Genetics Lab Report 3

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Dec 6, 2023

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Mossimo Tampol In this week’s lab, you are going to use FlyLab’s recombination feature to create a linkage map determining how far apart these genes are from each other on chromosome II. 1. Design a mating between a female with the purple mutation (eye color trait) and a male with the vestigial allele (wing size trait). a. What are the phenotypes of their offspring? - Both fly offspring have red eyes and normal wings (wild-type). b. What are their genotypes? - The genotypes of the offspring are all homozygous. 2. Select the F1 female for mating and design a new cross between her and a male fly showing both the purple and vestigial traits. a. Are the female’s mutant alleles in cis or trans arrangement? - The female’s mutant alleles are in trans arrangement. b. How do you know? - The recessive traits are located on the same gene. c. What are the phenotypes of the F2 offspring of this cross? - Of the F 2 offspring, 8 flies in total, half of them have the red wild-type eye color and the other half have purple. Of the red-eye-colored flies, 4 of them have normal wings and 4 of them have vestigial wings. This applies to the other half of purple-eye-colored flies as well. d. What are the genotypes of each of these phenotypes? It will be helpful to look at the offspring via the “Analyze” tab, instead of the “Crosses” tab. -

The genotypes of each phenotype are male and female EeSs, male and female Eess, male and female eeSs, and male and female eess. 3. Under the “Analyze” tab, please select the “Ignore sex of flies” box. a. List the total numbers of progeny of each genotype. - 62 were wild-type, 463 had wild-type red eyes but vestigial wings, 455 had purple eyes with wild-type wings, and 44 had both purple eyes and vestigial wings. b. Which of the genotypes from these F2 progeny are parental, and which are recombinant? - There are two recombination events resulting in the two recombinants. The flies that were strictly wild-type or mutant for having both purple eyes and vestigial wings act as the progeny, while the majority of the flies that had either purple eyes with wild-type wings or were wild-type for eyes with vestigial wings would be the recombinants.

4. Finally, what is the genetic distance between these two genes, in centimorgans? Show all calculations. - (463 + 455)/1024 x 100 = 89.64 cM Section II. Recombination in 3 genes, i.e., the 3-point cross When you consider three genes, by examining the segregation of different alleles into the offspring, you can determine the genes’ positions relative to each other by comparing the recombination frequencies of the different alleles. For example, if I am examining 3 genes, a, b, and c, and the recombination frequencies between them are: a-b: 3 cM and b-c: 2 cM, then we can draw these either as: 3 cM 2 cM 2 cM 3 cM a ------------- b -------- c OR a -- c --------- b The way to determine whether the genes are positioned on the chromosome as a-b-c or a-c-b is to look at the distance between a and c. In the examples above, that distance could be either 5 cM or 1 cM, depending on whether b or c is in the middle position. When we were looking at only 2 genes, there was only 1 location we were interested in for chiasmata to occur – between the 2 genes. However, now that there are 3 genes we are interested in, there are 2 possible locations for chiasmata to occur – between the first and second gene, and between the second and third. As a result, we also have the possibility now of having a chiasma form in each location at the same time. The result of

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this would be the first and third gene possessing alleles from the same parental chromosome, and the middle of the three genes having the allele from the other parental chromosome. This introduces an additional phenotype to the offspring. To determine the recombination frequency between genes in a three-point cross, we focus on just two genes at a time. Just as with 2-gene recombination, we need to look at offspring where the cis / trans arrangement of the offspring alleles does not match that of the parentals. The difference during a three-point cross is that you must consider more than 1 class of offspring with this change in allele arrangement. Review your lecture notes and be sure to read the relevant section of the chapter to better understand how to calculate recombination in a 3-point cross. For the following questions, you are going to examine the linkage of and gene order between the body color, eye color, and wing shape loci in Drosophila . 5. Design a mating between a female with these three traits: purple (eye color), black (body color), and curved (wing shape), and a wild-type male a. What are the phenotypes of their offspring? - Both fly offspring have red eyes and normal wings. b. What are their genotypes? - The genotype of both offsrping is WwYy. 6. Select the F1 female for mating and design a new cross between her and a male fly showing all three traits a. What are the phenotypes of the F 2 offspring of this cross? - Of the F 2 offspring, 16 flies in total, half of them have the wild-type body color and the other half are black. Of the wild-type body colored flies, 4 of them have

red eyes and 4 of them have purple. 2 of the red eyes and 2 of the purple eyes have curved wings. This applies to the other half of black-colored flies as well. b. What are the genotypes of each of these phenotypes? It will be helpful to look at the offspring via the “Analyze” tab, instead of the “Crosses” tab - The phenotypes of the offspring are +++, PR + + , + VG + , ++BL, PR VG + , PR + Bl, + VG BL , PR VG BL 7. Under the “Analyze” tab, please select the “Ignore sex of flies” box. a. List the total numbers of progeny of each genotype. - 393 flies with wild-type body color and eye color, 6 flies with wild-type body and purple eye mutation color, 73 flies with the wild-type eye color and curved wing mutation, 24 flies with the wild-type body and black body mutation, 32 flies with purple eye and curved wing mutations, 83 flies with the purple eye color and black body mutations, 4 flies with curved wing and black body mutation, and 390 flies with the black body color, purple eyes nad curved wing mutations. b. How many offspring are there in total? - 1005 c. Which of the genotypes from these F 2 progeny are parental, and which are recombinant? - The parental genotypes are the “+, PR, C, BL”. Everything else is a recombinant. There are two recombination events

8. What is the order of the three genes on the chromosome? - The order of the three genes on the chromosome goes BL, PR, and C. 9. Finally, what is the genetic distance between these two genes, in centimorgans? - 166/1005 x 100 = 16.52, 66/1005 x 100 = 6.57 BL ------- 16.52 ------- PR ------- 6.57 ------- C |-----------------------23.09 cM-----------------------|

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