Epidemiologic data on the population in the previous problem reve...

Question

Epidemiologic data on the population in the previous problem reveal that before the application of modern medical treatment, natural selection played a major role in shaping the frequencies of alleles. Heterozygous individuals have the highest relative fitness, and in comparison with heterozygotes, those who are βᴬβᴬ have a relative fitness of 82%, but only about 32% of those with SCD survived to reproduce. What are the estimated equilibrium frequencies of βᴬ and βˢ in this population?

Accepted Answer

Hello, everyone and welcome to today's video. So, cystic fibrosis is a genetic disease that is caused by a recessive allele, lowercase F homozygous dominant individuals do not have cystic fibrosis, but homozygous recessive individuals do have the disease, heterozygote are carriers of the disease and generally do not have symptoms. Although some may have mild symptoms. If 9% of the population is born with cystic fibrosis, what percentage of the population are carriers of the disease? Plus answer choice. A, we have 32% as answer choice B we have 48% as answer choice C we have 42% and as answer choice D we have 56%. So this problem is going to require a lot of steps. So let's try to break it down as much as possible so that we may solve the as efficiently as we can. So we're told that 9% of the population is born with cystic fibrosis. And we are told that cystic fibrosis is a recessive disease. So these 9% is going to be the frequency of the homozygosis recessive trait or genotype in the population. Now, if we look at the hardy Weinberg equation which is going to be P square plus two P Q plus Q square is going to be equal to one, these 9% is representing this Q square, which is the frequency of the homozygous recessive Geno type. So let's break this down a little bit more. We're going to take this Q square is equal to 0. or 9%. And we're going to find the square root of both sides. And we're going to obtain that Q is going to be equal to 0.3. Now this Q equals 0.3 is going to be the frequency of the recessive A in the population. If we take this Hardy Weinberg equation and we just obtain the other form of it, which is going to be P plus Q is equal to one, we can input this Q into the equation and solve for P. So it's going to be P is equal to one minus Q P is equal to one minus 0.3, P is equal to 0.7. So we know now that Q is equal to 0.3 and that P is equal to 0.7. The question is asking us for the frequency or the percentage of the population that are carriers. And we are told that these carriers are going to be heterozygote in the Hardy Weinberg equation. Two P Q is going to be the frequency of these heterozygote. So all we have to do really is to find these two P Q value, which is going to be two times P, which we said was 0.7 times Q, which we said was 0.3. And this is going to be equal to 42 or 0.42 which is the same as 42%. So this is the value or the frequency of the carriers of the disease in their population, which is represented by answer choice. C 42%. I really hope this video helped you and I hope to see you on the next one.

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