Genetics & Inheritance

In pea plants, the allele for tall (T) is dominant to dwarf (t). Two heterozygous tall plants (Tt) are crossed. What is the expected ratio of tall to dwarf offspring in the F₁ generation?

• A3 tall : 1 dwarf✓
• B1 tall : 1 dwarf
• C9 tall : 7 dwarf
• D1 tall : 2 dwarf

Answer: A monohybrid cross Tt × Tt gives genotypes 1 TT : 2 Tt : 1 tt. Both TT and Tt are phenotypically tall, so the phenotypic ratio is 3 tall : 1 dwarf.

In a dihybrid cross between two individuals heterozygous at both loci (RrYy × RrYy), with the genes unlinked and assorting independently, what phenotypic ratio is expected in the offspring?

• A1 : 1 : 1 : 1
• B9 : 3 : 3 : 1✓
• C3 : 1
• D12 : 3 : 1

Answer: For two unlinked heterozygous gene pairs, independent assortment produces the classic dihybrid ratio of 9 : 3 : 3 : 1 in the offspring phenotypes.

In the ABO blood group system, alleles Iᴬ and Iᴮ are codominant and both are dominant to the recessive allele i. A man with genotype IᴬIᴮ has children with a woman of genotype Iᴬi. What is the probability that a child has blood group O?

• A1/2
• B1/4
• C0✓
• D3/4

Answer: Group O requires genotype ii. The father (IᴬIᴮ) carries no i allele, so no child can be ii. The probability of group O is therefore 0.

Haemophilia is caused by a recessive allele on the X chromosome. A phenotypically normal woman whose father had haemophilia has children with an unaffected man. What proportion of their SONS is expected to have haemophilia?

• A0
• B1/4
• C1/3
• D1/2✓

Answer: The woman's father was XʰY, so she must be a carrier XᴴXʰ. The unaffected father is XᴴY. Sons receive Xᴴ or Xʰ from the mother with equal probability, so 1/2 of sons are XʰY (affected).

In a large randomly mating population at Hardy-Weinberg equilibrium, a recessive condition affects 1 in 10 000 individuals. What is the approximate frequency of carriers (heterozygotes)?

• AAbout 0.0198 (roughly 1 in 50)✓
• BAbout 0.01
• CAbout 0.0001
• DAbout 0.99

Answer: q² = 1/10 000 = 0.0001, so q = 0.01 and p = 0.99. Carrier frequency 2pq = 2 × 0.99 × 0.01 = 0.0198, roughly 1 in 50.

Which of the following is NOT one of the conditions required for a population to remain in Hardy-Weinberg equilibrium?

• ANo net migration into or out of the population
• BNatural selection is acting on the trait✓
• CRandom mating with respect to the trait
• DNo mutation altering allele frequencies

Answer: Hardy-Weinberg equilibrium assumes NO natural selection, along with no mutation, no migration, random mating, and a large population. Selection acting on the trait would change allele frequencies and break the equilibrium.

A heterozygous purple-flowered plant (Pp) is crossed with a white-flowered plant (pp), where P (purple) is dominant. This is an example of a test cross. What offspring ratio confirms the purple parent is heterozygous?

• AAll offspring purple
• B3 purple : 1 white
• C1 purple : 1 white✓
• DAll offspring white

Answer: A test cross with a homozygous recessive (pp) reveals the unknown genotype. Pp × pp gives 1 Pp (purple) : 1 pp (white), a 1 : 1 ratio, confirming the purple parent is heterozygous.

In Labrador retrievers, coat colour shows epistasis: gene E (with recessive e) controls pigment deposition, and a dog of genotype ee is yellow regardless of its B/b genotype. Gene B determines black (B) versus brown (b) only when pigment is deposited. What coat colour is a dog of genotype bbee?

• ABlack
• BBrown (chocolate)
• CYellow✓
• DWhite with no pigment cells

Answer: The ee genotype is epistatic: it prevents deposition of pigment in the coat, producing a yellow dog whatever the B/b alleles are. So bbee is yellow.

In shorthorn cattle, the alleles for red (Cᴿ) and white (Cᵂ) coat are codominant, and heterozygotes (CᴿCᵂ) are roan. Two roan cattle are crossed. What is the expected phenotypic ratio of the offspring?

• A1 red : 2 roan : 1 white✓
• B3 roan : 1 white
• CAll roan
• D1 red : 1 white

Answer: With codominance there is no dominance, so the genotypic ratio equals the phenotypic ratio: CᴿCᴿ (red) : CᴿCᵂ (roan) : CᵂCᵂ (white) = 1 : 2 : 1.

Red-green colour blindness is X-linked recessive. A colour-blind man (XᶜY) has children with a woman who is homozygous unaffected (XᴺXᴺ). Which statement about their children is correct?

• AHalf the sons will be colour blind
• BAll daughters will be carriers and all sons unaffected✓
• CAll daughters will be colour blind
• DHalf the daughters will be colour blind

Answer: Sons inherit the father's Y and the mother's Xᴺ, so all sons are XᴺY (unaffected). Daughters inherit the father's Xᶜ and the mother's Xᴺ, so all daughters are XᴺXᶜ carriers.