The rat poison Warfarin prevents blood from clotting Some ra

The rat poison Warfarin prevents blood from clotting. Some rats are resistant to Warfarin. Resistance controlled by a single gene. The normal function of this gene is to code for a protein involved in vitamin K metabolism. Homozygous recessive (rr) rats die of hemorrhages when exposed to Warfarin while RR rats need about 20 times as much vitamin K as other rats. RR rats need only 2 or 3 times as much vitamin K as do rats with the rr genotype. In one population, the genotypes have the following frequencies: a. What are the observed frequencies of the 2 alleles in this population? Be careful - you can\'t assume this population is in HW equilibrium; you need to make this problem look similar to the data in #3. b. What are the expected frequencies of each genotype if the population was in H-W equilibrium? Is this population in H-W equilibrium? Explain your decision.

Solution

5a) Genotype frequency: RR (p²) = 0.06, Rr (2pq) = 0.60, rr (q²) = 0.34

Rats with genotype rr will not survive because the allele R confers the resistance.

Rats with genotype RR also cannot survive or has less survival rate because they need 20 times more vitamin K which may not be available.

So the heterozygous genotype can only survive and that is alone taken into account to calculate allele frequenc

Allele frequency:

p = p² + 2pq/2 = 0 + 0.60/2 = 0.30

q = q²+ 2pq/2 = 0+0.60/2 = 0.30

5b) Expected genotype frequency

RR (p²) = 0.36 x 0.36 = 0.1296

rr (q²) = 0.64 x 0.64 = 0.4096

Rr (2pq) = 2 x 0.36 x 0.64 = 0.4608

Although genotype frequency equals to 1 this population is not in Hardy-Weinberg equilibrium because only the Rr genotype individuals can survive so the frequency of other genotypes will be less than the expected.

5c) This is an example of balanced polymorphism because both allele should be present to balance the requirement of Vitamin K and the allele R confers the resistance to warfarin.