Posted by Kellie on Friday, October 12, 2007 at 7:54pm.
Use the following genotypes to answer the question.
N= big nose
n= small nose
B= big nostril
b= small nostrl
H= hairy face
h= no hairy face
Z= zits on the nose
z= no zits on the nose
If two parents who are heterozygous for all the traits were 'crossed', what are the chance of getting a child with this genotype: NnBBHhXXZz?
Can someone do the problem then explain to me how you would do this? I've tried but i don't know if my answer is right
- Science:Biology - bobpursley, Friday, October 12, 2007 at 10:20pm
Lets do the sex gene first.
XX+ XY. The probabality of getting XX is 1/2
Nn+Nn = NN nn Nn nN so probability is 1/2
BB probability is 1/4 (BB, Bb, bB, bb)
Hh prob is 1/2
Zz prob is 1/2
Now multiply them all
(1/2)4 * (1/4)1
check my thinking and math.
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