IB DP Biology: Inheritance: alleles, monohybrid crosses, pedigrees and continuous variation (DP Biology)

This guide covers the first fifteen statements of section D3.2 (Inheritance) in the IB Biology guide, first assessment 2025, for DP Biology. All fifteen statements are SL and HL; none is marked additional higher level. Dihybrid crosses, linkage and chi-squared are taught separately. Revision notes are at /resources/ib-dp-biology-inheritance-alleles-monohybrid-crosses-pedigrees-and-revision-notes/ and questions at /resources/ib-dp-biology-inheritance-alleles-monohybrid-crosses-pedigrees-and-practice/. Chromosome numbers: diversity of organisms. Course hub: /boards/ib/ib-dp/biology/. Checklist: /checklists/ib/ib-dp/biology/. Find weak spots with the free diagnostics.

What this unit covers

Statement	You must be able to
D3.2.1	Explain inheritance as haploid gametes fusing to give a diploid zygote
D3.2.2	Describe crosses in flowering plants; use P, F1, F2 and Punnett grid
D3.2.3 to D3.2.5	Use genotype, phenotype, homozygous, heterozygous, dominant, recessive
D3.2.6	Explain phenotypic plasticity
D3.2.7	Describe phenylketonuria (PKU)
D3.2.8 to D3.2.10	Explain SNPs, multiple alleles, ABO, codominance, incomplete dominance
D3.2.11, D3.2.12	Predict sex and sex-linked inheritance, with haemophilia
D3.2.13	Deduce inheritance patterns from pedigrees
D3.2.14, D3.2.15	Interpret continuous variation; draw box-and-whisker plots

D3.2.1 Gametes and zygotes

Sexually reproducing eukaryotes make haploid gametes by meiosis. Two gametes fuse at fertilisation to form a diploid zygote. This is the means of inheritance in all eukaryotes with a sexual life cycle. A diploid cell holds two copies of each autosomal gene, one on each chromosome of a homologous pair. One copy came from each parent. A gamete holds only one copy.

Worked example. Pea plants have 2n = 14. A pollen nucleus has 7 chromosomes. An egg cell has 7. The zygote has 7 + 7 = 14, in seven homologous pairs. Each seed therefore carries two copies of every autosomal gene.

D3.2.2 Crosses in flowering plants

Use these terms exactly. The P generation is the parents. Their offspring are the F1 generation. Crossing F1 plants with each other gives the F2 generation. A Punnett grid shows the possible gamete combinations.

Pollen contains the male gametes. The female gametes are in the ovary, so a cross needs pollination: pollen must reach the stigma. Pea flowers hold male and female parts together, so a pea plant can self-pollinate and self-fertilise. A controlled cross must prevent that.

1. Choose two parent plants with different traits.
2. Remove the anthers from the flower of the female parent before they release pollen (emasculation).
3. Cover the flower to keep out stray pollen.
4. Transfer pollen from the chosen male parent to the stigma, for example with a small brush.
5. Cover the flower again, let seeds form, then collect and sow them to score the F1.

Such crosses are widely used to breed new crop and ornamental varieties.

Worked example. Pea pods are inflated (allele P) or constricted (allele p). A pure-breeding inflated plant is crossed with a pure-breeding constricted plant. The F1 are all Pp and all inflated. F1 plants are self-pollinated:

	P	p
P	PP	Pp
p	Pp	pp

The F2 is 3 inflated : 1 constricted. From 1,200 F2 plants you expect 900 inflated and 300 constricted.

D3.2.3 to D3.2.5 Genotype, phenotype and dominance

A gene is a heritable factor that controls a trait. An allele is one version of a gene. Pod shape is a gene; P and p are its alleles. The genotype is the combination of alleles an organism inherited. Two identical alleles make an organism homozygous; two different alleles make it heterozygous.

The phenotype is the set of observable traits. It results from the genotype and environmental factors. Human examples:

• Genotype only: ABO blood group.
• Environment only: a scar, or the language you speak.
• Both: adult height, because genes set a potential that nutrition and illness can limit; and PKU, where diet changes the outcome of the same genotype.

A dominant allele has the same effect on phenotype in a heterozygote as in a homozygote. A recessive allele shows its effect only in a homozygote. PP and Pp therefore give the same phenotype. A usual reason is that the dominant allele codes for a working polypeptide, often an enzyme. One working allele makes enough product. The recessive allele codes for a faulty polypeptide or none.

D3.2.6 Phenotypic plasticity

Phenotypic plasticity means an organism can develop traits that fit the conditions it meets. It works through varying gene expression, not through any change in genotype. The change may be reversible within one lifetime.

• Water crowfoot grows finely divided leaves when submerged and broad flat leaves in air. Same genotype; different gene expression.
• A person who moves to high altitude makes more red blood cells. The change reverses after returning to low altitude.

D3.2.7 Phenylketonuria

PKU is a recessive genetic condition. It is caused by a mutation in an autosomal gene for the enzyme that converts phenylalanine to tyrosine. In a person with two recessive alleles, phenylalanine builds up. Untreated, this damages brain development. A diet low in phenylalanine from early life prevents most of the harm. This changes the environment, not the genotype.

Two carriers (heterozygotes) have a 1 in 4 chance per child of an affected baby. A carrier makes enough working enzyme to be unaffected.

D3.2.8 SNPs and multiple alleles

A single-nucleotide polymorphism (SNP) is a position in the genome where individuals in a population differ by one base. Some SNPs change an amino acid and so an allele’s function. Many change nothing observable.

The gene pool is all the alleles of all genes in a population. A gene may have any number of alleles in the gene pool, but an individual inherits only two, one from each parent.

Worked example. A gene has 4 alleles in a population. The possible genotypes are 4 homozygous plus 6 heterozygous (each pair of different alleles), so 10. Any one person has two alleles only.

D3.2.9 and D3.2.10 ABO, codominance and incomplete dominance

The ABO gene has three alleles: Iᴬ, Iᴮ and i. Iᴬ and Iᴮ are each dominant to i, and codominant with each other.

Phenotype	Genotypes
Group A	IᴬIᴬ, Iᴬi
Group B	IᴮIᴮ, Iᴮi
Group AB	IᴬIᴮ
Group O	ii

With codominance, a heterozygote shows both alleles’ effects: the dual phenotype of group AB. With incomplete dominance, a heterozygote has a phenotype intermediate between the homozygotes. Marvel of Peru (Mirabilis jalapa, four o’clock flower): red (CᴿCᴿ) crossed with white (CᵂCᵂ) gives pink (CᴿCᵂ) in the F1. Pink × pink gives 1 red : 2 pink : 1 white, so the genotype ratio equals the phenotype ratio.

Worked example. A group A mother and a group B father have a group O child. The child is ii, so each parent gave i. The mother is Iᴬi and the father is Iᴮi. Their gametes: Iᴬ or i; Iᴮ or i. Offspring: IᴬIᴮ, Iᴬi, Iᴮi, ii, each 1/4, so groups AB, A, B and O are equally likely.

D3.2.11 and D3.2.12 Sex linkage and haemophilia

Human females are XX and males are XY. All eggs carry an X. A sperm carries X or Y, so the sperm’s sex chromosome decides whether the zygote develops male-typical or female-typical characteristics. The chance of a boy or a girl is 1/2 each. The X carries far more genes than the Y, so most X-linked genes have no matching allele on the Y.

Write alleles as superscripts on X. Haemophilia is a sex-linked recessive disorder in which blood clots poorly. Let Xᴴ be normal clotting and Xʰ be haemophilia.

• Males XᴴY are unaffected and XʰY are affected. A single recessive allele shows, so males are affected more often.
• Females XᴴXᴴ are unaffected, XᴴXʰ are unaffected carriers, and XʰXʰ are affected.

Worked example. A carrier woman (XᴴXʰ) and an unaffected man (XᴴY):

	Xᴴ	Y
Xᴴ	XᴴXᴴ	XᴴY
Xʰ	XᴴXʰ	XʰY

Daughters: half normal, half carriers. Sons: half normal, half affected. Of all children, 1/4 are affected. A father cannot pass an X-linked allele to a son, because he gives his son the Y.

D3.2.13 Pedigree charts

In a pedigree, squares are males, circles are females and shaded symbols are affected. A horizontal line joins partners and vertical lines lead to children. Generations are numbered I, II, III; individuals within them 1, 2, 3.

Clues:

• Two unaffected parents with an affected child: the condition is recessive, and both parents are carriers.
• Two affected parents with an unaffected child: the condition is dominant, and both parents are heterozygous.
• An unaffected father with an affected daughter rules out an X-linked recessive condition. A father with an X-linked recessive condition passes the allele to all his daughters.
• Mostly affected males, with the allele passed down through unaffected carrier women, suggests X-linked recessive.

Worked example. Parents I-1 and I-2 are unaffected. Their daughter II-2 is affected; their son II-3 is unaffected. The father is unaffected, yet his daughter is affected, so the allele is autosomal recessive (call it a). Both parents are Aa. II-3 is AA or Aa, with chances 1/3 and 2/3. If II-3 has a child with a woman known to be a carrier, P(affected child) = 2/3 × 1/4 = 1/6.

Nature of science: reading a pattern from some of the pedigree is inductive reasoning, generalising from some cases. Using that pattern to state the genotype of a particular person is deductive reasoning.

Many societies prohibit marriage between close relatives. Close relatives share recent ancestors, so both partners are more likely to carry the same rare recessive allele, and children are at greater risk of a recessive disorder.

D3.2.14 Continuous variation

Discrete variables fall into separate categories, such as ABO blood group. Continuous variables take any value in a range, such as height, and the population forms a smooth distribution. Continuous variation is caused by polygenic inheritance (many genes each with a small effect), by environmental factors, or both. Skin colour in humans is the example: several genes contribute to melanin production, and sunlight exposure alters it too.

Measures of central tendency:

• Mean: the sum divided by the number of values.
• Median: the middle value of the ordered data.
• Mode: the most frequent value.

D3.2.15 Box-and-whisker plots

A box plot displays six features of a data set: any outliers, the lowest value, the lower quartile (Q1), the median, the upper quartile (Q3) and the highest value. A value counts as an outlier when it lies further than 1.5 × IQR beyond either end of the box. The IQR is the upper quartile minus the lower quartile.

Worked example. Heights (cm) of 15 students, in order: 150, 153, 155, 156, 158, 158, 160, 161, 162, 163, 164, 165, 167, 168, 181.

• Median = 8th value = 161.
• Lower half (7 values): first quartile = 4th value = 156.
• Upper half (7 values): third quartile = 165.
• IQR = 165 - 156 = 9. Then 1.5 × 9 = 13.5.
• Upper limit = 165 + 13.5 = 178.5; lower limit = 156 - 13.5 = 142.5.
• 181 > 178.5, so 181 is an outlier and is plotted as a separate point.
• The whiskers reach the lowest value, 150, and the highest non-outlier, 168.
• Mean = 2,421 / 15 = 161.4 cm (to 1 decimal place); mode = 158 cm.

Quartile methods differ slightly between classes, so state yours. Here the median is left out of both halves.

Common errors

• Writing “gene” when you mean “allele”.
• Calling a dominant allele “stronger” or “more common”. Dominance describes phenotype only.
• Saying plasticity is caused by mutation. It is variable gene expression.
• Writing sex-linked alleles without the X, such as H and h on their own.

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).