2020-2021学年高一生物北师大版必修二随堂检测（7）基因的自由组合规律
※题型：单选题※知识点：单选题※试题难度：中等<br>下列杂交的组合中(遗传遵循自由组合定律),后代会出现两种表现型的是( )<br>A.AAbb<img src="https://picture2.zhihuicn.com/testImage/1/2020/10/19/52828785127BC1CD258531CDEA88FD52.png"><br>aaBB<br>B.AABb<img src="https://picture2.zhihuicn.com/testImage/1/2020/10/19/52828785127BC1CD258531CDEA88FD52.png"><br>aabb<br>C.AaBb<img src="https://picture2.zhihuicn.com/testImage/1/2020/10/19/52828785127BC1CD258531CDEA88FD52.png"><br>AABB<br>D.AaBB<img src="https://picture2.zhihuicn.com/testImage/1/2020/10/19/52828785127BC1CD258531CDEA88FD52.png"><br>AABb<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/f926492877244e81b7b3545668f6c72f/CB10E80A5A739A58F1312BA12F379467.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/0f28d3b25c5a4c2cbea30c1dba581963/390FD1F888E95F530EC29C938DDBD437.png"><br><br>※题型：单选题※知识点：单选题※试题难度：中等<br>在孟德尔进行的两对相对性状的遗传实验中,具有1:1:1:1比例关系的是( )<br>①杂种自交后代的性状分离比②杂种产生配子类型的比例<br>③杂种测交后代的表现型比例<br>④杂种自交后代的基因型比例<br>⑤杂种测交后代的基因型比例A.①②④ B.②③⑤ C.①③⑤ D.②④⑤<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/f7f7954e29ad4ddd861331fcb342e0ae/CB10E80A5A739A58F1312BA12F379467.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/efeaf1b8bd85450faffc1109f805c7fe/E45FB0185D83E65D71A52DB3712B7DE8.png"><br><br>※题型：单选题※知识点：单选题※试题难度：中等<br>大麦的黑稃(B)对白稃(b)为显性,刺芒(R)对光芒(r)为显性,这两对相对性状的遗传遵循自由组合定律。现有白稃刺芒品种甲和黑稃光芒品种乙,让大量的甲与乙杂交,F<sub>1</sub>中没有出现白稃光芒的个体,则( )<br>A.甲、乙都是纯合子<br>B.甲、乙都是杂合子<br>C.甲、乙都是纯合子,或者一个是纯合子一个是杂合子<br>D.条件不够,无法判断<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/0bebc2430fa94db1b1d4b93aeda1c3e2/F3FFF02E67737AECAA5F9376DCA7D618.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/38f3b91c539e43ea995c9c9de8e8e2b6/4B45FF73905BCCA336D89F872EB8EE7E.png"><br><br>※题型：单选题※知识点：单选题※试题难度：较易<br>下列有关孟德尔的两对相对性状的豌豆杂交实验的叙述,错误的是( )<br>A.F<sub>1</sub>自交后,F<sub>2</sub>出现绿圆和黄皱两种新性状组合<br>B.对F<sub>2</sub>每一对性状进行分析,比例都接近3︰1<br>C.F<sub>2</sub>的性状表现有4种,比例接近9︰3︰3︰1<br>D.F<sub>2</sub>的遗传因子组成有4种<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/c486da61fd80430182f1ff9a76d680dd/2BB966A0E0D0EF1A330D14E3ADEDA5C5.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/1a2f3fa067f4435ab704ee95b0c21dac/7EDE76A5F7945F1F71F742F80A484B7C.png"><br><br>※题型：单选题※知识点：单选题※试题难度：中等<br>科学家已经将控制豌豆7对相对性状的基因定位于豌豆的染色体上,结果如表所示。若要验证孟德尔自由组合定律,最适宜选取的性状组合是( )<br><tbody><br>基因所在染<br>色体编号<br>1号<br>4号<br>5号<br>7号<br>基因控制的相对性状<br>花的颜色子叶的颜色<br>花的位置、豆荚的形状、植株的高度<br>豆荚的颜色<br>种子的形状<br></tbody><br>A.花的颜色和子叶的颜色<br>B.豆荚的形状和植株的高度<br>C.花的位置和豆荚的形状<br>D.豆荚的颜色和种子的形状<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/0e985f92624b4bb899c186c3b1a73073/01A39E48B8A67989A23F6887F26FCC64.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/dc629c4c181541da98e216e675c8427d/B5B5653C874EF36853D59F2437D148F1.png"><br><br>※题型：单选题※知识点：单选题※试题难度：中等<br>下列属于孟德尔自由组合定律的演绎推理过程的是( )<br>A.F<sub>1</sub>杂合子自交产生四种性状表现,比例为9:3:3:1<br>B.由F<sub>2</sub>出现了“9:3:3:1”的性状分离比推测生物体产生配子时,成对的遗传因子彼此分离,不成对的遗传因子在形成配子时自由组合<br>C.若F<sub>1</sub>产生配子时成对的遗传因子分离,不成对的遗传因子自由组合,则测交后代会出现四种性状表现,且比例接近1:1:1:1<br>D.若F<sub>1</sub>产生配子时成对的遗传因子分离,不成对的遗传因子在形成配子时自由组合,则F<sub>2</sub>中有16种组合方式,9种遗传因子组成,4种性状表现<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/db66b226e5d845b8994286a8e6d2b573/06CA09E70C17EF8BA2D3B8C70D693953.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/98516da1f3984a08b2591d2f591c024f/559A26F66C93DC95372EE2942E361B58.png"><br><br>※题型：单选题※知识点：单选题※试题难度：中等<br>将基因型为AaBb植株的花粉粒和花瓣细胞同时在适宜的条件下进行离体培养遥通常情况下袁花粉粒和花瓣细胞发育成的幼苗的基因型不可能是( )<br>A.ab B.Ab C.aaBB D.AaBb<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/55e656aa41d9470dac130ceae4b1261d/F3FFF02E67737AECAA5F9376DCA7D618.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/87d2ea3dc4c64492ad2d990d4612f8e2/979369CB78AC8ED62A143F2725876FF8.png"><br><br>※题型：单选题※知识点：单选题※试题难度：中等<br>小麦的粒色受自由组合的两对基因R<sub>1</sub>和r<sub>1</sub>、R<sub>2</sub>和r<sub>2</sub>的控制。R<sub>1</sub>和R<sub>2</sub>决定红色,r<sub>1</sub>和r<sub>2</sub>决定白色,R对r为不完全显性,并有累加效应,所以麦粒的颜色随R的增加而逐渐加深。将红粒小麦(R<sub>1</sub>R<sub>1</sub>R<sub>2</sub>R<sub>2</sub>)与白粒小麦(r<sub>1</sub>r<sub>1</sub>r<sub>2</sub>r<sub>2</sub>)杂交得F<sub>1</sub>,F<sub>1</sub>自交得F<sub>2</sub>,则F<sub>2</sub>的表型有( )<br>A.4种 B.5种 C.9种 D.10种<br>答案<br><img src="https://picture2.zhihuicn.com/answer/1/9/2020/e90ab2d362db44c9b56fad3505e25102/76900E09FAB5EBA344B8EA84E844FF07.png"><br>解析<br><img src="https://picture2.zhihuicn.com/explain/1/9/2020/423bbd0592c14129b5cabe0ff8ea8fed/7D04C31E93A1E9BFBE9A633FBAF7A713.png"><br><br>
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