Question

A farmer plants transgenic Bt corn that is genetically modified to produce its own insecticide. Of the corn borer larvae feeding on these Bt crop plants, only 10 percent survive unless they have at least one copy of the dominant resistance allele \(B\) that confers resistance to the Bt insecticide. When the farmer first plants Bt corn, the frequency of the \(B\) resistance allele in the corn borer population is \(0.02 .\) What will be the frequency of the resistance allele after one generation of corn borers have fed on Bt corn?

Short answer

Answer: After one generation of feeding on Bt corn, the frequency of the resistance allele "B" in the corn borer population is approximately 0.1736 or 17.36%.

Step-by-step solution

Identify the initial allele frequencies

The problem states that the initial frequency of the resistance allele 'B' is 0.02. Since the 'B' allele is dominant, we are given that 10% of corn borer larvae survive if they have at least one copy of this allele. The remaining 90% of the larvae that do not have the resistance allele die off. Let 'b' represent the non-resistant allele. Given that there are only two alleles in the population, the frequency of the non-resistant allele 'b' is 1 - 0.02 = 0.98.

Determine initial genotype frequencies using Hardy-Weinberg principle

According to the Hardy-Weinberg principle, the genotype frequencies can be found using the equation: \(p^2 + 2pq + q^2 = 1\), where p represents the frequency of one allele and q represents the frequency of the other allele. In our case, p = B and q = b. \(BB\) frequency = \((0.02)^2 = 0.0004\) \(Bb\) frequency = \(2(0.02)(0.98) = 0.0392\) \(bb\) frequency = \((0.98)^2 = 0.9604\)

Evaluate the next generation's allele frequency after the selective pressure

In the next generation, only 10% of the non-resistant larvae (\(bb\)) survive after feeding on the Bt corn. So the new frequency of \(bb\) genotype is reduced to 10% of its initial value, which is 0.1 × 0.9604 = 0.09604. Now we can update the genotype frequencies to account for the selection pressure: New \(BB\) frequency = 0.0004 New \(Bb\) frequency = 0.0392 New \(bb\) frequency = 0.09604 The total frequencies should still sum up to 1, so we can normalize the frequencies: Normalized \(BB\) frequency = \(\frac{0.0004}{0.0004 + 0.0392 + 0.09604} = 0.0035\) Normalized \(Bb\) frequency = \(\frac{0.0392}{0.0004 + 0.0392 + 0.09604} = 0.3402\) Normalized \(bb\) frequency = \(\frac{0.09604}{0.0004 + 0.0392 + 0.09604} = 0.6563\)

Calculate the frequency of allele 'B' after selection

To find the frequency of resistance allele 'B' after one generation, we can use the genotype frequencies as follows: New 'B' frequency = (2 * Normalized \(BB\) frequency + Normalized \(Bb\) frequency) / 2 = \((2*0.0035+0.3402)/2 = 0.1736\) So, after one generation of corn borers feeding on Bt corn, the frequency of resistance allele 'B' in the population will be around 0.1736 or 17.36%.