Genetic inheritance, inherited disorders and sex determination – GCSE Biology exam questions with mark scheme

1Grade 4–55 marks

Cystic fibrosis is caused by a recessive allele, f. The dominant allele is F.

Two parents are both carriers of cystic fibrosis.

(a) Complete the Punnett square to show the possible genotypes of their children. [2]

	Father's gamete: ____	Father's gamete: ____
Mother's gamete: ____
Mother's gamete: ____

(b) Give the probability that a child of these parents will have cystic fibrosis. [1]

(d) Explain why the parents do not have cystic fibrosis. [1]

(a) gametes from both parents F and f(1)
(a) offspring genotypes FF, Ff, Ff, ff
allow fF for Ff
allow ecf from incorrect gametes(1)
(b) 0.25 / 25% / ¼ / 1 in 4
do not accept 1:4 or 1:3(1)
(c) 0.5 / 50% / ½ / 2 in 4 / 1 in 2
do not accept 1:1 or 2:4(1)
(d) they each have one dominant allele (F), which masks the recessive allele
allow they are heterozygous / Ff and the disorder needs two recessive alleles(1)
Answer: (a) FF, Ff, Ff, ff (b) 0.25 (c) 0.5 (d) Each has a dominant F allele.
Examiner tip: Probabilities written as ratios such as 1:4 are not accepted – use a fraction, decimal or percentage.

2Grade 8–9Higher5 marks

The diagram shows the inheritance of a genetic disorder in one family. The disorder is controlled by a single gene that is not on the sex chromosomes.

(a) Use evidence from the diagram to explain whether the disorder is caused by a dominant or a recessive allele. [2]

(b) Persons 1 and 2 plan to have another child. Give the probability that this child will have the disorder. [1]

(c) Person 5 does not have the disorder. Calculate the probability that person 5 is a carrier. Show your working, for example in a Punnett square. [2]

(a) recessive(1)
(a) because persons 1 and 2 do not have the disorder but their child, person 3, does (so 1 and 2 must both be carriers / heterozygous)
allow if it were dominant, at least one of persons 1 and 2 would have the disorder(1)
(b) 0.25 / ¼ / 25% / 1 in 4
do not accept 1:4 or 1:3(1)
(c) Punnett square Aa × Aa (or any letters) and person 5 cannot be aa, leaving AA, Aa, Aa(1)
(c) ⅔ / 0.67 / 67% / 2 in 3
allow 0.667 / 66.7%
do not accept 2:3 or 2:1
a correct answer with no working gains 2 marks(1)
Answer: (a) Recessive – unaffected parents 1 and 2 have an affected child 3 (b) 0.25 (c) ⅔
Examiner tip: In (c) many give ½: you must remove the affected genotype because person 5 is known to be unaffected.

3Grade 1–3Foundation4 marks

In humans, sex is determined by one pair of chromosomes.

(a) Give the sex chromosomes of a human male. [1]

(b) Draw a Punnett square to show how sex is inherited. [2]

(a) XY
allow X and Y(1)
(b) correct gametes: X and X from the mother and X and Y from the father
allow gametes in either row/column(1)
(b) offspring XX, XX, XY, XY (in the correct boxes)
allow ecf from incorrect gametes(1)
(c) 0.5 / 50% / ½ / 1 in 2 / 2 in 4
do not accept 50:50 or 1:1(1)
Answer: (a) XY (b) XX, XX, XY, XY (c) 0.5
Examiner tip: '50:50' is a ratio, not a probability, and does not score.

4Grade 1–3Foundation4 marks

(a) Draw one line from each genetic term to its correct description. [3]

Term	Description
Allele	A: the characteristic that can be seen in an organism
Genotype	B: a different version of the same gene
Homozygous	C: having two identical alleles of a gene
	D: a cell with half the normal number of chromosomes
	E: the combination of alleles an organism has

(b) Fur colour in mice is controlled by a single gene.

Which statement about most human characteristics is correct? Tick (✓) one box. [1]

☐ They are controlled by a single gene.
☐ They are controlled by several genes.
☐ They are controlled only by the environment.
☐ They are controlled by the sex chromosomes only.

(a) Allele → B (a different version of the same gene)(1)
(a) Genotype → E (the combination of alleles an organism has)(1)
(a) Homozygous → C (having two identical alleles of a gene)
if more than one line is drawn from a term, that term scores 0(1)
(b) They are controlled by several genes
do not award if more than one box is ticked(1)
Answer: (a) Allele – B; Genotype – E; Homozygous – C (b) controlled by several genes
Examiner tip: Genotype (the alleles) and phenotype (what you can see) are the most commonly swapped pair – learn them together.

5Grade 4–55 marks

Polydactyly is an inherited disorder in which a person has extra fingers or toes.

Polydactyly is caused by a dominant allele, D. The recessive allele is d.

A man with polydactyly has the genotype Dd. His partner does not have polydactyly.

(a) Give the genotype of the woman. [1]

(b) Complete the Punnett square to show the possible genotypes of their children. [2]

	Man's gamete: ____	Man's gamete: ____
Woman's gamete: ____
Woman's gamete: ____

(d) The couple have four children. None of the children has polydactyly.

Explain why this is possible. [1]

(a) dd(1)
(b) gametes: D and d from the man and d and d from the woman
allow gametes in either row or column(1)
(b) offspring Dd, Dd, dd, dd
allow ecf from incorrect gametes in (a)/(b)(1)
(c) 0.5 / ½ / 50% / 1 in 2 / 2 in 4
allow ecf from Punnett square
do not accept 1:1 or 50:50(1)
(d) each child / each fertilisation has the same (0.5) chance, independent of the other children
allow fertilisation is random / which sperm fertilises the egg is down to chance
allow probabilities only predict outcomes for large numbers / it is a small sample
ignore 'it is luck' unqualified(1)
Answer: (a) dd (b) Dd, Dd, dd, dd (c) 0.5 (d) Each fertilisation is random and independent, so every child has a 0.5 chance.
Examiner tip: A dominant disorder needs only one allele – so the unaffected parent must be homozygous recessive, dd.

6Grade 6–7Higher6 marks

In mice, the allele for black fur, B, is dominant to the allele for brown fur, b.

A scientist bred two mice that were both heterozygous for fur colour. The table shows the offspring from several litters.

Fur colour	Number of offspring
Black	47
Brown	13

(a) Calculate the ratio of black mice to brown mice.

Give your answer in the form x : 1, to 1 decimal place. [2]

(b) Calculate how many of the 60 offspring would be expected to be brown. [1]

(d) The scientist then crossed a black mouse of unknown genotype with a brown mouse. All 9 offspring were black.

What do these results suggest about the genotype of the black mouse? Explain your answer. [2]

(a) 47 ÷ 13 (= 3.615…)(1)
(a) 3.6 : 1
allow 3.6 alone
an answer of 3.6 : 1 with no working gains 2 marks(1)
(b) 15
(Bb × Bb gives 1 in 4 bb, ¼ × 60)(1)
(c) fertilisation is random / which gametes fuse is down to chance
allow sample size too small (for the ratio to be exact)
ignore errors in counting(1)
(d) (black mouse is probably) homozygous / BB(1)
(d) if it were Bb, about half the offspring would be expected to be brown (bb) / a brown mouse can only pass on b, so any b from the black parent would give brown offspring
do not penalise 'cannot be certain as only 9 offspring', but this alone does not gain the mark(1)
Answer: (a) 3.6 : 1 (b) 15 (c) fertilisation is random / small sample (d) BB – a Bb parent would be expected to produce brown (bb) offspring with a brown (bb) mouse.
Examiner tip: In (a) you must divide by the smaller number to get x : 1 – writing 47 : 13 is not in the form asked for.

7Grade 8–9Higher5 marks

Cystic fibrosis is caused by a recessive allele, f. The sex of a child is determined by the sex chromosomes.

(a) Give the number of chromosomes in a human sperm cell. [1]

(b) Two parents are both heterozygous (Ff) for cystic fibrosis.

Calculate the probability that their first child will be a girl who has cystic fibrosis. [2]

Give your answer as a fraction. [2]

(a) 23
do not accept 23 pairs(1)
(b) probability of a girl = ½ and probability of cystic fibrosis (ff) = ¼
may be shown in Punnett squares(1)
(b) ½ × ¼ = ⅛ / 0.125 / 12.5%
do not accept 1:8 or 1:7
correct answer with no working gains 2 marks(1)
(c) ¼ × ¼(1)
(c) ¹⁄₁₆
allow ecf from an incorrect probability of cystic fibrosis in (b), e.g. ½ × ½ = ¼ scores 1 mark
correct answer with no working gains 2 marks(1)
Answer: (a) 23 (b) ⅛ (c) 1/16
Examiner tip: Independent events are multiplied, not added – ¼ + ½ is a common error in (b).

8Grade 6–7Higher6 marks

The family tree shows the inheritance of polydactyly in one family.

Polydactyly is controlled by a single gene that is not on the sex chromosomes. Use D for the dominant allele and d for the recessive allele.

(a) Explain how the family tree shows that polydactyly is caused by a dominant allele. [2]

(b) Give the genotype of person 1. Explain your answer. [2]

Give the probability that this child will have polydactyly. Explain your answer. [2]

(a) persons 1 and 2 both have polydactyly but (their child) person 5 does not(1)
(a) so 1 and 2 must both carry a recessive allele that is masked / are heterozygous
allow if polydactyly were recessive, 1 and 2 would both be dd and all their children would have polydactyly(1)
(b) Dd(1)
(b) person 5 is dd / does not have polydactyly, so person 5 must have received a d allele from person 1
ignore 'person 1 has polydactyly so has a D allele' alone (it does not explain the d)(1)
(c) 0 / 0% / none
do not accept 'unlikely'(1)
(c) neither person 5 nor person 6 has polydactyly, so both are dd / have no D allele (to pass on)
allow all children would be dd(1)
Answer: (a) Affected parents 1 and 2 have an unaffected child, 5 (b) Dd – person 5 is dd so inherited d from person 1 (c) 0 – both parents are dd.
Examiner tip: With a dominant disorder, anyone without the disorder must be homozygous recessive – there are no 'carriers'.

9Grade 6–76 marks

A couple are both carriers of cystic fibrosis. They want to have a child who does not have cystic fibrosis.

Their doctor offers them IVF with embryo screening. Several embryos are produced in a laboratory. One cell is removed from each embryo and tested for the cystic fibrosis alleles. Only an embryo without cystic fibrosis is placed in the woman's uterus.

Evaluate the use of embryo screening by this couple.

You should include economic, social and ethical issues in your answer. [6]

For: prevents the child having cystic fibrosis / a lifelong disorder with suffering (e.g. thick mucus, lung infections)(1)
For: economic – saves money on lifelong medical treatment / care for a person with cystic fibrosis (for the NHS / family)(1)
For: social – reduces stress on the family / child can live a normal life; couple can have a biological child without the ¼ risk(1)
For: could choose a non-carrier embryo so the allele is not passed on to the next generation(1)
Against: economic – IVF and screening are expensive / may not be available on the NHS / only the rich could afford it(1)
Against: ethical – unused embryos (with cystic fibrosis) are destroyed / embryos have a right to life; religious objections(1)
Against: ethical – could lead to 'designer babies' / screening for non-medical characteristics(1)
Against: social – suggests people with cystic fibrosis are less valued / discrimination against people with disorders(1)
Against: removing a cell may damage the embryo / IVF has a low success rate / test could give a false result(1)
Judgement – a conclusion that weighs the points, e.g. screening is justified because cystic fibrosis causes severe suffering, provided it is only used for serious disorders(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, balanced evaluation that covers advantages and disadvantages, including economic, social and ethical issues, and reaches a reasoned judgement linked to the points made.
Level 2	3–4	Some advantages and disadvantages are described, covering at least two of economic, social and ethical issues; any conclusion is not fully supported.
Level 1	1–2	Simple statements of one or more relevant points, which may be one-sided or not linked to issues; no judgement.
No relevant content	0	Nothing worthy of credit.

Answer: Balanced evaluation: avoids a child with CF and long-term treatment costs, but IVF/screening is costly, involves destroying embryos and may lead to designer babies; reasoned conclusion.
Examiner tip: An 'evaluate' question needs a conclusion – a list of pros and cons with no judgement cannot reach Level 3.

10Grade 4–5Foundation5 marks

(a) How many pairs of chromosomes are there in a human body cell? [1]

(b) One pair of chromosomes in a body cell are the sex chromosomes. Give the sex chromosomes of a human female. [1]

The couple say their next baby is more likely to be a girl.

Give the probability that their next baby will be a girl. [1]

(d) Explain why the couple are wrong. [1]

(e) In one year, 412 babies were born in a hospital.

Calculate how many of these babies would be expected to be girls. [1]

(a) 23
do not accept 46(1)
(b) XX
allow X and X / two X chromosomes(1)
(c) 0.5 / ½ / 50% / 1 in 2
do not accept 50:50 or 1:1(1)
(d) the probability is the same (0.5) for every baby / is not affected by earlier children
allow each fertilisation is random / independent
allow it depends on whether the sperm carries an X or a Y chromosome (each time)(1)
(e) 206(1)
Answer: (a) 23 (b) XX (c) 0.5 (d) Each pregnancy is independent – the chance is always 0.5 (e) 206
Examiner tip: Previous children do not change the probability – every fertilisation is a fresh 0.5 chance.

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