Question

The Meritahti phenotype, on the planet Snnooozze is determined by one gene that has two alleles in the population. The alleles are: M (dominant allele) and m (recessive or wild-type allele). Previous studies have documented the following penetrance values: Pr(Meritahti phenotype | MM) = 0.67; Pr(Meritahti phenotype | Mm) = 0.52; Pr(Meritahti phenotype | mm) = 0.00. If a true breeding Snnooozzer who has the Meritahti phenotype mates with a true breeding Snnooozzer who has the wild-type phenotype, and two of their offspring mate, what is the most likely set of values for the F2 conditional-genotype frequencies (conditioned on having the Meritahti phenotype)?

A. Pr(MM genotype | Meritahti phenotype) = 0.60; Pr(Mm genotype | Meritahti phenotype) = 0.60; Pr(mm genotype | Meritahti phenotype) = 0.00.

B. Pr(MM genotype | Meritahti phenotype) = 0.40; Pr(Mm genotype | Meritahti phenotype) = 0.60; Pr(mm genotype | Meritahti phenotype) = 0.00.

C. Pr(MM genotype | Meritahti phenotype) = 0.25; Pr(Mm genotype | Meritahti phenotype) = 0.50; Pr(mm genotype | Meritahti phenotype) = 0.25.

D. Pr(MM genotype | Meritahti phenotype) = 0.14; Pr(Mm genotype | Meritahti phenotype) = 0.42; Pr(mm genotype | Meritahti phenotype) = 0.44.

E. Pr(MM genotype | Meritahti phenotype) = 0.00; Pr(Mm genotype | Meritahti phenotype) = 0.60; Pr(mm genotype | Meritahti phenotype) = 0.40.

Answer 1

Answer : B. Pr(MM genotype | Meritahti phenotype) = 0.40; Pr(Mm genotype | Meritahti phenotype) = 0.60; Pr(mm genotype | Meritahti phenotype) = 0.00.

Hardy-Weinberg equilibrium was given by Hardy and Weinberg, in 1908. According to the law, allelic frequency in a population remains constant if the population does not show any of these changes: genetic drift, mutation, migration, gene flow, and others. The sum of these frequencies is constant and equal to 1. Mathematically, the Hardy-Weinberg equation can be expressed as: (or) Let's assume, the frequency of M allele is p, the frequency of m allele is q, the frequency of MM genotype ispr , the frequency of Mrn is 2pq and the frequency of mm genotype is . The given genotype frequencies are as follows AA | p21-0.67 Aa (2pq) 0.52 2 So, the allelic frequencies are as foll oWs: 0.07-0.26 p 1-0.26-0.74 2pg (2x 0.26x0.74) 0.38 In the given case, a true breeding Snnooozzer who has the Meritahti phenotype (MM) mates with a true breeding Snnooozzer who has the wild-type phenotype (mm). It gives the offspring with the genotype (Mm). Now, two offspring (MmxMm) have matted to produce F2 progeny. The cross between (Mmx Mm) is as follows

Meritaht Wild (0.55) (0.07) (FI) Mm Mm Meritahti x Meritahti (0.31) (0.31) (0.2) (0.2) (0.2) (0.04) (0.04) F2) Mmmm (0.2 (0.04 (0.04) MM(0.04x1000) 40 or 0.4 □ Min = ((0.04+0.04)-0.02)x 1000) = 60 or 0.6 The frequency of Mm would be 0.6 due to of heterozygous advantage

Hence the correct option is

B. Pr(MM genotype | Meritahti phenotype) = 0.40;

Pr(Mm genotype | Meritahti phenotype) = 0.60;

Pr(mm genotype | Meritahti phenotype) = 0.00.