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laniyar laniyar
wrote...
Posts: 3
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11 years ago
(a)how many different phenotypic classes will result following selfing of this plant?

AabbCcDdee x AabbCcDdee
Reading these forums, I came up with:
n=the number of heterozygous genes
2^n = the number of phenotypic classes. So 2^3=8 phenotypic classes,right?

What if the cross was something like:
AabbCcDdee x AaBbccddEe

Would the n=3 or would n=5?
   
(b)   how many different genotypic classes will be obtained when this plant (AabbCcDdee) is used in a
test-cross?

How do you get (b)?

Would the testcross be:
AabbCcDdee x aaBbccddEe

Thanks

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Replies
wrote...
11 years ago
a. 1/2 x 1/2 x 1/2 x 1 x 1/4 = 1/32
Aa x AA = 1/2 Aa
Bb x BB = 1/2 BB
Cc x cc = 1/2 CC
dd x DD = 1 Dd
Ee x Ee = 1/4 EE

b. 0 chance!!
Cannot get CC in offspring, because only 1 parent has a C.
So the other parent is always going to provide c.
laniyar Author
wrote...
11 years ago
Those answers are wrong, but thanks for trying als2915.
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