Genetics: monohybrid crosses, epistasis and the Hardy–Weinberg equation

Set out every cross the same way

• Define the symbols explicitly: let A = allele for a purple flower, a = allele for a white flower.
• Write parental phenotypes, then genotypes, then gametes in circles.
• Draw the Punnett square, list offspring genotypes, then phenotypes with the ratio.
• Answer the actual question — if it asks for probability, give a fraction or percentage, not a ratio.

The variations examiners use

Codominance: both alleles are expressed, as in the AB blood group; use superscript notation such as I^A I^B. Multiple alleles: three or more forms exist in the population, though any individual still has two. Sex linkage: the gene is on the X chromosome, so write genotypes as X^H Y — males cannot be carriers, which explains why X-linked recessive conditions are more common in males.

Autosomal linkage: genes on the same chromosome are inherited together unless crossing over separates them, so the expected 9:3:3:1 dihybrid ratio is not seen and parental combinations are over-represented.

Epistasis: one gene masks the expression of another, giving modified ratios such as 9:7, 9:3:4 or 12:3:1. Identify which gene is epistatic before drawing anything.

Chi-squared, step by step

Calculate expected numbers from the predicted ratio applied to the actual total. Then chi-squared is the sum of (observed − expected) squared divided by expected. Degrees of freedom are the number of classes minus one. Compare with the critical value at p = 0.05.

The conclusion sentence must be written carefully: if the calculated value is less than the critical value, any difference between observed and expected is due to chance, so accept the null hypothesis. If it is greater, the difference is significant and something other than chance — linkage, selection, lethal alleles — is acting.

Hardy–Weinberg in practice

• p + q = 1, where p is the frequency of the dominant allele and q the recessive allele.
• p² + 2pq + q² = 1, giving the frequencies of homozygous dominant, heterozygous and homozygous recessive genotypes.
• Start from the recessive phenotype: its frequency is q², so q is its square root; then p = 1 − q, and carriers are 2pq.
• The assumptions — no mutation, no migration, no selection, random mating, a large population — must be stated when a question asks why the prediction might not hold.

Selection questions

Directional selection shifts the mean towards one extreme, typically after an environmental change, as in antibiotic resistance. Stabilising selection favours the mean and reduces variation, as in human birth mass. Every good answer follows the same chain: variation exists due to mutation, a selection pressure means some phenotypes survive and reproduce more, their alleles are passed on, and the allele frequency increases over generations.