Question

Consider a woman who is heterozygous for two traits: M+/M- for Maple Syrup Urine Disease on chromosome 19 and H+/H- for hemophilia on the X chromosome.

Which of the following gamete genotypes should this woman expect to produce?

1. M- H+

2. M+ M-

3. M+ H+

4. H+ H+

A. 1, 2 & 3

B. 1 & 3

C. 2 & 4

D. 4 only

E. All of 1, 2, 3 & 4 are correct.

Answer 1

We know that,

Maple syrup urine is an autosomal ressesive disease so while gamet (n) formation either M+ or M- will definately take part but not both(because 2 sets of same chromosome here chromosome 19 ,usually do not take part in gamet formation )

So from here two possibilities M+ or M-

Again, hemophilia is an  X linked (related to sex chromosomes) recessive disease.

According to the question the female is heterzygous for hemophilia that makes her the carrier(H+/H-). So we can see she can either give H+ or H- to the gamet.

So from here there are two possibilities either H+ or H-

Now , by combining the possibilities, gamate should contain both autosomal and sex chromosomal genes.

so the possibile genotypes may be:-

a. M+ H+............no.3

b. M+ H-

c. M- H+............no.1

d. M- H-

no.2 and 4 are not possible becase they represent genes either both from autosome or both from sex chromosome...so we can discard them straight away.

So only no.1 and no.3 are among the possibilities given here.

OPTION B  1&3 should be the answer.

I hope this was useful to you.