
Benjamin H. answered 12/01/21
Harvard Grad/Experienced Tutor in STEM, English, and Writing
Since both Eric and Janelle are carriers of a recessive allele, (let's call the healthy allele 'X' and the Tay-Sachs allele 'x') their genotypes are both Xx. Doing a punnet square, we can see that the distribution for offspring is %25 XX, 50% Xx, 25% xx. However, since Tay-Sachs is lethal in early childhood, we can assume that a healthy child is not xx. Since based on our Punnett square, their child is twice as likely to be a carrier (Xx) than a healthy non-carrier (XX), we can say there is a 33.333% chance that the child is not a carrier of Tay-Sachs.