Genetic Analysis 20.1 predicts the number of individuals expected...

Question

Genetic Analysis 20.1 predicts the number of individuals expected to have the blood group genotypes MM, MN, and NN. Perform a chi-square analysis using the number of people observed and expected in each blood-type category, and state whether the sample is in H-W equilibrium (see Section 2.5 for the chi-square formula and table).

Accepted Answer

Hello everyone and welcome to today's video. So a dominant A which is a case H is the cause of the genetic condition known as hunting tongue's disease. A person who inherits one copy of the H A will develop the disease later in life, typically between 30 and 50. So if a population is in Hardy Weinberg equilibrium for the H A and the frequency of unaffected individuals, which is lower case H and lower case H is 0.4. What is the frequency of the carriers in the population? So as answer choice A we have 0.90 as answer choice B, we have 0.80 as answer choice C we have 0.32. And as answer choice D we have 0.20. Well, in order to solve this problem, we really need to go back to the Hardy Weinberg equilibrium equation which is going to be P squared plus two P Q plus Q square, it is going to be equal to one. So we're told that the frequency of unaffected individuals which is going to be Homo Zygo recessive is 0.4 in the Hardy Weinberg equilibrium equation, this is going to be Q square. So we know that Q square is going to be equal to 0.4. If we take the square root of both sides, we know that Q is equal to 0.2. Now if we create another form of this equation which is going to be that P plus Q is going to be equal to one we can solve for P by inputting this Q value. So we know that B is going to be equal to one minus Q P is going to be equal to one minus 0.2 P is going to be equal to 0.8. So now we have both the values of Q which is 0.2 and the values of P which is 0.8. We are asked in the question for the frequency of carriers, these carriers are going to be Zygo because they're going, they're going to carry this dominant all. However, they're going to carry alongside a recessive allele, they're going to be heterozygous for the disease. Now, the frequency of these heterozygous individuals in the population is given in the hardy Weinberg equilibrium equation by two times B times Q. So all we need to do is solve for two times P times Q in order to solve this problem. So we have a two times P times Q. It is going to be equal to two times 0.8 times 0.2. So this is going to be equal to 0.32. That is going to be our frequency of carriers in our population, which is represented by answer choice C 0.32. I really hope this helped and I hope to see you on the next video.

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