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Edexcel GCSE
Biology
Unit B2: The Components of Life
Higher Tier
Additional Sample Assessment Material
Time: 1 hour
You must have:
Calculator, Ruler
Paper Reference
5BI2H/01
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Instructions
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• centrethe
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• Answer
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– use this as a guide as to how much time to spend on each question.
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•
S39593A
©2012 Edexcel Limited.
6/8/8/2/
*S39593A0116*
Turn over
Answer ALL questions
Some questions must be answered with a cross in a box . If you change your mind about an
answer, put a line through the box and then mark your new answer with a cross .
Blood
1 (a) Blood is described as a tissue.
State the meaning of the term tissue.
(1)
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(b)The photograph shows a blood smear from a healthy person.
Z
X
Y
(i)Name the two types of blood cells, X and Y, shown in the photograph.
(2)
X .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(ii) Complete the sentence by putting a cross ( ) in the box next to your
answer.
The function of the cells labelled X is to
A
engulf bacteria
B
produce antigens
C
transport carbon dioxide
D
transport oxygen
2
*S39593A0216*
(1)
(iii) Measure the diameter of the cell labelled Z.
diameter = . .................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . mm
(iv) The cell labelled Z is magnified × 900.
(1)
Calculate the actual diameter of the cell labelled Z.
actual diameter =
..................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(c) Name the liquid part of the blood.
(2)
µm
(1)
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(Total for Question 1 = 8 marks)
*S39593A0316*
3
Turn over
The embryo
2 (a) Cells in a human body can be haploid or diploid.
When an egg cell combines with a sperm cell a fertilised egg, known as a zygote,
is formed.
(i) Which row of the table gives correct information about the egg cell, the
sperm cell and the zygote?
(1)
Put a cross ( ) in the box next to your answer.
egg cell
sperm cell
zygote
A
diploid
diploid
haploid
B
diploid
haploid
diploid
C
haploid
diploid
haploid
D
haploid
haploid
diploid
(ii) The single cell of the zygote divides into two cells in 15 hours.
These cells continue to divide in this way to form an embryo.
The diagram shows an embryo with 32 cells.
Calculate the length of time it takes the single cell of the zygote to form an
embryo with 32 cells.
(2)
answer = .......................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . hours
4
*S39593A0416*
(iii) The cells of this embryo are genetically identical.
Explain how these genetically identical cells are produced.
(3)
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(b)The first embryonic stem cell trials took place in October 2010 to treat a patient.
This patient had been left partially paralysed by nerve damage after a spinal injury.
Explain why it is an advantage to use embryonic stem cells to treat spinal injuries.
(2)
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(Total for Question 2 = 8 marks)
*S39593A0516*
5
Turn over
Fat digestion
3 Lipase and bile are involved in the digestion of fat.
(a) Give the names of the two products of fat digestion by lipase.
(2)
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(b)State where bile is stored.
(1)
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(c) A student investigated the effect of bile on fat digestion.
The diagram shows the steps taken in this investigation.
Step 1: Pour some oil (liquid fat) on to water in a
test tube
Step 2: Shake the test tube and then measure the
diameters of the oil droplets
Step 3: Add bile and lipase to the oil and water
and shake the test tube
Step 4: After 20 minutes, measure the diameters
of the oil droplets
6
*S39593A0616*
(i) The student found that after 20 minutes the oil droplets were much smaller
and concluded that bile was needed for fat digestion.
A second student said that this conclusion may be incorrect.
Give a reason why the conclusion may be incorrect.
(1)
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(ii) Explain the effect of bile on oil droplets.
(3)
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(d)Explain how the structure of the small intestine enables efficient absorption of the
products of fat digestion.
(4)
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(Total for Question 3 = 11 marks)
*S39593A0716*
7
Turn over
Gas exchange
4 (a) Gas exchange in the air sacs of the lungs takes place in a similar way to gas
exchange between body cells and capillaries.
Each of these air sacs are surrounded by blood capillaries.
The diagram shows one air sac.
wall of
air sac
carbon
dioxide
oxygen
red
blood
cell
(i) Describe how oxygen is transported from the air sac into the surrounding
blood capillary.
(2)
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(ii) Complete the sentence by putting a cross ( ) in the box next to your answer.
The blood transports oxygen to body cells.
Oxygen is used by body cells when
A energy is released during respiration
B energy is released from carbon dioxide
C glucose is produced during respiration
D energy is taken in during respiration
8
*S39593A0816*
(1)
25
time / mins
35
_
15
_
5
_
_
heart rate
in beats per
minute
150–
135–
120–
105–
90–
75–
60–
45–
30–
15–
0–
0
_
(b)The graph shows how the heart rate of a person changes during and after aerobic
exercise.
_
45
(i) The volume of blood leaving the heart during one heart beat at 25 minutes is
0.07 dm3.
The person’s cardiac output can be calculated using the equation:
cardiac output = stroke volume × heart rate
Calculate the cardiac output of this person at 25 minutes.
(3)
cardiac output = ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(ii) Explain the trend shown by the graph between 5 and 25 minutes.
(3)
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*S39593A0916*
9
Turn over
(c) Anaerobic respiration takes place when the muscle cells are not supplied with
enough oxygen.
Give the word equation for anaerobic respiration.
(1)
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(Total for Question 4 = 10 marks)
10
*S39593A01016*
Transpiration
5 (a) Photograph A shows a Coleus plant that has wilted due to lack of water.
If this plant is given some water it will recover and stand upright again.
Photograph B shows the plant after it has been watered.
photograph A
photograph B
(i) Describe how a plant takes in water from the soil.
(2)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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(ii) The soil surrounding the roots of the Coleus plant contains water with a
higher concentration of dissolved mineral ions than the plant cell vacuole.
Which row of the table gives the method of transport of mineral ions into the
plant cell of this Coleus and the type of vessel which transports mineral ions
to its leaves?
Put a cross ( ) in the box next to your answer.
(1)
method of transport into the
plant cell
type of vessel which transports
mineral ions
A
active transport
phloem
B
active transport
xylem
C
diffusion
phloem
D
diffusion
xylem
*S39593A01116*
11
Turn over
(b)The water is transported from the roots up the plant by transpiration.
An investigation was carried out over a 12‑hour period to compare transpiration in
the Coleus plant with transpiration in another type of plant called a Begonia.
The photograph shows a Begonia plant.
The mass of each plant was recorded before and after the 12-hour period to find
out the effect of transpiration.
The investigation was repeated five times with the same plants.
The table shows the change in mass of each plant over each 12-hour period.
change in mass of plant / g
Coleus
Begonia
3.7
1.1
4.5
1.3
2.8
0.8
1.6
0.6
3.2
1.0
(i) Compare the water loss in the Coleus plant with the water loss in the Begonia
plant.
(2)
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12
*S39593A01216*
*(ii) The wilted Coleus plant was observed to recover more quickly than a wilted
Begonia plant when they were both watered.
Explain this observation.
(6)
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(Total for Question 5 = 11 marks)
*S39593A01316*
13
Turn over
Producing new cells
6 (a) The diagram shows the mass of DNA (m), before, during and after cell division in
one cell.
m
DNA replication
2m
cell division
½m
(i) Name the type of cell division taking place in this cell.
(1)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(ii) Complete the sentence by putting a cross ( ) in the box next to your answer.
A gene is a section of DNA that codes for
A an amino acid
B a base pair
C a protein
D a triplet code
14
*S39593A01416*
(1)
(b)Mutations can occur in genes.
Mutations can cause genetic disorders in humans.
Phenylketonuria (PKU) is a genetic disorder caused by a gene mutation.
People with PKU produce an inactive enzyme.
The normal base sequence and the mutated base sequence which can cause PKU
are shown below.
normal base sequence ............ C T C G G C C C T . . . . . . . . . . . .
mutated base sequence ............ C T T G G C C C T . . . . . . . . . . . .
(i) Describe how the changes that have occurred in the mutated base sequence
produce an inactive enzyme.
(2)
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*(ii) Explain how the mutated base sequence will result in an inactive enzyme
being produced during protein synthesis.
(6)
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*S39593A01516*
15
Turn over
(c) Explain how the shape of an enzyme can make it inactive.
(2)
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(Total for Question 6 = 12 marks)
total for paper = 60 marks
16
*S39593A01616*
Additional Sample Mark Scheme
GCSE Science 2011
GCSE
GCSE Biology (5BI2H/01)
Edexcel Limited. Registered in England and Wales No. 4496750
Registered Office: One90 High Holborn, London WC1V 7BH
General Marking Guidance
•
All candidates must receive the same treatment. Examiners must mark the
first candidate in exactly the same way as they mark the last.
•
Mark schemes should be applied positively. Candidates must be rewarded for
what they have shown they can do rather than penalised for omissions.
•
Examiners should mark according to the mark scheme not according to their
perception of where the grade boundaries may lie.
•
There is no ceiling on achievement. All marks on the mark scheme should be
used appropriately.
•
All the marks on the mark scheme are designed to be awarded. Examiners
should always award full marks if deserved, i.e. if the answer matches the
mark scheme. Examiners should also be prepared to award zero marks if the
candidate’s response is not worthy of credit according to the mark scheme.
•
For questions worth more than one mark, the answer column shows how
partial credit can be allocated. This has been done by the inclusion of part
marks eg (1).
•
Where some judgement is required, mark schemes will provide the principles
by which marks will be awarded and exemplification may be limited.
•
When examiners are in doubt regarding the application of the mark scheme
to a candidate’s response, the team leader must be consulted.
•
Crossed out work should be marked UNLESS the candidate has replaced it
with an alternative response.
Quality of Written Communication
Questions which involve the writing of continuous prose will expect candidates to:
•
Write legibly, with accurate spelling, grammar and punctuation in order to
make the meaning clear
•
Select and use a form and style of writing appropriate to purpose and to
complex subject matter
•
Organise information clearly and coherently, using specialist vocabulary
when appropriate.
Full marks will be awarded if the candidate has demonstrated the above abilities.
Questions where QWC is likely to be particularly important are indicated (QWC) in the mark
scheme, but this does not preclude others.
5BI2H/01
Additional Sample Mark Scheme
General Information
The following symbols are used in the mark schemes for all questions:
Symbol
eq
Meaning of symbol
Indicates that credit should be given for other
correct alternatives to a word or statement
/ oblique
Words or phrases separated by an oblique are
alternatives to each other
{ } curly brackets
( ) round brackets
Indicate the beginning and end of a list of
alternatives (separated by obliques) where
necessary to avoid confusion
Words inside round brackets are to aid
understanding of the marking point but are not
required to award the point
5BI2H/01
Additional Sample Mark Scheme
Question
Number
1(a)
Question
Number
1(b)(i)
Question
Number
1(b)(ii)
Answer
Acceptable answers
idea of (group of) cells that share a
similar role
Answer
(1)
Acceptable answers
•
X – white blood cell /
phagocyte (1)
•
Y – red blood cell /
erythrocyte (1)
Answer
Mark
Mark
(2)
Acceptable answers
Mark
A
(1)
Question
Number
1(b)(iii)
Answer
Acceptable answers
Mark
6 (mm)
(1)
Question
Number
1(b)(iv)
Answer
Acceptable answers
calculation (1)
6 x 1000
900
6 – 8 (µm)
Mark
6.7 / 7 (µm) (1)
Question
Number
1(c)
Answer
give full marks for correct answer,
no working shown
(2)
Acceptable answers
Mark
plasma
(1)
5BI2H/01
Additional Sample Mark Scheme
Question
Number
2(a)(i)
Answer
Acceptable answers
Mark
D
(1)
Question
Number
2(a)(ii)
Answer
Acceptable answers
Mark
correct calculation (1)
5 x 15
.
answer = 75 (hours) (1)
(2)
Question
Number
2(a)(iii)
Question
Number
2(b)
Answer
Acceptable answers
Mark
An explanation linking three of the
following points
•
mitosis (1)
•
(results in){chromosomes /DNA}
replication (1)
•
(and) separation of { sets of
chromosomes / eq} (1)
•
credit detail of separation
Answer
Reject references to meiosis
(3)
Acceptable answers
Mark
An explanation linking two of the
following points:
•
(stem cells) can differentiate
into any type of cell (1)
•
(so) can be used to
produce/repair/replace nerve
tissue (1)
•
easier to extract than adult
stem cells (1)
(2)
5BI2H/01
Additional Sample Mark Scheme
Question
Number
3(a)
Answer
Acceptable answers
•
fatty acid(s) (1)
•
glycerol (1)
Mark
(2)
Question
Number
3(b)
Answer
Acceptable answers
Mark
gall bladder
(1)
Question
Number
3(c)(i)
Answer
Acceptable answers
Mark
Any one of the following points
•
because {there was no negative control
/ no results for bile only} / eq (1)
•
more than one independent variable
(1)
(1)
Question
Number
3(c)(ii)
Answer
Acceptable answers
Mark
An explanation linking three of the following
points
•
reference to bile as an emulsifying
agent (1)
•
causing the oil droplets to break up
into smaller droplets (1)
•
resulting in a larger surface area of
oil(1)
•
by surrounding them / eq (1)
•
lipase activity increases (1)
(3)
5BI2H/01
Additional Sample Mark Scheme
Question
Number
3(d)
Answer
Acceptable answers
Mark
An explanation linking four of the
following points
•
absorption occurs in villi (1)
•
large number of villi /folding / length
of small intestine (1)
•
to increase the surface area (1)
•
single layer of cells (in villus) allows
faster absorption (1)
•
by diffusion (1)
•
credit correct reference to lacteals
(1)
(4)
5BI2H/01
Additional Sample Mark Scheme
Question
Number
4(a)
Question
Number
4(a)(ii)
Answer
Acceptable answers
Mark
A description linking the following points
•
high concentration (in air sacs) to
lower concentration ( in blood) (1)
•
by diffusion (1)
Answer
(2)
Acceptable answers
Mark
A
(1)
Question
Number
4(b)(i)
Answer
Acceptable answers
Mark
substitution (1)
150 x 0.07
evaluation (1)
10.5
unit (1)
dm3 per minute
Question
Number
4(b)(ii)
Question
Number
4(c)
Answer
(3)
Acceptable answers
Mark
An explanation linking three of the
following points
•
{heart / pulse} rate increases to
increase blood flow (1)
•
so more {oxygen / glucose}
delivered to {muscles / cells} (1)
•
for aerobic respiration (1)
•
increased blood flow removes
carbon dioxide faster (1)
Answer
Accept reference to muscles /
cells as site for aerobic
respiration
(3)
Acceptable answers
Mark
glucose → lactic acid
(1)
5BI2H/01
Additional Sample Mark Scheme
Question
Number
5(a)(i)
Answer
Mark
An explanation linking the following points
•
Question
Number
5(a)(ii)
Acceptable answers
•
by {osmosis / diffusion (of water)}
(1)
from an area where it is in high
concentration to an area where it is
in low concentration (1)
•
through the root hair cells (1)
Answer
(2)
Acceptable answers
Mark
D
(1)
Question
Number
5(b)(i)
Answer
Acceptable answers
Mark
Any two from the following points
•
more water lost from the Coleus / eq
(1)
•
ref to range of results for Coleus
being greater (1)
•
reference to mean for Coleus greater
(1)
(2)
5BI2H/01
Additional Sample Mark Scheme
Question
Number
QWC
*5(b)
(ii)
Level
1
0
1-2
2
3-4
3
5-6
Indicative Content
Mark
An explanation to include of the following points in a logical order
•
Coleus loses water from its leaves faster
•
therefore water drawn up Coleus faster
•
by transpiration
•
through the xylem
•
from the roots
•
details of transpiration given
•
Coleus could have a greater leaf surface area
•
because bigger leaves
•
Coleus could have more stomata
•
Coleus could have bigger stomata
•
more water taken in by osmosis
•
due to higher concentration in cells
(6)
•
reference to better root structure
No rewardable content
• the response recognises that there is a link between how much water is lost
and the recovery rate of the plants
• there is an attempt to suggest differences in the leaves of the two plants
• the answer communicates ideas using simple language and uses limited
scientific terminology
• spelling, punctuation and grammar are used with limited accuracy
• the response recognises that the Coleus plant loses water faster and that
this draws water up the plant faster, hence the faster recovery rate
• there is an attempt to describe transpiration and to suggest a reasonable
difference in the size of the leaves of the two plants
• there is some understanding of the reasoning behind each of the processes
• the answer communicates ideas showing some evidence of clarity and
organisation and uses scientific terminology appropriately
• spelling, punctuation and grammar are used with some accuracy
• the response gives a clear explanation of transpiration and links it to the
recovery of the Coleus
• more than one difference between the two plants is given, with the top
responses discussing differences relating to stomata
• the answer communicates ideas clearly and coherently uses a range of
scientific terminology accurately
• spelling, punctuation and grammar are used with few errors
5BI2H/01
Additional Sample Mark Scheme
Question
Number
6(a)(i)
Question
Number
6(a)(ii)
Answer
Acceptable answers
Mark
spelling must be correct
(1)
Acceptable answers
Mark
meiosis
Answer
C
(1)
Question
Number
6(b)(i)
Answer
Acceptable answers
Mark
A description including any two from
the following points
•
change in a base from C to T
(1)
•
(causes) change in one
{codon/triplet} of bases (1)
•
results in a different amino
acid (1)
(2)
5BI2H/01
Additional Sample Mark Scheme
Question
Number
6(b)
QWC
(ii)
Indicative content
An explanation linking some of the following points
•
•
•
•
•
•
•
•
•
•
Level
0
1
1-2
enzymes are proteins
mutation in DNA will result in different mRNA strand
during transcription
mRNA leaves the nucleus through the nuclear pore
the attachment of mRNA at the ribosomes
involvement of tRNA and amino acids
at the ribosome
which is translation
amino acid chain/peptide sequence altered
different protein formed/protein not folded correctly
•
•
•
•
3-4
•
•
3
•
5-6
•
•
•
Question
Number
6(c)
(6)
No rewardable material
•
2
Mark
an attempt is made to link the change in the base sequence to the
generation of a different amino acid
the answer communicates ideas using simple language and uses limited
scientific terminology
spelling, punctuation and grammar are used with limited accuracy
there is evidence that the mutation is linked to a change in the protein
there may be an attempt to explain transcription of the incorrect code
and translation of this although this explanation is likey to be unclear and
may have missing steps
the answer communicates ideas showing some evidence of clarity and
organisation and uses scientific terminology appropriately
spelling, punctuation and grammar are used with some accuracy
the response is likely to indicate the type of mutation and link this to a
change of codon/protein
the response will show good evidence of understanding that an incorrect
mRNA molecule is formed and translation by tRNA will result in an
incorrect amino acid being incorporated into the protein chain
the answer communicates ideas clearly and coherently uses a range of
scientific terminology accurately
spelling, punctuation and grammar are used with few errors
Answer
Acceptable answers
Mark
An explanation linking the following
points
• active site {different /
blocked / changed} (1)
•
substrate cannot bind /eq (1)
Accept reference to lock and key
(2)
5BI2H/01
Additional Sample Mark Scheme