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.
red-green colour blindness is X-linked recessive (allele b). A carrier mother, XBXb, has children with an affected father, XbY. The mother's gametes run down the side. Click the cell for a carrier daughter.
Two AaBb parents are crossed. Gene A gives yellow seeds (dominant) or green seeds; gene B gives purple flowers (dominant) or white flowers. Click the one cell for an offspring that is homozygous for yellow seeds and shows white flowers.
What is the net ATP yield of glycolysis per molecule of glucose?
In the intrinsic (mitochondrial) pathway of apoptosis, which event leads to caspase activation?
Answer: XB × Xb
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 · 2 ATP
Glycolysis uses 2 ATP and produces 4, a net gain of 2 ATP, plus 2 NADH and 2 pyruvate per glucose.
Answer: D · Release of cytochrome c from mitochondria into the cytosol
Cell stress lets pro-apoptotic Bcl-2 family proteins release cytochrome c. With Apaf-1 it forms the apoptosome, which activates caspase-9. Fas binding starts the extrinsic pathway.
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