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 unaffected father, XBY. The mother's gametes run down the side. Click the cell for an affected son.
♀ XBXb♂ XBY
- 1XB × XB
- 2XB × Y
- 3Xb × XB
- 4Xb × Y
2.GeneticsHard
In a population of 1000 in Hardy–Weinberg equilibrium, 9% show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 1000q² = 0.09
N =
3.Cell biologyMedium
Which cell junction anchors intermediate filaments and gives tissues such as skin mechanical strength?
- ATight junction
- BGap junction
- CDesmosome
- DPlasmodesma
4.Cell biologyHard
Cholera toxin modifies the alpha subunit of Gs so it cannot hydrolyse GTP. Why does this cause severe watery diarrhoea?
- AThe toxin destroys the villi
- BThe toxin blocks aquaporins in the colon
- CcAMP stays high in gut cells, driving chloride and water secretion
- DcAMP falls, so gut cells stop absorbing water and sodium from the lumen
3Solutions
1.Genetics
Answer: Xb × 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
XBXB, XBY, XbXB, XbY
Find the cell
Xb × Y = XbY
Answer
Cell Xb × Y → XbY
2.Genetics
Answer: N = 490
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 = 0.7
Homozygous dominant
p² = 0.49
N = 1000·0.49 = 490
Answer
N = 490
3.Cell biology
Answer: C · Desmosome
Why this is the answer
Desmosomes link the keratin intermediate filaments of neighbouring cells via cadherins, so tissues under stress do not tear apart.
Answer
C · Desmosome
4.Cell biology
Answer: C · cAMP stays high in gut cells, driving chloride and water secretion
Why this is the answer
Permanently active Gs keeps adenylyl cyclase on. High cAMP activates PKA, which opens CFTR chloride channels, and water follows into the gut lumen.
Answer
C · cAMP stays high in gut cells, driving chloride and water secretion
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