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Kry

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I'm having trouble figuring this out (been 3 years since I took genetics) for a friend. Does any one know how to solve it?

Labrador retrievers may be black, brown, or golden in color. While each color may breed true, many different outcomes occur if many litters are examined from a variety of matings, where the parents are not necessaryily true-breeding.

Propose a mode of inheritance that is consistent with these data, and indicate the corresponding genotypes of the parents in each mating. Indicate as well the genotypes of dogs that breed true for each color.



parents offspring

cross 1 black X brown all black
cross 2 black X brown 1/2 black 1/2 brown
cross 3 black X brown 3/4 black 1/4 brown
cross 4 black X golden all black
cross 5 black X golden 4/8 golden 3/8 black 1/8 brown
cross 6 black X golden 2/4 golden 1/4 black 1/4 brown
cross 7 brown X brown 3/4 brown 1/4 golden
cross 8 black X black 9/16 black 4/16 golden 3/16 brown
 

clowne

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This is just a guess, but I think two genes are involved (will call them A and B). If a dog is AABB or AaBb, it will be black. Brown is aabb, and golden is Aabb, AAbb, aaBb, or aaBb. Hope this helps. Disclaimer: I didn't spend much time on this, so it may be blatantly wrong.
 

Kry

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Thanks for trying. Your solution doesn't seem to fit for cross 3 or am I missing something?
 
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MErc44

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Originally posted by clowne
This is just a guess, but I think two genes are involved (will call them A and B). If a dog is AABB or AaBb, it will be black. Brown is aabb, and golden is Aabb, AAbb, aaBb, or aaBb. Hope this helps. Disclaimer: I didn't spend much time on this, so it may be blatantly wrong. [/QUOTE

I think you got it. I hate genetics
 

clowne

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Originally posted by Kry
Thanks for trying. Your solution doesn't seem to fit for cross 3 or am I missing something?

Maybe I guessed the wrong genotype for brown, you'll just have to play around with them and calculate the probabilities. I was just trying to guess at a starting point. Hope you figure it out.
 

PianoGirl04

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I've read in lots of textbooks that Labrador coat color is an example of epistasis (particularly recessive epistasis). So, looking at Cross 8 and it's 9:3:4 ratio, I came up with this:

A-B- = black (dominant at both)
A-bb = brown (dominant at A, recessive at b)
aaB- + aabb = golden (recessive at a --> b doesn't matter)

Hopefully that fits with your other crosses.
 

hale-bopp

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always liked genetics puzzles, but the only way I can make this one work is if in cross 3, the cross is actually Black x Black -> 3/4 black, 1/4 brown. (In which case, it would be AABb x A-Bb, using pianogirl's genotypes).

Since it is possible to cross black with anything and still get all black, it must be dominant at both of the genes.
If there are only two genes, the only way to get a 3:1 ratio with simple dominance is to have a cross like AABb x A-Bb (i.e. both heterozygous for one gene, and one homozygous dominant for the other or both homoxygous recessive for the other), in which case, both dogs would have to be black. This is the same pattern that we see in cross 7, except there, both have to be Aabb.

In that case, PianoGirl is right, and all of the rest of the crosses are possible, and the genotypes that breed true are black: AABB, brown: AAbb, golden: aa--.

I haven't tried to work out what would happen if there were three genes... maybe there is something there that would make it possible for cross 3 to stand as it is and make sense with the rest, but I think that the ratios all look like a two gene cross.

good luck to your friend.
 
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