Crossing Over in Flies, final challengeWhite (w) & Yellow (y) are both on the X chromosome and only 1.5 M.U. apart. Let us see if the data obtained fits with a sex-linked, linked gene (at 1.5 M.U.), recessive inheritance model.1. Determine expected frequencies of each individual progeny possibility based on the 1.5 M.U. distance. Do this by showing a punnet square for frequencies. This one is challenging since it involves two genes both on the X chromosome and we are going to assume that crossing over does not happen in males. However, we are still going to ignore the effects of multiple crossover events.a. (1.5 points) First, determine what gametes each adult of the F1 can produce (don’t calculate probabilities yet). You may need to consider sex, linkage, and lack of male crossing over to determine this. Remember this is class data so the P generation was affected males crossed to homozygous wild type females.(The F1 adults are X++/Y and X++/Xyw)(Note that a lack of crossover in males doesn’t actually come into play here since the male is hemizygous!) Students may represent this in slightly different ways. I think using superscripts work better to differentiate between the X and Y. I’m using parentheses in the square above just make it visually more clear that they are both on the X.b. (1 point) Which gametes represent the “parental” type? Which are “non-parental,” and so therefore are a result of crossing over?The X+ +, and Xy w are the parental types. The Xy + and X+ w are the non-parentals. They don’t have to specify the Y chromosome as either type, but if they do it counts as parental.c. (1.5 point) Construct the punnet square showing the genotype outcomes of this F1 sibling cross (don’t calculate probabilities yet).d. (1.5 points) Construct a punnet square showing the expected genotype probabilities/frequencies (based on 1.5 M.U.) and then calculate and show the phenotype frequencies. Remember how recombination frequency relates to the frequency of parental and non-parental gametes.Gamete probabilities are going to be ½ 1.5% for each of the recombinants/non-parental type (0.75% or 0.0075) and then ½ of 98.5% (49.25% or 0.4925) for the nonrecombinant/parental type. Filling out a square is easy from that point. They don’t need to have as many significant figures as indicated below (I just copied it from excel).Phenotypes are from summing together matching genotypesAll females are wt, so they can either list or ignore the various female mutants at this stage (if they list it, their probabilities should be 0).2. (1 point) Determine the expected number of flies (to the nearest hundreth) for each phenotype category.3. (1.5 points) You should now have enough information to run a Chi-square test to see if the data agrees with the linkage rates or not using an α of 0.05.Show your math, but otherwise you may make shortcuts in the process since you’ve completed one of these several times now. You may also need to refer to the note on p.131 of the lab manual.χ2 = 3.87 + 0.002592926 + 21.42437657 + 2.454444444+ 180.4544444=208.2df = k-1 = 5-1 =4 (note this is because 3 of the female types are not “expected”)The p-value is < 0.005 , therefore we must reject the null hypothesis.4. (1 points) Interpret your results. If your data did not match with the expected results, briefly explain why that might be the case.Since we rejected the null, we interpret that to mean that our results are not consistent with the genes being recessive, linked, and on the X chromosome at a distance of 1.5 M.U. apart. When we reject the null, the differences in the data must be due to some other factor. Anything reasonable is fine as an answer, but one possibility beyond biological mechanisms is also that our data is faulty in some way (sample size too small, incorrectly scored flies, etc.). There is also a list of reasons given from lecture that might apply as well. We also have the assumption of crossover only happening once.5. (1 point) Crossing over does happen at low rates in males due to non-standard mechanisms. In this specific problem, why would accounting for small amounts of crossing over in males not improve our model?Perhaps surprisingly (although once the students think about it I hope it’s not a surprise!), the fact that males have low rates of crossing over doesn’t matter in this case because the genes were X-linked: the males will never have a homologous pair to do crossover with in regard to these two specific genes.Name: ___________________________ Lab room, day, time: __________________TA name: _______________Crossing Over in Flies, final challengeOne more data set from this semester for the fly cross experiment (this tells you whatthe P generation and F1 generation crosses were like!) was selected and is provided below. You should use these data for the following questions.Female Malewild typeyellow body& white eyeyellow body (w/wild type eye)white eye (w/ wild type body)wild typeyellow body & white eyeyellow body (w/wild type eye)white eye (w/ wild type body)822 0 6 2 377 288 2 38White (w) & Yellow (y) are both on the X chromosome and only 1.5 M.U. apart. Let us see if the data obtained fits with a sex-linked, linked gene (at 1.5 M.U.), recessive inheritance model.1. Determine expected frequencies of each individual progeny possibility based on the 1.5 M.U. distance. Do this by showing a punnet square for frequencies. This one is challenging since it involves two genes both on the X chromosome and we are going to assume that crossing over does not happen in males. However, we arestill going to ignore the effects of multiple crossover events.a. (1.5 points) First, determine what gametes each adult of the F1 can produce (don’t calculate probabilities yet). You may need to consider sex, linkage, and lack of male crossing over to determine this. Remember this is class data so the P generation was affected males crossed to homozygous wild type females.(The F1 adults are X++/Y and X++/Xyw)X(+ +) X(y w) X(y +) X(+ w)X (+ +)Y(Note that a lack of crossover in males doesn’t actually come into play here since the male is hemizygous!) Students may represent this in slightly different ways. I think using superscripts work better to differentiate between the X and Y. I’m using parentheses in the square above just make it visually more clear that they are both on the X.b. (1 point) Which gametes represent the “parental” type? Which are “non-parental,” and