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
red-green colour blindness is X-linked recessive (allele b). A carrier mother, XBXb, has children with an affected father, XbY. The mother's gametes run down the side. Click the cell for a carrier daughter.
♀ XBXb♂ XbY
- 1XB × Y
- 2XB × Xb
- 3Xb × Y
- 4Xb × Xb
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
Why does inserting a single nucleotide into a coding sequence usually have a large effect?
- AIt always changes one amino acid only
- BIt stops the gene from being transcribed into messenger RNA
- CIt changes the sugar in the DNA backbone
- DIt shifts the reading frame of all following codons
4.Cell biologyHard
Knudson's "two-hit" hypothesis, based on retinoblastoma, explains that tumour suppressor genes:
- AAre only found in children
- BCause cancer when one allele is overactive
- CAre always inherited in mutated form from both the mother and the father
- DUsually need both alleles to be inactivated before a tumour develops
3Solutions
1.Genetics
Answer: XB × Xb
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
XBY, XBXb, XbY, XbXb
Find the cell
XB × Xb = XBXb
Answer
Cell XB × Xb → XBXb
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: D · It shifts the reading frame of all following codons
Why this is the answer
An insertion that is not a multiple of three causes a frameshift. Every codon downstream is read differently, often soon reaching a stop codon.
Answer
D · It shifts the reading frame of all following codons
4.Cell biology
Answer: D · Usually need both alleles to be inactivated before a tumour develops
Why this is the answer
Loss of one copy of RB1 is not enough; the second allele must also be lost. Children who inherit one mutant copy need only one more hit, so they often get bilateral tumours early.
Answer
D · Usually need both alleles to be inactivated before a tumour develops
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