· web viewcodes for black fur. these alleles are codominant. heterozygous females have ginger and...

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A-level Biology (7401/7402)Section 7 and 8 Practical Skills

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Time: 224

Marks: 192

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Q1.In cats, males are XY and females are XX. A gene on the X chromosome controls fur colour in cats. The allele G codes for ginger fur and the allele B codes for black fur. These alleles are codominant. Heterozygous females have ginger and black patches of fur and their phenotype is described as tortoiseshell.

(a) Explain what is meant by codominant alleles.

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(b) Male cats with a tortoiseshell phenotype do not usually occur. Explain why.

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(c) A tortoiseshell female was crossed with a black male. Use a genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.

Use XG to indicate the allele G on an X chromosome.Use XB to indicate the allele B on an X chromosome.

Genotypes of offspring .................................................................................

Phenotypes of offspring ................................................................................

Ratio of phenotypes ......................................................................................(3)

(d) Polydactyly in cats is an inherited condition in which cats have extra toes. The allele for polydactyly is dominant.

(i) In a population, 19% of cats had extra toes. Use the Hardy-Weinberg equation to calculate the frequency of the recessive allele for this gene in this population. Show your working.

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Answer = ............................(2)

(ii) Some cat breeders select for polydactyly. Describe how this would affect the frequencies of the homozygous genotypes for this gene in their breeding populations over time.

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(Total 8 marks)

Q2.(a) In fruit flies, the genes for body colour and wing length are linked. Explain what this means.

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A scientist investigated linkage between the genes for body colour and wing length. He carried out crosses between fruit flies with grey bodies and long wings and fruit flies with black bodies and short wings.

Figure 1 shows his crosses and the results.

• G represents the dominant allele for grey body and g represents the recessive allele for black body.

• N represents the dominant allele for long wings and n represents the recessive allele for short wings.

Figure 1

Phenotype of parents grey body,long wings ×

black body,short wings

Genotype of parents GGNN ggnn

Genotype of offspring GgNn

Phenotype of offspring all grey body, long wings

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These offspring were crossed with flies homozygous for black body and short wings.

The scientist’s results are shown in Figure 2.

Figure 2GgNn crossed with ggnn

Grey body, long wings

Black body,short wings

Grey body,short wings

Black body,long wings

Number of offspring 975 963 186 194

(b) Use your knowledge of gene linkage to explain these results.

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(c) If these genes were not linked, what ratio of phenotypes would the scientist have expected to obtain in the offspring?

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(d) Which statistical test could the scientist use to determine whether his observed results were significantly different from the expected results?

Give the reason for your choice of statistical test.

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(Total 8 marks)

Q3.(a) Explain what is meant by the term phenotype.

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(b) Tay-Sachs disease is a human inherited disorder. Sufferers of this disease often die during childhood. The allele for Tay-Sachs disease t, is recessive to allele T, present in unaffected individuals. The diagram shows the inheritance of Tay-Sachs in one family.

(i) Explain one piece of evidence from the diagram which proves that the allele for Tay-Sachs disease is recessive.

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(ii) Explain one piece of evidence from the diagram which proves that the allele for Tay-Sachs disease is not on the X chromosome.

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(c) (i) In a human population, one in every 1000 children born had Tay-Sachs disease. Use the Hardy-Weinberg equation to calculate the percentage of this population you would expect to be heterozygous for this gene. Show your working.

Answer = ...................................... %(3)

(ii) The actual percentage of heterozygotes is likely to be lower in future generations than the answer to part (c)(i). Explain why.

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(Total 10 marks)

Q4. Colour blindness is controlled by a gene on the X chromosome. The allele for colour blindness, Xb, is recessive to the allele for normal colour vision, XB . The gene controlling the presence of a white streak in the hair is not sex linked, with the allele for the presence of a white streak, H, being dominant to the allele for the absence of a white streak, h.

(a) Explain why colour blindness is more common in men than in women.

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(b) The diagram shows a family tree in which some of the individuals have colour blindness or have a white streak present in the hair.

(i) What are the genotypes of individuals 5 and 6?

Individual 5

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Individual 6

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(ii) Give the possible genotypes of the gametes produced by

individual 5;

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individual 6.

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(iii) What is the probability that the first child of individuals 5 and 6 will be a colour blind boy with a white streak in his hair? Show your working.

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Answer ............................................(2)

(Total 7 marks)

Q5.Oestrogen is a substance produced by the enzyme aromatase. In females, the main source of oestrogen is the ovaries but aromatase is produced by many other organs in the body, including the lungs. Oestrogen can stimulate the development of some lung tumours. In these tumours, binding of oestrogen to cell-surface receptors stimulates cell division.

Scientists investigated whether two drugs could prevent lung tumours in female mice. First, they removed the ovaries from these mice. They then injected the mice with a tumour-causing chemical found in tobacco twice a day for 4 weeks. The mice were then randomly allocated to one of four groups. Each group contained 10 mice.

• Group Q was given a placebo. This placebo did not contain either drug.• Group R was given the drug anastrozole. This inhibits the enzyme aromatase.• Group S was given the drug fulvestrant. This binds to oestrogen receptors.• Group T was given both anastrozole and fulvestrant.

The mice were given these drugs each week during weeks 5−15 of the investigation.

(a) The scientists removed the ovaries from the mice for the investigation. They also gave the mice injections of the substrate of aromatase each day.

Explain why these steps were necessary.

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(b) The scientists predicted that fulvestrant would be more effective when given with anastrozole than when given alone.

Use the information provided to suggest why they predicted this.

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At week 15, the lungs of the mice were removed and examined. The scientists then determined the number of tumours present and the mean tumour area for each group.

Figure 1 and Figure 2 show the scientists’ results.

Figure 1

Figure 2

(c) The scientists concluded that both drugs should be used together to reduce the risk of lung cancer in women exposed to tobacco products.

Do you agree? Explain your answer.(5)

(d) The scientists used tumour area as an indicator of tumour size.

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Explain why tumour area may not be the best indicator of tumour size and suggest a more reliable measurement.

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(e) The scientists repeated the investigation but this time they did not give the drugs until week 9.

Suggest why they gave the drugs at week 9, rather than at week 5.

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(f) Another group of scientists is currently using these drugs in human trials. However, the control group is not being given a placebo.

Suggest why a placebo is not being given and what is being given to this group instead.

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(Total 15 marks)

Q6. (a) Individuals in a population show phenotypic variation.

Give the two types of factor that cause this variation.

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(b) What is allopatric speciation?

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(Total 4 marks)

Q7. (a) What is sympatric speciation?

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Littorina saxatilis is a snail found on rocky seashores. It has a shell and a muscular foot that it uses to move and to attach to rocks. Crabs are predators of this snail. The crabs use their claws to break open the snails’ shells, or pull the snails from their shells.

Two forms of this snail are common in the UK.

Form T lives near the top of the shore. It lives in cracks in rock. Wave action is greatest near the top of the shore and there are very few crabs.

Form M lives on the middle shore. On the middle shore there are many crabs. Unlike form T, the snails of form M live on the open rock and not in cracks.

Forms T and M were produced by natural selection. The drawings show both forms of the snail.

The table shows features of these forms.

FeatureForm of Littorina saxatilis

T M

Size of shell Small Large

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Thickness of shell Thin Thick

Size of opening of shell Large Small

(b) Use this information to answer the following question.Give two differences between forms T and M.For each difference suggest how one environmental factor may have caused differential survival in the snail populations leading to this difference.

Difference 1 ................................................................................................

Suggestion ...................................................................................................

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Difference 2 .................................................................................................

Suggestion ...................................................................................................

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(c) Scientists placed male and female snails of forms T and M into an aquarium. They recorded how many form T males mated with form T females and how many mated with form M females.

The scientists found that the probability of a form T male mating with a form T female was greater than 90 %. They interpreted this result as evidence that speciation was taking place.

Explain why.

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(Total 8 marks)

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Q8. Warfarin is a substance which inhibits blood clotting. Rats which eat warfarin are killed due to internal bleeding. Some rats are resistant to warfarin as they have the allele WR.

Rats have three possible genotypes:

WRWR resistant to warfarinWRWS resistant to warfarinWSWS susceptible (not resistant) to warfarin.

In addition, rats with the genotype WRWR require very large amounts of vitamin K in their diets. If they do not receive this they will die within a few days due to internal bleeding.

(a) How can resistance suddenly appear in an isolated population of rats which has never before been exposed to warfarin?

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(b) A population of 240 rats was reared in a laboratory. They were all fed on a diet containing an adequate amount of vitamin K. In this population, 8 rats had the genotype WSWS, 176 had the genotype WRWS and 56 had the genotype WRWR.

(i) Use these figures to calculate the actual frequency of the allele WR in this population. Show your working.

Answer ..................................(2)

(ii) The diet of the rats was then changed to include only a small amount of vitamin K. The rats were also given warfarin. How many rats out of the population of 240 would be likely to die within a few days?

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(c) In a population of wild rats, 51% were resistant to warfarin.

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(i) Use the Hardy-Weinberg equation to estimate the percentage of rats in this population which would be heterozygous for warfarin resistance. Show your working.

Answer ....................................... %(3)

(ii) If all the susceptible rats in this population were killed by warfarin, more susceptible rats would appear in the next generation. Use a genetic diagram to explain how.

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(iii) The graph shows the change in the frequency of the WS allele in an area in which warfarin was regularly used. Describe and explain the shape of the curve.

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(iv) Give two assumptions that must be made when using the Hardy-Weinberg equation.

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(Total 15 marks)

Q9. (a) Explain the meaning of these ecological terms.

Population ....................................................................................................

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Community ...................................................................................................

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(b) Some students used the mark-release-recapture technique to estimate the size of a population of woodlice. They collected 77 woodlice and marked them before releasing them back into the same area. Later they collected 96 woodlice, 11 of which were marked.

(i) Give two conditions necessary for results from mark-release-recapture investigations to be valid.

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(ii) Calculate the number of woodlice in the area under investigation. Show your working.

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Answer ......................................................(2)

(c) Explain how you would use a quadrat to estimate the number of dandelion plants in a field measuring 100 m by 150 m.

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(d) Two similar species of birds (species A and species B) feed on slightly different sized insects and have slightly different temperature preferences. The diagram represents the response of each species to these factors.

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(i) Which of the numbered boxes describes conditions which represent

the niche of species A; .............

the niche of species B; .............

insects too small for species B and temperature too warm for species A; .............

insects too large for species A and temperature too cool for species B? .............(2)

(ii) These two species are thought to have evolved as a result of sympatric speciation. Suggest how this might have occurred.

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(Total 15 marks)

Q10. Sea otters were close to extinction at the start of the 20th century. Following a ban on hunting sea otters, the sizes of their populations began to increase. Scientists studied the frequencies of two alleles of a gene in one population of sea otters. The dominant allele, T, codes for an enzyme. The other allele, t, is recessive and does not produce a functional enzyme.

In a population of sea otters, the allele frequency for the recessive allele, t, was found to be 0.2.

(a) (i) Use the Hardy-Weinberg equation to calculate the percentage of homozygous recessive sea otters in this population. Show your working.

Answer ..................................... %(2)

(ii) What does the Hardy-Weinberg principle predict about the frequency of the t allele after another 10 generations?

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(b) Several years later, scientists repeated their study on this population. They found that the frequency of the recessive allele had decreased.

(i) A statistical test showed that the difference between the two frequencies of the t allele was significant at the P = 0.05 level.

Use the terms probability and chance to help explain what this means.

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(ii) What type of natural selection appears to have occurred in this population of sea otters? Explain how this type of selection led to a decrease in the frequency of the recessive allele.

Type of selection ................................................................................

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(Total 7 marks)

Q11.Malaria is a disease that is spread by insects called mosquitoes. In Africa, DDT is a pesticide used to kill mosquitoes, to try to control the spread of malaria.

Mosquitoes have a gene called KDR. Today, some mosquitoes have an allele of this gene, KDR minus, that gives them resistance to DDT. The other allele, KDR plus, does not give resistance.

Scientists investigated the frequency of the KDR minus allele in a population of mosquitoes in an African country over a period of 10 years.

The figure below shows the scientists’ results.

Year

(a) Use the Hardy–Weinberg equation to calculate the frequency of mosquitoes heterozygous for the KDR gene in this population in 2003.

Show your working.

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Frequency of heterozygotes in population in 2003 ...................................(2)

(b) Suggest an explanation for the results in the figure above.

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The KDR plus allele codes for the sodium ion channels found in neurones.

(c) When DDT binds to a sodium ion channel, the channel remains open all the time.Use this information to suggest how DDT kills insects.

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(d) Suggest how the KDR minus allele gives resistance to DDT.

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(Total 10 marks)

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Q12. Figure 1 shows sections through relaxed and contracted myofibrils of a skeletal muscle. The transverse sections are diagrams. The longitudinal sections are electron micrographs.

Figure 1

(a) (i) The electron micrographs are magnified 40 000 times.Calculate the length of band X in micrometres.Show your working.

Length of band X =..................................... µm(2)

(ii) Explain the difference in appearance between transverse sections A and C in Figure 1.

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(b) Explain what leads to the differences in appearance between the relaxed myofibril and the contracted myofibril.

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(c) Duchenne muscular dystrophy (DMD) is a condition caused by the recessive allele of a sex-linked gene. A couple have a son with DMD. They want to know the probability that they could produce another child with DMD. They consulted a genetic counsellor who produced a diagram showing the inheritance of DMD in this family.This is shown in Figure 2.

Figure 2

The couple who sought genetic counselling are persons 6 and 7.

(i) Give the evidence to show that DMD is caused by a recessive allele.

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(ii) Give the numbers of two people in Figure 2 who are definitely carriers of muscular dystrophy.

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(iii) Complete the genetic diagram to find the probability that the next child of couple 6 and 7 will be a son with muscular dystrophy. Use the following symbols:

XD = normal X chromosomeXd = X chromosome carrying the allele for muscular dystrophyY = normal Y chromosome

6 7

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Parental phenotypes Unaffected Unaffected

Parental genotypes .............. ..............

Gametes .............. ..............

Offspring genotypes .....................................................................

Offspring phenotypes .....................................................................

Probability of having a son with DMD ...................................................(4)

(d) DMD is caused by a deletion mutation in the gene for a muscle protein called dystrophin. A deletion is where part of the DNA sequence of a gene is lost. People in different families may inherit mutations in different regions of this gene.

Scientists isolated the dystrophin gene from DNA samples taken from children 10, 11 and 12. They cut the gene into fragments using an enzyme. The scientists then used two DNA probes to identify the presence or absence of two of these fragments, called F and G. This allowed them to find the number of copies of each fragment in the DNA of a single cell from each child.

The table shows their results.

ChildNumber of copies of gene fragment per cell

F G

10 (unaffected girl) 2 1

11 (unaffected girl) 2 2

12 (boy with DMD) 1 0

(i) The number of copies of gene fragments F and G shows that person 12 has DMD.Explain how.

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(ii) The number of copies of gene fragments F and G shows that person 12 is male.Explain how.

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(iii) The genetic counsellor examined the scientists' results. He concluded that person 10 is a carrier of DMD but her sister, 11, is not.

Describe and explain the evidence for this in the table.

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(e) Person 12 took part in a trial of a new technique to help people with DMD.

Doctors took muscle cells from person 12’s father and grew them in tissue culture.

They suspended samples of the cultured cells in salt solution and injected them into a muscle in person 12’s left leg. They injected an equal volume of salt solution into the corresponding muscle in his right leg. Person 12 was given drugs to suppress his immune system throughout the trial.

Four weeks later, the doctors removed a muscle sample from near the injection site in each leg. They treated these samples with fluorescent antibodies. These antibodies were specific for the polypeptide coded for by gene fragment G of the dystrophin gene.

The results are shown in the table.

Location andtreatment

Percentage of musclefibres labelled with

antibody

Left leg - injectedwith cultured cellssuspended in salt

6.8

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solution

Right leg - injectedwith salt solution

0.0

(i) Why was it necessary to treat person 12 with drugs to suppress his immune system?

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(ii) Explain why salt solution was injected into one leg and cultured cells suspended in salt solution into the other.

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(iii) This technique is at an early stage in its development. The doctors suggested that further investigations need to be carried out to assess its usefulness for treating people with DMD.

Explain why they made this suggestion.

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(Total 25 marks)

Q13. Cocaine is a highly addictive and illegal drug.

The release of the neurotransmitter dopamine in specific synapses in the brain leads to feelings of pleasure. Dopamine is removed from synapses by dopamine transporter proteins in the plasma membrane of neurones. Cocaine binds to the dopamine transporter protein.

Figure 1 shows a dopamine transporter protein and molecules of cocaine and dopamine.

Figure 1

(a) Using all of the information, suggest how cocaine leads to feelings of pleasure.

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(b) (i) Scientists isolated a mutated gene for the dopamine transporter protein.

Name one method that the scientists could have used to produce many copies of the mutated gene in the laboratory.

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(ii) Copies of the gene were then inserted into early embryos of mice. When these mice were born, samples of their DNA were tested using DNA probes to make sure that the mutated gene was present in the mice.

What is a DNA probe?

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(c) Figure 2 shows dopamine transporter proteins produced from the normal gene and from the mutated gene.

Figure 2

Explain how the mutation leads to the production of a protein that transports dopamine but is not affected by cocaine.

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(Total 9 marks)

Q14. A gene was broken into fragments using enzyme Z. The mixture of fragments produced was then separated by electrophoresis.

(a) What type of enzyme is enzyme Z?

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The table shows the number of base pairs present in the fragments.

Fragment Number of base pairs (× 103)

1 4.65

2 5.72

3 10.71

4 2.39

5 5.35

6 7.53

The diagram shows the electrophoresis gel used. The mixture of fragments was placed at the start point marked S and the process started. The boxes indicate the positions reached by the different fragments.

(b) Explain why base pairs are a suitable way of measuring the length of a piece of DNA.

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(c) (i) Write 6 above the appropriate box on the diagram to show the position you would expect fragment 6 to have reached.

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(ii) Explain how you arrived at your answer.

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(1)

(d) Enzyme Z recognises a particular sequence of bases in the gene. How many times does this sequence appear in the DNA of this gene?

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(Total 6 marks)

Q15. Taxol is a drug used to treat cancer. Research scientists investigated the effect of injecting taxol on the growth of tumours in mice. Some of the results are shown in Figure 1.

Figure 1

Number of days of treatment

Mean volume of tumour / mm3

Control group Group injected with taxol in saline

1 1 1

10 7 2

20 21 11

30 43 20

40 114 48

50 372 87

(a) Suggest how the scientists should have treated the control group.

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(b) Suggest and explain two factors which should be considered when deciding the number of mice to be used in this investigation.

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(c) The scientists measured the volume of the tumours. Explain the advantage of using volume rather than length to measure the growth of tumours.

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(d) The scientists concluded that taxol was effective in reducing the growth rate of the tumours over the 50 days of treatment. Use suitable calculations to support this conclusion.

(2)

(e) In cells, taxol disrupts spindle activity. Use this information to explain the results in the group that has been treated with taxol.

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(f) The research scientists then investigated the effect of a drug called OGF on the growth of tumours in mice. OGF and taxol were injected into different mice as separate treatments or as a combined treatment. Figure 2 and Figure 3 show the results from this second investigation.

Figure 2

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Figure 3

TreatmentMean volume of tumour

following 70 days treatment/mm3 (± standard deviation)

OGF 322 (± 28.3)

Taxol 207 (± 22.5)

OGF and taxol 190 (± 25.7)

Control 488 (± 32.4)

(i) What information does standard deviation give about the volume of the tumours in this investigation?

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(ii) Use Figure 2 and Figure 3 to evaluate the effectiveness of the two drugs when they are used separately and as a combined treatment.

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(4)(Total 15 marks)

Q16.(a) (i) A mutation of a tumour suppressor gene can result in the formation of a tumour.

Explain how.

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(ii) Not all mutations result in a change to the amino acid sequence of the encoded polypeptide.

Explain why.

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(b) Some cancer cells have a receptor protein in their cell-surface membrane that binds to a hormone called growth factor. This stimulates the cancer cells to divide.

Scientists have produced a monoclonal antibody that stops this stimulation.

Use your knowledge of monoclonal antibodies to suggest how this antibody stops the growth of a tumour.

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Q17. 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.

S (a) Explain the presence of bile in the colon.

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(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

bacteriaPercentage of

cancer patientsPercentage of

controls P

highhighlowlow

highlowhighlow

761374

98

3449

<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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(ii) Explain how the results indicate that other factors may be involved in causing colon cancer.

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S (c) Human cells contain genes that control their growth and division. One of these genes

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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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Q18.Cyanide is a poisonous substance. Cyanogenic clover plants produce cyanide when their tissues are damaged. The ability to produce cyanide is controlled by genes at loci on two different chromosomes. The dominant allele, A, of one gene controls the production of an enzyme which converts a precursor to linamarin. The dominant allele, L, of the second gene controls the production of an enzyme which converts linamarin to cyanide. This is summarised in the diagram.

(a) Acyanogenic clover plants cannot produce cyanide. Explain why a plant with the genotype aaLl cannot produce cyanide.

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(b) A clover plant has the genotype AaLl.

(i) Give the genotypes of the male gametes which this plant can produce.

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(ii) Explain how meiosis results in this plant producing gametes with these genotypes.

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(c) Two plants, heterozygous for both of these pairs of alleles, were crossed. What proportion of the plants produced from this cross would you expect to be acyanogenic but able to produce linamarin? Use a genetic diagram to explain your answer.

(3)

In an investigation, cyanogenic and acyanogenic plants were grown together in pots. Slugs were placed in each pot and records were kept of the number of leaves damaged by the feeding of the slugs over a period of 7 days. The results are shown in Table 1.

Table 1

Undamaged Damaged

Cyanogenic plants 160 120

Acyanogenic plants 88 192

(d) A x2 test was carried out on the results.

(i) Suggest the null hypothesis that was tested.

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(ii) x2 was calculated. When this value was looked up in a table, it was found to correspond to a probability of less than 0.05. What conclusion can you draw from this?

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A second investigation was carried out in a field of grass which had been undisturbed for many years. Table 2 shows the population density of slugs and the numbers of cyanogenic and acyanogenic clover plants at various places in the field.

Table 2

Population density of slugs

Number of acyanogenic clover

plants per m2

Number of cyanogenic clover

plants per m2

Very low 26 10

Low 17 26

High 0 10

Very high 0 5

(e) Explain the proportions of the two types of clover plant in different parts of the field.

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(Total 15 marks)

· web viewcodes for black fur. these alleles are codominant. heterozygous females have ginger and...

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