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 unaffected 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% of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.01
f =
3.Cell biologyMedium
Cyclin-dependent kinases (CDKs) that drive the cell cycle become active when they bind to:
- ACyclins
- BHistones
- CTubulins
- DRibosomes
4.Cell biologyHard
In E. coli, under which conditions is the lac operon most strongly transcribed?
- ABoth lactose and glucose absent
- BLactose present and glucose absent
- CBoth lactose and glucose present
- DLactose absent and glucose present
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 = 0.18
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 = √0.01 = 0.1
p = 1 − 0.1 = 0.9
Carriers
2pq = 2·0.9·0.1 = 0.18
Answer
f = 0.18
3.Cell biology
Answer: A · Cyclins
Why this is the answer
CDKs are only active when bound to cyclins, whose levels rise and fall through the cycle. Active cyclin–CDK complexes phosphorylate targets that push the cell forward.
Answer
A · Cyclins
4.Cell biology
Answer: B · Lactose present and glucose absent
Why this is the answer
Lactose (as allolactose) removes the repressor. Low glucose raises cAMP, which activates CAP to boost transcription. Both are needed for maximal expression.
Answer
B · Lactose present and glucose absent
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