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, round seeds?
AaBb × aaBb
P =
2.GeneticsHard
Two AaBb parents are crossed. Gene A gives tall plants (dominant) or short plants; gene B gives round seeds (dominant) or wrinkled seeds. Click the one cell for an offspring that is shows both short plants and wrinkled seeds.
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
During which phase of mitosis do sister chromatids separate?
- AAnaphase
- BMetaphase
- CTelophase
- DProphase
4.Cell biologyHard
Knudson's "two-hit" hypothesis, based on retinoblastoma, explains that tumour suppressor genes:
- AUsually need both alleles to be inactivated before a tumour develops
- BAre always inherited in mutated form from both the mother and the father
- CAre only found in children
- DCause cancer when one allele is overactive
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(short) = 0.5
P(round seeds) = 0.75
Multiply
P = 0.5·0.75 = 0.375
Answer
P = 0.375
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: A · Anaphase
Why this is the answer
In anaphase cohesin is cleaved and chromatids are pulled to opposite poles by spindle microtubules.
Answer
A · Anaphase
4.Cell biology
Answer: A · Usually need both alleles to be inactivated before a tumour develops
Why this is the answer
Loss of one copy of RB1 is not enough; the second allele must also be lost. Children who inherit one mutant copy need only one more hit, so they often get bilateral tumours early.
Answer
A · Usually need both alleles to be inactivated before a tumour develops
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