题目内容

【题目】某XY型性别决定的植物花色遗传中，常染色体基因D决定花色素的生成，X染色体上的基因H和h分别使花色呈现橙色和黄色，不能生成色素时花为白色。现将纯合白花雄株和纯合黄花雌株杂交，后代中有橙花植株。请回答下列问题：

（1）F1中雌株的基因型及雄株的表现型分别为 。让F1 雌雄株杂交得到F2，F2中橙花∶黄花∶白花比例为 。F2代中黄花植株的基因型有__________种。

（2）请利用纯合品系，设计杂交方案，探究 F2 中白花雌株的基因型：

①让F2 白花雌株与纯合 杂交；

②观察并统计后代的表现型，预期白花雌株的基因型。

如果 ；

如果 。

【答案】

（1）DdXHXh、黄花 3∶3∶2 4

（2）①黄花雄株

②子代中橙花:黄花＝1∶1，则白花雌株的基因型为ddXHXh

子代中全为黄花，则白花雌株的基因型为ddXhXh

【解析】

（1）依题意可知：纯合白花雄株和纯合黄花雌株的基因型分别为ddXHY或ddXhY、DDXhXh，二者杂交，后代中有橙花植株，说明亲本纯合白花雄株的基因型为ddXHY，进而推知：F1中雌株和雄株的基因型分别为DdXHXh、DdXhY，雄株的表现型为黄花。F1雌雄株杂交所得到的F2中，其基因型及其比例为DD ∶Dd∶dd＝1 ∶2∶1，XHXh ∶XhXh ∶XHY ∶XhY＝1∶1 ∶1 ∶1,表现型为橙花（1DDXHXh、2DdXHXh、1DDXHY、2DdXHY）∶黄花（1DDXhXh、2DdXhXh、1DDXhY、2DdXhY）∶白花（1ddXHXh、1ddXhXh、1ddXHY、1ddXhY）＝3∶3∶2。F2代中黄花植株的基因型有4种。

（2）①利用纯合品系，设计杂交方案，探究 F2 中白花雌株的基因型，可让该白花雌株与纯合黄花雄株（DDXhY）杂交，观察并统计后代的表现型。

②若F2 中白花雌株的基因型为ddXHXh，则子代的基因型为DdXHXh ∶DdXhXh∶DdXHY∶DdXhY＝1∶1∶1∶1，表现型为橙花∶黄花＝1∶1；若F2 中白花雌株的基因型为ddXhXh，则子代的基因型为DdXhXh∶DdXhY＝1∶1，表现型均为黄花。

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