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OCR GCSE Combined Science

Topic questions Genetics and inheritance Monohybrid inheritance

Monohybrid inheritance

18 questions · 89 marks · about 1.5 hours

Monohybrid inheritance questions

OCR J250 Monohybrid inheritance 18 questions

Biology Paper 2 June 2025 Foundation tier, Question 9 1 mark

The diagram shows a genetic cross for inheriting cystic fibrosis.

Cystic fibrosis is a recessive inherited condition.

Figure for question 9

What percentage of the offspring will have cystic fibrosis?

A 0%

B 25%

C 75%

D 100%

Question 9, shown as an image
[1]
Reveal mark scheme Mark scheme for question 9

Biology Paper 2 June 2024 Foundation tier, Question 11 9 marks

Inheritance is passing genetic material from one generation to the next.

(a)
Figure for question 11
[2]
Reveal mark scheme Mark scheme for question 11(a)
(b)

Which terms describe sex cells in humans?

Tick (✓) two boxes.

Question 11, shown as an image
[2]
Reveal mark scheme Mark scheme for question 11(b)
(c)

Peas develop inside pods.

Figure for question 11

Pea pods can be green or yellow.

(c)(i)

A pea plant with green pods is crossed with a pea plant with yellow pods. All the pods on the new plant are green.

Complete the sentences about the pea pods.

Use words from the list.

dominant haploid heterozygous

homozygous recessive

The results of this cross show that the allele for green is........ and the

allele for yellow is........

The genotype of the new plant is Gg. This means it is........ for pod

colour.

[3]
Reveal mark scheme Mark scheme for question 11(c)(i)
(c)(ii)

The new plants with green pods are then crossed.

What is the probability that a plant from this cross will have yellow pods?

Complete the genetic diagram to explain your answer.

Question 11, shown as an image
Figure for question 11
[2]
Reveal mark scheme Mark scheme for question 11(c)(ii)

Biology Paper 8 June 2024 Higher tier, Question 12 8 marks

Genetic material is required for inheritance in eukaryotic and prokaryotic cells.

(a)

Compare the genetic material found in eukaryotic and prokaryotic cells.

[2]
Reveal mark scheme Mark scheme for question 12(a)
(b)

Peas develop inside pods.

Figure for question 12
(b)(i)

Pea pods are either green or yellow. The allele for green is dominant over the allele for yellow.

The table shows some information about three different pea pods.

Complete the table.

PhenotypeGenotypeDescription of genotype
yellowgghomozygous .........................................
greenGGhomozygous dominant
............................Gg.........................................
[2]
Reveal mark scheme Mark scheme for question 12(b)(i)
(b)(ii)

A gardener crosses a plant that has green pods with a plant that has yellow pods. The gardener thinks both these plants have a homozygous genotype.

The result is 15 new plants with green pods and 3 plants with yellow pods.

Show that the gardener’s green plant did not have a homozygous genotype by completing the two genetic diagrams.

Use letters G and g for the alleles.

If the green plant has a homozygous genotype:

Question 12, shown as an image
Figure for question 12

If the green plant does not have a homozygous genotype:

Question 12, shown as an image
Figure for question 12
[3]
Reveal mark scheme Mark scheme for question 12(b)(ii)
(b)(iii)

The ratio the gardener achieved was 5 green to 1 yellow.

Suggest why this does not match either of the offspring ratios.

[1]
Reveal mark scheme Mark scheme for question 12(b)(iii)

Biology Paper 2 June 2023 Foundation tier, Question 16 12 marks

A gardener grows a species of plant that has either red or white flowers. The colour of the flowers is controlled by a single pair of alleles, R and r.

The gardener crosses a plant that is homozygous for red flowers with a plant that is homozygous for white flowers. The diagram shows the results.

Figure for question 16
(a)(i)

Explain why the offspring in the diagram are all red.

[2]
Reveal mark scheme Mark scheme for question 16(a)(i)
(a)(ii)

The offspring are all heterozygous for flower colour.

What is meant by the term heterozygous ?

[1]
Reveal mark scheme Mark scheme for question 16(a)(ii)
(b)
Figure for question 16
[3]
Reveal mark scheme Mark scheme for question 16(b)
(c)

The gardener grows another species of plant that can produce flowers of five different colours.

The gardener counts the number of plants for each flower colour.

The graph shows their results.

Figure for question 16
(c)(i)

The gardener counted 97 plants.

Calculate the percentage of these plants that have blue flowers.

Give your answer to the nearest whole number.

[3]
Reveal mark scheme Mark scheme for question 16(c)(i)
(c)(ii)

What evidence is there in the graph that flower colour is an example of discontinuous variation in this species?

[1]
Reveal mark scheme Mark scheme for question 16(c)(ii)
(d)

The male gametes of a plant are called pollen.

The sentences in the text box are about pollen. There are two words in the sentences that are not correct.

Question 16, shown as an image

Pollen cells are haploid. This means that the pollen cells of a plant have twice

the number of chromosomes as the leaf cells of the plant.

Pollen cells are formed by a type of cell division called mitosis.

[2]
Reveal mark scheme Mark scheme for question 16(d)

Biology Paper 8 June 2023 Higher tier, Question 11 12 marks

A gardener grows a species of plant that has either red or white flowers. The colour of the flowers is controlled by a single pair of alleles, R and r.

The gardener crosses a plant that is homozygous for red flowers with a plant that is homozygous for white flowers. The diagram shows the results.

Figure for question 11
(a)(i)

Explain why the offspring in the diagram are all red.

[2]
Reveal mark scheme Mark scheme for question 11(a)(i)
(a)(ii)

The offspring are all heterozygous for flower colour.

What is meant by the term heterozygous ?

[1]
Reveal mark scheme Mark scheme for question 11(a)(ii)
(b)

The gardener crosses one of these offspring plants with a plant that is homozygous for white flowers.

Predict the probability that the next generation of plants will have white flowers.

Complete the genetic diagram to explain your answer.

Figure for question 11
[3]
Reveal mark scheme Mark scheme for question 11(b)
(c)

The gardener grows another species of plant that can produce flowers of five different colours.

The gardener counts the number of plants for each flower colour.

The graph shows their results.

Figure for question 11
(c)(i)

The gardener counted 97 plants.

Calculate the percentage of these plants that have blue flowers.

Give your answer to the nearest whole number.

[3]
Reveal mark scheme Mark scheme for question 11(c)(i)
(c)(ii)

What evidence is there in the graph that flower colour is an example of discontinuous variation in this species?

[1]
Reveal mark scheme Mark scheme for question 11(c)(ii)
(d)

The male gametes of a plant are called pollen.

The sentences in the text box are about pollen. There are two words in the sentences that are not correct.

Question 11, shown as an image

Pollen cells are haploid. This means that the pollen cells of a plant have twice

the number of chromosomes as the leaf cells of the plant.

Pollen cells are formed by a type of cell division called mitosis.

[2]
Reveal mark scheme Mark scheme for question 11(d)

Biology Paper 2 June 2022 Foundation tier, Question 4 1 mark

The diagram shows how a condition caused by a recessive gene mutation is inherited.

Which child, A, B, C or D, will inherit and develop the condition?

Figure for question 4
Question 4, shown as an image
[1]
Reveal mark scheme Mark scheme for question 4

Biology Paper 2 June 2022 Foundation tier, Question 9, also on the Higher paper 1 mark

The diagram shows a genetic cross for seed shape in peas.

Figure for question 9

Which prediction about the offspring is most likely?

A All the offspring will be heterozygous for seed shape.

B All the offspring will be homozygous for seed shape.

C The ratio of heterozygous to homozygous offspring will be 1: 1.

D The ratio of heterozygous to homozygous offspring will be 3: 1.

Question 9, shown as an image
[1]
Reveal mark scheme Mark scheme for question 9

Biology Paper 8 June 2022 Higher tier, Question 1, also on the Foundation paper 1 mark

The diagram shows a genetic cross for seed shape in peas.

Figure for question 1

Which prediction about the offspring is most likely?

A All the offspring will be heterozygous for seed shape.

B All the offspring will be homozygous for seed shape.

C The ratio of heterozygous to homozygous offspring will be 1:1.

D The ratio of heterozygous to homozygous offspring will be 3:1.

Question 1, shown as an image
[1]
Reveal mark scheme Mark scheme for question 1

Biology Paper 8 June 2022 Higher tier, Question 15 10 marks

(a)

Some forms of breast cancer are linked to specific mutated genes called BRCA.

BRCA is a dominant allele.

Inheriting the BRCA allele means in both males and females, the risk of developing breast cancer is affected. It does not mean a person has been diagnosed with cancer.

(a)(i)

A male who is heterozygous for BRCA has children with a female who is homozygous recessive.

How are they affected by the BRCA allele?

Put ticks (✓) in the correct boxes.

Figure for question 15
[1]
Reveal mark scheme Mark scheme for question 15(a)(i)
(a)(ii)

What is the percentage probability of them having a child with an increased risk of developing breast cancer?

Use a labelled genetic diagram to explain your answer.

Use the letter B for the dominant BRCA allele.

[3]
Reveal mark scheme Mark scheme for question 15(a)(ii)
(b)
Figure for question 15
[6]
Reveal mark scheme Mark scheme for question 15(b)

Biology Paper 2 November 2021 Foundation tier, Question 13 11 marks

Figure for question 13
(a)(i)

Complete Table 13.1 by choosing words from this list to identify the genetic term used to describe cell A and cell B.

diploid genotype haploid phenotype

[2]
Reveal mark scheme Mark scheme for question 13(a)(i)
(a)(ii)

Give an explanation for each choice of word in Table 13.1.

Figure for question 13
[2]
Reveal mark scheme Mark scheme for question 13(a)(ii)
(b)
Figure for question 13
[2]
Reveal mark scheme Mark scheme for question 13(b)
(c)

Huntington’s disease can cause death.

Fig. 13.3 shows the occurrence of HD by age when symptoms first appear.

Figure for question 13

What is the modal age range for the data shown in Fig. 13.3 ?

[1]
Reveal mark scheme Mark scheme for question 13(c)
(d)

Early studies in the UK showed one in 15 000 have the HD allele. There are 66 000 000 people in the UK.

Calculate the number of people in the UK with the HD allele.

[2]
Reveal mark scheme Mark scheme for question 13(d)
(e)

Read the information about HD in the box.

• The HD allele is a dominant allele that can lead to death.

• Alleles that cause death rarely persist in populations.

• About 10% of people with HD acquire the HD allele from a newly formed mutation and not through inheriting it from their parents.

Use this information and the data in Fig. 13.3 to suggest why the HD allele still persists in the human population.

[2]
Reveal mark scheme Mark scheme for question 13(e)

Biology Paper 8 November 2021 Higher tier, Question 5 1 mark

A flower homozygous dominant for colour is crossed with a homozygous recessive flower.

What is the predicted percentage of heterozygous offspring?

A 25%

B 50%

C 75%

D 100%

Question 5, shown as an image
[1]
Reveal mark scheme Mark scheme for question 5

Biology Paper 8 November 2020 Higher tier, Question 1 1 mark

A scientist crossed a white flower with a purple flower. The white flower is homozygous recessive. The purple flower is heterozygous.

The scientist uses this Punnett square to predict the expected ratio of offspring.

Pp
pPppp
pPppp

What is the expected ratio of phenotypes in the offspring?

A 1 purple: 1 white

B 1 purple: 2 white

C 2 purple: 1 white

D 4 purple: 0 white

Question 1, shown as an image
[1]
Reveal mark scheme Mark scheme for question 1

Biology Paper 8 November 2020 Higher tier, Question 7 1 mark

The phenotype of rabbits can be black or white fur. The allele for black fur is dominant.

Which genetic cross identifies the genotype of a rabbit with black fur (rabbit X) as Bb?

A Cross rabbit X with a black rabbit. All offspring have black fur.

B Cross rabbit X with a black rabbit. 50% of offspring have black fur and 50% have white fur.

C Cross rabbit X with a white rabbit. All offspring have white fur.

D Cross rabbit X with a white rabbit. 50% of offspring have black fur and 50% have white fur.

Question 7, shown as an image
[1]
Reveal mark scheme Mark scheme for question 7

Biology Paper 2 June 2019 Foundation tier, Question 6 1 mark

Huntington’s disease is a single gene disorder caused by a dominant allele (H).

The Punnett square shows the genetic cross between two heterozygous parents.

Hh
HHHHh
hHhhh

What is the probability of these parents having a child with Huntington’s disease?

A 0 in 4

B 1 in 4

C 2 in 4

D 3 in 4

Question 6, shown as an image
[1]
Reveal mark scheme Mark scheme for question 6

Biology Paper 8 June 2019 Higher tier, Question 3 1 mark

A genetic cross occurs between a heterozygous individual and a homozygous recessive individual.

What is the probability of these individuals having offspring that are homozygous recessive?

A 0 in 4

B 1 in 4

C 2 in 4

D 3 in 4

Question 3, shown as an image
[1]
Reveal mark scheme Mark scheme for question 3

Biology Paper 2 June 2018 Foundation tier, Question 14 10 marks

(a)

Some children can inherit a rare kidney condition called ARPKD. It is caused by a recessive allele.

A pregnant woman has ARPKD. The baby’s father is heterozygous for ARPKD.

What is the probability of their baby having ARPKD?

Complete the genetic diagram to explain your answer.

Use the letters:

• R for the dominant allele

• r for the recessive allele.

Figure for question 14
[2]
Reveal mark scheme Mark scheme for question 14(a)
(b)(i)

It is estimated that 1 in 70 people in the UK is a carrier of ARPKD.

The estimated population of the UK in 2011 was 63.2 million.

Estimate the number of people in the UK in 2011 who were carriers.

[2]
Reveal mark scheme Mark scheme for question 14(b)(i)
(b)(ii)

Explain why this number is only an estimate and not the real number of carriers.

[1]
Reveal mark scheme Mark scheme for question 14(b)(ii)
(c)
Figure for question 14
[2]
Reveal mark scheme Mark scheme for question 14(c)
(d)

The kidneys in people with ARPKD have not developed normally.

Stem cells could be used to repair the damaged kidneys.

Discuss reasons for and against using stem cells to treat ARPKD.

[3]
Reveal mark scheme Mark scheme for question 14(d)

Biology Paper 8 June 2018 Higher tier, Question 14 7 marks

(a)
Figure for question 14
[1]
Reveal mark scheme Mark scheme for question 14(a)
(b)

Type 1 diabetes is a disorder caused by lack of insulin production in the pancreas. Type 1 diabetes has a genetic link.

In humans the haploid number of chromosomes is 23. Insulin is a hormone controlled by a single gene found on chromosome number 11.

(b)(i)

How many chromosomes are there in a single human pancreas cell?

[1]
Reveal mark scheme Mark scheme for question 14(b)(i)
(b)(ii)

Some individuals have a copy of the gene that would cause diabetes but do not have the disorder.

Suggest a reason why.

[1]
Reveal mark scheme Mark scheme for question 14(b)(ii)
(c)

Some people can inherit a rare condition called ADPKD that results in kidney disease.

The PKD1 gene provides instructions for making a protein called polycystin-1. Polycystin-1 controls kidney cell differentiation.

In ADPKD, polycystin-1 does not work properly.

(c)(i)

What name is given to a change in the PKD1 gene that causes ADPKD?

[1]
Reveal mark scheme Mark scheme for question 14(c)(i)
(c)(ii)

Explain why the ADPKD version of polycystin-1 prevents the kidney working properly.

[1]
Reveal mark scheme Mark scheme for question 14(c)(ii)
(c)(iii)

A pregnant woman is heterozygous for ADPKD. The baby’s father does not have ADPKD and is homozygous recessive.

ADPKD is caused by a dominant allele (D).

What is the probability of their baby having ADPKD?

Use a labelled genetic diagram to explain your answer.

[2]
Reveal mark scheme Mark scheme for question 14(c)(iii)

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