I) (14 pts) You have discovered a new species of worms for which the haploid num
ID: 95024 • Letter: I
Question
I) (14 pts) You have discovered a new species of worms for which the haploid number is three. You isolated a dumpy worm and want to determine the linkage group of the gene dpy-1. The dpy-1 mutation is recessive. Using recessive marker mutations for each chromosome i.e., marker 1 is on chromosome I, marker 2 is on chromosome II, marker 3 is on chromosome III), you set-up 3 crosses as follows: Cross : marker mutation 1 x dpy-1 Cross 2: marker mutation 2 x dpy-1 Cross 3: marker mutation 3 x dpy-1 Fl progeny for the different crosses are allowed to mate, and you select the animals carrying the marker mutation from this cross. You find the following results: Of 300 marker mutation 1 animals, 200 are dpy-1 Of 300 marker mutation 2 animals, 150 are dpy-1 Of 300 marker mutation 3 animals, 15 are dpy-1 a) On which chromosome is dpy-1? You now wish to determine the position of dpy-1 on the chromosome. You have identified several other recessive mutations on that chromosome. These other mutations include eyeless and hairy. You have a female worm that is heterozygous for eyeless, hairy, and dpy-1 and testcross the animal (i.e., mate the het to an animal homozygous for eyeless, hairy, dpy-1). Among the 1000 F1 progeny from the testcross, the following phenotypes were observed: Dpy-1, eyeless, hairy wild type hairy, dpy-1 eyeless eyeless, hairy dpy-1 eyeless, dpy-1 hairy 450 470 30 40 34 36 b) What is the order of genes on the chromosome and the map distances between the genes? 2) (12 pts) You were taking the subway to class one day and captured several family members of an unknown rodent species on the platform. You gave them to your genetics professor to analyze, and she performed the following crosses. The F2 progeny are the result of crossing the Fl progeny to each other.Explanation / Answer
1)
a) dpy-1 is on chromosome 1.
b) This is a three point cross.
Parental genotypes are present in largest numbers. So, the parental genotypes are Dpy-1,eyeless, hairy and wild type. The double recombinants are present inleast number. Here, eyeless, dpy-1 and hairless are double recombinants. Double recombination pushes the middle allele over the non sister chromatids. Therefore, the gene order is eyeless, hairy, dpy-1.
Distance between eyeless, hairy = 100 x (34+30+4+6)/1070 = 6.92 cM
Distance between hairy,dpy-1 = 100 X (40+36+4+6)/1070 = 8.04 cM
The gene map is,
eyeless hairy dpy-1
I-------------I---------------------I
6.92 cM 8.04cM