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 1 in 2500 of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.0004
f =
3.Cell biologyMedium
In which direction do DNA and RNA polymerases add new nucleotides?
- A5′ to 3′
- BIn both directions equally
- CFrom the centromere outwards
- D3′ to 5′
4.Cell biologyHard
Cyanide is rapidly lethal because it inhibits which component of the cell?
- AHexokinase in glycolysis
- BCytochrome c oxidase (complex IV)
- CThe Na⁺/K⁺ pump directly
- DATP synthase's rotor
3Solutions
1.Genetics
Answer: P = 0.1875
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) = 0.75
P(wrinkled seeds) = 0.25
Multiply
P = 0.75·0.25 = 0.1875
Answer
P = 0.1875
2.Genetics
Answer: f = 0.0392
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 = √0.0004 = 0.02
p = 1 − 0.02 = 0.98
Carriers
2pq = 2·0.98·0.02 = 0.0392
Answer
f = 0.0392
3.Cell biology
Answer: A · 5′ to 3′
Why this is the answer
Polymerases only add nucleotides to the 3′ end of the growing strand, so synthesis always runs 5′ to 3′. This causes the leading and lagging strands.
Answer
A · 5′ to 3′
4.Cell biology
Answer: B · Cytochrome c oxidase (complex IV)
Why this is the answer
Cyanide binds cytochrome c oxidase and stops electron transfer to oxygen. The proton gradient collapses and aerobic ATP production stops.
Answer
B · Cytochrome c oxidase (complex IV)
ExamProblems · examproblems.com/courses/biology