Question

The Huntington’s Disease (HD) gene (HTT) is located on chromosome 4, and the age of onset of the condition is closely related to the repeat-expansion size within this gene. Recent research shows that a particular variant on chromosome 15 acts as a modifier for the condition. Someone with a repeat expansion in their HTT gene as well as this modifier variant will develop HD symptoms approximately six years earlier than someone without this modifier variant. A woman with Huntington’s disease has one copy of this modifier variant, and has a child with an unaffected man, who also has one copy of this modifier variant.

a) What are the chances their child has inherited the HD expansion, but not the modifier variant?

b) What are the chances their child has inherited the HD expansion and at least one copy of the modifier variant?

Answer 1

The gene for HD, which is an autosomal dominant condition, is on chromosome 4 and the gene for the modifier is ion chromosome 15. So they are not linked and would be inherited independently. So if the woman has a genotype of Hh for HD and Mm for the modifier variant ( so she is HhMm) and the man is hhMm ( since he does not have HD but has one copy of the modifier). the possible gametes are HM, Hm, hM, hm( woman), and hM, hm( man).

	hm	hM
HM	HhMm	HhMM
hM	hhMm	hhMM
Hm	Hhmm	HhMm
hm	hhmm	hhMm

a) So, chances that their child has inherited HD expansion and not the variant is the probability of Hhmm offspring, which is 1/8 ( red-colored) or 0.125.

b) there are 3 instances when the gene for HD expansion (H) and the modifier (M, one or two copies) are present (green colored). So, the probability for HD gene along with one or two copies of the modifier ( at least one copy includes instances of one or more copies of the modifier) is 3/8 0r 0.375.