OCR A Level Biology A: Variation, genetic diagrams and chi-squared (H420)

This guide covers section 6.1.2(a) to (d), Patterns of inheritance, of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. It teaches variation, genetic diagrams, linkage, epistasis, the chi-squared test, and continuous and discontinuous variation. Use it with the revision notes and the practice questions. See also the OCR A Level Biology hub, the checklist, the free diagnostics, variation, adaptation and natural selection and the practical skills revision notes.

What this unit covers

Spec reference	What you must be able to do
6.1.2(a)(i)	Explain genetic and environmental contributions to phenotypic variation
6.1.2(a)(ii)	Explain how sexual reproduction gives genetic variation
6.1.2(b)(i)	Use genetic diagrams: monogenic, dihybrid, multiple alleles, sex linkage, codominance
6.1.2(b)(ii)	Use phenotypic ratios to identify linkage and epistasis
6.1.2(c)	Use the chi-squared test on observed and expected results
6.1.2(d)	Explain the genetic basis of continuous and discontinuous variation

Genetic and environmental variation, 6.1.2(a)

Phenotype, the observable characteristics of an organism, depends on genotype (the alleles it carries) and on the environment in which it develops. Most characteristics are affected by both.

Genetic contribution. Alleles are inherited, so differences in alleles cause differences in phenotype. New alleles arise by mutation. Sexual reproduction then shuffles existing alleles in three ways:

• Crossing over in prophase I of meiosis exchanges sections of non-sister chromatids within a bivalent, so chromatids carry new allele combinations.
• Independent assortment: each homologous pair lines up at the equator in metaphase I independently of the others, so maternal and paternal chromosomes are distributed randomly into the gametes.
• Random fusion of gametes at fertilisation: which sperm reaches which egg is a matter of chance.

Worked example. A garden pea has 2n = 14, so n = 7 pairs of homologous chromosomes. Ignoring crossing over, one plant makes 2⁷ = 128 different gametes, and two parents give 128 × 128 = 16 384 combinations at fertilisation.

Environmental contribution. Two examples are specified.

• Diet in animals. Genetically similar animals on a richer diet reach a greater body mass than those on a poor diet.
• Etiolation in plants. A seedling grown in the dark has a long, thin, pale stem, small unexpanded leaves and a yellow colour, because chlorophyll formation needs light. The extra stem growth carries the shoot tip towards light. Cuttings from one plant are genetically identical, so differences between light and dark grown cuttings are environmental.

Genes and environment can interact. In Himalayan rabbits the allele codes for an enzyme that works only at lower temperatures, so dark pigment forms only on the cooler ears, paws, nose and tail.

To separate the causes, compare genetically identical individuals (clones, identical twins) in different environments.

Genetic diagrams, 6.1.2(b)(i)

Terms: an allele is a version of a gene; a locus is a gene’s fixed place on a chromosome. Homozygous means two identical alleles, heterozygous two different alleles. A dominant allele is expressed in the heterozygote; a recessive allele only when no dominant allele is present. Codominant alleles are both expressed in the heterozygote. A carrier has one copy of a recessive allele and does not show the condition.

A full genetic diagram gives: a key defining the symbols, parental phenotypes, parental genotypes, gametes (circled or listed), offspring genotypes (a table is clearest), offspring phenotypes and the ratio.

Monogenic inheritance

A monogenic cross follows one gene. In mice, allele C gives pigmented fur and c gives albino. Two pigmented mice have an albino pup, so both parents must be Cc.

	C	c
C	CC	Cc
c	Cc	cc

Phenotypic ratio 3 pigmented : 1 albino. A test cross mates an unknown-genotype individual showing the dominant phenotype with a homozygous recessive partner. Recessive offspring show the unknown parent was heterozygous (ratio 1 : 1); all dominant offspring suggest homozygous.

Codominance

In some cattle, red coat (Cᴿ) and white coat (Cᵂ) are codominant, and the heterozygote is roan with both red and white hairs. Roan × roan gives CᴿCᴿ : CᴿCᵂ : CᵂCᵂ in the ratio 1 : 2 : 1, which is 1 red : 2 roan : 1 white. Genotype ratio and phenotype ratio match, a clue to codominance.

Multiple alleles

Multiple alleles means three or more versions of one gene exist in a population; each diploid individual still carries just two. In rabbit coat colour, take four alleles in the dominance order Cᶠ (full colour) > Cᶜ (chinchilla) > Cʰ (Himalayan) > Cᵃ (albino). These labels are for this example only.

Cross a full-colour rabbit CᶠCʰ with a chinchilla rabbit CᶜCᵃ. Gametes: Cᶠ or Cʰ; Cᶜ or Cᵃ. Offspring: CᶠCᶜ, CᶠCᵃ (both full colour), CᶜCʰ (chinchilla), CʰCᵃ (Himalayan). Ratio 2 full colour : 1 chinchilla : 1 Himalayan.

Dihybrid inheritance

A dihybrid cross tracks two genes on separate chromosomes that assort independently. In maize, purple kernels (P) are dominant to yellow (p), and smooth (S) is dominant to wrinkled (s). Cross two PpSs plants. Each makes four gametes in equal numbers: PS, Ps, pS, ps. The 16 boxes give:

Phenotype	Genotypes	Boxes
purple smooth	P_S_	9
purple wrinkled	P_ss	3
yellow smooth	ppS_	3
yellow wrinkled	ppss	1

From 640 kernels you would expect 360 : 120 : 120 : 40. Using probability, purple wrinkled is (3/4) × (1/4), which is 3/16. Test crossing PpSs with ppss gives purple smooth : purple wrinkled : yellow smooth : yellow wrinkled in the ratio 1 : 1 : 1 : 1 when the genes assort independently.

Sex linkage

A sex-linked gene lies on the X chromosome, where the Y has no matching locus. A male (XY) has one allele, so a recessive allele is always expressed. Red-green colour blindness is X-linked recessive: Xᴺ is normal vision, Xⁿ is colour blindness.

A carrier woman (XᴺXⁿ) has children with a man with normal vision (XᴺY).

	Xᴺ	Xⁿ
Xᴺ	XᴺXᴺ	XᴺXⁿ
Y	XᴺY	XⁿY

Daughters all have normal vision (half are carriers). Half the sons are colour-blind. The probability that a child is colour-blind is 1/4, but for a son it is 1/2. A colour-blind father passes Xⁿ to every daughter and never to a son.

Linkage and epistasis from ratios, 6.1.2(b)(ii)

An unexpected ratio is information. The method is to compare the observed ratio with the expected ratio, then explain the difference.

Autosomal linkage. Two genes on one autosome are usually inherited as a unit. Alleles on one chromosome are written together, for example PL/pl. In a dihybrid test cross, independent assortment gives 1 : 1 : 1 : 1. With linkage, the two parental phenotype classes are much commoner than the two recombinant classes. Recombinants arise only when crossing over happens between the two loci.

Worked example. A sweet pea plant PL/pl (purple flowers and long pollen both dominant) is test-crossed with ppll. Of 480 offspring: 214 purple long, 22 purple round, 24 red long, 220 red round. The parental classes (purple long, red round) total 434; recombinants total 46, which is 46/480 = 9.6 per cent of offspring. Expected under independent assortment is 120 in each class, giving χ² = 313.8 (see below), so the genes are linked.

Sex linkage. Suspect it when the ratio differs between males and females, or when reciprocal crosses differ. For an X-linked recessive allele, an affected male × homozygous normal female gives no affected children, while an affected female × normal male gives affected sons and carrier daughters.

Epistasis. The allele at one locus masks or modifies the expression of alleles at a second locus. Two heterozygous parents give a modified version of 9 : 3 : 3 : 1, with 16 as the total.

Worked example. In Labrador retrievers, B (black) is dominant to b (brown), and gene E/e controls whether pigment is deposited in the coat: ee dogs are yellow whatever their B/b alleles. BbEe × BbEe gives 9 B_E_ (black), 3 bbE_ (brown), and 3 B_ee plus 1 bbee (4 yellow). The ratio is 9 black : 3 brown : 4 yellow, with the last two classes merged because ee masks B/b. This is recessive epistasis.

Ratio from two heterozygotes	Interpretation
9 : 3 : 3 : 1	two genes, independent assortment, no interaction
9 : 3 : 4	recessive epistasis, as above
12 : 3 : 1	dominant epistasis: one dominant allele hides the other gene
9 : 7	both dominant alleles needed for the phenotype
15 : 1	either dominant allele is enough for the phenotype

The chi-squared test, 6.1.2(c)

Observed counts rarely match the expected ratio exactly, because fertilisation is random. The chi-squared test judges whether the difference between observed and expected results is significant. The formula will be provided in assessments:

χ² = Σ (fo − fe)² / fe

where fo is the observed frequency and fe the expected frequency in each category.

1. State the null hypothesis: there is no significant difference between the observed and expected results.
2. Work out expected values: share the total number of offspring in the expected ratio.
3. Calculate (fo − fe)² / fe for each category and add them.
4. Degrees of freedom = number of categories − 1.
5. Read the critical value at p = 0.05 for those degrees of freedom: 3.84 for 1, 5.99 for 2, 7.81 for 3.
6. A χ² below the critical value means no significant difference: chance may explain it, so do not reject the null hypothesis. If χ² is equal to or greater than it, the difference is significant, so reject the null hypothesis and look for a cause, such as linkage or epistasis.

Use raw counts, not percentages or proportions, and only for categories (discontinuous data).

Worked example. Maize, PpSs × PpSs, expected 9 : 3 : 3 : 1. Observed from 560 kernels: 322, 118, 86, 34.

Class	fo	fe	(fo − fe)² / fe
purple smooth	322	315	49/315 = 0.156
purple wrinkled	118	105	169/105 = 1.610
yellow smooth	86	105	361/105 = 3.438
yellow wrinkled	34	35	1/35 = 0.029

χ² = 5.23. With 3 degrees of freedom the critical value is 7.81. As 5.23 < 7.81, the difference is not significant, so the results are consistent with 9 : 3 : 3 : 1.

Worked example (epistasis). Labradors, expected 9 : 3 : 4 from 280 puppies. Observed 150 black, 62 brown, 68 yellow. Expected 157.5, 52.5, 70. χ² = 0.357 + 1.719 + 0.057 = 2.13. With 2 degrees of freedom (critical 5.99) the difference is not significant, so the data fit recessive epistasis.

Continuous and discontinuous variation, 6.1.2(d)

	Discontinuous	Continuous
Phenotypes	distinct categories, no intermediates	a range, with all values between the extremes
Genes	one gene, or a small number of genes	many genes, each with a small additive effect (polygenic)
Environment	little or no effect	often a large effect
Graph	bar chart	histogram; often a bell-shaped (normal) curve
Examples	ABO blood group, pea seed shape	height, body mass, leaf length

Why many genes give a smooth curve. Suppose two genes each have an allele that adds one unit to a trait (A, B) and an allele that adds nothing (a, b). Cross AaBb × AaBb: count the capital letters in each of the 16 offspring. The numbers with 0, 1, 2, 3 and 4 capital letters are in the ratio 1 : 4 : 6 : 4 : 1, giving five classes with most offspring in the middle. With three such genes there are seven classes (1 : 6 : 15 : 20 : 15 : 6 : 1). In general, n genes give 2n + 1 classes. With many genes, and environmental effects, the steps blur into a bell-shaped curve.

Common errors

• Writing gametes with two alleles of the same gene. Each gamete has one.
• Treating Xⁿ as inherited from the father by a son.
• Using percentages in chi-squared, or forgetting to subtract 1 for degrees of freedom.
• Claiming that a low χ² proves the expected ratio.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: section 6.1.2(a) to (d), Patterns of inheritance.