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Answer :
The given question is asking about the calculation of individuals of each genotype that would be expected if the population had not evolved. Suppose you observed 455 individual cockroaches in the infested apartment after 5 years of exposure to hydramethylnon glucose bait. If the population had not evolved, how many individuals of each genotype would you expect .
The number of individuals of each genotype that would be expected if the population had not evolved can be calculated as follows: Firstly, we need to assume that cockroach population is diploid, and the locus governing resistance is the only polymorphic locus in the genome of the cockroach species. Let R represent the dominant allele conferring resistance to the insecticide and r represent the recessive non-resistant allele.
Secondly, we need to assume that resistance is caused by a single gene that is codominantly expressed. This implies that individuals homozygous for the resistance allele (RR) and those that are heterozygous (Rr) are both resistant to the hydramethylnon glucose bait. Thus, their genotypes are indistinguishable based on the phenotypes observed. On the other hand, individuals homozygous for the non-resistant allele (rr) are non-resistant to the hydramethylnon glucose bait and are easy to differentiate from the resistant individuals.
Therefore, we expect to see only two phenotypes in the population: resistant and non-resistant. Finally, we will assume that the population is in Hardy-Weinberg equilibrium. This is a null hypothesis which stipulates that in a large, randomly-mating population, the frequency of alleles at a locus will remain constant across generations as long as there is no migration, mutation, natural selection, or genetic drift operating on the population. Since we know the total number of cockroaches in the population (N = 455), we can infer the number of individuals with each phenotype as follows: Let p represent the frequency of the R allele and q represent the frequency of the r allele.
Because there are only two alleles at the locus, p + q = 1. In addition, because the population is diploid, the frequency of each genotype can be calculated using the binomial expansion of[tex](p + q)² = p² + 2pq + q²[/tex]. Since the RR and Rr genotypes are indistinguishable,
= frequency of RR genotype2pq = frequency of Rr genotypeq²
= frequency of rr genotype Assuming the population is in Hardy-Weinberg equilibrium, we can estimate the frequency of the r allele as the square root of the frequency of the rr genotype:
q = √(frequency of rr genotype) = √(number of non-resistant cockroaches/N) = √(75/455) = 0.3109Using the equation p + q = 1, we can calculate the frequency of the R allele:
p = 1 - q = 1 - 0.3109 = 0.6891Therefore, we would expect the following number of individuals of each genotype:
RR = p²N = (0.6891)²(455) = 208.5 ≈ 209Rr
= 2pqN = 2(0.6891)(0.3109)(455) = 167.0 ≈ 167rr = q²N
= (0.3109)²(455) = 75.0 ≈ 75Hence, if the population had not evolved, we would expect to see 209 RR individuals, 167 Rr individuals, and 75 rr individuals in a population of 455 cockroaches exposed to hydramethylnon glucose bait.
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