)
Complementary pairs: A-T and G-C (NB: Letters are acceptable here because they
have been linked to the base names in the first part of the answer)
Q24: Hydrogen bonds between the complementary base pairs.
Q25: The hydrogen bonds that hold them together are weak bonds (compared to those
that hold the sugar phosphate backbone together).
Q26: Double helix
Q27: Antiparallel
Q28: 3 : deoxyribose;5 : phosphate.
Q29: Helicase unwinds and unzips the DNA molecule.
Q30: DNA polymerase
Q31: It allows DNA polymerase to start replication.
Q32: The deoxyribose / 3 end of the strand.
Q33: The 3 strand / leading strand.
Q34: The enzyme ligase joins them.
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5 Gene expression in human cells
Gene expression through protein synthesis: Questions (page 80)
Q1: a) True
Q2: Extracellular
Q3: ADP
Q4: Transcription and translation
Q5: 48
Structure and functions of RNA: Questions (page 83)
Q6:
Q7:
RNA
DNA
Structure
not a double helix
double helix
Preferred form
single-stranded
double-stranded
Number of types
>1
1
Present in
the cytoplasm and the
nucleus
the nucleus
Nitrogenous bases
adenine, uracil, guanine,
cytosine
adenine, thymine,
guanine, cytosine
Q8: mRNA, tRNA and rRNA
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Q9: mRNA copies the code from the DNA molecule and carries it out to the ribosomes
where the proteins are synthesised.
tRNA molecules are found in the cytoplasm. They attach to specific amino acids and
bring them to the ribosomes where the amino acids are joined together.
Q10: c) both DNA and RNA
Q11: rRNA and proteins
Transcription of DNA into mRNA: Questions (page 90)
Q12: Translation
Q13: Helicase unwinds the DNA double helix and unzips the DNA molecule.
RNA polymerase binds the RNA nucleotides together.
Q14: b) DNA → RNA
Q15: d) GACUG
Q16: The mRNA molecule must start at the 5 end.
Q17: Addition of the 5 cap, joining of the exons, and removal of introns.
Q18: RNA splicing
Translation of mRNA into a polypeptide: Questions (page 96)
Q19:
1. DNA
2. tRNA
3. Hydrogen bonds
4. Amino acid
5. Start
6. Small subunit
7. mRNA
8. Peptide bonds
9. Ribosome subunits disassociation
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ANSWERS: TOPIC 5
Q20:
Single gene, several proteins: Question (page 100)
Q21: One gene can be expressed as many proteins as a result of RNA splicing
and post-translational modification. Different mRNA molecules are produced from the
same primary transcript depending on which RNA segments are treated as exons and
introns. Protein structure can be altered post-translation by:
cutting and combining polypeptide chains, e.g. the formation of insulin from proinsulin;
adding phosphate or carbohydrate groups to the protein.
Extended response question: Transcription and post-translational modification
(PTM) (page 103)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of ten marks can be gained.
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ANSWERS: TOPIC 5
A) Transcription (maximum of 6 marks):
1.
2.
3.
4.
5.
6.
7.
8.
Transcription is the formation of a mRNA molecule on a DNA template.
DNA unwinds, and strands separate by the action of helicase.
RNA nucleotides attach to exposed bases of DNA.
DNA A pairs with RNA U, DNA T - RNA A, DNA G - RNA C and DNAC - RNA
G.
RNA polymerase joins nucleotides (to each other).
This produces the primary transcript.
Which contains introns and exons.
Exons are protein coding sections of mRNA and introns are non-coding
sections.
B) Post-translational modification (PTM) (maximum of 4 marks):
i
ii
iii
iv
v
vi
It takes place after the polypeptide has been completed on the ribosome.
It allows several proteins to be formed from one gene.
It may involve enzyme cutting and combining of polypeptide chains.
e.g. insulin from pro-insulin.
It may involve adding of phosphate or carbohydrate groups.
Addition of phosphate enables enzymes / receptors to be switched 'on' and
'off'.
End of Topic 5 test (page 104)
Q22: The individual's phenotype is determined by the proteins produced by gene
expression. These proteins are generated from the DNA in the nucleus of each cell,
but in any one cell only a fraction of the genome is expressed. Intracellular factors can
influence gene expression, e.g. phosphofructokinase is activated by high levels of
ADP. Extracellular factors may also have an effect, e.g. low blood blood sugar levels
activate the production of glucagon by the alpha cells of the pancreas. Gene expression
has two stages, transcription in the nucleus and translation at the ribosome, both of
which may be used to control the process. The code of a gene consists of a series
of triplets of bases in the DNA. A triplet of bases, of which there are four types, can
represent 64 different instructions.
Q23:
RNA:
Phosphate, base, ...
Adenine, cytosine, guanine, ...
G-C, ...
Messenger, transfer, ...
mRNA:
tRNA:
rRNA:
Anticodon:
single strand of nucleic acid
ribose sugar
uracil
A-U
ribosomal
nucleus to cytoplasm
amino acid to ribosome
reads mRNA code
3 exposed bases on tRNA
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Q24:
Transcription:
DNA triplet code to mRNA.
RNA polymerase:
DNA unwound and unzipped.
RNA base uracil:
DNA base adenine pairs with it.
mRNA codon:
DNA triplet of bases matches with it.
Primary transcript:
first RNA strand formed from DNA.
Exons:
protein-coding regions of mRNA.
Introns:
non-coding regions of mRNA.
RNA splicing:
joining up of exons.
Mature transcript:
final form of mRNA.
Q25:
1. Translation takes place at the ribosome.
2. Ribosomes consist of rRNA and proteins.
3. At the ribosome, rRNA reads the incoming mRNA.
4. When a start codon is reached, the ribosome starts attaching tRNA bases to the
mRNA.
5. mRNA has sets of three bases called codons and tRNA has matching sets called
anticodons.
6. Each tRNA molecule carries a specific amino acid.
7. Enzymes in ribosome join the amino acids by peptide bonds.
8. tRNA disengages and goes off to find another of its amino acids.
9. When a stop codon is reached, the ribosome diassociates and the polypeptide is
released.
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Q26:
The body contains many more
proteins than there are
genes.
One gene must be able to code for
several proteins.
One primary transcript can produce
several different mRNA molecules.
Alternative RNA splicing treats
different RNA segments as
exons and introns.
Post-translational modification may
involve adding
carbohydrate groups.
Cutting and combining of
polypeptides is another form of
post-translational modification.
Q27: Gene expression produces the proteins which determine phenotype.
Q28: Phosphofructokinase is activated / by high levels of ADP. (or other suitable)
Q29: Low blood sugar levels activate the (alpha cells of the islets of Langerhans in the)
pancreas / to produce glucagon.
Q30: In any nucleotide there are 4 possible bases, so with three nucleotides there are
4 x 4 x 4 = 64 possible combinations.
Q31: A-U, G-C, T-A
Q32: messenger RNA, transfer RNA, ribosomal RNA
Q33: The transfer of the gene information from the triplet code on the DNA template to
mRNA.
Q34: Introns are removed and / the exons are joined together.
Q35: Joining of the exons together.
Q36: Base pairing between different parts of the mRNA strand fold the molecule,
exposing three bases at one end (the anticodon) / and an attachment point for an amino
acid at the other end.
Q37: Any one amino may be coded for by more than one anticodon.
Q38: The ribosome disassociates / and the chain of amino acids is released as a
polypeptide.
Q39: Different mRNA segments being treated as exons and introns.
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6 Genes and proteins in health and disease
Proteins: Questions (page 112)
Q1: Carbon, hydrogen, nitrogen, oxygen.
Q2: Amino acids; 20.
Q3: Peptide
Q4: Polypeptide
Q5: The arrangement of amino acids in polypeptide chains;
the association of more than one polypeptide chain in the mature protein;
the inclusion of non-protein structures, e.g. phosphates, metal ions, and carbohydrates.
Different levels of protein structure: Question (page 117)
Q6:
Level of structure
Description
Primary:
the sequence of amino acids.
Secondary:
held by weak hydrogen bonds between amino acids.
Tertiary:
held by strong (disulphide) bonds between amino acids.
Quaternary:
two or more polypeptide chains associated together.
Functions of proteins: Questions (page 119)
Q7:
Type of protein
Example
structural
collagen, elastin
contractile
actin and myosin in muscle cells
hormones
insulin
receptors
insulin receptor in liver cells (forming part
of the structure of the plasma membrane)
transport proteins
transporter of glucose into the cell
defence proteins
immunoglobulins
enzymes
lipase, pepsin, maltase
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Treatment of bacteria with UV light: Question (page 121)
Q8: As the length of time for which the bacteria are exposed to UV increases, so the
number of bacteria killed increases.
Mutations and genetic disorders: Questions (page 121)
Q9:
An agent which increases the rate of mutation.
Q10: Radiation, e.g. X-rays, UV light; chemicals, e.g. benzene, arsenic.
Effects of gene mutations on amino acid sequences (page 125)
Q11: 4 amino acids have been changed; yes, a deletion mutation is a frameshift
mutation.
Q12: 4 amino acids have been changed; yes, an insertion mutation is a frameshift
mutation.
Q13: 1 amino acid has been changed; no, a substitution mutation is a point mutation,
not a frameshift mutation.
Gene mutations: Questions (page 129)
Q14: a) one nucleotide
Q15: b) nonsense
Q16: c) splice-site
Q17: a) missense
Q18: b) sometimes
Q19: b) after it
Q20: b) after it
Q21: a) 0
Q22: d) all of these
Q23:
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Q24:
Q25:
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Q26: If three nucleotides are involved, only a single amino acid will be affected.
Q27: A section of a gene with a repeating series of triplets of bases.
Q28: Deletion or insertion mutations.
Q29: Huntingdon's disease / Fragile X syndrome.
The effect of mutations on the structure and function of proteins: Question (page
134)
Q30:
Inherited disease
Beta thalassemia
Cystic fibrosis
Cause
splice-site, autosomal recessive
Duchenne muscular dystrophy (DMD)
nonsense, X-linked recessive
Phenylketonuria (PKU)
missense, autosomal recessive
missense, codominant
Sickle-cell disease
Tay-Sachs syndrome
frameshift deletion, autosomal recessive
frameshift insertion, autosomal recessive
Chromosome structure mutations: Question (page 137)
Q31:
Deletion and duplication are examples of
structural
abnormality.
Chromosome mutations occur during
cell division.
Deletion is loss of a segment of chromosome
during
meiosis I.
The condition Cri-du-Chat is caused by a
deletion.
Extra chromosomal material will be present
because of
duplication.
Duplication of these genes is the cause of many
cancers
oncogenes.
The most common effect of an inversion mutation
is
reduced fertility.
A gene from chromosome 22 fusing a gene on
chromosome 9
translocation.
A cancer of stem cells in the bone marrow
myeloid leukaemia.
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Extended response question: Proteins and gene mutation (page 140)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of seventeen marks can be gained.
A) Give an account of the structure and function of proteins.
a) Protein structure (maximum of 4 marks):
1.
2.
3.
4.
5.
6.
Proteins consist of amino acids joined together (in chains).
The amino acids are joined by strong peptide bonds. . .
. . .to produce the primary structure of the polypeptide.
Further bonding between amino acids. . .
. . .produces the secondary and tertiary structures.
Folding of the polypeptide chains forms the three-dimensional shape of
the protein.
b) Protein function (maximum of 4 marks):
I
II
III
IV
V
Some proteins are enzymes + named example (e.g. the digestive
enzyme amylase).
Some proteins are hormones + named example (e.g. insulin).
Some proteins are antibodies which help the body fight infections.
Some proteins transport substances + named example (e.g.
haemoglobin).
Some proteins provide structure + named example (e.g. collagen).
B) Give an account of gene mutation. (maximum of 9 marks)
i
A gene mutation is a change in the base type or sequence in a gene.
ii
In a substitution (mutation) one base is replaced by another.
iii If substitution produces a new stop codon it is a nonsense mutation.
iv If substitution affects introns and exons it is a splice-site mutation.
v
If substitution changes one mRNA codon it is a missense mutation.
vi In a deletion (mutation) a base is removed.
vii In an insertion (mutation) a base is added.
viii A change to a single nucleotide / base is a point mutation.
ix Insertion and deletion (mutations) can potentially cause frameshift mutation.
x
A frameshift mutation alters all the triplets following it.
xi If any of these mutations occurs in a protein-coding gene, then the protein
produced may be altered (or not produced at all).
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ANSWERS: TOPIC 6
End of Topic 6 test (page 141)
Q32: Proteins are composed of chains of amino acids joined together by peptide
bonds to form a polypeptide. They are held in complex three-dimensional shapes
by hydrogen bonds and interactions between individual amino acids. In addition,
several polypeptide chains may associate together and non-protein groups such as
carbohydrates and metal ions may be bound into the final protein structure. Proteins
have many functions including acting as enzymes (e.g. phosphofructokinase),
hormones (e.g insulin), transport (e.g. haemoglobin), and structural components (e.g.
collagen).
Q33:
Term
Point mutation
Definition
single base involved
Substitution
Missense
one base replaced
Nonsense
Splice-site
new stop codon
Deletion
Insertion
Frameshift
one codon replaced
new exon-intron codons
removal of a base
addition of a base
base sequence changed
Q34:
Of what are chromosome structural mutations the result?
(all of these) addition,
removal, reversal
Which type of mutation results from the loss of a segment of
chromosome?
deletion
Which codons are potentially altered by structural mutations?
(all of these) exons,
introns, stop codons
Which type of mutation causes Cri-du-Chat?
Which type of mutation is associated with breast cancer?
deletion
duplication
Which type of mutation causes chronic myeloid leukaemia?
translocation
Which type of mutation usually has not noticeable effect?
inversion
What is the likely effect of substantial changes to the
chromosome?
lethal effect
Q35: Peptide
Q36: A chain of amino acids
Q37: They hold the polypeptides in complex shapes.
Q38: Carbohydrates and metal ions
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7 Human genomics
Sequencing DNA: Questions (page 146)
Q1: 3
Q2: DNA sequence variations that occur when a single nucleotide in the genome
sequence is altered.
Q3: Coding regions
Bioinformatics: Questions (page 147)
Q4: Computer science, statistics, mathematics and engineering.
Q5: Coding sequences similar to known genes, start codons, or sequences lacking
stop codons.
Q6: They identify base sequences that correspond to the amino acid sequence of the
protein.
Systematics: Questions (page 148)
Q7: c) Homologous
Q8:
Scientists can work out the number of differences in an amino acid sequence between
two species.
They can then calculate the time since the two species diverged from a common
ancestor.
If there are lots of differences between the maps, it can be deduced that the species
diverged longer ago than if there are only a few differences.
Personalised medicine: Question (page 150)
Q9: Any four from:
• production of more 'powerful' medicines;
• production of better and safer drugs more quickly;
• more accurate methods of determining drug dosages;
• advanced screening for a particular disease;
• allows a person to make adequate lifestyle and environmental changes;
• pharmaceutical companies will be able to discover potential therapies more easily.
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ANSWERS: TOPIC 7
Amplification and detection of DNA sequences: Question (page 155)
Q10: After the mixing of the components in the PCR process, three further stages
follow. Heating to 95 ◦ C which denatures the DNA by separating its two strands.
The temperature is then dropped to 54 ◦ C, which allows primers to bind to their
complementary sequences on the separated strands.
Primers are short DNA
sequences of about 20 bases that bind on either side of the target sequence. This
stage is known as annealing. The temperature is then raised again to 72 ◦ C in the
extension stage. Taq polymerase adds nucleotides to the 3 ends of the primers and
extends them into new complementary strands. Continued cycles of heating and cooling
cause the DNA to be doubled repeatedly, so that in one hour, 20 cycles will amplify the
target DNA more than 1 million times.
DNA probes: Questions (page 156)
Q11: c) identify a particular sequence of DNA.
Q12: b) The pentose sugar ring of a gene probe is labelled with radioactive
14 C.
Medical and forensic applications: Questions (page 159)
Q13:
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Q14:
1.
DNA extracted from blood sample.
2.
DNA digested with a restriction enzyme.
3.
Gel electrophoresis.
4.
5.
Blotting.
6.
Autoradiography.
Hybridisation with labelled probe.
Q15: c) Suspect 3
Q16: There is the possibility that the suspect's DNA fingerprint matches that left at the
scene of the crime purely by chance.
Q17: By increasing the number of probes hybridised to the DNA blot.
Extended response question: The PCR process (page 163)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of eight marks can be gained.
1. The components of the PCR (template DNA, primers, nucleotides and the Taq
DNA polymerase) are mixed together.
2. Heating to 90 - 95 ◦ C. . .
3. . . .separates (denatures) the double-stranded DNA (containing the target DNA
sequence).
4. Annealing at 54 ◦ C. . .
5. . . .allows the primers to bind (by forming hydrogen bonds) to their complementary
sequences on the separated strands
6. Primers are short DNA sequences (each about 20 bases long),
7. that bind at either side of the target sequence (one on each of the complementary
strands of the target DNA).
8. Extension at 72◦ C.
9. Taq polymerase now adds nucleotides to the 3 ends of the primers and extends
them into new complementary strands.
10. Repeated cycles multiply the target DNA exponentially. . .
11. . . .since each new double strand separates to become two templates for further
synthesis.
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End of Topic 7 test (page 164)
Q18:
The sequence of bases can be
determined for individual genes by
using
DNA sequencing.
The use of computer technology to
identify DNA sequences is known
as
bioinformatics.
Comparison of the human genome
with genomes of other species is
known as
systematics.
An understanding of the genetic
component of the risk of disease is
required for
personalised medicine.
A technique for the amplification of
DNA in vitro is known as
polymerase chain reaction (PCR).
Short sequences of
single-stranded DNA that are
complementary to specific target
sequences are known as
primers.
Short single stranded fragments of
DNA are known as
DNA probes.
Comparison of highly variable
repetitive sequences of DNA is
known as
DNA profiling.
Q19:
1. Temperature of the reaction adjusted to 95 ◦ C.
2. Penetration of the double-stranded DNA.
3. Temperature of the reaction adjusted to 55 ◦ C.
4. Annealing of the primers to the single stranded DNA.
5. Temperature of the reaction adjusted to 72 ◦ C.
6. Synthesis of DNA by the enzyme DNA polymerase.
Q20: a) the full complement of DNA possessed by the organism.
Q21: b) The enzyme is relatively stable at high temperatures.
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Q22: During the process of PCR, extension occurs at 72 ◦ C.
Q23: During the process of PCR, DNA primers bind to DNA single strands during
annealing.
Q24: During the process of PCR, heating separates the DNA into single strands.
Q25: b) 8
Q26: a) Primers allow DNA probes to attach to single stranded DNA.
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ANSWERS: TOPIC 8
8 Cell metabolism
Metabolism: Question (page 169)
Q1:
All the chemical reactions of the body:
metabolism
The chemical on which an enzyme acts:
substrate
Chemical produced by an enzyme's action:
product
Chemicals involved in cells reactions:
metabolites
Anabolic and catabolic pathways: Question (page 171)
Q2:
Anabolic
Catabolic
Biosynthesis
Breakdown
Requires energy input
May release energy
ATP → ADP + Pi
ADP → Pi → ATP
DNA polymerase
Glycogen phosphorylase
Reversible and irreversible pathways: Questions (page 174)
Q3: Reversible reactions only change the arrangement of atoms in a molecule,
whereas irreversible reactions add phosphate.
Reversible reactions do not require ATP to release energy, whereas irreversible
reactions require the input of energy from ATP breakdown.
Q4: The presence of the substrate will cause the reaction to run from left to right.
Build-up of the substrate will cause the reaction to run from right to left.
Q5: The pentose phosphate pathway takes glucose 6-phosphate and produces
NADPH and a 5-carbon (pentose) sugar.
Control by regulation of enzyme production: Questions (page 177)
Q6:
Mutation of the gene (for an enzyme in a metabolic pathway).
Q7:
The enzyme for converting phenylalanine to tyrosine is absent.
Q8:
Extracellular come from outside the cell; intracellular are formed within the cell.
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Q9:
i
An example of an extracellular signal molecule is any hormone, e.g. estrogen,
cytokines;
ii
An example of an intracellular signal molecule is the proteins in the cascade that
are initiated by cytokines attaching to their receptors in the cell membrane. (names
definitely not needed)
Enzyme regulation: Question (page 177)
Q10:
Concentration of end-product:
negative feedback
Post-translational modification:
glycogen synthase
Local environment:
changing pH
Localisation within the cell:
Golgi apparatus
Reversible reactions: Question (page 182)
Q11:
1. The presence of the substrate will cause the reaction to go from left to right.
2. Build up of the product will cause the reaction to be reversed.
Multi-enzyme complexes: Questions (page 182)
Q12: Any of:
• RNA/DNA synthesis;
• glycolysis;
• TCA cycle.
Q13:
1. The product of one reaction is immediately available for the next.
2. The product of one reaction is presented so that it fits immediately into the active
site of the next enzyme in the chain.
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Competitive and non-competitive inhibition of enzymes: Question (page 185)
Q14:
Extended response question: The control of the action of enzymes (page 189)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of ten marks can be gained.
1. Enzymes that are in constant use are continuously present.
2. The presence of the substrate may initiate a reaction.
3. The build-up of the product may reverse the direction of a reaction.
4. The active site is the part of the enzyme molecule to which the substrate
(temporarily) binds.
5. Competitive inhibitors resemble the substrate and compete with it for the active
site.
6. Increased substrate concentration reduces the effect of competitive inhibition.
7. Competitive inhibition is reversible.
8. Non-competitive inhibitors bind to allosteric sites. . .
9. . . .and alter the shape of the sites, preventing the substrate from binding to them.
10. Enzyme activators bind to allosteric sites and alter the shape of the active site to
fit the substrate.
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ANSWERS: TOPIC 8
307
11. Feedback inhibition involves non-competitive inhibition.
12. Products from reactions late in metabolic pathway act as non-competitive inhibitors
to enzymes earlier in the pathway.
End of Topic 8 test (page 189)
Q15:
Biosynthetic means
building up complex molecules.
Anabolic pathways
require energy input.
Catabolic pathways
usually release energy.
Steps which require energy are
irreversible.
Reactions which can run both ways are
reversible.
Alternative routes
by-pass steps in a pathway.
Q16:
Regulation of the activity of key enzymes
enables control of
metabolic pathways.
Extracellular signal molecules regulate
hormones.
Protein-protein interactions are set off by
intracellular signal
molecules.
Some enzymes are always present in cells as a
result of
continuously
expressed genes.
A reaction may be initiated by the presence of
substrate.
A reaction may be reversed by the build-up of
product.
Q17: The active site is the point on the enzyme molecule to which the substrate
molecule(s) temporarily binds. In many cases, the active site does not exactly fit the
substrate and it is the substrate molecule itself which causes the necessary change in
the conformation of the enzyme molecule. This is called induced fit.
The active site orientates reactants, lowers activation energy, and releases products
with low affinity for the active site.
Many metabolic reactions are reversible and the presence of a substrate or relative
concentration of a product will determine their direction.
Glycolysis is an example of metabolic pathway controlled by a metabolon, or multienzyme complex.
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ANSWERS: TOPIC 8
Q18:
Sites where activators may bind:
allosteric.
Inhibitors bind to sites away from the active site:
non-competitive.
Inhibitors which resemble substrate:
competitive.
Increase reduces competitive inhibition:
substrate
concentration.
Competitive inhibition:
reversible.
End-products inhibit earlier enzymes:
feedback control.
Q19: Anabolic pathways build up more complex molecules from simpler ones and
require an input of energy. Catabolic pathways break down complex molecules to
simpler ones and generally release energy.
Q20: The presence of the substrate and the relative concentration of the product of the
reaction.
Q21: The active site orientates reactants, lowers the activation energy of the transition
state and releases products with a low affinity for the active site.
Q22: Feedback control acts through end products, inhibiting an enzyme which
catalyses a reaction early in the pathway.
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ANSWERS: TOPIC 9
9 Cellular respiration I: Glycolysis
Introduction: Questions (page 194)
Q1: It provides the energy to drive them.
Q2: Anabolic
Q3: Catabolic
Summary of glucose breakdown: Question (page 196)
Q4:
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ANSWERS: TOPIC 9
What is ATP: Question (page 198)
Q5: ATP consists of a ribose sugar molecule with adenine attached at 1 carbon and
three phosphates attached in a chain at 5 carbon.
ATP and energy transport: Question (page 199)
Q6: ATP can be readily converted to ADP and P i to release energy, and reconverted
to store energy.
ATP and phosphorylation: Question (page 200)
Q7:
The addition of a phosphate group to a molecule.
Q8:
Glucose 6-phosphate
Q9:
It switches them 'on' or 'off'.
Glycolysis: Questions (page 201)
Q10: Cytoplasm
Q11: The product of each reaction is readily available as the substrate for the next
enzyme in the chain.
Q12: Two molecules of pyruvate, two molecules of ATP, and two hydrogen ions on
carrier molecules (as NADH).
The energy investment stage: Questions (page 203)
Q13: Two molecules of ATP are used up in it.
Q14: ATP contributes the phosphate which attached to the substrate molecule.
Q15: The phosphorylation reactions involve the addition of phosphate to the molecule,
whereas the other reactions re-arrange the existing atoms in the molecule.
The energy investment stage: Questions (page 205)
Q16: 2
Q17: 2
Q18: Phosphorylation
Q19: The phosphate comes directly from the substrate.
Q20: It can enter the mitochondrion, form Acetyl CoA, and enter the Krebs Cycle / Citric
Acid Cycle / Tricarboxylic Acid (TCA) Cycle.
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ANSWERS: TOPIC 9
Q21: It is converted to lactic acid.
Q22: Alanine and fatty acids.
Feedback inhibition and regulation of the glycolytic pathway: Questions (page
206)
Q23: An enzyme's activity is reduced by an increase in the concentration of a product
from a later step in the pathway.
Q24: Damaging metabolites don't build-up and resources, such as ATP, are not wasted.
Q25: ATP inhibits phosphofructokinase by binding to an allosteric site and reducing its
affinity for its sustrate.
Synchronisation of the rates of glycolysis and the citric acid cycle: Questions
(page 207)
Q26: When the cell is not using much energy, the concentration of ADP will be low as
ATP is not being broken down. Consequently, the phosphorylation steps of the pay-off
stage will be slowed. And vice versa.
Q27: Glycolysis supplies the pyruvate which is the fuel for the citric acid cycle. If
glycolysis runs too slowly, the cell will not get the energy it needs from the citric acid
cycle; if it runs to fast, energy will be wasted.
Extended answer question: Feedback inhibition (page 210)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of seven marks can be gained.
1. Feedback inhibition in a metabolic pathway occurs when an enzyme's activity is
reduced. . .
2. . . .by an increase in the concentration of a product from a later step in the pathway.
3. In the energy investment stage of glycolysis. . .
4. . . .phosphofructokinase is inhibited by ATP. . .
5. . . .which is released by glycolysis and the cytochrome system.
6. If the ATP concentration rises in the cell, phosphofructokinase will become less
active. . .
7. . . .and the production of ATP will be reduced, or vice versa.
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ANSWERS: TOPIC 9
8. The production of ATP will be matched to the cell's need for energy. . .
9. . . .and resources will be conserved.
End of Topic 9 test (page 210)
Q28: Respiration provides energy for all the cell's anabolic reactions, movement,
and heat release. In addition, intermediate compounds may be used in many other
metabolic pathways. During aerobic respiration, glucose is broken down, oxygen is
used up, and carbon dioxide and water are released. hydrogen ions and electrons are
removed by dehydrogenase enzymes. As a result, ATP is formed and made available
to the cell.
Q29:
ATP is hydrolysed to
ADP and P i
ATP is formed by
the addition of P i to ADP.
Glycolysis and the cytochrome system
generate ATP.
Energy for anabolic processes is
provided by ATP.
Glucose to glucose 1-phosphate occurs
by
phosphorylation.
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313
Q30:
Glycolysis takes place
in the cytoplasm.
Glycolysis of one glucose molecule
yields two molecules of pyruvate.
The energy investment stage
uses up two molecules of ATP.
The energy pay-off stage
forms two molecules of ATP per
pyruvate.
The energy investment stage
contains
two irreversible phosphorylations.
Glucose 6-phosphate is formed
by phosphorylation of glucose.
Glucose 6-phosphate and pyruvate
can join several metabolic pathways.
The product of phosphofructokinase
can only proceed to glycolysis.
Transfer of phosphate directly to ADP
substrate-level phosphorylation.
In the absence of oxygen
pyruvate can be converted to lactic
acid.
Q31: Glycolysis and the citric acid cycle are synchronised by the feedback inhibition
of phosphofructokinase in the energy investment stage and the supply of ADP to the
phosphorylation reactions in the energy pay-off stage. Feedback inhibition occurs
when an enzyme's activity is reduced by an increase in the concentration of a product
from a a later step in the pathway. Phosphofructokinase in the energy investment
stage of glycolysis is inhibited by high levels of ATP. This will reduce the production of
pyruvate, and therefore of ATP, when the cell's need for energy is low.
Q32: The respiration pathways provide intermediate compounds which may be used in
several other metabolic pathways.
Q33: One molecule of glucose yields two molecules of pyruvate, two molecules of ATP
and two hydrogen ions as 2 NADH.
Q34: These ATP molecules are formed by the transfer of phosphate directly from the
substrate to ADP.
Q35: They are synchronised by the feedback inhibition of phosphofructokinase in the
energy investment stage, and the supply of ADP to the phosphorylation reactions in the
energy pay-off stage.
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ANSWERS: TOPIC 10
10 Cellular Respiration II: Citric acid cycle
Mitochondria: Question (page 216)
Q1:
Glycolysis
multi-enzyme complex in cytoplasm
Pyruvate to acetyl-coA
mitochondrial matrix
Mitochondrial matrix
citric acid cycle
Electron transport chain
cristae of mitochondrion
The steps of the citric acid cycle: Questions (page 219)
Q2:
Once
Q3:
Twice
Q4:
Acetyl coenzyme A / acetyl-coA
Q5:
Oxaloacetate
Q6:
ATP, hydrogen ions, carbon dioxide.
Q7:
It regenerates the oxaloacetate which joins to the acetyl group to form citrate.
The enzymes and carriers of the citric acid cycle: Questions (page 221)
Q8:
A different enzyme
Q9:
Dehydrogenases
Q10: NAD / NAD + and FAD
Q11: NADH and FADH 2
Q12: To the electron transport chain on the cristae (of the mitochrondrion).
Alternative substrates for respiration: Questions (page 224)
Q13: Glycogen
Q14: It is readily interconvertible with glucose.
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ANSWERS: TOPIC 10
Q15:
1. Conversion to glucose.
2. Conversion to intermediates in glycolysis.
Q16: They are converted to intermediates in glycolysis or in the citric acid cycle.
Q17: Liver
Q18: It is converted to an intermediate in glycolysis.
Q19: They are converted to acetyl-coA.
Extended answer question: Enzymes and coenzymes (page 226)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of six marks can be gained.
1. Each step in the citric acid cycle is mediated by a different enzyme.
2. Dehydrogenase enzymes remove hydrogen and electrons.
3. Coenzymes NAD / NAD + and FAD accept hydrogen and electrons.
4. NAD / NAD+ accepts hydrogen to form NADH.
5. FAD accepts hydrogen to form FADH 2.
6. NADH / FADH 2 / reduced coenzymes carry hydrogen and electrons to the electron
transport chain. . .
7. . . .which results in the generation of ATP.
End of Topic 10 test (page 226)
Q20: The mitochondrion consists of an outer boundary membrane, an inner membrane
which is folded to form cristae, and the central matrix. Pyruvate enters the matrix of
the mitochondrion, is converted to acetate, and attached to co-enzyme A to form acetyl
co-enzyme A (acetyl-coA). Acetyl-coA passes the acetate into the Citric Acid cycle in
the matrix of the mitochondrion. Hydrogen and high energy electrons are released
from the citric acid cycle and are taken by carrier molecules to the electron transport
chain on the cristae of the inner membrane of the mitochondrion.
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ANSWERS: TOPIC 10
Q21:
Pyruvate is produced by
glycolysis.
Pyruvate is broken down to form
an acetyl group.
Acetyl-coA passes the acetyl group to
oxaloacetate.
Oxaloacetate combines with the
acetyl group to form
citrate.
The citric acid cycle directly
generates
ATP.
Hydrogen is removed from the citric
acid cycle by
dehydrogenases.
Hydrogen is transferred from the citric
acid cycle to
coenzymes.
High energy electrons are carried by
NAD+ and FAD.
NADH and FADH 2 carry hydrogen to
the
electron transport chain.
The electron transport chain is
located on the
cristae.
Q22:
A storage carbohydrate
glycogen
A sugar which can be converted into
a glycolysis intermediate:
fructose
Some can be converted to citric acid
cycle intermediates:
amino acids
Formed from glycerol and fatty acids:
lipids
Converted into glucose in the liver:
glycerol
Converted into acetyl-coA:
fatty acids
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ANSWERS: TOPIC 10
Q23: It results in the direct generation of ATP, the release of carbon dioxide, the transfer
of hydrogen ions to carriers, and the regeneration of oxaloacetate.
Q24: Coenzymes NAD + and FAD accept hydrogen ions and electrons to form NADH
and FADH 2 . These then release the high-energy electrons to the electron transport
chain on the cristae.
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ANSWERS: TOPIC 11
11 Cellular Respiration III: Electron transport system
The electron transport chain: Questions (page 233)
Q1:
Hydrogen ions and high energy electrons.
Q2:
The inner mitochondrial membrane.
Q3:
Protein
Q4:
To act as enzymes to pass electrons from donors to acceptors.
Q5: To pump hydrogen ions across the inner membrane into the intermembrane
space.
Q6:
Oxygen
Q7:
Water
ATP synthesis: Questions (page 235)
Q8:
Energy
Q9:
They are pumped across the inner membrane into the intermembrane space.
Q10: ATP synthase
Q11: The return flow of H + ions from the intermembrane space into the matrix.
Q12: ATP is formed from ADP and P i .
Q13: The electron transport chain.
Extended answer question: The electron transport chain (page 237)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of eight marks can be gained.
1. NADH and FADH 2 carry H+ ions and high energy electrons. . .
2. . . .from glycolysis and the citric acid cycle.
3. At the inner mitochondrial membrane, NADH and FADH 2 pass the hydrogen ions
and high energy electrons. . .
4. . . .to carriers in the electron transport chain.
5. The electron transport chain is a collection of proteins attached to the inner
mitochondrial membrane.
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ANSWERS: TOPIC 11
319
6. As the high energy electrons released from NADH and FADH 2 pass along the
chain, . . .
7. . . .they release their energy in stages.
8. This energy is used to pump H + ions into the intermembrane space.
9. The final electron acceptor is oxygen.
10. The (negatively charged) oxygen combines with two H + ions to form water.
End of Topic 11 test (page 237)
Q14: The electron transport chain is a collection of proteins attached to the inner
mitochondrial membrane. Coenzymes NAD + and FAD collect hydrogen ions and high
energy electrons from glycolysis and the citric acid cycle, forming NADH and FADH 2 .
They travel to the inner mitochondrial membrane and pass the hydrogen ions and
high energy electrons to carriers in the electron transport chain. As the high energy
electrons released from NADH and FADH 2 pass along the chain of electron acceptors
and donors, they release energy which is used by enzymes to pump hydrogen ions
(H+ ) across the inner mitochondrial membrane into the intermembrane space. The final
electron acceptor is oxygen which combines with hydrogen ions to form water.
Q15:
Energy is released as electrons pass
along
the electron transport chain.
Enzymes pump H+ ions into
the intermembrane space.
The return flow of H + ions into the
matrix rotates
part of ATP synthase.
ATP synthase catalyses
the formation of ATP.
The electron transport chain
produces most of
the ATP from aerobic respiration.
Q16: The high energy electrons / released by NADH and FADH 2 / pass along a chain
of transporter proteins (attached to the inner mitochondrial membrane). / As they are
passed from protein to protein, they release their energy in stages. / This energy is
used to pump hydrogen ions (H + ) / across the inner mitochondrial membrane into the
intermembrane space.
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ANSWERS: TOPIC 12
12 Energy systems in muscle cells
Creatine phosphate: Questions (page 243)
Q1:
2
Q2:
Muscle, brain
Q3:
They have high and variable energy demands.
Q4:
ATP, creatine
Q5:
10
Q6:
When energy demands are low, ATP from respiration is used.
Lactic acid metabolism: Questions (page 245)
Q7: There is not sufficient oxygen to provide all of the cell's energy needs from the
electron transport chain.
Q8:
Pyruvate/pyruvic acid
Q9:
NAD+
Q10: Fatigue
Q11:
i
Glucose
ii
Pyruvate/pyruvic acid
Q12: ATP
Q13: Aerobic respiration requires an additional oxygen supply (to provide this ATP).
Types of skeletal muscle fibres: Questions (page 246)
Q14: Slow and fast twitch.
Q15: 2
Q16: Slow twitch / Type 1
Q17: Fast twitch / Type 2
Slow twitch (Type 1) fibres: Questions (page 248)
Q18: Slow, but sustained
Q19: Endurance
Q20: Aerobic respiration
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ANSWERS: TOPIC 12
321
Q21:
Level of sustained exercise
Source of respiratory substrate
Light
Blood glucose
Moderate
Fat and glycogen
High
Glycogen
Q22: High levels of myoglobin to store oxygen.
High numbers of mitochondria for aerobic respiration.
High density of capillaries to give maximum oxygen supply.
Fast twitch (Type 2) fibres: Questions (page 250)
Q23: Fast for short periods.
Q24: Power
Q25: Creatine phosphate and glycolysis / anaerobic respiration.
Q26: Creatine phosphate and glycogen.
Q27: Lower levels of myoglobin as oxygen does not have to be stored.
Lower numbers of mitochondria as there is much less aerobic respiration.
Lower density of capillaries as less oxygen has to be supplied.
Extended answer question: Creatine phosphate in muscles (page 252)
Answer scheme
Each line represents a point worth one mark. The concept may be expressed in other
words. Words which are bracketed are not essential. Alternative answers are separated
by a solidus (/); if both such answers are given, only a single mark is allocated. In
checking the answer, the number of the point being allocated a mark should be written
on the answer paper. A maximum of eight marks can be gained.
1. Muscle cells do not contain any reserves of ATP which will fuel more than 2s of
activity.
2. Creatine phosphate is an energy reserve found in cells with high and variable
energy demands. . .
3. . . .such as muscle / brain.
4. Creatine phosphate is used by fast twitch muscle fibres.
5. Creatine phosphate breaks down anaerobically. . .
6. . . .to release energy and phosphate. . .
7. . . .which is used to convert ADP and P i to ATP at a fast rate.
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ANSWERS: TOPIC 12
8. There is only sufficient creatine phosphate to fuel strenuous muscle activity for a
maximum of 10s.
9. When energy demands are low, creatine phosphate reserves are restored. . .
10. . . .using ATP provided by aerobic respiration.
End of Topic 12 test (page 252)
Q28: Muscle cells do not contain any reserve of ATP which will fuel more than 2s of
activity. Creatine phosphate is an energy reserve found in cells with high and variable
energy demands, such as the muscles and the brain. There is only sufficient creatine
phosphate to fuel strenuous muscle activity for a maximum of 10s.
Creatine phosphate breaks down anaerobically to release energy and phosphate
which is used to convert ADP and P i to ATP at a fast rate. When energy demands
are low, creatine phosphate reserves are restored using ATP provided by aerobic
respiration.
Q29:
During vigorous exercise, muscle
cells cannot support
the electron transport chain.
Without oxygen, pyruvate is
converted to
lactate.
Conversion of pyruvate involves
transferring hydrogen from
NADH.
The use of NADH to form lactate
regenerates
NAD+ .
NAD+ is required to maintain ATP
production by
glycolysis.
Accumulation of lactate causes
muscle fatigue.
Lactate is reconverted to pyruvate in
the
muscles.
Lactate is reconverted to glucose in
the
liver.
Reconversion of lactate requires ATP
from
aerobic respiration.
Reconversion of lactate causes
oxygen debt.
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ANSWERS: TOPIC 12
323
Q30:
There are two types of muscle fibre,
namely
slow and fast twitch.
All skeletal muscles contain
both types of muscle fibre.
Athletes show patterns of muscle
fibres that
reflect their events.
Slow twitch fibres contract slowly but
sustain contraction longer.
Slow twitch fibres are good for
endurance events.
Slow twitch fibres rely on
aerobic respiration.
A large blood supply and many
mitochondria are linked to
a high concentration of myoglobin.
Slow twitch fibres use fuel stores of
glycogen and fat.
Fast twitch fibres contract quickly but
for
short periods only.
Fast twitch fibres are good for
power events.
Fast twitch fibres generate ATP by
creatine phosphate and glycolysis.
Few mitochondria and a low blood
supply are linked to
little myoglobin.
Q31: Creatine phosphate is found in cells with high and variable energy demands, such
are muscles and the brain.
Q32: For 2 seconds the muscle cells rely on stored ATP; for the next 10 seconds
creatine phosphate; and then lactic acid metabolism.
Q33: Slow twitch fibres are dark red / because they have a high concentration of
myoglobin / which stores oxygen for aerobic respiration. / Fast twitch fibres are pale
pink / because they contain low concentrations of myoglobin / as they mainly rely on
anaerobic forms of respiration / namely, creatine phosphate and lactic acid metabolism.
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ANSWERS: TOPIC 13
13 End of unit test
End of Unit 1 test (page 256)
Q1:
Cellular differentiation means that
cells
develop more specialised functions.
Muscle cells can only
express genes characteristic of their
cell type.
Embryonic stem cells can
develop into any cell type.
Tissue stem cells can
produce a limited range of cell types.
Germline cells are able to
produce gametes.
Mitosis provides the ability to
maintain the chromosome number.
Mutations in germline cells
affect all cells of offspring.
Skin epidermis can be used to
exemplify epithelial tissue.
Stem cell research lets scientists
understand cell growth and
differentiation.
The Human Tissue Authority has a
remit to
regulate cell research in the UK.
Cancer cells do not
respond to normal regulatory signals.
A tumour is benign if it does not
invade neighbouring tissues.
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ANSWERS: TOPIC 13
325
Q2:
A nucleotide is composed of
deoxyribose,
phosphate and a nitrogenous base.
The sugar-phosphate backbone is
held together by
strong covalent bonds.
The base pairings of DNA are
adenine and thymine,
guanine and cytosine.
At the 3 and 5 ends of the DNA
strands are
deoxyribose and phosphate.
The enzyme which joins fragments of
DNA together is
ligase.
The formation of mRNA from a DNA
template is
transcription.
The base in RNA which replaces
thymine is
uracil.
The first strand of RNA formed from
DNA is
the primary transcript.
The protein-coding regions of mRNA
are
exons.
The polypeptide is released when
reaching
a stop codon.
One primary transcript can produce
several different mRNA molecules.
Post-translational modification may
involve adding
carbohydrate groups.
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ANSWERS: TOPIC 13
Q3:
Pathways that need an energy input
are called
anabolic.
The active site lowers
activation energy.
Glycolysis is catalysed by
a multi-enzyme complex.
The product of glycolysis is
pyruvate.
The product of phosphofructokinase
only enters
glycolysis.
Acetyl-coA passes the acetyl group to
oxaloacetate.
A substrate that is converted into a
glycolysis intermediate is
fructose.
Enzymes pump H+ ions into
the intermembrane space.
High energy electrons are carried by
NAD+ and FAD.
Q4: Without oxygen, pyruvate is converted to lactate.
The use of NADH to form lactate regenerates NAD + .
Lactate is reconverted to glucose in the liver.
Creatine phosphate reserves are restored by aerobic respiration.
Slow twitch fibres are good for endurance events.
Fast twitch fibres generate ATP by creatine phosphate and glycolysis.
Q5:
Mitosis
Q6:
They only express genes that produce proteins characteristic of that type of cell.
Q7: Epithelial - covers the organ surfaces.
Connective - gives shape to organs / supports organs.
Q8:
Any two from:
• divide excessively;
• do not respond to regulatory signals;
• fail to attach to each other.
Q9: Mutations to a somatic cell will affect only the tissue in which it occurs.
Mutations to a germline cell will affect all cells in any offspring produced from that cell.
Q10: Deoxyribose sugar, phosphate, base.
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ANSWERS: TOPIC 13
Q11: Hydrogen
Q12: The two strands run in opposite directions. / One strand ends with deoxyribose at
the 3 end / and the other ends with phosphate at the 5 end.
Q13: The base bonding to adenine (A) is thymine (T), not uracil (U).
Q14: DNA polymerase
Q15: A primer
Q16: One strand is replicated continuously, The other strand is replicated in fragments
which are joined together by ligase.
Q17: Changes to genes or chromosomes.
Q18: Proteins are not expressed / proteins do not function correctly / proteins are faulty.
Q19: a) Deletion
Q20: b) Duplication
Q21: c) Translocation
Q22: Outcome: often lethal.
Explanation: mutation results in loss of genes / breaks up groups of genes which work
together.
Q23: Anabolic: biosynthetic - requires an energy input. or
Catabolic: breakdown of molecules - releases energy and building blocks.
Q24: Any two from:
• orientates reactants (so that they react);
• lowers the activation energy of the transition state;
• release of products (with low affinity for active site).
Q25: Example: glycolysis (or other suitable).
Advantage: product from one enzyme is substrate for the next / presented in correct
position / orientation for next active site.
Q26: Meaning: the end-product binds to an enzyme which catalyses a reaction early in
the pathway.
Example: ATP / citrate inhibits the action of phosphofructokinase. Threonine deaminase
inhibited by isoleucine (in transamination pathway).
Q27: Pyruvate / pyruvic acid
Q28: Pyruvate is broken down to form an acetyl group. It is combined with coenzyme A
to form acetyl coenzyme A / acetyl-coA.
Pyruvate to acetyl-coA. (only 1 mark)
Q29: Insufficient oxygen reaching muscle for aerobic respiration to supply required ATP,
e.g. during a power event such as sprinting (example needed).
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ANSWERS: TOPIC 13
Q30: The athletes have an oxygen debt / lactic acid has built up in their muscles. Lactic
acid must be reconverted to pyruvate / pyruvic acid in the muscles. or
Lactic acid must be reconverted to glucose in the liver. Reconversion of pyruvate /
pyruvic acid requires ATP from aerobic respiration.
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