Question

Question 35 (1 point) A patient with Huntington's disease, but heterozygous for the trait, and another patient with HD, also heterozygous, have a child. Assuming the child lives to maturity and no cure is found, what are the chances that the child will develop HD by the age of 45? a) 75% Ob) 100% O c) 0% O d) 50% Question 36 (1 point) In pea plants, green (G) seeds are dominant to yellow (g) seeds. In a genetic cross of two pea plants that are heterozygous for the seed colour trait, what percentage of the offspring will have the yellow seed colour? O a) 100% Ob) 25% O c) 50% O d) 75%

Answer 1

35. Answer: a)

Representation:

H = Normal allele = Dominant

h = mutated allele for Huntinton disease

Since both parnts are heterozygous, their genotype = Hh

Cross: Hh x Hh

	H	h
H	HH	Hh
h	Hh	hh

Since Huntington disease is inherited as autosomal dominant manner, the dominant heterozygote and homozygote will give rise to disease phenotype.

Thus, probability of child may have Huntinton disease = 75% (HH= 25% + Hh = 50%)

36. Answer: b)

Representation:

G= dominant

g= recessive

Green = genotype GG/Gg

Yellow = genotype gg

Cross:

	G	g
G	GG	Gg
g	Gg	gg

Green = 75% (GG/Gg)

Yellow = 25% (gg)