Author: Philip Colls
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Mutations and Gene expression booklet revision questions   NAME:
Q1. (a)     There are different types of gene mutation.
Put a tick (✓) in the box next to the statement which describes incorrectly the effect of the mutation in an exon of a gene.
 
A substitution may not result in a change to the encoded amino acid. |
An inversion will result in a change in the number of DNA bases. |
A deletion will result in a frame shift. |
An addition will result in a frame shift. |
(1)
(b)     Describe how alterations to tumour suppressor genes can lead to the development of tumours.
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(c)     A type of malignant tumour cell divides every 8 hours. Starting with one of these cells, how many tumour cells will be present after 4 weeks? Assume none of these cells will die. Give your answer in standard form.

Answer = ____________________                     (2)
(Total 6 marks)
Q2. (a)     What name is used for the non-coding sections of a gene?
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          Figure 1 shows a DNA base sequence. It also shows the effect of two mutations on this base sequence. Figure 2 shows DNA triplets that code for different amino acids.
Figure 1
 
Original DNA base sequence | A | T | T | G | G | C | G | T | G | T | C | T
Amino acid sequence | | | |
Mutation 1 DNA base sequence | A | T | T | G | G | A | G | T | G | T | C | T
Mutation 2 DNA base sequence | A | T | T | G | G | C | C | T | G | T | C | T
Figure 2
 
DNA triplets | Amino acid
GGT, GGC, GGA, GGG | Gly
GTT, GTA, GTG, GTC | Val
ATC, ATT, ATA | Ile
TCC, TCT, TCA, TCG | Ser
CTC, CTT, CTA, CTG | Leu
(b)     Complete Figure 1 to show the sequence of amino acids coded for by the original DNA base sequence.
(1)
(c)     Some gene mutations affect the amino acid sequence. Some mutations do not.
Use the information from Figure 1 and Figure 2 to explain
(i)      whether mutation 1 affects the amino acid sequence
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(ii)     how mutation 2 could lead to the formation of a non-functional enzyme.
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(d)     Gene mutations occur spontaneously.
(i)      During which part of the cell cycle are gene mutations most likely to occur?
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(ii)     Suggest an explanation for your answer.
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(Total 9 marks)
Q3. (a)     Define what is meant by epigenetics.
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(b)     In eukaryotes, transcription of target genes can be stimulated or inhibited when specific transcriptional factors move from the cytoplasm into the nucleus.
Oestrogen, methyl groups and acetyl groups are control factors that can play a role in initiating transcription.
Complete the table to show features of these control factors.
Put a tick (✔) in the box if the control factor shows the feature.
 
| Feature
Control factor | Binds with DNA | Binds with protein
Oestrogen | |
Methyl groups | |
Acetyl groups | |
(2)
(c)     Explain how increased methylation could lead to cancer.
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(d)     Give one way in which benign tumours differ from malignant tumours.
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(1)
(Total 8 marks)
Q4.
Phenylketonuria is a disease caused by mutations of the gene coding for the enzyme PAH. The table shows part of the DNA base sequence coding for PAH. It also shows a mutation of this sequence which leads to the production of non-functioning PAH.
 
DNA base sequence coding for PAH | C | A | G | T | T | C | G | C | T | A | C | G
DNA base sequence coding for non-functioning PAH | C | A | G | T | T | C | C | C | T | A | C | G
(a)     (i)      What is the maximum number of amino acids for which this base sequence could code?
 
(1)
(ii)     Explain how this mutation leads to the formation of non-functioning PAH.
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(3)

PAH catalyses a reaction at the start of two enzyme-controlled pathways.
The diagram shows these pathways.
 
(b)     Use the information in the diagram to give two symptoms you might expect to be visible in a person who produces non-functioning PAH.
1. _________________________________________________________________
2. _________________________________________________________________
(2)
(c)     One mutation causing phenylketonuria was originally only found in one population in central Asia. It is now found in many different populations across Asia. Suggest how the spread of this mutation may have occurred.
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(1)
(Total 7 marks)
Q5.
One hypothesis for the cause of cancer of the colon (large intestine) is that Clostridium bacteria present in the gut can convert bile steroids into cancer-causing substances.
(a)     Explain the presence of bile in the colon.
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(2)

(b)     The concentrations of bile steroids and numbers of Clostridium bacteria were measured in people with colon cancer and in controls without colon cancer. The table shows the results.
 
Concentration of bile steroids | Number of Clostridium bacteria | Percentage of cancer patients | Percentage of controls | P

high
high
low
low

high
low
high
low

76
13
7
4

9
8
34
49

<0.01 <0.01 <0.01 <0.01

A statistical test showed there was a significant difference between the cancer patients and the controls in each of the four categories.
(i)      Explain how the results could be used to support the hypothesis that Clostridium bacteria convert bile steroids into substances which cause colon cancer.
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(2)
(ii)     Explain how the results indicate that other factors may be involved in causing colon cancer.
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(1)
(c)     Human cells contain genes that control their growth and division. One of these genes codes for a protein that prevents cell division. The substances formed from bile steroids by Clostridium bacteria may cause gene mutation. Describe and explain how these substances could cause colon cancer.
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Q6.
Alzheimer’s disease (AD) is a non-reversible brain disorder that develops over a
number of years. At the start of 2014 the number of Americans with AD was
estimated to be 5.4 million. Every 30 seconds another person in America
develops AD.
5        In the brain of a person with AD there is a lower concentration of acetylcholine.
This affects communication between nerve cells and initially results in memory
loss and confusion. Some of the symptoms of AD that are associated with
communication between nerve cells are reduced by taking the drug donepezil.
Donepezil inhibits the enzyme acetylcholinesterase.
10      A gene mutation called E280A found on chromosome 14 causes early-onset AD
at a mean age of 49 years. The age at which the E280A mutation is expressed
to cause AD varies.
Yaramul is a town in a historically isolated region of the Andes Mountains. The
population of this town has the highest frequency of the E280A mutation in the
15      world. The origin of the E280A mutation in this population has been traced back
to a common ancestor in the 17th century. Natural selection has not reduced
the frequency of the E280A mutation in the population.
This autosomal dominant mutation involves a change in triplet 280 from GAA to
GCA. Scientists analysed chromosome 14 from 102 individuals from Yaramul.
20      They recorded a sample size of 204 and detected 75 E280A mutations but only
74 potential AD cases. The scientists identified individuals with the mutation by
whole genome sequencing. They had decided that a DNA probe would not be a
suitable method to detect the E280A mutation.
(a)     Assuming no one with AD died in 2014, calculate the annual percentage increase in AD cases in America for 2014 (lines 2–4).
Answer = ____________________ %
(2)
(b)     Explain how donepezil could improve communication between nerve cells (lines 7–9).
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(3)
(c)     Suggest and explain two reasons why there is a high frequency of the E280A mutation in Yaramul (lines 13–15).
1. _________________________________________________________________
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2. _________________________________________________________________
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(2)
(d)     Explain why natural selection has not reduced the frequency of the E280A mutation in the population (lines 16–17).
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(e)     The age at which the E280A mutation is expressed to cause AD can vary (lines 11–12).
Suggest and explain one reason for this.
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(f)      One scientific study which analysed chromosome 14 involved 102 individuals. The scientists recorded a sample size of 204. In this sample they detected 75 E280A mutations but only 74 potential AD cases (lines 19–21).
Suggest explanations for the figures the scientists recorded.
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(2)
(g)     Suggest why a DNA probe for the mutated triplet was not considered a suitable method for detection of the E280A mutation (lines 22–23).
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(2)
(Total 15 marks)
Q7.
(a)     Each year, a few people with type I diabetes are given a pancreas transplant. Pancreas transplants are not used to treat people with type II diabetes.
Give two reasons why pancreas transplants are not used for the treatment of type II diabetes.
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2. _________________________________________________________________
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(2)
(b)     The pancreas produces the hormone insulin.
Put a tick (✔) in the box next to the statement which describes incorrectly the action of insulin.
 
Activates enzymes involved in the conversion of glucose to glycogen. |
Controls the uptake of glucose by regulating the inclusion of channel proteins in the surface membranes of target cells. |
Attaches to receptors on the surfaces of target cells. |
Activates enzymes involved in the conversion of glycerol to glucose. |
(1)

(c)     Scientists investigated the use of induced pluripotent stem cells (iPS cells) to treat type I diabetes in mice. The scientists used four transcription factors to reprogramme skin cells to form iPS cells. The scientists then stimulated the in vitro differentiation of iPS cells into pancreatic cells.
The scientists set up three experimental groups:
•   Group A – 30 mice with type I diabetes received pancreatic cell transplants derived from iPS cells.
•   Group B – 30 mice with type I diabetes were left untreated.
•   Group C – 30 mice without diabetes were left untreated.
The scientists measured the blood glucose concentration of all the mice on a weekly basis for 12 weeks.
The results the scientists obtained are shown in the graph.
 
Suggest how transcription factors can reprogramme cells to form iPS cells.
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(2)

(d)     Using all the information provided, evaluate the use of iPS cells to treat type I diabetes in humans.
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(4)
(Total 9 marks)
Q8.
Read the following passage.
Complete achromatopsia is a form of complete colour blindness. It is caused
by having only rods and no functional cone cells. People with complete
achromatopsia have difficulty in seeing detail. Complete achromatopsia is
caused by an autosomal recessive allele and is usually very rare in
populations with only one in 40 000 being affected. However on the Pacific              5
island of Pingelap ten percent of the population are affected.
One form of red-green colour blindness is caused by a sex-linked recessive
allele which affects more men than women. People with this red-green
colour blindness are unable to distinguish between red and green, and also
between other colours. They have green-sensitive cones but the                             10
photoreceptive pigment they contain does not function.
Scientists investigated the use of gene therapy to correct red-green colour
blindness in monkeys. They injected viruses containing the gene for the
green-sensitive pigment directly into the eyes of the monkeys. Although the
monkeys maintained two years of colour vision, there is debate on whether            15
this form of gene therapy is worthwhile. No clinical trials of this procedure
have been carried out on humans. Current research into the treatment of
red-green colour blindness involves the use of induced pluripotent stem cells
(iPS cells). The use of iPS cells could have advantages over the use of gene
therapy.                                                                                                                       20
Use the information in the passage and your own knowledge to answer the following questions.

(a)     People with complete achromatopsia have difficulty in seeing detail (lines 2–3).
Explain why.
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(3)
(b)     Ten percent of the population on the Pacific island of Pingelap are affected by complete achromatopsia (lines 3–6).
Use the Hardy-Weinberg equation to calculate the percentage of this population who are heterozygous for this disorder.
Show your working.
Answer = ____________________ %
(2)
(c)     Red-green colour blindness affects more men than women (lines 7–8).
Explain why.
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(2)
(d)     People with red-green colour blindness are unable to distinguish between red and green, and also between other colours (lines 8–10).
Explain why.
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(e)     Current research into the treatment of red-green colour blindness involves the use of induced pluripotent stem cells (iPS cells) (lines 17–19).
Suggest how iPS cells could correct red-green colour blindness.
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(f)     The use of iPS cells could have advantages over the use of gene therapy to correct red-green colour blindness (lines 19–20).
Using the information from the passage, suggest and explain reasons why.
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Q9. Scientists found a correlation between prostate cancer and exposure to cadmium ions.
The scientists investigated the effects of cadmium ions on cells from a human prostate gland.
They grew a culture of these cells in liquid growth medium and removed samples at intervals.
For each sample they measured
•    how much DNA was not methylated,
•    the activity of the enzyme methyltransferase.
Methyltransferase is an enzyme that adds methyl groups to some of the bases in DNA.
The addition of a methyl group is called methylation.
(a)     The scientists set up another culture as a control.
Describe how the scientists would have set up a control experiment for this investigation.
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(b)     Figures 1 and 2 show the scientists’ results.
Figure 1                                                                        Figure 2
 
(i)      The scientists expressed their results as percentages of the control values.
Suggest why.
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(ii)     Use information from Figure 1 to describe how exposure to cadmium ions affected the methylation of DNA.
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(iii)     Use information from Figure 2 to suggest what caused the change to the DNA shown in Figure 1.
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(c)     Prostate gland cells contain a tumour suppressor gene called p16.
During the investigation, the scientists also measured the amount of  p16 protein produced. Figure 3 shows their results.
Figure 3
 
The scientists found that the promoter DNA of the p16 gene had become methylated. The promoter is the sequence of bases where the enzyme RNA-polymerase binds to a DNA molecule. Explain how methylation of the promoter sequence of the p16 gene could cause the changes shown in Figure 3.
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(d)     Each week of the investigation, the scientists took samples of the cadmium-treated prostate cells from the laboratory cultures. They injected these cells into mice and monitored the mice for the growth of tumours. It was only the samples taken in the tenth week that caused tumours to begin to grow in the mice.
Use information from Figures 1, 2 and 3 to suggest why.
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Q10.
(a)     What is meant by a genome?
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(1)
Chromatin immunoprecipitation is one method to determine where a transcription factor binds to DNA. The principle behind this procedure is shown in the chart.

(b)     Explain why the antibody binds to the transcription factor.
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(2)
(c)     Use the chart to explain what ‘precipitated DNA’ consists of.
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(1)
Soybeans are used in a number of processed foods. However, soybeans contain a protein known as P34 that causes an allergic response in some people. Scientists have created transgenic soybeans that produce single-stranded cDNA, which prevents transcription of the P34 gene. They used recombinant plasmids as vectors to transform soybean cells. After they had screened these cells for production of the P34 protein, they cultured the transformed cells to form soybean plants.
(d)     Suggest how single-stranded cDNA could prevent transcription of the P34 gene.
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(e)     Describe the roles of two named types of enzymes used to insert DNA fragments into plasmids.
Type of enzyme _____________________________________________________
Role _______________________________________________________________
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Type of enzyme _____________________________________________________
Role _______________________________________________________________
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(f)      The soybean cells were screened for the presence of the P34 protein. This process involved the use of gel electrophoresis to separate proteins extracted from soybean cells. Suggest two features of the structure of different proteins that enable them to be separated by gel electrophoresis.
1. _________________________________________________________________
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2. _________________________________________________________________
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(2)  (Total 9 marks)
Q11.
Scientists have investigated the use of different types of stem cell to treat damage to the heart after a myocardial infarction. During a myocardial infarction, a number of different cell types in the heart die. This includes cardiomyocytes which are heart-muscle cells.
Embryonic pluripotent stem cells (ESCs) can divide and differentiate into a wide range of different cell types.
(a)     Using the information given, suggest one reason why ESCs might be suitable to treat damage to the heart.
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(1)
(b)     ESCs have not yet been used to treat people who have had a myocardial infarction. This is because of concern that the use of ESCs might lead to more harm to the person. One way that ESCs might lead to more harm is by differentiating into the wrong types of cells.
Suggest one other way that putting ESCs into a person’s heart might lead to more harm to the person.
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(2)
(c)     Transplants of cardiomyocytes have been shown to increase the repair of heart tissue damaged by myocardial infarction.
One group of scientists investigated the hypothesis that these transplants work by stimulating growth of new blood vessels into damaged heart tissues. They obtained three groups of mice, A, B and C that had suffered myocardial infarctions.
•        Group A were operated on but no transplant was given.
•        Group B were operated on and given transplants containing cardiomyocytes and two other types of heart cell.
•        Group C were operated on and given transplants containing the two other types of heart cells but no cardiomyocytes.
After a suitable time, the scientists measured the mean number of capillaries per mm^2 in sections taken from areas of the hearts of the mice affected by myocardial infarction.
Their results are shown in the graph below. The bars show ±2 standard deviations, which includes 95.4% of the data.
 
Group A was a control group. Explain two ways in which Group A acts as a control.
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2. _________________________________________________________________
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(2)
(d)     What can you conclude from these data about the stimulation by cardiomyocytes on growth of new blood vessels into damaged heart tissues?
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(3)

(e)     Suggest how the growth of new blood vessels into damaged heart tissues could increase the rate of repair of tissues.
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(f)      The scientists used an optical microscope to measure the number of capillaries in thin sections cut from samples of heart muscle.
Describe the method they would have used to find the mean number of capillaries per mm^2.
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___________________________________________________(4)(Total 15 marks)
Q12. (a)     Explain how the structure of DNA is related to its functions.

(6)
Scientists investigated three genes, C, D and E, involved in controlling cell division.
They studied the effect of mutations in these genes on the risk of developing lung cancer.
The scientists analysed genes C, D and E from healthy people and people with lung cancer.
•        If a person had a normal allele for a gene, they used the symbol N.
•        If a person had two mutant alleles for a gene, they used the symbol M.
They used their data to calculate the risk of developing lung cancer for people with different combinations of N and M alleles of the genes. A risk value of 1.00 indicates no increased risk. The following table shows the scientists’ results.

Gene C

Gene D

Gene E

Risk of
developing
lung cancer

N | N | N | 1.00
M | N | N | 1.30
N | N | M | 1.78
N | M | N | 1.45

N = at least one copy of the normal allele is present
M = two copies of the mutant allele are present

(b)     What do these data suggest about the relative importance of the mutant alleles of genes C, D and E on increasing the risk of developing lung cancer? Explain your answer.
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(3)

Chemotherapy is the use of a drug to treat cancer. The drug kills dividing cells.
The figure below shows the number of healthy cells and cancer cells in the blood of a patient receiving chemotherapy. The arrows labelled F to I show when the drug was given to the patient.
 
                                    Time / days
(c)     Calculate the rate at which healthy cells were killed between days 42 and 46.

 
____________________ cells killed per unit volume of blood per day
(1)
(d)     Describe similarities and differences in the response of healthy cells and cancer cells to the drug between times F and G.
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(e)     More cancer cells could be destroyed if the drug was given more frequently.
Suggest why the drug was not given more frequently.
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(2)
(Total 15 marks)

Mark schemes
Q1.
(a)     Box 2.
An inversion will result in a change in the number of DNA bases.
Reject if more than one box with tick. Ignore crossed-out ticks
1
(b)     1.      (Increased) methylation (of tumour suppressor genes);
Accept abnormal methylation or hypermethylation
Ignore decreased acetylation of histones
2.      Mutation (in tumour suppressor genes);
3.      Tumour suppressor genes are not transcribed/expressed
OR
Amino acid sequence/primary structure altered;
Accept mRNA for transcription/transcribed
Accept tertiary structure altered
Accept different amino acid
Ignore reference to protein not being formed
4.      (Results in) rapid/uncontrollable cell division;
Accept cell division cannot be regulated
Ignore growth
3 max
(c)     1.      Correct answer of 1.9/1.93 × 10^25 = 2 marks;;
Accept 2 × 10^25 = 2 marks
Ignore any numbers after 1.93
2.      Incorrect answer but shows 84 = 1 mark
OR
28 × 3 = 1 mark
OR
Incorrect answer but shows 672 divided by 8 = 1 mark;
2
[6]
Q2.
(a)     Introns;
1
(b)     Ile Gly Val Ser;
1
(c)     (i)      Has no effect / same amino acid (sequence) / same
primary structure;
Q Reject same amino acid formed or produced.
1
Glycine named as same amino acid;
1
It still codes for glycine = two marks.
(ii)     Leu replaces Val / change in amino acid (sequence) / primary structure;
Change in hydrogen / ionic bonds which alters tertiary structure / active site;
Q Different amino acid formed or produced negates first marking point.
Substrate cannot bind / no longer complementary / 
no enzyme-substrate complexes form;
Active site changed must be clear for third marking point but does not need reference to shape.
3
(d)     (i)      Interphase / S / synthesis (phase);
1
(ii)     DNA / gene replication / synthesis occurs / longest stage;
Allow ‘genetic information’ = DNA.
Allow ‘copied’ or ‘formed’ = replication / synthesis
1
[9]
Q3.
(a)     1.      Heritable changes in gene function;
2.      Without changes to the base sequence of DNA;
2
(b)
 
Control element | Binds with DNA | Binds with protein
Oestrogen | | ✔
Methyl groups | ✔ |
Acetyl groups | | ✔
1 mark for each correct column.
Accept both boxes ticked in oestrogen row.
2
(c)     1.      Methyl groups (could be) added to (both copies of) a tumour suppressor gene;
2.      The transcription of tumour suppressor genes is inhibited;
3.      Leading to uncontrolled cell division.
3
(d)     Cells of benign tumours cannot spread to other parts of the body / metastasise;
OR
Cells of benign tumours cannot invade neighbouring tissues.
Accept answers clearly in the context of malignant tumours.
1
[8]
Q4.
(a)     (i)      4;
1
(ii)     1.      Change in amino acid / (sequence of) amino acids / primary structure;
1. Reject = different amino acids are 'formed'
2.      Change in hydrogen / ionic / disulphide bonds alters tertiary structure / active site (of enzyme);
2. Alters 3D structure on its own is not enough for this marking point.
3.      Substrate not complementary / cannot bind (to enzyme / active site) / no enzyme- substrate complexes form;
3
(b)     1.      Lack of skin pigment / pale / light skin / albino;
2.      Lack of coordination / muscles action affected;
2 max
(c)     Founder effect / colonies split off / migration / interbreeding;
Allow description of interbreeding e.g. reproduction between individuals from different populations
1
[7]
Q5.
(a)     secreted by the liver / storage / release from gall bladder into the duodenum / small intestine;
bile passes unchanged from small intestine to colon;
2
(b)     (i)      chance alone has not caused the difference (between the two patients types);
high steroid high bacteria (significantly) higher percentage of cancer patients / low steroids low bacteria (significantly) higher percentage of control patients;
2
(ii)     some patients with low levels of one / both factor(s) have cancer;
1
(c)     change in code / base sequence / structure of gene;
addition / deletion / substitution;
mRNA / transcription changed;
gene product / protein structure / amino acid sequence changed / different protein;
loss of function;
uncontrolled cell division;
4 max
[9]
Q6.
(a)     1.      Correct answer of 19.4 / 19.41%
OR
19.47 / 19.5% = 2 marks;
2.      Incorrect answer but shows increase of
1,048,320 OR 1,051,200 = one mark;
Accept: 19.46% for one mark.
2
(b)     1.      Less / no acetylcholine broken down;
2.      Acetylcholine attaches to receptors;
3.      (More) Na^+ enter to reach threshold / for depolarisation / action potential / impulse;
1.      Accept: more acetylcholine present / remains.
1 and 2. Accept: remains attached for longer = 2 marks.
3.      Must be sodium ions.
3
(c)     1.      Isolated so inbreeding / low genetic diversity / small gene pool;
2.      Allele inherited (through generations) from (common) ancestor;
1.      Ignore: Founder effect.
1.      Accept: no interbreeding with other populations.
1.      Reject: interbreeding within the population.
2
(d)     1.      AD / symptoms develops late / at 49;
2.      Have already reproduced;
Note: ‘It’ is not equivalent to AD / symptom as the question stem relates to the mutation.
2
(e)     1.      Epigenetics / environment / named factor e.g. stress, alcohol, toxins, diet, exercise, smoking;
2.      methylation (of genes)
OR
acetylation (of histones);
1.      Ignore: gender and lifestyle.
2.      If further details are provided the context must be correct e.g. increased methylation or decreased acetylation inhibit gene expression / transcription.
2
(f)      1.      One person was homozygous dominant / has two dominant alleles = 2 marks;
2.      For one mark has two alleles / chromosomes;
1.      Accept; homozygous dominant genotype e.g. ‘one person has AA’ for 2 marks.
2.      Accept: is diploid or has two copies of the gene.
2
(g)     1.      (GCA / triplet) is common / found in other places;
2.      Would not know if it was the mutation / allele / gene
OR
Produces ‘false positives’
1.      Accept: Probe will bind elsewhere.
2
[15]
Q7.
(a)     1.      (Usually)Type II produce insulin;
2.      Cells / receptors less sensitive / responsive (to insulin)
OR
Faulty (insulin) receptors;
3.      (Treated / controlled by) diet / exercise;
2.      Accept: cells / receptors do not respond.
2.      Accept: ‘fewer receptors’
3.      Accept: (Treated / controlled by) weight loss / medication / drugs.
3.      Ignore: diabetes is caused by diet / exercise.
2 max
(b)     Tick in box 4
1
(c)     1.      Attach to gene / DNA / promoter region;
2.      Stimulate / inhibit transcription / RNA polymerase;
Note: Genes being expressed / inhibited or switched on / off is not enough on its own.
2
(d)     1.      (Effective as) group A / with iPS / treated lower than group B / with diabetes;
2.      (Effective as) group A similar to group C / without diabetes;
3.      (Investigation) done on mice not humans;
4.      Only shows results for 12 weeks / short-time period / long-term effects not known;
Ignore: Only one study / not repeated / sample size.
2.      Accept: ‘healthy’ or ‘normal’ or control for group C.
4
[9]
Q8.
(a)     1.      No (functional) cones
OR
Only rods;
2.      Cones are connected to a single neurone
OR
Several rods connected to a single neurone;
Accept correct reference to retinal convergence
Accept ‘bipolar/nerve cell’ for neurone
Accept ‘many’ 2 or more for ‘several’
3.      (Cones) Separate (sets of) impulses to brain
OR
(Rods) Single (set of) impulse/s to brain;
Accept ‘optic nerve’ for brain
Reject ‘signals’, ‘messages’ for ‘impulses’
Accept ‘action potential’
3
(b)     1.      Correct answer in range 42 – 44% = 2 marks;;
2.      Incorrect answer but shows that understanding that 2pq = heterozygous/carriers = 1 mark;
Accept 1 – (p^2 + q^2)
Accept understanding of 2pq by using calculation involving 2 × two different numbers
2
(c)     1.      (Gene/allele) is on the X chromosome;
2.      Females require two alleles/females can be heterozygous/carriers and males require one allele;
Reference to allele is essential but only required once
Reference to females and males required
Reject dominant allele
2
(d)     1.      Green sensitive pigment/cones non-functional
OR
Cones that detect green light non-functional;
2.      Three different types of pigment/cone;
3.      Other/different colours (‘seen’) due to stimulation of more than one cone/pigment;
1, 2 and 3. Reject reference to ‘green cones’/ ‘blue cones / ‘red cones’ but once only
1, 2 and 3. Reject reference to ‘green pigment’/ ‘blue pigment/ ‘red pigment but once only and only if ‘green cones’ etc, (see above) has not been rejected.
3
(e)     1.      (iPS cells) divide;
2.      (iPS cells) develop/differentiate into (green sensitive) cones;
Accept ‘produce’/’specialise’ ‘turn in to’ / ‘genes switched on’ / ’turned on’ for ‘develop’ but ignore ‘grow’
Reject develop into ‘green cones’/blue’ cones’/’red cones’
Ignore develop/differentiate into (blue/red sensitive) cones;
Reject reference to develop in to ‘green pigment’/ ‘blue pigment/ ‘red pigment
2
(f)      1.      (Use of iPS cells) long-term;
Accept ‘gene therapy short-term’ or ‘only two years’
Accept ‘permanent’
2.      (Use of iPS cells) less chance of rejection/immune response;
3.      (Use of iPS cells) single treatment;
Accept ‘gene therapy ‘regular/frequent treatment’’
4.      Harm/side effects from using viruses (in gene therapy);
3 max
[15]
Q9.
(a)     No cadmium;
Other conditions same as cadmium-treated group;
2
(b)     (i)      As a measure of the effect due to cadmium /
to make a comparison;
1
(ii)     Becoming more methylated;
Ignore later slight decrease/no change
1
(iii)     Production of more methyltransferase enzyme /
increased activity of transferase;
Extra incorrect relevant information - cancel
1
(c)     RNA-polymerase could not bind (to DNA / to promoter);
mRNA of p16 could not be made / no transcription of p16 gene;
2
(d)     Any four from:
1.  Cadmium causes expression of methyltransferase gene /
     increased activity transferase (from 2 to 3 weeks in);
2.  Methyl groups on to promoter / p16 gene / suppressor (gene);
3.  (p16) normally suppresses tumour growth;
4.  p16 protein / p16 expression falls after 4 weeks / after methylation;
5.  Tumour formation occurs (after 10 weeks) after p16 falls /
     after suppressor gene activity falls;
4 max
[11]
Q10.
(a)     (All) the DNA in a cell/organism;
Accept
‘(all) the ‘genes’/alleles’ ‘genetic material/code’ in a cell/organism/ person’
‘the total number of DNA bases in a cell/organism’
Reject all the DNA/ genes within a species
1
(b)     1.      (Transcriptional factor/antibody) has a specific/tertiary structure/shape;
Accept (antibody) has a specific variable region
Accept (transcription factor/antibody) has a specific binding site
Reject active site but only once.
2.      Complementary (shape/structure);
Reject active site but only once.
2
(c)     DNA, transcription factor and antibody;
Accept Nucleotides for DNA
Ignore ‘reference to chemicals’
1
(d)     Binds to P34 gene/DNA/mRNA
OR
Binds to transcription factor gene/DNA
OR
Binds to promoter;
Reject binds to transcription factor
1
(e)     1.      Restriction (endonuclease/enzyme) to cut plasmid/vector;
2.      Ligase joins gene/DNA to plasmid/vector;
2
(f)     1.      Mass/number of amino acids/polypeptides;
Accept weight for mass
Ignore density/size
Accept length of polypeptide/amino acid chain
Accept primary structure /sequence of amino acids.
Accept tertiary structure
2.      Charge;
3.      R groups (differ);
2 max
[9]
Q11.
(a)     1.      (ESCs) can replace any type of (heart) cell;
Accept named type of cell, e.g. heart muscle cell
1
(b)     1.      Might divide out of control;
2.      Leading to tumour / cancer;
2
(c)     1.      Shows the effects of surgery;
2.      Allows effects of transplants / treatment to be seen;
Allow in either order
2
(d)     1.      Other cell types cause some increase but most of increase due to cardiomyocytes;
2.      Large SD, so some not much increase / no better than control;
3.      Overlap of SDs indicates / suggests no significant difference;
3
(e)     1.      Greater blood supply (to damaged areas);
2.      Bringing more oxygen / glucose for respiration;
3.      Brings more amino acids for protein synthesis;
4.      For cell repair / mitosis / division;
3 max
(f)     1.      Measure diameter of field of view and calculate area;
2.      Using micrometer slide and eyepiece graticule;
Accept descriptions
3.      Count number of capillaries in large number of fields of view and calculate mean;
4.      Select fields of view randomly
4
[15]
Q12.
(a)     1.      Sugar-phosphate (backbone) / double stranded / helix so provides strength / stability / protects bases / protects hydrogen bonds;
Must be a direct link / obvious to get the mark
Neutral: reference to histones
2.      Long / large molecule so can store lots of information;
3.      Helix / coiled so compact;
Accept: can store in a small amount of space for ‘compact’
4.      Base sequence allows information to be stored / base sequence codes for amino acids / protein;
Accept: base sequence allows transcription
5.      Double stranded so replication can occur semi-conservatively / strands can act as templates / complementary base pairing / A-T and G-C so accurate replication / identical copies can be made;
6.      (Weak) hydrogen bonds for replication / unzipping / strand separation / many hydrogen bonds so stable / strong;
Accept: 'H-bonds' for ‘hydrogen bonds’
6
(b)     1.      (Mutation) in E produces highest risk / 1.78;
2.      (Mutation) in D produces next highest risk / 1.45;
3.      (Mutation) in C produces least risk / 1.30;
Must be stated directly and not implied
E > D > C = 3 marks
Accept: values of 0.78, 0.45 and 0.30 for MP1, MP2 and MP3 respectively
If no mark is awarded, a principle mark can be given for the idea that all mutant alleles increase the risk
3
(c)     180;
1
(d)     (Similarities):
1.      Same / similar pattern / both decrease, stay the same then increase;
2.      Number of cells stays the same for same length of time;
Ignore: wrong days stated
(Differences):
(Per unit volume of blood)
3.      Greater / faster decrease in number of healthy cells / more healthy cells killed / healthy cells killed faster;
Accept: converse for cancer cells
Accept: greater percentage decrease in number of cancer cells / greater proportion of cancer cells killed
4.      Greater / faster increase in number of healthy cells / more healthy cells replaced / divide / healthy cells replaced / divide faster;
Accept: converse for cancer cells
For differences, statements made must be comparative
3 max
(e)     1.      More / too many healthy cells killed;
2.      (So) will take time to replace / increase in number;
Neutral: will take time to ‘repair’
3.      Person may die / have side effects;
2 max
[15]