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
Tall (A) is dominant over short (a), and round seeds (B) over wrinkled (b). The genes are on different chromosomes. For the cross AaBb × Aabb, what is the probability P that an offspring is tall, round seeds?
AaBb × Aabb
P =
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 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
- CThe phase in which DNA is copied
- DA resting state in which cells have left the cycle and stopped dividing
4.Cell biologyHard
A mutation in the lac repressor stops it binding to the operator. What is the result?
- AThe lac genes are expressed only when glucose is present in the medium
- BThe lac genes are expressed all the time, even without lactose
- CThe bacteria can no longer take up glucose
- DThe lac genes can never be expressed
3Solutions
1.Genetics
Answer: P = 0.375
Mendelian genetics: each parent passes on one allele of each gene, chosen at random. A Punnett square lists every combination of gametes, each equally likely. A dominant allele (capital letter) shows whenever it is present; a recessive one only in the homozygous form. Genes on different chromosomes are inherited independently, so their probabilities multiply.
Each gene on its own
P(tall) = 0.75
P(round seeds) = 0.5
Multiply
P = 0.75·0.5 = 0.375
Answer
P = 0.375
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: D · 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
D · A resting state in which cells have left the cycle and stopped dividing
4.Cell biology
Answer: B · The lac genes are expressed all the time, even without lactose
Why this is the answer
If the repressor cannot bind the operator, RNA polymerase is never blocked, so the operon is constitutively expressed.
Answer
B · The lac genes are expressed all the time, even without lactose
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