Question

Coco Sex-Linked Problems (Alloles Located on the X Chromosome) of these problems, you do that humanae have one chrome and Ych males have two X chrom e They comecome does not carry the gene found on the X chromom genes that determine 11. Inhumans, the condition f cing (H) dominates the condition for onclotting the Both alleles are linked to the chrome made hemophilia maries a woman who is a carries for this condition in this specta carrier is a woman who has an alle for normal Mood costing and a lle for hemophilia y a what the chance he will be mal for blood clotting? 12. For this problem, use the information given in problem 11. A male who has normal blood clotting marries a woman who is a carrier for hemophilia. What are the chances that they will have a son who is normal for blood clotting? ..::TRE EFFFFFFFFF 13. Color Lindness is a condition in which a person cannot distinguish specific colors from one another. For example, the person may not be able to distinguish red from green or blue from yellow. However, the person is able to distinguish some colors. Because color blind people are not blind and they can see some colors, many people prefer to use the term color deficient. In humans, the condition for normal vision dominates color blindness. Both alleles are linked to the X chromosome. A color-blind male marries a color-blind female. If they have a daughter, what are the chances she will have normal vision?

Answer 1

11) Let, XH = X chromosome containing normal blood clotting allele (H), Xh = X chromosome containing hemophilic allele (h)

Now, genotype of hemophiliac man will be XhY & genotype of carrier female will be XHXh. The cross between them is shown in below Punnett square.

Gametes	Xh	Y
XH	XHXh (Carrier Female)	XHY (Normal male)
Xh	XHXh (Carrier Female)	XhY (Affected male)

From the above Punnett square, we find that if they have a son probability that he will be normal for blood clotting is 1/2 (Here, we ignore probability for female. Thus, we will consider only XHY & XhY genotype).

12) Now, genotype of man with normal blood clotting will be XHY & genotype of carrier female will be XHXh. The cross between them is shown in below Punnett square.

Gametes	XH	Y
XH	XHXH (Normal Female)	XHY (Normal male)
Xh	XHXh (Carrier Female)	XhY (Affected male)

From the above Punnett square, we find that if they have a son probability that he will be normal for blood clotting is 1/2 (Here, we ignore probability for female. Thus, we will consider only XHY & XhY genotype).

13) Let, X = normal X chromosome, Xc = X chromosome containing color blind allele

Now, genotype of color-blind male will be XcY & genotype of color-blind female will be XcXc. The cross between them is shown in below Punnett square.

Gametes	Xc	Y
Xc	XcXc (Color blind female)	XcY (Color blind male)
Xc	XcXc (Color blind female)	XcY (Color blind male)

From the above Punnett square, we find that if they have a daughter probability that she will have normal vision is 0 (Here, we ignore probability for male. Thus, we will consider only XcXc genotypes).