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 tall plants (dominant) or short plants; gene B gives yellow seeds (dominant) or green seeds. Click the one cell for an offspring that is shows short plants and is homozygous for yellow 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
What is crossing over during meiosis?
- AFusion of two gametes
- BExchange of genetic material between homologous chromosomes in prophase I
- CSeparation of sister chromatids as they move to opposite poles in anaphase II
- DCopying of DNA before division
4.Cell biologyHard
A mutation in the lac repressor stops it binding to the operator. What is the result?
- AThe lac genes are expressed all the time, even without lactose
- BThe lac genes can never be expressed
- CThe lac genes are expressed only when glucose is present in the medium
- DThe bacteria can no longer take up glucose
3Solutions
1.Genetics
Answer: P = 0.1875
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(wrinkled seeds) = 0.25
Multiply
P = 0.75·0.25 = 0.1875
Answer
P = 0.1875
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: B · Exchange of genetic material between homologous chromosomes in prophase I
Why this is the answer
Non-sister chromatids of homologous chromosomes exchange segments at chiasmata, increasing genetic variation.
Answer
B · Exchange of genetic material between homologous chromosomes in prophase I
4.Cell biology
Answer: A · 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
A · The lac genes are expressed all the time, even without lactose
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