Population genetics

Cleft lip and palate (CLP) is a split in the upper lip and/or the roof of the mouth. CLP is present from birth.

CLP can be caused by various genetic and environmental factors.

CDH1 is a gene involved in lip and mouth development in a fetus before a baby is born.

Scientists hypothesised that CLP is caused by a ‘two-hit’ model:

• a ‘genetic hit’ through loss of CDH1 function

• an ‘environmental hit’ leading to CDH1 methylation caused when inflammation affects a fetus.

Inflammation can result from the mother having an infection or diabetes.

Scientists investigated the ‘two-hit’ model using human cells grown in vitro.

They:

• grew normal cells and cells with a loss of CDH1 function

• treated some of each type of cell with a drug to cause inflammation and left some of each type of cell untreated

• measured the percentage CDH1 methylation of the cells.

Figure 4 shows the scientists’ results.

The frequency of live births with CLP is 1 in 700

Despite knowing this frequency, the Hardy–Weinberg equation cannot be used to estimate any allele frequencies for CLP.

Suggest two reasons why.

Describe one effect methylation could have on CDH1

Suggest and explain three reasons why the data in Figure 4 might not provide evidence for the ‘two-hit’ model.

In fruit flies, a gene for body colour has a dominant allele for grey body, G, and a recessive allele for black body, g.

Explain how you would determine if the genotype of a grey fly is homozygous or heterozygous for body colour.

In fruit flies, males are XY and females are XX.

A cross between a grey-bodied male fly and a black-bodied female fly produced some black-bodied females. Explain how this shows that the gene for body colour is not sex-linked.

A population of fruit flies contained 19% grey-bodied flies.

Use the Hardy–Weinberg equation to calculate the percentage of flies heterozygous for gene G.

Show your working.

In fruit flies, a gene for wing shape has a dominant allele for curly wings, R, and a recessive allele for normal wings, r. The alleles for this gene are on a different pair of chromosomes from the gene for body colour.

Fruit flies that are homozygous dominant for curly wings do not survive beyond the embryo stage.

A curly-winged fly, homozygous for grey body colour was crossed with a curly-winged, black-bodied fly.

Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected to develop into adults from this cross.

Phenotypes of parents Curly-winged, grey-bodied, × Curly-winged, black-bodied

Genotypes of offspring

Phenotypes of offspring

Ratio of offspring

Define the biological term population.

Ecologists monitored the biodiversity of birds in a protected woodland.

They recorded the number of bird species breeding in the woodland on the same day, every year for a 30-year period.

Figure 4 shows their results.

Figure 4

Figure 4 not reproduced here due to third-party copyright restrictions

The figure is taken from the following website:

Figure 1a. Anne E. Magurran, Stephen R. Baillie, Stephen T. Buckland, Jan McP. Dick, David A. Elston, E. Marian Scott, Rognvald I. Smith, Paul J. Somerfield, Allan D. Watt, Long-term datasets in biodiversity research and monitoring: assessing change in ecological communities through time, Trends in Ecology & Evolution, Volume 25, Issue 10, 2010, Pages 574-582, ISSN 0169-5347, https://doi.org/10.1016/j.tree.2010.06.016.

Which statement correctly represents the species richness of the woodland?

Tick () one box.

A student studying the information in this question concluded that all woodlands should be protected to increase the biodiversity of birds.

Evaluate the student’s conclusion.

It is not possible to extrapolate the data for the number of bird species recorded breeding each year beyond 1979.

Explain why.

In this woodland, the ecologists measured the similarity in the bird community by comparing each year to 1949 using an index of similarity.

This index ranges from 1.0 for total similarity to 0.0 for total dissimilarity.

Figure 5 shows their results.

Figure 5

Figure 5 not reproduced here due to third-party copyright restrictions

The figure is taken from the following website:

Figure 1b, Anne E. Magurran, Stephen R. Baillie, Stephen T. Buckland, Jan McP. Dick, David A. Elston, E. Marian Scott, Rognvald I. Smith, Paul J. Somerfield, Allan D. Watt, Long-term datasets in biodiversity research and monitoring: assessing change in ecological communities through time, Trends in Ecology & Evolution, Volume 25, Issue 10, 2010, Pages 574-582, ISSN 0169-5347, https://doi.org/10.1016/j.tree.2010.06.016.

Suggest how the changes in the index of similarity in the bird community provide evidence for the process of succession.

In Figure 5, the index of similarity for the last 10 years remains fairly constant.

Name the stage of a succession this represents.

Suggest one reason why the index of similarity is not absolutely constant.

The Living Planet Index (LPI) is an index designed to monitor the state of the world’s biodiversity.

The LPI is arbitrarily scaled to be 1.0 in 1970, the baseline year.

Figure 6 shows the LPI from the Living Planet Report, 2008. The dotted lines represent ± 2 standard deviations from the mean, which includes over 95% of the data.

Figure 7 shows an alternative version of Figure 6 published on a news website.

Figure 6 Figure 7

Figures 6 and 7 not reproduced here due to third-party copyright restrictions

The figures are taken from the following website:

Figure 2a and 2b, Anne E. Magurran, Stephen R. Baillie, Stephen T. Buckland, Jan McP. Dick, David A. Elston, E. Marian Scott, Rognvald I. Smith, Paul J. Somerfield, Allan D. Watt, Long-term datasets in biodiversity research and monitoring: assessing change in ecological communities through time, Trends in Ecology & Evolution, Volume 25, Issue 10, 2010, Pages 574-582, ISSN 0169-5347, https://doi.org/10.1016/j.tree.2010.06.016.

The news website published the headline:

The LPI shows human activities cause significant decrease in biodiversity.

Suggest three reasons why this headline is not valid.

Use all the information provided.

Turn over for Section B

Section B

one question.

You are advised to spend no more than 45 minutes on this section.

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 female.

The two alleles, F and f of a different gene, which is not sex-linked, interact with the gene controlling fur colour. The allele F is dominant and stops the formation of pigment in the fur, resulting in white fur. The allele f is recessive and has no effect on fur colour.

Name the type of interaction between the two genes affecting fur colour.

What phenotype would a cat with the following genotype have?

Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.

Phenotypes of parents X^GX^GFf x X^BYff

Ratio of offspring phenotypes

In a population, 36% of cats had the F allele and had white fur.

Use the Hardy–Weinberg equation to calculate the frequency of the f allele in this population.

Show your working.

Give two types of cell that can stimulate an immune response.

An autoimmune disease causes the immune system to attack healthy body tissues. Scientists investigated the immune responses of healthy mice and mice with autoimmune disease.

The chemical OXA causes an immune response in mice and can make their skin swell. Mice had olive oil applied to their left ear and OXA in olive oil applied to their right ear.

The immune response was recorded in two ways:

• the cellular response by measuring the mean increase in ear thickness 24 hours after exposure to OXA

• the humoral response by measuring the mean concentration of anti-OXA antibody in blood 14 days after exposure to OXA.

Table 2 shows the results of this investigation. The values in the brackets show ± 2 standard deviations. A value of ± 2 standard deviations from the mean includes over 95% of the data.

Suggest and explain one reason why olive oil was applied to the left ear of the mice.

Plot a suitable graph for mean increase in ear thickness for each group of mice in Table 2.

Include the data for ± 2 standard deviations on your graph.

What can you conclude about the effects of autoimmune disease on the cellular response and the humoral response in male and female mice?

Use the data to justify your conclusions.

Some studies have shown that in humans, oestrogen has the opposite effect on two different autoimmune diseases. Oestrogen:

• accelerates the progression of systemic lupus erythematosus (SLE)

• prevents the progression of rheumatoid arthritis (RA).

The scientists investigated the effect of oestrogen on the immune response in healthy mice and mice with autoimmune disease.

Table 3 shows the scientists’ results.

A student studying these data made the following conclusions.

1. In humans, SLE is caused by an overproduction of antibodies.

2. In humans, RA is caused by an overproduction of cytotoxic T cells (T_C cells).

Evaluate the student’s conclusions.

In mice, one type of autoimmune disease is inherited as a dominant allele. Would the Hardy–Weinberg principle hold true for a population of mice, some of which had this autoimmune disease?

Explain your answer.

In humans, the ABO blood groups and Rhesus blood groups are under genetic control. The inheritance of the ABO blood groups is controlled by three alleles of a single gene, I^A, I^B and I^O. The alleles I^A and I^B are codominant, and the allele I^O is recessive to I^A and recessive to I^B.

There are four ABO phenotypes, A, B, AB and O.

The gene for the Rhesus blood groups has two alleles. The allele for Rhesus positive, R, is dominant to the allele for Rhesus negative, r.

The genes for the ABO and Rhesus blood groups are not sex-linked and are not on the same chromosome.

Figure 3 shows the phenotypes in a family tree for the ABO and Rhesus blood groups.

Give the genotypes of the ABO blood groups for individuals 1 and 2.

Do not include the genotypes for the Rhesus blood groups in your answer.

Explain one piece of evidence from Figure 3 that the allele for Rhesus positive is dominant.

Calculate the probability of individuals 1 and 2 producing a Rhesus positive son with blood group A (individual 3). You can assume that individual 1 is heterozygous for the Rhesus blood group.

Show your working.

Scientists determined the frequencies of the ABO alleles and ABO phenotypes in a large population. They then used a statistical test to determine if the frequencies of the four phenotypes differed significantly from the frequencies expected according to the Hardy–Weinberg equation.

The frequencies of the I^A and I^O alleles were 0.15 and 0.65. What is the frequency of

Name the statistical test you should use to determine if the observed frequencies of the four phenotypes differed significantly from the frequencies expected according to the Hardy–Weinberg equation.

State how many degrees of freedom should apply.

Number of degrees of freedom

The scientists concluded that the observed frequencies of the four phenotypes differed significantly from the expected frequencies. Use your knowledge of the Hardy–Weinberg principle to suggest two reasons why.

In one species of squirrel, Sciurus carolinensis, fur colour is controlled by one gene, with two codominant alleles. C^G represents the allele for grey fur colour, and C^B represents the allele for black fur colour.

Table 1 shows the three possible phenotypes.

In a population of 34 S. carolinensis, 2 had black fur.

Use the Hardy–Weinberg equation to estimate how many squirrels in this population had brown-black fur. Show your working.

The actual number of squirrels in this population that had brown-black fur was 16.

Use all of the information to calculate the actual frequency of the C^G allele.

Do not use the Hardy–Weinberg equation in your calculation.

Give your answer to 2 decimal places.

S. carolinensis were first introduced to the UK from North America in the 1870s. They are now widely distributed across the UK.

S. carolinensis from both North America and the UK show exactly the same genotypic and phenotypic variation. An identical mutation causing black fur has also been found in several other species closely related to S. carolinensis.

Use this information to deduce which one of the following conclusions is most likely true.

Tick () one box.

The mutation that caused the C^B allele was due to a 24 base-pair deletion from the C^G allele.

The protein coded for by the C^B allele is 306 amino acids long.

Calculate the percentage reduction in size of the protein coded for by the C^B allele compared with the protein coded for by the C^G allele.

Give your answer to 3 significant figures and show your working.

In S. carolinensis, fur colour depends on the distribution and relative amounts of light pigments and dark pigments in the hairs of the fur. Figure 1 shows how the protein produced from the C^G allele can result in the production of a light pigment or a dark pigment.

The deletion mutation in the C^B allele results in the production of a receptor protein that does not have glutamic acid. The lack of glutamic acid in the receptor protein has the same effect as αMSH leaving the receptor protein.

Use Figure 1 and this information to suggest why S. carolinensis with the genotype C^BC^B have black fur rather than grey fur.

Mutation is one cause of genetic variation in organisms.

Give two other causes of genetic variation.

In a species of flowering plant, the T allele for tallness is dominant to the t allele for dwarfness. In the same species, two alleles C^R (red) and C^W (white) code for the colour of flowers. When homozygous red-flowered plants were crossed with homozygous white-flowered plants, all the offspring had pink flowers.

Name the relationship between the two alleles that code for flower colour.

A dwarf, pink-flowered plant was crossed with a heterozygous tall, white-flowered plant.

Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.

Phenotypes of parents: Dwarf, pink-flowered Tall, white-flowered

Genotypes of offspring:

Phenotypes of offspring:

Ratio of phenotypes:

A population of this species of plant contained 9% of red-flowered plants.

Use the Hardy–Weinberg equation to calculate the percentage of pink-flowered plants in this population.

Show your working.

In fruit flies, a gene for body colour has a dominant allele for grey body, G, and a recessive allele for black body, g.

A gene for eye colour has a dominant allele for red eyes, R, and a recessive allele for white eyes, r, and is located on the X chromosome.

Figure 4 shows the phenotypes of fruit flies over four generations.

Give the full genotype of the fly numbered 6 in Figure 4.

Give one piece of evidence from Figure 4 to show that the allele for grey body colour is dominant.

Explain one piece of evidence from Figure 4 to show that the gene for body colour is not on the X chromosome.

A population of fruit flies contained 64% grey-bodied flies. Use the Hardy–Weinberg equation to calculate the percentage of flies heterozygous for gene G.

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.

People with complete achromatopsia have difficulty in seeing detail (lines 2–3).

Explain why.

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.

Red-green colour blindness affects more men than women (lines 7–8).

Explain why.

People with red-green colour blindness are unable to distinguish between red and green, and also between other colours (lines 8–10).

Explain why.

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.

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.

What is meant by the term phenotype?

The inheritance of fruit colour in summer squash plants is controlled by two genes, A and B. Each gene has two alleles.

Figure 7 shows the interaction of these two genes in controlling fruit colour in summer squash plants.

Name the type of gene interaction shown in Figure 7.

What fruit colour would you expect the following genotypes to have?

Genes A and B are not linked.

Complete the genetic diagram to show all the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.

Genotypes of parents aabb × AaBb

Genotypes of offspring

Phenotypes of offspring

Ratio of phenotypes

A population of summer squash plants produced only green and yellow fruit. The percentage of plants producing yellow fruit in this population was 36%.

Use the Hardy−Weinberg equation to calculate the percentage of plants that were heterozygous for gene B.

What is a gene pool?

Lord Howe Island in the Tasman Sea possesses two species of palm tree which have arisen via sympatric speciation. The two species diverged from each other after the island was formed 6.5 million years ago. The flowering times of the two species are different.

Using this information, suggest how these two species of palm tree arose by sympatric speciation.