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

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

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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c. 1/2

Explanation:

Cross:

Parents      X  

Gametes   X  

Offspring 2 , 2

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

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