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 affected father, XhY. The mother's gametes run down the side. Click the cell for a carrier daughter.
Two AaBb parents are crossed. Gene A gives purple flowers (dominant) or white flowers; gene B gives yellow seeds (dominant) or green seeds. Click the one cell for an offspring that is homozygous for purple flowers and shows green seeds.
What is the net ATP yield of glycolysis per molecule of glucose?
A disease is passed from an affected mother to all her children but never from an affected father. Where is the mutation most likely located?
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: 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: B · 2 ATP
Glycolysis uses 2 ATP and produces 4, a net gain of 2 ATP, plus 2 NADH and 2 pyruvate per glucose.
Answer: C · In mitochondrial DNA
Mitochondria, and their DNA, are inherited almost only from the egg. Mitochondrial disorders therefore follow maternal inheritance.
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