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Population genetics problem If you have these data:Individuals are homozygous dominant for insertion of PV92 -90Individuals are heterozygous for insertion of PV92...

Question

Population genetics problem If you have these data:Individuals are homozygous dominant for insertion of PV92 -90Individuals are heterozygous for insertion of PV92-60How many individuals are homozygous recessive - 45Calculate the following:a) frequency of observed genotypesb) determine the total number of alleles observedc) Calculate the total number of alleles (2N)d) Determine the allele frequencye) Calculate the expected genotype frequencyf) Calculate the expected number of each genotype.g) P

Population genetics problem If you have these data: Individuals are homozygous dominant for insertion of PV92 -90 Individuals are heterozygous for insertion of PV92-60 How many individuals are homozygous recessive - 45 Calculate the following: a) frequency of observed genotypes b) determine the total number of alleles observed c) Calculate the total number of alleles (2N) d) Determine the allele frequency e) Calculate the expected genotype frequency f) Calculate the expected number of each genotype. g) Perform the chi square test h) Determine the degrees of freedom i) Compare the results with the chi square distribution table using 0.5 as the alpha value. j) Interpret your results.



Answers

Consider 10 populations that have the genotype frequencies shown in the following table: a. Which of the populations are in Hardy-Weinberg equilibrium? b. What are $p$ and $q$ in each population? c. In population $10,$ the $A \rightarrow a$ mutation rate is discovered to be $5 \times 10^{-6}$. What must be the fitness of the $a / a$ phenotype if the population is at equilibrium? d. In population $6,$ the $a$ allele is deleterious; furthermore, the $A$ allele is incompletely dominant; so $A / A$ is perfectly fit, $A / a$ has a fitness of $0.8,$ and a/a has a fitness of $0.6 .$ If there is no mutation, what will $p$ and $q$ be in the next generation?

In this question, we're testing whether the births are uniformly distributed across seasons. So I have 120 observations of students births according to the season that they're born. That's 100 20 and for party were asked, What is the expected number of births in each season, assuming that there is a uniform distribution? So since there are four seasons, we expect 1/4 of the births to be in each season. So 1/4 of 120 for each season, so 1/4 of 100 20 is 30. So we're expecting 30 berths in each season for Part B were asked to compute the chi squared statistic. So you recall that this is the sum of so this will be the some of of the observed values minus C expected values squared, divided by the expected value. So we could make another column for this table and calculate this parameter for each season. So for the 1st 1 would do that just as an example. So observed is 25. That's the 25 birth in the winter, minus the expected, which is 30 squared over 30 and that comes out to 25/30 so that would be the 1st 1 And then you might do three more so that you have one for each season, and then you could some them. So I used a calculator to calculate the chi squared statistic for me gave me a value of 1.93 Now moving on to see, whereas how many degrees of freedom does this chi squared statistic have that is equal to end minus one degrees of freedom and minus one, and we have four categories, so it's four minus one gives us three degrees of freedom for D, Using a significance level of 0.5 we're to look up a critical value. So we have way have Alfa 0.5 and we also have our degrees of freedom. So what? These two things We could go to the table and check. So Alfa is the area in the upper tail, so 0.5 and we have three degrees of freedom. So are critical. Value is 7.815 and finally for E using the critical value were supposed to conclude about the no hypothesis. But the question didn't explicitly ask for the no hypothesis, but no hypothesis is that birther evenly distributed across season. And so we're comparing two values are chi squared statistic and the Critical Valley. And because they're chi squared, statistic is less than the critical value. We we failed to reject the null hypothesis, and we can say that we do not have sufficient evidence to believe that births are not distributed evenly across season.

The drawing shown is a sample list of Gina types at Locus A for 10 individuals to calculate the observed frequency of recess civil eels or, in this case, the lower case a weenie dicked solve. For Q and use the following equation. I am going to use the highlighting tool to show you where I'm getting these numbers from your home. Ozai Ghous Recess Civil eels need to be counted first and then multiplied by two highlighted in yellow. There are four homos, I guess. Recessive alleles. You're hetero Ziegesar Leal's need to be counted next. In this case, it will be your lower and uppercase A's or one dominant Alil and one recess of Alil. I have highlighted them in blue, and we have four. We then divide this by N, which is the total number of individuals. They're 10, and I have highlighted them in red. We know that two times four is eight eight plus four is 12, two times 10 equals 20 and 12, divided by 20 equals 200.6. So, using the equation, we have determined that the frequency of the recess of Lille is also equal to Q, which is 0.6 now we will solve for the frequency of the dominant Alil, also known as P. This is solved by using the following equation. I will highlight the information again so you know where all of the numbers air coming from. First, we need to count the Homo Zika's dominant A Leal's and insert them into the equation. This is going to be highlighted in red. There are two of them total. Then we're going to count the hetero Ziegesar Leal's, which are going to be highlighted in yellow. There are four and lastly, the total number of individuals is also needed to solve the equation. This is going to be highlighted in blue. There are 10 now. We saw the equation two times two is four. Four plus four is eight. Two times 10 is 20 on finally eight, divided by 20 is 200.4. This tells us the frequency of the Homo dominant trait, or P, is point. For now, we're going to solve for the observed frequency of Homo zegas recessive genes. Using the same information, we will count the total number of Homo zegas recess of the Leal's. This is going to be highlighted in yellow. There are four, then we're going to divide it by the total number of individuals. This will be highlighted in blue. There are 10 four. Divided by 10 is 100.4, so the frequency of Homo Zika's recessive is 0.4. Next, we will calculate the frequency for hetero Ziegesar, Leo's. This is solved by counting the total number of heterocyclic individuals. This is highlighted in green and there are four. Then we divide that by the total number of individuals, which is highlighted in blue, there are 10 four. Divided by 10 is 100.4, So the hetero Zika's frequency is 0.4. Now we will solve for the Homo Zegas dominant frequency. We need to count the total number of Homo Zegas dominant genes there, too, and they're highlighted in yellow. We then needed to divide that by the total number of individuals, which is 10 again. It's highlighted in blue two. Divided by 10 is 100.2, so the Homo Zika's dominant frequency is point to. If we use the hardy Weinberg expectation, weaken solve for the frequency of the next generation. Using the following equations, we will start with the Homo Zika's dominant genotype. I will highlight the numbers that we need in yellow. This could be solved by P. Squared P is 0.4. So 0.4 squared is 0.16 This means 0.16 is the Ho Mose, I guess. Dominant frequency. Now we will solve for the hedgerows, I guess. One. This will be highlighted in green. The equation needed is two times p. Times Q. This means we need to multiply two times 20.6 times 0.4. This equals 0.48 So 0.48 is your hetero Zika's frequency. Lastly, we will calculate the Ho Mose, I guess. Recessive frequency. This is highlighted in red, thick way Asian needed is Q squared. So 0.6 squared equals 0.36 This is your home owes, I guess recessive frequency for the Hardy Weinberg expectations toe hold True, certain conditions must be followed if they are not met. Some of the observed Gina types may differ. The expectations are as followed. There could be no mutations. There could be no selection among Gino types. There can be no gene flow. The population size is infinite and the mating is random. If these conditions air held, we should expect to see the frequencies we calculated

So, um, this is a little bit of a confusing question. Our goal is to figure out how common the double header is. I go big, a little a big be that will be will be when the population is in a linkage equilibrium. So, as we says, they threw questions, starts off the game. Meats are thes frequencies, and I'm just going to scribble them down here. So this is point former, This is 0.0.1. Uh, this is 0.1, and this is point for All right, So this is the comedic frequencies, and there's think it's just equilibrium right now, which means that a and B or not, um, sort of traveling independently in recombination that there because they're close together, say, on the crossover, somebody like that, um, they are traveling together. And so that's why we have this thing on, said Afghan means here. But if you take a moment and look and let's divide them in half here, uh, Big A's makeup 0.5 of the Grammys Little A's make up 25. The gambits. Do you see that? So point foreign 0.0.1. So half the game meats are big A or 0.5. The game meats, our Big A and then half of the game meats are little A and the same you'll notice, is true for Bigby and Little B. So point for Post 0.0.1 is 25. So half of the game meats have big being them, and then the same thing. Half of the game means have little being them. That doesn't have to be that way. That's that's how this question was set up so that we would hopefully pay attention to it. So that's just the frequency of the of the A. Leal's in his population. Now the question says, eventually the population will make it to legalistic delivery are linkage, equilibrium. And then we're interested in how common this doubleheaders I go. It will be, But we can do this in two steps. We can say, how frequent, How common will the headers? I go to a big a big A B and how come, well, the headers like a big, be little BB, though. And of course, if it's in Hardy Weinberg equilibrium, it'll be two p. Q. And so that's just too Times 0.5, which is the frequency, say, of Big a and then frequency of Q would be awesome 0.5. And that's, Ah, frequency of little A. And so, if you do that, of course, um, 2.5 times 25.25 times two is 0.5. And we could do exactly the same math for big be little B and get exactly the same notion. And so what this tells us is that after the linkage, equilibrium occurs right half of the time we will see headers. I goes Bigby little be in in the population of unit face. So then the question comes, How common will the doubleheaders? I got a big a big, a little a big be little b become in the population. Well, that's simply the probability of these two independent events happening. So remember us multiply when you have independent events. Does he just most by 0.55 point five, and so that would be 0.0 to 5. All right, so what does that mean? It means that eventually, when these gammy its end up in a population that's in think it's equilibrium, right? The probability of getting a double headers. I go big, a little a baby little bit is simply the product of these independent probability. So 0.5 times 0.5 and so that gives us a probably of 0.25 or 25%. That's how common doubleheaders. I guess we'll be in the population no.

Tricky problem here. So, um, we're basically told about an ensemble dominant alil that has, ah, very negative effects on individuals who possess it. Um, and it apparently is mean maintained only by mutations from the non dominant excessive into the dominant forms that seem street was going on here. Um, this is not the easiest question. I think there's some assumptions you have to make in order to get this one. But let's just go through this. The formula, that's, um we're going to need is this one, which is the one that connects mutation rates to selection. And the only difference here is we're gonna actually resolving for P instead of Q because we're working with the dominant rather than a recessive alil. Um, and so we have to go through. And so this is what we're look shooting for. We're trying to find this number. The the mutation rate, Um, s is the selection. And remember that we're told in the question that there is a 70. No, there's a 30% um, sort of success rate that there's a detriment of 70%. Eso selection is 700.7. In this question. This is the degree of dominance, little age story of dominance, and we're not really told with the degree of dominance is. But presumably the home is highest dominance die. And so, if that's the case, um, we're going to assign the value of one Teoh the degree of dominance value. And so we have to find que in order to get new. Um, so remember that we have headers. I got's, um, The question says that the number of infected individuals is off, or the proportion effective individuals is four times 10 the minus six. The assumption going is that those were the headers I got said that that's because the home is, I guess diamonds were dying. So that means that this number equals two p you know we want to solve for Q. But we can't fall for P two as well, because, you know they can have to solve for two variables at the same time here. Um, so what we're going to do is say that Q is approximately one, and again that makes sense because we've been told this auto sort of dominant has very negative effects. Selection should be acting very strongly against it, so that should be a very rare alil in the population. And so if he's make this one, then weaken solve for P. Ah, and so p is simply going to be for times. End of man six divided by two. Right, So we'll take this to write this by two. And so that will give us two times 10 to the minus six. All right, so now what we're gonna do is use this equation again, but we're going to solve for P instead of Q because it's again. It's a dominant A Leo. So we just have toe, uh, plug our numbers in and solve from you. So let's just solve from you first. So new equals Ah, H s p. All right. And so, uh, then U equals begin one. That's a number we just assigned to age. Assuming that that the Leo ah is very bad. 0.7 and then this number right here to times 10 to the minus six. So if you do that, you get the mutation rate that's maintaining this khalil In the population that's 1 24 times 10 to the minus six


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