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 plants in their offspring?

• 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 (roughly 1 in 100 people)
• CAbout 0.0001 (roughly 1 in 10 000)
• DAbout 0.99 (roughly 99 out of 100)

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 (roan dominant)
• CAll roan (no white)
• D1 red : 1 white only

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

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.

What is an allele?

• AA different version of the same chromosome
• BA sequence of three bases in mRNA
• CA different version of the same gene✓
• DA section of DNA that codes for a protein

Answer: Alleles are alternative forms of a gene occupying the same locus on homologous chromosomes. A gene is the section of DNA itself; the triplet in mRNA is a codon.

What is the expected phenotypic ratio from a monohybrid cross between two heterozygotes?

• A1 dominant to 1 recessive
• B9 dominant to 7 recessive
• C1 dominant to 2 recessive
• D3 dominant to 1 recessive✓

Answer: Crossing Aa × Aa gives genotypes AA, Aa, Aa and aa. The three carrying at least one dominant allele share a phenotype, giving 3:1. The 9:3:3:1 ratio belongs to a dihybrid cross.

Why are males more often affected by X-linked recessive conditions?

• AThey have only one Y chromosome, so one allele is enough
• BThey inherit two copies of the recessive allele from parents
• CThey have only one X chromosome, so one allele is enough✓
• DThe allele is carried on the Y chromosome they inherit

Answer: With a single X, a male has no second copy to mask a recessive allele, so one faulty allele produces the condition. A female would need it on both X chromosomes, which is far less likely — haemophilia follows exactly this pattern.

What is codominance?

• ANeither allele is expressed in the phenotype at all
• BOne allele completely masks the effect of the other
• CThe phenotype is intermediate between the two alleles
• DBoth alleles are fully expressed in the phenotype✓

Answer: In codominance both alleles show, as in the AB blood group where both A and B antigens appear. An intermediate phenotype is incomplete dominance, a different thing entirely.

What does epistasis describe?

• AOne allele masking the expression of another allele
• BOne gene masking the expression of another gene✓
• CTwo genes located close together on one chromosome
• DTwo alleles of a gene being expressed together

Answer: In epistasis the alleles at one locus alter the effect of a completely different locus — for example a gene that blocks pigment production regardless of which colour alleles are present. Masking between alleles at the SAME locus is ordinary dominance.

What does the Hardy-Weinberg principle assume?

• ANo selection, mutation, migration or genetic drift✓
• BStrong selection but no mutation or genetic drift
• CRandom mutation but no selection or migration
• DConstant migration but no selection or mutation

Answer: The principle predicts allele frequencies in a population that is large, randomly mating and free from selection, mutation, migration and drift. Real populations rarely meet all the conditions, which is why a departure from the predicted frequencies is itself informative.

In the Hardy-Weinberg equations, what does 2pq represent?

• AThe frequency of homozygous dominant individuals
• BThe frequency of homozygous recessive individuals
• CThe frequency of heterozygous individuals✓
• DThe frequency of the dominant allele in the population

Answer: p² is homozygous dominant, q² homozygous recessive and 2pq heterozygous, with p² + 2pq + q² = 1. Allele frequencies themselves are p and q, which sum to 1.

Why does linkage reduce the variety produced by independent segregation?

• AGenes on one chromosome are always inherited separately
• BLinked genes cannot undergo crossing over at any point
• CLinked genes are located on homologous chromosomes
• DGenes on one chromosome tend to be inherited together✓

Answer: Independent segregation shuffles whole chromosomes, so alleles on the same chromosome usually travel as a unit. Crossing over can separate them, and the closer two loci lie, the less often that happens — which is how gene maps are built.

What is the purpose of a test cross?

• ATo find whether an organism showing the recessive phenotype is heterozygous
• BTo find the number of chromosomes present in a somatic cell
• CTo find whether an organism showing the dominant phenotype is heterozygous✓
• DTo find which genes are linked together on one chromosome

Answer: Crossing the unknown with a homozygous recessive is decisive: any recessive offspring proves the parent carried a recessive allele and was heterozygous. An organism showing the recessive phenotype must already be homozygous.