Inheritance

People with the inherited blood disorder haemophilia A (HA) have a reduced ability to clot blood. One variant of HA is due to an inversion mutation in the F8 gene. This mutation causes the production of a non-functional form of the protein, factor VIII. Factor VIII is not an enzyme.

Explain how an inversion mutation can cause the production of non-functional factor VIII.

Currently, the most effective treatment for HA is weekly or monthly injections of factor VIII. Scientists investigated the effectiveness of a gene therapy product, Roctavian, as a treatment for HA. Roctavian contains a viral vector that has the DNA coding for factor VIII.

The scientists:

• selected 112 male volunteers with severe HA, who were at least 18 years of age

• treated each volunteer with a single injection of Roctavian into their blood

• recorded the number of bleeding episodes (internal bleeding) in these volunteers over a two-year period (set A results)

• compared these data with the number of bleeding episodes per year in the volunteers before receiving Roctavian (set B results).

Table 1 shows the scientists’ results.

A value of ±2 × SD from the mean includes over 95% of the data.

Use all the information provided to evaluate the use of Roctavian as a treatment for HA.

Haemophilia C is an autosomal recessive condition.

Two parents are both heterozygous for the haemophilia C allele.

Calculate the probability of these parents producing a son who has haemophilia C.

Explain what is meant by the term phenotype.

In genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios.

Suggest two reasons why.

Do not refer to sex-linkage or autosomal linkage in your answer.

In sweet pea plants, the F allele for purple flowers is dominant to the f allele for red flowers. The L allele for long pollen is dominant to the l allele for round pollen.

A scientist carried out a genetic cross between a plant with purple flowers and long pollen (heterozygous for both genes) and a plant with red flowers and round pollen.

Table 2 shows the results of this cross.

Use your knowledge of autosomal linkage to explain the results in Table 2.

Your answer should include:

• an explanation of what is meant by autosomal linkage

• a comment about the relative proportions of the different genotypes of gametes produced in this cross.

Cystic fibrosis (CF) is a recessive inherited disorder caused by a defective CFTR gene.

Table 6 shows the number of people with CF and the estimated population of the UK in 2020 and 2022.

Calculate how many more people had CF than was expected in 2022.

You can assume the relative frequency of CF in a population remains constant.

Show your working.

Additional number of people

CF can be caused by a three-base deletion mutation in the CFTR gene. This deletion

creates a recessive allele called F508del.

DNA probes and gel electrophoresis were used to screen two parents and their two children for the F508del allele. One of the children was diagnosed with CF.

Figure 5 shows the results.

What is a DNA probe?

The identity of each family member can be determined from the relative position of the bands shown in Figure 5.

Use information in Figure 5 to:

• complete Table 7

• explain how you identified the child with CF.

You can use F to represent the dominant CFTR allele and f to represent the recessive F508del CFTR allele.

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

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.

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 fruit flies, males have the sex chromosomes XY and the females have XX. In fruit flies, a gene for eye colour is carried on the X chromosome. The allele for red eyes, R, is dominant to the allele for white eyes, r.

Male fruit flies are more likely than female fruit flies to have white eyes.

Explain why.

A female fruit fly with white eyes was crossed with a male fruit fly with red eyes to produce a large number of offspring.

In fruit flies, the genes for body colour and for wing development are not on the sex chromosomes. The allele for grey body colour, G, is dominant to the allele for black body colour, g. The allele for long wings, L, is dominant to the allele for short wings, l.

A geneticist carried out a cross between fruit flies with grey bodies and long wings (heterozygous for both genes) and fruit flies with black bodies and short wings.

Table 2 shows the results of this cross.

Explain the results in Table 2.

The first generation of a population of fruit flies had 50 females.

Calculate how many female fruit flies would be produced from this population in the fifth generation.

You can assume:

• each female produces 400 offspring each generation

• half the offspring produced each generation are female

• there is no immigration or emigration

• no flies die before reproducing.

Show your working.

Give your answer in standard form.

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.

In genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios.

Suggest two reasons why.

In tomato plants, the genes for height and for the type of leaf are on the same homologous pair of chromosomes. The allele T, for a tall plant, is dominant to the allele t, for a dwarf plant. The allele M, for normal leaves, is dominant to the allele m for mottled leaves.

A biologist carried out crosses between parent plants heterozygous for both genes and examined the offspring produced. The position of the two alleles for both genes was the same in each parent plant as shown in Figure 7. The phenotypes and number of offspring produced are shown in Table 3.

What would be the genotype of the offspring with dwarf plants and mottled leaves?

Use the information provided to explain the results in Table 3.

Complete Table 4 to show the expected ratio of phenotypes if the same cross had been carried out but the genes for height of plant and for the type of leaf were on different homologous pairs of chromosomes.

Lactose is the main sugar in milk and is hydrolysed by the enzyme lactase. Lactase is essential to newborn mammals as milk is their only source of food. Most mammals stop producing lactase when they start feeding on other food sources. Humans are an exception to this because some continue to produce lactase as adults. The ability to continue producing lactase is known as lactase persistence (LP) and is controlled by a dominant allele. A number of hypotheses based on different selection pressures have been put forward to explain LP in humans.

One hypothesis for LP in humans suggests that the selective pressure was related to some human populations farming cattle as a source of milk.

Describe how farming cattle as a source of milk could have led to an increase in LP.

Use the information provided to explain why the number of people showing LP would rapidly increase once selection for this condition had been established.

Lactase persistence is caused by a mutation in DNA. This mutation does not occur in the gene coding for lactase.

Suggest and explain how this mutation causes LP.

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.

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.

In the brain of a person with AD there is a lower concentration of acetylcholine. 5 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.

A gene mutation called E280A found on chromosome 14 causes early-onset AD 10 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 world. The origin of the E280A mutation in this population has been traced back 15 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. They recorded a sample size of 204 and detected 75 E280A mutations but only 20 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.

Assuming no one with AD died in 2014, calculate the annual percentage increase in AD cases in America for 2014 (lines 2–4).

Explain how donepezil could improve communication between nerve cells (lines 7–9).

Suggest and explain two reasons why there is a high frequency of the E280A mutation in Yaramul (lines 13–15).

Explain why natural selection has not reduced the frequency of the E280A mutation in the population (lines 16–17).

The age at which the E280A mutation is expressed to cause AD can vary (lines 11 – 12).

Suggest and explain one reason for this.

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.

Suggest why a DNA probe for the mutated triplet was not considered a suitable method for detection of the E280A mutation (lines 22–23).