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(2 pts.) Determine the ultimate frequency of the allele with random mating when the relative fitnesses of AA_ Aa, and aa are: 1.00 , 1.02,1.02 Pa Freauertj A F(AA )...

Question

(2 pts.) Determine the ultimate frequency of the allele with random mating when the relative fitnesses of AA_ Aa, and aa are: 1.00 , 1.02,1.02 Pa Freauertj A F(AA ) VzF(A ^) ?wa 2P4 WK +9wa4 (b) [.00, 1.01 [.02(c) 1.00.0.98,0.96(d) |.00,1.02,0.98

(2 pts.) Determine the ultimate frequency of the allele with random mating when the relative fitnesses of AA_ Aa, and aa are: 1.00 , 1.02,1.02 Pa Freauertj A F(AA ) VzF(A ^) ?wa 2P4 WK +9wa4 (b) [.00, 1.01 [.02 (c) 1.00.0.98,0.96 (d) |.00,1.02,0.98



Answers

In a population with two alleles, $B$ and $b$, the allele frequency of $b$ is $0.4 . B$ is dominant to $b .$ What is the frequency of individuals with the dominant phenotype if the population is in HardyWeinberg equilibrium?
a. 0.16
b. 0.36
c. 0.48
d. 0.84

So we're going to be running through hardy's equations that he has given to show how we can check and figure out what the genotype frequency is going to be. So we're going to go ahead and start with the equation that we would use which is P squared plus two PQ plus Q squared equals one. So we're gonna next give some values here. So P Is going to equal in this case .3 and Q is going to equal 0.7 in this example. So those are the values we're going to start off with. So first we're going to start figuring what P squared is going to be. So P squared equals 0.3 Squared which would give us .09. Next we're going to figure out what cues value is going to be which is Q squared. So Q squared equals 0.7 squared. And that value gives us .49. So now that we know what P squared and cubed squared are, we need to determine what Our other value two PQ is. So we're gonna go two P. Cute equals this value here. So we're gonna set that equation. So too times P. s. value is .3 from when we started Up here we know that that's what the value of P is Q is .7. So it's times 0.7 and that will calculate and give us the value of 0.42. So now to double check your math and make sure that we're doing this right, remember that this equation that we just determine equals one. Let's add up all of our values that we got. So we'll go .42 right? Plus we're going to add in our P. Which was .09 and then also at an RQ, Which is .49. So when we add all of these up, they give us a grand total of one. So we know that that is correct.

Hello Today we're going to solve a problem. Number seven and a population with to Aly ls Capital B and lower his be the Alil frequency of lower case V is point for uppercase B is dominant toe lower case be What is the frequency of individuals with the dominant FINA type? If the population is in Hardy Weinberg equilibrium. So first we want to determine the frequency of the dominant alil off Pete so we can use the second Hardy Weinberg equation, which is P plus Q equals one. In this case, P is our dominant uppercase be alil. So if we say sorry about that, if we say P plus q equals one in this case, we're going to sub uppercase B and lower case be equals one. So B plus 0.4 equals one. Which means that upper case be should be 0.61 minus point for this 0.0.6. So now we want to determine the frequency of individuals with the dominant FINA type. Okay, so remember, the dominant FINA type can be two things. It could be all individuals with capital B Capital B, but it could also be Capital B lower case be okay, Those would all those air the to gina types that would give us the FINA type of the dominant Oh, Leo. So we can plug our new numbers into the first Hardy Weinberg equation on our screen so it would look a little bit like this. Okay, we're gonna do P squared. Plus two p. Q plus, Q squared is equal toe one. So in this case, we're gonna do 0.6 squared plus two times 0.6 times 0.4 plus zero point for squared equals one. So if we resume solving, we now find that the frequency of P squared is 0.36 The frequency of two p Q is 20.48 and the frequency of Q squared is 0.16 Now, remember, in this question, we're looking for where these FINA types. So that would be these two right here. So if we add those two numbers together 20.36 plus point for eight, we get our final answer of 80.84 I hope that helped

Hello Today we're going to solve problem number seven and a population with two Leal's capitally and lower wispy the Alil frequency off Lower case beat His point for upper case B is dominant. A lower case B What is the frequency of individuals with the dominant FINA type if the population is in Hardy Weinberg equilibrium. So first we want to determine the frequency off the dominant alil off Pete so we can use the second Hardy Weinberg equation, which is P plus Q equals one. In this case, P is are dominant upper case be olio. So we say Sorry about that. If we say P plus q equals one. In this case, we're going to sub uppercase B and lower case being equals one. So being plus 0.4 equals one, which means the upper case be should be 0.61 minus point form is 0.0.6. So now we want to determine the frequency of individuals with the dominant Fino type. Okay, so remember, the dominant FINA type can be two things. It could be all individuals with capital B capital being, but it could also be capital be lower case be okay. Those would all those air the to Gina types That would give us the FINA type of the dominant A Leo. So we can plug our new members into the first hardy wine working equation on our screen, so it would look a little bit like this. Okay, we're gonna do P squared. Plus two p. Q plus Q square, Busy little one. So in this case, we're gonna do 0.6 squared plus two times 0.6 times 0.4 plus 0.4 squared equals one. So if we resume solving, we find that the frequency of P squared is 0.36 The frequency of to pick you is 0.48 and the frequency of queues, Where does 0.16? Now, remember, in this question, we're looking for where these fina types. So that would be these two right here. So if we add those two numbers together when 36 plus put for a we get our final answer of point a four, I hope

Here we have a question regarding Hardy Weinberg equilibrium. So we are given the low frequency of Alil Capital B, and that is equal to 0.7. Now, what we have to do to solve this question is to look at our to Hardy Weinberg equations. And for our first part, we need to determine the olio frequency of lower case. Be so what we would do. We take our 0.7, we plug it into the first equation, and we see that 0.7 plus lower case B equals one. So now salt for B and we see that B equals 0.3. So now, knowing these two facts were able to use the second equation to determine the olio frequencies of the hetero zygotes, the Homo Zegas dominance and the homo Saugus recesses. So all we have to do is we simply plug in for our hetero zygotes. We know that this would be a capital being a lower case, being being one dominant only on one recessive Leo, we plug in Teoh, find the little frequency of this. So a little frequency we plug in from our second equation this term right here so we'll bring it down to be so that's the legal frequency of Capital B, which is 0.7 times the olio frequency of lower case be, which is 0.3. So we get to time 0.7 times 0.3, which equals 0.42 So that will be the frequency of our hetero Zongo's. Right now we have to figure out the frequency for our home owes, I guess. Dominance. Okay, Our homeless, I guess dominance will be to capital these and, uh, to figure this one out, we would just simply once again plug in the term from our second party Weinberg equation into Ah, plug it in right here. So we find that b squared equals our olio frequency. So we take the illegal frequency of B capital B, we square it, and that is equal to 0.49 So 0.49 will be the frequency for our homo Zegas dominance. And then we need to find the frequency for our homeless. I guess Recesses actually do that right next to it. So our home owes, I guess recessive, remember, Since their recessive, they will have to recess Civil wheels since they're homeless, I guess, um And now we just take this term right here and we bring it down and we find that B squared is equal to our legal frequency for lower case be, which we found to be 0.3. And we're just going to square that and we get 0.9 as our answer for our home was, I guess recessive olio. Ah, frequency. So are three answers would be 0.42 0.49 and 0.9 And the, um, the Gino types for those would be as follows. We have capital be lower case be than to capital these where home was August dominance. And to lower case be for our homeless, I guess processes.


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