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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 don't know if my answer is right

  • Science:Biology -

    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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