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 short, wrinkled seeds?
AaBb × aabb
P =
2.GeneticsHard
A recessive condition affects 9% of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.09
f =
3.Cell biologyMedium
What directly drives ATP synthase to make ATP in mitochondria?
- AProtons flowing back into the matrix down their gradient
- BGlucose binding to ATP synthase
- COxygen diffusing into the matrix
- DElectrons from NADH flowing directly through ATP synthase
4.Cell biologyHard
In the most common cystic fibrosis mutation (ΔF508), the CFTR protein is made but is missing from the cell membrane. Why?
- AThe gene is not transcribed at all
- BThe mRNA cannot leave the nucleus
- CThe protein is secreted out of the cell instead of being inserted
- DThe misfolded protein is held in the ER and degraded
3Solutions
1.Genetics
Answer: P = 0.25
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(short) = 0.5
P(wrinkled seeds) = 0.5
Multiply
P = 0.5·0.5 = 0.25
Answer
P = 0.25
2.Genetics
Answer: f = 0.42
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 = √0.09 = 0.3
p = 1 − 0.3 = 0.7
Carriers
2pq = 2·0.7·0.3 = 0.42
Answer
f = 0.42
3.Cell biology
Answer: A · 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
A · Protons flowing back into the matrix down their gradient
4.Cell biology
Answer: D · The misfolded protein is held in the ER and degraded
Why this is the answer
Loss of one phenylalanine makes CFTR misfold. ER quality control retains it and sends it for proteasomal degradation, so little reaches the membrane.
Answer
D · The misfolded protein is held in the ER and degraded
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