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 short, wrinkled seeds?
AaBb × aabb
P =
2.GeneticsHard
Two AaBb parents are crossed. Gene A gives yellow seeds (dominant) or green seeds; gene B gives purple flowers (dominant) or white flowers. Click the one cell for an offspring that is shows green 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
A competitive inhibitor binds the active site of an enzyme. What can overcome its effect?
- ARemoving the enzyme's cofactor
- BRaising the inhibitor concentration
- CRaising the substrate concentration
- DLowering the temperature to 0 °C
4.Cell biologyHard
Okazaki fragments are formed on:
- AThe lagging strand during DNA replication
- BThe leading strand
- CmRNA during translation
- DtRNA during transcription
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(short) = 0.5
P(wrinkled seeds) = 0.5
Multiply
P = 0.5·0.5 = 0.25
Answer
P = 0.25
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: C · Raising the substrate concentration
Why this is the answer
Because substrate and inhibitor compete for the active site, enough substrate can outcompete the inhibitor. Vmax is unchanged but the apparent Km rises.
Answer
C · Raising the substrate concentration
4.Cell biology
Answer: A · The lagging strand during DNA replication
Why this is the answer
The lagging strand is synthesised discontinuously in short fragments away from the fork, then joined by ligase.
Answer
A · The lagging strand during DNA replication
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