AQA A-level Biology: Inheritance and Chi-squared (7402)

This guide covers AQA A-level Biology (7402) section 3.7.1, Inheritance, from the AQA AS and A-level Biology specification for first teaching 2015. The specification marks 3.7.1 as A-level only (not AS). It sits in topic 7, so it falls in the content assessed by A-level Paper 2 (topics 5 to 8) and Paper 3 (topics 1 to 8). The chi-squared test is linked to maths skill 1.9. Condense it with the revision notes, then test yourself with the practice questions. Course pages: AQA A-level Biology hub, the printable checklist and the free diagnostics. Meiosis is covered in mutation and meiosis.

What the specification asks for

Statement in 3.7.1	What you must be able to do
Genotype and phenotype	Define both; explain that phenotype also reflects interaction with the environment
Alleles	Know a gene can have many alleles; use dominant, recessive, codominant, homozygous, heterozygous
Monohybrid and dihybrid crosses	Interpret or predict results with fully labelled genetic diagrams
Sex linkage, autosomal linkage, multiple alleles, epistasis	Use each in a cross and explain why the ratio is not the simple one
Chi-squared test	Compare observed phenotypic ratios with expected ratios and judge the fit

No required practical is listed in 3.7.1. The specification suggests crosses of Drosophila or Fast Plant as a way to investigate genetic ratios, and the use of probability (maths skill 1.4) and of phenotypic ratios (maths skill 0.3).

Key terms

• Genotype: the genetic constitution of an organism, the alleles it carries.
• Phenotype: the expression of that genetic constitution and its interaction with the environment.
• Locus: the position of a gene on a chromosome. A diploid organism has two alleles at each locus.
• Homozygous: two identical alleles at a locus. Heterozygous: two different alleles.
• Dominant: an allele that is expressed in the heterozygote. Recessive: expressed only when no dominant allele is present. Codominant: both alleles are expressed in the phenotype of the heterozygote.

The same genotype can give different phenotypes. Many hydrangea varieties are genetically identical when cloned, yet their flower colour changes with the pH of the soil.

Setting out a genetic diagram

The specification asks for fully labelled genetic diagrams, so include every stage every time:

1. Key: state what each letter means.
2. Parental phenotypes and parental genotypes.
3. Gametes, written as separate letters or pairs, circled or listed.
4. Offspring genotypes (a grid for crosses with more than two gamete types).
5. Offspring phenotypes and the ratio.

Use the same letter for both alleles of one gene (capital for dominant), and for codominance use a capital letter with superscripts, such as Cᴿ and Cᵂ.

Monohybrid crosses

A monohybrid cross follows one gene. In fruit flies, normal wings (V) are dominant to vestigial wings (v). Two heterozygous flies are crossed.

Gametes	V	v
V	VV normal	Vv normal
v	Vv normal	vv vestigial

The offspring ratio is 3 normal : 1 vestigial. Of 240 offspring you would expect 180 normal and 60 vestigial. Real results seldom match exactly because fertilisation is random.

Test cross. To find the genotype of a normal-winged fly, cross it with a vestigial (vv) fly. Any vestigial offspring show the fly was Vv (ratio 1 : 1). If all are normal it was probably VV, but only a large sample makes that safe.

Codominance. In a flower species, red petals (Cᴿ) and white petals (Cᵂ) are codominant, and heterozygotes have red-and-white patches. Crossing two patched plants gives CᴿCᴿ red : 2 CᴿCᵂ patched : CᵂCᵂ white, a 1 : 2 : 1 genotype and phenotype ratio. From 200 offspring expect 50 : 100 : 50. A 1 : 2 : 1 phenotype ratio is the sign of codominance.

Multiple alleles

A gene can have more than two alleles in a population, but each individual carries only two. In the ABO blood group gene, Iᴬ and Iᴮ are codominant with each other, and the recessive allele i is masked by either.

Worked example. A woman of group A and a man of group B have a child of group O. The child is ii, so each parent carries i. The parents must be Iᴬi and Iᴮi. Gametes: Iᴬ or i from the mother, Iᴮ or i from the father. Offspring: IᴬIᴮ (AB), Iᴬi (A), Iᴮi (B), ii (O), each with probability 1/4. The probability that their next child is group AB is 1/4. The probability that their next two children are both group O is 1/4 × 1/4 = 1/16. Multiply probabilities of independent events and add probabilities of alternative outcomes.

Dihybrid crosses

A dihybrid cross follows two genes on different chromosomes, which assort independently in meiosis. In a tomato variety, assume purple stem (A) is dominant to green (a) and tall (D) is dominant to dwarf (d). Cross two plants that are AaDd.

Each parent makes four gamete types in equal numbers: AD, Ad, aD, ad. The 16-box grid gives:

Phenotype	Boxes	Ratio
Purple tall	9	9
Purple dwarf	3	3
Green tall	3	3
Green dwarf	1	1

From 320 offspring expect 180 : 60 : 60 : 20. The quick route is probability: purple dwarf needs purple (3/4) and dwarf (1/4), so 3/4 × 1/4 = 3/16. A test cross of AaDd with aadd gives 1 : 1 : 1 : 1 if the genes assort independently. A result that departs from 9 : 3 : 3 : 1 suggests autosomal linkage or epistasis.

Sex linkage

Sex linkage means the gene is on a sex chromosome. In humans and fruit flies, females are XX and males XY. The X carries many genes with no equivalent on the much smaller Y, so males have one allele and express whichever it is. Write the allele on the X as a superscript, for example Xᴴ and Xʰ.

Worked example. Duchenne muscular dystrophy is X-linked recessive. A carrier mother (XᴰXᵈ) and an unaffected father (XᴰY) have children.

Gametes	Xᴰ	Y
Xᴰ	XᴰXᴰ unaffected girl	XᴰY unaffected boy
Xᵈ	XᴰXᵈ carrier girl	XᵈY affected boy

The probability of an affected child is 1/4. Among sons it is 1/2, and no daughters are affected.

Codominant sex linkage. In cats the X carries alleles for black (Xᴮ) and orange (Xᴼ) fur, and they are codominant. Tortoiseshell cats are nearly always female, XᴮXᴼ. Crossing a tortoiseshell female with a black male (XᴮY) gives XᴮXᴮ black females, XᴮXᴼ tortoiseshell females, XᴮY black males and XᴼY orange males, 1 : 1 : 1 : 1.

Birds. In birds the pattern is reversed: males are ZZ and females ZW. A barred-plumage allele Zᴮ is dominant to non-barred Zᵇ. Cross a non-barred male (ZᵇZᵇ) with a barred female (ZᴮW). Sons are ZᴮZᵇ barred and daughters are ZᵇW non-barred, so chicks can be sexed by plumage.

Autosomal linkage

Genes on the same chromosome (autosomal linkage) are inherited together unless crossing over separates them. The alleles on one chromosome are written together, for example GN/gn.

Worked example. A female fruit fly carries grey body (G, dominant) and normal wings (N, dominant) on one chromosome and black body (g) and vestigial wings (n) on its homologue. She is crossed with a black, vestigial male (gn/gn), which can only pass on gn. Of 1000 offspring:

Phenotype	Observed
Grey, normal	412
Black, vestigial	388
Grey, vestigial	106
Black, normal	94

Independent assortment predicts 250 of each. Instead, the two parental combinations are far commoner. The grey-vestigial and black-normal flies are recombinants made by crossing over between the loci in the mother’s meiosis. There are 200 of them, 20% of the offspring. The closer two loci are, the fewer recombinants appear. A chi-squared test on these figures (below) confirms linkage.

Epistasis

Epistasis is when the allele of one gene masks or suppresses the expression of the alleles of a different gene at another locus. The result is a modified dihybrid ratio.

Recessive epistasis (9 : 3 : 4). Assume onion bulb colour needs a pigment allele C. Bulbs with C are red if R is also present and yellow if rr. Bulbs without C (cc) are white whatever the R locus has. Cross CcRr × CcRr: 9 C_R_ red, 3 C_rr yellow, 3 ccR_ and 1 ccrr white, so 9 red : 3 yellow : 4 white. From 160 offspring expect 90 : 30 : 40.

Dominant epistasis (12 : 3 : 1). In a squash variety, assume W (white fruit) masks both colours. W_ gives white (9 + 3 = 12), wwY_ yellow (3), wwyy green (1). Expect 96 : 24 : 8 from 128.

Complementary action (9 : 7). Two enzymes are needed in a pathway. Pigment appears only with a functional allele of each gene, so 9 coloured : 7 white.

To decode an epistasis question, find the total of the ratio (16 for two heterozygous parents), identify which classes have merged, and explain which gene masks the other. Say that expression at one locus depends on the alleles at the other.

The chi-squared test

The chi-squared (χ²) test compares observed counts with expected counts in categories, and the specification uses it to judge the goodness of fit of observed phenotypic ratios. Use it for counts of individuals in separate categories, never for percentages, means or continuous data. The other tests named in maths skill 1.9 are the Student’s t-test (means) and the correlation coefficient.

1. Null hypothesis: chance alone explains any gap between the observed and expected counts.
2. Expected numbers: share the total in the expected ratio.
3. Calculate: χ² = Σ (O − E)² / E, summed over every category.
4. Degrees of freedom: number of categories − 1.
5. Compare with the critical value for those degrees of freedom at p = 0.05. The critical values are 3.84 (1), 5.99 (2), 7.81 (3) and 9.49 (4). Use the critical value given in the question or a table.
6. Conclude: if χ² is less than the critical value, the difference is not significant, so you fail to reject the null hypothesis; the results fit the expected ratio. If χ² is equal to or greater, reject the null hypothesis.

Worked example 1 (dihybrid). Expected 9 : 3 : 3 : 1. Observed from 320 plants: 172 purple tall, 68 purple dwarf, 54 green tall, 26 green dwarf. Expected: 180, 60, 60, 20.

χ² = (172−180)²/180 + (68−60)²/60 + (54−60)²/60 + (26−20)²/20 = 0.356 + 1.067 + 0.600 + 1.800 = 3.82

Degrees of freedom = 3, critical value 7.81. Since 3.82 < 7.81, the difference is not significant at p = 0.05 and the data fit 9 : 3 : 3 : 1.

Worked example 2 (monohybrid). Expected 3 : 1, observed 118 and 42 (n = 160). Expected 120 and 40. χ² = 4/120 + 4/40 = 0.133, with 1 degree of freedom (critical 3.84). Not significant.

Worked example 3 (linkage). The fly data above against 1 : 1 : 1 : 1 gives χ² = 361 with 3 degrees of freedom, far above 7.81. Reject the null hypothesis: the genes do not assort independently, which is evidence for linkage.

Worked example 4 (epistasis). Observed 83 red, 36 yellow, 41 white against 9 : 3 : 4 (expected 90, 30, 40): χ² = 0.544 + 1.200 + 0.025 = 1.77; 2 degrees of freedom (critical 5.99). Not significant, so the data are consistent with recessive epistasis.

Do not write that a non-significant result proves the ratio. Say the difference could be due to chance and the results are consistent with the ratio.

Common errors

• Writing the genotype of gametes as pairs of the same gene (Aa) instead of one allele each.
• Using a 9 : 3 : 3 : 1 grid for linked genes.
• Forgetting that a male has one X allele, so X-linked recessive traits need no second copy.
• Using observed totals or percentages as the expected values in a chi-squared test.
• Using degrees of freedom = number of individuals − 1.

Official syllabus

AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, section 3.7.1 (version 1.5), published by AQA.