为研究裂翅基因在染色体上的分布情况，将裂翅果蝇进行杂交实验，结果如下图所示
(裂翅(A)对非裂翅(a)为显性)。
根据实验结果分析，这对等位基因的分布有两种可能：
(1)如果这对等位基因位于常染色体上，则父本的基因型为_____，F1中的裂翅的基因型为_____。
(2)如果这对等位基因于X染色体上，则母本的基因型为_____，F1中裂翅的基因型为_____。
(3)上述两种情况下，亲本一定为杂合子的是_____(父本／母本)。
正确答案及解析
正确答案
解析
(1)an；Aa
(2)XAxa；XAxa、XAY(3)母本
(1)显性个体与隐性个体的杂交后代显、隐性之比为1：1，符合测交后代的表现型比例，如果等位基因位于常染色体，则为AaXaa，父本基因型为aa。F1裂翅基因型为Aa。
(2)若亲本基因型为XAxa,Xay，其杂交后代符合题意1：1：1：1的情况。
(3)由(1)(2)可知母本一定是杂合子。
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