OxfordAQA International GCSE Biology: Genetic variation, inheritance, DNA and disorders (9201) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.5.3 and 3.5.4 of the OxfordAQA International GCSE Biology (9201) specification, Version 5.2, for exams May/June 2018 onwards. The qualification is untiered. Revise with the study guide or revision notes first. Course pages: the OxfordAQA IGCSE Biology hub and the checklist.

Questions

1. Define each term.

(a) Allele [1] (b) Homozygous [1] (c) Genotype [1] (d) Dominant allele [1]

2. A gardener takes cuttings from one plant. Six are grown in fertilised compost and six in unfertilised compost. After eight weeks the mean height is 58 cm with fertiliser and 41 cm without.

(a) Explain what the results show about the causes of variation in height. [2] (b) Calculate how much taller the fertilised plants are, as a percentage of the unfertilised mean. Give your answer to 1 decimal place. [2] (c) Suggest why the gardener used cuttings, not seeds. [1]

3. (a) Draw a genetic diagram to show how sex is determined in humans. [3] (b) A couple have three children. Calculate the probability that two are boys and one is a girl. [2]

4. In a species of moth, grey wings (G) are dominant to white wings (g). Two heterozygous moths are crossed.

(a) Give the gametes, the offspring genotypes and the expected phenotype ratio. [3] (b) Calculate the number of white moths expected among 64 offspring. [1] (c) The 64 offspring include 17 white moths. A student says this disproves the model. Evaluate the student’s claim. [2]

5. A grey moth of unknown genotype is crossed with a white moth. Of 76 offspring, 38 are white and 38 are grey.

(a) Deduce the genotype of the grey parent and explain your answer. [2] (b) State the offspring expected if the grey parent had been GG. [1]

6. The table shows a family in which polydactyly (extra fingers or toes) is caused by a dominant allele, D.

Person	Relationship	Phenotype
1, 2	Parents	Both polydactyly
3	Daughter	No polydactyly
4	Son	Polydactyly
5	Person 4’s child with an unaffected partner	No polydactyly

(a) Explain how persons 1 to 3 show that the allele is dominant. [2] (b) Give the genotypes of persons 1, 2 and 3. [2] (c) Deduce the genotype of person 4. [2]

7. Two parents without cystic fibrosis have a child who has it. Use C for the dominant allele and c for the recessive allele.

(a) Explain why the allele for cystic fibrosis is recessive, and give the parents’ genotypes. [2] (b) Complete a genetic diagram and state the probability that their next child has cystic fibrosis. [2] (c) Their unaffected daughter wants to know the probability that she is a carrier. Calculate it. [2]

8. Sickle cell anaemia is caused by a recessive allele. The table shows five families in which both parents are carriers.

Family	Children	Children with sickle cell anaemia
A	4	1
B	3	0
C	2	1
D	5	1
E	2	0

(a) Calculate the percentage of all these children who have sickle cell anaemia. Give 1 decimal place. [2] (b) State the number you would expect to be affected. [1] (c) Explain why the observed result differs from the expected one. [2]

9. Down’s syndrome is caused by an extra chromosome.

(a) State the number of chromosomes in a body cell of a person with Down’s syndrome. [1] (b) State the number in a gamete that causes it. [1]

The table gives illustrative figures for cases per 10 000 births at different maternal ages.

Maternal age (years)	25	35	40	45
Cases per 10 000 births	8	28	95	310

(c) Describe the pattern, and calculate how many times higher the figure is at 40 than at 25. Give 1 decimal place. [3]

10. (a) Describe the structure of DNA. [2] (b) Explain how a gene is linked to a protein. [2] (c) A protein has 220 amino acids. Calculate the minimum number of bases in the coding sequence. [2]

11. Mendel crossed pure-breeding yellow-seeded pea plants with pure-breeding green-seeded plants. All the first generation had yellow seeds. When these plants were allowed to self-pollinate, the next generation had 6022 yellow and 2001 green seeds. Yellow (Y) is dominant to green (y).

(a) Explain why all of the first generation had yellow seeds. [2] (b) Complete a genetic diagram for the second generation. [2] (c) Calculate the ratio yellow : green as x : 1, to 1 decimal place. [2] (d) Give one reason why it mattered that Mendel counted so many seeds, and one reason why his explanation was not linked to chromosomes in his lifetime. [2]

Answers

1. (a) An alternative version of the same gene [1]. (b) Both chromosomes of a pair carry the same allele [1]. (c) The combination of alleles an individual has [1]. (d) An allele that controls the characteristic even when present on only one chromosome [1]. [4] Examiner insight: Short definitions gain credit when they compare the two alleles; “a gene” alone for allele usually does not.

2. (a) The cuttings are genetically identical [1], so the difference in height is environmental (the fertiliser) [1]. [2] (b) 58 - 41 = 17 [1]; 17 ÷ 41 × 100 = 41.5% [1]. [2] (c) Seeds would differ genetically, so genes could also affect height [1]. [1] Examiner insight: In a percentage question, state which value you divided by; dividing by 58 gives a different answer.

3. (a) Parents XX and XY [1]; gametes X and X from the mother, X and Y from the father [1]; offspring XX, XX, XY, XY giving 1 female : 1 male [1]. [3] (b) Three orders (BBG, BGB, GBB) [1], each (1/2)³ = 1/8, so 3/8 [1]. [2] Examiner insight: Count the orders as well as multiplying; (1/2)³ alone answers a different question.

4. (a) Gametes G and g from each parent [1]; offspring GG, Gg, Gg, gg [1]; 3 grey : 1 white [1]. [3] (b) 64 × 1/4 = 16 [1]. [1] (c) 17 is close to 16 expected [1]; fertilisation is random, so small differences are normal and the claim is not supported [1]. [2] Examiner insight: Evaluate answers do best when they use the numbers (17 against 16) and not just the word “chance”.

5. (a) Gg [1]; white offspring are gg, so each received a g from the grey parent [1]. [2] (b) All offspring grey (Gg) [1]. [1] Examiner insight: Explain using the offspring: a recessive phenotype shows the alleles it must have received.

6. (a) Persons 1 and 2 both show polydactyly [1] but have an unaffected child, so they carry the recessive allele and the characteristic shows with one dominant copy [1]. [2] (b) Persons 1 and 2: Dd [1]; person 3: dd [1]. [2] (c) Person 5 is dd, so received d from person 4 [1]; person 4 is polydactyly, so Dd [1]. [2] Examiner insight: Genotype answers are clearest when you name the evidence person by person.

7. (a) Neither parent shows the disorder yet a child does, so the allele is masked in them [1]; both parents are Cc [1]. [2] (b) Cc × Cc gives CC, Cc, Cc, cc [1]; probability 1/4 [1]. [2] (c) Unaffected outcomes: CC, Cc, Cc [1]; two of the three are carriers, so 2/3 [1]. [2] Examiner insight: A common slip is answering 1/2, which forgets that the daughter is already known to be unaffected.

8. (a) Total children 4 + 3 + 2 + 5 + 2 = 16 and affected 3 [1]; 3 ÷ 16 × 100 = 18.8% [1]. [2] (b) 16 × 1/4 = 4 [1]. [1] (c) Fertilisation is random [1]; with only 16 children a small difference (3 against 4) is expected [1]. [2] Examiner insight: Mention sample size when asked to explain a difference from a prediction.

9. (a) 47 [1]. (b) 24 [1]. [2] (c) The number of cases rises as maternal age increases [1]; 95 ÷ 8 [1] = 11.9 times [1]. [3] Examiner insight: Describe the trend using at least one pair of values from the table.

10. (a) Two very long strands [1] twisted into a double helix, containing four different bases [1]. [2] (b) A gene is a small section of DNA [1] whose base sequence sets the sequence of amino acids that are joined into one specific protein [1]. [2] (c) 220 × 3 [1] = 660 bases [1]. [2] Examiner insight: Show the multiplication by three; a bare number gives no way to recognise the idea.

11. (a) Pure-breeding parents are YY and yy, so every offspring is Yy [1]; yellow is dominant so all show yellow [1]. [2] (b) Yy × Yy, gametes Y, y and Y, y [1]; YY, Yy, Yy, yy giving 3 yellow : 1 green [1]. [2] (c) 6022 ÷ 2001 [1] = 3.0 : 1 [1]. [2] (d) Large numbers reduce the effect of chance, so ratios show clearly [1]; chromosomes and their link to inheritance had not yet been discovered (later scientists linked them) [1]. [2] Examiner insight: Link the large sample to chance variation to gain credit; “more accurate” is vague.

Where marks are usually lost

• Defining allele as “a gene” without saying “different form”.
• Writing a percentage change relative to the wrong base value.
• Leaving out the gametes when constructing a genetic diagram.
• Answering with a ratio such as 3 : 1 when the question wants 3/4 or 75%.
• Forgetting to condition on the unaffected child when finding carrier probability.
• Using a total of all offspring when only a subset applies.
• Giving a bare number for calculations on DNA bases.
• Calling Down’s syndrome an inherited allele problem.

Next steps

• Revise with the revision notes or the study guide.
• Course pages: the OxfordAQA IGCSE Biology hub and the checklist.
• Take the free 10-minute diagnostics.
• Want teacher support? Book a free trial class.

Official syllabus

OxfordAQA International GCSE Biology (9201), Version 5.2, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.5.3 Genetic variation and 3.5.4 Genetic disorders.