两株高茎豌豆杂交,F1中既有高茎又有矮茎,选择F1中高茎豌豆让其全部自交,则自交后代形状分离比为( )

2025-03-24 09:57:29
推荐回答(3个)
回答1:

依题意得知高茎为显性,所以F1中高茎豌豆有1/3AA,2/3Aa,自交后代中出现矮茎概率为2/3*1/4=1/6,高茎为1-1/6=5/6,故分离比为高茎:矮茎=5:1.

回答2:

根据题意亲本高茎豌豆为杂合子,基因型为Dd和Dd,F1的基因型为DD、Dd、dd,比例为1:2:1,其中高茎有DD和Dd,比例为1/3、2/3,让F1中高茎豌豆让其全部自交即1/3(DDxDD),2/3(DdxDd),DDxDDr的后代全为DD,占子代的1/3,DdxDd的后代为DD、Dd、dd,占子代的比例分别为2/3*1/4=1/6、2/3*1/2=1/3、2/3*1/4=1/6,所以自交后代的性状分离比为(1/3+1/6+1/3)/(1/6)=5:1。

回答3:

因为F1中既有高茎又有矮茎,所以亲本一定是两个杂合,也就是Aa和Aa。
而在F1中就会有高茎豌豆基因型AA:Aa=1:2,即为AA占F1中高茎的1/3、Aa则占2/3
让后让AA和Aa分别自交,AA的后代只有AA,即在F2中有1/3的AA;Aa的后代则有AA:Aa:aa=1:2:1,也就是F2中有1/4的AA、2/4的Aa、1/4的aa。
所以性状分离比=F2中的显性基因型:F2中的隐性基因型=[1/3+2/3(1/4+2/4)]:2/3*1/4=5:1

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