A man has extra digits (six fingers on each hand and six toes on each foot). His wife and their daughter have the normal number of digits (five fingers on each hand and five toes on each foot.) Having extra digits is a dominant trait. The couple's second child has extra digits. What is the probability that their next (third) child will have extra digits?

a. 3/4
b. 1/16
c. 1/2
d. 1/8

Polydactyly, the presence of extra digits on hands and feet, possesses an autosomal dominant mode of inheritance.
To calculate the probability of the third child being polydactyl, we must first figure out the genotypes of the parents.
Since polydactyly is a dominant trait, the mother must be homozygous recessive as she does not have extra digits.
Since the first daughter is normal, the father is heterozygous for the allele. If he would have been homozygous dominant, all the children would be affected.

Cross:

Parents [tex]M^{D} M^{d}[/tex] X [tex]W^{d}W^{d}[/tex]

Gametes [tex]M^{D}, M^{d}[/tex] X [tex]W^{d}, W^{d}[/tex]

Offspring 2 [tex]M^{D} W^{d}[/tex], 2 [tex]M^{d}W^{d}[/tex]

Probability of polydactyl child = 2/4 or 1/2 or 50%

batolisis batolisis · Answer 2 · 2021-11-25T16:48:51+01:00

The probability that the third child will have extra digits = 1/2

Assume :

X = dominant trait ( i.e. having extra digits )

y = recessive trait

Since the first child does not have extra digits it simply means that the wife is h0m0zygous recessive while the man is heterozygous dominant

i.e. Father = Xy

Mother = yy

Crossing the parents using punnett square

X y

y Xy yy

Number of children with five digits ( yy ) = 2

Number of children with six digits ( Xy ) = 2

Total = 4

∴ probability of a child having six digits = 2 / 4 = 1/2

Hence we can conclude that The probability that the third child will have extra digits = 1/2 .

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