Crossing Over in Flies1. Sepia & Ebony have a published distance of 44.7 M.U. For that data, run a Chi-square test to see if the data obtained fits with the published distance. The following questions will walk you through that processa. (1.5 point) First, determine expected frequencies of each individual progeny possibility based on the 44.7 M.U. distance. Do this by showing a punnet square for frequencies. Think about what M.U.s mean and how that affects gamete probabilities. Assume that crossing over occurs equally between males and females.Gamete probabilities are going to be ½ 44.7% for each of the recombinants/non-parental type (22.35% or 0.2235) and then ½ of 55.3% (27.65% or 0.2765) for the nonrecombinant/parental type. They just need to then fill out a square. They don’t need to have as many significant figures as indicated below (I just copied it from excel).b. (1 points) From your punnet square, calculate the expected frequencies of each phenotype based on a 44.7 M.U. distance.They should add together the individual frequencies for each phenotype to get this answer.c. (1 points) Determine the expected number of flies of each category (round to the nearest hundredth of a fly).2. (1 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. Make sure you follow all the steps!Answers should generally be along these lines, but can vary slightly.χ2 = 46.12 + 16.66 + 63.84 + 0.21 = 126.83df = k-1 = 4-1 =3The p-value is < 0.005 , therefore we must reject the null hypothesis.3. (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 a distance of 44.7 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 in the 11/4 lecture that might apply as well. Then, of course, there is the issue addressed in the next set of questions that male flies do not actually experience crossing over.4. We do know that crossing over in male Drosophila is naturally very rare.a. (1.5 point) Determine how this will affect the offspring rates by repeating the calculations you did for question 1a but assume that crossing over only occurs in females.Their Punnet square will end up fairly different because they should assume that crossing over will not happen in males. This means they will only generate two kinds of gametes (the parental type). They don’t need to include the Ab or aB lines from the male (so their table can just be a 2x4 instead of a 4x4). The probabilities of the female gametes are the same as in question 6a, but the male can only produce two possibilities of equal probability (0.5 each).b. (1 points) Repeat parts 1b and 1c using the Punnet square you made for part a of this question.d. (1 points) Interpret your results. Also compare your results with those calculated from assuming crossing over happens equally in males and females. Why might they be the same or different? Even if you reached the same conclusion, explain if one seemed more accurate or not.Since we rejected the null, we interpret that to mean that our results are not consistent with the genes being a distance of 44.7 M.U. apart. In both cases the null was rejected, likely due to the various factors discussed in question 3. However, the total X2 value when we assumed crossing over did not occur in males was smaller (126.64 vs. 64.90). While it still easily rejects the null, the fact that the differences in the actual data and the model were smaller does indicate that it was somewhat better at describing the pattern seen (another way to look at this is that the smaller the X2 value is, the larger the exact p-value is. The chart the students use doesn’t give exact p-values, however).Name: ___________________________ Lab room, day, time: __________________TA name: _______________Crossing Over in FliesData from the fly cross experiment were selected. An associated excel file is supplied, but fly counts are listed below in questions 1c and 4c. (The data is similar to that given for homework 9, but that assignment erroneously used last semester’s data. The calculated crossover frequencies have been supplied as a point of comparison). Linked genes:1. Sepia & Ebony have a published distance of 44.7 M.U. For that data, run a Chi-square test to see if the data obtained fits with the published distance. The following questions will walk you through that processa. (1.5 point) First, determine expected frequencies of each individual progeny possibility based on the 44.7 M.U. distance. Do this by showing apunnet square for frequencies. Think about what M.U.s mean and how thataffects gamete probabilities. Assume that crossing over occurs equally between males and females.Gamete probabilities are going to be ½ 44.7% for each of the recombinants/non-parental type (22.35% or 0.2235) and then ½ of 55.3% (27.65% or 0.2765) for the nonrecombinant/parental type. They just need to then fill out a square. They don’t need to have as many significant figures as indicated below (I just copied it from excel).Gamete + + se e + e se +Gamete Probability 0.2765 0.2765 0.2235 0.2235+ + 0.27650.076452250.07645225 0.06179775 0.06179775se e 0.27650.076452250.07645225 0.06179775 0.06179775+ e 0.22350.061797750.06179775 0.04995225 0.04995225se + 0.22350.061797750.06179775 0.04995225 0.04995225b. (1 points) From your punnet square, calculate the expected frequencies of each phenotype based on a 44.7 M.U. distance.They should add together the individual frequencies for each phenotype to get this answer.Wild type Ebony SepiaSepia and ebonyPercentage 0.5764522 0.1735477 0.17354775 0.07645225s 5 5c. (1 points) Determine the expected number of flies of each category (round to the nearest hundredth of a fly).WT Ebony (body)Sepia (eye) Sepia&Ebony Total (contributing)Observed727 117 66 71981Expected565.50 170.25 170.25 75.00 981.002. (1 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. Make sure you follow all the