IB DP Biology: Inheritance: alleles, monohybrid crosses, pedigrees and continuous variation (DP Biology) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover D3.2.1 to D3.2.15 (Inheritance, first part) of the IB Biology guide, first assessment 2025, for DP Biology. All content is SL and HL. Learn the content in the study guide and the revision notes. Course hub: /boards/ib/ib-dp/biology/. Checklist: /checklists/ib/ib-dp/biology/.

Questions

1. Tomato plants have 2n = 24. (a) State the number of chromosomes in a pollen nucleus and in a zygote. [2] (b) Explain why a tomato zygote has two copies of each autosomal gene. [1]

2. Describe how you would make a controlled cross between a pure-breeding red-fruited tomato plant and a pure-breeding yellow-fruited one, given that each flower can self-pollinate. [4]

3. Seed colour in a plant is controlled by a gene with alleles G (green, dominant) and g (yellow). (a) Define the term homozygous. [1] (b) Explain why plants of genotype GG and Gg have the same phenotype. [3]

4. (a) For each trait, state whether it is due to genotype only, environment only, or both: ABO blood group; the number of scars on a person’s knee; body mass. [3] (b) Two genetically identical hydrangea cuttings are grown in soils of different acidity and produce different flower colours. Explain this using phenotypic plasticity. [2]

5. Two unaffected parents have a child with phenylketonuria (PKU). Use A and a for the alleles.

(a) State the genotype of both parents. [1] (b) Calculate the probability that their next child has PKU. [2] (c) Explain why phenylalanine builds up in a child with PKU. [2] (d) Explain why a low-phenylalanine diet changes the phenotype but not the genotype. [1] (e) Explain why marriage between close relatives is prohibited in many societies. [2]

6. (a) Define single-nucleotide polymorphism. [1] (b) A gene has five alleles in a population. State how many alleles one person has, and why. [2] (c) Calculate the number of possible genotypes. [2]

7. A woman of blood group A and a man of group B have a child of group O.

(a) Deduce the genotypes of both parents, using Iᴬ, Iᴮ and i. [2] (b) Find the probability that their next child is group AB. [2] (c) Explain why group AB shows codominance. [2]

8. In Mirabilis jalapa (marvel of Peru), red flowers (CᴿCᴿ) crossed with white (CᵂCᵂ) give pink flowers. Two pink plants are crossed and produce 320 seedlings.

(a) Predict the expected number of seedlings of each colour. [3] (b) Distinguish between incomplete dominance and codominance. [2]

9. A man with haemophilia (XʰY) and a carrier woman (XᴴXʰ) have children.

(a) Draw a Punnett grid and state the probability that a daughter has haemophilia. [3] (b) Explain why X-linked recessive disorders are more common in males. [2] (c) Explain why the man cannot pass his haemophilia allele to a son. [1] (d) Explain how sperm determine the sex of the zygote. [2]

10. In a family, I-1 (male) and I-2 (female) are both affected by a condition. Their children are II-1 (male, affected), II-2 (female, unaffected) and II-3 (female, affected). II-3 and an unaffected man II-4, from outside the family, have two children: III-1 (male, unaffected) and III-2 (female, affected).

(a) Deduce that the condition is autosomal dominant. [3] (b) Using D and d, state the genotypes of I-1 and II-3. [2] (c) II-3 and II-4 have a third child. Calculate the probability it is affected. [2] (d) Identify which step above used inductive reasoning and which used deductive reasoning. [2]

11. Shoot heights (cm) of 12 seedlings: 23, 27, 27, 27, 29, 30, 31, 33, 34, 35, 36, 52.

(a) Find the median, first quartile, third quartile and interquartile range. [3] (b) Determine whether 52 is an outlier. [2] (c) State the values at which the whiskers end. [1] (d) Calculate the mean and state the mode. [2] (e) Explain why shoot height shows continuous variation, unlike ABO blood group. [2]

Answers

1. (a) Pollen nucleus 12 [1]; zygote 24 [1]. (b) One copy of each chromosome comes from each parent, so the zygote has homologous pairs [1]. [3] Examiner insight: the pollen nucleus is haploid and the zygote diploid; give both numbers and do not swap them.

2. Remove the anthers from the red-fruited flower before they release pollen [1]. Cover the flower to keep out foreign pollen [1]. Transfer pollen from the yellow-fruited plant to the stigma [1]. Collect the seeds and grow them to score the F1 [1]. [4] Examiner insight: name the part removed and say why; “stop it self-pollinating” without emasculation is incomplete.

3. (a) Having two identical alleles of a gene [1]. (b) Gg is green [1]. G codes for a functional polypeptide, such as an enzyme [1]. One copy makes enough product, so Gg matches GG [1]. [4] Examiner insight: “dominant masks recessive” is a restatement of the term; an explanation needs the working polypeptide.

4. (a) ABO: genotype only [1]; scars: environment only [1]; body mass: both [1]. (b) The genotype is identical, so the difference is caused by the environment altering gene expression [1]. Plasticity lets the plant develop a form matched to its conditions, with no change to the DNA [1]. [5] Examiner insight: “a mutation occurred” is a tempting wrong answer; state that the genotype is unchanged.

5. (a) Both parents are Aa [1]. (b) Gametes A or a from each parent [1]; the child is aa with probability 1/4 [1]. (c) The enzyme that converts phenylalanine to tyrosine is non-functional [1], so phenylalanine accumulates [1]. (d) The diet reduces phenylalanine intake, an environmental change, while the child’s alleles stay aa [1]. (e) Close relatives share recent ancestors [1], so both partners are more likely to carry the same rare recessive allele [1]. [8] Examiner insight: in “calculate”, show the gametes or grid before the fraction.

6. (a) A position in the genome where individuals differ by a single base [1]. (b) Two [1]; a diploid person inherits one allele from each parent [1]. (c) Genotypes = 5 × 6 ÷ 2 [1] = 15 [1]. [5] Examiner insight: show the method, and remember that homozygous genotypes count as well as heterozygous ones.

7. (a) Child ii, so each parent gave i [1]. Mother Iᴬi, father Iᴮi [1]. (b) Gametes Iᴬ or i; Iᴮ or i [1]. IᴬIᴮ is one of four equally likely outcomes, so 1/4 [1]. (c) The heterozygote has both alleles [1] and both are expressed, giving a dual phenotype [1]. [6] Examiner insight: write genotypes with the superscripts; a plain “A” and “B” does not show the alleles.

8. (a) Cross CᴿCᵂ × CᴿCᵂ gives 1 red : 2 pink : 1 white [1]. Red = 80 and white = 80 [1], pink = 160 [1]. (b) In incomplete dominance the heterozygote has an intermediate phenotype [1]; in codominance it shows both alleles’ phenotypes at once [1]. [5] Examiner insight: give expected counts, not just the ratio, when the question supplies a total.

9. (a) Gametes: father Xʰ or Y; mother Xᴴ or Xʰ [1]. Daughters are XᴴXʰ and XʰXʰ [1]. Half of daughters are affected: 1/2 [1]. (b) Males have one X [1], so one recessive allele gives the disorder; a female needs two [1]. (c) He gives his son the Y, not an X [1]. (d) Eggs always carry X [1]; a sperm carrying X gives a girl, and one carrying Y gives male-typical characteristics [1]. [8] Examiner insight: “males are affected more” needs the reason that males have a single X.

10. (a) Affected parents I-1 and I-2 have an unaffected daughter [1], so the condition is dominant and both parents carry a recessive allele [1]. Affected father I-1 has an unaffected daughter, so it is not X-linked dominant, as all his daughters would be affected: autosomal [1]. (b) I-1 is Dd, since his daughter II-2 is dd and received d from him [1]. II-3 is Dd, since III-1 is dd and received d from her [1]. (c) Dd × dd: gametes D or d; d [1]. Probability = 1/2 [1]. (d) Concluding autosomal dominant from this family’s pattern is inductive [1]; using that rule to assign II-3’s genotype is deductive [1]. [9] Examiner insight: ruling out the alternatives is rewarded; naming the pattern alone is less convincing.

11. (a) Median = (30 + 31) ÷ 2 = 30.5 [1]. First quartile = median of the lower six = (27 + 27) ÷ 2 = 27; third quartile = median of the upper six = (34 + 35) ÷ 2 = 34.5 [1]. IQR = 34.5 − 27 = 7.5 [1]. (b) Upper limit = 34.5 + 1.5 × 7.5 = 45.75 [1]; 52 exceeds it, so 52 is an outlier [1]. (c) 23 and 36 [1]. (d) Mean = 384 ÷ 12 = 32.0 cm [1]; mode = 27 cm [1]. (e) Height can take any value in a range, while ABO falls into four categories [1]. Many genes and the environment, such as nutrition, contribute [1]. [10] Examiner insight: show the limit calculation; stating “52 is far from the others” is not a test.

Where marks are usually lost

• Saying a dominant allele “overpowers” the recessive instead of describing the polypeptide.
• Writing genotypes such as H and h without the X.
• Treating plasticity as a mutation or a change of genotype.
• Giving a ratio when a probability, a count or a percentage is asked for.
• Forgetting that a male passes his Y, not an X, to his sons.
• Naming an outlier without calculating the 1.5 × IQR limits.
• Saying a pedigree is “recessive” without ruling out other patterns.
• Confusing codominance with incomplete dominance.

Next steps

• Revise with the revision notes and study guide.
• Visit the IB Biology hub and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization, section D3.2 Inheritance (statements D3.2.1 to D3.2.15).