Question

5.) You have discovered mutants in two new gene mutants in Drosophila. One mutant, curl, has small, curly, nonfunctional wings. The other mutant, big, has enormous eyes. You decide to test if these two new genes are on chromosome #2, so you perform a three-gene linkage testcross using a gene that you know is on chromosome #2, trp, a mutant that cannot make its own tryptophan (and so must get it from its food). In all three cases, these mutant alleles are recessive to their wild-type counterparts. After doing the appropriate crosses, you obtain a triple mutant, and you cross it with a triple-hybrid (appearing wild-type for all three genes, but the offspring of a triple-mutant). Here are the results of this cross (showing phenotype only). [N.B. the categories are not necessarily listed/organized by “recombinant type" so check that first!] 537 curl, big, trp 519 curft, bigt, trp 91 curl, big, trp 82 curl, big, trp* 81 curl, big, trp* 77 curl*, big, trp 4 curl, big, trp* 3 curl*, big, trp a.) Are one, both, or none of these two genes on chromosome #2? How do you know? b.) Identify the "category types" and which flies are in which categories. c.) Which gene is in the middle of these three? How do you know? d.) Draw a linkage map, with distances in CM, for these three genes. e.) In the map you drew, there are two possible distances for the two outside genes. Which is more accurate and why? f.) Calculate the crossover interference that is observed in this experiment.

Answer 1

This is the question of linkage and crossover which is there is bsc standards.

The question simply means that there are 3 genes having different traits .

Mutant alleles are recessive to wild type.

Here first two results having frequency 537 and 519 are parental one and Non cross over gametes whereas the last two are double cross over one because there are less chances Or double cross over hence frequency is less.

Now I am providing you an answer .
Iloil 7 Atur recognising middle geven big, mee tep big malt tript им , - - 91 - 82 curt tept Ist and 3rd NON curl, big, tip CROES OVER curlt, bigt, topt cure, bigt, trp curl, big, trpt curtt, bigt, trpt curtt, bigt, trg DOUBLE Scurt, bigt, topt CROSS I wilt, big, top OVER. big - 81 - 77 cult tept curt tip bigt big curl curt tept tre M 11394 Total 57 → Non cross over is equal to 537+519 -1056 I 1056 > sau As the non coose over gamete more than 50% and the recombinant gametes are less than 50%. Therefou the genes. are linked 8 are on chromosome 2. & NON CROSS OVER in curl, big, trp o curtt, bigt, topt DOUBLE CROSS OVER :- cure, bigt, topt & curtt, big, big. By comparing non cross oves with double crossover we understand thant (curl) is the central gene becaus it is always the middle gene which get Displaced in Double cross over. big wil tip Dco big curl tap big cult to Non cross cross over big will trapt لطمه

takes place between over are → chromosomal map.. let us ignore the zeid locê, ef messing big & uirl, the cross over gametes Yuigt udstre + curt biau tep sco Leig curl topt big curl top These gamites are present in a combinations the frequency big + curl top- bigt anul trpt - 4 big werlt tipt -81 189 x 10o = 13.558 1394 13.558 CM. big curlt top - 3 I 189 Therefore the distance between big & curl is 13.558 cm) let us ignore the Ist locê, if crossing over takes places between curt & fep , the voss ovel & the gamets are. bigt coolt tept sa big wie tap so sco sig uuel top? big cure topt big+ cuelt top - 77 big wilt top 3 topta trpt - 4 big 166 x 100 211.9D8% 1394 2 11.908cm at wil top T66 Therefore the sistance between cull & top is 11.908 CM. CHROMOSOM AL MAP ON CHROMOSOMEⓇ : 13.5580M 11.9080M cul tip. OOM 25.466 cm

4. Interference double croes over. Percentage of expected 0.1356 0.1191 0.0161 0.0161 x 10o 1.610y 189 x 166 = 1394 1394 sco so of big & ut ure & top PRODUCT RUE , DUO. of sco observed Double Percentage of cross over. 7 x 100 1394 = 0.502 % Coefficient of co-incidence z observed so espected DOO 0.502 2 1.610 0.3118 Inferference - [1- coefficient of co-incidence] - [1- 0.31187 2 0.6882