作业帮一道生物题 有关于遗传疾病的问题
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一道生物题 有关于遗传疾病的问题
Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.
Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.
Would you mind if I answer in English,'cause I haven't used Chinese for a while.Ok,let's start.
First,the answer is 1/16.
Let's use A&a respectively to stand for the dominant and recessive alleles of the sickle-cell disease.According to the problem discribed,we could easily deduce that the genotype of both Karen and Steve's parents are "Aa".That means that both Karen and Steve are 50% "a" allele carrier (since Karen and Steve must not be aa based on their phenotypes,so they are 25%AA,50%Aa).So they each have 25% to give "a" allele during mating (50% time 50%).So the final probability for them to have a sickle-cell-diseased child is 25% times 25% （or (1/4)*(1/4)）,equals to 1/16.
再问： sorry, but answer is 1/9
再答： Sorry, I did not read the problem set really carefully. Yes, the answer should be 1/9. Karen and Steve is 25%AA, 50%Aa, so they each has 1/3 probability to provide "a" allele containing gamete during mating. And the final probability for their child to catch the disease is (1/3)*(1/3)=1/9. Thanks for informing me. If you have any further concerns, please feel free to contact me.
再问： why they each has 1/3 probability to provide "a" allele containing gamete during mating？
再答： Sorry, I am pretty much sure I've made another mistake. They both are 1/3AA, 2/3Aa, so that is 100% total. Just now I said 25%, 50%, that is only 75%. I was influenced by my first answer. Ok, this time it should be fine. Since they have 2/3Aa which is inherited from their parents, they each has a probability of (1/2)*(2/3)=1/3 to produce a "a" containing gamete. Thus, the final probability is 1/9. Thanks.
First,the answer is 1/16.
Let's use A&a respectively to stand for the dominant and recessive alleles of the sickle-cell disease.According to the problem discribed,we could easily deduce that the genotype of both Karen and Steve's parents are "Aa".That means that both Karen and Steve are 50% "a" allele carrier (since Karen and Steve must not be aa based on their phenotypes,so they are 25%AA,50%Aa).So they each have 25% to give "a" allele during mating (50% time 50%).So the final probability for them to have a sickle-cell-diseased child is 25% times 25% （or (1/4)*(1/4)）,equals to 1/16.
再问： sorry, but answer is 1/9
再答： Sorry, I did not read the problem set really carefully. Yes, the answer should be 1/9. Karen and Steve is 25%AA, 50%Aa, so they each has 1/3 probability to provide "a" allele containing gamete during mating. And the final probability for their child to catch the disease is (1/3)*(1/3)=1/9. Thanks for informing me. If you have any further concerns, please feel free to contact me.
再问： why they each has 1/3 probability to provide "a" allele containing gamete during mating？
再答： Sorry, I am pretty much sure I've made another mistake. They both are 1/3AA, 2/3Aa, so that is 100% total. Just now I said 25%, 50%, that is only 75%. I was influenced by my first answer. Ok, this time it should be fine. Since they have 2/3Aa which is inherited from their parents, they each has a probability of (1/2)*(2/3)=1/3 to produce a "a" containing gamete. Thus, the final probability is 1/9. Thanks.