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 affected son.
Two AaBb parents are crossed. Gene A gives yellow seeds (dominant) or green seeds; gene B gives round seeds (dominant) or wrinkled seeds. Click the one cell for an offspring that is shows both green seeds and wrinkled seeds.
Apoptosis is:
Knudson's "two-hit" hypothesis, based on retinoblastoma, explains that tumour suppressor genes:
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: ab × ab
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.
A homozygous genotype needs the same gamete from both parents, so only one cell in the 16 fits.
Answer: A · Programmed cell death
Apoptosis removes damaged or unneeded cells in a controlled way via caspases; necrosis is uncontrolled death with inflammation.
Answer: A · Usually need both alleles to be inactivated before a tumour develops
Loss of one copy of RB1 is not enough; the second allele must also be lost. Children who inherit one mutant copy need only one more hit, so they often get bilateral tumours early.
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