IB Biology 2025 · Theme D · Continuity and change

D3.2 Inheritance. Practice questions with markscheme.

109 original IB-style questions on D3.2, written from the 2025 guide: 54 multiple-choice, 35 short-answer, 13 data-based, 3 extended-response parts, 2 drawing, 2 labelling. Below are two ready-made practice papers built from them, one at standard level (22 marks) and one at higher level (22 marks); switch between them above the paper. Hand one out as a class quiz or homework, or sit it yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.

What the guide asks for

15 statements at SL and HL, 6 additional higher level.

D3.2.1SL / HL Production of haploid gametes in parents and their fusion to form a diploid zygote as the means of inheritance
D3.2.2SL / HL Methods for conducting genetic crosses in flowering plants
D3.2.3SL / HL Genotype as the combination of alleles inherited by an organism
D3.2.4SL / HL Phenotype as the observable traits of an organism resulting from genotype and environmental factors
D3.2.5SL / HL Effects of dominant and recessive alleles on phenotype
D3.2.6SL / HL Phenotypic plasticity as the capacity to develop traits suited to the environment experienced by an organism, by varying patterns of gene expression
D3.2.7SL / HL Phenylketonuria as an example of a human disease due to a recessive allele
D3.2.8SL / HL Single-nucleotide polymorphisms and multiple alleles in gene pools
D3.2.9SL / HL ABO blood groups as an example of multiple alleles
D3.2.10SL / HL Incomplete dominance and codominance
D3.2.11SL / HL Sex determination in humans and inheritance of genes on sex chromosomes
D3.2.12SL / HL Haemophilia as an example of a sex-linked genetic disorder
D3.2.13SL / HL Pedigree charts to deduce patterns of inheritance of genetic disorders
D3.2.14SL / HL Continuous variation due to polygenic inheritance and/or environmental factors
D3.2.15SL / HL Box-and-whisker plots to represent data for a continuous variable such as student height
D3.2.16HL Segregation and independent assortment of unlinked genes in meiosis
D3.2.17HL Punnett grids for predicting genotypic and phenotypic ratios in dihybrid crosses involving pairs of unlinked autosomal genes
D3.2.18HL Loci of human genes and their polypeptide products
D3.2.19HL Autosomal gene linkage
D3.2.20HL Recombinants in crosses involving two linked or unlinked genes
D3.2.21HL Use of a chi-squared test on data from dihybrid crosses

In the bank for D3.2

54 multiple-choice
35 short-answer
13 data-based
3 extended-response parts
2 drawing
2 labelling
28 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

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Biology · topic quiz

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35 minutes22 marks

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Covers D3.2 Inheritance · SL

Name:

In the pedigree, I-1 (blood group A) and I-2 (blood group B) already have one daughter, II-1, with blood group O. What are the possible blood groups of a further child of I-1 and I-2?

[1]

O only
A or B only
AB only
O, A, B or AB

Soybean (Glycine max) flowers contain both anthers and an ovary, and each flower normally pollinates itself before it opens. Purple flower colour (P) is dominant to white flower colour (p). A breeder crossed true-breeding white-flowered plants, used as the female parent, with true-breeding purple-flowered plants. The anthers of each white flower were to be removed while still immature, and pollen from a purple-flowered plant was then brushed onto its stigma. The seeds were collected and grown as the F1 generation.

(a)In one batch of white flowers, the anthers were removed only after they had released their pollen. Predict, with a reason, the flower colours of the F1 plants grown from the seeds of this batch.[2]

(b)A second breeder made the reciprocal cross, using purple-flowered plants as the female parent and white-flowered plants as the pollen donor, and made the same mistake with one batch of flowers. Predict whether the mistake can be detected from the flower colours of the F1 plants, giving a reason.[2]

(c)The second breeder lets every F1 plant self-pollinate and grows a large number of seeds from each one. Predict the flower colours among the offspring of an F1 plant that came from the intended cross, and among the offspring of an F1 plant that came from the mistake.[2]

The pedigree chart shows a condition caused by a dominant allele. Use Xᴰ for the allele causing the condition and Xᵈ for the normal allele.

(a)State the genotype of individual I-1.[1]

(b)Explain how the children of I-1 and I-2 show that the allele is on the X chromosome rather than on an autosome.[2]

(c)Deduce, with a reason, the genotype of individual II-2.[2]

In rabbits (Oryctolagus cuniculus), coat colour is controlled by one gene with several alleles. Three of them are Cᶠ, which gives full colour, Cᵍ, which gives a chinchilla (silver-grey) coat, and Cᵃ, which gives an albino (all white) coat. A breeder recorded the offspring of two crosses over several litters.

Cross	Parents	Offspring
1	chinchilla, from a line that has bred true for many generations × albino	17 chinchilla
2	full colour × chinchilla	13 full colour, 7 chinchilla, 6 albino

(a)Deduce, with reasons, the order of dominance of the three alleles.[3]

(b)Deduce the genotypes of the two parents in cross 2.[2]

(c)Each full-colour offspring of cross 2 carries Cᶠ with either Cᵍ or Cᵃ. Outline a cross that would show which of these a particular full-colour offspring carries, and the results expected in each case.[2]

The pedigree chart shows the inheritance of haemophilia, an X-linked recessive condition, in one family. Squares are males, circles are females; filled symbols are affected, and a dot indicates a confirmed carrier. Identify: (i) the parent from whom II-4 inherited her haemophilia allele; (ii) the females in generation II who could pass the haemophilia allele to a son; (iii) one male in generation II who cannot pass the haemophilia allele to any of his children.

[3]

II.

III.

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One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. D3.2.9 [1]

D: their daughter's group O shows both parents carry the i allele, so I-1 is Iᴬi and I-2 is Iᴮi; crossing Iᴬi × Iᴮi can produce IᴬIᴮ (AB), Iᴬi (A), Iᴮi (B) or ii (O);

2. D3.2.2, D3.2.5

(a) [2]
- white-flowered plants appear in the F1 (some or all of the plants, with any others purple-flowered);
- the white-flowered plants grew from seeds formed by self-fertilization, when the flower's own pollen reached its stigma (pp × pp gives pp);
- the purple-flowered plants grew from seeds of the intended cross, which are all Pp;
- no plant from the intended cross can be white, because every one of its seeds receives P from the purple (PP) parent, OWTTE;
Accept a mixture of purple and white plants, or all white plants, when justified. Award [1 max] for a correct prediction with no reason.
(b) [2]
- it cannot be detected: all of the F1 plants are purple-flowered;
- seeds formed by self-fertilization are PP, while seeds from the intended cross are Pp;
- PP and Pp plants have the same phenotype because P is dominant / one P allele is enough to give purple flowers, OWTTE;
(c) [2]
- an F1 plant from the intended cross (Pp): about 3 purple-flowered : 1 white-flowered offspring;
- an F1 plant from the mistake (PP): all purple-flowered offspring;
- so any F1 plant with white-flowered offspring is a true hybrid, and the plants with only purple-flowered offspring came from the mistake, OWTTE;
Both predictions are needed for [2].

3. D3.2.11, D3.2.13

(a) [1]
- XᴰY;
(b) [2]
- a father passes his X chromosome to all of his daughters and his Y chromosome to all of his sons;
- both daughters (II-2 and II-4) are affected and neither son (II-3 or II-5) is, which would not be expected if the allele were on an autosome / an autosomal dominant father would affect sons and daughters equally;
(c) [2]
- XᴰXᵈ / heterozygous;
- she is affected so carries Xᴰ, and two of her children (III-3 and III-4) are unaffected so she must also carry Xᵈ;

4. D3.2.5, D3.2.8

(a) [3]
- Cᵍ is dominant to Cᵃ: every offspring of cross 1 receives Cᵃ from the albino parent, yet all are chinchilla;
- Cᶠ is dominant to Cᵃ: the full-colour parent in cross 2 must carry Cᵃ, since it has albino offspring, yet it has full colour;
- Cᶠ is dominant to Cᵍ: half of the offspring of cross 2 are full colour (2 : 1 : 1, not 1 : 2 : 1), so the CᶠCᵍ offspring, a quarter of the total, must show full colour;
- order: Cᶠ > Cᵍ > Cᵃ;
The order alone, with no reasons, scores [1].
(b) [2]
- full-colour parent CᶠCᵃ;
- chinchilla parent CᵍCᵃ;
- both must carry Cᵃ, because the albino offspring (CᵃCᵃ) received Cᵃ from each parent;
(c) [2]
- cross it with an albino (CᵃCᵃ) rabbit;
- if it is CᶠCᵍ: about half full colour and half chinchilla offspring, and no albinos;
- if it is CᶠCᵃ: about half full colour and half albino offspring, and no chinchillas;
- count the offspring of several litters, since a small litter could lack one class by chance, OWTTE;
Award the second mark only if the expected results for both genotypes are given.

5. D3.2.12 [3]

(i) I-2: her father I-1 is unaffected (XᴴY) so gave her Xᴴ, so her Xʰ must have come from her mother;
(ii) II-1 and II-4, both needed: II-4 is a known carrier, and II-1 may be one, as her carrier mother I-2 passed her either Xᴴ or Xʰ; a son's X always comes from his mother;
(iii) II-2 / II-5: an unaffected male is XᴴY, so he passes Xᴴ to his daughters and Y to his sons;

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Biology · topic quiz

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Covers D3.2 Inheritance · HL

Name:

The figure shows a breeding programme in a plant for two linked genes, A and B. The alleles A and B are dominant. Each pair of vertical bars is a pair of homologous chromosomes, with the alleles carried at the two loci shown beside them. True-breeding AABB plants were crossed with aabb plants, and the F1 plants were then crossed with aabb plants. Which row shows the first stage at which crossing over between the two loci can produce gametes with a combination of alleles not carried by either P-generation plant, and the first generation that contains plants with recombinant phenotypes?

[1]

Row	First stage at which crossing over can produce recombinant gametes	First generation with recombinant phenotypes
A.	stage 1	F1
B.	stage 1	offspring of the test cross
C.	stage 3	offspring of the test cross
D.	stage 3	F1

Row D
Row B
Row C
Row A

The pedigree chart shows a rare sex-linked recessive condition. Assume that I-2 and the two people who married into the family, II-1 and II-4, do not carry the allele. III-2 and III-4 each have a son with an unaffected man. What is the probability that the son of III-2 is affected, and what is the probability that the son of III-4 is affected?

[1]

0 and ½
¼ and ¼
¼ and ½
0 and ¼

In snapdragons, crossing red-flowered with white-flowered plants gives all pink offspring. What pattern of inheritance is this?

[1]

Incomplete dominance
Complete dominance
Codominance
Sex linkage

Human height shows continuous variation. Which explanation is correct?

[1]

Height is controlled by one gene with two alleles
Height is controlled by a single gene with many alleles
Height is determined entirely by environmental factors
Height is influenced by many genes and by the environment

A person has blood group AB. What does this indicate about the ABO alleles they carry?

[1]

They carry the Iᴬ and Iᴮ alleles
They carry two copies of the i allele
They carry three alleles of the gene
They are homozygous for the Iᴬ allele

The pedigree chart shows a condition controlled by a single gene. Assume that the condition is caused by a recessive allele of an autosomal gene. What is the probability that the next child of II-3 and II-4 will be affected?

[1]

In the pedigree, I-1 (XᴴY) and I-2 (XᴴXʰ, a known carrier) have an unaffected daughter, II-1. What is the probability that II-1 is a carrier of the haemophilia allele?

[1]

Adult haemoglobin is a protein built from two α-globin and two β-globin polypeptides. A student used an online database to find one gene coding for each polypeptide. In a locus such as 16p13.3, the first number is the chromosome, p is its short arm and 13.3 is a band on that arm. What can be concluded from the data?

[1]

Gene	Polypeptide product	Locus
HBA1	α-globin	16p13.3
HBB	β-globin	11p15.4

Each polypeptide is translated at the locus of its gene, and the two are then joined together
Genes coding for the polypeptides of one protein can have loci on different chromosomes
Genes coding for the polypeptides of one protein must be next to each other on a chromosome
One gene codes for both polypeptides, and its sequence is split between two chromosomes

Genes A and B are on the same chromosome, close together. What is the expected result of a dihybrid test cross (AaBb × aabb)?

[1]

A 9:3:3:1 ratio of the four possible phenotypes
Mostly parental combinations, with a few recombinants
A 1:1:1:1 ratio of the four possible phenotypes
Only recombinant phenotype combinations in the offspring

10.

Explain the cause of phenylketonuria (PKU) and why early diagnosis of the condition is important.

[3]

11.

Sickle-cell anaemia is caused by a recessive allele of the haemoglobin β-chain gene. Explain how two healthy parents can have a child with sickle-cell anaemia.

[3]

12.

Skin colour can be measured with a reflectometer as a melanin index (arbitrary units): the higher the index, the darker the skin. The melanin index of the inner upper arm was measured in 120 adults from one population. The median was 41, the first quartile 36 and the third quartile 48. The lowest value recorded was 14 and the highest was 70.

(a)Calculate the interquartile range (IQR) of the melanin index.[1]

(b)Using the rule that a value is an outlier if it lies more than 1.5 × IQR beyond the nearer quartile, deduce whether the lowest and highest values should be plotted as outliers.[2]

(c)Explain why the melanin index was measured on the inner upper arm rather than on the face or the back of the hand.[2]

(d)Explain why these data are displayed as a box-and-whisker plot rather than as a bar chart of categories.[2]

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1. D3.2.20 [1]

C: each P plant has the same alleles on both homologues (A with B on both, or a with b on both), so crossing over in its meiosis at stage 1 exchanges identical alleles and makes no new combination; the F1 carries A with B on one homologue and a with b on the other, so a crossover between the loci in its meiosis at stage 3 gives Ab and aB gametes; the F1 plants themselves grew from parental AB and ab gametes and are all AaBb, so recombinant phenotypes first appear among the offspring of the test cross;

2. D3.2.11, D3.2.13 [1]

D: II-2 received a Y chromosome, not an X, from his affected father, so neither he nor his daughter III-2 carries the allele (0); II-3 received her father's X, so she is certainly a carrier; III-4 has a ½ chance of receiving that X from her and, if she is a carrier, a ½ chance of passing it to a son: ½ × ½ = ¼;

3. D3.2.10 [1]

A: intermediate heterozygote = incomplete dominance; codominance would show both parental phenotypes together (as in AB blood);

4. D3.2.14 [1]

D: polygenic inheritance plus environmental influence produces a continuous (normal) distribution, unlike single-gene discrete traits;

5. D3.2.9 [1]

A: Iᴬ and Iᴮ are codominant (both antigens made); i (O) is recessive; ABO also illustrates multiple alleles;

6. D3.2.13 [1]

A: II-3 is affected, so aa; II-4 is unaffected but has an affected son III-2, so he must be Aa; aa × Aa gives Aa and aa in a 1 : 1 ratio, so the probability is ½;

7. D3.2.12, D3.2.13 [1]

C: II-1 always receives Xᴴ from her father; from her mother she receives Xᴴ or Xʰ with equal probability, so there is a ½ chance she is XᴴXʰ;

8. D3.2.18 [1]

B: HBA1 is on chromosome 16 and HBB on chromosome 11, so the genes for two polypeptides of one protein lie on different chromosomes (and are inherited independently); A fails because each polypeptide has its own gene, C is contradicted by the loci, and D fails because translation happens at ribosomes in the cytoplasm, not at the gene;

9. D3.2.19 [1]

B: linked genes do not assort independently: parental combinations dominate; recombinant frequency reflects crossover distance;

10. D3.2.7 [3 max]

PKU is due to a recessive allele, so only homozygotes are affected (the parents are usually unaffected carriers);
the enzyme that converts phenylalanine to tyrosine (phenylalanine hydroxylase) is missing/non-functional;
phenylalanine (from dietary protein) accumulates in the blood/tissues, damaging the developing brain (impairing cognitive development);
(newborn screening allows) a diet low in phenylalanine from birth, which prevents the accumulation and allows (near-)normal development, OWTTE;

11. D3.2.5 [3 max]

both parents are carriers/heterozygous (HbᴬHbˢ), healthy because the normal allele is dominant / one normal allele makes enough normal haemoglobin;
each parent passes Hbˢ with probability ½;
(so) a child has a ¼ chance of being homozygous HbˢHbˢ;
homozygotes make only sickle haemoglobin, causing the disease, OWTTE;

12. D3.2.14, D3.2.15

(a) [1]
- 12 (48 − 36);
(b) [2]
- 1.5 × IQR = 18, so the limits are 36 − 18 = 18 and 48 + 18 = 66;
- both 14 (below 18) and 70 (above 66) are outliers, so each is plotted as a separate point;
(c) [2]
- exposure to sunlight (UV) increases melanin production / causes tanning, an effect of the environment on phenotype;
- the inner upper arm is rarely exposed to the sun, so its colour reflects genotype rather than differences in sun exposure, OWTTE;
(d) [2 max]
- melanin index is a continuous variable that can take any value within a range, so there are no separate categories to plot as bars;
- a box-and-whisker plot summarizes the centre and spread of continuous data (median, quartiles, minimum, maximum and outliers);
- a bar chart suits a discrete variable with a few distinct classes, OWTTE;

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