IB Biology · Theme D Continuity and change · Organisms
D3.2 Inheritance
Each parent gives one allele of every gene through a haploid gamete; the zygote gets two. Dominance, codominance, multiple alleles and sex linkage set how those alleles show in the phenotype. At HL, two genes at once: independent assortment, linkage, recombinants and the chi-squared test.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Biology guide (first assessment 2025, updated May 2026 for 2028).
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D3.2.1 Haploid gametes fuse to make a diploid zygote
A haploid gamete carries one set of chromosomes: one copy of each autosomal gene.
Two gametes fuse into a diploid zygote: two copies of each gene, one per parent.
Every body cell made by mitosis carries the same two copies.
This pattern is shared by all eukaryotes with a sexual life cycle: plants, fungi, animals.
Students often think a parent passes on its whole genotype. In fact each gamete carries one allele of each gene.
Students often think only animals have gametes and zygotes. In fact plants and fungi with sexual life cycles do too.
D3.2.2 How a plant cross is done
The P generation is crossed; offspring are the F1; F1 crossed or selfed give the F2.
Male gametes are in pollen, female gametes in the ovary, so pollination is needed.
The breeder moves pollen to the chosen stigma, often removing the recipient's anthers first.
Peas make both gametes on one plant, so they can self-pollinate and self-fertilise.
A Punnett grid lists one parent's gametes across the top and the other's down the side; the cells give offspring genotypes and ratios. Such crosses breed new crop and ornamental varieties.
Students often think the F2 comes from crossing F1 back to the parents. In fact F2 comes from crossing F1 with each other or selfing them.
Students often think a cross needs two separate plants. In fact a pea flower can fertilise itself.
D3.2.3 Genotype: the alleles you inherited
The genotype is the combination of alleles an organism inherited, written like Tt.
A gene is DNA at a locus; an allele is one version of it.
Homozygous means two copies of the same allele: TT or tt.
Heterozygous means two different alleles, Tt, giving two gamete types in equal numbers.
Students often use gene and allele as one word. In fact a gene is the locus; alleles are its alternative versions.
Students often think heterozygous means having two different genes. In fact it means two different alleles of one gene.
D3.2.4 Phenotype: genotype plus environment
The phenotype is the observable traits: structure, physiology, behaviour.
Some traits are genotype only, such as ABO blood group.
Some are environment only, such as scars or the language first spoken.
Many result from interaction, such as height or skin colour.
Students often think the same genotype must give the same phenotype. In fact environment shapes many traits.
Students often think acquired traits can be inherited. In fact they do not change the genotype, so they are not passed on.
D3.2.5 Why one dominant allele is enough
A dominant allele gives the same phenotype in homozygotes and heterozygotes.
Typically it codes for a functional protein, and one copy makes enough.
A recessive allele shows only when homozygous; often it codes for no working protein.
In a heterozygote the recessive allele is still present and can be passed on.
Students often think the dominant allele destroys the recessive one. In fact it is still there; the dominant allele's protein simply masks it.
Students often think the dominant phenotype must be the common one. In fact dominance is about the heterozygote, not frequency.
D3.2.6 Phenotypic plasticity: adjusting without changing genes
Phenotypic plasticity is developing traits suited to the environment experienced.
It works by changing gene expression, not the genotype.
The change may be reversible in the individual's lifetime, as with skin darkening in sunlight.
Students often think the environment alters the genes. In fact it alters which genes are expressed.
Students often think a plastic change is permanent and inherited. In fact it may reverse, and it does not pass through gametes.
D3.2.7 PKU: a recessive condition
Phenylketonuria is caused by a recessive allele of an autosomal gene.
The gene codes for the enzyme converting phenylalanine to tyrosine.
Homozygotes lack the enzyme, so phenylalanine builds up and can damage the developing brain.
Two carriers are unaffected; each child has a 1 in 4 chance of PKU.
Students often think PKU comes from eating too much phenylalanine. In fact it comes from lacking the enzyme that breaks it down.
Students often think 3:1 means exactly one in four children is affected. In fact each child independently has a 1 in 4 chance.
D3.2.8 Many alleles in the gene pool, two in each person
The gene pool is all alleles of all genes in an interbreeding population.
A SNP is a position where one nucleotide differs between people; most alleles differ this way.
A gene can have multiple alleles in the gene pool.
A diploid individual inherits only two of them.
Students often think every gene has exactly two alleles. In fact any number can exist in the population.
Students often think a person can carry several alleles of one gene. In fact a person carries at most two.
D3.2.9 ABO blood groups: three alleles
The blood group gene has three alleles: IA, IB and i.
IA and IB are codominant with each other; both are dominant to i.
IAIA or IAi give A; IBIB or IBi give B; IAIB gives AB; ii gives O.
Students often think IA is dominant to IB. In fact they are codominant, so IAIB is group AB.
Students often think A and B parents cannot have an O child. In fact two heterozygotes can each pass on i.
D3.2.10 Codominance versus incomplete dominance
In codominance the heterozygote shows both phenotypes at once: a dual phenotype.
Blood group AB (IAIB) has both A and B antigens.
In incomplete dominance the heterozygote is intermediate between the homozygotes.
In Mirabilis jalapa, red × white gives pink; pink × pink gives red, pink, white in 1:2:1.
Students often treat the two as the same or swap the labels. In fact codominance is both together; incomplete dominance is in between.
Students often think pink flowers have a merged pink allele. In fact they carry one red and one white allele, and both reappear.
D3.2.11 Sex determination: the sperm decides
Humans have one pair of sex chromosomes: females XX, males XY.
Every egg carries an X; a sperm carries X or Y, so the sperm decides.
The X chromosome carries far more genes than the small Y.
Students often think the mother decides the sex. In fact the egg is always X; the sperm brings X or Y.
Students often think X and Y carry matching genes. In fact most X genes have no partner on the Y.
D3.2.12 Haemophilia is X-linked
A sex-linked gene is on a sex chromosome, almost always the X.
Write alleles as superscripts: XH, Xh; males are XHY or XhY.
A male with one Xh has haemophilia; there is no second X to mask it.
An XHXh female is an unaffected carrier; XhXh is rare.
Students often think haemophilia passes from father to son. In fact a son gets his father's Y; the X goes to daughters.
Students often think a carrier female is mildly affected. In fact her dominant allele supplies the clotting factor.
D3.2.13 Reading a pedigree, and inductive versus deductive reasoning
From 2028 this is supplied in the Biology data booklet — you need to recognise and interpret it, not reproduce it from memory.
An affected child of unaffected parents shows a recessive condition and two heterozygous parents.
An affected daughter with an unaffected father rules out X-linked recessive inheritance.
Close relatives often share a recessive allele from a common ancestor, raising homozygous children.
Finding the pattern from cases is inductive; applying it to one genotype is deductive.
Students often think a condition that skips a generation must be dominant. In fact skipping is the mark of a recessive allele carried silently.
Students often think cousin marriages create new mutations. In fact they raise the chance of pairing an allele already in the family.
D3.2.14 Continuous variation, polygenes and averages
Continuous variation takes any value in a range, such as skin colour or height.
Polygenic inheritance: several genes each add a small effect, giving many grades.
Environment adds to it: sunlight affects skin colour.
A discrete variable falls into separate classes, such as ABO blood group.
The mean is total divided by count; the median is the middle value in order; the mode is the most frequent value.
Students often think skin colour varies because one gene has many alleles. In fact several genes plus the environment produce the gradient.
Students often think blood group is continuous because there are several groups. In fact it is discrete: four classes, no intermediates.
D3.2.15 Box-and-whisker plots
From 2028 this is supplied in the Biology data booklet — you need to recognise and interpret it, not reproduce it from memory.
The plot shows minimum, first quartile, median, third quartile, maximum and outliers.
The box runs from first to third quartile with a line at the median.
The IQR is third quartile minus first quartile: the middle half.
An outlier lies more than 1.5 × IQR beyond the first or third quartile.
Students often measure outliers from the median. In fact the rule uses 1.5 × IQR beyond the quartiles.
Students often think the box shows the full range. In fact it shows the interquartile range; whiskers reach the non-outlier extremes.
D3.2.16 Segregation and independent assortment come from meiosis HL
Segregation: a gene's two alleles part in anaphase I; Aa gives A and a equally.
Independent assortment: each bivalent orients at random in metaphase I, independently of the others.
So AaBb with unlinked genes gives AB, Ab, aB and ab gametes in equal proportions.
Students often place independent assortment in anaphase II or blame crossing over. In fact it is the random orientation of bivalents in metaphase I.
Students often think alleles from one grandparent stay together. In fact unlinked genes combine at random.
D3.2.17 Dihybrid ratios: 9:3:3:1 and 1:1:1:1 HL
AaBb × AaBb for unlinked genes gives a 9:3:3:1 phenotypic ratio.
Multiply the two 3:1 ratios: 9/16, 3/16, 3/16 and 1/16.
A test cross AaBb × aabb gives 1:1:1:1, one class per gamete type.
This is Mendel's second law; it fails for linked genes, so biological laws have exceptions.
Students often think 9:3:3:1 is a genotype ratio. In fact it is phenotypic; there are nine genotypes behind it.
Students often think every dihybrid cross gives 9:3:3:1. In fact only unlinked double heterozygotes crossed together do.
D3.2.18 Loci: where a gene sits HL
A locus is a gene's fixed position on a particular chromosome.
A database records each gene's chromosome, position and polypeptide product.
Genes on different chromosomes are unlinked; genes close together on one chromosome are linked.
Students often think different alleles sit at different loci. In fact the locus is fixed; only the allele there varies.
D3.2.19 Linked genes travel together HL
Autosomal linkage: two genes with loci on the same autosome.
The allele combination on each chromosome passes on as a unit.
Only crossing over between the loci in prophase I separates them; closer loci, less often.
Show the alleles beside vertical lines representing the homologous chromosomes.
Students often confuse linkage with sex linkage. In fact autosomal linkage is about two genes sharing an autosome.
Students often think linked genes can never recombine. In fact crossing over between the loci makes recombinants, at a rate set by distance.
D3.2.20 Spotting recombinants in a test cross HL
A recombinant carries a combination of alleles unlike either parental combination.
In a test cross the recessive parent adds nothing visible, so each offspring reveals one gamete.
For unlinked genes, recombinants are half the offspring, by independent assortment.
Linked genes: recombinants are the two less frequent classes, made by crossing over.
Students often call any offspring unlike its parents a recombinant. In fact it is one whose allele combination differs from the heterozygous parent's two original combinations.
D3.2.21 Chi-squared: do the numbers fit the ratio? HL
The null hypothesis: no significant difference between observed and expected; differences are chance.
Expected numbers are the total split in the predicted ratio, such as 9/16 and 3/16.
Sum (observed − expected)² ÷ expected; degrees of freedom = classes minus one.
If chi-squared exceeds the critical value at p = 0.05, reject the null hypothesis.
The F2 counted is a sample standing for the population; in other experiments the sample is the repeated measurements.
Students often think a large chi-squared means a good fit. In fact a large value means the observed numbers differ from expected.
Students often make the null hypothesis the linkage claim. In fact the null hypothesis is that there is no significant difference.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement correctly describes the means of inheritance in eukaryotes that have a sexual life cycle?
Answer and reasoning
Each parent produces haploid gametes by meiosis, and their fusion forms a diploid zygote with two copies of each autosomal gene — Meiosis halves the chromosome number so that each gamete carries one copy of each autosomal gene; fertilization restores the diploid number, giving the zygote two copies, one from each parent. This pattern is common to all sexually reproducing eukaryotes.
Each parent's gametes carry both of its alleles of every gene, so the zygote's traits blend the two parental genotypes — A student who thinks a parent passes on its whole genotype picks this. Gametes are haploid and carry only one allele of each gene, chosen at random in meiosis, and the alleles remain separate in the zygote rather than blending.
Haploid gametes fusing to form a zygote is the pattern in animals, while plants pass on their genes through pollen and seeds instead — A student who files plant reproduction as a separate process picks this. Pollen contains male gametes and the ovule contains the female gamete; their fusion forms a diploid zygote, so plants follow the same haploid–diploid pattern as animals.
Each gamete carries the parent's single copy of each gene, so the zygote formed at fertilization has one copy of each gene — A student who pictures a gene as a single object in the cell picks this. A diploid parent has two copies of each autosomal gene and puts one into each gamete; the zygote therefore has two copies, one on each homologous chromosome.
2 A pea plant has the genotype Tt for the gene controlling stem height. Which statement about this plant is correct?
Answer and reasoning
It has two separate genes controlling stem height, called T and t — A student who uses gene and allele as synonyms picks this. There is one gene for stem height; T and t are alleles, different versions of that gene, and the genotype is the combination of alleles inherited.
It is heterozygous because its phenotype differs from both its parents — A student who attaches 'hetero' to the parents rather than to the alleles picks this. Heterozygous describes the two alleles of a gene being different, whatever the parents looked like; a Tt plant is tall, like a TT plant.
It is heterozygous, since it has two different alleles of one gene — T and t are two alleles, alternative forms of the one gene for stem height, occupying the same locus on the two homologous chromosomes. Having two different alleles is the definition of heterozygous.
It carries only the T allele in its body cells, with t only in its gametes — A student who thinks a cell holds a single copy of each gene picks this. Every diploid body cell of the plant carries both T and t; meiosis then places one or the other into each haploid gamete.
3 In peas the allele T for tall stems is dominant to the allele t for dwarf stems, and TT and Tt plants are equally tall. What is the reason that the heterozygote is not shorter than the homozygous dominant plant?
Answer and reasoning
One T allele makes enough functional protein for full stem growth, so the t allele has no observable effect — This is why a homozygous-dominant genotype and a heterozygous genotype produce the same phenotype: the dominant allele codes for a functional protein and a single copy makes enough of it, while the recessive allele contributes a non-functional or absent product.
The T allele switches off the t allele in the heterozygote, so the plant effectively has two working T alleles — A student who imagines dominance as one allele suppressing the other picks this. The t allele is not switched off or altered; it simply makes a product that does not contribute, and one T allele is sufficient on its own.
Tall plants are more common than dwarf plants, so the tall allele is the stronger one and takes over — A student who equates dominance with frequency or strength picks this. Dominance is defined by the phenotype of the heterozygote, and has nothing to do with how common an allele is in the population.
The heterozygote is in fact slightly shorter, because a single T allele makes half as much of the protein as two — A student who reasons that one allele must give half the effect picks this, but the stem says the plants are equally tall. With complete dominance one allele produces enough protein for the full phenotype; a reduced amount does not reduce the trait.
4 PKU usually appears in a child whose parents are both unaffected. Which statement correctly describes phenylketonuria (PKU)?
Answer and reasoning
It is caused by a dominant allele of an autosomal gene, which is why the condition is expressed in every person who carries it — A student who reasons that a condition that is expressed must be dominant picks this. PKU appears only in people homozygous for the mutant allele; heterozygous carriers are unaffected, which is the signature of a recessive allele.
It is caused by a recessive allele that prevents the body making phenylalanine, so affected people need extra phenylalanine in their diet — A student who takes the name to mean a shortage of phenylalanine picks this. The missing enzyme converts phenylalanine to tyrosine, so phenylalanine accumulates and the treatment is a diet low in phenylalanine, not rich in it.
It is caused by a recessive allele of an autosomal gene coding for the enzyme that converts phenylalanine to tyrosine — PKU is a recessive genetic condition due to mutation in an autosomal gene for the enzyme needed to convert phenylalanine to tyrosine. A homozygous recessive person lacks the functional enzyme, so phenylalanine accumulates.
It is caused by a recessive allele on the X chromosome, so it is expressed in males far more often than in females — A student who assumes that any recessive disorder appearing in the children of unaffected parents is sex-linked picks this. The PKU gene is autosomal, so the condition is equally frequent in males and females.
5 A woman with blood group A whose genotype is IAi has children with a man of blood group B whose genotype is IBi. Which blood groups are possible in their children, and with what probabilities?
Answer and reasoning
Groups A, B and AB, each 1 in 3, but not group O, as neither parent is group O — A student who thinks a child's blood group must match a parent's picks this. Both parents carry the recessive allele i, and an ii child, group O, is produced in 1 in 4 fertilizations.
Groups AB, A, B and O are all possible, each with a probability of 1 in 4 — The Punnett grid for IAi × IBi gives IAIB (AB), IAi (A), IBi (B) and ii (O), one cell each. IA and IB are codominant and both dominant to i, so all four ABO groups are possible with equal probability.
Groups A and AB only, each 1 in 2, because the IA allele is dominant to IB — A student who forces the three alleles into a dominant/recessive hierarchy picks this. IA and IB are codominant, so IAIB is group AB, and IBi is group B; the option also ignores the ii children.
Groups A and B only, each 1 in 2, since each parent passes on its own group's allele — A student who treats blood group as a unit trait that a child takes from one parent or the other picks this. Each parent is heterozygous and passes on either of its two alleles with equal probability, so four genotypes, IAIB, IAi, IBi and ii, are possible, giving all four blood groups.
6 Which statement about sex determination and the sex chromosomes in humans is correct?
Answer and reasoning
The sex chromosome carried by the sperm determines the sex of the zygote, and the X chromosome carries far more genes than the Y — All eggs carry an X chromosome, whereas a sperm carries either an X or a Y, so the sperm decides whether the zygote is XX or XY. The Y chromosome is much smaller and carries far fewer genes than the X.
The sex chromosome carried by the egg determines the sex of the zygote, and the X chromosome carries far more genes than the Y does — A student who assumes the mother decides the child's characteristics picks this. Every egg carries an X, so the egg cannot vary the outcome; it is the X or Y in the sperm that determines the sex of the zygote.
The sex chromosome carried by the sperm determines the sex of the zygote, and the X and Y chromosomes carry the same genes — A student who treats X and Y as a fully homologous pair picks this. The Y is much smaller than the X and carries relatively few genes; most X-linked genes have no counterpart on the Y, so males have a single copy of them.
Both parents contribute equally to determining the sex of the zygote, and the Y chromosome carries far more genes than the X — A student who reasons that both parents contribute a sex chromosome and so both decide picks this. Only the sperm varies between X and Y, and it is the X, not the Y, that carries far more genes.
7 A geneticist examines the pedigrees of many families in which a disorder occurs and concludes that it is inherited as an autosomal recessive condition. She then uses this conclusion to state that a particular unaffected couple in one pedigree, who have an affected son, must both be heterozygous. Which description of the two steps of reasoning is correct?
Answer and reasoning
The first step is inductive, a general conclusion from observed cases; the second is deductive, applying that conclusion to specific individuals — Scientists draw general conclusions by inductive reasoning when they base a theory on observations of some but not all cases, as in inferring the pattern of inheritance from pedigrees. Using that theory to deduce the genotypes of particular individuals is deductive reasoning.
The first step is deductive, because it produces a theory; the second is inductive, because it uses the observations recorded in the pedigree — A student who has the two terms reversed picks this. Producing a general theory from particular observations is induction; working from the theory to a conclusion about specific people is deduction, even though the pedigree is consulted in both.
Both steps are inductive, because both rely entirely on observations recorded in pedigree charts rather than on any general theory — A student who labels any reasoning from data as inductive picks this. The second step starts from the general conclusion already reached and applies it to a specific case, which is deductive reasoning.
The second step is invalid, because unaffected parents who have an affected child show that the disorder must be dominant — A student who associates a disorder appearing in a child with dominance picks this. An affected child of two unaffected parents is the signature of a recessive allele carried by both, so the deduction that they are heterozygous is sound.
8 For a class's heights, the first quartile is 150 cm, the median is 158 cm, the third quartile is 166 cm, and the minimum and maximum are 130 cm and 192 cm. Above what height would a data point be categorized as an outlier on a box-and-whisker plot?
Answer and reasoning
190 cm — The interquartile range is Q3 − Q1 = 166 − 150 = 16 cm, so 1.5 × IQR = 24 cm. A point is an outlier if it is more than 1.5 × IQR above the third quartile: 166 + 24 = 190 cm. The maximum of 192 cm is therefore an outlier and would be plotted separately.
182 cm — A student who measures 1.5 × IQR from the median rather than from the third quartile gets 158 + 24 = 182 cm. The rule uses the quartiles: the upper limit is Q3 + 1.5 × IQR.
174 cm — A student who adds 1.5 × IQR to the first quartile instead of the third gets 150 + 24 = 174 cm. The upper limit is measured from the third quartile, and the lower limit from the first.
259 cm — A student who takes the range (192 − 130 = 62 cm) as the interquartile range gets 1.5 × 62 = 93 and 166 + 93 = 259 cm. The IQR is the length of the box, Q3 − Q1 = 16 cm, not the full range of the data.
9 Two pea plants heterozygous for two unlinked genes (RrYy) are crossed. Why is the expected phenotypic ratio among their offspring 9:3:3:1? HL
Answer and reasoning
The 16 cells of the Punnett grid contain four genotypes in the proportions 9:3:3:1, and each of these genotypes gives a different phenotype — A student who reads 9:3:3:1 as a genotypic ratio picks this. The grid contains nine different genotypes; 9:3:3:1 appears only when they are grouped by phenotype, the 9 class alone containing four genotypes.
Each parent makes four types of gamete and these combine with the single gamete type of the other parent in equal numbers, giving 9:3:3:1 — A student who has confused the two dihybrid crosses picks this. Four gamete types meeting a single gamete type describes the test cross with a double homozygous recessive, which gives 1:1:1:1; here both parents make four gamete types.
Each gene independently gives 3 dominant : 1 recessive, and multiplying the probabilities gives 9/16, 3/16, 3/16 and 1/16 for the four phenotype combinations — Because the genes assort independently, the chance of round is 3/4 and of yellow 3/4, so round yellow is 3/4 × 3/4 = 9/16; round green and wrinkled yellow are each 3/4 × 1/4 = 3/16; wrinkled green is 1/4 × 1/4 = 1/16. The same result comes from a 16-cell Punnett grid of the four gamete types from each parent.
Alleles inherited together stay together, so each parent makes only RY and ry gametes and these combine to give the four phenotype classes — A student who thinks maternal and paternal alleles travel as sets picks this. Two gamete types from each parent would give only two phenotypes (round yellow and wrinkled green) in a 3:1 pattern, from three genotypes; independent assortment produces four gamete types, which is what generates the four phenotype classes in the 9:3:3:1 ratio.
10 In a fruit fly the loci for body colour and wing shape are on the same autosome. A fly inherited one chromosome carrying the alleles G and W and the homologous chromosome carrying g and w, shown as G W and g w beside two vertical lines. Why does this fly not produce GW, Gw, gW and gw gametes in equal numbers? HL
Answer and reasoning
The two genes are permanently joined together on the chromosome, so only GW and gw gametes can ever be formed by this fly — A student who takes 'linked' to mean inseparable picks this. Crossing over between the loci exchanges segments between non-sister chromatids, so recombinant Gw and gW gametes are produced, though at less than 50%.
The alleles on each chromosome are passed on together unless crossing over between the loci separates them, which occurs in only some meioses — Linked alleles fail to assort independently because they are on the same DNA molecule. GW and gw gametes are the majority; Gw and gW recombinant gametes arise only when a chiasma forms between the two loci in prophase I, and the closer the loci the less often that happens.
Because the genes are linked they must be on the X chromosome, so only female flies are able to produce the recombinant gametes — A student who equates linkage with sex linkage picks this. The stem states the loci are on an autosome; autosomal linkage has nothing to do with the sex chromosomes or the sex of the fly.
Because the fly received G and W from one parent and g and w from the other, the alleles from each parent stay together as a set in its gametes — A student who thinks alleles from the same parent travel as a set picks this. Alleles stay together here because the two loci are on the same chromosome, not because of their parental origin; for unlinked genes the parental combinations are broken up by independent assortment, and even for linked genes crossing over produces some recombinants.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
19 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A breeder wants to cross a tall pea plant with a dwarf pea plant to obtain an F1 generation. Which procedure is required, and why?
Answer and reasoning
Place pollen from one plant on the stigma of the other, where it fuses with the female gamete, so there is no need to remove the anthers first — A student who thinks the pollen grain is the gamete and that fertilization happens on the stigma picks this. The female gametes are in the ovary, and unless the anthers are removed the flower's own pollen may fertilize them, spoiling the cross.
Remove the anthers from flowers of one plant, then transfer pollen from the other plant to their stigmas, because a pea flower would otherwise self-pollinate — Peas produce male and female gametes on the same plant and normally self-pollinate. To make a cross the breeder must prevent selfing by removing the anthers and then carry out pollination by hand, transferring pollen (which contains the male gametes) to the stigma so that a pollen tube can reach the female gametes in the ovary.
Grow the two plants next to each other, because a pea flower cannot be fertilized by its own pollen and so has to be cross-pollinated by insects — A student who assumes an organism cannot fertilize itself picks this. Peas readily self-pollinate and self-fertilize, which is exactly why the breeder has to intervene to obtain a cross between two different plants.
Allow the tall plant to self-pollinate and cross its seedlings with the dwarf plant, because the F1 is produced from a parent's own offspring — A student who has confused the generation labels picks this. The F1 generation is the direct offspring of the P generation cross between the tall and dwarf plants; no intermediate selfing is involved.
2 Which list gives a human trait due to genotype only, then a trait due to environment only, then a trait due to interaction between genotype and environment, in that order?
Answer and reasoning
Adult height; a facial scar; skin colour — A student who thinks inherited traits are fixed by the genes alone files height as genotype only. Height is strongly influenced by nutrition and health during growth, so it is an interaction trait, not a genotype-only one.
ABO blood group; adult height; a facial scar — A student who treats any observable trait as part of the heritable phenotype puts the scar with the traits that involve genotype. A scar is due to the environment only, and height, listed here as environment only, has a clear genetic component.
ABO blood group; skin colour; adult height — A student who credits sunlight with all variation in skin colour puts it as environment only. Skin colour depends on several genes as well as on exposure to sunlight, so it is an interaction trait, like height.
ABO blood group; a scar; adult height — ABO blood group is set entirely by the alleles inherited; a scar has no genetic basis; adult height depends on the alleles inherited and on environmental factors such as nutrition during growth, so it results from interaction between genotype and environment.
3 A person's skin darkens after several weeks in strong sunlight and returns to its previous colour over the winter. How is this change best explained?
Answer and reasoning
It is phenotypic plasticity: ultraviolet light causes mutations in the melanin genes, which produce extra pigment until the mutations are repaired in winter — A student who thinks a new trait needs a change in the genes picks this. Phenotypic plasticity is not due to changes in genotype; the base sequences of the alleles are unaltered and only their expression changes.
Phenotypic plasticity: sunlight alters the expression of genes for melanin production, and the change reverses because the genotype is unchanged — Phenotypic plasticity is the capacity to develop traits suited to the environment experienced by varying patterns of gene expression. The alleles are exactly as inherited, so when the environmental signal is removed the expression pattern, and the trait, can revert.
It is an adaptation acquired during life, and because it is adaptive it will be permanent and passed on to the person's children — A student who applies the evolutionary sense of adaptation to an individual picks this. The stem itself shows the change reverses, and since the genotype is unchanged the tan cannot be inherited by offspring.
It is a trait due to environment only, so it forms no part of the person's phenotype and involves no change in gene expression — A student who thinks only inherited traits count as phenotype picks this. Phenotype includes all observable traits however they arise, and a tan is produced by increased expression of the genes for melanin synthesis in response to light.
4 A man and a woman are both unaffected carriers of the PKU allele. They already have one child with PKU. What is the probability that their next child will have PKU?
Answer and reasoning
0.50 — A student who treats the 2 in 4 heterozygous carriers as the children who 'have' PKU picks this. Carriers are unaffected, because one functional allele produces enough enzyme; only the 1 in 4 homozygous recessive (pp) children have PKU.
0.75 — A student who believes PKU is caused by a dominant allele expects three quarters of the children of two heterozygotes to be affected. PKU is recessive, so only the homozygous recessive quarter is affected.
0.00 — A student who reads a 3:1 ratio as a quota picks this, thinking the family has already had its one affected child in four. Each fertilization is independent; the probability is 1 in 4 for every child regardless of earlier children.
0.25 — Both parents are heterozygous (Pp × Pp). The Punnett grid gives PP, Pp, pP and pp, so each fertilization has a 1 in 4 chance of producing a homozygous recessive child with PKU. Each child is an independent event, so the earlier affected child makes no difference.
5 A gene in a human population has eleven known alleles, which differ from one another at single-nucleotide polymorphisms (SNPs). Which statement is correct?
Answer and reasoning
All eleven alleles can exist in the gene pool, but any one individual inherits at most two of them — Any number of alleles of a gene can exist in a gene pool because mutations, including SNPs, have arisen in different lineages. A diploid individual has two copies of the locus, one from each parent, and so carries at most two different alleles.
Only two of the eleven can be true alleles, one dominant and one recessive; the rest must be alleles of other genes — A student who has generalised from introductory two-allele crosses picks this. Multiple alleles of one gene are common; the ABO gene has three and many human genes have far more.
An individual can carry several of the eleven alleles at once, since each SNP creates a separate copy of the gene — A student who confuses multiple alleles in a population with multiple alleles in a person picks this. A SNP is a variation at one nucleotide position within the same locus; it does not add copies, and an individual still has just two.
The eleven alleles are found at eleven different loci, since each allele has a different base sequence — A student who thinks different forms of a gene must sit in different places picks this. All alleles of a gene occupy the same locus; they differ in base sequence, not in position.
6 In the four o'clock plant Mirabilis jalapa, a cross between a red-flowered plant and a white-flowered plant gives only pink-flowered offspring. Two of these pink plants are crossed. What phenotypes are expected in their offspring?
Answer and reasoning
Pink only, because the red and white alleles have merged into a pink allele — A student who thinks the intermediate phenotype means the alleles have blended picks this. Alleles are discrete: the pink plants still carry one red and one white allele, and red and white reappear in the next generation.
Pink and white in a 3:1 ratio, because pink is dominant to white — A student who judges the phenotype seen in the heterozygote to be the dominant one picks this. Neither allele is dominant; pink exists only in heterozygotes, so a pink × pink cross also gives red homozygotes.
Red, pink and white flowers in a 1:2:1 ratio, as both alleles persist — Flower colour shows incomplete dominance: the heterozygote is intermediate. Each pink plant (CRCW) produces CR and CW gametes in equal numbers, so the cross gives CRCR red, CRCW pink and CWCW white in a 1:2:1 ratio, recovering both parental phenotypes.
Flowers with red and white patches, because both alleles are expressed — A student who has merged codominance with incomplete dominance picks this. A dual phenotype with both traits shown separately is codominance; in Mirabilis jalapa the heterozygote is uniformly pink, an intermediate phenotype.
7 In the ABO system the red blood cells of a person with genotype IAIB carry both the A antigen and the B antigen. In Mirabilis jalapa a heterozygote for the flower-colour alleles has pink petals. What do these observations show about the two patterns of inheritance?
Answer and reasoning
Both are examples of the same pattern, in which the alleles are mixed in the heterozygote, so AB and pink are both intermediate phenotypes — A student who files both as 'blending' picks this. Group AB is not intermediate between A and B: both antigens are present in full, which is a dual phenotype. Pink petals are intermediate, with neither allele fully expressed.
Blood group AB shows that IA is dominant to IB, while pink petals show that the red allele is dominant to the white allele — A student who expects every pair of alleles to show simple dominance picks this. If IA were dominant to IB the cells would carry only the A antigen, and if red were dominant the heterozygote would be red, not pink.
Pink is codominance, since both alleles contribute to the colour; AB is incomplete dominance, because the phenotype is between A and B — A student who has the two terms reversed picks this. Codominance means both phenotypes appear separately and completely, as with the A and B antigens; a uniformly pink flower is a single intermediate phenotype, which is incomplete dominance.
Blood group AB is codominance, since both alleles are fully expressed together; pink is incomplete dominance, since the heterozygote is intermediate — The two patterns differ at the phenotypic level. In codominance the heterozygote has a dual phenotype, both antigens present; in incomplete dominance the heterozygote has an intermediate phenotype, neither red nor white. The AB blood type and Mirabilis jalapa are the guide's examples of each.
8 A woman who is a carrier of haemophilia (XHXh) has children with a man who does not have haemophilia (XHY). Their first child is a boy. What is the probability that he has haemophilia?
Answer and reasoning
0.25 — A student who takes the 1 in 4 from the whole Punnett grid (XHXH, XHXh, XHY, XhY) and applies it to a son picks this. Among sons only, the two possibilities are XHY and XhY, so the probability is 1 in 2.
0.50 — A son receives his Y chromosome from his father and his single X from his mother. The mother passes XH or Xh with equal probability, so half of sons on average are XhY and have haemophilia.
1.00 — A student who believes a carrier mother passes the condition to all of her sons picks this. She has one XH and one Xh and passes either with equal probability, so only half of her sons are expected to be affected.
0.00 — A student who thinks haemophilia passes from father to son, and notes that this father is unaffected, picks this. A son inherits his X chromosome from his mother, so a carrier mother can have affected sons whatever the father's genotype.
9 Why is haemophilia far more common in males than in females?
Answer and reasoning
The allele is carried on the Y chromosome, so that a father passes the condition directly to each one of his sons — A student who expects a male-biased condition to pass from father to son picks this. The gene is on the X chromosome; a father gives his Y to sons, so sons inherit the haemophilia allele only from their mother.
Females who inherit the allele can only ever be carriers, because the female body cannot develop the condition — A student who believes females cannot have haemophilia picks this. A female with two Xh alleles does have haemophilia; the condition is rare in females because inheriting Xh from both parents is much less likely.
Males have one X chromosome, so a single recessive allele produces the condition, whereas a female needs two — The haemophilia allele is recessive and on the X chromosome. A male XhY has no second X to carry a dominant allele, so he has haemophilia; a female must be XhXh, which requires an affected father and a carrier or affected mother and is therefore rare.
The Y chromosome carries a second copy of the gene that is more often mutated than the copy on the X — A student who thinks the X and Y carry the same genes picks this. The Y chromosome carries far fewer genes than the X and has no copy of the haemophilia gene, which is precisely why males have only one copy.
10 A man and a woman who are both unaffected by a genetic disorder have three children: an affected daughter, an unaffected son and an unaffected daughter. The affected daughter later has four children with an unaffected man, and all four are unaffected. What can be deduced about the disorder?
Answer and reasoning
It is autosomal dominant, because the daughter shows the disorder while her parents are merely carriers — A student who links 'appears in the child' with 'dominant' picks this. A dominant allele is expressed in everyone who carries it, so an affected child would have an affected parent; carriers who show nothing are heterozygous for a recessive allele.
It is X-linked recessive, since the disorder skipped a generation and none of the affected daughter's children show it — A student who treats any skipped generation as a sign of sex linkage picks this. An X-linked recessive disorder in a daughter requires her father to carry, and therefore show, the allele; her father is unaffected, so the gene must be autosomal.
It arose as a new mutation in the daughter, because parents who do not show a disorder cannot pass it on — A student who thinks the recessive allele is eliminated in an unaffected heterozygote picks this. The allele is intact in each carrier parent and is placed in half their gametes; the daughter received it from both, so no new mutation is needed.
It is autosomal recessive, and both parents of the affected daughter must be heterozygous carriers — Unaffected parents with an affected child shows the allele is recessive and that both parents carry it. An affected daughter must have received a recessive allele from her father; had the gene been X-linked her father would have been affected, so the gene is autosomal. Her unaffected children are expected if her partner is homozygous dominant.
11 Many societies prohibit marriage between close relatives. What is the genetic basis for this prohibition?
Answer and reasoning
Reproduction between close relatives causes new mutations, producing genetic disorders that were not present in the family beforehand — A student who reaches for mutation as the cause of any genetic harm picks this. No new mutation is involved; the risk comes from existing recessive alleles being inherited from both sides of the family.
Close relatives are more likely to carry the same recessive allele inherited from a common ancestor, so their children are more likely to be homozygous for it — Everyone carries some harmful recessive alleles that have no effect in a heterozygote. Two relatives may both have inherited the same one from a shared ancestor, so the chance that a child receives it from both parents, and so has the disorder, is raised.
Recessive alleles that were switched off in each parent become active again when the same allele arrives from both sides of the family — A student who imagines the dominant allele switching off the recessive one picks this. The recessive allele is never inactivated; it simply has no observable effect in a heterozygote, and it is expressed when a child is homozygous for it.
Children of close relatives may inherit more than two alleles of a gene, which disrupts the normal dominant and recessive pattern — A student who confuses multiple alleles in a population with multiple alleles in a person picks this. Any child inherits exactly two alleles of each autosomal gene, one from each parent, whether or not the parents are related.
12 Human skin colour varies continuously across a population, whereas ABO blood group falls into four separate classes. Which explanation of this difference is correct?
Answer and reasoning
Skin colour is controlled by one gene with a very large number of alleles, whereas the blood group gene has only three alleles — A student who confuses multiple alleles with polygenic inheritance picks this. However many alleles a gene has, an individual carries only two, so one gene gives a small number of phenotypes; continuous variation needs many genes acting together.
Skin colour is determined by the environment only, with no genetic contribution, whereas blood group is determined by genotype only — A student who credits sunlight with all the variation picks this. Skin colour has a strong genetic component from several genes; the environment adds to it, which is why it is an example of interaction between genotype and environment.
Skin colour depends on several genes and on sunlight, giving many grades; blood group depends on one gene with discrete alleles — Continuous variation is due to polygenic inheritance and/or environmental factors. Several genes each add to the amount of melanin and exposure to sunlight alters it further, so skin colour is a continuous variable; ABO blood group depends on the two alleles of one gene and is a discrete variable with four categories.
Both are continuous variables, but blood group is recorded in four categories only because the test used cannot detect intermediate values — A student who judges by how data are recorded rather than by whether intermediates exist picks this. There are no intermediate blood groups: a person's red cells carry the A antigen, the B antigen, both or neither, so blood group is a discrete variable.
13 The heights in centimetres of nine students are: 150, 152, 152, 155, 158, 160, 163, 170 and 175. What are the mean, median and mode of these heights?
Answer and reasoning
Mean 158 cm, median 159.4 cm, and mode 152 cm — A student who thinks the median is the calculated average and the mean is the middle value swaps the two. The mean is the sum divided by the number of values, 159.4; the median is the middle value in order, 158.
Mean of 159.4 cm, median 152 cm, mode 158 cm — A student who takes the mode to be the middle value and the median to be the most frequent value swaps those two. The median is the middle value, 158; the mode is the value that occurs most often, 152.
Mean 158.6 cm, with median 158 cm and mode 152 cm — A student who treats the highest and lowest values as outliers and discards both before calculating picks this: (1435 − 175 − 150) / 7 = 158.6. A value is an outlier only if it lies more than 1.5 × IQR beyond a quartile; by that rule none of these nine values is an outlier, and the mean of the full data set is 159.4 cm.
Mean 159.4 cm, median 158 cm, mode 152 cm — The nine values sum to 1435, and 1435 / 9 = 159.4 (to one decimal place). The values are already in order, so the median is the fifth value, 158. The only value that appears twice is 152, so it is the mode.
14 A plant is heterozygous for two genes on different chromosomes (AaBb). Which event in meiosis determines which allele of the first gene ends up in a gamete together with which allele of the second gene? HL
Answer and reasoning
Crossing over between the two genes during prophase I, which exchanges alleles between chromosomes to make the four combinations — A student who credits crossing over with all reshuffling picks this. Crossing over exchanges segments between homologues of the same pair; it cannot move an allele of gene A onto the chromosome carrying gene B, which is a different chromosome.
The random orientation of each bivalent on the equator at metaphase I, which is independent of the orientation of the other bivalent — Which homologue of the A pair faces a given pole has no influence on which homologue of the B pair faces it. The alleles then segregate at anaphase I, so the four combinations AB, Ab, aB and ab are produced in equal proportions: this is independent assortment, the basis of dihybrid ratios.
The separation of sister chromatids at anaphase II, which sends the two alleles of each gene into two different gametes — A student who places the shuffling at the most memorable separation event picks this. Sister chromatids carry identical alleles (apart from crossing over); the alleles of a heterozygote are separated when homologues part at anaphase I, and the combinations are fixed by orientation at metaphase I.
The parental origin of the chromosomes, so that alleles from the plant's mother are packaged together and those from its father together — A student who thinks maternal and paternal chromosomes move as sets picks this. Each bivalent orients independently, so a gamete receives a random mixture of maternal and paternal chromosomes and all four allele combinations are equally likely.
15 Mendel's second law predicts a 9:3:3:1 ratio in the F2 of a dihybrid cross, yet a cross involving two genes close together on the same chromosome gives a ratio far from this. What does this show about the law? HL
Answer and reasoning
The law is universal, so the unexpected ratio must be the result of new mutations or of errors made in scoring the offspring — A student who has learned the law without its conditions picks this. The deviation is real and repeatable: genes close together on one chromosome are inherited together, and no scoring error or mutation is needed to explain it.
The law still applies because all genes assort independently; the unexpected ratio must arise from unequal survival of the different offspring — A student who believes every gene assorts independently picks this. Genes on the same chromosome do not assort independently, because their alleles travel together unless crossing over separates them, so the ratio changes with no difference in survival.
The law applies only to autosomal genes, so this deviation shows that the two genes must be located on one of the two sex chromosomes — A student who equates linkage with sex linkage picks this. Two genes close together on any chromosome, including an autosome, are linked; the stem says nothing about sex chromosomes, and autosomal linkage is the common case.
The law applies only when genes are on different chromosomes or far enough apart to recombine freely; all biological laws have exceptions — Mendel's second law applies only if genes are on different chromosomes or are far enough apart on one chromosome for recombination rates to reach 50%. Linked genes are a genuine exception, and students should recognize that there are exceptions to all biological laws under certain conditions.
16 In a genome database, a student finds that a human gene has its locus on chromosome 11 and that a second gene has its locus a few thousand base pairs away on the same chromosome; each record names a different polypeptide product. Which conclusion is correct? HL
Answer and reasoning
The two genes are linked, because their loci are in close proximity on the same chromosome, and each codes for its own polypeptide — A locus is the specific position of a gene on a chromosome, and a database record gives the chromosome, the position and the polypeptide product. Two genes whose loci are close together on the same chromosome are linked and their alleles tend to be inherited together.
The two records must describe alleles of one gene, because different sequences at nearby positions are alternative forms of the same gene — A student who thinks alleles sit at different positions picks this. All alleles of a gene share one locus; two loci with two different polypeptide products are two different genes, however close together they lie.
The two genes are not linked, because gene linkage applies only to genes that are carried on the X chromosome — A student who equates linkage with sex linkage picks this. Linkage means two loci are on the same chromosome; chromosome 11 is an autosome, so this is autosomal linkage.
The positions given are true only for the person whose genome was sequenced, since the locus of a gene varies between people — A student who thinks gene position varies between individuals picks this. The locus of a human gene is the same in every person; people differ in which alleles they carry at that locus, not in where it is.
17 A fly heterozygous for two linked genes (AaBb), which inherited the alleles A and B on one chromosome and a and b on the other, is test-crossed with an aabb fly. The offspring are 42 AaBb, 38 aabb, 11 Aabb and 9 aaBb. Which offspring are the recombinants, and what produced them? HL
Answer and reasoning
The 42 AaBb and 38 aabb, because these offspring differ in genotype from the homozygous recessive parent — A student who reads 'recombinant' as 'different from a parent' picks this. Recombinants have new combinations of the heterozygote's alleles; AaBb received the parental AB chromosome and aabb the parental ab chromosome, so both are parental classes.
None of them, because two genes that are linked cannot give rise to any new combinations of their alleles — A student who thinks linkage is absolute picks this. Crossing over between the loci in some meioses produces Ab and aB gametes, and these recombinants are visible here as the 11 Aabb and 9 aaBb offspring.
The 11 Aabb and 9 aaBb, produced by crossing over between the two loci in the heterozygous parent — The aabb parent contributes only ab gametes, so each offspring genotype reveals the gamete from the heterozygote. AB and ab gametes give the parental classes (42 and 38); Ab and aB gametes, formed only when crossing over occurs between the loci, give the two smaller recombinant classes (11 and 9, a recombination frequency of 20%).
All four classes, since a test cross of a double heterozygote gives four recombinant types in a 1:1:1:1 ratio — A student who applies the unlinked test-cross ratio to every dihybrid picks this. The 1:1:1:1 ratio requires independent assortment; here the two classes of 42 and 38 far exceed the classes of 11 and 9, which is the pattern of linked genes with a minority of recombinants.
18 A cross between two double heterozygotes gives 160 F2 offspring: 95 show both dominant traits, 28 show the first dominant and the second recessive, 30 show the first recessive and the second dominant, and 7 show both recessive traits. The critical value of chi-squared at p = 0.05 is 7.815 for 3 degrees of freedom and 9.488 for 4 degrees of freedom. What is the outcome of a chi-squared test against a 9:3:3:1 ratio? HL
Answer and reasoning
Expected numbers are 90, 30, 30 and 10; chi-squared is about 1.3, below 7.815, so the difference is significant and the two genes are linked — A student who thinks a value below the critical value indicates a significant difference picks this. Significance requires chi-squared to exceed the critical value; 1.3 is far below 7.815, so the data do not differ significantly from 9:3:3:1.
Expected numbers are 9, 3, 3 and 1; chi-squared is far above 9.488 for 4 degrees of freedom, so the null hypothesis must be rejected — A student who uses the ratio itself as the expected numbers and sets degrees of freedom equal to the number of classes picks this. Expected numbers must be scaled to the sample total of 160, and four classes give 3 degrees of freedom, not 4.
Expected numbers are 90, 30, 30 and 10; chi-squared is about 1.3, below 7.815, so the null hypothesis that the genes are linked is accepted — A student who takes the null hypothesis to be the interesting claim picks this. The null hypothesis states there is no significant difference from the expected ratio, meaning the genes assort independently; a small chi-squared retains that hypothesis and gives no support for linkage.
Expected numbers are 90, 30, 30 and 10; chi-squared is about 1.3, below 7.815, so the null hypothesis is not rejected and the data fit 9:3:3:1 — Expected = 160 × 9/16, 3/16, 3/16, 1/16 = 90, 30, 30, 10. Chi-squared = 25/90 + 4/30 + 0/30 + 9/10 = 0.28 + 0.13 + 0 + 0.90 = 1.31. With four classes there are 3 degrees of freedom; 1.31 is below 7.815, so the difference is not significant at p = 0.05 and the observed results are consistent with independent assortment.
19 A student carries out a chi-squared test on the F2 offspring of a dihybrid cross. Which statement about the test is correct? HL
Answer and reasoning
The null hypothesis is that observed and expected numbers do not differ significantly, and the F2 counted is a sample of the population — The null hypothesis states there is no difference beyond chance between observed and expected results. Statistical testing uses a sample to represent a population: here the sample is the F2 generation that was counted, just as in many experiments the sample is a set of replicated measurements.
The null hypothesis is that the two genes are linked, so that the observed numbers will differ from those expected from the predicted ratio — A student who expects the hypothesis under test to be the interesting claim picks this. The null hypothesis is the hypothesis of no difference; linkage, which would produce a difference, belongs to the alternative hypothesis.
If the calculated chi-squared value is below the critical value at p = 0.05, the null hypothesis has been proved to be true — A student who upgrades 'not rejected' to 'proved' picks this. A small chi-squared shows the data are consistent with the expected ratio; it does not prove the null hypothesis, and a larger sample could still reveal a difference.
The number of degrees of freedom equals the number of phenotype classes, so a cross that produces four classes has four degrees of freedom — A student who sets degrees of freedom to the most obvious number picks this. Degrees of freedom are the number of classes minus one, because once the total is fixed only three of the four class sizes are free to vary, so the critical value is read for 3.
That was your twenty minutes. Real practice on D3.2 is past-paper questions marked against the mark scheme.
What the exam asks of D3.2
Paper 1A asks you to pick genotypes, blood groups or the outcome of a single cross, and to classify a variable as discrete or continuous. Paper 1B gives a pedigree or a box-and-whisker plot to interpret, or dihybrid counts for a chi-squared calculation. Paper 2 uses *deduce* and *predict*: write the gametes, draw the Punnett grid, state the ratio. HL questions use *explain* for why linked genes break the expected ratio, and *evaluate* a chi-squared result: state the hypothesis, compare with the critical value, conclude.
Compiled from the IB Biology guide (first assessment 2025, updated May 2026 for 2028) and our question bank · Specialist review in progress. How these pages are made ·
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