One parent has a dominant gene for a
straight thumb and a recessive gene for a
bent thumb. The other parent has two
recessive genes for a bent thumb. Let T
represent straight thumbs and t represent
bent thumbs. What is the probability, or
possibility, that one of their children will
have a bent thumb?
The probability of a child with a bent thumb
is out of 4, or
%.

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monica789412 monica789412 · Answer 1 · 2020-03-30T07:53:26+02:00

The probability of a child with a bent thumb is out of 4 is 50%.

Explanation:

Information given:

dominant gene = straight thumb (genotype TT, Tt)

recessive trait = bent thumb (genotype tt)

from the question genotype of parents can be deduced:

Parent 1 = Tt

parent 2 = tt

from the Punnet square:

T t

t Tt tt

From the results of Punnet square it is seen that 50% of progeny will be straight thumb and 50% of progeny will be bent thump.

phenotype ratio and genotype ratio = 1:1

There will be 50 % probability that one of their offspring will be with bent thumb.

progeny with bend thump percent = [tex]\frac{progeny with bent thumb}{total number of offspring}[/tex] x100

= 50%.