Awoman is phenotypically normal, but her father had the x-linked recessive condition of red-green colorblindness. she married a man with normal vision. what is the probability that their first child will be a daughter with normal vision? round your answer properly to 3 decimal digits.

Considering that the father of the woman in question had the X-linked recessive condition, we can deduce that her mother had the normal phenotype (since exercise tells us that the woman was normal, so she received the mother's normal X, and X' affected from her father). So the genotypes are these:

  x    X'Y

      X'X   <---------- woman's genotype

Obs.: We are not sure of the genotype of the woman 's mother, so we put the symbol (?) We know that at least one of the genes she passed on to her daughter is normal, but we don't know if she had both normal Xs or if she had one of them affected.

Now this woman has married a normal man. We know that normal men can only have one genotype, which is XY. Making the crossing we have the following probabilities:

X'X      x       XY

          X            Y

X'      X'X          X'Y

X      XX           XY

So the probability of this couple having a daughter with the normal phenotype is 50%, since for the woman to manifest the disease she needs to have both Xs affected.

it is a learned trait

explanation:

geographic isolation