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

These notes condense 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. For fuller explanations, read the study guide, then test yourself with the practice questions. Tick off the unit on the checklist, see the OCR A Level Biology hub, and use the free diagnostics to check recall.

Variation, 6.1.2(a)

Source	Key points
Genetic	different alleles; new alleles arise by mutation
Environmental	diet in animals (body mass, size); light in plants (etiolation: long, pale stems, small leaves, no chlorophyll)
Both	phenotype = genotype acting in an environment; for example the temperature-sensitive enzyme in Himalayan rabbits

How sexual reproduction creates genetic variation

• Crossing over in prophase I: non-sister chromatids in a bivalent swap alleles.
• Independent assortment in metaphase I: random distribution of maternal and paternal chromosomes. For n pairs there are 2ⁿ gamete types (ignoring crossing over).
• Random fusion of gametes: chance decides which gametes meet.

Test of the idea. Genetically identical organisms (clones) in different environments show environmental variation only.

Genetic diagram method, 6.1.2(b)(i)

1. Key: choose a letter, capital for dominant; define each allele and phenotype.
2. Parental phenotypes, then genotypes.
3. Gametes carry one allele per gene.
4. Offspring genotypes (table or grid).
5. Offspring phenotypes and the ratio.

Type	Notation and key points
Monogenic	Aa × Aa gives AA, Aa, Aa and aa (3 dominant : 1 recessive); test cross Aa × aa gives 1 : 1
Codominance	both alleles expressed in the heterozygote, written as superscripts such as Cᴿ and Cᵂ; Cᴿ Cᵂ × Cᴿ Cᵂ gives 1 : 2 : 1 for genotype and phenotype
Multiple alleles	three or more alleles in the population, two in each individual; state the dominance order
Dihybrid	AaBb × AaBb gives 9 : 3 : 3 : 1; test cross AaBb × aabb gives 1 : 1 : 1 : 1
Sex linkage	allele on X; males XY have one allele; write alleles as superscripts on X, such as Xᴰ and Xᵈ, and always write the Y

Small reminder (X-linked recessive). Cross an affected mother XᵈXᵈ with an unaffected father XᴰY.

	Xᵈ	Xᵈ
Xᴰ	XᴰXᵈ	XᴰXᵈ
Y	XᵈY	XᵈY

All daughters are unaffected carriers, and all sons are affected. A son always gets his X from his mother.

Small reminder (dihybrid gametes). AaBb makes four gamete types in equal numbers: AB, Ab, aB, ab. Two such parents give a 4 × 4 grid of 16 boxes. If the genes are linked, write the chromosomes instead (AB/ab) and expect mainly AB and ab gametes.

Small reminder (probabilities). For independent genes, multiply. Aabb from AaBb × AaBb: 1/2 × 1/4 = 1/8 for Aa and bb together.

Using ratios, 6.1.2(b)(ii)

Observed ratio	Likely explanation
3 : 1	monogenic cross, heterozygote × heterozygote
1 : 1	test cross of a heterozygote
9 : 3 : 3 : 1	two unlinked genes, no interaction
1 : 1 : 1 : 1	test cross, unlinked genes
two large classes, two small	autosomal linkage; large classes are the parental combinations, small ones the recombinants
different in males and females, or reciprocal crosses differ	sex linkage
9 : 3 : 4	recessive epistasis
12 : 3 : 1	dominant epistasis
9 : 7	both dominant alleles needed (complementary)
15 : 1	either dominant allele sufficient

Linkage in steps. (1) Write the expected 1 : 1 : 1 : 1. (2) Identify parental and recombinant classes: parental classes are the biggest. (3) Recombinants come from crossing over between the loci, in prophase I. (4) The closer the loci, the fewer the recombinants.

Complementary example (9 : 7). A flower pigment forms from a colourless precursor in two steps, each needing an enzyme coded by a dominant allele (A and B). Plants lacking either enzyme are white, so only A_B_ (9 of 16) are coloured and the other 7 are white.

Epistasis in steps. (1) Add the ratio numbers to 16. (2) See which classes have merged. (3) Name the gene whose allele masks the other, and state the allele that does the masking. (4) Give the enzyme pathway reason if the question describes one.

Reminder. Of 144 offspring from a 9 : 3 : 4 cross you expect 81, 27 and 36. Of 320 from a 12 : 3 : 1 cross you expect 240, 60 and 20.

Chi-squared, 6.1.2(c)

The formula is given: χ² = Σ (fo − fe)² / fe. Learn the steps and the interpretation.

Step	What to write
1	Null hypothesis: no significant difference between observed and expected results
2	Expected counts from the ratio and the total
3	Add (fo − fe)² / fe over every class
4	Degrees of freedom: number of classes minus one
5	Look up the critical value at p = 0.05
6	Conclusion in words, linked to the context

Degrees of freedom	Critical value at p = 0.05
1	3.84
2	5.99
3	7.81

Small worked reminder. Observed 100 and 20 against 3 : 1. Total 120, expected 90 and 30. χ² = 100/90 + 100/30 = 1.11 + 3.33 = 4.44. One degree of freedom; 4.44 is greater than 3.84, so the difference is significant: reject the null hypothesis. The cross does not fit 3 : 1, so something other than chance, such as lethal genotypes or linkage, may be involved.

Interpretation wording. Below the critical value: the difference is not significant; it could be due to chance; the results fit the expected ratio. At or above: the difference is significant; reject the null hypothesis; chance is unlikely to explain it.

Rules. Use counts, never percentages. Categories only. Do not stop at “reject”: say what the evidence suggests about the genes.

Continuous and discontinuous variation, 6.1.2(d)

	Discontinuous	Continuous
Genes	one or a few; alleles have large, distinct effects	many genes (polygenic), each a small additive effect
Environment	little effect	usually large effect
Distribution	separate categories	range; bell-shaped curve
Graph	bar chart	histogram or line graph of frequency
Examples	ABO blood group, round or wrinkled pea seeds	height, mass, leaf length

Two additive genes give five classes, 1 : 4 : 6 : 4 : 1; n genes give 2n + 1 classes.

Must-know distinctions

• Genotype is the alleles; phenotype is the characteristics expressed.
• Codominant (both alleles expressed) versus incompletely dominant (blended): the specification names codominance.
• Linkage (genes on one chromosome, parental classes commonest) versus epistasis (one gene masks another because their products interact, giving a modified 9 : 3 : 3 : 1).
• Sex linkage (gene on the X) versus autosomal linkage (two genes on one autosome).
• Significant (reject the null hypothesis) versus not significant (do not reject it).

Quick self-test

1. State three ways sexual reproduction produces genetic variation.
2. An organism has 2n = 8. How many different gametes can independent assortment produce?
3. Give the genotype ratio and the phenotype ratio for Aa × Aa with complete dominance.
4. Two parents with normal colour vision have a colour-blind son (X-linked recessive). State both parents’ genotypes.
5. A dihybrid test cross gives 48 : 2 : 3 : 47 instead of 1 : 1 : 1 : 1. What does this show, and what percentage of offspring are recombinants?
6. A 3 : 1 cross gives 62 and 18 offspring. Calculate χ² and state the conclusion at p = 0.05.
7. How many degrees of freedom apply to a test with four phenotype classes, and what is the critical value at p = 0.05?
8. How many phenotype classes do three genes with equal additive effects give?
9. Which type of variation is best shown as a bar chart, and why?
10. Why can identical twins reared apart differ in height?

Answers

1. Crossing over, independent assortment, random fusion of gametes.
2. 2⁴ = 16.
3. Genotype 1 AA : 2 Aa : 1 aa; phenotype 3 dominant : 1 recessive.
4. Mother XᴺXⁿ, father XᴺY.
5. Autosomal linkage: the parental classes (48 and 47) are far commoner than the recombinants. Recombinants are 2 + 3 = 5 out of 100, which is 5 per cent.
6. Total 80, expected 60 and 20. χ² = 4/60 + 4/20 = 0.067 + 0.200 = 0.27. One degree of freedom, critical value 3.84. Not significant: the results fit 3 : 1.
7. 3; 7.81.
8. 2 × 3 + 1 = 7.
9. Discontinuous variation, because the phenotypes fall into separate categories.
10. They share genes, but differences in diet and environment affect height, so the environmental contribution differs.

Where marks are usually lost

• Genetic diagrams with no key, no gametes row or gametes carrying two alleles of one gene.
• Writing a sex-linked cross without the Y, or giving the male an allele for the second X.
• Stating a ratio without saying which phenotype goes with which number.
• Calling a 9 : 3 : 4 or 12 : 3 : 1 result “a dihybrid ratio” without naming epistasis and the gene doing the masking.
• Explaining linkage without naming crossing over as the source of the few recombinants.
• Chi-squared on percentages, a wrong expected total or degrees of freedom equal to the number of classes.
• Concluding “the ratio is proved” or “the hypothesis is accepted”; use the language of significance and chance.
• Explaining variation using only genes, or only environment, when the question gives both.
• Describing continuous variation as “controlled by environment”; it is controlled by many genes, with the environment adding to the effect.

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.