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
A recessive condition affects 1 in 400 of a population in Hardy–Weinberg equilibrium. What fraction f of the population are carriers (heterozygous)?
q² = 0.0025
f =
3.Cell biologyMedium
What is the G0 phase of the cell cycle?
- AA resting state in which cells have left the cycle and stopped dividing
- BThe phase in which DNA is copied
- CThe final stage of mitosis
- DA phase found only in cancer cells that have stopped responding to signals
4.Cell biologyHard
Where in the cell does the electron transport chain of aerobic respiration take place?
- AOuter mitochondrial membrane
- BCytoplasm
- CInner mitochondrial membrane
- DMitochondrial matrix
3Solutions
1.Genetics
Answer: P = 0.25
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) = 0.5
P(round seeds) = 0.5
Multiply
P = 0.5·0.5 = 0.25
Answer
P = 0.25
2.Genetics
Answer: f = 0.095
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 = 1 − 0.05 = 0.95
Carriers
2pq = 2·0.95·0.05 = 0.095
Answer
f = 0.095
3.Cell biology
Answer: A · 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
A · A resting state in which cells have left the cycle and stopped dividing
4.Cell biology
Answer: C · Inner mitochondrial membrane
Why this is the answer
Complexes of the ETC sit in the inner membrane, pumping protons into the intermembrane space; ATP synthase uses the gradient.
Answer
C · Inner mitochondrial membrane
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