Question

1) Researchers cross parental fruit flies with mutations in an eye color gene (either normal Red or mutant Bone) and a body bristle gene (either normal Wild-type or mutant Broken-Bristle). All of the first generation of flies had unbroken bristles and Red eyes. They mated the F1 flies to a fly that was Bone and Broken-Bristle (a test-cross). They examined the phenotypes of the resulting offspring and found four classes of flies. They counted the number of flies in each class. Based on these data, answer the following questions. Bone Eye Bone Eye Red Eye Red Eye Broken-Bristle WT Bristle Broken-Bristle WT Bristle 15 flies 67 flies 61 flies 17 flies 160 flies A. What were the phenotypes of each of the initial parental flies? (don't worry about which was male and which was female) Parent 1: Parent 2: B. What is the distance between the Bone and Broken-Bristle genes? Show your calculations for determining this interval. C. The data indicates that the Bone and Broken-Bristle genes are on the same chromosome. If they had been on different chromosomes, if you had examined 160 flies as in the original experiment, how many flies of each class would you have expected to see? Bone Eye Broken-Bristle = Bone Eye WT Bristle = Red Eye Broken-Bristle = Red Eye WT Bristle =

.

.

Please provide an explanation foreach part, and I will give a like. Thank you!!!

Answer 1

Lets, Normal wild-type eye-Red colon - R mutant-tyke eye - bone wlos - Wild-type boustle B mutant broken-brustleb en F1 flies are unbroken (wild-tyle) bristles and Red eyes. Thus, fl shies genotyle could be Given assume 2 → Robb assume 3 RIBB assume a → RR BB RR BB assume assume 4 - RR Bb Assume-1 genotyke (Rs 3b) testcours with Bone and buis broken-boustle (Gobb). Then, be Robb & gabb parent enotype enotype RB Rb sob ob RB Rb a B ob ab/RGB) Robb 036 Iosbl Red Eye Redeye 1 Bone eye Bone eye WT boustle Broken-bristle WT brustle booken bristle Thus, four phenotype offsprings will be produced. • Now, if Assume-2 genotyke (RRIBB) testcross, Then PA RROB & Gabb gamete RB ab RaBb (Red eye wt bristle) - only one phenotybe offspring will be produce ol.

-) If Assume-3 genotyke (RGBB) testcrost. The RABB X as zamete RB 5 Rosh (Red eye (Bone eye WT bristle 1 wr bristle) - Just like assume -3, assume & genoty be w two phenoty be offsprings. ssume a neusty he will have - As, the given situation, test cross proence tour phenotyke offsprings. Thus, fl genotyle will be RGBb. FA - Rabb phenotybe can be obtain only when Red ege WT bristle -tque breeding (RR BB) coross with bone eye broken bristle - toque breeding (orot bb). P1 RRBB X nabb V game te RB зь FLV Rabb Thus, phenotype of cutia parents will be Parent & Red eye wild-type Boustle Parenta - Bone eye broken boistle 02.

B) Distance between Bone and Borcken-bruntle gener. Now, given in the test corons f 1 fly x bone eye-broken-bristle Rabb & nubb same as Assume a corolt in page 1). Bone Eye Broken Bristle - 15 flies Bone eye wt Bristle - 67 flieg Red Eye Broken Bristle - 61 flies. Red Eye WT Bristle – 14 flies The one larger in population are actually parental (Nou-crossover) ty be. Thus, Bone eye wt boustle (67 flies) and Red Eye Broken Bristle 16 thes) are parental types. while, Bone-eye Boroken-boristle and Red-eye wt bristle are non-parental (Recombinant) tybes. - Hence, distance between Bone and Broken- ? bristlegenes No. of Recombinant Xrou Total No. oi progenies 215 +/7 Vio 760 2 3 2 x 100 16o 2 20 map vort 03

1 e c) If Bone and broken- borustle genes un had been on different chromosomes. Then the natio of all phenotybe offsprings will be Thus, Number of each flies class out of 160 flics would be 40. Therefore, Bone - Eye Broken. Bruitle - 40 flies Bone - Eye WT-Bristle - 40 flies Red-eye Broken Boristle = 40 flies Red-Eye WT - Borstle = 40 fries 160 flies