Q28 — Epistasis Between Two Hair-Pigmentation Genes

Theoretical B Real exam question — full text reproduced under IBO's CC BY-NC-SA 4.0 license

Gene Interactions: Epistasis & Complementation Analysis

In a hypothetical human population genes A and B control hair pigmentation, where hair color can be either black or white. Both genes A and B are autosomal, have two alleles, exert complete dominance, and are unlinked. In addition, these genes follow the Hardy-Weinberg principle and the frequency of the dominant allele of gene A is 0.4.

In this question, you are given a genetic pedigree of individuals from this population, where black squares and circles represent black-haired individuals. Genotypes of some individuals are shown as well. Note, that the genotype of individual IV-8 for gene A is unknown.

A five-generation pedigree with genotype labels (e.g. Aabb, AaBb, aaBb) shown under several individuals, black-filled shapes marking black-haired individuals, ending in individual V-1 whose phenotype is being asked about. Figure 1. Pedigree of the hypothetical population, generations I-V, with known genotypes labeled under the relevant individuals.

On your answer sheet, indicate “T” for true statements and “F” for false ones.

A. There is an epistatic relationship between genes A and B.

B. The probability that double heterozygous parents will have a child with white hair is 7/16.

C. The probability that individual V-1 will have white hair is 0.35.

D. Now, assume that you don't know whether gene A is autosomal or X-linked, but you know that it is not located on chromosome Y. In this case, based on the pedigree with the indicated genotypes alone, gene A can be located in every human chromosome except chromosome Y.

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Question reproduced from IBO 2024, Theoretical Exam Part B, licensed under CC BY-NC-SA 4.0 — attributed to the International Biology Olympiad. Open the full exam PDF