2018-2022英语周报5答案

32.【答案】(1)深棕色眼基因(se)与无眼基因(ey)位于非同源染色体上(2分)(2)野生型:细毛:无眼:细毛无眼=9:3:3:1(2分)3号(1分)(3)12(2分)(4)深棕色眼雌(1分)全为野生型(1分)野生型:深棕色眼=1:1(1分)全为深棕色眼(1分)【关键能力】本题考查考生对基因自由组合定律的理解能力、分祈能力和综合运用能力以及获取信息的能力。【解题思路】(1)由于深棕色眼基因(se)位于3号染色体上,无眼基因(ey)位于4号染色体上,3号4号染色体为非同源染色体,故深棕色眼与无眼不是一对相对性状。(2)据甲组、乙组杂交实验中F2的表现型及比例分析,细毛基因(h)与短翅基因(dy)位于非同源染色体上,其遗传遵循基因的自由组合定律,细毛基因(h)与深棕色眼基因(se)位于3号染色体上。基因se与基因ey位于非同源染色体上,故丙组杂交实验中的表现型及比例为野生型:细毛:无眼:细毛无眼=9:3:3:1。(3)只考虑体毛和翅形性状,乙组杂交实验F2中细毛雄果蝇的基因型为 hhDyDy,甲组杂交实验F2中细毛雌果蝇的基因型为/4 hhDy Dy2/4 hhDydy、1/4 hhdydy,二者杂交,后代中纯合子( hhDy Dy)的概率为1/4+2/4×1/2=1/2。(4)鉴定无眼突变体X雄果蝇的眼色基因组成时,若选择短翅突变体与X杂交,则后代都是野生型,可以选择深棕色眼雌果蝇与X雄果蝇杂交。若X雄果蝇的基因型为 eyeySeSe,则eyeySeSe x EyEysese→F1: EyeySese(野生型);若X雄果蝇的基因型为eyeySese,则 eyeySese x EyEysese→F1: EyeySese(野生型): Eyeysese(深棕色眼)=1:1;若X雄果蝇的基因型为 eyeysese,则 eyeysese xEyEysese→F1: Eyeysese(深棕色眼)。

第二节书面表达【写作指导】类型:通知审题时态:一般将来时和一般现在时为主人称:第三人称为主丰富校生活Para.1h陈述举办竞赛的目的增强学生食品安全意识国作思时间:竞賽时间和报名截止时间Pa2赛的时间地点和方式地点学校大礼堂报告厅芋方式:问答Para.3呼吁同学们积极参加【佳作展台】Noticecontest concerning the subject is to be held in our scho. a knowledgeAimed at raising students' awareness of food safetyThe contest is scheduled in the school hall from 2: 00 pm to 4: 00 pmon November 27th. It mainly consists of two rounds. First, all competitorsare required to answer five questions about food safety. The top 20competitors will have the opportunity to compete in the second roundwhere they will be asked another five more challenging questions. If youare interested, please sign up at the Student Union office before November20thLooking forward to your active participation【亮点梳理】亮点词汇: aimed at, awareness, schedule,berequired to do, sign up亮点结构:分词短语作状语( Aimed at raising studIents'awareness ofod safety), where引导的非限制性定语从句( where they will be askedanother five more challenging

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