If the initial allele frequencies are p = 0.5 and q = 0.5 and allele a is a lethal recessive, what will be the frequencies after 1, 5, 10, 25, 100, and 1000 generations?

Video transcript

Everyone, let's take a look at this question together, assuming that a population is in hardy Weinberg equilibrium. What is the value of P. If a recessive trait has a frequency of 5 to 1300 in the population. So let's recall what we know about hardy Weinberg equilibrium. To figure out what the value of P is, if we have that recessive trade frequency of 5 to 1300 so we know that the hardy Weinberg equilibrium formula looks like this where we have P squared plus two PQ plus Q squared equals one. Where P squared is the homesickness dominant? This PQ is the hetero ziggy's frequency and Q squared is the home recessive frequency of the population. And we also know that P plus Q equals one. And since we're trying to find the value of P that means that we have P equals one minus Q. And so looking at the recessive trait frequency, we know that Q squared equals 5 to 1300. Or in other terms five divided by 1300 which means that Q squared equals 0.34. And because we have the square we have to take the square root of both sides and we end up with Q equals 0.62. And because we're trying to find the value of P and we have Q. We plug in Q into this equation and we end up with P equals one minus 0.62 or P equals 0.938, which is answer choice A the correct answer because when this population is in hardy Weinberg equilibrium and we have that Q squared value of 5 to 1300 we do the math and we end up with q squared equals 0.34 and take the square root, so Q 0.62. And that gives us a p value of 0.938. Or answer choice A. The correct answer. I hope you found this video to be helpful. Thank you and goodbye.