Re: [問題] 請問一題Genetics Questions消失

看板Biology (生物學)作者xiaoa時間11年前 (2014/11/16 21:17)推噓0(0推 0噓 0→)

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※ 引述《runnial (呵呵)》之銘言： : 【問題】：Sickle cell disease (SCD) is found in numerous populations whose ancestral : homes are in the malaria belt of Africa and Asia. SCD is an autosomal : recessive disorder that results from homozygosity for a mutant β-globin gene : allele. Data on one affected population indicates that approximately 4 in 100 : newborn infants have SCD. : Epidemiologic data on the population in the previous problem reveal that : before the application of modern medical treatment, natural selection played : a major role in shaping the frequencies of alleles. Heterozygous individuals : have the highest relative fitness; and in comparison to heterozygotes, those : who are βAβA have a relative fitness of 91 percent, but only about 25 : percent of those with SCD survived to reproduce. : What is the estimated equilibrium frequency of βA in this population? : What is the estimated equilibrium frequency of βS in this population? : 【問題起因】：mastering genetics : 【個人看法】： : q^2=0.25-->q=0.5 ( βS in this population) : 2pq=0.91-->p=0.91/2*0.5 (βA in this population) : 答案是錯的 : 有人知道怎們做嗎?謝謝我沒有算過這種題目, 我在推文裡說的也不對 (25% 存活至生育與 SCD患者在pop.裡的比例並無直接關係) 這題目我一看完第一個想法是 4% 開平方根號(因為每100個小孩有4個有SCD) 而事實是, 這個題目不能用哈溫定律算我懶得動筆算了, 但我可以告訴你幾個方向 1. SCD患者(βSβS) 只有25% 能存活到生育意思若是pop中的 4% SCD患者, 平均只有1%會活到生小孩 2. βAβA 的 relative fitness, 是混型βAβS的 91% 意思是 βAβA 的平均存活率 x 生育率, 是βAβS的 91% (若是我來算, 我會嘗試假設三種人的生育率都相同, 差別只是存活率) 3. 粗估初生嬰兒有4%是 SCD患者(這條我不確定重不重要, 上面算完後拿來當複查吧 orz) 4. βS = 1 - βA(應該沒有其他的allele了吧...) -- Nothing is true, everything is permitted -- the Assassin's Creed -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 203.106.156.178 ※ 文章網址: http://www.ptt.cc/bbs/Biology/M.1416143852.A.EC5.html