Question

一道高一生物题，有答案，求分析

答得好有追加，帮帮忙吧
Ⅰ．“翻翅平衡致死系”个体的体细胞中一条2号染色体上有一显性基因Cy（curly， 翻翅），这是纯合致死的；另一条2号染色体上有另一显性基因S（star， 星状眼），也是纯合致死的。
⑴“翻翅平衡致死系”雌雄个体相互交配，其遗传图解如下，请补充完善该图解：
Cy + × Cy +
+ S + S
翻翅星状眼 翻翅星状眼

F1
⑵由上述遗传图解，可以得出的结论是：_____。
Ⅱ．现有一批经诱变处理的雄果蝇，遗传学家用翻翅作标志性状，利用上述平衡致死系设计了如下的杂交实验，通过三代杂交实验，就可以判断出这批雄果蝇2号染色体的其他位置上是否带有隐性突变或隐性致死突变基因，用M和N表示要鉴定的野生型果蝇的两条2号染色体，如下：

⑴将诱变处理后的雄果蝇与“翻翅平衡致死系”的雌蝇交配，得F1，F1有_____种类型的个体（不考虑雌雄差异）；
⑵在F1中选取翻翅雄蝇，再与“翻翅平衡致死系”的雌蝇单对交配，分别饲养，得到F2，F2有_____种类型的个体（不考虑雌雄差异）；
⑶在F2中选取翻翅雌雄个体相互交配，得到F3。预期结果和结论：
①若F3中有1/3左右的野生型果蝇，则说明___。
②若F3中只有翻翅果蝇，则说明___________。
③若F3中除翻翅果蝇外，还有1/3左右的突变型，则说明___________________。
⑷若该隐性突变发生在X染色体上，位于X与Y的非同源区段上，则采用上述的方案，应在_____（F1、F2、F3）代，就可得到结果。
答案：Ⅰ．⑴ ；⑵“翻翅平衡致死系”的雌雄个体交配，其后代总是翻翅星状眼的个体，能非常稳定地遗传。Ⅱ．⑴四；⑵三；⑶最初第2染色体上不带致死基因//最初第2染色体含有致死基因//最初第2染色体上会有隐性可见突变基因；⑷F2。

Answer 1

I（1）Cy cy Ss × Cy cySs
FI 基因为CYcySs 因为翻翅星状眼是纯合子显性致死 翻翅平衡致死系的雌雄个体交配得到的CYCYSS和cycyss全部死亡
II(1)四种基因为CYCYSS CYcySS CYCYss cycyss
（2）即答案(1）前3种CYCYSS CYcySS CYCYss 与CYCYSS 交配得
CYCYSS CYcySS CYCYSs三种

Answer 2

飞盛大富翁

Answer 3

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