Question

Cystic fibrosis (CF) is caused by a recessive allele. A couple who do not have CF have a child that has CF. What is the probability that if they have another child, that the child will be heterozygous for CF?

1/2 or 50%

1/4 or 25%

3/4 or 75%

0%

Answer #1

Let the recessive allele causing CF be represented by X and non-recessive allele be Y.

Let us consider the couple. Since both do not have CF, it figures that one of their genomes contains X while the other contains Y.

Hence the genome of the couple should be XY and XY respectively.

Now the possible genome combinations when they have a child is (XX, XY, YX, YY)

Only in case of XX would the child have CF otherwise not.

The child will be heterozygous for CF in the case of XY and YX i.e inherited different form of a particular gene from each parent.

The required probability is thus 2/4 = 1/2 =
**50%**

**If satisfied,
please do upvote! Let me know in the comments if anything is
unclear. I will reply ASAP**

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