Mandy :) Posted January 25, 2011 Share Posted January 25, 2011 In humans, the ability to taste PTC is dominant over non-tasting. Two heterozygous tasters have three children. What is the probability that one or more of the children will be non-tasters? a. 0.684 b. 0.578 c. 0.251 d. 0.422 e. 0.004 I assumed that the two heterozygous parents would have a 1/4 chance of passing on the recessive, non-tasting gene, aa. With three children, I would think to use the product rule, thus multiplying (1/4)(1/4)(1/4)= 1/64 = 0.0156 Please help explain what I have done wrong. Thank you. Link to comment Share on other sites More sharing options...
niharika Posted April 9, 2011 Share Posted April 9, 2011 it should be .251.(1/4) what you did is correct when he asks about probability of first three children being non tasters.but he is asking only about one or more children.he gave no. of children just to confuse you. Link to comment Share on other sites More sharing options...
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