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.

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1Formulas and methods

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

2Practice problems

1.GeneticsMedium

haemophilia is X-linked recessive (allele h). A carrier mother, X^HX^h, has children with an unaffected father, X^HY. The mother's gametes run down the side. Click the cell for a carrier daughter.

♀ X^HX^h♂ X^HY

1X^h × Y
2X^h × X^H
3X^H × Y
4X^H × X^H

2.GeneticsHard

In a population of 5000 in Hardy–Weinberg equilibrium, 1 in 1600 show a recessive trait. How many individuals N are expected to be homozygous dominant?

N = 5000q² = 0.000625

N =

3.Cell biologyMedium

A competitive inhibitor binds the active site of an enzyme. What can overcome its effect?

ALowering the temperature to 0 °C
BRaising the inhibitor concentration
CRaising the substrate concentration
DRemoving the enzyme's cofactor

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: X^h × X^H

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

X^hY, X^hX^H, X^HY, X^HX^H

Find the cell

X^h × X^H = X^hX^H

Answer

Cell X^h × X^H → X^hX^H

2.Genetics

Answer: N = 4753

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 = √0.000625 = 0.025

p = 0.975

Homozygous dominant

p² = 0.9506

N = 5000·0.9506 = 4753

Answer

N = 4753

3.Cell biology

Answer: C · Raising the substrate concentration

Why this is the answer

Because substrate and inhibitor compete for the active site, enough substrate can outcompete the inhibitor. Vmax is unchanged but the apparent Km rises.

Answer

C · Raising the substrate concentration

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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