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 an unaffected son.
In a population of 5000 in Hardy–Weinberg equilibrium, 1 in 10000 show a recessive trait. How many individuals N are expected to be homozygous dominant?
Cyclin-dependent kinases (CDKs) that drive the cell cycle become active when they bind to:
How can one human gene produce several different proteins?
Answer: XH × Y
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: N = 4901
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: A · Cyclins
CDKs are only active when bound to cyclins, whose levels rise and fall through the cycle. Active cyclin–CDK complexes phosphorylate targets that push the cell forward.
Answer: C · By alternative splicing of its pre-mRNA
Alternative splicing joins different combinations of exons from the same pre-mRNA, producing related proteins from one gene.
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