Download AM 05 Biology Paper 1

Index No.:______________
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MATRICULATION AND SECONDARY EDUCATION CERTIFICATE EXAMINATIONS BOARD
UNIVERSITY OF MALTA, MSIDA
MATRICULATION EXAMINATION
ADVANCED LEVEL
MAY 2016
SUBJECT:
PAPER NUMBER:
DATE:
TIME:
BIOLOGY
I
5th May 2016
4.00 p.m. to 7.05 p.m.
Directions to Candidates
•
Write your index number in the space at the top left-hand corner of this page.
•
Answer ALL questions. Write all your answers in the spaces provided in this booklet.
•
The mark allocation is indicated at the end of each question. Marks allocated to parts of questions are
also indicated.
•
You are reminded of the necessity for good English and orderly presentation in your answers.
•
In calculations you are advised to show all the steps in your working, giving your answer at each stage.
•
The use of electronic calculators is permitted.
For examiners’ use only:
Question
1
2
3
4
5
6
7
8
9
10
Total
10
10
10
8
14
10
8
10
10
10
100
Score
Maximum
© The MATSEC Examinations Board reserves all rights on the examination questions in all examination papers set by the said Board.
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1
This question is about taxonomy.
1.1
Why is it advantageous to have a system of classification for organisms?
It helps us clarify the evolutionary relationships among organisms.
Accept ANY ONE of the following:
It helps us remember the organisms and their characteristics.
It lets us communicate clearly regarding the identity of the organisms being
studied.
It provides stable names to organisms in order to avoid ambiguity of
common names.
1 mark
1 mark
[two marks]
1.2
Define the term species.
Accept ANY of the following options:
Option 1
A species is one of the basic units of biological classification
and a taxonomic rank
Option 2
A species is defined as a group of similar organisms
Which are capable of interbreeding/ exchange genes to
produce fertile offspring
1 mark
1 mark
1 mark
1 mark
[two marks]
Euphorbia melitensis is an endemic flowering plant found in the Maltese islands.
1.3
Identify TWO major taxonomical groups within the plant kingdom for Euphorbia melitensis.
Tracheophyta
1 mark
Magnoliophyta / Angiosperm
1 mark
Dicot; Magnoliopsida (class); Malpighiales(order); Euphorbaceae (family) can also be
acceped
[two marks]
1.4
Euphorbia melitensis is the scientific name of this plant. Mention TWO advantages of using
the binomial system for naming organisms.
Accept TWO of the following:
Universal in the scientific community/eliminates use of several common
names.
No misunderstanding of the organisms being studied/attributed to a
particular species.
Gain information about plant
1 mark each
MAX 2
marks
[two marks]
1.5
What do the following represent in the taxonomic rank?
1.5.i. Euphorbia:
Genus
1 mark
1.5.ii. melitensis:
species
1 mark
[two marks]
[Total: Ten marks]
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2.
This question is about biomolecules.
Figure 1 represents the molecular structure of ATP.
Figure 1: Molecular structure of ATP
(Source: http://chemwiki.ucdavis.edu/@api/deki/files/25707/800px-ATP_chemical_structure.png?revision=1)
2.1
2.2
What does ATP stand for?
Adenosine Triphosphate
1 mark
Reject if answer is not
complete
[one mark]
Mention THREE reasons why ATP is an optimal energy currency in living organisms.
Accept any THREE of the following:
Gives off small packets of energy
Small and water-soluble so can move around the cell
Immediate energy donor
Rapid turnover
1 mark each
reason –
MAX 3
marks
[three marks]
Proteins are made up of amino acids. Twenty amino acids make up polypeptides and proteins
in animals.
2.3
What is an amino acid?
Accept ANY of the following:
An organic molecule consisting of the basic amino group (NH2), a hydrogen
atom, an acidic carboxylic group (COOH) and an organic side group (R)
which is attached to the carbon atom.
A biochemical with the following formula: NH2CHRCOOH.
The basic chemical unit that makes up proteins.
An amino acid is the chemical which is coded for by a triplet codon on the
DNA molecule.
1 mark
[one mark]
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2.4
Name the type of reaction that occurs when two amino acids form a dipeptide.
Condensation reaction
1 mark
Reject dehydration reaction
[one mark]
2.5
Valine and Glycine are two of the twenty amino acids found in animals. Their molecular
structure is represented below in Figure 2.
Valine
Glycine
Figure 2: Molecular structure of valine and glycine
In the space below, draw the molecular structure of the dipeptide formed between these two
molecules.
Correct R-groups
Amino and carboxylic acid ends are showing
Water is eliminated
Peptide bond drawn correctly.
2.6.i Describe the primary structure of a protein.
The linear sequence of amino acids that form the molecule.
2.6.ii What is the importance of the primary structure of a protein?
Accept any of the following:
The primary structure is the basis of identity of the protein.
Forms the basis of all other proteins
0.5 marks
0.5 marks
0.5 marks
0.5 marks
[two marks]
1 mark
[one mark]
1 mark
[one mark]
[Total: Ten marks]
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3
This question is about gaseous exchange.
3.1
List TWO characteristics of a respiratory surface.
Accept any TWO of the following
Large surface area
One cell thick / thin
An extensive blood network
Concentration gradient present
Moist
1 mark
each –
MAX 2
marks
[two marks]
3.2
How does the countercurrent system found in the gills of fish provide efficient gaseous
exchange?
Water and blood flow in opposite directions.
1 mark
Concentration gradient maintained all along length of lamellae /Equilibrium is
1 mark
not reached
Water always next to blood with a lower concentration of oxygen.
1 mark
Accept diagram if clearly annotated with above points
[three marks]
Neoparamoeba perurans is a parasite which causes amoebic gill disease in fish. This disease
causes the lamellae to thicken and to fuse together.
3.3 Explain why these two effects give rise to a reduction in the efficiency of gaseous exchange
in fish.
Thicker lamellae mean a bigger diffusion distance.
Lamellae fuse therefore the surface is reduced/ Reduced surface area to
volume ratio.
1 mark
1 mark
[two marks]
3.4
Use the information in Table 1 to calculate the volume of water that would have to pass over
the gills every hour to provide a fish, weighing 0.6 kg, with the required oxygen. Show your
working.
Table 1: Hypothetical features of gaseous exchange in fish at rest
Oxygen required by fish
85 cm3kg-1hour-1
Volume of oxygen absorbed by the gills from each Litre of water
8 cm3
Oxygen required by fish = 0.6 x 85 = 51
Volume = 51/8
Correct answer 6.375L
1 mark
1 mark
1 mark
Deduct 0.5 marks if units are
not given or incorrect
[three marks]
[Total: Ten marks]
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4.
4.1
This question is about hormonal control.
Which hormones are responsible for the regulation of blood glucose in the body?
Insulin
1 mark
glucagon
1 mark
[two marks]
An experiment was carried out to investigate how the blood glucose level fluctuates
following ingestion of glucose.
At the start of the experiment, each member of a group of volunteers ingested a syrup
containing 50 g of glucose. The concentration of glucose was determined in blood samples at
intervals over a period of 2 hours. The results shown in the graph next page (Figure 3) are
mean values for the group of volunteers.
Mean blood glucose
concentration / mmolL-1
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
0
20
40
60
80
100
120
140
Time / minutes
Figure 3: Graph showing mean blood glucose concentration versus time
4.2
Describe and explain the changes in the concentration of glucose during the following time
intervals.
4.2.i
0 to 30 minutes
Blood glucose level increases
Peaks at 7.5 mMol/L
Due to absorption of glucose from gut/ enters blood stream
4.2.ii 30 to 120 minutes
Concentration decreases
Rise of glucose stimulates insulin secretion
Uptake by liver/muscle/taken into cells/ Change into glycogen
1 mark
1 mark
1 mark
[three marks]
1 mark
1 mark
1 mark
[three marks]
[Total: Eight marks]
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5.
This question is about genetics.
5.1
Explain what is meant by the following terms:
5.1.i
Allele:
One of the forms of a gene that are found on the same loci of a chromosome.
1 mark
[one mark]
5.1.ii Dominant:
An allele that will always show in the phenotype if present
5.1.iii Heterozygous:
Two different alleles of a gene
In a particular locus
1 mark
[one mark]
0.5 mark
0.5 mark
[one mark]
5.1.iv Genotype:
Accept ANY of the following:
Alleles present in an organism
Particular alleles of a gene
Gene constitution
1 mark
[one mark]
5.2
Melopsittacus undulatus, commonly known as the budgerigar, is a small parrot which can
have green, blue, yellow or white feathers.
Feather colour is controlled by two genes, B/b or Y/y according to the following genotypes:
Colour of bird
Genotype
Green
At least one dominant B allele and one dominant Y allele
Blue
At least one dominant allele B but are homozygous recessive for y
Yellow
At least one dominant allele Y but are homozygous recessive for b
White
Homozygous recessive for both b and y
5.2.i
If two green-feathered budgerigars which are heterozygous at both loci were crossed, what is
the probability of producing offspring with blue feathers?
A genetic diagram should be included.
BY
BY
BBYY
By
BBYy
bY
BbYY
by
BbYy
Probability = 3/16 or 18.75%
By
BBYy
BByy
BbYy
Bbyy
Parents BbYy x BbYy
Gametes BY By bY by
Correct Punnett square (or genetic diagram)
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bY
ByYY
BbYy
bbYY
bbYy
by
BbYy
Bbyy
bbYy
bbyy
1 mark
1 mark
3 marks
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Correct probability
1 mark
[six marks]
5.2.ii Two blue budgerigars were crossed and over the years, they produced 22 offspring, five of
which were white. Deduce the most likely genotypes for the two blue birds.
Blue birds can be BByy or Bbyy
1 mark
Since ~ 25% were white, b must have been present in both parents
1 mark
Correct punnett square (or genetic diagram)/explanation
1 mark
Therefore the parents were Bbyy and Bbyy
1 mark
[four marks]
[Total: Fourteen marks]
6
This question is about biotechnology.
6.1
Define biotechnology.
Accept ANY of the following:
The application of organisms, biological systems or biological processes to
manufacture certain services.
Manipulation of DNA to produce genetically modified products
1 mark
[one mark]
6.2
List the FOUR main stages that are employed in genetic engineering using bacteria
Gene extraction
1 mark
Recombinant DNA formation
1 mark
Introduction of vector DNA into host cell
1 mark
Screening/DNA proofing
1 mark
DNA amplification/ cloning can also be accepted
[four marks]
6.3
The following are techniques used in analysing DNA organisation. Give the function of
each:
Gel electrophoresis
Separation of DNA fragments
1 mark
Southern blotting
Detection
of
fragments/identification
of
1 mark
genes/visualisation of DNA fragments
[two marks]
PCR is an important tool used in biotechnology.
6.4
6.5
What does PCR stand for?
Polymerase chain reaction
1 mark
Reject if answer is not complete
[one mark]
Explain the function of PCR in biotechnology.
PCR enables one to amplify (to generate a large number of copies)
a specific fragment of DNA
1 mark
1 mark
[two marks]
[Total: Ten marks]
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7
This question is about reproduction.
7.1
Briefly mention the main advantages and disadvantages of asexual reproduction.
Advantages:
Accept any two:
1 mark each
1 parent needed
– MAX 2
No energy waste in mating
marks
No energy wasted in gamete production
Fast/efficient
Disadvantages:
1 mark
Slow rate of generation of genetic variation (relies only on mutations)
Low genetic variation within population results in susceptibility
1 mark
[four marks]
7.2.
The following hormones are involved in the control of the human menstrual cycle. For each
hormone, state where it is produced and the organ/s on which it acts.
Oestrogen:
ovary / granulosa cells/ ovarian/ Graafian
0.5 marks
follicle
affects ovary, uterus and anterior pituitary
0.5 marks
Reject if organs in bold
are not mentioned
gland
Progesterone:
ovary / corpus luteum
0.5 marks
affects the lining of uterus and pituitary gland
0.5 marks
Reject if organs in bold
are not mentioned
[two marks]
The placenta plays an important role in pregnancy.
7.3.
7.3.i
State ONE important role of the placenta in the:
protection of the embryo.
Accept any of the following:
Prevents certain harmful pathogens from passing to the embryo.
Releases progesterone to prevent the wall to contract.
Protects against the high blood pressure.
Passage of antibodies from the mother which gives the embryo a way of
defence.
Passage of waste from embryo to mother – avoids intoxication of embryo
1 mark
[one mark]
7.3.ii nutrition of the embryo.
Accept any of the following:
It transfers nutrients, food and necessary vitamins from the blood of the
mother to the embryo.
Responsible for oxygen exchange between mother and embryo.
1 mark
[one mark]
[Total: Eight marks]
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8
This question is about excretion and osmoregulation.
The water balance of the body in different situations was investigated. The results were
recorded on three separate days, namely, a warm day, a cold day and a day during which the
individual did heavy exercise (Figure 5). The total water intake by the body was the same as
the water output in each situation.
Figure 5: Water balance for 3 different days – a warm day, a cold day and a day
whereby heavy exercise was performed.
8.1 Explain what is meant by oxidation of food and how it produces water in the body.
Oxidation of food - during cellular respiration, organic molecules are
1 mark
broken down to release energy OR addition of oxygen to food molecules
to release energy or produce water
During the final stage of electron transport chain oxygen accepts hydrogen
1 mark
and electrons forming water OR by cytochrome oxidase
[two marks]
8.2.i
Describe the differences in water output between the cold day and the day of heavy exercise.
On a cold day, the volume of urine was more than that on the day of heavy
1 mark
exercise.
Sweat was higher on heavy exercise day
1 mark
More water lost from lungs on the heavy exercise day
1 mark
[three marks]
8.2.ii Explain what is causing these differences
Cool temperatures mean less sweating.
Sweat due to body heat generation and thermoregulatory mechanisms to let
out heat.
since more heavy breathing to provide more oxygen for activity
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0.5 mark
0.5 mark
1 mark
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[two marks]
8.3
Explain why more water is produced by the oxidation of food on the day of heavy exercise.
A lot of food being respired.
1 mark
More energy for the exercise.
1 mark
Respiration of fats therefore more water in the oxidation chain.
1 mark
[three marks]
[Total: Ten marks]
9
This question is about the immune system.
9.1
Explain the term disease.
A particular abnormal condition, a disorder of a structure
function/impairment of the normal state of the individual
that is a response to a pathogen, the environment or genetic disposition
.
or
1 mark
1 mark
[two marks]
An individual can acquire two types of immunity. Table 6 represents the different types of
immunity.
Table 6:
Different types of immunity
Immunity
Type of immunity Production of memory
acquired
cells / yes or no
Injection with an antibody against Artificial Passive
No
specific disease- causing organism
Injection with a live, weakened Artificial Active
Yes
disease-causing organism
0.5 marks each point
9.2
Complete Table 6.
[two marks]
When a pathogen enters the body, a primary immune response occurs. This can include the
production of antibodies.
9.3
9.4
Define the term antibody.
Accept any of the following:
Option 1 A molecule that is made by an animal in response to the presence
of a pathogen/ substances
known as antigens
Option 2
a Y-shaped protein with 2 heavy chains and 2 light chains (held by
disulphide bridges)
Bind to epitope
1 mark
1 mark
1 mark
1 mark
[two marks]
Describe the events in the immune response which lead to antibodies being produced.
Antigen binds to receptor or membrane of B-cells/ B-lymphocytes
1 mark
T helper cell or T lymphocyte, activate the B-cells
1 mark
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Killer T cells release cytokines to stimulate the B-cells
Plasma cells form and produce antibodies
1 mark
1 mark
[four marks]
[Total: Ten marks]
10
This question is about evolution.
10.1
Explain what is meant by biological evolution.
Biological evolution involves a change in the gene pool (allele and genotype
frequencies) of a population from one generation to the next.
Small differences accumulate with each generation and can, over time, cause
substantial changes in the organisms.
1 mark
1 mark
[two marks]
10.2
An important factor for natural selection to occur is the variation in the phenotypes of the
population. Explain.
Variation means the presence of different characteristics within a population
1 mark
Results in different survival rates
1 mark
More successful phenotypes are passed on to next generation
1 mark
[three marks]
10.3
Briefly differentiate between sympatric and allopatric speciation.
Allopatric speciation is speciation that results when a population is separated
by a physical barrier. It is also referred to as geographic speciation.
Sympatric speciation is speciation that occurs without physical separation of
members of the population but is the product of divergent evolution/
specialisation.
1 mark
1 mark
[two marks]
10.4
Briefly explain how geographic isolation can lead to the evolution of a new species.
Populations separated by physical barrier therefore no mixing of gene pools/
Separated population experiences founder effect and there is no mixing of
gene pools
Due to different selection pressures, become adapted to local environment
Mutation in one group adds to changes in allele frequency
1 mark
1 mark
1 mark
[three marks]
[Total: Ten marks]
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