Genetics
- Monohybrid cross (Aa × Aa)
- genotypes 1 : 2 : 1, phenotypes 3 : 1
- Dihybrid cross
- phenotypes 9 : 3 : 3 : 1
- Hardy–Weinberg
- p + q = 1, p² + 2pq + q² = 1
- Independent genes
- multiply the probabilities
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
haemophilia is X-linked recessive (allele h). A carrier mother, XHXh, has children with an unaffected father, XHY. The mother's gametes run down the side. Click the cell for a carrier daughter.
A recessive condition affects 16% of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
Which enzyme makes the second messenger cyclic AMP when activated by a Gs protein?
Bacteria moved to a much colder environment change the fatty acids in their membranes. Which change keeps the membrane fluid?
Answer: Xh × XH
Punnett square: one parent's gametes run along the top, the other's down the side. Each cell joins the gamete above it with the gamete beside it, and every cell is equally likely. A recessive trait only shows when both alleles are recessive; for an X-linked trait, sons get their only X from the mother.
Answer: f = 0.48
Hardy–Weinberg: in a large, randomly mating population with no selection, the allele frequencies p (dominant) and q (recessive), p + q = 1, give genotype frequencies p², 2pq and q².
Answer: B · Adenylyl cyclase
Activated Gs stimulates adenylyl cyclase to convert ATP into cAMP. cAMP activates protein kinase A, and phosphodiesterase breaks cAMP down.
Answer: A · More unsaturated fatty acids
Double bonds in unsaturated fatty acids create kinks that prevent tight packing, keeping the membrane fluid at low temperatures.
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