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Ok this is what I ended up doing chance of father being carrier 2/3. Chance of mother being carrier 1/100
2/3*1/100= .00666
Then I multiplied the chance that if both parents are carriers that the recessive gene will be the one carried on (1/4)
So .00666 * (1/4)= .00166 or 1/600
Can OP confirm what the right answer is?
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father's brother is affected .so chance left 3 . 1 chance is to be normal , 2 chance is to be carrier .so 2 out of 3 =2/3 is to be carrier .hope realize
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drangel you know father is 2/3 because you know for sure he does not have the disease. In this case that it's a recessive gene that knocks out the possibility completely of the father being aa. Meaning the father could only be AA, Aa, or aA. So the probability of him being a carrier is 2/3
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where did super6818 get these questions of genetics??.. plz can u tell me i wanna practice it too
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in the hardy weinburg equation is the p always 1??
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ccc is correct explained by rvusmle
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i have not been agree with you. check the carrier frequency .it would be probability of mother being carrier.
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The correct answer is C. The risk that the husband is a carrier is 2/3, and the risk that his wife is a carrier is 1/100. If they are both carriers, the risk to a child is 1/4. The overall risk to a child is therefore (2/3) (1/100) (1/4) = 1/600.
