AP Biology Help?

115. If a human recessive X-linked characteristic occurred with a 10 percent frequency, what would its frequency be in males and females?

123. If an X-linked recessive gene is expressed in 4 percent of the men, what proportion of women would express the recessive trait?

a. galactosemia

b. Turner syndrome

c. AIDS

d. hemophilia

e. Down syndrome

D 137. For this disorder, both a phenotypic cure and a genotypic cure are potentially possible.

By the way, the D before question 137 means difficult, not Hemophilia.

115.
Since it's recessive and X-linked, the frequency of appearance would be the same as the frequency of expression for males.

For females, the frequency of expression would be the probability that a female has the characteristic on both X-chromosomes. Using basic probability: 10% * 10% = 1%

123.
Same as above for the most part. 4% is frequency of the gene. Thus for women the rate is based on having it on both X chromorsomes. 4% * 4% = 0.16%

137.
Ugh... making me remember stuff I learned 15 years ago... ;-)
galactosemia - inability to digest galactose. Gene therapy could lead to proper expression of galactase. It'd only need to be expressed in the liver so it's somewhat feasible to solve genotypically.

Turner Syndrome - missing X chromosome in women. Good luck inserting an extra chromosome in all the necessary cells... Phenotypic cure is hormone treatment.

AIDS - Complete removal of the HIV genome from the infected person is extremely difficult (due to integration in the host DNA) but possible.

Hemophilia - X-linked gene that has either a missing or deficient blood clotting factor (I'l be damned if I recall which ones they are...). About as difficult to cure genotypically as Turner Syndrome.

Down Syndrome - extra chromosome. Even if you could remove all the extra chromosomes, the body has already developed incorrectly.

Not 100% sure, but I'd say C) AIDS is the best answer.