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第二节参考范文Th[文]e girl's mother told Kelly that they were moving to another city. After that they could no longer run thefairy garden. She had reached an agreement[章] with the community[来] that planned to turn it into a park for local resi-dent[自]s to relax in, But her daughter did not want to leave the garden, and she had cried several times. She waseven more unwilling[知] to leave her dear fairy. Her mother wanted Kelly to meet her daughter personall[嘛]y. but shewasnt sure how to make her daughter happy.The[答] girI's mother preferred[案] Kelly to act as a fairy to say goodbye to her daughter.[网] Kelly accepted thegirls mother's request. Then they agreed on a time and a place to meet. On the day of the meeting, Kellyappe[文]ared in front of the girl dressed like a fairy. The little girl was surprised[章] and delighted[来] to see her. theyjoine[自]d hands and started dancing. She chatted excitedly[知] with Kelly for a long time, Kelly gave her many giftsthat[嘛] she liked. The girl made Kelly promise that Kelly would visit her often.
23.(除标注外,[答]每空2分,共10分[案])(1)线性(1分[网])全为心形叶(1分[文](2)选择染色体缺[章]失个体与戟形叶个体[来]进行正反交(正交:[自]A0aad;反交:[知]A0dxaa),观[嘛]察后代的性状表现;[答]若正交后代全为心形[案]叶,反交后代心形叶[网]:戟形叶=1:1,[文]则染色体缺失会使雌[章]配子致死;若正交后[来]代心形叶:戟形叶=[自]1:1,反交后代全[知]为心形叶,则染色体[嘛]缺失会使雄配子致死[答]。(4分)3)出现[案]四种表现型,比例为[网]9:3:3:1出现[文]两种表现型,比例为[章]3:1;或出现三种[来]表现型,比例为1:[自]2:1【解析】本题[知]考查基因自由组合定[嘛]律的应用、相关表现[答]型的计算、遗传实验[案]设计。(1)由图可[网]知,图中A、B、C[文]基因在5号染色体上[章]呈线性排列。由于含[来]缺失染色体的一种配[自]子(不确定是雄配子[知]还是雌配子)致死,[嘛]所以只考虑烟草的叶[答]形,根据上述分析可[案]知,该株烟草自交后[网]代的基因型为AA、[文]AO,均表现为心形[章]叶。(2)为了确定[来]染色体缺失致死的配[自]子是雄配子还是雌配[知]子,应选择染色体缺[嘛]失个体与戟形叶个体[答]进行正反交(正交:[案]A0ad;反交:A[网]0dxa),并观察[文]后代的性状表现。若[章]正交后代全为心形叶[来],反交后代心形叶:[自]戟形叶=1:1,则[知]染色体缺失会使雌配[嘛]子致死;若正交后代[答]心形叶:戟形叶=1[案]:1,反交后代全为[网]心形叶,则染色体缺[文]失会使雄配子致死。[章](3)两对基因若位[来]于两对同源染色体上[自](即D、d基因不位[知]于5号染色体上),[嘛]遗传时遵循自由组合[答]定律,让基因型为B[案]bDd的植株自交,[网]后代应出现四种表现[文]型,比例为9:3:[章]3:1;若两对基因[来]位于一对同源染色体[自]上,若不考虑交叉互[知]换,则BbDd的植[嘛]株会产生两种数量相[答]等的配子,可能是B[案]D:bd=1:1,[网]也可能是BdbD=[文]1:1,若为前者,[章]后代会出现两种表现[来]型,比例为3:1:[自]若为后者,则后代会[知]出现三种表现型,比[嘛]例为1:2:1。
