组成细胞的元素和化合物 水和无机物训练题

组成细胞的元素和化合物 水和无机物训练题 班级: 姓名: 1((2013年高考海南卷)关于生物体内有机化合物所含元素的叙述，错误的是( ) A(叶绿素含有镁元素 B(血红蛋白含有铁元素 C(脱氧核糖含有磷元素 D(胰岛素含有碳元素 2((2013年高考江苏卷)下列关于生物体与水分的关系，叙述正确的是( ) A(贮藏中的种子不含水分，以保持休眠状态 B(水从根系向地上部分的运输与细胞壁无关 C(适应高渗环境的动物可排出体内多余的盐 D(缺水时，动物体的正反馈调节能促使机体减少水的散失 3((2011年高考上海卷)生长在含盐量高、干旱土壤中的盐生植物，通过在液泡中贮存 ，，大量的Na而促进细胞吸收水分，该现象说明液泡内Na参与( ) A(调节渗透压 B(组成体内化合物 C(维持正常pH D(提供能量 4((2012年高考安徽卷)某同学以新鲜洋葱鳞片叶内表皮为材料，经不同处理和染色剂 染色，用高倍显微镜观察。下列描述正确的是( ) A(经吡罗红甲基绿染色，可观察到红色的细胞核 B(经吡罗红甲基绿染色，可观察到绿色的细胞质 C(经健那绿染色，可观察到蓝绿色颗粒状的线粒体 D(经苏丹?染色，可观察到橘黄色颗粒状的蛋白质 5(一种植物和一种哺乳动物体内细胞的某些元素含量(占细胞干重的质量分数:%)如表 所示，下列有关叙述正确的是( ) 元素 C H O N P Ca S 植物 43.57 6.24 44.43 1.46 0.20 0.23 0.17 动物 55.99 7.46 14.62 9.33 3.11 4.67 0.78 A.C的含量说明有机物是干物质的主要成分 B(这两种生物体内所含的化学元素的种类差异很大 C(N、S含量说明动物组织含蛋白质较多，若该动物血钙过高则会发生肌肉抽搐 D(经测定该植物某有机物含C、H、O、N、S，此化合物可能携带氨基酸进入核糖体 6(组成生物的化学元素在生物体中起重要作用，下列关于几种元素与光合作用关系的 叙述中，正确的是( ) of door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of door and window frames would preferably be conducted upon completion of the construction of the main structure, door and window door installation should be at the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500mm elevations in advance, as door and window level control line. Exterior door and window installation will pop up on the walls a uniform elevation control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doors and Windows. Check the embedded specification of the number, location and planting method, aluminum window has no damage, deformation, repair, correction or replacement should be carried out promptly. Check for corrosion materials, sealing materials and cleaning materials with the design requirements and specifications. 2, installation procedure 1) door and window frame installation according to the location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and window frames size is greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window frames edge 15mm position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete formwork for concrete pouring. After the base strength. Door frame mounting hole place to wedge A(C是组成糖类的基本元素，在光合作用中C元素从CO先后经C、C形成(CHO) 2352B(N是叶绿素的组成元素之一，没有N，植物就不能进行光合作用 C(O是构成有机物的基本元素之一，光合作用制造的有机物中的氧来自于水 D(P是构成ATP的必需元素，光合作用中光反应和暗反应过程中均有ATP的合成 7((2014年山东淄博模拟)下图横坐标表示细胞中的几种化合物，纵坐标表示每种成分 在细胞鲜重中的含量，以下按图中????顺序排列的是( ) A(水、蛋白质、糖类、脂质 B(蛋白质、糖类、脂质、水 C(水、蛋白质、脂质、糖类 D(蛋白质、水、脂质、糖类 8(下列关于水的叙述中，正确的是( ) A(若不同种生物细胞中的自由水和结合水的比值相同，则它们的代谢强度相同 B(在有氧呼吸过程中，线粒体既消耗水又产生水 C(酵母菌的有氧呼吸和无氧呼吸均不产生水 D(线粒体、核糖体、中心体在其生理活动中都可以产生水 9(关于水在人体内的生理作用，下列叙述正确的是( ) A(膝跳反射的兴奋传导离不开水 B(基因表达中的翻译过程没有水生成 C(用于留种的晒干种子中不含自由水 D(结合水的比例越高，人体细胞的代谢活动越旺盛 10(结合下列曲线，分析有关无机物在生物体内含量的说法，错误的是( ) A(曲线?可表示人一生中体内自由水与结合水的比值随年龄变化的曲线 B(曲线?可以表示细胞呼吸速率随自由水与结合水比值的变化 according to the location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and ng materials with the design requirements and specifications. 2, installation procedure 1) door and window frame installationcleanition, repair, correction or replacement should be carried out promptly. Check for corrosion materials, sealing materials and Windows. Check the embedded specification of the number, location and planting method, aluminum window has no damage, deformas and uniform elevation control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doorevations in advance, as door and window level control line. Exterior door and window installation will pop up on the walls a mm elshould be at the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500 es would preferably be conducted upon completion of the construction of the main structure, door and window door installationr and window framof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of dook for concrete pouring. After the base strength. Door frame mounting hole place to wedgeormworframes edge 15mm position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete fow frames size is greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window wind2 编制:苗秀丽 审核:高二生物组 2015年3月 C(曲线?可以表示一粒新鲜的玉米种子在烘箱中被烘干的过程中，其内无机盐的相对含量变化 D(曲线?可以表示人从幼年到成年，体内水相对含量的变化 11((2014年临沂模拟)下列有关无机盐的叙述，正确的是( ) A(人体的血浆渗透压由无机盐维持 B(细胞中的无机盐主要以化合物形式存在 C(细胞进行脂肪、DNA、RNA等物质合成时都需要磷酸盐 D(K是构成细胞的大量元素，对维持细胞正常功能有重要作用 12(下列对无机盐的相关叙述，正确的是( ) A(豆芽是大豆浸水萌发的产物，根据这一现象说明豆芽的生长不需要无机盐 B(哺乳动物血液中钙离子含量超标时，会经常出现抽搐现象 C(硼元素能促进花粉的萌发和花粉管的伸长，故甘蔗地缺硼会严重影响其产量 D(细胞中的无机盐有两种存在形式，绝大多数以离子形式存在，部分以化合态存在 二、非选择题 13((2014年浙江萧山测试)互花米草是一种滩涂草本盐沼植物，对水淹的生态环境有很强的适应能力，曾被称为“保滩护堤、促淤造陆的最佳植物”。互花米草在引入我国后，出现生长蔓延、破坏生态环境的严重问题。为了探究互花米草对水淹生态环境的适应机制，进一步为调控的研究提供依据，某科研小组以“探究互花米草体内自由水与结合水的比值和潮汐水淹时间的关系”为课题，设计了实验。 (1)写出科研小组的实验思路(培养的具体过程和测量方法不作要求): (2)下图曲线甲是科研小组对互花米草处理50天后的结果，请回答: ?自由水与结合水的比值和植物的代谢能力及抗逆性有着密切的关系。根据实验结果可知，互花米草在水淹时间为________h/d的环境下，抗逆性最强;在水淹时间为________h/d的环境下，代谢活动最旺盛。 ?在实验结束时水淹时间为24 h/d的实验组互花米草长势最差，甚至死亡，导致该现象发生的主要原因是 _________________________________________________________________ _________________________________________________________________ cretem position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete formwork for conis greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window frames edge 15m s sizethe location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and window frameith the design requirements and specifications. 2, installation procedure 1) door and window frame installation according to als worrection or replacement should be carried out promptly. Check for corrosion materials, sealing materials and cleaning materithe embedded specification of the number, location and planting method, aluminum window has no damage, deformation, repair, c Check ion control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doors and Windows.vance, as door and window level control line. Exterior door and window installation will pop up on the walls a uniform elevatin ad the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500mm elevationsrably be conducted upon completion of the construction of the main structure, door and window door installation should be at r and window frames would prefeof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doopouring. After the base strength. Door frame mounting hole place to wedge3 ，?该实验小组同时测定K吸收速率与水淹时间的关系，如曲线乙所示。由图可知，互 ，花米草吸收K的方式是_______ _ 。 14(由于无土栽培作物易于管理，优质高产，因此该项技术广泛应用于现代农业，同时无土栽培也是确定一种元素是否是植物所必需的元素的最常用的方法。 (1)无土栽培所用营养液中的无机盐在植物体内的作用是________和________。植物因种类和生长发育阶段不同对无机盐的需求也不同，所以应视具体情况调整________________，供作物_______ _ 性吸收无机盐离子。 (2)为了验证镍为植物生活所必需的元素，请完成下列问题。 ?材料用具:完全营养液甲、缺镍的营养液乙(其他元素均具备)、适当的容器和固定材料、长势相似的玉米幼苗、含镍的无机盐等。 ?方法步骤(完善表格): 项目 A组(20株长势相似的玉米幼苗) B组(a.____) 处理 放入缺镍的营养液乙中培养 b.______ 培养 相同且适宜的条件下培养数天 观察 c.______________________ 预期结果 d.____________ e.______ ?从科学研究的严谨角度出发，为进一步证实A、B两组玉米生长状况的差异是由于镍元素供应不同引起的，还应增加的实验步骤及结果是: ________________________________________________________________________ ________________________________________________________________________。 k for concrete pouring. After the base strength. Door frame mounting hole place to wedgeormworframes edge 15mm position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete fow frames size is greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window wind according to the location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and ng materials with the design requirements and specifications. 2, installation procedure 1) door and window frame installationcleanition, repair, correction or replacement should be carried out promptly. Check for corrosion materials, sealing materials and Windows. Check the embedded specification of the number, location and planting method, aluminum window has no damage, deformas and uniform elevation control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doorevations in advance, as door and window level control line. Exterior door and window installation will pop up on the walls a mm elshould be at the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500 es would preferably be conducted upon completion of the construction of the main structure, door and window door installationr and window framof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doo4 编制:苗秀丽 审核:高二生物组 2015年3月 答案 1 C 2 C 3 A 4(C 5( A 6( B 7( D 8( B 9( A 10( C 11( D 12( D 13( 解析:(1)实验设计要遵循单一变量原则、对照原则～无关变量相同且适宜～本实验自变量是水淹时间～因变量是自由水与结合水的比值。根据50天后的实验结果图示可知～实验应分为5组～每天分别给予0、3、6、12、24小时的水淹处理。(2)根据实验结果可知～互花米草在水淹时间为3 h/d的环境下～自由水与结合水的比值最小～抗逆性最强,在水淹时间为0 h/d的环境下～自由，水与结合水的比值最大～代谢活动最旺盛。互花米草对K的吸收是主动运输过程～需消耗细胞内的能量。 答案:(1)将互花米草平均分为5份，每天分别给予0、3、6、12、24小时的水淹处理。50天后，测量每组互花米草整株的自由水和结合水的含量。 (2)?3 0 ?长期水淹导致互花米草进行无氧呼吸造成供能不足，酒精中毒 ?主动运输 14( 解析:(1)解答本题需要考虑清楚组成细胞的化合物中无机盐的作用以及细胞对无机盐的吸收具有选择性。题中无土栽培所用营养液中无机盐的作用是构成细胞的组成成分和调节植物的生命活动。从题中给出的信息“植物因种类和生长发育阶段不同对无机盐的需求也不同”～可以看出植物对不同无机盐的吸收根据需要具有选择性～所以应根据不同作物、不同生长发育时期对无机盐离子不同需求的具体情况～在配制营养液时随时调整无机盐的组成和比例～以便供作物选择性地吸收无机盐离子。 (2)实验设计要遵循对照原则和单一变量原则。A组玉米幼苗放在缺镍的营养液乙中培养～则B组幼苗应放在等量的完全营养液甲中培养。本题要确定不能生长的一组确实是因缺镍导致的～故要进行二次实验～应在缺镍的营养液乙中加入一定量的含镍的无机盐～一段时间后看玉米幼苗能否恢复正常生长。 答案:(1)细胞的组成成分 调节植物的生命活动 营养液中无机盐的组成和比例(营养液的配方) 选择 (2)?a.20株同A组长势相似的玉米幼苗 b(放入等量的完全营养液甲中培养 c(玉米幼苗的生长状况 d(玉米幼苗生长不正常 e(玉米幼苗生长正常 ?在缺镍的营养液乙中加入一定量的镍元素(或含镍的无机盐)，一段时间后玉米幼苗恢复正常生长(或症状消失)，则镍是植物生活所必需的元素 答案 1 C 2 C 3 A 4(C 5( A 6( B 7( D 8( B 9( A 10( C 11( D 12( D 13( 解析:(1)实验设计要遵循单一变量原则、对照原则～无关变量相同且适宜～本实验自变量是水淹时间～因变量是自由水与结合水的比值。根据50 is greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window frames edge 15m s sizethe location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and window frameith the design requirements and specifications. 2, installation procedure 1) door and window frame installation according to als worrection or replacement should be carried out promptly. Check for corrosion materials, sealing materials and cleaning materithe embedded specification of the number, location and planting method, aluminum window has no damage, deformation, repair, c Check ion control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doors and Windows.vance, as door and window level control line. Exterior door and window installation will pop up on the walls a uniform elevatin ad the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500mm elevationsrably be conducted upon completion of the construction of the main structure, door and window door installation should be at r and window frames would prefeof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doopouring. After the base strength. Door frame mounting hole place to wedge cretem position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete formwork for con5 天后的实验结果图示可知～实验应分为5组～每天分别给予0、3、6、12、24小时的水淹处理。(2)根据实验结果可知～互花米草在水淹时间为3 h/d的环境下～自由水与结合水的比值最小～抗逆性最强,在水淹时间为0 h/d的环境下～自由，水与结合水的比值最大～代谢活动最旺盛。互花米草对K的吸收是主动运输过程～需消耗细胞内的能量。 答案:(1)将互花米草平均分为5份，每天分别给予0、3、6、12、24小时的水淹处理。50天后，测量每组互花米草整株的自由水和结合水的含量。 (2)?3 0 ?长期水淹导致互花米草进行无氧呼吸造成供能不足，酒精中毒 ?主动运输 14( 解析:(1)解答本题需要考虑清楚组成细胞的化合物中无机盐的作用以及细胞对无机盐的吸收具有选择性。题中无土栽培所用营养液中无机盐的作用是构成细胞的组成成分和调节植物的生命活动。从题中给出的信息“植物因种类和生长发育阶段不同对无机盐的需求也不同”～可以看出植物对不同无机盐的吸收根据需要具有选择性～所以应根据不同作物、不同生长发育时期对无机盐离子不同需求的具体情况～在配制营养液时随时调整无机盐的组成和比例～以便供作物选择性地吸收无机盐离子。 (2)实验设计要遵循对照原则和单一变量原则。A组玉米幼苗放在缺镍的营养液乙中培养～则B组幼苗应放在等量的完全营养液甲中培养。本题要确定不能生长的一组确实是因缺镍导致的～故要进行二次实验～应在缺镍的营养液乙中加入一定量的含镍的无机盐～一段时间后看玉米幼苗能否恢复正常生长。 答案:(1)细胞的组成成分 调节植物的生命活动 营养液中无机盐的组成和比例(营养液的配方) 选择 (2)?a.20株同A组长势相似的玉米幼苗 b(放入等量的完全营养液甲中培养 (玉米幼苗生长不正常 e(玉米幼苗生长正常 ?c(玉米幼苗的生长状况 d 在缺镍的营养液乙中加入一定量的镍元素(或含镍的无机盐)，一段时间后玉米幼苗恢复正常生长(或症状消失)，则镍是植物生活所必需的元素 答案 1 C 2 C 3 A 4(C 5( A 6( B 7( D 8( B 9( A 10( C 11( D 12( D 13( 解析:(1)实验设计要遵循单一变量原则、对照原则～无关变量相同且适宜～本实验自变量是水淹时间～因变量是自由水与结合水的比值。根据50天后的实验结果图示可知～实验应分为5组～每天分别给予0、3、6、12、24小时的水淹处理。(2)根据实验结果可知～互花米草在水淹时间为3 h/d的环境下～自由水与结合水的比值最小～抗逆性最强,在水淹时间为0 h/d的环境下～自由，水与结合水的比值最大～代谢活动最旺盛。互花米草对K的吸收是主动运输过程～需消耗细胞内的能量。 k for concrete pouring. After the base strength. Door frame mounting hole place to wedgeormworframes edge 15mm position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete fow frames size is greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window wind according to the location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and ng materials with the design requirements and specifications. 2, installation procedure 1) door and window frame installationcleanition, repair, correction or replacement should be carried out promptly. Check for corrosion materials, sealing materials and Windows. Check the embedded specification of the number, location and planting method, aluminum window has no damage, deformas and uniform elevation control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doorevations in advance, as door and window level control line. Exterior door and window installation will pop up on the walls a mm elshould be at the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500 es would preferably be conducted upon completion of the construction of the main structure, door and window door installationr and window framof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doo6 编制:苗秀丽 审核:高二生物组 2015年3月 答案:(1)将互花米草平均分为5份，每天分别给予0、3、6、12、24小时的水淹处理。50天后，测量每组互花米草整株的自由水和结合水的含量。 (2)?3 0 ?长期水淹导致互花米草进行无氧呼吸造成供能不足，酒精中毒 ?主动运输 14( 解析:(1)解答本题需要考虑清楚组成细胞的化合物中无机盐的作用以及细胞对无机盐的吸收具有选择性。题中无土栽培所用营养液中无机盐的作用是构成细胞的组成成分和调节植物的生命活动。从题中给出的信息“植物因种类和生长发育阶段不同对无机盐的需求也不同”～可以看出植物对不同无机盐的吸收根据需要具有选择性～所以应根据不同作物、不同生长发育时期对无机盐离子不同需求的具体情况～在配制营养液时随时调整无机盐的组成和比例～以便供作物选择性地吸收无机盐离子。 (2)实验设计要遵循对照原则和单一变量原则。A组玉米幼苗放在缺镍的营养液乙中培养～则B组幼苗应放在等量的完全营养液甲中培养。本题要确定不能生长的一组确实是因缺镍导致的～故要进行二次实验～应在缺镍的营养液乙中加入一定量的含镍的无机盐～一段时间后看玉米幼苗能否恢复正常生长。 答案:(1)细胞的组成成分 调节植物的生命活动 营养液中无机盐的组成和比例(营养液的配方) 选择 (2)?a.20株同A组长势相似的玉米幼苗 b(放入等量的完全营养液甲中培养 c(玉米幼苗的生长状况 d(玉米幼苗生长不正常 e(玉米幼苗生长正常 ?在缺镍的营养液乙中加入一定量的镍元素(或含镍的无机盐)，一段时间后玉米幼苗恢复正常生长(或症状消失)，则镍是植物生活所必需的元素 答案 1 C 2 C 3 A 4(C 5( A 6( B 7( D 8( B 9( A 10( C 11( D 12( D 13( 解析:(1)实验设计要遵循单一变量原则、对照原则～无关变量相同且适宜～本实验自变量是水淹时间～因变量是自由水与结合水的比值。根据50 、3、6、12、24天后的实验结果图示可知～实验应分为5组～每天分别给予0小时的水淹处理。(2)根据实验结果可知～互花米草在水淹时间为3 h/d的环境下～自由水与结合水的比值最小～抗逆性最强,在水淹时间为0 h/d的环境下～自由，水与结合水的比值最大～代谢活动最旺盛。互花米草对K的吸收是主动运输过程～需消耗细胞内的能量。 答案:(1)将互花米草平均分为5份，每天分别给予0、3、6、12、24小时的水淹处理。50天后，测量每组互花米草整株的自由水和结合水的含量。 (2)?3 0 ?长期水淹导致互花米草进行无氧呼吸造成供能不足，酒精中毒 ?主动运输 s sizethe location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and window frameith the design requirements and specifications. 2, installation procedure 1) door and window frame installation according to als worrection or replacement should be carried out promptly. Check for corrosion materials, sealing materials and cleaning materithe embedded specification of the number, location and planting method, aluminum window has no damage, deformation, repair, c Check ion control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doors and Windows.vance, as door and window level control line. Exterior door and window installation will pop up on the walls a uniform elevatin ad the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500mm elevationsrably be conducted upon completion of the construction of the main structure, door and window door installation should be at r and window frames would prefeof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doopouring. After the base strength. Door frame mounting hole place to wedge cretem position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete formwork for conis greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window frames edge 15m7 14( 解析:(1)解答本题需要考虑清楚组成细胞的化合物中无机盐的作用以及细胞对无机盐的吸收具有选择性。题中无土栽培所用营养液中无机盐的作用是构成细胞的组成成分和调节植物的生命活动。从题中给出的信息“植物因种类和生长发育阶段不同对无机盐的需求也不同”～可以看出植物对不同无机盐的吸收根据需要具有选择性～所以应根据不同作物、不同生长发育时期对无机盐离子不同需求的具体情况～在配制营养液时随时调整无机盐的组成和比例～以便供作物选择性地吸收无机盐离子。 (2)实验设计要遵循对照原则和单一变量原则。A组玉米幼苗放在缺镍的营养液乙中培养～则B组幼苗应放在等量的完全营养液甲中培养。本题要确定不能生长的一组确实是因缺镍导致的～故要进行二次实验～应在缺镍的营养液乙中加入一定量的含镍的无机盐～一段时间后看玉米幼苗能否恢复正常生长。 答案:(1)细胞的组成成分 调节植物的生命活动 营养液中无机盐的组成和比例(营养液的配方) 选择 (2)?a.20株同A组长势相似的玉米幼苗 b(放入等量的完全营养液甲中培养 c(玉米幼苗的生长状况 d(玉米幼苗生长不正常 e(玉米幼苗生长正常 ?在缺镍的营养液乙中加入一定量的镍元素(或含镍的无机盐)，一段时间后玉米幼苗恢复正常生长(或症状消失)，则镍是植物生活所必需的元素 课时考点训练 题组一、组成细胞的元素和无机物 1((2013年高考海南卷)关于生物体内有机化合物所含元素的叙述，错误的是( ) A(叶绿素含有镁元素 B(血红蛋白含有铁元素 C(脱氧核糖含有磷元素 D(胰岛素含有碳元素 解析:脱氧核糖为五碳糖～元素组成为C、H、O～故C项错误。 答案:C 2((2013年高考重庆卷)下列有关细胞物质组成的叙述，正确的是( ) A(在人体活细胞中氢原子的数目最多 B(DNA和RNA分子的碱基组成相同 C(多糖在细胞中不与其他分子相结合 D(蛋白质区别于脂质的特有元素是氮 解析:本题主要考查细胞物质组成的相关知识。人体活细胞中含量最多的化合物是HO～2因此氢原子的数目最多,DNA和RNA分子的碱基组成不完全相同～DNA分子的碱基组成为A、T、C、G～而RNA分子的碱基组成为A、U、C、G,多糖在细胞中可与蛋白质结合形成糖蛋白,蛋白质区别于脂质的特有元素是硫。故A项正确。 答案:A 3((2013年高考江苏卷)下列关于生物体与水分的关系，叙述正确的是( ) A(贮藏中的种子不含水分，以保持休眠状态 B(水从根系向地上部分的运输与细胞壁无关 C(适应高渗环境的动物可排出体内多余的盐 D(缺水时，动物体的正反馈调节能促使机体减少水的散失 解析:本题从不同的角度考查了生物体内水的有关知识。贮藏的种子中自由水含量低～ k for concrete pouring. After the base strength. Door frame mounting hole place to wedgeormworframes edge 15mm position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete fow frames size is greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window wind according to the location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and ng materials with the design requirements and specifications. 2, installation procedure 1) door and window frame installationcleanition, repair, correction or replacement should be carried out promptly. Check for corrosion materials, sealing materials and Windows. Check the embedded specification of the number, location and planting method, aluminum window has no damage, deformas and uniform elevation control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doorevations in advance, as door and window level control line. Exterior door and window installation will pop up on the walls a mm elshould be at the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500 es would preferably be conducted upon completion of the construction of the main structure, door and window door installationr and window framof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doo8 编制:苗秀丽 审核:高二生物组 2015年3月 以保持休眠状态,水沿着由死亡细胞的细胞壁组成的导管从植物的根系向地上部分运输,缺水时～动物体的负反馈调节能促使机体减少水的散失,适应高渗环境的动物体液渗透压相对较高～但仍要排出体内多余的盐～以维持内环境稳态。 答案:C 4((2011年高考上海卷)生长在含盐量高、干旱土壤中的盐生植物，通过在液泡中贮存，，大量的Na而促进细胞吸收水分，该现象说明液泡内Na参与( ) A(调节渗透压 B(组成体内化合物 C(维持正常pH D(提供能量 解析:液泡内的各种矿质离子能够参与调节细胞液的渗透压～盐生植物液泡中贮存大量，的Na～增大了细胞液的渗透压～吸水能力增强。 答案:A 7((2012年高考安徽卷)某同学以新鲜洋葱鳞片叶内表皮为材料，经不同处理和染色剂染色，用高倍显微镜观察。下列描述正确的是( ) A(经吡罗红甲基绿染色，可观察到红色的细胞核 B(经吡罗红甲基绿染色，可观察到绿色的细胞质 C(经健那绿染色，可观察到蓝绿色颗粒状的线粒体 D(经苏丹?染色，可观察到橘黄色颗粒状的蛋白质 解析:本题主要考查细胞内DNA、RNA、蛋白质等物质及线粒体结构的鉴定。吡罗红甲基绿染色剂用于鉴定细胞中DNA和RNA的分布～甲基绿使DNA呈现绿色～吡罗红使RNA呈现红色。DNA主要分布在细胞核中～RNA主要分布在细胞质中～故A、B不正确。线粒体经健那绿染色呈蓝绿色～故C正确。经苏丹?染色～可观察到橘黄色颗粒状的脂肪～而不是蛋白质～故D不正确。 答案:C 课时高效训练 一、选择题 1(一种植物和一种哺乳动物体内细胞的某些元素含量(占细胞干重的质量分数:%)如表所示，下列有关叙述正确的是( ) 元素 C H O N P Ca S 植物 43.57 6.24 44.43 1.46 0.20 0.23 0.17 动物 55.99 7.46 14.62 9.33 3.11 4.67 0.78 A.C的含量说明有机物是干物质的主要成分 B(这两种生物体内所含的化学元素的种类差异很大 C(N、S含量说明动物组织含蛋白质较多，若该动物血钙过高则会发生肌肉抽搐 D(经测定该植物某有机物含C、H、O、N、S，此化合物可能携带氨基酸进入核糖体 解析:有机物都以碳链为基本骨架～故C在有机物中所占比例很大～表中动植物体内C含量较高～说明干物质的主要成分是有机物～A正确。由表可知～这两种生物体内所含元素的种类相同～但元素的含量差异较大～B错误。动物组织中蛋白质含量较多～故N、S含量也较多～肌肉抽搐是血钙过低引起的～C错误。携带氨基酸进入核糖体的物质是tRNA～RNA中不含S但含P～D错误。 答案:A 2(组成生物的化学元素在生物体中起重要作用，下列关于几种元素与光合作用关系的叙述中，正确的是( ) A(C是组成糖类的基本元素，在光合作用中C元素从CO先后经C、C形成(CHO) 2352 B(N是叶绿素的组成元素之一，没有N，植物就不能进行光合作用 C(O是构成有机物的基本元素之一，光合作用制造的有机物中的氧来自于水 D(P是构成ATP的必需元素，光合作用中光反应和暗反应过程中均有ATP的合成 解析:光合作用中C元素的流经途径是CO?C?C，(CHO),N是叶绿素的组成元2352 素之一～而叶绿素是植物进行光合作用的必需条件,光合作用中HO中的氧全部形成氧气～2 有机物中的氧则来自CO,光合作用光反应产生ATP～暗反应消耗ATP。 2 答案:B 3((2014年山东淄博模拟)下图横坐标表示细胞中的几种化合物，纵坐标表示每种成分 the embedded specification of the number, location and planting method, aluminum window has no damage, deformation, repair, c Check ion control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doors and Windows.vance, as door and window level control line. Exterior door and window installation will pop up on the walls a uniform elevatin ad the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500mm elevationsrably be conducted upon completion of the construction of the main structure, door and window door installation should be at r and window frames would prefeof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doopouring. After the base strength. Door frame mounting hole place to wedge cretem position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete formwork for conis greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window frames edge 15m s sizethe location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and window frameith the design requirements and specifications. 2, installation procedure 1) door and window frame installation according to als worrection or replacement should be carried out promptly. Check for corrosion materials, sealing materials and cleaning materi9 在细胞鲜重中的含量，以下按图中????顺序排列的是( ) A(水、蛋白质、糖类、脂质 B(蛋白质、糖类、脂质、水 C(水、蛋白质、脂质、糖类 D(蛋白质、水、脂质、糖类 解析:在细胞鲜重中水的含量最多～其次是蛋白质～脂质～糖类和核酸含量较少。 答案:D 4(下列关于水的叙述中，正确的是( ) A(若不同种生物细胞中的自由水和结合水的比值相同，则它们的代谢强度相同 B(在有氧呼吸过程中，线粒体既消耗水又产生水 C(酵母菌的有氧呼吸和无氧呼吸均不产生水 D(线粒体、核糖体、中心体在其生理活动中都可以产生水 解析:细胞代谢的强度不仅与自由水和结合水的比值有关～还与温度等因素有关～A错误,有氧呼吸第二阶段丙酮酸在水的参与下彻底氧化分解～第三阶段[H]与氧结合生成水～B正确,有氧呼吸过程有水产生～C错误,中心体的功能是形成纺锤体～无水产生～D错误。 答案:B 5(关于水在人体内的生理作用，下列叙述正确的是( ) A(膝跳反射的兴奋传导离不开水 B(基因表达中的翻译过程没有水生成 C(用于留种的晒干种子中不含自由水 D(结合水的比例越高，人体细胞的代谢活动越旺盛 解析:任何生命活动都离不开水～膝跳反射是一种反射活动～这个过程离不开水的参与。基因表达中的翻译过程也就是蛋白质的合成过程～即脱水缩合过程～有水的生成。用于留种的晒干种子中虽然失去了大部分的自由水～但是仍然含有少量的自由水。自由水的比例越高～人体细胞的代谢活动越旺盛。 答案:A 6(结合下列曲线，分析有关无机物在生物体内含量的说法，错误的是( ) A(曲线?可表示人一生中体内自由水与结合水的比值随年龄变化的曲线 B(曲线?可以表示细胞呼吸速率随自由水与结合水比值的变化 C(曲线?可以表示一粒新鲜的玉米种子在烘箱中被烘干的过程中，其内无机盐的相对含量变化 D(曲线?可以表示人从幼年到成年，体内水相对含量的变化 解析:一般来说～细胞内的结合水含量是相对稳定的～但自由水的含量变化较大,衰老的细胞中自由水含量减少～细胞内自由水与结合水的比值也将减小,自由水是细胞内的良好溶剂～细胞内许多化学反应都需要有水的参与～所以细胞内自由水比值升高后～代谢增强～细胞呼吸速率会加强,玉米种子被烘干的过程中所含水分越来越少～其内的无机盐相对含量逐渐增加～最后达到一恒定值,人体衰老的特征之一就是水的含量减少～幼儿体内水的含量远远高于成年人体内水的含量。答案:C 7((2014年临沂模拟)下列有关无机盐的叙述，正确的是( ) A(人体的血浆渗透压由无机盐维持 B(细胞中的无机盐主要以化合物形式存在 C(细胞进行脂肪、DNA、RNA等物质合成时都需要磷酸盐 D(K是构成细胞的大量元素，对维持细胞正常功能有重要作用 frames edge 15mm position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete fow frames size is greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window wind according to the location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and ng materials with the design requirements and specifications. 2, installation procedure 1) door and window frame installationcleanition, repair, correction or replacement should be carried out promptly. Check for corrosion materials, sealing materials and Windows. Check the embedded specification of the number, location and planting method, aluminum window has no damage, deformas and uniform elevation control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doorevations in advance, as door and window level control line. Exterior door and window installation will pop up on the walls a mm elshould be at the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500 es would preferably be conducted upon completion of the construction of the main structure, door and window door installationr and window framof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of dook for concrete pouring. After the base strength. Door frame mounting hole place to wedgeormwor10 编制:苗秀丽 审核:高二生物组 2015年3月 解析:人体血浆渗透压的大小主要与无机盐、蛋白质含量有关～A错误,细胞中的无机盐主要以离子形式存在～B错误,脂肪的构成元素只含有C、H、O～不含P～其合成时不需要磷酸盐～C错误。 答案:D 8(下列对无机盐的相关叙述，正确的是( ) A(豆芽是大豆浸水萌发的产物，根据这一现象说明豆芽的生长不需要无机盐 B(哺乳动物血液中钙离子含量超标时，会经常出现抽搐现象 C(硼元素能促进花粉的萌发和花粉管的伸长，故甘蔗地缺硼会严重影响其产量 D(细胞中的无机盐有两种存在形式，绝大多数以离子形式存在，部分以化合态存在 解析:豆芽的生长离不开无机盐～没有N和P等元素～DNA就不能合成～A错误,哺乳动物抽搐的原因是血钙含量低～B错误,甘蔗收获的是茎～故缺硼对甘蔗的产量影响不大～C错误。答案:D orrection or replacement should be carried out promptly. Check for corrosion materials, sealing materials and cleaning materithe embedded specification of the number, location and planting method, aluminum window has no damage, deformation, repair, c Check ion control line and vertical line from bottom to top, elevation and vertical deviation is controlled with doors and Windows.vance, as door and window level control line. Exterior door and window installation will pop up on the walls a uniform elevatin ad the end of the indoor and outdoor decoration, so as to avoid civil installation damage to each other. Popup +500mm elevationsrably be conducted upon completion of the construction of the main structure, door and window door installation should be at r and window frames would prefeof door and window boxes, the jambs set aside in advance to check geometry meets the design requirements. Installation of doopouring. After the base strength. Door frame mounting hole place to wedge cretem position and the surface is rough. If the thickness is greater than 30mm, should be of fine stone concrete formwork for conis greater than 15mm, first with cement mortar plastering the wall size on all sides from the door and window frames edge 15m s sizethe location of the doors and Windows installed wire, when doors and Windows opening to the edge of the door and window frameith the design requirements and specifications. 2, installation procedure 1) door and window frame installation according to als w11