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 thi
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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 thi](/uploads/image/z/14459978-2-8.jpg?t=%E4%B8%80%E9%81%93%E7%94%9F%E7%89%A9%E9%A2%98+%E6%9C%89%E5%85%B3%E4%BA%8E%E9%81%97%E4%BC%A0%E7%96%BE%E7%97%85%E7%9A%84%E9%97%AE%E9%A2%98Karen+and+Steve+each+have+a+sibling+with+sickle-cell+disease.+Neither+Karen+nor+Steve+nor+any+of+their+parents+have+the+disease%2C+and+none+of+them+have+been+tested+to+reveal+sickle-cell+trait.+Based+on+thi)
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 thi
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.
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 thi
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.
2/3*2/3*1/4=1/9