Answer:
Genotypic ratio of offsprings will be: 1 (Hh) : 1 (hh)
Explanation:
This question involves a single gene coding for the possession or not of Hitchhiker's thumb in humans. The allele that codes for Hitchhiker's thumb (H) is dominant over the allele for no hitchhiker's thumb (h).
Based on this question, if woman who does not have hitchhiker's thumb (hh) marries a man who is heterozygous for hitchhiker's thumb (Hh) i.e. hh × Hh, the following gametes will be produced by each parent:
hh - h and h
Hh - H and h
Using these gametes in a punnet square (see attached image), the offsprings with the following genotypic ratio will be likely produced:
1 (Hh) : 1 (hh)
N.B:
- Two (2) of them were Hh i.e. with Hitchhiker's thumb
- The other two were hh i.e. no hitchhiker's thumb.
Genotype is the allelic combination of the trait. The the probable genotypic ratio (Hh:hh) of their children is 1:1.
Hitchhiker's thumb:
It is a trait in which person have very flexible thumb that can bent backward more than the normal range of motion.
Given here,
Hitchhiker's thumb (H) - Dominant trait
No hitchhiker's thumb (h) - Recessive trait
Mother- hh - recessive
Father - Hh - Heterozygous
h h
H Hh Hh
h hh hh
From punnet square, the 50% of child have heterogygous genotype while 50% have recessive.
Therefore, the the probable genotypic ratio (Hh:hh) of their children is 1:1.
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