Inheritance: past paper questions
Real questions on inheritance
The newest 20 of 166 written questions, plus 14 multiple-choice. Open one to see it on the real paper.
- 9700/22 · October/November 2025 · Q3a[3]
Suggest differences between gene LCT and gene MCM6, other than their locations in different positions on the same chromosome. (a) LCT and MCM6 are located on the same chromosome in humans.
See it on the paper → - 9700/22 · October/November 2025 · Q3c[1]
A mutation in a regulatory intron of MCM6 allows lactase synthesis to continue. This is known as lactase persistence. In this mutation, the number of nucleotides in the intron remains the same, but one of the nucleotides is different to the original nucleotide. State the type of mutation that is the cause of lactase pe...
See it on the paper → - 9700/41 · May/June 2025 · Q3a[2]
key 1 female without the disease 2 male without the disease 3 4 5 6 female with the disease male with the disease Fig. 3.1 X-linked protoporphyria is described as a sex-linked disease. Use the information in Fig. 3.1 to make one other conclusion about the pattern of inheritance of X-linked protoporphyria. Give evidence...
See it on the paper → - 9700/42 · May/June 2025 · Q2[6]
When organisms reproduce, they pass on their alleles to the next generation. There are many factors that can affect how allele frequencies change over time in a population. Explain how genetic drift and the founder effect may affect allele frequencies in populations.
See it on the paper → - 9700/42 · May/June 2025 · Q3a(i)[5]
The development of awns is controlled by two genes: gene A/a and gene B/b. The genes are present on different autosomes. The presence of either dominant allele A or dominant allele B results in rice plants producing grains with awns. A cross was carried out between a rice plant that is homozygous dominant for the two g...
See it on the paper → - 9700/42 · May/June 2025 · Q3a(ii)[1]
Deduce the type of inheritance shown by the ratio of the offspring phenotypes stated in 3(a)(i).
See it on the paper → - 9700/42 · May/June 2025 · Q3c(i)[4]
Explain the principles used by farmers to produce rice plant grains with no awns. (i)
See it on the paper → - 9700/42 · May/June 2025 · Q3c(ii)[3]
The normal allele for the gene An‑1 codes for a transcription factor that has a role in awn development and in the number of grains of rice produced. When the transcription factor is present there is: • • an increase in the expression of genes involved in awn development (positive regulation) a decrease in the expressi...
See it on the paper → - 9700/42 · May/June 2025 · Q4a[3]
In most species of plants and animals, the cell that is formed as a result of fertilisation is diploid and contains homologous chromosomes. Explain why the cell that is formed as a result of fertilisation is a diploid cell and contains homologous chromosomes.
See it on the paper → - 9700/42 · May/June 2025 · Q4b[2]
reason stage in meiosis (b) State the name of the stage in meiosis when reduction division occurs and explain a reason for your choice.
See it on the paper → - 9700/42 · May/June 2025 · Q4c(i)[2]
Fig. 4. Identify the stages of meiosis shown in Fig. 4.1 and Fig. 4.2.
See it on the paper → - 9700/42 · May/June 2025 · Q4c(ii)[1]
The cells formed at the end of meiosis in a Lilium pollen mother cell each have 12 chromosomes. State the number of sister chromatids found in a Lilium pollen mother cell at the start of meiosis.
See it on the paper → - 9700/43 · May/June 2025 · Q2b(i)[3]
Outline how natural selection differs from selective breeding.
See it on the paper → - 9700/43 · May/June 2025 · Q2b(ii)[2]
Selective breeding is used to produce uniform varieties of maize. The maize plants in a crop ripen at the same time and are the same height. The advantage of this is that harvesting is easy and quick. The disadvantage is that farmers must buy new seeds each year. Explain why farmers must buy new seeds each year.
See it on the paper → - 9700/42 · February/March 2025 · Q4a[3]
Construct a genetic diagram of a monohybrid cross to show how two parents who do not have hereditary haemochromatosis can produce a child with the disease. Use the following symbols: H = normal HFE allele h = mutant HFE allele.
See it on the paper → - 9700/42 · February/March 2025 · Q4b[2]
Some scientists believe that the C282Y mutation may have first occurred in Ireland. Scientists sequenced DNA obtained from two human fossil skeletons in Ireland. One of the fossils was 5200 years old and the other was 4000 years old. The scientists concluded that: the human living 4000 years ago did have the C282Y muta...
See it on the paper → - 9700/42 · February/March 2025 · Q4c(i)[4]
Suggest how the C282Y allele of HFE has been maintained in European populations.
See it on the paper → - 9700/42 · February/March 2025 · Q4c(ii)[3]
Suggest explanations for the differences in the percentage of people in different European countries who have one C282Y allele of the HFE gene, as shown in Fig. 4.1.
See it on the paper → - 9700/42 · February/March 2025 · Q4d[1]
Biological databases contain DNA sequence data from a large number of different people. Table 4.1 shows three of these databases and the percentage of people in each database who have one C282Y allele of the HFE gene. Table 4.1 database percentage of people who have one C282Y allele of the HFE gene database A 2.6 datab...
See it on the paper → - 9700/42 · February/March 2025 · Q4e[2]
In one population consisting of 2501 people, there were 9 people who were homozygous recessive for the HFE gene. Use equation 1 and equation 2 of the Hardy–Weinberg principle to calculate the number of people in the population who are heterozygous for the HFE gene. equation 1: p + q = 1 equation 2: p 2 + 2pq + q 2 = 1...
See it on the paper →
What examiners said
Most candidates recognised that both parents needed to be heterozygous for the C282Y allele and used a genetic diagram to show how a child with hereditary haemochromatosis could be produced. Several responses omitted to state which of the offspring genotypes would result in a child that could develop haemochromatosis, often only stating that there was a 25% probability of any child from this cros...
— Examiner report, February/March 2025, 9700/42 Q4(a) (Cambridge)
Stronger responses typically referred to the use of bioinformatics or information available in biological databases to locate the C282Y allele in DNA sequences obtained from the fossil skeletons. Weaker responses often considered alternative techniques such as microarrays or gel electrophoresis, despite information in the question clearly stating that the data analysed were obtained from DNA sequ...
— Examiner report, February/March 2025, 9700/42 Q4(b) (Cambridge)
This was a challenging question. Although many responses recognised that maintenance of the C282Y allele would require individuals with the mutation to survive and reproduce, few appreciated that this could be linked to the selection pressure for low-iron diets. Strong responses considered the role of directional selection and recognised that heterozygotes were not at a disadvantage.
— Examiner report, February/March 2025, 9700/42 Q4(c)(i) (Cambridge)
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