Genetics and evolution

ZIMSEC A Level · Biology · Form 6

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Inheritance

Monohybrid crosses give 3 : 1 phenotype ratios (heterozygote × heterozygote) and 1 : 1 (test cross).
Dihybrid crosses with unlinked genes give 9 : 3 : 3 : 1.
Codominance: both alleles expressed (e.g. roan cattle; blood group AB).
Multiple alleles: ABO blood groups (Iᴬ, Iᴮ, Iᴼ).
Sex linkage: genes on the X chromosome (haemophilia, red–green colour blindness) are more often expressed in males.
Linkage: genes on the same chromosome tend to be inherited together unless separated by crossing over.

The chi-squared test

χ² = Σ(O − E)²/E compares observed and expected results. Degrees of freedom = number of classes − 1. If χ² is below the critical value at p = 0.05, the difference is due to chance.

Population genetics

Hardy–Weinberg: p + q = 1 and p² + 2pq + q² = 1, for a large, randomly mating population without selection, mutation or migration.

Worked example: In a population, 16% show a recessive phenotype. Find the frequency of heterozygotes.

q² = 0.16 → q = 0.4, p = 0.6; heterozygotes 2pq = 2 × 0.6 × 0.4 = 0.48 (48%)

Evolution

Natural selection: variation, overproduction, competition, survival of the best adapted, inheritance of advantageous alleles.
Types: stabilising, directional, disruptive.
Speciation: allopatric (geographical isolation) and sympatric (e.g. behavioural or polyploidy).
Examples: antibiotic resistance in bacteria; insecticide resistance in mosquitoes.

Exam tip: In Hardy–Weinberg problems, always start from q², the frequency of the recessive phenotype.

Check yourself

What ratio results from a dihybrid cross of two heterozygotes (unlinked genes)?
Why are sex-linked conditions more common in males?
State two conditions for Hardy–Weinberg equilibrium.

Show answers

9 : 3 : 3 : 1.
Males have only one X chromosome, so a single recessive allele is expressed.
Any two: large population, random mating, no selection, no mutation, no migration.

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