Anyone care to shed light on this?
Hi Kyle!
I found the answer by looking at a similar question online, this is how they solved:
You are correct that the parents must be both Aa to produce an aa child with a probability of 1/4.
But what we don’t know is the genotype of the parents
they could be aa, aA, Aa, or AA
we need to find the probability that the parents are healthy carriers of the trait
(there is a flaw in the question, as we need to assume that the parents must be healthy to solve, and ignore the possibility of having an aa child with Aa x aa or aa x aa.
in other questions i found online this is stated explicitly)
With this in mind the parents now have 1/3 chance of being AA and 2/3 of being Aa
P(both parents are healthy carriers) = 2/3 * 2/3 = 4/9
P(aa child from healthy carriers) = 4/9 * 1/4 = 4/36 = 1/9
Hope this helps!!
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That’s where I got thrown off because I couldn’t decide how to calculate if both or just one of the parents was a carrier, or neither Very Very helpful, tricky question - Thank you!
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