【生物】河南省信阳高级中学2012-2013学年高一下学期6月月考试题

C项，F1产生配子时成对遗传因子分离，这是属于假说内容，“测交后代会出现两种性状比接近1:1”这是根据假说进行演绎推理，属于“演绎”过程，故C项正确。

A项、B项和D项，是根据实验结果后做出的假说过程，不属于演绎过程，故A项、B项和D 项错误。

综上所述，本题正确答案为C。

3.一杂合子Dd植株自交时，含有隐性基因的配子的花粉有25%的死亡率，则自交后代的基因型比例为（）A.1:1:1B.4:4:1C.4:7:3D.1:2:1分析:基因频率的计算：（1）在种群中一对等位基因的频率之和等于1，基因型频率之和也等于1；（2）一个等位基因的频率=该等位基因纯合子的频率+1/2杂合子的频率。

解析：Dd植株中雌配子有1/2D+1/2d，雄配子d有25%的致死，说明雄配子是1/2D+1/2×3/4d，也就是雄配子中有4/7D+3/7d。

所以后代各种基因型的频率：故后代各种基因型所占的比例为：DD：Dd：dd=4：7：3故选：C。

点评：本题考查基因频率的变化，要求考生识记基因频率和基因型频率的概念，掌握相关计算方法，能根据基因型频率计算相关基因频率，然后计算后代的基因型频率。

属于考纲识记和理解层次的考查。

4.豌豆的红花与白花是一对相对性状（分别由A、a基因控制），现有一批基因型为AA与Aa的红花豌豆，两者的数量之比是3：1，自然状态下其子代中基因型为AA、Aa、aa的数量之比为（）A. 13：2： 1B. 49：14： 1C. 1：2：1 D. 9：6：1分析：1、豌豆是自花传粉植物，而且是闭花受粉，所以AA的个体后代都是AA，Aa 的个体后代会出现性状分离。

2、基因分离定律的实质：在杂合子的细胞中，位于一对同源染色体上的等位基因，具有一定的独立性；生物体在进行减数分裂形成配子时，等位基因会随着同源染色体的分开而分离，分别进入到两个配子中，独立地随配子遗传给后代。

解答：根据题意分析可知：AA：Aa=3：1，AA占3/4，Aa占1/4，豌豆自然状态下通过自花传粉繁殖，子代AA占3/4+1/4*1/4=13/16，,Aa占1/4*1/2=2/16，aa占1/4*1/4=1/16。

河南省鹤壁市淇滨高级中学【最新】高一6月月考生物试题学校:___________姓名：___________班级：___________考号：___________一、单选题1．下列性状中，不属于相对性状的是（）A．高鼻梁与塌鼻梁B．卷发与直发C．五指与多指D．眼大与眼角上翘2．遗传因子组成为AaBb（这两对遗传因子独立遗传）的水稻自交，自交后代中两对遗传因子都是纯合子的个数占总数的A．4/16 B．2/16 C．6/16 D．8/163．减数分裂过程中，染色体数目减半是由于（）A．着丝点的分裂B．同源染色体的分离C．DNA的复制D．同源染色体的联会4．一条染色单体含有一个双链的DNA分子，那么，四分体时期的一条染色体含有( ) A．四个双链DNA分子B．二个双链DNA分子C．二个单链DNA分子D．一个双链DNA分子5．如图所示为某高等动物精子形成过程中不同时期的细胞，其正确的排列顺序应为A．④①③⑤②B．④③①⑤②C．①④③⑤②D．①④⑤③②6．如图表示某动物的一个正在分裂的细胞，请判断下列说法正确的是()A．该细胞是次级精母细胞或次级卵母细胞B．该细胞中1与2；3与4是同源染色体C．该细胞中有两对姐妹染色单体，1与2,3与4D．该细胞中，如果1是Y染色体，则2也是Y染色体，3与4是常染色体7．卵巢中的5个卵原细胞经过一次减数分裂，形成的结果A．20个卵细胞B．10个卵细胞和10个极体C．5个卵细胞和15个极体D．5个卵细胞和5个极体8．人类精子中含有的染色体是（）A．44+XY B．23+X或23+YC．22+X或22+Y D．44+XX9．下列关于基因与染色体关系的叙述，正确的是：A．染色体是基因的主要载体B．一条染色体上只有一个基因C．等位基因位于同一条染色体上D．非等位基因位于两条染色体上10．最早证实“基因在染色体上”的实验是（）A．孟德尔的豌豆杂交实验B．萨顿蝗虫细胞观察实验C．摩尔根的果蝇杂交实验D．现代分子生物学技术印证11．抗维生素D佝偻病是由位于X染色体的显性致病基因决定的一种遗传病，这种遗传病的特点之一是( )A．男患者与女患者结婚，其女儿正常B．男患者与正常女子结婚，其子女均正常C．女患者与正常男子结婚，儿子正常女儿患病D．患者的正常子女不携带该患者传递的致病基因12．已知一色觉正常女孩的祖母是红绿色盲，这个女孩携带红绿色盲基因的可能性是（）A．25% B．50%C．75% D．100%13．下图是一个血友病遗传系谱图，从图可以看出患者7的致病基因来自（）A．1 B．4 C．1和3 D．1和414．在噬菌体侵染细菌的实验中，下列对噬菌体衣壳蛋白质合成的描述，正确的是：（）A．氨基酸原料和酶来自细菌B．氨基酸原料和酶来自噬菌体C．氨基酸原料来自细菌，酶来自噬菌体D．氨基酸原料来自噬菌体，酶来自细菌15．格里菲思（F. Griffith）用肺炎双球菌在小鼠身上进行了著名的转化实验，此实验结果（）A．证明了DNA是遗传物质B．证明了RNA是遗传物质C．证明了蛋白质是遗传物质D．没有具体证明哪一种物质是遗传物质16．下列关于DNA复制的叙述，正确的是A．DNA分子在解旋酶的作用下，水解成脱氧核苷酸B．在全部解旋之后才开始进行DNA复制C．解旋后以一条母链为模板合成两条新的子链D．复制后，每个新DNA分子中含有一条母链和一条子链17．DNA分子的双链在复制时解旋，从氢键连接处分开的碱基是（）A．鸟嘌呤与尿嘧啶B．鸟嘌呤与胞嘧啶C．腺嘌呤与尿嘧啶D．鸟嘌呤与胸腺嘧啶18．DNA分子的一条链中（A+C）/（T+G）=0.4，则在其互补链中及整个DNA中该比例分别是（）A．0.4，0.4 B．2.5，0.4C．2.5，1.0 D．0.6，1.019．用15N标记的细菌DNA，在14N的培养基上连续分裂两次后，其后代中含有14N标记的新个体与含有15N标记的新个体的比值为（）A．3：1 B．2：1 C．1：1 D．7：120．如图所示转录、翻译过程中，所涉及的碱基种类和核苷酸种类分别有几种（）A．5、2 B．6、2 C．5、7 D．5、521．DNA分子模板链上的碱基序列携带的遗传信息最终翻译成的氨基酸如下表所示，则图所示的转运RNA所携带的氨基酸是(反密码子从携带氨基酸的一端开始读码)（）A．丙氨酸B．苏氨酸C．半胱氨酸D．赖氨酸22．一条多肽链中有氨基酸1000个，作为合成该多肽链模板的mRNA分子和用来转录成mRNA的DNA分子，至少含有碱基的数量分别是（）A．3000个和3000个B．1000个和2000个C．2000个和4000个D．3000个和6000个23．下列关于基因、性状以及二者关系的叙述，正确的是()A．基因在染色体上呈线性排列，基因的前端有起始密码子，末端有终止密码子B．基因能够通过复制实现遗传信息在亲代和子代之间的传递C．性状受基因的控制，基因发生突变，该基因控制的性状也必定改变D．通过控制酶的合成从而直接控制性状，是基因控制性状的途径之一24．现代生物进化理论认为，决定生物进化方向的是（）A．基因重组B．自然选择C．染色体变异D．基因突变25．下列变异中，不属于染色体结构变异的是：A．非同源染色体之间相互交换片段B．染色体缺失片段C．染色体中DNA的一个碱基发生改变D．染色体增加片段26．基因工程的正确操作步骤是（）①目的基因与运载体结合②将目的基因导入受体细胞③目的基因的检测与鉴定④提取目的基因A．③④②①B．②④①③C．③④①②D．④①②③27．在基因工程操作过程中，DNA连接酶的作用是（）A．将任意两个DNA分子连接起来B．将具有相同黏性末端的DNA分子连接，包括DNA分子的基本骨架和碱基对之间的氢键C．只连接具有相同黏性末端的DNA分子的基本骨架，即磷酸二酯键D．只连接具有相同黏性末端的DNA分子碱基对之间的氢键28．新物种形成最显著的标志是（）A．具有一定的形态结构B．改变了基因频率C．产生了地理隔离D．产生了生殖隔离29．在一个种群中随机抽出一定数量的个体，其中基因类型AA的个体占24％，基因型为Aa的个体占72％，基因型为aa的个体占4％，那么基因A和基因a的频率分别是（）A．24％、72％B．36％、64％C．57％、43％D．60％、40％二、综合题30．图1表示细胞分裂的不同时期染色体数与核DNA数比例的变化关系，图2表示同种动物细胞分裂不同时期的图像。

河南省信阳高级中学2020-2021学年下学期高一年级6月月考生物试卷一、选择题（每题2分，共50分）1．下列有关ATiRNA是一种小分子RNA，某miRNA能抑制W基因控制的蛋白质（W 蛋白）的合成。

某真核细胞内形成该miRNA及其发挥作用的过程示意图如下，下列叙述正确的是（）A．miRNA基因转录时，RNA聚合酶与该基因的起始密码子相结合B．W基因转录形成的mRNA在细胞核内加工后，进入细胞质用于翻译C．miRNA与W基因mRNA结合遵循碱基互补配对原则，即A与T、C与G配对D．miRNA抑制W蛋白的合成是通过双链结构的miRNA直接与W基因的mRNA结合所致18．家系图中属于常染色体隐性遗传、Y染色体遗传、X染色体显性遗传、X染色体隐性遗传的依次是（）A．③①②④B．②④①③C．①④②③D．①④③②19．直翅果蝇经紫外线照射后出现一种突变体，表现型为翻翅，已知直翅和翻翅这对相对性状完全显性，其控制基因位于常染色体上，且翻翅基因纯合致死（胚胎期）。

选择翻翅个体进行交配，F1中翻翅和直翅个体的数量比为2：1。

下列有关叙述错误的是（）A．紫外线照射使果蝇的直翅基因结构发生了改变B．果蝇的翻翅对直翅为显性C．F1中翻翅基因频率为1 3D．F1果蝇自由交配，F2中直翅个体所占比例为4 920．下列关于人类探索遗传奥秘历程中的科学方法及技术的叙述，错误的是（）A．孟德尔在研究豌豆杂交实验时，运用了假说—演绎法B．萨顿根据基因和染色体的行为存在平行关系，类比推理出基因位于染色体上C．格里菲思利用肺炎双球菌研究遗传物质时，运用了放射性同位素示踪技术D．沃森和克里克研究DNA分子结构时，运用了建构物理模型的方法21．甲、乙、丙个体的基因组成分别为、和，若三者都产生了Ab、aB的配子，则下列个体及产生上述配子的机理对应一致的是（）A．甲——同源染色体分离；乙——染色体变异；丙——基因重组B．甲——染色体变异；乙——交叉互换；丙——同源染色体分离C．甲——同源染色体分离；乙——交叉互换；丙——基因重组D．甲——交叉互换；乙——染色体变异；丙——交叉互换22．二倍体水毛茛黄花基因q1中丢失3个相邻碱基对后形成基因q2，导致其编码的蛋白质中氨基酸序列发生了改变，下列叙述正确的是（）A．正常情况下q1和q2可存在于同一个配子中B．利用光学显微镜可观测到q2的长度较q1短C．突变后翻译时碱基互补配对原则发生了改变D．突变后水毛茛的花色性状不一定发生改变23．人类遗传病往往给患者和家庭带来身心两方面的持久影响。

河南省信阳高级中学北湖校区2023-2024学年高一下期06月测试（A）英语试题分值：150分时间：120分钟命题人：冯凡审题人：丁睿、李瑶注意事项：1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上一、单词汉译英、英译汉(共20小题；每小题1.5分，共30分)1.a adj.令人钦佩的；极其出色的2.r adj.不寻常的；显著的；非凡的3.c adv.创造性地；有创造力地4.e vt.建立，设立5.p adj.大有希望的6.W n.知识，学问；智慧7.i adj.令人钦佩的，给人印象深刻的8.a n.认识；意识9.s adj.很严重的10.e vi.参加，参与11.career n.12.intelligent adj.mittee n.14.staff n.15.despite prep.16.independence n.17.novelist n.18.reputation n.19.depression n.20.biography n.二、阅读理解(共两节，共20小题，每小题2.5分，满分50分)第一节(共15小题；每小题2.5分，共37.5分)AIn today's classroom,homework is completed not only with the pen and paper,but also with computers and cell phones.How to make the best use of such digital products is an important question for teachers and students.The following top digital portfolio platforms(作品集平台)offer a broad range of functions.Most importantly,they provide a way to share students' work with pride.Google SitesCreating a digital portfolio could not be any easier than Google Sites makes it.It lets students quickly put in contents,such as text,images,videos,maps,and much e one of the provided six themes or create a custom one,and then publish it as a public or restricted-view(限制浏览的)site.Seesaw for SchoolsDesigned for education,Seesaw for Schools provides a platform through which students complete and share school projects.By recording the process of gradually improving their schoolwork,kids gain a sense of mastery(熟练)and pride,so it is popular among them.Plus,parents can be involved too—just download the Seesaw Family app.ArtsoniaArtsonia is like a dream come true for art-minded students:a free,safe,educational space through which students show their digital creativity.Friends and family can view,comment on and buy objects.Just celebrate the artistic skills with Artsonia!EdublogsOne of the oldest and best-known web platforms for education,Edublogs makes it easy for teachers and students to start building a free WordPress platform.The free plan offers some storage space,class management tools,and no advertising.What impresses the users most is that a set of educators guide them and answer their questions on this site.2yWhich is true about Google Sites?A.It offers students online publishing rights for free.B.It provides students seven themes.C.It allows students to fill in contents in different forms.D.I lets students learn from others' work.22.Which platform can help students track their progressA.Google.C.Artsonia B.Seesaw for Schools.blogs23.Which of the following make Edublogs different from other platforms?A.It requires little chargeB.It is a platform for art-minded studentsC.It can store large amounts of information.D.It provides the guidance from specialistsB1 was siting in between my parents on a 15-hour flight from Toronto to Seoul, where I intended to discuss a book project.The hotel room I reserved had enough space for us all.So when my parents proposed it as a half-joke that they would come,too,I didn't think it made no sense.I was only 8 when we began our immigration(移民)journey.Korea was their country.They wouldn't need me there as they did in Canada.But the first week proved to be tough. We were upside down with the time change.My father announced that the first thing they wanted to do was visit my grandma's grave(坟墓).We had discussed visiting a few relatives,but going to grandma's grave had never come up.It was like: family duties upset my work schedule.“You guys go,”I refused. While my father showered,my mother took me aside.“Your dad has always been counting the days for the moment when he can show your grandma how well you have grown up.”I was deeply moved. So I decided to accompany them.As we approached the graveyard,I gathered some colored wildflowers from the parking lot and tied them with a long piece of grass. My parents got busy weeding around the headstone.“Your name is on the back"my father said.“See here?"I looked,and there was my Korean name carved beside those of my siblings and cousins. It felt odd to see our names on the headstone all of us,the living and dead, connected.I saw a link between generations.I didn't know how to tell them that the trip was amazing.I realized how I was closely associated with them. We don't belong to languages or countries. My grandma died only four months before we moved to Canada, when she was too weak to make the trip.I hope she knows that we did take her,and that all we have is each other.24. What was the main reason for the author's trip to Seoul？A.To visit grandmas graveyard.C.To work on a project. B. To have a family reunion.D.To accompany his parents25.why was the author initially unwilling to visit his grandma's grave?A.He was exhausted from the flight C.He liked visiting other relatives better.B.He had work arrangement.D.He preferred exploring alone.26.Why lid the author's father want to take him to grandma's grave?A.To introduce the author to relatives.C.To observe a family ceremony. B.To fulfil a long-awaited wish.D.To help with weeding.27.What did the author realize during the trip?A. Never forget our home country.C.The world is one big family.B.Family is a bridge to our future.D.Home is the bond among people. CThe elderly residents(居民)in care homes in London are being given hens to look after to stop them feeling lonely.The project was dreamed up by a local charity(慈普组织)to reduce loneliness and improve elderly people's wellbeing,It is also being used to help patients suffering dementia,a serious illness of the mind. Staff in care homes have reported a reduction in the use of medicine where hens are in use.Among those taking part in the project is 80-year-old Ruth Xavier.She said:“I used to keep hens when I was younger and had to prepare their breakfast each morning before I went to school.”“I like the project a lot.I am down there in my wheelchair in the morning letting the hens out and down there again at night to see they've gone to bed.'"It's good to have a different focus.People have been bringing their children in to see the hens and residents come and sit outside to watch them.I'm enjoying the creative activities,and it feels great to have done something useful.”There are now 700 elderly people looking after hens in 20 care homes in the North East,and the charity has been given financial support to roll it out countrywide.Wendy Wilson,extra care manager at 60 Penfold Street,one of the first to embark on the project,said:“Residents really welcome the idea of the project and the creative sessions.We are looking forward to the benefits and fun the project can bring to people here.”Lynn Lewis,director of Notting Hill Pathways,said:“We are happy to be taking part in the project.It will really help connect our residents through a shared interest and creative activities.”28.What is the purpose of the project?A.To ensure harmony in care homes.B.To provide part-time jobs for the aged.C.To raise money for medical research.D.To promote the elderly people's welfare(幸福).29.How has the project affected Ruth Xavier?A..She has learned new life skills.B.She has gained a sense of achievement.C.She has recovered her memory.D.She has developed a strong personality.30.What do the underlined words“embark on”mean in paragraph 7?A.improve.B.oppose.C.beginD.hesitate,31 What can we learn about the project from the last two paragraphsA.It is well received. C.It is highly profitable.B.It needs to be more creative.D.It takes ages to see the results. DMassive heat waves and droughts are already becoming a threat to farmers.Over the next three decades,California's San Joaquin Valley alone could lose up to 535,00 acres of farmland as a result of the shortage of water supplies.Jennifer Brophy,an assistant professor of bioenginering at Stanford,is working on methods she hopes will be used to change crop species so they can survive severe conditions.She is developing new genetic(基因的)engineering techniques that can help plants grow in a variety of different conditions.By changing the genome(基因组)of both crops and soil bacteria,she thinks it may be possible to help plants survive droughts.To make plants respond to extreme climate,Brophy is building what she calls“genetic circuits(回路).”In addition to changing the genes within plant cells,this method also change show and when those genes are activated(激活).If the plant senses a specific sugar,it can express one protein(蛋白质);if it senses another signal,it'll express a different protein,starting an entirely separate range of events.If both signals are there,the plant may be able to express something else entirely.“Using circuits,you can have plants respond in new ways to all these different inputs,”she says.Humans can learn from the weather forecast whether it's going to be really hot or really dry in the next couple of weeks so that we can plan accordingly.“A plant doesn't necessarily know what's coming.It just knows if it's hot or mild right now,”says Brophy.This can lead to problems when weather becomes unstable.A plant that usually flowers in spring may flower in winter if there are a few unseasonably warm days.When temperatures fall again,the flowers die and ruin a year of crops.“It'd be great to be able to communicate with plants to tell them,'hey,you should wait on that flowering,”she adds.Controlling plants' growth at this level is a promising future,but actually putting it into practice still has a long way to go.At the moment,Brophy is testing the concept in the lab using a small weedy plant called Arabidopsis,trying to figure out how to activate certain genes on command.32.What problem are farmers faced with according to the text?A.The shortage of strong labour force.B.The water pollution caused by factories.C.The pressure of feeding a growing population.D.The harmful effect of climate change on farming.33.What is the expected advantage of“genetic circuits”?A.Increasing farmlandB.Controlling plants' growthC.Telling the weather in advance.D.Allowing plants to give instructions independently.34.What do we know about Brophy s research?A.It is still in its early stages. C.It has led to serious problems.B.It has been put into practice D.It is funded by the government35.What's the text mainly about?A.A scientific breakthrough in bioengineering.B.A potential battle against global warming.C.A new type of plant resistant to heat waves.D.Away to help crops survive severe conditions第二节七选五(共5小题；每小题2.5分，共12.5分)Studies have shown that walking among trees makes us relaxed. 36 There's plenty of research showing the link between being in nature and being happy,but most of the research has focused on adults.37 Explaining the motivation behind the study,researchers note that with environmental issues such as global warming,more studies are focusing on the relationships between humans and nature to find solutions to these problems,and the future of the planet lies in the hands of children and their actions.However,few studies focus on this aspect.For the study,researchers worked with 296 children between the ages of 9 and 12.To measure how connected they were to nature,the kids were asked how much they enjoy activities like seeing wildflowers and wild animals,hearing sounds of nature and touching animals and plants. 38 They also found that the more concerned children were about the environment and nature,the more likely they were to be happy.So exposure to nature is related to active behaviors and happiness of children.What does that mean to adults?As we all know,children usually tend to follow adults',especially their parents' behaviors.That's to say,when adults are outside and appreciate nature,kids learn by example. 39Researchers say that children need role models who can gently guide them to nature with excitement and an attitude of a lifelong learner. 40 What's more important is that they spend time together with children by exploring a fun and safe environment.A.It's so good going for a walk around trees.B Even just smelling the trees helps reduce anxietyC.And if parents enjoy being close to nature,so will their children.D.Parents don't have to be experts in environmental science or nature studies.E.They found that the kids connected to nature were more likely to behave actively.F.In a new study researchers wanted to se if kids get the same benefits from being outdoors.G.Researchers will study more about the connection between children's happiness and nature.第三节完形填空(共15小题；每小题1分，满分15分)Like many young people,Jessica wants to travel the globe.Unlike most of them,this25-year-old is doing it 41 .She and her husband have spent the last two years traveling the world,stopping everywhere from Paris to Singapore.It might sound like one long,expensive 42 ,but the couple has an unusual way to make their travel 43 .They're part of a new form of the 44 economy:an online group of house sitters(代管人).Throughout their no-cost stays in 45 homes,they feed pets and water plants in thehomeowner's 46 .It's not all sightseeing.The two travelers carefully 47 their trips,scheduling their days around the pets that are sometimes difficult to 48 .But house sitting also offers a level of 49 they can't find in a hotel.“It's like 50 at a friend's house,”Jessica says.The couple has a high 51 rate in getting accepted as house sitters and they always go beyond the homeowner's 52 .For Jessica,that means 53 plenty of pictures of happy pets,keeping the house 54 and leaving a nice small gift before heading to the next house.“You want to make the homeowner feel that they made the right 55 ,”she says.41. A. indoors42. A. game43. A. safe44. A. local45. A. strangers46. A. favor47. A. plan48. A. buy49. A. support50. A. cooking51. A. success52. A. budget53. A. admiring54. A. clean55. A. guessB. onlineB. serviceB. busyB. privateB. parentsB. defenseB. explainB. transport B. comfortB. stayingB. survivalB. abilitiesB. donatingB. openB. decisionC. single-handed C. vacationC. helpfulC. sharingC. co-workers C. honor C. compareC. chooseC. controlC. waitingC. growthC. expectations C. sendingC. simpleC. responseD. full-timeD. procedure D. affordable D. agricultural D. neighborsD. absenceD. completeD. pleaseD. attentionD. studyingD. unemployment D. understanding D. borrowing D. emptyD. impression第Ⅱ卷(非选择题，共55分)第一部分语法填空(共10小题;每小题1.5分,共15分)阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式(注意:答案请写在答题卡上,写在本卷无效)。

河南省信阳市高级中学校本部高一生物月考试题含解析一、选择题（本题共40小题，每小题1.5分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

）1. 某正常女人，其父母色觉正常、弟弟色盲。

如果该女人和一个正常的男人结婚，生一个色盲儿子的概率是( )A．1/2 B．1/8 C．1/4D．1/16参考答案：B略2. 动物生理学家为了确定下丘脑在体温调节中的作用，做了以下实验：刺激下丘脑的前部，发现实验动物有出汗的现象；而刺激下丘脑的后部，实验动物出现寒战现象，据此分析，下列叙述正确的是（）A.下丘脑前部是散热中枢所在地，下丘脑后部是产热中枢所在地B.下丘脑前部是产热中枢所在地，下丘脑后部是散热中枢所在地C.温度感受器只分布在皮肤D.下丘脑的活动仍然受大脑皮层的控制参考答案：A3. 细胞膜的一定流动性是动植物细胞完成多种生理功能的结构基础，下列哪项与它关系最小（）A．肌细胞排出二氧化碳 B．白细胞吞噬细菌C．动物细胞细胞分裂 D．植物细胞质壁分离参考答案：A解析：肌细胞排出二氧化碳的方式是自由扩散，与细胞膜的流动性无关。

4. 马歇尔和沃伦因引起胃溃疡的幽门螺杆菌的开创性研究成果，获得了2005年诺贝尔生理学或医学奖．请问幽门螺杆菌与蓝藻的共同特点是（）A．是营腐生或寄生的异养生物B ．没有核膜，有染色体和核糖体C．遗传物质主要位于拟核D．细胞膜的组成和结构完全相同参考答案：C【考点】27：原核细胞和真核细胞的形态和结构的异同．【分析】幽门螺杆菌与蓝藻都属于原核生物，真核细胞和原核细胞的比较：类别原核细胞真核细胞细胞大小较小（一般1～10um）较大（1～100um）细胞核无成形的细胞核，无核膜、核仁、染色体，只有拟核有成形的细胞核，有核膜、核仁和染色体细胞质只有核糖体，没有其它复杂的细胞器有核糖体、线粒体等，植物细胞还有叶绿体等细胞壁细细胞壁主要成分是肽聚糖细胞壁的主要成分是纤维素和果胶增殖方式二分裂有丝分裂、无丝分裂、减数分裂可遗传变异来源基因突变基因突变、基因重组、染色体变异共性都含有细胞膜、核糖体，都含有DNA和RNA两种核酸等B、幽门螺杆菌与蓝藻都属于原核生物，两者都没有核膜和染色体，有核糖体，B错误；C、幽门螺杆菌与蓝藻都属于原核生物，两者的遗传物质主要位于拟核，C正确；D、幽门螺杆菌与蓝藻的细胞膜的组成和结构相似，而不是完全相同，D错误．故选：C．5. 豌豆灰种皮（G）对白种皮（g）为显性，黄子叶（Y）对绿子叶（y）为显性，每对性状的杂合子（F1）自交后代为F2，下列叙述正确的是( )①F1代上结的种子，灰种皮比白种皮接近3：1 ②F1代上结的种子，黄子叶比绿子叶接近3：1 ③F1代上结的种子，胚的基因型有9种④F2代上结的种子，灰种皮比白种皮接近3：1A.②③B. ②③④C.③④ D. ①②④参考答案：B略6. 在生物实验室内模拟生物体DNA复制所必需的条件是①酶类②游离四种脱氧核苷酸③ATP④DNA分子⑤mRNA⑥tRNA⑦适宜的温度⑧适宜的pH值A. ①②③④⑤⑥B. ②③④⑤⑥⑦C. ①②③⑤⑦⑧D. ①②③④⑦⑧参考答案：D试题分析：DNA复制需要模板、原料、能量和酶，模板是DNA分子，原料是游离的四种脱氧核苷酸，能量是ATP，酶需要适宜的温度和PH，故需要①②③④⑦⑧，D正确。

河南省信阳高级中学北湖校区2023-2024学年高一下期06月测试（A）生物试题考试时间:75分钟注意事项:1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上一、单项单选题:本题共16小题,每小题3分,共48分。

每小题只有一个选项符合题目要求。

1．下列关于科学家及其研究成果的叙述，错误的是A．沃森和克里克制作的DNA分子双螺旋结构模型，属于物理模型B．艾弗里利用“减法原理”进行肺炎链球菌体外转化实验，证明DNA是遗传物质C．摩尔根以果蝇为材料，利用假说–演绎法，证明基因位于染色体上D．梅塞尔森和斯塔尔利用放射性同位素示踪技术，证明DNA的复制是半保留复制2．孟德尔科学地解释了其豌豆杂交实验的结果，后人将其归纳为孟德尔第一、第二定律。

图中A/a与B/b两对等位基因独立遗传。

下列相关叙述错误的是A．③⑥过程可反映自由组合定律的实质B．①②过程可反映分离定律的实质C．⑥过程中双显性个体中能稳定遗传的占1/9D．④⑤过程中可以发生基因重组3．科学地选择实验材料是科学研究取得成功的重要保障之一、豌豆和果蝇均是研究遗传学的常用实验材料，下列不属于两者共同优点的是A．便于进行人工杂交实验B．具有易于区分的相对性状C．易于区分雌雄，便于研究伴性遗传D．子代数目多，利于实验结果的分析4．如图为真核细胞中遗传信息的传递和表达过程。

相关叙述正确的是A．①②过程中碱基配对情况相同C．①②过程所需要的酶相同B．③中核糖体的移动方向是由左向右D．②③过程发生的场所相同5．下图为中心法则示意图。

下列相关叙述正确的是A．病毒的遗传信息都会沿着④或⑤流动B．有丝分裂前的间期会发生过程①②③C．除了③过程外，其他4个过程所需原料相同D．过程①④所发生的碱基互补配对方式相同6．如果某个动物产生了一个新的基因，那么这个基因的形成可能是由于A．基因重组B．基因突变C．染色体变异D．出芽生殖7．下图中，a、b、c表示某生物细胞中一条染色体上相邻的三个基因，m、n为基因间的间隔序列。

河南省信阳市罗山县高级中学高一生物月考试卷含解析一、选择题（本题共40小题，每小题1.5分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

）1. 放在30%蔗糖溶液中会发生质壁分离的细胞是A．人的口腔上皮细胞 B．洋葱根尖生长点细胞C．洋葱表皮细胞 D．干种子细胞参考答案：C2. 细胞膜上与细胞的识别、免疫反应、信息传递和血型决定有密切关系的化合物是（）A．糖蛋白B．磷脂C．脂肪D．核酸参考答案：A3. 植物细胞吸水不会涨破，这是由于存在着( )A．细胞膜 B．叶绿体 C．液泡 D．细胞壁参考答案：D4. 现有一个性染色体为XXY的男孩。

①若男孩色觉正常，其母色盲，其父正常；②若男孩色觉正常，其父色盲，其母正常；③若男孩色盲，其父正常，其母色盲。

则①②③中此染色体变异分别发生在什么生殖细胞 ( ) A．精子、卵、不确定 B．精子、不确定、卵C．卵、精子、不确定 D．精子、不确定、精子参考答案：C5. 如图是酵母菌呼吸作用实验示意图，相关叙述正确的是（）A. 条件X下葡萄糖中能量的去向有三处B. 条件Y下，葡萄糖在线粒体中被分解，并产生CO2和水C. 试剂甲为溴麝香草酚蓝水溶液D. 物质a产生的场所为线粒体基质参考答案：A根据产物酒精判断条件X为无氧，无氧呼吸过程中葡萄糖中的能量一部分储存在酒精中，一部分储存在ATP中，大部分以热能形式散失，A正确；线粒体不能利用葡萄糖，B错误；试剂甲为酸性重铬酸钾溶液，C错误；图中无氧呼吸产生CO2的场所为细胞质基质，有氧呼吸产生CO2的场所为线粒体基质，D错误。

【考点定位】细胞呼吸、探究酵母菌的呼吸方式【名师点睛】解决本题需要知道酵母菌在有氧条件下进行有氧呼吸，产物是CO2和水；在无氧条件下进行无氧呼吸，产物是CO2和酒精；同时需要知道鉴定CO2可用溴麝香草酚蓝水溶液，鉴定酒精需要酸性的重铬酸钾溶液。

6. 以下关于遗传和减数分裂的叙述中，正确的是（）A．在减数分裂的过程中，非同源染色体间的部分交换属于基因重组B．在减数分裂的过程中，基因A和a的分离只能发生在减数第一次分裂的后期C．在正常减数分裂过程中，基因和染色体的行为是一致的D．一个基因型为AaBb的精原细胞经减数分裂后形成4种精子参考答案：C【考点】61：细胞的减数分裂．【分析】减数分裂过程：（1）减数第一次分裂间期：染色体的复制；（2）减数第一次分裂：①前期：联会，同源染色体上的非姐妹染色单体交叉互换；②中期：同源染色体成对的排列在赤道板上；③后期：同源染色体分离，非同源染色体自由组合；④末期：细胞质分裂．（3）减数第二次分裂过程：①前期：核膜、核仁逐渐解体消失，出现纺锤体和染色体；②中期：染色体形态固定、数目清晰；③后期：着丝点分裂，姐妹染色单体分开成为染色体，并均匀地移向两极；④末期：核膜、核仁重建、纺锤体和染色体消失．【解答】解：A、在减数分裂的过程中，非同源染色体间的部分交换属于染色体结构变异中的易位，A 错误；B、在减数分裂的过程中，等位基因A和a的分离一般发生在减数第一次分裂的后期，若发生基因突变或交叉互换，也可发生在减数第二次分裂后期，B错误；C、由于基因位于染色体上，因此在正常减数分裂过程中，基因和染色体的行为是一致的，C正确；D、一个基因型为AaBb的精原细胞经减数分裂后形成4个精子，但只有2种，D错误．故选：C．【点评】本题考查细胞的减数分裂，要求考生识记细胞减数分裂不同时期的特点，掌握减数分裂过程中染色体行为和数目变化规律，能结合所学的知识准确判断各选项．7. 下表中有关人体细胞化合物的各项内容，正确的是( )参考答案：C8. 下列各项中均属于真核生物的一组是A. 绿藻、小球藻、颤藻、黑藻B. 大肠杆菌、酵母菌、乳酸菌、破伤风芽孢杆菌C. 洋葱、伞藻、蝾螈、刀豆D. 蓝球藻、SARS病毒、Rous肉瘤病毒、香菇参考答案：C【详解】细菌（如杆菌、球菌、螺旋菌、乳酸菌等）、蓝藻（如颤藻、蓝球藻、念珠藻、发菜等）等都是原核生物；绿藻、小球藻、黑藻、伞藻、香菇、酵母菌、洋葱、蝾螈、刀豆等都是真核生物；所以选C。

命题人：郸城一高赵瑞—、选择题（前25小题每小题1分、后10小题每小题2分共45分）1．下列是几个放射性同位素示踪实验，对其结果的叙述不正确的是 ( ) A．要得到含32P的噬菌体，必须先用含32P的培养基培养细菌B．给水稻提供14CO2，则14C的转移途径大致是：14CO2→14C3→（14CH2O）C．给水稻提供14CO2，则其根细胞在缺氧状态有可能出现14C2H5OHD．小白鼠吸入18O2后，红细胞呼吸作用会产生少量的H218O2．1861年巴斯德发现，利用酵母菌酿酒的时候，如果发酵容器存在氧气，会导致酒精产生停止，这就是所谓的巴斯德效应。

直接决定“巴斯德效应”发生与否的反应及其场所是( )A．酒精+O2→丙酮酸细胞质基质 B．丙酮酸+ O2→CO2 线粒体基质C．H2O→O2+[H] 类囊体膜 D．[H]+O2→H2O 线粒体内膜3.如图表示人体内氢元素随化合物代谢转移过程，下列分析合理的是（）A．①过程发生在核糖体中，水中的H只来自—NH2B．在缺氧的情况下③过程中不会发生脱氢反应C．M物质是丙酮酸，④过程不会发生在线粒体中D．在O2充足的情况下，②③过程发生于线粒体中4.鉴别一个进行有丝分裂的细胞是植物细胞还是动物细胞，最可靠的方法是检查它的（）A．DNA的自我复制 B．是否出现中心体C．是否出现纺锤体 D．细胞质分裂的方式5．对于人类的单基因遗传病①、②、③、④，在被调查的若干家庭中发病情况如右表。

据此推断，最可能属于伴性遗传的是（注：每类家庭人数200～300人，表中+为发现病症表现者，-为正常表现者）A．①③④B．②③C．②③④D．②④6.小麦的粒色受不连锁的两对基因R1和r1、R2和r2控制。

R1和R2决定红色，r1和r2决定白色，R对r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（R1R1R2R2）与白粒（r1r1r2r2）杂交得F1，F1自交得F2，则F2的表现型有（）A．10种 B．9种 C．5种 D．4种7.下图中的图1表示细胞分裂的不同时期与每条染色体上DNA含量变化的关系，图2表示处于细胞分裂不同时期的细胞图像。

信阳高级中学2012-2013 学年高一下学期 6 月月考（理）本试卷分第Ⅰ卷( 选择题 ) 和第Ⅱ卷 ( 非选择题 ) 两部分，共150 分 . 考试时间120 分钟 .第Ⅰ卷（选择题共60分）注意事项：1．答第Ⅰ卷前，考生务必将机读卡上的姓名、学号用黑色字迹的签字笔填写，用2B 铅笔将学号对应的信息点涂黑．2．每小题选出答案后，将答题卡上对应题目的答案选中涂满涂黑，黑度以盖住框内字母为准，修改时用橡皮擦除干净，在试卷上作答无效。

一、选择题：本大题共12 小题，每小题 5 分，共 60 分，在每小题给出的的四个选项中，只有一项是符合题目要求的．1．已知集合A={-1,0,1},B={x|x=|a+1|,a∈ A} ,则A∩B=（）A.{0}B.{1}C.{0,1}D.{0,1,2}2 若点(a,9)在函数y=3x的图象上,则tan的值为（）A ． 0B ．C1 D ．．3.从分别写有 ABCDE 的 5 张卡片中任取两张，两字母恰好相连的概率（）A0.2B0.4C0.3D0.74. 设函数f (x)log a x(0 a 1)的定义域为 [ m,n]( m n) ,值域为 [0,1] ,若 n m 的最小值为1,则实数 a 的值为（）3A ．1B ．1或2C．2D ．2或3 4433345.将区间 [0,1] 内的随机数转化为[-2,6] 内的均匀随机数，需实施的变换为()A.a=a1 8B.a=a 1 *8+2C.a=a1 *8-2D.a=a1 *66．已知向量a, b 满足 | a |1,b(1,3) ,且a a b ,则 a 与 b的夹角为（）A .60B . 90C. 120 D . 1507．如右图所示的程序框图的输出值y ∈ (1,2] ，则输入值 x∈（）A． [-log 23,-1) ∪ (1,3]B． (-1,-log 32]∪ [1,2)C． [-1,-log32]∪(1,2] D. (-log 23,-1]∪ [1,3)8. 已知函数① y sin x cos x , ② y 2 2 sin x cos x , 则下列结论正确的是（）A ．两个函数的图象均关于点(,0) 成中心对称4B ．①的纵坐标不变 , 横坐标扩大为原来的2 倍, 再向右平移个单位即得②4C ．两个函数在区间 (, ) 上都是单调递增函数4 4D ．两个函数的最小正周期相同9. 已知函数 y sin( x ) ( 0) 的部分图象如右图所示, 设 P 是图象的最高点 , A, B 是图象与 x 轴的交点 , 则 tan APB （）A ． 10B ． 8C ．8D ．47 710 设三棱柱的侧棱垂直于底面, 所有棱长都为 a , 顶点都在一个球面上, 则该球的表面积为 （）A ． a2B ． 7a2C ．11a 2D ． 5 a 23311. 已知 P(x,y) 是直线 kxy 40(k0) 上一动点 ,PA,PB 是圆 C: x 2y 2 2y0 的两条切线 , A ． B 是切点 , 若四边形 PACB 的最小面积是 2, 则 k 的值为 ( )A ． 3B ．21C ． 2 2D ．221x 1 , x( , 2)12. 设函数 f (x) 1, 则函数 F ( x) xf ( x)1的零点的个数（）f ( x 2), x [2,)2A ． 4B ． 5C ． 6D ．7二、填空题（每题 5分共 20分。