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Edexcel a level biology
B question paper 1 june
2024 salters Nuffield +
mark scheme

Please check the examination details below before entering your candidate information
Candidate surname

Centre Number

Other names

Candidate Number

Pearson Edexcel Level 3 GCE

Wednesday 5 June 2024
Afternoon (Time: 1 hour 45 minutes)

Paper
reference

Biology B

9BI0/01
 

Advanced

PAPER 1: Advanced Biochemistry, Microbiology and

Genetics

You must have:
Scientific calculator, HB pencil, ruler

Total Marks

Instructions

Use black ink or ball‑point pen
...

Answer all questions
...


Information

The total mark for this paper is 90
...

In question(s) marked with an asterisk (*), marks will be awarded for your ability to
• structure
your answer logically showing how the points that you make are related
or follow on from each other where appropriate
...

•
Try to answer every question
...

Turn over

P78663A

©2024 Pearson Education Ltd
...



(a) Which box in each row of the table shows where endothelial cells and valves
are found?

Features of
blood vessels

(2)

Type of blood vessel
both capillaries
and veins

capillaries
only

veins only

neither capillaries
nor veins

Endothelial cells
Valves



(b) Nutrients and oxygen pass into cells from tissue fluid
...


(2)


...



...



...


(Total for Question 1 = 4 marks)

2

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Some questions must be answered with a cross in a box
...


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Answer ALL questions
...



(a) Sucrose is made from glucose and one other molecule
...


(1)

(3)


...



...



...



...



...




(a) Explain why viruses have to be cultured with ‘appropriate cells’
...



...



...



...

8

7

Log10 number
of viruses

6

5

4

0

20

40

60

80

100

Time after adding viruses to the cells / min

4

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3 Viruses can replicate using the lytic cycle
...


(3)


...



...



...



...



...


Give your answer to three significant figures
...

(Total for Question 3 = 7 marks)

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5

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(a) Glucose consists of the elements of carbon, hydrogen and oxygen
...


(2)

Carbon 
...

Oxygen
...
It is needed for photolysis and for the
excitation of electrons in the photosystems
...
This molecule is used by plants to
produce glucose and other organic molecules
...


Electricity generated from solar panels was used to convert carbon dioxide and
water into the organic compound acetate
...



(i) Explain why these plants were able to grow in the dark
...



...



...



...



...




(ii) The scientists hope that this technique will be able to produce
plant‑based food
...

Give a reason for your answer
...



...


(Total for Question 4 = 7 marks)

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7

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(a) In one year, 245 000 people were diagnosed with septicaemia and 49 735 of these
people died as a result
...
3%
B 25
...
3%
D
79
...
 coli) and
Staphylococcus aureus (S
...





E
...
 aureus is Gram positive
...
Both bacteria have peptidoglycan (murein) in their cell walls
...

S
...



3
...
 coli is thinner than that of S
...

A
2 only
B
3 only
C
1 and 2 only
D
1 and 3 only

8

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5 Septicaemia is a life‑threatening condition that can arise from bacterial infection
...
 coli and S
...


E
...

Which of the following statements about the toxins produced by these
bacteria are correct?
1
...
 coli produces endotoxins
...
The toxins produced by S
...




3
...
 aureus are components of their cell wall
...
The antigens they contain
stimulate only a weak immune response
...
 coli or S
...


The bacterial antigens are held in a mesh that contains chemicals to attract
tissue macrophages
...



(i) The graph shows the number of T helper cells in the lymph nodes that drain
the part of the body where the vaccine was implanted
...


(3)


...



...



...



...



...
 coli and S
...


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(ii) In part of this investigation, mice had the vaccine implanted 35 days before
being infected with E
...


A control group that had not received the vaccine was also infected
with E
...

The levels of antibody in the blood of both groups of mice were determined
...

6
mice with vaccine implanted
4
Concentration
of antibody / a
...

2

control group of mice

0
Time after infected with E
...


(3)


...



...



...



...



...


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(c) Gel electrophoresis can be used to analyse the DNA of a person
...


Position 1
Position 2

Position 3

Position 4
Position 5
Mother


Father

Daughter

Son 1

Son 2

Analyse the information to explain the bands found at positions 1 to 5 in this family
...



...



...



...



...



...



...


C
A
C
U
G
U
AA

A
G
U
G
A
C
A
U

uracil replaced with cytosine

A C
A
C
A
A
UUC U C
A AUCG
C
AAG A G
C
C
A
A
UU
U U A G CA
AG A
U A
U A
A U
A U
C G
A
C
A
U
UUU


(i) Name this type of mutation
...




(ii) Explain how this mutation could affect the structure of the tRNALys molecule
shown in the diagram
...



...



...



...


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(iii) Explain how this mutation could affect the role of the tRNALys molecule
...



...



...



...


(Source: © Andrew_Howe/Getty Images)



These birds all have mitochondria in their red blood cells
...




(i) Describe how the remaining ATP is produced
...



...



...



...


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(ii) In the mitochondria, ATP is synthesised on the inner
mitochondrial membrane
...
This is called leaked respiration
...


(3)


...



...



...



...



...




Length from head to tail / cm

Wing span / cm

Mass / g

Coal tit

11

19

11

Blue tit

12

18

11

Great tit

14

24

18

(i) Calculate the magnification of the photograph of the blue tit
...


(1)

Answer
...


Give your answer to one decimal place
...


18

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Species of bird

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(c) The generation of heat by red blood cell mitochondria in these birds in autumn
and winter was investigated
...
7 °C and in winter 7
...

The flowchart shows some of the stages in this investigation
...
They
were ringed, measured and a blood sample taken before they were released

Samples of intact red blood cells and red blood cell mitochondria were
separated from the blood

Leaked respiration and the volume of mitochondria were measured



(i) Explain why well‑stocked bird feeders containing nuts were available
throughout this investigation
...



...



...



...
4
Mean rate of
respiration / a
...

0
...
0

Season

20

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*(ii) The graphs show the results of this investigation
...
4
Mean rate of leaked
respiration / a
...

0
...
0

Au
t

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Graph 3: leaked respiration

Season
Explain the results of this investigation
...


(6)


...



...



...



...



...



...


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21

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...



...


(Total for Question 7 = 15 marks)

22

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8 Changes must occur to mammalian sperm cells as they pass through the female
reproductive tract, if successful fertilisation is to take place
...




The acrosome reaction (AR) must follow capacitation but not too soon nor too late
...


(3)


...



...



...



...



...


Protein G‑P results in the stimulation of capacitation and protein G results in the
inhibition of the AR
...

Protein G is located in the head region of the sperm
...

Complete the table to show which form of protein is present and which is absent
...


Presence of protein G
Event

Head region

Midpiece and
flagellum

(4)

Presence of protein G‑P
Head region

Midpiece and
flagellum

Just before capacitation
During capacitation
Just before the AR



(c) In some infertile males, the gene coding for protein G is methylated
...


(3)


...



...



...



...



...


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(d) Scientists have used ‘knockout’ mice to investigate the effect of protein G
on fertility
...


...



...



...



...


Describe how this investigation should be designed to confirm that protein G
affects fertility
...



...



...



...



...



...



...

Yolk
sac

Structures responsible for
synthesising globin subunits

Liver

Spleen

Bone marrow

100
gamma

80
Percentage of
globin subunits
synthesised (%)

beta

60
40
epsilon

20
0

delta
0

2

4

6

8

10

12

14

birth
Time after fertilisation / months


(i) Explain why there is no synthesis of globin subunits in the first few days
after fertilisation
...



...



...



...



...


26

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9 The structure of haemoglobin changes as the embryo develops into a fetus and after
the child is born
...


(4)


...



...



...



...



...



...


(iii) Calculate the rate of decrease in percentage of gamma globin subunit
synthesis from 7
...

Give your answer to two decimal places with appropriate units
...




*P78663A02732*

27

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(b) Sickle cell disease (SCD) is an inherited disorder
...

Bone marrow stem cells removed from patient

Stem cells genetically modified by turning the BCL11A gene off

Genetically modified stem cells injected back into the same patient



(i) There are approximately 67 million people in the UK, 15 000 of these people
are affected by SCD
...

Express your answer in whole numbers
...


28

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A new treatment for SCD is being developed that involves genetic modification
...

Analyse all the information in this question to discuss how this approach
could provide a cure for SCD and the ethical issues surrounding
this treatment
...


(6)


...



...



...



...



...



...



...



...



...


(Total for Question 9 = 17 marks)
TOTAL FOR PAPER = 90 MARKS

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29

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31

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Mark Scheme (Results)
Summer 2024
Pearson Edexcel Advanced Level GCE
In Biology B (9BI0)
Paper 1: Advanced Biochemistry,
Microbiology and Genetics

Edexcel and BTEC Qualifications
Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body
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For further information visit our qualifications websites at
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Alternatively, you can get in touch with us using the details
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Our aim is to help everyone progress
in their lives through education
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We’ve been involved in education for over 150 years, and by
working across 70 countries, in 100 languages, we have built an international reputation for our
commitment to high standards and raising achievement through innovation in education
...
pearson
...
Examiners
must mark the first candidate in exactly the same way as they
mark the last
...
Candidates must
be rewarded for what they have shown they can do rather
than penalised for omissions
...


•

There is no ceiling on achievement
...


•

All the marks on the mark scheme are designed to be
awarded
...
e
...
Examiners should also be prepared to award zero
marks if the candidate’s response is not worthy of credit
according to the mark scheme
...


•

When examiners are in doubt regarding the application of the
mark scheme to a candidate’s response, the team leader
must be consulted
...
7

•

correct answer given to 3 sig figs (1)

10 000 000 – 501 187 = 9 498 813 = 9 500 000
more particles /or
10 000 000 ÷ 501 187 = 19
...
0 x
7 – 5
...
3 = 20
...
797, 0
...
7
B is incorrect because (245 000 – 49735) ÷ 245 000 = 0
...
797 × 100 = 79
...
797, 0
...
7

(1)

Question
Number
5(b)(i)

Answer

Additional Guidance

Mark

The only correct answer is D
A is incorrect because both have peptidoglycan in their cell wall (1) and a Gram
negative bacteria has a thinner cell wall (3) S
...
aureus is gram positive so
statement 2 is incorrect

Question
Number
5(b)(ii)

Answer

Additional Guidance

Mark

The only correct answer is A
B is incorrect because S aureus releases exotoxins whilst it is alive
C is incorrect because S aureus releases exotoxins whilst it is alive
D is incorrect because S aureus releases exotoxins whilst it is alive

(1)

Question
Number
5(c)

Answer

Additional Guidance

Mark

The only correct answer is A
B is incorrect because a vaccine is artificial and the antigens contained stimulate
an immune response
C is incorrect because a vaccine is artificial and the antigens contained stimulate
an immune response

(1)

D is incorrect because a vaccine is artificial and the antigens contained stimulate
an immune response

Question
Number
5(d)(i)

Answer
An explanation that makes reference to three of the following:

Additional Guidance

Mark

IGNORE descriptions of the immune response not
linked to time

•

few T helper cells at start as delay whilst antigen is being
presented (1)

•

some T helper cells present as they have been activated and are
beginning to divide (1)

ACCEPT description of clonal selection linked to
increase

•

large number of T helper cells by day 2 due to clonal expansion
(1)

ACCEPT large number due division by mitosis

•

drop in number after 21 days as T helper cells have {died /
moved out of the lymph nodes} (1)

(3)

Question
Number
5(d)(ii)

Answer

Additional Guidance

Mark

An explanation that makes reference to three of the following:
•

no antibodies in control mice as they had not been exposed to
antigen before (1)

•

Rapid production of antibodies in vaccinated mice due to
secondary immune response (1)

•

immediate increase in antibodies in vaccinated mice as they had
memory cells present as a result of the vaccine (1)

•

therefore rapid increase in plasma cells to release antibodies (1)

Question
Number
6(a)

Answer

Mouse must undergo primary immune response
(which takes time)

(3)
ACCEPT converse

Additional Guidance

Mark

The only correct answer is C
A is incorrect because there is no cytoplasm in mitochondria
B is incorrect because mitochondrial DNA is found in the matrix
D is incorrect because mitochondria do not have a nucleus

(1)

Question
Number
6(b)

Answer

Additional Guidance

Mark

The only correct answer is A
B is incorrect because circular DNA has two more phosphodiester bonds than linear
DNA with the same number of pentoses
C is incorrect because mitochondrial DNA is circular
(1)
D is incorrect because mitochondrial DNA is circular

Question
Number
6(c)

Answer
An explanation that makes reference to four of the following:

Additional Guidance

Mark

IGNORE references to how the bands form
Must refer to DNA in band at least once

•

position 1 : only the daughter has inherited this DNA from the
mother (1)

•

position 2 : (genes located in) this DNA is common to all people
(1)

•

position 3 : daughter and son 1 have inherited this DNA from
their father but son 2 has not (1)

•

position 4 : this could be mitochondrial DNA as it is found in the
mother and all children (1)

•

position 5 : this could be the mitochondrial DNA of the father as
it is not present in any of the children (1)

(4)

Question
number
6(d)(i)

Answer

•

Additional Guidance
ACCEPT transition

substitution (1)

Mark

(1)

DO NOT ACCEPT insertion / addition / deletion
/ subtraction / chromosome mutation /
frameshift / inversion / duplication/
replacement

Question
Number
6(d)(ii)

Answer

Additional Guidance

Mark

An explanation that makes reference to two of the following:
• the cytosine will not bind with the adenine (1)
• therefore the hydrogen bonds will not form (1)
• so {the left hand loop will open up / other bonds may form} (1)

(2)
ACCEPT tRNA will be a different shape

Question
Number
6(d)(iii)

Answer

Additional Guidance

Mark

An explanation that makes reference to two of the following:
• the tRNA may not be able to bind with the ribosome (1)
• and therefore not hold the amino acid in place (1)
OR
• the amino acid may not be able to bind to the tRNA (1)
• and therefore {this amino acid cannot be / fewer of these amino
acids} brought to the ribosome (1)
OR
• the anticodon may not be able to bind to the codon on the mRNA
(1)

(2)

• and therefore the amino acid will not be held in place (1)
Question
Number
7(a)(i)

Answer

Additional Guidance

Mark

A description that makes reference to two of the following:
• during glycolysis (1)
• glucose is converted to pyruvate (1)
• by substrate level phosphorylation (1)

ACCEPT anaerobic respiration
(2)

Question
Number
7(a)(ii)

Answer

Additional Guidance

An explanation that makes reference to the following:
• because the proton gradient would be shallower (1)

ACCEPT electrochemical/chemiosmotic gradient

• so protons would flow through the ATP synthase more slowly (1)

IGNORE fewer protons

• therefore less energy for the phosphorylation of ADP (1)

ACCEPT less ADP + Pi converted to ATP

Question
Number
7(b)(i)

Answer
• length from head to tail in photo in cm ÷ 12 (1)

Question
Number
7(b)(ii)

Answer

(3)

Additional Guidance
40-42 mm
40 = 0
...
34 x 42= 0
...
8 : 1 (1)

Question
Number
7(c)(i)

Mark

Answer

Additional Guidance

Mark

An explanation that makes reference to two of the following:
• To attract the birds to be caught (1)
• so that respiration would not be limited by energy source (1)
• to make the investigation valid (1)

so that all birds had plenty to eat
(2)

Question Indicative content
Number
*7(c)(ii) Graph 1 (volume of mitochondria)
•
•
•
•
•
•
•
•
•

Mark

great tits have larger volume of mitochondria than blue tits
blue tits have more mitochondria than coal tits
error bars do not overlap between autumn and winter so differences are significant
because bigger birds need to produce more ATP for flying
coal tits are smaller so need less ATP
all three birds increase their number of mitochondria in the winter
by division
because more heat needs to be generated to keep warm in the colder season
greater volume needed in winter because more leaked respiration

Graph 2 (respiration producing ATP)
•
•
•
•
•
•
•
•
•
•

coal tits respire faster than blue tits and great tits
because they have fewer mitochondria
no significant difference between blue tits and great tits
there is no difference in rate of respiration in great tits between autumn and winter
the rate of respiration decreases in winter in blue tits and coal tits
because they need to {produce more heat / switch to leaked respiration}
as they have a larger surface area : volume ratio
and therefore lose more body heat
if switch to leaked respiration is made there is a smaller proton gradient
therefore less respiration to make ATP

Graph 3 (leaked respiration)
• Rate of leaked respiration is faster in all three species in winter
• rate of leaked respiration is faster in coal tits than blue tits
• rate of leaked respiration is similar in great tits than blue tits
• there are no error bars so cannot say if any differences are significant
• because leaked respiration produces more heat

(6)

•
•

needed for coal tits because they have a larger surface area : volume ratio
and therefore lose more body heat

Additional guidance
Level 0
Level 1

Level 2

Level 3

0
1-2

3-4

5-6

No awardable content
An explanation may be attempted but with limited interpretation or
analysis of the scientific information and with a focus on mainly just one
piece of scientific information
...

An explanation will be given, with occasional evidence of analysis,
interpretation and/or evaluation of both pieces of scientific information
...

An explanation is made that is supported throughout by sustained
application of relevant evidence of analysis, interpretation and/or
evaluation of both pieces of scientific information
...


Simple descriptions of the data
1 mark = a description relating to one graph
2 marks = a description for three graphs
Some explanation of the data
3 marks = explanation for one of the graphs
4 marks = explanation for two of the graphs
Detailed explanation of the data
5 marks = explanation of two graphs and discussion of
error bars
6 marks = explanation of all three graphs and both size and
seasonal variation discussed

Question
Number
8(a)

Answer

Additional Guidance

Mark

An explanation that makes reference to the following:
•

if AR is too early then there will not be the enzymes available for
fertilisation (1)

•

and it will not be able to digest through the zona pellucida/jelly
coat (1)

•

if AR is too late then {the egg/sperm (cell) may have died / the
sperm have swam past the egg (cell)} (1)

Question
Number
8(b)

Answer

Event
Just before capacitation
During capacitation
Just before the AR

Presence of protein G
Mid piece and
Head region
flagellum
✓

✓




(3)

Additional Guidance
Presence of protein G-P
Mid piece and
Head region
flagellum

✓

✓
✓
✓

Mark

(4)

Question
Number
8(c)

Answer

Question
Number
8(d)(ii)

Mark

An explanation that makes reference to the following:
•

DNA methylation {silences / switches off} a gene (1)

•

Because is a form of epigenetic modification (1)

•

therefore the gene cannot be transcribed / therefore no protein
G will be produced (1)

ACCEPT Transcription factors cannot bind to DNA

without protein G the timings of capacitation and AR will be
wrong (1)

ACCEPT If AR not inhibited, AR will happen too
soon

•

Question
Number
8(d)(i)

Additional Guidance

Answer

(3)

Additional Guidance

Mark

An answer that makes reference to the following:
•

genetically-modified (laboratory) mice (1)

•

that have had the gene coding for protein G {inactivated /
replaced} (1)

ACCEPT genes coding for protein G prevented from
being expressed

Answer

Additional Guidance

(2)

Mark

A description that makes reference to three of the following:
•

all female mice would have to be fertile (1)

•

fertile males used as a control (1)

•

pregnancy rate measured (1)

•

Same species/age of mouse/diet (1)

ACCEPT Control group that have not had gene
modified
ACCEPT Number of offspring produced

(3)

Question
Number
9(a)(i)

Question
Number
9(a)(ii)

Answer

Additional Guidance

Mark

An explanation that makes reference to three of the following:
•

because the embryo is increasing in cell number (1)

ACCEPT Mitosis is occurring

•

all the cells are unspecialised (1)

ACCEPT Cells have not differentiated/are
differentiating

•

and have no genes switched on to produce the globin subunits (1)

•

no yolk sac has developed yet (1)
Answer

IGNORE genes haven’t been switched off

Additional Guidance

(3)

Mark

An answer that makes reference to four of the following:
•

the structure responsible for synthesising the globin subunits
changes with time after fertilisation (1)

•

the {embryo / fetus} contains gamma globin and either epsilon or
beta globin (and some delta) (1)

•

the baby has increasing levels of beta (and delta globin) and
decreasing levels of gamma globin (1)

•

yolk sac is responsible for synthesis of epsilon globin (1)

•

the liver synthesises mainly gamma globin (1)

ACCEPT Liver synthesises gamma and beta

•

bone marrow synthesises both components of adult haemoglobin
(1)

ACCEPT Bone marrow synthesises beta haemoglobin

ACCEPT Mainly gamma made before birth and
mainly beta after birth

(4)

Question
Number
9(a)(iii)

Answer

Additional Guidance

•

correct two values read from graph, subtracted and divided by
2
...
4 = 21
...
67 month-1 /month = 2 marks
21
...
6666666666666 : 1

Question Indicative content
Number
*9(b)(ii)
(1)Globin subunits and sickle cell disease :
• sickle cell anaemia is caused by a mutation
• in the gene coding for the beta globin subunit
• which causes the red blood cells change shape
• and reduces the blood supply to cells of the body
• fetal haemoglobin contains gamma globin which is switched to beta globin at birth
(2)BCL11A gene :
• switch due to the presence of the gene product from the BCL11A gene
• if BCL11A switched off there will be no transcription
• therefore BCL11A will not be produced
• and therefore gamma globin will still be produced (fetal)
• which is not defective
• therefore red blood cells will no longer be sickle shaped
• haemoglobin affinity would be higher
(3)Method used :
• bone marrow stem cells used as these are the cells that produce haemoglobin in the fetus and the baby
• stem cells will divide by mitosis
• to produce genetically-identical cells
• that will all contain the modified BCL11A gene
• advantage of using own stem cells is that they will not cause an immune response when returned to body
• as there will not be any (foreign) antigens for the immune system to recognise
• and own stem cells will not be rejected
• wont need immunosuppressants
(4)Ethics:
• new technology so little evidence on effect
• e
...
that this will work in the long term
• e
...
there are no unanticipated changes
• relevant comment about cost
• changes will not be inherited
• painful to extract cells from patient

Mark

(6)

•
•
•
•
•
•

may become cancerous
Cells taken from patient not embryo
Treatment would save lives
If young, they cannot consent
Who will the treatment be available to
Don’t know long term effects of GM

Additional guidance
Level 0
Level 1

Level 2

Level 3

0
1-2

3-4

5-6

No awardable content
Demonstrates isolated elements of biological knowledge and
understanding to the given context with generalised comments made
...
The
discussion will contain basic information with some attempt made to link
knowledge and understanding to the given context
...
Consequences are
discussed which are occasionally supported through linkage to a range of
scientific ideas, processes, techniques and procedures
...

Demonstrates comprehensive knowledge and understanding by selecting
and applying relevant biological facts / concepts
...
The discussion
shows a well-developed and sustained line of scientific reasoning which is
clear and logically structured
...
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