Patterns of Inheritance (OCR A Level Biology): Exam Questions

When studying variation, it is sometimes impractical to analyse DNA.

A student was investigating variation between a number of students in their school. They recorded the frequency of students that could and could not roll their tongue.

The results are shown in the table.

(i) Represent the data in the table as a bar chart on the grid provided below.

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(ii) Since 1940, people have believed that the ability to roll the tongue is caused by a single gene with two alleles.

R is dominant and allows tongue-rolling.

r is recessive and does not allow tongue-rolling.

The genotype of students who can roll their tongue could be either RR or Rr.

In the results shown in the table opposite

• the total number of students who could roll their tongue = 171

• the total number of students who could not roll their tongue = 77.

The Hardy–Weinberg principle allows us to estimate the proportion of each genotype.

Use the Hardy–Weinberg principle to estimate the proportion of heterozygous individuals in the school survey in the table.

Use the equations:

p² + 2pq + q² = 1
p + q = 1

proportion = .......................................................... [3]

(iii) The Hardy–Weinberg principle might not give an accurate estimate of the proportion of genotypes for the results of the student’s investigation.

The population of students varies from year to year and so cannot be said to be stable.

State two other reasons why it might be inappropriate to use the Hardy–Weinberg principle to estimate allele frequencies for the results in the table.

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The colour of onion bulbs is determined by two genes, A/a and B/b.

• A is a dominant allele and codes for the production of a red pigment.

• Onion bulbs that are homozygous for the recessive allele, a, produce no pigment and are white.

• B is a dominant allele that inhibits the expression of allele A.

• The recessive allele, b, allows the production of the red pigment.

A white onion plant was cross-pollinated with a red onion plant. All 15 offspring had the genotype AaBb.

(i) Identify the following:

The genotype of the white onion plant .....................
The genotype of the red onion plant ..........................
The phenotype of the offspring ....................................

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(ii) State the type of gene interaction shown by the genes A/a and B/b.

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(iii) Suggest how allele B inhibits the expression of allele A.

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Agammaglobulinemia and Vici syndrome are both genetic diseases.

Agammaglobulinemia results in a lack of mature B lymphocytes in a person’s blood.

(i) Suggest and explain one symptom of agammaglobulinemia.

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(ii) Fig. 4 shows the inheritance pattern of agammaglobulinemia in a family.

What conclusions can you draw about the location and nature of the allele responsible for causing agammaglobulinemia? Explain your conclusions.

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Vici syndrome is a genetic disease that shows a recessive inheritance pattern. The allele responsible for Vici syndrome is found on chromosome 18.

(i) Two carriers of Vici syndrome have six children.

Calculate how many of the six children you would expect to:

• have Vici syndrome

• be carriers of Vici syndrome.

Vici syndrome ...........................
Carriers ........................ [1]

(ii) A daughter of these parents and a male carrier of Vici syndrome have a child.

Calculate the probability of the child having Vici syndrome.

Answer = ................... [1]

DNA profiling can be used to analyse the risk of inheriting conditions such as agammaglobulinemia and Vici syndrome. 

(i) To produce a DNA profile, DNA first needs to be purified.

Explain why a protease enzyme is added to the mixture during the DNA purification process.

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(ii) DNA samples can be amplified using the polymerase chain reaction (PCR).

In theory, how many fragments of DNA might be present after 12 cycles of PCR?

Assume one DNA fragment was present at the beginning of the PCR process. Represent your answer as a log₁₀ value.

.......................................... fragments [2]

(iii) Suggest why the figure you calculated in (ii) may not be achieved in practice.

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(iv) State the name of the enzyme used in PCR to synthesise new DNA strands.

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(v) DNA fragments are separated to produce a DNA profile using electrophoresis.

A student wrote the following description of the electrophoresis procedure:

We will set up an agarose gel plate and place the DNA samples in the wells at the cathode. Voltage will be passed through the gel for one minute. The gel will then be placed in purified water and we will be able to see the banding pattern of each DNA sample.

Describe two changes you would make to the student’s procedure and explain how these changes would improve electrophoresis.

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The sweet pea plant has been used to study inheritance since the nineteenth century. The seeds of the sweet pea can vary in colour and shape.

The gene that controls colour has two alleles:

• Y is dominant and produces yellow seeds.

• y is recessive and produces green seeds.

The gene that controls shape has two alleles:

• R is dominant and produces round seeds.

• r is recessive and produces wrinkled seeds.

In the nineteenth century, Gregor Mendel crossed a pea plant that was heterozygous for both seed colour and shape with a pea plant that had green and wrinkled seeds.

(i) List the gametes that would be produced by a sweet pea plant that was heterozygous for both seed colour and shape.

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(ii) List the genotypes of the offspring that were produced from Mendel’s cross and state the corresponding phenotypes.

genotypes ........................................................................ phenotypes ......................................................................

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When Mendel crossed two pea plants that were heterozygous for both seed colour and shape, the ratio of phenotypes in the offspring was:

• 9 yellow round

• 3 green round

• 3 yellow wrinkled

• 1 green wrinkled

Some students tried to recreate this investigation using a modern variety of plant that showed the same phenotypic variation in seed colour and shape.a)

The students crossed two of the modern plants that were heterozygous for both seed colour and shape. The results of this cross were:

• 58 yellow and round

• 31 green and round

• 21 yellow and wrinkled

• 2 green and wrinkled

The students used the chi-squared test to compare their data to the expected 9:3:3:1 ratio.

(i) Use the chi-squared formula χ² = to calculate the χ² value for these data.

You may use the table below for working out.

χ² = .............................[3]

 Table 17 shows a χ² probability table.

Table 17

(ii) After analysing the results, the students stated that the inheritance of the seed colour and shape in their investigation was different from that in Mendel’s investigation.

Using Table 17, discuss whether the results of the investigation and the chi-squared test support the students’ statement.

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(iii) A ratio that is different from the expected 9:3:3:1, in a cross such as this, can be the result of epistasis.

Suggest and explain one reason, other than epistasis, why the phenotype ratio might not be 9:3:3:1.

Suggestion .......................................................
Explanation  .....................................................

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The yellow colour in peas is the result of an enzyme that breaks down chlorophyll, which is green.

• The Y allele codes for the production of an enzyme that breaks down chlorophyll.

• The y allele is the result of a mutation in the Y allele.

• The y allele codes for an inactive form of this enzyme.

(i) Outline how the Y allele codes for the production of this enzyme and explain why the y allele codes for an enzyme with a different primary structure.

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(ii) With reference to the proteins coded for by the seed colour gene, explain why the y allele is recessive.

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Tigers, Panthera tigris, are predatory mammals. They have evolved striped patterns on their fur (as shown in Fig. 3.1a), which provide camouflage in their habitats.

Fig 3.1a

(i) Adaptations can be divided into three types.

State the type of adaptation represented by the tiger’s stripes.

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(ii) Describe and explain how a tiger with striped fur may have evolved from a non-striped ancestor.

In your answer you should discuss the different types of genes that might be involved in the creation of the striped pattern in the tiger’s fur.

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One subspecies of tiger is the Bengal tiger. One in 10000 Bengal tiger births results in a white Bengal tiger.

White Bengal tigers (shown in Fig. 3.1b) have black stripes but lack orange fur.

Fig 3.1b

The allele that causes white fur is recessive and is a result of a mutation to a gene called SLC45A2.

According to the Hardy-Weinberg principle, the following equations can be used to estimate allele frequency within a population:

p² + 2pq + q² = 1

p + q = 1

Use the Hardy-Weinberg equations to calculate the percentage of Bengal tigers that are heterozygous for the SLC45A2 gene.

Give your answer to one significant figure.

Show your working.

Answer: .................... %

In domesticated, farmed pigs, the following two traits have been studied:

• The allele for curly tail, T, is dominant to the allele for straight tail, t.

• The allele for pink skin (dermis), D, is dominant to the allele for black skin, d.

(i) Draw a genetic diagram to show the results of crossing pigs that are heterozygous for both traits, tail and skin. Use the letters given above.

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(ii) Describe in words how this phenotypic ratio might be different if the two genes were autosomally linked.

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A pig farmer crossed one group of pigs, heterozygous for both traits, with another group homozygous recessive for both traits. The farmer expected to get roughly equal numbers of each of the four possible mixtures of tail and skin phenotype. The results that actually occurred are shown in Table 17.2.

Table 17.2

(i) The farmer thought from these results that the two genes might be autosomally linked.

Calculate x² (You may wish to use Table 17.2 to write figures for steps in your calculation process.)

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(ii) The farmer had concluded that the genes are linked.

Use your calculation and Table 17.3 to justify whether the farmer’s conclusion can be supported or not.

Table 17.3

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