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A change in membrane potential
Answer:
Allele frequency
Normal allele 
Mutant r allele 
Genotype frequency
Homozygous normal bugs 
Homozygous mutant bug 
Heterozygous normal bug with one mutant r allele and one normal allele 
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 
From this we can evaluate the frequency of mutant "r" allele.
Thus, 

As per Hardy-Weinberg first equilibrium equation, 
Substituting the value of q in above equation, we get

Thus, the frequency of homozygous normal bug is equal to

As per Hardy-Weinberg second equilibrium equation-

Substituting all the available values we get -

Allele frequency
Normal allele 
Mutant r allele 
Genotype frequency
Homozygous normal bugs 
Homozygous mutant bug 
Heterozygous normal bug with one mutant r allele and one normal allele 
This may have been written wrong, but the first three are all possible. EE or Ee will show as a fish having two eyes, and ee will show as having only one. the percentage of chance for each is 75% for two eyes, and 25% for one.