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

red-green colour blindness is X-linked recessive (allele b). A carrier mother, X^BX^b, has children with an unaffected father, X^BY. The mother's gametes run down the side. Click the cell for a carrier daughter.

♀ X^BX^b♂ X^BY

1X^b × Y
2X^b × X^B
3X^B × Y
4X^B × X^B

2.GeneticsHard

In a population of 1000 in Hardy–Weinberg equilibrium, 1 in 400 show a recessive trait. How many individuals N are expected to be homozygous dominant?

N = 1000q² = 0.0025

N =

3.Cell biologyMedium

What role does cholesterol play in animal cell membranes?

AIt makes the membrane completely rigid at every temperature
BIt helps keep membrane fluidity stable across temperatures
CIt forms channels for ions
DIt is the main carrier of glucose

4.Cell biologyHard

Which checkpoint ensures all chromosomes are attached to the spindle before anaphase begins?

AThe spindle assembly (metaphase) checkpoint
BThe G2/M checkpoint
CThe S-phase checkpoint
DThe G1/S checkpoint

3Solutions

1.Genetics

Answer: X^b × X^B

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^bY, X^bX^B, X^BY, X^BX^B

Find the cell

X^b × X^B = X^bX^B

Answer

Cell X^b × X^B → X^bX^B

2.Genetics

Answer: N = 902.5

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 = √0.0025 = 0.05

p = 0.95

Homozygous dominant

p² = 0.9025

N = 1000·0.9025 = 902.5

Answer

N = 902.5

3.Cell biology

Answer: B · It helps keep membrane fluidity stable across temperatures

Why this is the answer

Cholesterol reduces fluidity at high temperatures and prevents tight packing at low temperatures, buffering membrane fluidity.

Answer

B · It helps keep membrane fluidity stable across temperatures

4.Cell biology

Answer: A · The spindle assembly (metaphase) checkpoint

Why this is the answer

The spindle checkpoint delays anaphase until every kinetochore is attached, preventing unequal chromosome distribution.

Answer

A · The spindle assembly (metaphase) checkpoint

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