Question

7. Pyrethrums are chemicals used to kill insects. In bed bugs, the mutant-type allele, r, confers resistance, but only in homozygous individuals. Assume the populations are in Hardy-Weinberg proportions before pyrethrum is sprayed and that there are two alleles at this gene. a. In 1955, pyrethrum was sprayed and killed 99% of the bed bug population. What were the genotype frequencies and allele frequencies in 1955 immediately before the spraying occurred

Answer 1

Answer:

Allele frequency

Normal allele  $= 0.9$

Mutant r allele $= 0.1$

Genotype frequency

Homozygous normal bugs $= 0.81$

Homozygous mutant bug $= 0.01$

Heterozygous normal bug with one mutant r allele and one normal allele $= 0.18$

Explanation:

It is given that 99% of the bugs were killed after the spray of  pyrethrum. This suggests that 1% of the bugs that were not killed must be homozygous for the mutant type allele "r"

Thus, the frequency of homozygous "rr" species i.e $q^2 = 0.01$

From this we can evaluate the frequency of mutant "r" allele.

Thus, $q = \sqrt{0.01}$

$q = 0.1$

As per Hardy-Weinberg first equilibrium equation, $p + q = 1$

Substituting the value of q in above equation, we get

$p = 1 - q\\p = 1 - 0.1\\p = 0.9$

Thus, the frequency of homozygous normal bug is equal to

$p^2 = 0.9^2 = 0.81$

As per Hardy-Weinberg second equilibrium equation-

$p^2 + q^2 + 2pq = 1$

Substituting all the available values we get -

$0.81 + 0.01+ 2pq = 1\\2pq = 0.18$

Allele frequency

Normal allele  $= 0.9$

Mutant r allele $= 0.1$

Genotype frequency

Homozygous normal bugs $= 0.81$

Homozygous mutant bug $= 0.01$

Heterozygous normal bug with one mutant r allele and one normal allele $= 0.18$