7.生物学的一些实验中需要借助特定的颜色来观察实验现象或物质鉴定。下列叙述错误的是(A.黑藻叶肉细胞中存在的叶绿体有助于观察液泡B.检测花生子叶中的脂肪,需用苏丹Ⅲ染色后再用50%酒精洗去浮色C观察洋葱根尖细胞的染色体,用龙胆紫染色后再用清水洗去多余染液D用洋葱鳞片叶内表皮观察质壁分离现象时,可用红墨水对蔗糖溶液进行染色8我国宋代著作《种艺必用》中记载:“凡嫁接矮果及花,用好黄泥晒干,筛过,以小便浸之。又晒干,再浸之。又晒又浸,凡十余次。以泥封树枝…则根生。”结合所学知识,对促进枝条生根起重要作用的物质及其作用最可能是(A.小便中的无机盐,营养B.小便中的有机物,调节C黄泥中的无机盐,营养D黄泥中的有机物,调节9某遗传病患者表现有智力低下、身材矮小和性器官发育不良等症状,经诊断得到该患者的染色体组型图如右图所¥【自XR示。下列判断错误的是()A.该病为染色体数目异常病,其核型可表示为2如-1=45t男斯1011B.该病可能是由于其母亲因减数分裂异常产生了不含X染131415161718色体的卵细胞引起C.该病人到达育龄时减数分裂产生正常配子的概率为50%196D对孕期妇女进行绒毛细胞检查可减少该病患儿的出生10.下列关于植物细胞工程的说法,错误的是()A.可以用植物病毒作载体将目的基因导入植物细胞B进行农杆菌转化时往往需要检测受体细胞对农杆菌的敏感性C,尽量选择性状优良、细胞全能性表达充分的基因型作为外植体D再分化形成的丛状苗等培养物进行重新切割或接种到其它培养基之前必须要消毒11.红树林生态系统分布在热带亚热带海岸的潮间带,图1为某红树林生态系统的结构示意图,图2为该生态系统部分营养级之间的能量流动示意图。下列说法错误的是(半红树植物红树植物、伴生植物高潮线低潮线生产者初级消费者图2图1红树林生态系统结构示意图A.从低潮线到陆地分布的物种具有明显的差异,体现了群落的水结构B红树林具有抗风消浪、固碳储碳的作用,体现了生物多样性的间接使用价值C.红树植物的板状根和支柱根有助于适应滩涂淤泥周期性海水浸淹造成的无氧环境D.第一营养级到第二营养级的传递效率为(A2+B2+C2+D2)/(A1+B1+C1+D)*100%生物选考试题卷第2页(共8页)
水位高、面积大。9.A解析:由于M湖区降水主要受季节性西风影响,其水量最少的时段降水量少,强发量大,岁明受西风影响弱,其时副热带高气压带较强,受其影响,东亚季风气候区暖混的东南以势力增强,喜爱季风区降水增多,气候趋向于暖湿。[命题意图]以全球气候变化为素材,考查地理环境整体性和差异性一直是高考试愿中的“网红”或题组以内蒙古M湖泊沉积岩岩芯的性状变化为切入点,综合考查气候变化对渐治面称及元系%慢牙的影响。注重基础知识、基本原理和基本技能的考查,关注区减认知,综合愿维和地理实易力答学系素养的考查。10.B解析:依据图中三个横剖面的河床海拔高度可推知自上游到下游的顺序是甲,乙两,乙处何岸陡峭,河床中只有崩塌堆积物,应是瀑布溯源侵蚀形成的,11.A解析:岩层上硬下软,下层侵蚀速度快,上层失去支撑,发生朔塌,爆布保特着能崖状态且向上善移动。断层切断河流会形成瀑布,但不一定使瀑布保持陡崖状态:流量只影响期源侵焕的速度,不影响瀑布状态:地壳一直上升,河流就会整体下切侵蚀,不单独影响瀑布12.A解析:越江河河口段河床一直没有泥沙淤积,说明河口段河床落差大,水流急,下切浸睡速度使泥沙难以在此淤积。[命题意图]本题组以嘉陵江支流越江河低坑瀑布附近三个河谷横制面为餐景,考查流水作用,考查学生的综合思维等学科素养13.C解析:C集团在俄罗斯投资建厂:生产高端产品,主要是为了占领俄罗斯市场,甚至将市场扩大至欧洲。14.D解析:文字信息显示,该整车制造厂的生产工艺和工装设备先进,制造水一流,主要目的是提高产品质量。[命题意图]我国知名企业拓展海外市场是当前的热点话题:工业活动也是近年来高考试题中的然点之一。该题组以长城汽车集团在俄罗斯设立工厂为切入点,综合考查工业活动的相关知识。侧重对区域认知、综合思维等学科素养的考查,考查的基本能力是获取和解读地遵信息的能力、满动和运用地理知识的能力:15.B解析:图中信息显示,西太洋台风主要源地有两个中心,一个是菲律宾以东洋面,一个是南善。台风形成后向西或西北方向移动,南海形成的台风影响台高的几率较小,形成于台湾乌附近的台风较少,且台风在源地时一般势力较小,在移动过程中逐渐加强,形成强台风。因此可推新,影响台湾的台风多来自菲律宾以东洋面。神州智达地理,调研卷丁答差算?面(止A
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(生物参考答案)广东省2023届高考..文件预览广东省2023届高考综合能力测试(三)生物学参考答案一、单项选择题:共16小题,第1一12小题每小题2分,第13一16小题每小题4分,共40分。在每小题给出的四个选项中,只有一个选项符合题目要求,选错或不答得0分。题号1235676答案DA0DCCB题号910111213141516答案6DB0D二、非选择题:本大题共5小题,共60分。17.(11分,除说明外,每空2分)(1)减小CO2吸收减少,暗反应减弱(2)减缓高浓度C02处理能减缓气孔开度减小程度,有利于吸收C02,提高光合速率(3)不同意(1分),水淹胁迫时水稻与空气接触减少,细胞中O2含量下降,有氧呼吸速率减小:CO2浓度增大,有氧呼吸也受抑制。(2分)18.(11分,除说明外,每空2分)(1)低(1分)降低(1分)降低(1分)(2)先用等量蒸馏水喂养4周;再用不含二甲双胍的羟甲基纤维素钠溶液灌胃与A组对比确定造模是否成功,与C组对比,确定二甲双胍对甲状腺激素的影响(3)促进甲状腺激素的合成(4)测定三组大鼠血清中T$H的含量19.(12分,除说明外,每空2分)(1)蚕、鱼类、人类(1点1分,答出2点及以上得2分)桑树、粮食作物等植物通过光合作用固定的太阳能(2)生态位垂直(1分)阳光(光能)(1分)(3)底泥中含有丰富的有机物,被微生物分解后可产生无机盐和C02,供给桑树和粮食作物利用(4)获得了更多产品,增加收益:减少化肥等使用,减少了环境污染20.(14分,除说明外,每空2分)(1)常(1分)HhZBZ必(1分)(2)42/3(3)不同意。(1分)纯合黑羽(♀)和纯合白羽()Bb基因型的分别为ZW和ZZ,得到的F1中雌性鹌鹑全为白羽,与正交实验的F1表型不一致。(2分)(4)实验思路:选亲本中黑羽雄性个体与该白羽个体杂交,观察并统计后代表型及比例。(2分)
9.细菌病毒(即噬菌体)可通过两种途径(如图4所示,原噬菌体状态和裂解状态)复制,这两种生长状态是由病毒编码的cI和Co两个转录调控蛋白控制的。在原噬菌体状态,cI表达;在裂解状态,C和表达。当细菌用紫外线轻度照射后,cI蛋白被降解。下列相关说法,正确的是、cI蛋白质Cro蛋白质Cro基因c基因Cro基因原噬菌没有Cro基因转录体状态没有c基因转录裂解状态图4A.两种基因表达时,均需要大量的脱氧核苷酸作为原材料B.由图可推测,除了调控其他基因表达外,c是Cro基因转录的阻抑物,Cro蛋白是cI基因的激活物C.紫外线轻度照射后,Co基因转录会增强D.噬菌体在细菌中的繁殖过程可表示为:吸附→注入→组装→合成→释放10.研究发现,有高血压病史或过度劳累等原因可导致内皮素(ET,21肽激素)增高,而ET能促进黑色素细胞的增殖,并产生大量黑色素;ET拮抗剂可以竞争黑色素细胞膜上的ET受体,使ET作用减弱。下列相关叙述正确的是A.黑色素细胞内黑色素积累增多,可能诱导癌变B.可通过口服或涂抹ET于面部,实现皮肤美白C.ET拮抗剂与ET受体结合后,能抑制黑色素细胞产生黑色素D.控制合成ET和ET受体的基因只在黑色素细胞中选择性表达11.橙子辅导发现年轻胃癌患者有更多的CDH1基因突变;而老年胃癌患者往往有更多的TP53、NOTCH1和FAT4基因突变。下列叙述正确的是A.无致癌因子时,癌变便不会发生B.基因突变只发生在生物个体发育的成熟时期C.基因TP53可能属于原癌基因或抑癌基因D.癌细胞膜上的脂蛋白减少,导致细胞易分散和转移12.研究人员用三种不同颜色的荧光物质分别给基因型为eGg的雄牛基因e、G、g进行标记,再观察其分裂过程,下列叙述正确的是A.在有丝分裂前期,细胞中有3种颜色4个荧光点B.在有丝分裂后期,细胞中有3种颜色4个荧光点C.一个初级精母细胞中可能有3种颜色4个荧光点D.一个次级精母细胞中可能有3种颜色4个荧光点生物·第3页(共8页)
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公式项和为S,且2Sw2+有分行销汽车以某速度(未知妇速行驶机然B微候得特,经反应时间(未知)后采取制动措施,使车开始做一碱连直线运动,气术知时速度减为0,此时汽车与障碍物的距离1一1.5人司机发规腾码物到汽车止,对应的图像如图所示。。知汽车做匀诚速直线运动的笔以前进速度大小e=m。求:3答案解析网答案解析网1)汽车做减速直线运动。2司机的饭应时间答案解析与龙解析网小程片微信搜素关注答案解析网物块静A糙台上方向成0角、F知物块与台间的动摩2可0.取重@速度头小g条指利答案解析网暖点间的距管案解析网Mmg
回lk9s@①80062%☐112:36解得=3m/s对A、B第一次碰撞的过程,根据动量守恒定律有mw=mu+2m2(1分)对该过程,根据机械能守恒定律有4/5【高三物理·参考答案第3页(共5页)】2005C·HEN·名mw=m2+号×m2(1分)解得=1m/s(1分)2=4m/s。(1分)(3)A、B第一次碰撞后,A向右做匀加速直线运动,C向右做匀减速直线运动,假设B滑到C右端前,A、C已达到共同速度,设该共同速度的大小为vc,根据动量守恒定律有U十m6=2muc(1分)解得c=2m/s设从A、B第一次碰撞到A、C达到共同速度的时间为1,根据动量定理有ngb=muc-m(1分)解得t=0.2s在该段时间1内,A、B、C向右运动的距离分别为x1=边士些h=0,3m,m=4=0.8m,a=+Y1=0.5m1分)22因为x一x©

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