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
Biology: Genetics and Cell Biology: cram pack
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
1Formulas and methods
2Practice problems
1.GeneticsMedium
Tall (A) is dominant over short (a), and round seeds (B) over wrinkled (b). The genes are on different chromosomes. For the cross AaBb × aaBb, what is the probability P that an offspring is tall, wrinkled seeds?
AaBb × aaBb
P =
2.GeneticsHard
A recessive condition affects 4% of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.04
f =
3.Cell biologyMedium
What directly drives ATP synthase to make ATP in mitochondria?
- AElectrons from NADH flowing directly through ATP synthase
- BProtons flowing back into the matrix down their gradient
- CGlucose binding to ATP synthase
- DOxygen diffusing into the matrix
4.Cell biologyHard
Bacteria moved to a much colder environment change the fatty acids in their membranes. Which change keeps the membrane fluid?
- AFewer membrane proteins
- BMore saturated fatty acids
- CLonger fatty acid tails
- DMore unsaturated fatty acids
3Solutions
1.Genetics
Answer: P = 0.125
Mendelian genetics: each parent passes on one allele of each gene, chosen at random. A Punnett square lists every combination of gametes, each equally likely. A dominant allele (capital letter) shows whenever it is present; a recessive one only in the homozygous form. Genes on different chromosomes are inherited independently, so their probabilities multiply.
Each gene on its own
P(tall) = 0.5
P(wrinkled seeds) = 0.25
Multiply
P = 0.5·0.25 = 0.125
Answer
P = 0.125
2.Genetics
Answer: f = 0.32
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.
Allele frequencies
q = √0.04 = 0.2
p = 1 − 0.2 = 0.8
Carriers
2pq = 2·0.8·0.2 = 0.32
Answer
f = 0.32
3.Cell biology
Answer: B · Protons flowing back into the matrix down their gradient
Why this is the answer
The electron transport chain pumps protons into the intermembrane space. Their flow back through ATP synthase turns it and powers ATP synthesis.
Answer
B · Protons flowing back into the matrix down their gradient
4.Cell biology
Answer: D · More unsaturated fatty acids
Why this is the answer
Double bonds in unsaturated fatty acids create kinks that prevent tight packing, keeping the membrane fluid at low temperatures.
Answer
D · More unsaturated fatty acids
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