Question

Warfarin is a blood anticoagulant used for rat control during and after World War II. Initially highly successful, the effectiveness of the rodenticide gradually diminished due to the evolution of resistance among some target populations. Among Norway rats in Great Britain, resistance results from an otherwise harmful mutation WARF-R in a gene in which the normal, sensitive allele may be denoted WARF-S. In the absence of warfarin, the relative fitness of SS, SR, and RR genotypes have been estimated as 1.00, 0.77, and 0.46, respectively. In the presence of warfarin, the relative fitness values have been estimated as 0.68, 1.00, and 0.37, respectively (May 1985).

A) If the initial allele frequency of R is 0.4 in this population, calculate the expected frequency of R and mean fitness in the next generation.

B) According to the fitness scheme above, what type of selection is this?

Answer 1

WT=WARF-S (S) MUTANT = WARF-R(R)

3 genotypes : SS(homozygous wt) SR(heterozygous) RR(homozygous mutant)

relative fitness(w) is given relative fitness=numberof progenies with specific genotype/max. number of progenies

without warfin:

w(SS)=1 w(SR) =0.77 w(RR)=0.46  

with warfin:

w(SS)=0.68 w(SR)=1 w(RR)=0.37

A)allele frequency(q) of R= 0.4 allele frequency(p) of S= 1-0.4=0.6

=(0.6*0.6)0.68+2*0.6*0.4*1+0.4*0.4*0.37=0.2448+0.48+0.0592=0.784

w of S=(0.36*0.68)/0.784=0.24/0.784=0.306

w of R=0.0592/0.784=0.075

B)initially the fitness of homozygous SS was more thenn it is shifted to heterozygous in presence of warfin avoiding both the extreme genotypes. so it is a stabilising selection