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
haemophilia is X-linked recessive (allele h). A carrier mother, XHXh, has children with an unaffected father, XHY. The mother's gametes run down the side. Click the cell for an affected son.
♀ XHXh♂ XHY
- 1XH × Y
- 2XH × XH
- 3Xh × Y
- 4Xh × XH
2.GeneticsHard
A recessive condition affects 1 in 10000 of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.0001
f =
3.Cell biologyMedium
Why does DNA replication need primase?
- APrimase proofreads the new DNA
- BPrimase joins the Okazaki fragments together once their RNA primers have been removed
- CPrimase unwinds the DNA double helix
- DDNA polymerase can only extend an existing strand, so a short RNA primer is needed
4.Cell biologyHard
Mitochondria have their own DNA, but where are most mitochondrial proteins made?
- AIn the Golgi apparatus
- BOn cytosolic ribosomes from nuclear genes, then imported
- CIn the rough ER and delivered by vesicles
- DInside the mitochondrion from mitochondrial genes
3Solutions
1.Genetics
Answer: Xh × 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
XHY, XHXH, XhY, XhXH
Find the cell
Xh × Y = XhY
Answer
Cell Xh × Y → XhY
2.Genetics
Answer: f = 0.0198
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 = √0.0001 = 0.01
p = 1 − 0.01 = 0.99
Carriers
2pq = 2·0.99·0.01 = 0.0198
Answer
f = 0.0198
3.Cell biology
Answer: D · DNA polymerase can only extend an existing strand, so a short RNA primer is needed
Why this is the answer
Primase makes a short RNA primer with a free 3′ end. DNA polymerase extends it; the primer is later removed and replaced with DNA.
Answer
D · DNA polymerase can only extend an existing strand, so a short RNA primer is needed
4.Cell biology
Answer: B · On cytosolic ribosomes from nuclear genes, then imported
Why this is the answer
Most mitochondrial proteins are encoded in the nucleus, made in the cytosol and imported through the TOM and TIM translocases. Human mtDNA encodes only 13 proteins.
Answer
B · On cytosolic ribosomes from nuclear genes, then imported
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