Question

a woman is a healthy carrier of the cystic fibrosis gene (a recessive trait) and her husband is an unaffected male. what is the probability of an unaffected child from this pairing

Answer 1

The probability of having an unaffected child from the pairing = 0.5

Assumption

Let ;  X = Dominant trait

        y = recessive trait

The woman been a healthy carrier will have an allele combination of ;  Xy

while the unaffected male will have  = XX

Their offsprings will have the following allele combinations

         X         X

X      XX       XX

y       Xy       Xy

Number of allele combinations = 4

Number of carrier = 2

Number of unaffected = 2

∴ probability of  having an unaffected child = 2/4  = 0.5

Hence we can conclude that The probability of having an unaffected child from the pairing = 0.5

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