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

XY= man

XX= woman

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 dont 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}

or .0156

check my thinking and math.

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