PLEASE SHOW ALL WORK A non-hemophiliac man who is blind from aniridia (see previ

Question

PLEASE SHOW ALL WORK

A non-hemophiliac man who is blind from aniridia (see previous problem) and whose mother is not blind marries a non-hemophiliac woman who is not blind and whose father has hemophilia. What kinds of children might they have and in what proportions? A normal-visioned man marries a normal-visioned woman whose father is color-blind. They have two daughters who grow up and marry. The first daughter has five sons, all normal-visioned. The second daughter has two normal-visioned daughters and a color-blind son. Diagram the family history in a pedigree and indicate the genotypes of all family members.

Explanation / Answer

9. ANS:

(hhBB) – Non – haemophiliac man

(hhbb) – Non-haemophiliac man

According to the Punnett Square and on crossing, we get

hhBB X hhbb ----------------> hhbb , hhBB, hhBB, BBhh which means

25% non-hemophilic and non-blind

50% non-hemophilic and Blind

25% non-hemophilic and Blind.

10. ANS:

Let XC(Capital C)represent an X chromosome carrying the dominant normal allele; Xc (Small c) represents an X chromosome carrying the recessive colorblind allele. Y represents the Y chromosome, which lacks this gene (and all sex-linked genes that we will discuss).

The man with normal vision is XC Y; his father doesn't matter (because he inherited Y from his father) The woman's father was XcY, so the woman got the colorblind allele from him and is XC Xc .

Now, the cross is XC Xc H XC Y. So:

for daughters, ½ XC XC , normal; ½ XC Xc, normal but carriers

for sons, ½ XCY, normal; ½ XcY, colorblind

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