Black body b is recessive to wild type body (+) and cinnabar eye color cn is rec
ID: 255224 • Letter: B
Question
Black body b is recessive to wild type body (+) and cinnabar eye color cn is recessive to wild type eye (). You cross + +/b cn x +/b cn What offspring genotypes and frequencies should you get from this cross if the genes are on the same chromosome, and there is no recombination between the two loci (0 cM)? (choose all that apply) all ++/b cn all b cn/b cn O 25% ++/++, 5096 ++/b cn, 25% b cn / b cn all+cn/b+ QUESTION 2 What phenotypes will be seen in the above non-recombinant offspring? (choose all that apply) wild-type O black body, wt eye O wt body, cinnabar eye black body, cinnabar eye QUESTION 3 If the genes are 8cM apart what what is the total frequency of recombinant gametes produced by the mother? o 16% O 10% o 8% 2% QUESTION 4 What kinds of gametes can be produced by the/b cn fathers? (choose all that apply remember there is no recombination in Drosophila males) ?? + cnExplanation / Answer
Answer:
1). D). 25% ++/++, 50% ++/b cn, 25% b cn / b cn
++ / b cn x ++ / b cn
++
b cn
++
++/++
++/b cn
b cn
++/b cn
b cn / b cn
Explanation: As there is no recombination, all progeny are parental genotypes. ++/b cn
2).
B). Black body, wt eye and
C). wt body, cinnabar eye
3). C). 8%
Explanation: Distance between the genes (cM) = (%) Percentage of recombination frequency
4).
A). ++
B). b cn
Explanation: As there is no recombination, only non-recombinat gametes are formed.
++
b cn
++
++/++
++/b cn
b cn
++/b cn
b cn / b cn
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