OxfordAQA International AS and A-level Biology: Inheritance and genetic crosses (9610) – Revision Notes

Quick recall for sections 3.3.6.1 and 3.3.6.2 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1 (Unit 3, listed as International A-level only). For full explanations and worked genetic diagrams, use the study guide. Then try the practice questions. Course pages: OxfordAQA A-level Biology hub, printable checklist and free diagnostics.

3.3.6.1 Definitions

Term	Meaning
Genotype	The alleles an organism carries (its genetic constitution)
Phenotype	What is expressed: the genotype’s effect plus the environment’s
Allele	A version of a gene
Locus	Where a gene sits on a chromosome
Homozygous	Both alleles at the locus are the same
Heterozygous	The two alleles at the locus differ
Dominant	Shows in the heterozygote’s phenotype
Recessive	Hidden in a heterozygote; shows only without a dominant allele
Codominant	Each allele shows in the heterozygote’s phenotype

A population can hold many alleles of a gene. A diploid individual carries two.

Environment can change the phenotype of a fixed genotype. Example: a plant with alleles for tallness grows shorter if short of nitrates.

Method: a fully labelled genetic diagram

1. Key (letters, dominance).
2. Parental phenotypes and genotypes.
3. Gametes: one allele (or one allele of each gene) per gamete.
4. Offspring genotypes (use a grid if there are more than two gamete types).
5. Phenotype for each genotype, then the ratio.

Ratios to recognise

Parents	Offspring ratio	Clue
Aa × Aa (complete dominance)	3 : 1	Both parents show the dominant trait and some offspring do not
Aa × aa (test cross)	1 : 1	Used to find an unknown genotype
Codominant heterozygotes × each other	1 : 2 : 1	Heterozygote has its own phenotype
AaBb × AaBb, independent genes	9 : 3 : 3 : 1	Dihybrid
AaBb × aabb, independent genes	1 : 1 : 1 : 1	Dihybrid test cross
Linked genes in a test cross	Two large and two small classes	Recombinants are the small classes
Recessive epistasis	9 : 3 : 4	Two classes merged
Dominant epistasis	12 : 3 : 1	Two classes merged
Complementary genes	9 : 7	Both genes needed

Check that the total is 16 for a two-gene cross between double heterozygotes.

Multiple alleles

• Possible genotypes = n(n + 1)/2 for n alleles: 6 for three alleles, 10 for four.
• ABO: Iᴬ and Iᴮ codominant, i recessive. Group O is ii; group AB is IᴬIᴮ.
• Dominance series: write the order (for example a1 > a2 > a3) and apply it to each genotype before drawing the grid.

Sex-linkage

• Humans and Drosophila: XX female, XY male. Birds: ZZ male, ZW female.
• The sex with one copy (XY male; ZW female) expresses whichever allele it has.
• X-linked recessive: an affected daughter has an affected father and a carrier or affected mother. A son gets his X from his mother only.
• X-linked dominant: an affected father passes it to all daughters and no sons.
• Reciprocal crosses give different results for a sex-linked gene and the same for an autosomal gene.
• Carrier mother × unaffected father (X-linked recessive): half the sons affected, no daughters affected, half the daughters carriers.

Autosomal linkage

• Linked genes lie on the same chromosome and are inherited together.
• Parental combinations are the commonest classes; recombinants arise by crossing over in meiosis.
• Recombination frequency = recombinant offspring ÷ total offspring × 100.
• Loci that are close together give few recombinants. Independent assortment would give 50% recombinants in a test cross, because all four classes would be equal.

Epistasis

• One gene masks or modifies the expression of another gene at a different locus.
• Dominance acts between alleles of the same gene; epistasis acts between genes.
• State the biochemical idea where you can: for example, an enzyme that makes a precursor is missing, so a later gene has nothing to act on.

The chi-squared test

Step	Action
1	State the null hypothesis: any difference from the expected ratio is due to chance
2	Work out expected counts from the ratio
3	χ² = Σ (O − E)² / E
4	Degrees of freedom = categories − 1
5	Critical value at p = 0.05: df 1 is 3.84, df 2 is 5.99, df 3 is 7.81, df 4 is 9.49
6	Result below the critical value: not significant, null hypothesis stands. Equal or above: significant, reject the null hypothesis

Use counts, never percentages. Include every category. A significant result means the ratio does not fit; it does not say why.

Reminder. Fowl blue × blue, 80 chicks, observed 17 black : 45 blue : 18 splashed white. Expected 20 : 40 : 20. χ² = 0.45 + 0.625 + 0.2 = 1.275, so 1.28. With df = 2 the critical value is 5.99, so the result is not significant and fits 1 : 2 : 1.

Worked reminders

Sex-linked cross. Assume an X-linked recessive allele Xᵃ causes a condition. Carrier mother XᴬXᵃ × unaffected father XᴬY.

• Gametes: mother Xᴬ or Xᵃ; father Xᴬ or Y.
• Offspring: XᴬXᴬ unaffected girl, XᴬXᵃ carrier girl, XᴬY unaffected boy, XᵃY affected boy.
• Probability a child is affected: 1/4. Probability a son is affected: 1/2.

Dihybrid expectation. AaBb × AaBb with 480 offspring, independent genes: 9/16 × 480 = 270 double dominant, 3/16 × 480 = 90 for each single-recessive class, 1/16 × 480 = 30 double recessive.

Linkage. Test cross of a double heterozygote CD / cd with cd / cd. Parental classes (CD and cd) are large; recombinant classes (Cd and cD) are small and about equal, because crossing over gives each reciprocal product equally often.

Epistasis. If a pp genotype blocks pigment, every pp plant is white whether it is RR, Rr or rr. In PpRr × PpRr, four of the 16 boxes (one ppRR, two ppRr and one pprr) merge into one white class, so the ratio is 9 coloured with R : 3 coloured with rr : 4 white.

Model answer shapes

• A question asks why a dihybrid ratio is unusual: name the pattern (linkage or epistasis), say what the observed classes show, link it to the mechanism.
• Explain a sex-linked result: state which sex has the single allele, say sons or daughters inherit from which parent, and compare with an autosomal prediction.
• Interpret a chi-squared result: give χ², degrees of freedom, critical value, comparison, then a statement about the null hypothesis in the context of the cross.

Must-know distinctions

• Genotype versus phenotype. Codominance versus incomplete dominance: the second is not named in the specification, so describe what you see.
• Linkage versus sex-linkage: linkage is two genes on one chromosome; sex-linkage is a gene on a sex chromosome.
• Epistasis versus dominance.
• Recombinant versus parental offspring classes.

Quick self-test

1. Define phenotype.
2. A heterozygote is crossed with a homozygous recessive. What ratio do you expect?
3. A species has four alleles of one gene. How many genotypes are possible?
4. 160 offspring from a cross expected to give 9 coloured : 7 white are 103 coloured and 57 white. Calculate χ² and say if it fits.
5. A test cross gives 24, 27, 25 and 24 in four classes. Test against 1 : 1 : 1 : 1.
6. A carrier woman has a child with an unaffected man (X-linked recessive). What is the probability that a son is affected?
7. A test cross gives 500 offspring, of which 60 are recombinants. Give the recombination frequency.
8. State why X-linked recessive conditions are commoner in males.
9. A two-gene cross gives 12 : 3 : 1. Name the pattern.
10. Give the degrees of freedom for a test on a 9 : 3 : 4 ratio.

Answers

1. The expression of the genotype and its interaction with the environment.
2. 1 : 1.
3. 4 × 5 / 2 = 10.
4. Expected 90 and 70. χ² = (13²/90) + (13²/70) = 1.878 + 2.414 = 4.29. One degree of freedom, critical value 3.84. Greater than 3.84, so significant: the data do not fit 9 : 7.
5. Total 100, expected 25 each. χ² = 0.04 + 0.16 + 0 + 0.04 = 0.24. Three degrees of freedom, critical value 7.81. Not significant: consistent with independent assortment.
6. 1/2.
7. 60/500 × 100 = 12%.
8. A male has one X, so a single recessive allele is expressed. A female needs two.
9. Dominant epistasis.
10. 3 categories − 1 = 2.

Where marks are usually lost

• Leaving out the key, so the examiner cannot tell which allele is dominant.
• Putting a whole genotype in a gamete.
• Giving a ratio without the phenotypes it refers to.
• Treating a recombination frequency as an expected count rather than the proportion of recombinants.
• Applying chi-squared to percentages.
• Using the total number of individuals instead of the number of categories − 1.
• Writing “the null hypothesis is accepted” or “proved” when the wording should be “not rejected”.
• Not saying what the significant difference means for the cross, for example linkage or epistasis.

Official syllabus

Source: OxfordAQA, International AS and A-level Biology (9610), Version 5.1, International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, sections 3.3.6.1 and 3.3.6.2.