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 affected father, XbY. The mother's gametes run down the side. Click the cell for a carrier daughter.
♀ XBXb♂ XbY
- 1Xb × Xb
- 2Xb × Y
- 3XB × Xb
- 4XB × Y
2.GeneticsHard
In a population of 500 in Hardy–Weinberg equilibrium, 1 in 400 show a recessive trait. How many individuals N are expected to be homozygous dominant?
N = 500q² = 0.0025
N =
3.Cell biologyMedium
What is the G0 phase of the cell cycle?
- AThe final stage of mitosis
- BA phase found only in cancer cells that have stopped responding to signals
- CA resting state in which cells have left the cycle and stopped dividing
- DThe phase in which DNA is copied
4.Cell biologyHard
Why is DNA replication called semi-conservative?
- AThe new DNA is completely new
- BMutations are conserved in half of the daughter cells after every division
- COnly half of the DNA is copied
- DEach new double helix contains one original strand and one new strand
3Solutions
1.Genetics
Answer: XB × Xb
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
XbXb, XbY, XBXb, XBY
Find the cell
XB × Xb = XBXb
Answer
Cell XB × Xb → XBXb
2.Genetics
Answer: N = 451.3
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 = 500·0.9025 = 451.3
Answer
N = 451.3
3.Cell biology
Answer: C · A resting state in which cells have left the cycle and stopped dividing
Why this is the answer
Cells in G0 have exited the cycle and do not prepare to divide. Some can return to G1 when stimulated, such as liver cells after injury.
Answer
C · A resting state in which cells have left the cycle and stopped dividing
4.Cell biology
Answer: D · Each new double helix contains one original strand and one new strand
Why this is the answer
Meselson and Stahl showed that after one replication the DNA had intermediate density, consistent with one old and one new strand.
Answer
D · Each new double helix contains one original strand and one new strand
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