If a woman who is a nonPTC taster aa with heterozygous hitch

If a woman who is a non-PTC taster (aa) with heterozygous hitchhikers thumb (Bb) has children with a man who is a heterozygous PTC taster (Aa) with straight thumbs (bb), what is the probability of them having each of the following types of children? a. How many PTC taster, Hitchhikers thumb? b. How many PTC taster, straight thumb? c. How many Non-PTC taster, Hitchhikers thumb? d. How many Non- PTC taster, straight thumb?

Solution

Answer:

Woman= aaBb

Man= Aabb

Let’s do punnett square

aB

ab

Ab

AaBb

Aabb

ab

aaBb

aabb

PTC taster and Hitchhiker thumb (AaBb)= 1

PTC taster and straight thumb (Aabb)=1

non-PTC taster and Hitchhiker thumb (aaBb)= 1

	aB	ab
Ab	AaBb	Aabb
ab	aaBb	aabb