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 an unaffected son.

♀ X^HX^h♂ X^HY

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

2.GeneticsHard

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

N = 1000q² = 0.01

N =

3.Cell biologyMedium

Which enzyme unwinds the DNA double helix during replication?

APrimase
BHelicase
CDNA polymerase
DDNA ligase

4.Cell biologyHard

Why can translation in bacteria begin before transcription of the mRNA is finished?

AThere is no nuclear envelope separating DNA from ribosomes
BBacteria do not use mRNA
CBacterial mRNA is spliced faster
DBacterial ribosomes can read the DNA template directly in the cytoplasm

3Solutions

1.Genetics

Answer: X^H × 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

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

Find the cell

X^H × Y = X^HY

Answer

Cell X^H × Y → X^HY

2.Genetics

Answer: N = 810

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 = √0.01 = 0.1

p = 0.9

Homozygous dominant

p² = 0.81

N = 1000·0.81 = 810

Answer

N = 810

3.Cell biology

Answer: B · Helicase

Why this is the answer

Helicase separates strands at the replication fork; primase lays primers, polymerase extends them and ligase joins fragments.

Answer

B · Helicase

4.Cell biology

Answer: A · There is no nuclear envelope separating DNA from ribosomes

Why this is the answer

In prokaryotes DNA lies in the cytoplasm, so ribosomes can attach to mRNA while it is still being made. In eukaryotes mRNA must be processed and exported first.

Answer

A · There is no nuclear envelope separating DNA from ribosomes

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