The following phenotypes are seen after mating: 400 tall/blue 200 tall/green 200 short/blue 400 short/green If the tall/green and short/blue phenotypes are recombinant, what is the linkage distance for these genes?
A. 3.0 LMU
B. 6.0 LMU
C. 33.3 LMU
D. 66.7 LMU
@Abhisar

Question

The following phenotypes are seen after mating: 400 tall/blue 200 tall/green 200 short/blue 400 short/green If the tall/green and short/blue phenotypes are recombinant, what is the linkage distance for these genes?
	A.	3.0 LMU
	B.	6.0 LMU
	C.	33.3 LMU
	D.	66.7 LMU
@Abhisar

anonymous · Answer

@Abhisar do you know the answer

abhisar · Answer

No i am not sure about this one !

anonymous · Answer

Do you know anything about it

abhisar · Answer

Not very much !
::(

anonymous · Answer

Tbh I think its B or D

abhisar · Answer

Why ?

abhisar · Answer

A user asked me this question before, sadly i didn't knew it that time too. I never got the time to research about it. http://openstudy.com/study#/updates/53ac3e52e4b0ffdda15f3a77

miracrown · Answer

there is a formula to use for linkage distances

abhisar · Answer

I know that linkage distances are expressed in %age recombinant values. I am not sure how to do this variant of question.

abhisar · Answer

I think @mrdoldum may be able to help !

abhisar · Answer

@Miracrown do hv any idea about this one ?

miracrown · Answer

1 LMU = 1% 
we have to add up that least of the 4
then divide that by the total

miracrown · Answer

so in this case
200+200/total
then multiply that by 100 for the %

miracrown · Answer

in other words, recominants/total x 100 = LMU

miracrown · Answer

and also the total is all the offspring added together :)

miracrown · Answer

so it should be like this:

$$\frac{ 400 }{ 1200 } = 0.33333333333 \space then \space \times 100= 33.3333333333$$

abhisar · Answer

Good job @Miracrown !