A woman with normal vision has a son. The woman’s maternal grandfather is color-blind. What is the probability that the son is colorblind, if his father is normal? Assume there are no other color-blind individuals in the pedigree.A. 1/2B. 1/3C. 1/4D. 1/6E. 1/8

Respuesta :

Answer:

= [tex]\frac{1}{4}[/tex]

Explanation:

Given that the woman’s maternal grandfather is color-blind, then it is possible that the woman is a carrier of color-blind allele.

Then probability that the woman is a carrier of color-blind allele is = [tex]\frac{1}{2}[/tex]  

So if a carrier woman cross with a normal man

Let say the gene of the carrier woman is: [tex]XX^c[/tex]

and the normal man = [tex]XY[/tex]

so if we have a cross between  [tex]XY[/tex] and [tex]XX^c[/tex]; we have:

                     [tex]X[/tex]                     [tex]Y[/tex]

[tex]X[/tex]                   [tex]XX[/tex]                 [tex]XY[/tex]

[tex]X^c[/tex]                 [tex]XX^c[/tex]                 [tex]X^cY[/tex]

we can see the offspring to be :

[tex]XX[/tex] =normal unaffected female

[tex]XY[/tex] = normal unaffected male

[tex]XX^c[/tex] = carrier female for color-blind

[tex]X^cY[/tex] = carrier male for color-bind

Then probability that of  inheriting defective allele = [tex]\frac{2}{4}[/tex] = [tex]\frac{1}{2}[/tex]

Let's not forget that the probability that the woman is a carrier of color-blind allele is = [tex]\frac{1}{2}[/tex]  

∴  the probability that the son is colorblind = [tex]\frac{1}{2}*\frac{1}{2}[/tex]

= [tex]\frac{1}{4}[/tex]