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    【题目】在家鼠中短尾（T）对正常尾（t）为显性。一只短尾鼠与一只正常鼠交配，后代中正常尾与短尾比例相同；而短尾类型相互交配，子代中有一类型死亡，能存活的短尾与正常尾之比为2∶1，则不能存活类型的遗传因子组成可能是 （ ）

    A．TT B．Tt C．tt D．TT或Tt

    【答案】A

    【解析】根据一只短尾鼠与一只正常鼠交配，后代中正常尾与短尾比例相同，即后代正常尾∶短尾=1∶1可知短尾亲本的遗传因子组成为Tt，子代正常尾遗传因子组成为tt，短尾遗传因子组成为Tt。又因为短尾类型相交配，子代中有一类型死亡，能存活的短尾与正常尾之比为2∶1，如果亲本有纯合子，则后代不会出现性状分离，所以该实验中亲本短尾遗传因子组成全为Tt，他们之间杂交的后代遗传因子组成比例应为TT∶Tt∶tt=1∶2∶1，由于第二代杂交结果是短尾与正常之比是2∶1，则可以推出不能存活类型的遗传因子组成为TT。

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