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
Duchenne muscular dystrophy is X-linked recessive (allele d). A carrier mother, XDXd, has children with an affected father, XdY. The mother's gametes run down the side. Click the cell for an affected son.
♀ XDXd♂ XdY
- 1Xd × Y
- 2Xd × Xd
- 3XD × Y
- 4XD × Xd
2.GeneticsHard
A recessive condition affects 1 in 400 of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.0025
f =
3.Cell biologyMedium
In which direction do DNA and RNA polymerases add new nucleotides?
- A3′ to 5′
- B5′ to 3′
- CFrom the centromere outwards
- DIn both directions equally
4.Cell biologyHard
How does a cell mark a damaged or short-lived protein for destruction by the proteasome?
- ABy attaching a chain of ubiquitin molecules
- BBy removing all its phosphate groups
- CBy adding a signal sequence for the ER
- DBy adding mannose-6-phosphate
3Solutions
1.Genetics
Answer: Xd × Y
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.
Fill the square
XdY, XdXd, XDY, XDXd
Find the cell
Xd × Y = XdY
Answer
Cell Xd × Y → XdY
2.Genetics
Answer: f = 0.095
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 = √0.0025 = 0.05
p = 1 − 0.05 = 0.95
Carriers
2pq = 2·0.95·0.05 = 0.095
Answer
f = 0.095
3.Cell biology
Answer: B · 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
B · 5′ to 3′
4.Cell biology
Answer: A · By attaching a chain of ubiquitin molecules
Why this is the answer
Enzymes attach a polyubiquitin chain to the target protein. The proteasome recognises this chain, unfolds the protein and breaks it into short peptides.
Answer
A · By attaching a chain of ubiquitin molecules
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