1. White eyes can e due to the combination of two mutations which eliminate both pigment types normally found in the Drosophila eye. Assume for this question that the two mutations are autosomal, recessive, and not linked.a. (0.5) What are the genotypes of the P generation? Recall how the cross was set up in lab. Based on the above information, you should be able to determine which two genes are involved.Since they are unlinked, this should be brown with scarlet (brown is on chromosome 2, scarlet on 3)P: ♀ +/+; +/+; +/+; +/+ x ♂ +/Y; bw/bw ; st/st ; +/+; +/+)b. (0.5) Write the genotypes of the expected F1 progeny.F1: ♀/♂ +/bw ; +/stc. (1) Write the Punnet square showing the genotypes expected in the F2 progeny. (Be careful here! Remember that maternal and paternal contributions are noted specifically even if certain genotypes are ultimately equivalent.)A Punnet square is best here since the maternal/paternal chromosome tracking gets complicated. Paternal gametes are typically shown on the row across the top with maternal along the left column, but as long as they specify it can be done the other way.♀ ♂+ ; +bw ; ++ ; stbw ; st+ ; ++/+ ; +/++/bw ; + /++/+ ; +/st+/bw ; +/stbw ; +bw/+ ; +/+bw/bw ; +/+bw/+ ; +/stbw/bw ; +/st+ ; st+/+ ; st/++/bw ; st/++/+ ; st/st+/bw ; st/stbw ; stbw/+ ; st/+bw/bw ; st/+bw/+ ; st/stbw/bw ; st/std. (0.5) What are the phenotypes AND phenotypic ratios expected in the F2 progeny?The phenotypes will be 9 brick red eyes (wild type) : 3 bright red eyes : 3 brown eyes :1 white eyes. (They can also use percentages for their answer.)e. (0.5) Write the combined null hypothesis-prediction statement for this pattern.Something along the lines of the following:If scarlet and brown are recessive, autosomal, and unlinked genes, then the F2 progeny will not differ statistically from an expected phenotypic ratio of 9 brick red eyes (wild type) : 3 bright red eyes : 3 brown eyes :1 white eyes.Special: You have one of three possible gene combinations that result in white-eyed flies. Based on the above information, make a note in your lab manual of all possible eye phenotypes you will need to look for in your F2 generation. They may not all be present in your particular line of flies, but you need to know what to look for. (This satisfies the pre-lab assignment referenced on p. 25 of the lab manual.)Name: ___________________________ Lab room, day, time: __________________TA name: _______________Patterns of inheritance:This semester you are working to figure out the pattern of inheritance of 2 different lines of flies. This task and its associated statistical tests are much easier if you can recognize or reason through the patterns you are seeing. Last week you were asked toconsider the outcomes in the F2 generation and you will do the same this week. For the following questions you will get the “safest” results by doing the punnet square rather than just guessing. When in doubt, write it out.1. White eyes can e due to the combination of two mutations which eliminate both pigment types normally found in the Drosophila eye. Assume for this question that the two mutations are autosomal, recessive, and not linked.a. (0.5) What are the genotypes of the P generation? Recall how the cross was set up in lab. Based on the above information, you should be able to determine which two genes are involved.Since they are unlinked, this should be brown with scarlet (brown is on chromosome 2, scarlet on 3)P: ♀ +/+ ; +/+ x ♂ bw/bw ; st/st (they can also write out all the chromosomes, as: P: ♀ +/+; +/+; +/+; +/+ x ♂ +/Y; bw/bw ; st/st ; +/+; +/+)b. (0.5) Write the genotypes of the expected F1 progeny.F1: ♀/♂ +/bw ; +/stc. (1) Write the Punnet square showing the genotypes expected in the F2 progeny. (Be careful here! Remember that maternal and paternal contributions are noted specifically even if certain genotypes are ultimately equivalent.)A Punnet square is best here since the maternal/paternal chromosome tracking gets complicated. Paternal gametes are typically shown on the row across the top with maternal along the left column, but as long as they specify it can be done the other way.♀ \ ♂ + ; + bw ; + + ; st bw ; st+ ; + +/+ ; +/+ +/bw ; + /+ +/+ ; +/st +/bw ; +/stbw ; + bw/+ ; +/+ bw/bw ; +/+ bw/+ ; +/st bw/bw ; +/st+ ; st +/+ ; st/+ +/bw ; st/+ +/+ ; st/st +/bw ; st/stbw ; st bw/+ ; st/+ bw/bw ; st/+ bw/+ ; st/st bw/bw ; st/std. (0.5) What are the phenotypes AND phenotypic ratios expected in the F2 progeny?The phenotypes will be 9 brick red eyes (wild type) : 3 bright red eyes : 3 brown eyes :1 white eyes. (They can also use percentages for their answer.)e. (0.5) Write the combined null hypothesis-prediction statement for this pattern.Something along the lines of the following:If scarlet and brown are recessive, autosomal, and unlinked genes, then the F2 progeny will not differ statistically from an expected phenotypic ratio of 9 brick red eyes (wild type) : 3 bright red eyes : 3 brown eyes :1 white eyes.Special: You have one of three possible gene combinations that result in white-eyed flies. Based on the above information, make a note in your lab manual of all possible eye phenotypes you will need to look for in your F2 generation. They may not all be present in your particular line of flies, but you need to know what to look for. (This satisfies the pre-lab assignment referenced on p. 25 of the lab manual.)PV92Read Appendix 2 of the lab manual (p. 119 - 124 and 135). Follow the example provided for running a Chi-square test on PV92 data (p. 125-126). You may also want to preview the use of Chi-square with flies (p. 127-134).2. Using the total class data (which will not be available until after the last lab on Thursday night), determine the following. Show your work.a. (0.5) What are the actual allele frequencies?Total data was 35 +/+, 18 +/-, 92 -/-(35 x 2) + 18 = 88 “+” alleles.88/(145*2) = 0.3034 = f “+” allele(92*2)+18 = 202 “-“ alleles 201/290 = 0.6966 = f “-“ alleleThe can also calculate one frequency and get the other by subtracting it from 1.Their answers can be to fewer decimal places, so rounding will cause some small variability.b. (0.5) What are the actual genotype frequencies?They should not do H-W for this part.f +/+ = 35/(35+18+92) = 0.2414f +/- = 18/145 = 0.1241f -/- = 92/145 = 0.6345They may also calculate 2 and subtract those from 1 to get the third.c. (1) Calculate the expected genotype frequencies