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, wrinkled seeds?
AaBb × aaBb
P =
2.GeneticsHard
Two AaBb parents are crossed. Gene A gives round seeds (dominant) or wrinkled seeds; gene B gives purple flowers (dominant) or white flowers. Click the one cell for an offspring that is shows wrinkled seeds and is homozygous for purple flowers.
AaBb × AaBb
- 1aB × aB
- 2aB × AB
- 3aB × ab
- 4aB × Ab
- 5Ab × aB
- 6Ab × AB
- 7Ab × ab
- 8Ab × Ab
- 9ab × aB
- 10ab × AB
- 11ab × ab
- 12ab × Ab
- 13AB × aB
- 14AB × AB
- 15AB × ab
- 16AB × Ab
3.Cell biologyMedium
In which direction do DNA and RNA polymerases add new nucleotides?
- A3′ to 5′
- BIn both directions equally
- CFrom the centromere outwards
- D5′ to 3′
4.Cell biologyHard
In the intrinsic (mitochondrial) pathway of apoptosis, which event leads to caspase activation?
- ABinding of a death ligand to the Fas receptor on the cell surface
- BEntry of glucose into the mitochondria
- CFusion of lysosomes with the nucleus
- DRelease of cytochrome c from mitochondria into the cytosol
3Solutions
1.Genetics
Answer: P = 0.125
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(wrinkled seeds) = 0.25
Multiply
P = 0.5·0.25 = 0.125
Answer
P = 0.125
2.Genetics
Answer: aB × aB
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.
Genotype asked for
aaBB
Gametes
A homozygous genotype needs the same gamete from both parents, so only one cell in the 16 fits.
aB × aB = aaBB
Answer
Cell aB × aB → aaBB
3.Cell biology
Answer: D · 5′ to 3′
Why this is the answer
Polymerases only add nucleotides to the 3′ end of the growing strand, so synthesis always runs 5′ to 3′. This causes the leading and lagging strands.
Answer
D · 5′ to 3′
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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