× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
8
5
z
4
n
4
t
4
k
3
x
3
r
3
m
3
j
3
c
3
l
3
New Topic  
kiki-star kiki-star
wrote...
Posts: 5
Rep: 0 0
10 years ago
C elegans mutant was found that lives much longer than wildtype worms.  This homozygous mutant was crossed to a pure breeding dpy mutant that had a normal lifespan. Note that the pure breeding long lived mutant was wildtype for Dpy.

L= long life span,  N = normal lifespan,  dpy+= wildtype bodysize allele, dpy- = Dpy mutant.

F1: all long lived, normal body size x tester
F2: 286 long lived, normal body size; 11 long lived, Dpy; 290 normal lifespan,  Dpy; 13 normal lifespan,  normal body size.

Write put the parental and F1 genotypes above. Then show your work to indicate the relative position of the longevity gene (L) compared to the dpy gene.

Attempt:
I'm not sure if the genotypes I have are right:
Parents are LL dpy+, dpy+ × NN, dpy- dpy-
F1 are all LN dpy+dpy-

And I don't know how to do the second part
Read 498 times
1 Reply

Related Topics

Replies
wrote...
Donated
Valued Member
10 years ago
Parental Genotypes:    Mutant: LL; dpy+ dpy+   Cross: NN; dpy dpy
F1 Genotypes:      LN; dpy+ dpy
The 11 long lived/Dpy and 13 normal lifespan/Dpy+ are recombinants, therefore 24/600 are recombinants.  24/600 = 0.04 = 4% recombination = 4 map unit distance between gene loci:

--------Lifespan gene---------------------Dpy------------
                                  4 m.u.
Pretty fly for a SciGuy
New Topic      
Explore
Post your homework questions and get free online help from our incredible volunteers
  282 People Browsing
 123 Signed Up Today
Related Images
  
 646
  
 278
  
 6380
Your Opinion