# Genetics

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#### chiddler

##### Full Member
10+ Year Member
B = Black
b = brown

A=normal pigment
a = albino

Experiment 1:
A female hamster with BBAA is crossed with a male hamster bbaa

Experiment 2:
Female offspring from the cross in experiment 1 were backcrossed to the bbaa parent. The distribution of offspring is as follows: black (66), brown (34), white (100).

What is the frequency of recombination between the two loci?
----------------
Lets see...

Experiment 1: BBAA * bbaa = BbAa

Experiment 2: then BbAa * bbaa = 25% each of BbAa, Bbaa, bbAa, bbaa. This means that half are white, 1/4 are black, 1/4 are brown.

So they must be linked since they do not display the above distribution. Knowing now that they are linked, how do I proceed?

B = Black
b = brown

A=normal pigment
a = albino

Experiment 1:
A female hamster with BBAA is crossed with a male hamster bbaa

Experiment 2:
Female offspring from the cross in experiment 1 were backcrossed to the bbaa parent. The distribution of offspring is as follows: black (66), brown (34), white (100).

What is the frequency of recombination between the two loci?
----------------
Lets see...

Experiment 1: BBAA * bbaa = BbAa

Experiment 2: then BbAa * bbaa = 25% each of BbAa, Bbaa, bbAa, bbaa. This means that half are white, 1/4 are black, 1/4 are brown.

So they must be linked since they do not display the above distribution. Knowing now that they are linked, how do I proceed?

Recombinant individuals are the ones that have phenotypes different from their parents. Since the parents were albino (aabb) and Black (BbAa) then only the brown individuals are recombinant, hence 34.

I'm just confused for the fact that the formula for frequency of recombination is actually # of recombinant individuals/total number of offspring.. so shouldn't it be 34 / (66+34+100) ?

nevermind. thinking.

incredible solution. not extraordinary, but just I never would have thought of that incredible.

there are 34 brown which are recombinant. we can also assume the other recombinant possibility is present but is lost among the albinos. Therefore, there are two times the recombinants than there are brown animals and it is 34*2 / 200.

Last edited:
nevermind. thinking.

incredible solution. not extraordinary, but just I never would have thought of that incredible.

there are 34 brown which are recombinant. we can also assume the other recombinant possibility is present but is lost among the albinos. Therefore, there are two times the recombinants than there are brown animals and it is 34*2 / 200.

wait, I'm confused. So you're just assuming that there are 34*2 recombinants?

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wait, I'm confused. So you're just assuming that there are 34*2 recombinants?

yes. because the other half are hidden due to the albino trait.

wait, I'm confused. So you're just assuming that there are 34*2 recombinants?

wow, that's a huge assumption. I'm not saying it's invalid just quite a leap. Interesting. If this is from TPRH, can you let me know of which passage no this pertains?

Question is flawed because no offspring are white pigmented.

wow, that's a huge assumption. I'm not saying it's invalid just quite a leap. Interesting. If this is from tprh, can you let me know of which passage no this pertains?

33

lol

TBR has a nearly identical passage as TPR's

yes. because the other half are hidden due to the albino trait.

wow.. I wouldn't have known to make this assumption!
I guess this is the case because there is epistasis going on, right? Otherwise we would have had other 34 indiv with recombinant phenotype..