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?

Video transcript

Hi, everybody. Let's take a look at this practice problem together given that the initial frequency of the capital B resistance A lil in the corn borer population is 0.3. What would be the frequency of the recessive genotype? Little B, little B? So in the corn borer population, the question tells us that the recess of A lil is little B. Therefore capital B is the dominant a little. So how do we use the frequency of the dominant A will to solve for the frequency of the recessive genotype? Recall that we will use the Hardy Weinberg equation on the screen. Let's review it. We've got P squared plus two PQ plus Q squared equals one where P is the frequency of the dominant A lil and Q is the frequency of the recessive a little. And also note that P plus Q equals one or the frequency of both alleles has to equal 100%. And then in the equation, we have P squared, P squared represents the frequency of the homocide dominant genotype. So for our question, that is capital B, capital B, Then we have two PQ And two PQ represents the frequency of the Hetero Ziggy's genotype. So for us that is capital B little B, that means Q squared represents the frequency of the Homo zegas recessive genotype, Little B Little B. Okay. Thus far, we've reviewed the Hardy Weinberg Equation. But how do we use it? Recall that in the question, we were told that the frequency of the dominant lil is 0.3, that means we know that p equals 0.3. So there's really just two steps at this point that we're going to use to solve our answer. The first step is we're going to determine que the frequency of the recessive A little using P plus Q equals one. Then we will solve for Q square, the frequency of the Homo Zegas recessive genotype. So let's solve step one, we have P plus Q equals one, 0.3 plus Q equals one. Therefore Q equals 0.7. So this is the frequency of the recessive A little little B. Then we've got step two and step two is Q squared equals 0.7 squared equals 0.49. And this is the frequency of the homesickness recessive genotype. Little B little B. So 0.49 is the correct answer, which means d option D is our correct answer. Alright, everyone. I hope you found this helpful and I'll see you soon for the next practice problem.