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 × Y
- 2Xb × XB
- 3XB × Y
- 4XB × XB
2.GeneticsHard
A recessive condition affects 9% of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.09
f =
3.Cell biologyMedium
What does it mean that the genetic code is "degenerate" (redundant)?
- AMost amino acids are coded by more than one codon
- BEach codon codes for several amino acids
- CSome codons are never used
- DThe code differs completely between species
4.Cell biologyHard
In the intrinsic (mitochondrial) pathway of apoptosis, which event leads to caspase activation?
- AFusion of lysosomes with the nucleus
- BEntry of glucose into the mitochondria
- CBinding of a death ligand to the Fas receptor on the cell surface
- DRelease of cytochrome c from mitochondria into the cytosol
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
XbY, XbXB, XBY, XBXB
Find the cell
Xb × Y = XbY
Answer
Cell Xb × Y → XbY
2.Genetics
Answer: f = 0.42
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 = 1 − 0.3 = 0.7
Carriers
2pq = 2·0.7·0.3 = 0.42
Answer
f = 0.42
3.Cell biology
Answer: A · Most amino acids are coded by more than one codon
Why this is the answer
Several codons can code for the same amino acid, often differing in the third base. This is why many point mutations are silent.
Answer
A · Most amino acids are coded by more than one codon
4.Cell biology
Answer: D · Release of cytochrome c from mitochondria into the cytosol
Why this is the answer
Cell stress lets pro-apoptotic Bcl-2 family proteins release cytochrome c. With Apaf-1 it forms the apoptosome, which activates caspase-9. Fas binding starts the extrinsic pathway.
Answer
D · Release of cytochrome c from mitochondria into the cytosol
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