英语周报高考综合第30期b版答案

32答案】(除注明外,每空1分,共1分动,:(1)遵循A纯合致死(2分)1,m(0比欲(2)aaBB100%(3)红花:黄花:蓝花:白花=2:2:1:1但2分)4)宽叶x有(x)的雄配子具有致死效应(2分(【解析】(1)根据题干信息“红花和黄花中均含B基因”分析题图,可知推断F1红花植株的基因型为AaBb,正常情况下,基因型为AaBb的雌雄个体杂交,后代的性状分离比应为9:3:3:1,将图中F2的分离比与正常情况下的分离比进行比较,可推断AA显性纯合有致死作用。进一步判断等位基因Ma和B/b的遗传遵循基因的自由组合定律。(2)由此可进一步判断图中蓝花植铁的因型全为A的,黄花的因型为B或aB,其中只有基因型为aB的赏花与蓝自千花株(xah)杂交,后代才会出现两种类型的植株由此确定P中黄花植株的基因型是anB(3)分析题图,F2中蓝花植株的基因型是Aahb,黄花的基因型为aBB(1/3)或aBb(2/3),若让F2中的黄花与蓝花之间随机受粉,则后代的表现型及比例是红花:黄花:蓝花:白花=2:21:1(其体计算可先求各自产生的配子类型及概率,再利用棋盘法进行分析求算,如下表,再利用镇盘法进行分析求算,如下表)黄花植株的配子蓝花植株的配子I/ab1/2Ab2/6ABb红花。L6Aab蓝花1/2ab2/6aBb黄花1/6abb白花(4)根据题中“两株宽叶植株杂交,F表现型及比例为宽叶雌株:宽叶雄株:窄叶雄株=2:1:1可推断基因D控制的性状是宽叶,由于后代雌雄性状分离比不同,可判断该基因位于X染色体上。若让F中窄叶雄株(X“Y)与亲代中的母本(X"x)杂交,正常情况下,窄叶雄株(XY)产生xY两种雄配子,XX产生、N两种雌配子,若后代中全为雄株,则可能的原因是含d(或X“)雄配子有致死效应。

【参考范文】One day, Spotty returned from his daily walk with a broken leg. Appearing exhausted, he came to myroom and sat near me, the leg bleeding. I called my mother and she quickly tied a bandage around his leg andgave him food to eat. I was very upset. But the next day, Spotty followed me wherever I went happily asusual though he limped a bit. After this incident my relation with Spotty became more intense. I reallyadmired him a lot for his courageAlmost a year later, one midnight we heard Spotty barking breathlessly. We rushed out and saw himbarking continuously, heading somewhere. After some time Spotty became quiet. I patted him on his backand came inside. The next morning, my heart skipped a beat when I didn't see Spotty. I searched for him ineach and every corner but in vain. And this time he had gone and would never come back. I cried and waitedfor him. But there were no signs of him and I only say him looking at me with his sparking eyes in my dream

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