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
Duchenne muscular dystrophy is X-linked recessive (allele d). A carrier mother, XDXd, has children with an affected father, XdY. The mother's gametes run down the side. Click the cell for an unaffected son.
♀ XDXd♂ XdY
- 1Xd × Xd
- 2Xd × Y
- 3XD × Xd
- 4XD × Y
2.GeneticsHard
In a population of 500 in Hardy–Weinberg equilibrium, 1 in 10000 show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 500q² = 0.0001
N =
3.Cell biologyMedium
Which enzyme unwinds the DNA double helix during replication?
- APrimase
- BDNA polymerase
- CDNA ligase
- DHelicase
4.Cell biologyHard
In the most common cystic fibrosis mutation (ΔF508), the CFTR protein is made but is missing from the cell membrane. Why?
- AThe protein is secreted out of the cell instead of being inserted
- BThe mRNA cannot leave the nucleus
- CThe gene is not transcribed at all
- DThe misfolded protein is held in the ER and degraded
3Solutions
1.Genetics
Answer: XD × 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
XdXd, XdY, XDXd, XDY
Find the cell
XD × Y = XDY
Answer
Cell XD × Y → XDY
2.Genetics
Answer: N = 490.1
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 = √0.0001 = 0.01
p = 0.99
Homozygous dominant
p² = 0.9801
N = 500·0.9801 = 490.1
Answer
N = 490.1
3.Cell biology
Answer: D · Helicase
Why this is the answer
Helicase separates strands at the replication fork; primase lays primers, polymerase extends them and ligase joins fragments.
Answer
D · Helicase
4.Cell biology
Answer: D · The misfolded protein is held in the ER and degraded
Why this is the answer
Loss of one phenylalanine makes CFTR misfold. ER quality control retains it and sends it for proteasomal degradation, so little reaches the membrane.
Answer
D · The misfolded protein is held in the ER and degraded
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