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

This guide covers sections 3.3.6.1 (Principles) and 3.3.6.2 (Patterns of inheritance) of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1. Both sit in Unit 3, Populations and genes, which the specification lists as International A-level only. The revision notes condense the ideas and the practice questions let you check them. Course pages: the OxfordAQA A-level Biology hub, the printable checklist and the free diagnostics.

What the specification asks for

Section	What you must be able to do
3.3.6.1	Define genotype and phenotype; explain that phenotype also reflects interaction with the environment
3.3.6.1	Use the terms allele, dominant, recessive, codominant, homozygous and heterozygous; know a gene can have many alleles
3.3.6.2	Use fully labelled genetic diagrams to interpret or predict monohybrid and dihybrid crosses with dominant, recessive and codominant alleles
3.3.6.2	Do the same for sex-linkage, autosomal linkage, multiple alleles and epistasis
3.3.6.2	Use the chi-squared test to decide whether observed ratios differ significantly from expected ratios

The scenarios below are written for this page. Where an allele pattern is assumed for teaching, the text says “assume”.

3.3.6.1 Principles

Genotype means the alleles an organism carries, its genetic constitution. Phenotype is how that constitution is expressed, together with the effect of the environment on it. Genotype sets what is possible; environment influences what appears.

Example: a child with two recessive alleles for phenylketonuria cannot break down the amino acid phenylalanine properly. The genotype is the same whatever the diet, but a diet low in phenylalanine from birth gives a very different phenotype from an unrestricted diet.

An allele is one version of a gene. Each gene has a fixed place on a chromosome, its locus. Because organisms are diploid, each carries two alleles at a locus, one from each parent, and is:

• homozygous: two identical alleles (AA or aa)
• heterozygous: two different alleles (Aa)

A dominant allele shows in the phenotype of a heterozygote. A recessive allele shows only in a homozygote or when the other allele cannot mask it. Codominant alleles both contribute to the phenotype of a heterozygote. A population can hold many alleles of one gene, but each individual carries at most two.

How to set out a genetic diagram

A fully labelled diagram has these stages. Leave none out.

1. Key: what each letter means, and which allele is dominant.
2. Parental phenotypes and genotypes.
3. Gametes, shown as separate letters (one allele each, because meiosis separates the pair).
4. Offspring genotypes, using a grid if the parents make more than two kinds of gamete.
5. The phenotype of each genotype, then the overall ratio.

Use one letter per gene, capital for dominant. For codominance or multiple alleles use a base letter with superscripts, such as Fᴮ and Fᵂ.

Monohybrid crosses

A monohybrid cross follows one gene. Assume that in the fast-growing plant Brassica rapa purple stem (P) is dominant to green stem (p). Cross two heterozygous plants.

Gametes	P	p
P	PP purple	Pp purple
p	Pp purple	pp green

Ratio 3 purple : 1 green. A test cross crosses a purple plant of unknown genotype with a green plant (pp). If any offspring are green, the plant was Pp (expect 1 : 1). If all are purple, it was probably PP, though only a large sample makes that safe.

Codominance. In Andalusian fowl, assume black feathers (Fᴮ) and splashed-white feathers (Fᵂ) are codominant, and the heterozygote is blue-grey. Blue × blue gives FᴮFᴮ black : 2 FᴮFᵂ blue : FᵂFᵂ splashed white. A 1 : 2 : 1 phenotype ratio signals codominance, because the heterozygote is visibly different. Blue × splashed white gives 1 blue : 1 white.

Multiple alleles

One gene controls ABO blood group and has three alleles: Iᴬ, Iᴮ and i. The first two are codominant with each other, and i is recessive to both. This gives six genotypes and four phenotypes.

Worked example. A man of group AB has children with a woman of group A whose mother was group O. The woman must be Iᴬi (she inherited i from her mother).

Gametes	Iᴬ	Iᴮ
Iᴬ	IᴬIᴬ A	IᴬIᴮ AB
i	Iᴬi A	Iᴮi B

Ratio 2 group A : 1 group AB : 1 group B. Group O is impossible, because the father has no i allele. Probability of group B is 1/4; of a child that is not group A is 1/2.

Dominance series. Assume a beetle’s wing-case colour has three alleles, Cᵈ (dark) > Cᵐ (mid) > Cˡ (light), each dominant to those after it. Cross CᵈCˡ × CᵐCˡ: offspring CᵈCᵐ dark, CᵈCˡ dark, CᵐCˡ mid, CˡCˡ light, so 2 dark : 1 mid : 1 light. The number of possible genotypes for n alleles is n(n + 1)/2, which is 6 for three alleles.

Dihybrid crosses

A dihybrid cross follows two genes. If they are on different chromosomes they assort independently at meiosis. Assume in Brassica rapa that hairy leaves (H) are dominant to hairless (h) and that this gene is on a different chromosome from the stem colour gene.

Cross PpHh × PpHh. Each parent makes four gamete types in equal numbers: PH, Ph, pH, ph. The 16-box grid gives:

Phenotype	Boxes
Purple, hairy	9
Purple, hairless	3
Green, hairy	3
Green, hairless	1

The quick route is probability. Purple and hairless is 3/4 × 1/4 = 3/16. From 320 offspring expect 180 : 60 : 60 : 20. If the genes assort independently, a test cross PpHh × pphh gives four equally sized classes. If one gene is codominant, its heterozygote forms an extra class: a 1 : 2 : 1 gene crossed with a 3 : 1 gene gives six phenotypes in the ratio 3 : 6 : 3 : 1 : 2 : 1.

Sex-linkage

A sex-linked gene lies on a sex chromosome. In humans and Drosophila the female has two X chromosomes and the male an X and a Y. The X carries many genes that the short Y lacks, so a male has a single allele at each of them and it is always expressed. Write alleles as superscripts on X.

Worked example. In Drosophila assume red eyes (Xᴿ) are dominant to white (Xʷ). Cross a white-eyed female with a red-eyed male.

Gametes	Xʷ	Xʷ
Xᴿ	XᴿXʷ red female	XᴿXʷ red female
Y	XʷY white male	XʷY white male

All daughters are red-eyed and all sons white-eyed. In an autosomal cross, sons and daughters would not differ. Crossing the F1 flies gives 1 red female : 1 white female : 1 red male : 1 white male.

Birds. In birds males are ZZ and females ZW, so the female has one allele. Assume in chickens late feathering (Zᴷ) is dominant to early (Zᵏ). A heterozygous male (ZᴷZᵏ) × early-feathering female (ZᵏW) gives ZᴷZᵏ late male, ZᵏZᵏ early male, ZᴷW late female and ZᵏW early female, 1 : 1 : 1 : 1.

X-linked dominant. An affected father passes the allele to every daughter and to no son, because sons receive his Y.

Autosomal linkage

Genes lying on the same chromosome are linked, and their alleles tend to travel together. Alleles on one chromosome are written together (CSh / csh). Linked genes do not assort independently, so results depart from the unlinked ratios.

Crossing over between non-sister chromatids in prophase I (meiosis I) can separate linked alleles, producing recombinants. The closer two loci lie, the less often crossing over falls between them, so the fewer recombinants.

Worked example. Assume in maize C (coloured kernel) and Sh (full kernel) are dominant to c and sh, and the loci are linked. A plant CSh / csh is test-crossed with csh / csh, which can only give csh gametes. Of 1200 offspring:

Phenotype	Observed	Expected if unlinked
Coloured, full	520	300
Colourless, shrunken	536	300
Coloured, shrunken	72	300
Colourless, full	72	300

The parental combinations are far commoner than 1 : 1 : 1 : 1, so the genes are linked. The two small classes are recombinants: (72 + 72)/1200 = 0.12, a 12% recombination frequency.

Epistasis

Epistasis occurs when a gene at one locus masks or alters how a gene at another locus is expressed. It alters the 9 : 3 : 3 : 1 ratio by merging classes. The total stays 16.

Recessive epistasis, 9 : 3 : 4. Assume in a monkeyflower that a pigment needs allele P. With P, R gives red petals and rr gives pink. The genotype pp gives white whatever the R locus holds. PpRr × PpRr gives 9 P_R_ red : 3 P_rr pink : 4 white (3 ppR_ + 1 pprr). From 144 offspring expect 81 : 27 : 36.

Dominant epistasis, 12 : 3 : 1. Assume in oats an allele I gives black husks whatever the second gene holds. ii with Y gives grey and iiyy gives white. IiYy × IiYy gives 12 black (9 + 3) : 3 grey : 1 white.

Complementary genes, 9 : 7. If two enzymes act in sequence to make a pigment, a flower needs a functional allele of each gene. Only A_B_ (9/16) is coloured. The other 7/16 lack one enzyme and are white.

To decode a ratio: add the numbers, check whether it totals 16, find which classes merged, then state which gene masks which and why.

The chi-squared test

The chi-squared (χ²) test decides whether the difference between observed and expected counts is significant. Use it for counts in separate categories, not for percentages, means or measurements.

1. Null hypothesis: chance alone accounts for any gap between observed and expected results.
2. Expected counts: divide the total according to the expected ratio.
3. χ² = Σ (O − E)² / E, adding one term for each category.
4. Degrees of freedom (df) = categories − 1.
5. Find the critical value for p = 0.05: df 1 is 3.84, df 2 is 5.99, df 3 is 7.81 and df 4 is 9.49.
6. A calculated χ² below the critical value means no significant difference, so the null hypothesis stands. A value equal to or above it means a significant difference, and you reject the null hypothesis.

Example 1 (monohybrid). Heterozygous purple plants are crossed. Of 210 offspring, 148 are purple and 62 green. Expected 3 : 1 gives 157.5 and 52.5.

χ² = (148 − 157.5)²/157.5 + (62 − 52.5)²/52.5 = 0.573 + 1.719 = 2.29. With df = 1 the critical value is 3.84. The result is below it, so the offspring fit 3 : 1.

Example 2 (dihybrid). Observed from 320 offspring: 171, 66, 55, 28. Expected 180, 60, 60, 20.

χ² = 0.45 + 0.60 + 0.417 + 3.2 = 4.67. With df = 3 the critical value is 7.81. The result is below it, so the offspring fit 9 : 3 : 3 : 1.

Example 3 (linkage). Using the maize data against 1 : 1 : 1 : 1 (expected 300 each): χ² = 161.3 + 185.7 + 173.3 + 173.3 = 693.5. This is far above 7.81, so the null hypothesis is rejected: the two genes do not assort independently.

Example 4 (epistasis). For the monkeyflower, observed 78 red, 31 pink, 35 white; expected 81, 27, 36. χ² = 0.111 + 0.593 + 0.028 = 0.73. With df = 2 the critical value is 5.99, so the flowers fit 9 : 3 : 4.

A significant result shows the data do not fit the expected ratio. It does not prove which alternative explains them.

Common errors

• Writing two letters in a gamete, or leaving gametes out of the diagram.
• Calling a 1 : 2 : 1 ratio dominance. It indicates codominance.
• Writing the X-linked allele without X, or giving a male two alleles.
• Putting percentages into χ² instead of counts, or skipping a category.
• Using the number of categories, not categories − 1, as degrees of freedom.
• Saying “accept the hypothesis” when you should say you do not reject the null hypothesis.
• Describing epistasis as “dominance”. Dominance is between alleles of one gene; epistasis is between genes.

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.