Loading... 以下,均认为母代为$P$,第$n$代子代为$F_n$。 ## 显性杂合子自交 杂合子$Aa$连续自交,在$F_n$中(淘汰$aa$个体),基因型比例为:$Aa$占$\cfrac{1}{2^{n-1}}$,$AA$占$\cfrac{2^{n-1}-1}{2^n}$,$aa$占$\cfrac{2^{n-1}-1}{2^n}$。 ## 显性杂合子自交(淘汰隐性个体) 杂合子$Aa$连续自交,并逐代淘汰隐性个体,在$F_n$中(淘汰$aa$个体),基因型比例为:$Aa$占$\cfrac{2}{2^n+1}$,$AA$占$\cfrac{2^n-1}{2^n+1}$,$aa$占$0$。<hr class="content-copyright" style="margin-top:50px" /><blockquote class="content-copyright" style="font-style:normal"><p class="content-copyright">版权属于:淡淡的路瑶</p><p class="content-copyright">本文链接:<a class="content-copyright" href="https://www.starroad.top/archives/737.html">https://www.starroad.top/archives/737.html</a></p><p class="content-copyright">如果文章出现任何问题,请下方评论,紧急情况请发邮件。谢谢支持!</p></blockquote> Last modification:March 20th, 2020 at 09:44 am © 允许规范转载 Support 如果您觉得我的文章有用,给颗糖糖吧~ ×Close Appreciate the author Sweeping payments Pay by AliPay Pay by WeChat
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必修二?同高一