What if Mike and Stephanie have a child together and that child does not show the condition. Given this new information, what is the probability that both Mike and Stephanie are carriers for this rare autosomal recessive condition at the same time?

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∻Answer∻ ⚹ The probability that both Mike and Stephanie are carriers for the rare autosomal recessive condition, given that their child does not show the condition, is 2/3 for each parent, so the combined probability is 4/9. ⚹ ∻Solution∻ ‖ a ⋮ Autosomal recessive inheritance: In this pattern, both parents must be carriers (heterozygous) for a child to have a chance of exhibiting the condition (homozygous recessive). ‖ ‖ b ⋮ Prior probability: Initially, without any information about the child, the probability of each parent being a carrier is 2/3, assuming that they do not exhibit the condition themselves. ‖ ‖ c ⋮ Child's genotype: Since the child does not exhibit the condition, the child is either not a carrier (homozygous dominant) or a carrier (heterozygous), but not affected (homozygous recessive). ‖ ‖ d ⋮ Calculation of combined probability: The probability that both parents are carriers given that their child is not affected is calculated using the formula $P( ext{both carriers}| ext{child not affected}) = \frac{P( ext{child not affected}| ext{both carriers}) imes P( ext{both carriers})}{P( ext{child not affected})}$. ‖ ‖ e ⋮ Probability of child not affected if both carriers: If both parents are carriers, there is a 1/4 chance the child will be affected and a 3/4 chance the child will not be affected. ‖ ‖ f ⋮ Probability of both parents being carriers: The probability that both parents are carriers is $(2/3) imes (2/3) = 4/9$. ‖ ‖ g ⋮ Probability of child not affected: The probability that the child is not affected is the sum of the probabilities of the child being homozygous dominant or heterozygous, which is $1 - \frac{1}{4} = \frac{3}{4}$ if both parents are carriers. ‖ ‖ h ⋮ Final calculation: Using the formula from step d, the probability is $P( ext{both carriers}| ext{child not affected}) = \frac{\frac{3}{4} imes \frac{4}{9}}{\frac{3}{4}} = \frac{4}{9}$. ‖ ∻Key Concept∻ ⚹ Conditional probability in genetics ⚹ ∻Explanation∻ ⚹ The probability that both parents are carriers of an autosomal recessive condition given that their child is unaffected is determined using the principles of conditional probability and Mendelian inheritance. ⚹◊What is the formula to calculate the probability of both parents being carriers for a rare autosomal recessive condition if their child does not exhibit the condition?⍭ Generate me a similar question◊

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