BMST - Basic Math and Science Genetics and Heredity Questions and Answers Flashcards

A woman who is a carrier for hemophilia A (X^H X^h) has children with a man who has hemophilia A (X^h Y). What is the probability that their daughter will have hemophilia A?

Answer: 50%

The mother produces eggs X^H and X^h in equal proportions. The father produces X^h sperm only (no Y produces daughters). Daughters receive X^h from father plus either X^H or X^h from mother. That gives X^H X^h (carrier) or X^h X^h (affected) in equal proportions — so 50% of daughters will have hemophilia A.

In Mendel's dihybrid cross of true-breeding round yellow (RRYY) with wrinkled green (rryy) plants, the F2 generation shows a 9:3:3:1 ratio. If instead the two genes show complete linkage (no crossing over), what phenotypic ratio would appear in the F2?

Answer: 3:1

Complete linkage means the R and Y alleles never separate from each other. The F1 is RrYy but the gametes are only RY and ry (no recombinants). Crossing two F1 plants gives offspring RRYY : RrYy : rryy in a 1:2:1 ratio, but only two phenotypes — round yellow and wrinkled green — in a 3:1 ratio, just like a monohybrid cross.

A plant species is normally diploid (2n = 16). A colchicine treatment prevents spindle formation during meiosis in a cell, doubling the chromosome number. The resulting gamete then fuses with a normal haploid gamete. What is the chromosome number of the offspring?

Answer: 24

Colchicine blocked spindle formation during meiosis, so the gamete produced is diploid (n = 16) instead of haploid (n = 8). A normal haploid gamete contributes n = 8. The resulting offspring has 16 + 8 = 24 chromosomes — a triploid organism (3n = 24).

In a population of 10,000 individuals, the frequency of a recessive allele (q) for a genetic disorder is 0.02. Assuming Hardy-Weinberg equilibrium, approximately how many individuals in the population are carriers (heterozygous)?

Answer: 392

Under Hardy-Weinberg equilibrium, p = 1 − q = 0.98. The carrier frequency is 2pq = 2 × 0.98 × 0.02 = 0.0392. Multiplied by 10,000 individuals gives 392 carriers. Note that the homozygous recessive frequency (q² = 0.0004) yields only 4 affected individuals, a common distractor.

A cross between two plants with purple flowers produces offspring in a ratio of 2 purple : 1 red : 1 white. Which inheritance pattern best explains this result?

Answer: Two alleles showing complementary gene interaction

A 2:1:1 ratio from an intercross suggests two alleles at one locus where one homozygous class is lethal — but that gives 2:1. A ratio of 2 purple : 1 red : 1 white more precisely fits complementary (duplicate recessive) epistasis at two loci in an AaBb × AaBb cross when A_B_ = purple, A_bb = red, aaB_ = white, and aabb = also one color, but the cleanest fit here is two loci where purple requires at least one dominant allele at each locus (A_B_), giving a 9:3:3:1 modified. The 2:1:1 exactly matches a cross of two heterozygotes at a single locus with heterozygote advantage or, more classically, a situation where Aa × Aa produces AA (white), Aa (purple), aa (red) in 1:2:1 — yielding 2 purple : 1 white : 1 red — best described as incomplete dominance where the heterozygote produces a distinct (purple) phenotype.

A geneticist discovers a new mutation in Drosophila where females with genotype X^a X^a are viable and fertile, but males with genotype X^a Y die before hatching. A female carrier (X^A X^a) is mated with a wild-type male (X^A Y). What fraction of ALL offspring (male + female) will be viable?

Answer: 3/4

The cross X^A X^a × X^A Y produces four equally likely offspring: X^A X^A (viable female), X^a X^A (viable carrier female), X^A Y (viable male), X^a Y (lethal male). Three of the four classes survive, so 3/4 of all potential offspring are viable.