某基因型为AaBbDdEe的植物个体,四对基因的位置如右图所示,预测其自交后代表现型的种类,最少和最多是多少
例出是哪几种
推荐回答（3个）
从给定信息来说，A/B性状对，以及D/E性状对，有可能紧密连锁，也有可能不连锁，当紧密连锁时，后代表现型种类最少，不连锁时，表现型种类最多。
连锁时（最少）：
配子类型：AbDE, Abde, aBDE, aBde
-
合子类型：AAbbDDEE, AAbbDdEe, AaBbDDEE,AaBbDdEe,
AAbbddee, AaBbDdEe, AaBbddee,
aaBBDDEE, aaBBDdEe,
aaBBddee
（这是所有组合方式，你可以将相同的合并起来）
-
表现型： AbDE, ABDE, Abde, ABde, aBDE, aBde
这里认为AA和Aa的表现型一致，即完全显性性状。
由此你看到了，最少有6种表现型；
不连锁时（最多）：
配子类型，不连锁嘛，就是完全随机组合，各种可能性都有，所以不用这么麻烦去计算了
-
表现型：A两种*B两种*D两种*E两种=2*2*2*2=16种可能组合，
所以最多有16种表现型。
其实前面连锁时的可能性也可以用组合的可能性来计算，只是反正位点也不多，都列出来也没浪费多少时间。
btw：想想还是给你说一下吧，虽然麻烦些。对于计算最少可能性，我们有前提假设是A/B连锁，D/E连锁，那么，我们可以把A/B看做一个基因座，D/E看做一个基因座，A/B基因座具有3个等位基因型AAbb/AabB/aaBB，对应Ab/A/B/aB三种表现型；类似地，D/E具有3个等位基因型DDEE/ddee/DdEe，对应DE/de两种表现型；由于A/B基因座和D/E基因座独立分离并决定表现型，因此，最少的表现型为3*2=6种。
最少4种 AbDE aBde Abde aBDE
最多16种 AbDE aBde Abde aBDE abDE ABde ABDE abde AbDe aBdE aBDe AbdE abDe ABdE ABDe abdE
9种，AAbbDDEE AaBbDDEE AaBbDdEe AAbbDdEe AAbbddee AaBbddee aaBBDDEE aaBBDdEe aaBBddee
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