ccna3 第2单元章节测试答案
ccna第二章答案
1. Which statements correctly identify the role of intermediary devices in the network? (Choose three.)determine pathways for data 确定数据通路initiate data communicationsretime and retransmit data signals 重分发originate the flow of datamanage data flows 管理数据final termination point for data flow2. Select the statements that are correct concerning network protocols. (Choose three.)define the structure of layer specific PDU'sdictate how to accomplish layer functionsoutline the functions necessary for communications between layers limit the need for hardware compatibilityrequire layer dependent encapsulationseliminate standardization among vendors3. What are two functions of encapsulation? (Choose two.)tracks delay between end devicesenables consistent network paths for communicationallows modification of the original data before transmissionidentifies pieces of data as part of the same communicationensures that data pieces can be directed to the correct receiving end device4. What is a primary function of the trailer information added by the data link layer encapsulation?supports error detection 支持错误检测‘ensures ordered arrival of dataprovides delivery to correct destinationidentifies the devices on the local networkassists intermediary devices with processing and path selection5. Which statements correctly identify the role of intermediary devices in the network? (Choose three.)determine pathways for datainitiate data communicationsretime and retransmit data signalsoriginate the flow of datamanage data flowsfinal termination point for data flow6. What is a PDU?corruption of a frame during transmissiondata reassembled at the destinationretransmitted packets due to lost communicationa layer specific encapsulation 一个具体的封装层7. Which characteristic correctly refers to end devices in a network?manage data flowsoriginate data flow 原数据流retime and retransmit data signalsdetermine pathways for data8.Refer to the exhibit. "Cell A" at IP address 10.0.0.34 has established an IP session with "IP Phone 1" at IP address 172.16.1.103. Based upon the graphic, which device type best describes the function of wireless device "Cell A?"the destination devicean end devicean intermediate devicea media device9.Refer to the exhibit. Which three labels correctly identify the network types for the network segments that are shown?(Choose three.)Network A -- WANNetwork B -- WANNetwork C -- LANNetwork B -- MANNetwork C -- WANNetwork A -- LAN10. Which three statements best describe a Local Area Network (LAN)? (Choose three.)A LAN is usually in a single geographical area.The network is administered by a single organization.The connection between segments in the LAN is usually through a leased connection.The security and access control of the network are controlled by a service provider.A LAN provides network services and access to applications for users within a common organization.Each end of the network is generally connected to a Telecommunication Service Provider (TSP).11.Refer to the exhibit. Which networking term describes the data interleaving process represented in the graphic?pipingPDUstreamingmultiplexingencapsulation12. What is the primary purpose of Layer 4 port assignment?to identify devices on the local mediato identify the hops between source and destinationto identify to the intermediary devices the best path through the network to identify the source and destination end devices that are communicating to identify the processes or services that are communicating within the end devices13. What device is considered an intermediary device?1．file serve2.IP phone ptop 4.printer 5.switch14.Refer to the exhibit. Which term correctly identifies the device type that is included in the area B? sourceendtransferintermediary 中间设备15.Refer to the exhibit. What type of network is shown?WANMANLANWLAN16. Which layer encapsulates the segment into packets?physicaldata linknetwork 网络层封装的是数据包transport17. What can be identified by examining the network layer header?the destination device on the local mediathe destination host address 网络层头部封装目的主机的地址the bits that will be transferred over the mediathe source application or process creating the data18. What is the purpose of the TCP/IP Network Access layer?path determination and packet switchingdata presentationreliability, flow control, and error detectionnetwork media control 接入层控制网络媒体the division of segments into packets19. During the encapsulation process, what occurs at the data link layer? No address is added.The logical address is added.The physical address is added. 数据链路层添加一个物理地址The process port number is added.20.Refer to the exhibit. Which set of devices contains only end devices? A, C, DB, E, G, HC, D, G, H, I, JD, E, F, H, I, JE, F, H, I, J21. What is the proper order of the layers of the OSI model from the highest layer to the lowest layer?4physical, network, application, data link, presentation, session, transportapplication, physical, session, transport, network, data link, presentationapplication, presentation, physical, session, data link, transport, networkapplication, presentation, session, transport, network, data link, physicalpresentation, data link, session, transport, network, physical, application22. Which two layers of the OSI model have the same functions as the TCP/IP model Network Access Layer? (Choose two.)34NetworkTransportPhysicalData Link TCP/IP 工作在第一层和第二层Session。
必修三第2章综合检测以及答案
第2章综合检测A．激素X是促性腺激素，激素Y为雌性激素一、选择题 B．激素Y到达靶细胞后，其跨膜运输方式是主动运输1．如图为反射弧结构示意图，下列有关说法不正确的是( ) C．该生理过程中存在反馈调节D．长期注射激素Y会导致性腺衰退5．下图为人体甲状腺激素分泌的调节示意图，图中①②③为激素。
下列叙述不正确的是( ) A．由abcde组成了一个完整的反射弧B．若从③处剪断神经纤维，刺激①处，效应器仍能产生反应A．含量很少的①经过分级调节作用，可明显增加③的分泌C．图中②的结构决定了神经元之间的兴奋传递只能是单向的B．激素②是调节甲状腺细胞分泌功能的主要激素D．若从①处剪断神经纤维，刺激③处，效应器仍能产生反应C．血中③的浓度过低时，对下丘脑和垂体的促进作用减弱2．下图为某一传出神经元与肌细胞形成的突触。
下列说法错误的是( ) D．③几乎作用于全身的靶细胞，促进其细胞代谢6．如图表示人体和人体细胞内某些信息传递机制的模式图，图示中箭头表示信息的传递方向。
下列有关叙述中，正确的是( ) A．①的形成与高尔基体有关 A．如果该图表示反射弧，则其中的信息是以局部电流的形式传导的B．参与突触形成的肌细胞膜面积增大有利于神经递质的作用B．如果该图中的a为下丘脑、b为垂体、c为甲状腺，则c分泌的甲状腺激素增加到一定程度后，C．④兴奋时，其两侧的电位表现为内正外负对a分泌d，b分泌e具有抑制作用D．②释放的神经递质一定会使肌细胞兴奋C．如果该图表示细胞中遗传信息的传递过程，则过程d只发生于细胞核中 3.如图表示突触的亚显微结构，下列说法正确的是( ) D．如果该图表示细胞免疫过程，则a为效应T细胞，b为靶细胞，c为抗体7．如图所示，①②分别代表不同的细胞，a代表物质，下列各项不符合该示意图的是( ) ＋A．①中内容物使b兴奋时，兴奋部位的膜对Na通透性减小A．①为前一神经元、②为后一神经元、a为神经递质B．②处的液体为组织液，③一定是一个神经元的树突膜B．①为垂体细胞、②为甲状腺细胞、a 为促甲状腺激素释放激素C．在a中发生电信号→化学信号的转变，信息传递需要能量C．①为下丘脑细胞、②为肾小管细胞、a为抗利尿激素D．当兴奋沿b神经元传导时，其膜内电流方向与兴奋传导方向相D．①为甲状腺细胞、②为下丘脑细胞、a为甲状腺激素反 8．细胞间和细胞内的信息传递过程中，需要受体对信号的识别。
ccna第二章练习答案.docx
yy1 ・ Refer to the exhibit・ Identify the devices labeled A, B, C, and D in the network physical documentation. 匚A=bridge, B=switch, C=router, D=hubK A=bridge, B=hub, C=router, D=switchC A=bridge, B=router, C=hub, D=switchc A=hub, B=bridge, C=router, D=switch2 The central hub has malfunctioned in the network. As a result, the entire network is down. Which type of physical network topology is implemented?匚busE starC ring匚mesh3. A switch has failed in the network. As a result, only one segment of the network is down. Which type of physical network topology is implemented?C busC ring匚starE extended star4. Which three features apply to LAN connections? (Choose three.)厂operate using serial in terfacesV make network connection using a hubV limited to operation over small geographic areas厂 provide part-time connectivity to remote servicesV typically operate under local administrative control厂provide lower bandwidth services compared to WANs5. What is one advantage of defining network communication by the seven layers of the OSI model? 【:It in creases the ban dwidth of a n etwork.E It makes networking easier to learn and understand・It eliminates many protocol restrictions.匚It increases the throughput of a network.c It reduces the need for testing network connectivity.6. What makes it easier for different networking vendors to design software and hardware that willin teroperate?E OSI modelc proprietary desig ns匚IP addressing schemec standard logical topologies匚standard physical topologies7. Which term describes the process of adding headers to data as it moves down OSI layers?匚division匚encoding匚separation匚segmentationE encapsulation8. What is the term used to describe the transport layer protocol data unit?匚bits c packetsE segments匚framesc data streams9. Which of the following are layers of the TCP/IP model? (Choose three.)▽ Applicati onr PhysicalV Internet0 Network Access厂Prese ntation10. Which of the following are data link layer encapsulation details? (Choose two.)0 A header and trailer are added・厂Data is con verted into packets ・V Packets are packaged into frames・厂Frames are divided into segments.厂Packets are changed into bits for Internet travel.11. Which layer of the OSI model provides network services to processes in electronic mail and file transfer programs?匚data link匚transport匚n etworkE application12. Which two features apply to WAN connections? (Choose two.)V operate using serial interfacesP make network connection using a hub厂limited to operation over small geographic areasI- typically operate under local administrative control▽provide lower bandwidth services compared to LANs13. Which of the following are ways that bandwidth is commonly measured? (Choose three.)厂GHzps▽kbpsV Mbps厂Nbps厂MHzpsV Gbps14. Refer to the following list. Choose the correct order of data encapsulation when a device sends information. segments bits packetsdata frames匚1 -3-5-4-2C 2- 1-3-5-4C 2-4-3-5-1匚4-3-1-2-5E 4-1 -3-5-2C 3-5-1-2-415. Which of the following are factors that determine throughput? (Choose two.)厂types of passwords used on servers厂type of Layer 3 protocol usedV n etwork topologywidth of the network cable0 number of users on the network16. Refer to the exhibit・ Which column shows the correct sequenee of OSI model layers?E D17. Which layer of the OSI model provides connectivity and path selection between two end systems where routing occurs?c physical layer匚data link layerE network layer匚transport layer18. Which best describes the function of the physical layer?E Defines the electrical and functional specifications for the link between end systems・匚Provides reliable transit of data across a physical link.匚Provides connectivity and path selection between two end systems・匚Concerned with physical addressing, network topology and media access.。
CCNA02-final3
1Which of the following are required when adding a network to the OSPF routing process conf iguration? (Choose three.)network addressloopback addressautonomous system numbersubnet maskwildcard maskarea ID2Using default settings, what is the next step in the router boot sequence after t he IOS loads from flash?Perf orm the POST routine.Search for a backup IOS in ROM.Load the bootstrap program f rom ROM.Load the running-config file from RAM.Locate and load the startup-config file from NVRAM.3Refer to the exhibit. What is the most efficient summarization of the routes attached to router R1?198.18.0.0/16198.18.48.0/21198.18.32.0/22198.18.48.0/23198.18.49.0/23198.18.52.0/224Refer to the exhibit. When troubleshooting a network, it is important to interpret the output of various router commands. On the basis of the exhibit, which three statem (Choose three.)The missing information f or Blank 1 is the command show ip route .The missing information f or Blank 1 is the command debug ip route .The missing information f or Blank 2 is the number 100.The missing information f or Blank 2 is the number 120.The missing information f or Blank 3 is the letter R.The missing information f or Blank 3 is the letter C.5Refer to the exhibit. Packets destined to which two networks will require the router to perform a recursive lookup? (Choose t wo.)10.0.0.0/864.100.0.0/16128.107.0.0/16172.16.40.0/24192.168.1.0/24192.168.2.0/246When presented with multiple valid routes to a destination, what criteria does a router use to determine which routes to add to the routing table?The router selects the routes with the best metric. All routes that hav e the same best metric are added to the routing table.The router f irst selects routes with the lowest administrative distance. The resulting routes are then prioritized by metric and the routes with the best metric arouting table.The router selects the routes with the lowest administrativ e distance. All routes with the same lowest administrative distanc e are added to the routing table.The router installs all routes in the routing table but uses the route with the best metric most when load balancing.7Refer to the exhibit. All routers in the network are running RIPv2 and EIGRP with default routing protocol settings and have interfaces configured with the bandwidths t exhibit. Which protocol will be used and how will traffic between the Router1 LAN and Router5 LAN be routed through the network?RIPv2 will load balance across both paths between Router1 and Router5.EIGRP will load balance across both paths between Router1 and Router5.RIPv2 traffic will use the path Router1, Router2, Router5 because it has the least hops.EIGRP traffic will use the path Router1, Router3, Router4, Router5 because it has the best metric.8Of the listed routing protocols, which two will propogate LSPs to all neighbors? (Choose two.)IS-ISEIGRPOSPFRIPv1RIPv29Refer to the exhibit. Routers 1 and 2 are directly connected over a serial link. Pings are f ailing between the two routers. W hat change by the administrator will correctSet the encapsulation on both routers to PPP.Decrease the bandwidth on Serial 0/1/0 on router 2 to 1544.Change the cable that connects the routers to a crossover cable.Change the IP address on Serial 0/1/0 on router 2 to 192.168.0.1/30.10Refer to the exhibit. Which three statements are true of the routing table for Router1? (Choose three.)The route to network 172.16.0.0 has an AD of 156160.Network 192.168.0.16 can best be reached using FastEthernet0/0.The AD of EIGRP routes has been manually changed to a value other than the default value.Router1 is running both the EIGRP and OSPF routing process.Network 172.17.0.0 can only be reached using a default route.No default route has been conf igured.11Which statement correctly describes a feasible successor in EIGRP?It is a primary route that is stored in the routing table.It is a backup route that is stored in the routing table.It is a primary route that is stored in the topology table.It is a backup route that is stored in the topology table.12Refer to the exhibit. A ping from R1 to 10.1.1.2 is successful, but a ping from R1 to 192.168.2.0 fails. What is the cause of this problem?There is no gateway of last resort at R1.The serial interf ace between the two routers is down.A default route is not configured on R1.The static route for 192.168.2.0 is incorrectly conf igured.13Refer to the exhibit. A network administrator has run the show interface command. The output of this command is displayed. What is the first step that is required toSwitch the cable with a known working cable.Issue the no shutdown command on the interface.Configure the interf ace as a loopback interface.Set the encapsulation for the interface.14Which two locations can be the source of the Cisco IOS that is used by a router during the bootup process? (Choose two.)flash memoryRAMNVRAMTFTP serverconfiguration register15All routers in a network are configured in a single OSPF area with the same priority value. No loopback interface has been set on any of the routers. Which secondrouters use to determine the router ID?The highest MAC address among the active interfaces of the network will be used.There will be no router ID until a loopback interf ace is configured.The highest IP address among the activ e FastEthernet interfaces that are running OSPF will be used.The highest IP address among the activ e interfaces will be used.16Refer to the exhibit. A network administrator has conf igured OSPF using the following command:network 192.168.1.32 0.0.0.31 area 0Which router interf ace will participate in OSPF?FastEthernet0/0FastEthernet0/1Serial0/0/0Serial0/0/117Refer to the exhibit. All routers are properly configured with default configurations and are running the OSPF routing protocol. The network is f ully converged. A host on th with a host on the 192.168.2.0/24 network.Which path will be used to transmit the data?The data will be transmitted v ia R3-R2.The data will be transmitted v ia R3-R1-R2.The traffic will be load-balanced between two paths — one via R3-R2, and the other via R3-R1-R2.The data will be transmitted v ia R3-R2, and the other path via R3-R1-R2 will be retained as the backup path.18Which two situations require the use of a link-state protocol? (Choose two.)Fast convergence of the network is critical.The network is very large.The network administrator has limited knowledge to configure and troubleshoot routing protocols.The network is a flat network.The capacity of the router is low.19Refer to the exhibit. What is the meaning of the highlighted value 120?It is the metric that is calculated by the routing protocol.It is the v alue that is used by the DUAL algorithm to determine the bandwidth for the link.It is the administrative distance of the routing protocol.It is the hold-down time, measured in seconds, before the next update.20In a lab test environment, a router has learned about network 172.16.1.0 through four different dynamic routing processes. Which route will be used to reach this nD 172.16.1.0/24 [90/2195456] via 192.168.200.1, 00:00:09, Serial0/0/0O 172.16.1.0/24 [110/1012] via 192.168.200.1, 00:00:22, Serial0/0/0R 172.16.1.0/24 [120/1] via 192.168.200.1, 00:00:17, Serial0/0/0I 172.16.1.0/24 [100/1192] via 192.168.200.1, 00:00:09, Serial0/0/021Which statement is true about the RIPv1 protocol?It is a link-state routing protocol.It excludes subnet information from the routing updates.It uses the DUAL algorithm to insert backup routes into the topology table.It uses classless routing as the default method on the router.22Refer to the exhibit. The 10.4.0.0 network f ails. What mechanism prevents R2 f rom receiving false update information regarding the 10.4.0.0 network?split horizonhold-down timersroute poisoningtriggered updates23Which statement is true about the metrics used by routing protocols?A metric is a value used by a particular routing protocol to compare paths to remote networks.A common metric is used by all routing protocols.The metric with the highest value is installed in the routing table.The router may use only one parameter at a time to calculate the metric.24Refer to the exhibit. Both routers are using the RIPv2 routing protocol and static routes are undef ined. R1 can ping 192.168.2.1 and 10.1.1.2, but is unable to ping 192.16 What is the reason for the ping f ailure?The serial interf ace between two routers is down.R2 is not forwarding the routing updates.The 192.168.4.0 network is not included in the RIP configuration of R2.RIPv1 needs to be configured.25Which statement correctly describes a feature of RIP?RIP is a link-state routing protocol.RIP uses only one metric—hop count— for path selection.Adv ertised routes with hop counts greater than 10 are unreachable.Messages are broadcast every 10 seconds.26Refer to the exhibit. When the show cdp neighbors command is issued f rom router C, which devices will be displayed in the output?B, DA, B, DD, SWH-2SWH-1, A, BSWH-1, SWH-2A, B, D, SWH-1, SWH-227Refer to the exhibit. Although R2 is conf igured correctly, host A is unable to access the Internet. What are two static routes that can be c onfigured on R1, either of which w (Choose two.)ip route 0.0.0.0 0.0.0.0 Fa0/0ip route 0.0.0.0 0.0.0.0 Fa0/1ip route 0.0.0.0 0.0.0.0 10.1.1.1ip route 0.0.0.0 0.0.0.0 10.1.1.2ip route 209.165.202.0 255.255.255.0 10.1.1.128Which component is typically used to connect the WIC interface of a router to a CSU/DSU?V.35 cableRJ-45 adaptercrossover cablestraight-through cable29Refer to the exhibit. Which combination of IP address and subnet mask can be used on the serial interf ace of Router2 in order to put the interf ace in the same networ interface of Router1?IP 172.16.0.18, subnet mask 255.255.255.0IP 172.16.32.15, subnet mask 255.255.255.240IP 172.16.0.18, subnet mask 255.255.255.252IP 172.16.32.18, subnet mask 255.255.255.25230Refer to the exhibit. Which host has a combination of IP address and subnet mask on the same network as Fa0/0 of Router1?host Ahost Bhost Chost D31Refer to the exhibit. OSPF is used f or the routing protocol and all interf aces are configured with the correct IP addresses and subnet masks. During testing, it is foun unable to form an adjacency with R2. What is the cause of this problem?Both routers have been configured with incorrect router IDs.Both routers have been configured in different OSPF areas.Both routers have been configured with an incorrect network type.Both routers have been configured with different hello and dead intervals.32You hav e been asked to explain converged networks to a trainee. How would you accurately describe a converged network?A network is converged when all routers have f ormed an adjacency.A network is converged immediately after a topology change has occurred.A network is converged when all routers flush the unreachable networks from their routing tables.A network is converged after all routers share the same information, calculate best paths, and update their routing tables.33 A network administrator is in charge of two separate networks that share a single building. What dev ice will be required to connect the two networks and add a comthe Internet that can be shared?hubrouteraccess pointEthernet switch34Which port should a terminal emulator be connected to in order to access a router without network connectivity?T1serialconsoleFastEthernet35Refer to the exhibit. What needs to be done to allow these two routers to connect successf ully?Add a clock rate to S0/0/0 on R1.Add an interface description to S0/0/1 on R2.Change the serial interface on R2 to S0/0/0 so that it matches R1.Change the IP address of S0/0/1 on R2 so that it is in the same subnet as R1.36Refer to the exhibit. Based on the partial output in the exhibit, why can users establish a console connection to this router without entering a password?The login command was not entered on the console line.The enable password should be an enable secret password.No username and password combination has been configured.Console connections cannot be conf igured to require users to provide passwords.37Refer to the exhibit. When a static IP address is being configured on the host, what address should be used for the default gateway?10.1.1.110.1.1.2172.16.1.1192.168.1.138Refer to the exhibit. The entire 192.168.1.0 network has been allocated to address hosts in the diagram. Utilizing VLSM with contiguous address blocks, which set of pref ixes could be used to create an addressing solution with a minimum waste of IP addresses?39What is a function of a router?It extends the Layer 2 broadcast domain.It eliminates collisions among PCs on the same local network.It provides connectivity among PCs on the same physical segment.It forwards packets to different IP networks based on Layer 3 inf ormation.40Refer to the exhibit. Which solution provides the most efficient use of router resources for forwarding traffic between BR and HQ?RIPRIPv2EIGRPstatic routes41Refer to the exhibit. A network administrator configures a static route on router R1 to reach the 192.168.1.0/24 network. Which IP address should be used as the next ip route command?192.168.1.1192.168.2.1192.135.250.1192.135.250.242Refer to the exhibit. The network is conf igured with RIPv2. However, network administrators notice that communication cannot be successfully completed f rom one L network administrator issues the show ip route command on the HQ router. Based on the output, what should be done to correct the problem?Disable the load balancing feature of RIPv2.Issue the no auto-summary command for RIPv2.Replace RIPv2 with EIGRP which supports VLSM.Make sure that the network statements include the correct subnet mask.43Which protocol is used by EIGRP to deliver and receive update packets?FTPRTPTCPTFTPUDP44Refer to the exhibit. What OSPF network statements are required for the router B to adv ertise the three networks that are attached?router ospf 1network 10.0.0.0 0.0.0.255 area 0router ospf 1network 10.1.1.0 0.3.255.255 area 0network 10.10.1.0 0.255.255.255 area 0network 10.20.1.0 0.255.255.255 area 0router ospf 1network 10.1.1.0 0.0.0.3 area 0network 10.10.1.0 0.0.255.255 area 0network 10.20.1.0 0.0.255.255 area 0router ospf 1network 10.1.1.0 0.0.0.3 area 0network 10.10.1.0 0.0.0.255 area 0network 10.20.1.0 0.0.0.255 area 045Which two v alues are used by default to calculate a metric in EIGRP? (Choose two.) loaddelayreliabilityhop countbandwidth46Refer to the exhibit. Which statement is true concerning the routing configuration?Using dy namic routing instead of static routing would hav e required f ewer configuration steps.The 10.1.1.0/24 and 10.1.2.0/24 routes have adjacent boundaries and should be summarized.Packets routed to the R2 Fast Ethernet interface require two routing table lookups.The static route will not work correctly.47Refer to the exhibit. On the basis of the show running-config output, which option correctly ref lects the routes that will be listed in the R2 routing table?48Refer to the exhibit. Which command should be issued to configure the address of the router interface that is attached to this computer?Router(config-if)# ip address 192.168.2.1 255.255.255.0Router(config-if)# ip address 192.168.2.2 255.255.255.0Router(config-if)# ip address 192.168.0.1 255.255.255.0Router(config-if)# ip address 192.168.0.2 255.255.255.049Refer to the exhibit. Which route will be removed from the routing table if manual EIGRP summarization is disabled on the Serial0/0/0 interface of Router3?0.0.0.0/0172.16.0.0/16172.16.1.0/24172.16.3.0/3050Refer to the exhibit. All interfaces are addressed and f unctioning correctly. The network administrator runs the tracert command on host A. Which two f acts could be output of this command? (Choose two.)The gateway for Host A is missing or improperly conf igured.The gateway for Host B is missing or improperly conf igured.The entry for 192.168.1.0/24 is missing f rom the routing table of R1.The entry for 192.168.1.0/24 is missing f rom the routing table of R2.The entry for 192.168.2.0/24 is missing f rom the routing table of R1.The entry for 192.168.2.0/24 is missing f rom the routing table of R2.。
2023-2024学年高中生物浙科版选修3第2章 克隆技术单元测试(含答案解析)
2023-2024学年浙科版高中生物单元测试班级 __________ 姓名 __________ 考号 __________一、选择题（本大题共计17小题每题3分共计51分）1.下列同时属于基因工程和植物细胞工程实际应用的是（）A. 培育抗病毒植物B. 生产脱毒植株C. 制作“生物导弹”D. 培育单倍体【答案】A【解析】解培育抗病毒植物需要将抗病毒的相应目的基因经过转基因技术导入植物细胞再经过植物组织培养得到抗病毒植株是基因工程和植物细胞工程的应用故选 A2.小鼠克隆胚胎着床后胎盘发育显著异常可能是与克隆胚胎中H3K27me3印记基因过度表达有关 H3K27me3印记基因敲除极大提高了体细胞克隆的成功率 H3K27me3印记基因敲除的实验组克隆小鼠的体重与受精卵来源的小鼠一致而对照组克隆小鼠的体重显著高于受精卵来源的小鼠实验组克隆小鼠的胎盘直径和重量也显著低于对照组克隆小鼠而与受精卵来源的小鼠一致下列相关分析错误的是（）A. 克隆胚胎过大可能是克隆胚胎着床后成活率低的原因B. H3K27me3印记基因过度表达抑制胎盘的发育C. 克隆胚胎过大的原因可能是克隆胚胎的胎盘过大D. H3K27me3印记基因的表达产物可能影响早期胚胎滋养层细胞的发育【答案】B【解析】解 A.克隆胚胎过大可能是克隆胚胎着床后成活率低的原因 A正确B.根据实验结果 H3K27me3印记基因过度表达能够促进胎盘的发育 B错误C.克隆胚胎过大的原因可能是克隆胚胎的胎盘过大 C正确D.H3K27me3印记基因的表达产物可能影响早期胚胎滋养层细胞的发育 D正确故选 B3.对人工种子正确的解释是（）A. 利用植物组织培养技术获得的种子B. 种皮能够自我生成的种子C. 通过科学手段人工合成的种子D. 人工种子发芽率低【答案】A【解析】解 A.人工种子是通过植物组织培养得到的胚状体、不定芽、顶芽和腋芽等为材料通过人工薄膜包装得到的种子 A正确B.人工种皮是化学方法合成的不能生成种子 B错误C.人工种子是通过植物组织培养获得的种子不是人工合成 C错误D.人工种子中含有不定芽、顶芽和腋芽可以直接栽培不涉及发芽率 D错误故选 A4.下列属于微型繁殖的是（）A. 用豌豆自交获得较多的子代B. 用枣树茎的形成层繁殖C. 用红薯的块根繁殖D. 用嫩柳条扦插繁殖【答案】B【解析】解用于快速繁殖优良品种的植物组织培养技术叫作植物的微型繁殖技术也叫快速繁殖技术A.用豌豆自交获得较多的子代属于自然繁殖过程不需要植物组织培养技术 A错误B.用枣树茎的形成层繁殖需要采用植物组织培养技术属于微型繁殖 B正确C.用红薯的块根繁殖属于营养繁殖不需要植物组织培养技术 C错误D.用嫩柳条扦插繁殖属于营养繁殖不需要植物组织培养技术 D错误故选 B5.人绒毛膜促性腺激素（HCG）是女性怀孕后胎盘滋养层细胞分泌的一种糖蛋白制备抗HCG单克隆抗体可用于早孕的诊断如图是抗HCG单克隆抗体制备流程示意图有关说法错误的是（）A. 早孕的诊断试剂盒之所以准确灵敏是因为单克隆抗体的特异性强、纯度高B. ②过程通过选择培养基筛选出的细胞都能产生抗HCG抗体C. ③过程经抗体检测呈阴性的杂交瘤细胞既能迅速大量增殖又能产生抗体D. ④过程能在体内进行也可在体外进行【答案】B【解析】解 A.早孕的诊断试剂盒之所以准确灵敏是因为单克隆抗体的特异性强、纯度高 A正确B.②过程通过选择培养基筛选出的细胞可能有骨髓瘤细胞自身融合的细胞不一定都能产生抗HCG抗体 B错误C.③过程经抗体检测呈阴性的杂交瘤细胞既能迅速大量增殖又能产生抗体 C正确D.④大规模培养过程能在体内进行也可在体外进行 D正确故选 B6.如图示一人工种子下列与人工种子培育生产过程有关的叙述中不正确的是（）A. 人工种子一般用离体的植物细胞通过组织培养技术获得B. 胚状体是由愈伤组织分化而成离体细胞只有形成愈伤组织才能表现出全能性C. 同一批次的人工种子可以保证具有相同的基因型D. 胚状体是由未分化的、具有分裂能力的细胞构成【答案】D【解析】A、人工种子是以植物组织培养得到的胚状体不定芽、顶芽和腋芽等为材料经过人工薄膜包装得到的种子 A正确B、胚状体是由愈伤组织分化形成的离体细胞都能表现出全能性 B正确C、同一批次的人工种子由同一离体的植物组织、器官经过脱分化和再分化形成所以基因型相同 C正确D、胚状体是由愈伤组织分化形成的但具有分裂能力的细胞 D错误7.下列不属于克隆的一项是（）A. 将某个DNA片段利用PCR技术进行扩增B. 将草莓的匍匐茎压入泥土中长出新的草莓植株C. 将某种瘤细胞在体外培养繁殖成一个细胞系D. 将小鼠骨髓瘤细胞与经过免疫的脾细胞融合成杂交瘤细胞【答案】D【解析】解克隆技术指从众多的基因或细胞群体中通过无性繁殖和选择获得目的基因或特定类型细胞的技术操作可从三个层面上理解（1）分子水平上的克隆指基因克隆（2）细胞水平上的克隆指细胞克隆（3）个体水平的克隆指个体的无性生殖A.将某个DNA片段利用PCR技术进行扩增是从分子水平上进行的克隆 A不符合题意B.将草莓的匍匐茎压入泥土中长出新的草莓植株属于无性生殖是个体水平的克隆 B不符合题意C.将某种瘤细胞在体外培养繁殖成一个细胞系属于细胞培养是细胞水平的克隆 C不符合题意D.将小鼠骨髓瘤细胞与经过免疫的脾细胞融合成的杂交瘤细胞其遗传物质发生了改变不属于克隆 D符合题意故选 D8.下列不属于单克隆抗体的应用的是（）A. 作为诊断试剂B. 用于治疗疾病C. 作为基因探针D. 用于运载药物【答案】C【解析】A、单克隆抗体最广泛的用途是作为诊断试剂 A错误B、单克隆抗体可用于治疗疾病 B错误C、基因探针是指用放射性同位素（或荧光分子）标记的核酸片段单克隆抗体不能作为基因探针 C正确D、单克隆抗体可用于运载药物如”生物导弹“ D错误9.对人工种子正确的解释是（）A. 植物体细胞杂交技术获得的种子B. 同一批次的人工种子基因型不同C. 人工种子胚状体是由未分化的、具有分裂能力的细胞构成D. 利用植物组织培养技术获得的种子【答案】D【解析】解 A、人工种子是以植物组织培养得到的胚状体不定芽、顶芽和腋芽等为材料经过人工薄膜包装得到的种子 A错误B、同一批次的人工种子由同一离体的植物组织、器官经过脱分化和再分化形成所以基因型相同 B错误C 、胚状体是由愈伤组织分化形成的但具有分裂能力的细胞 C错误D、人工种子是通过植物组织培养得到的胚状体、不定芽、顶芽和腋芽等为材料通过人工薄膜包装得到的种子 D正确．故选 D．10.与“多利”羊等克隆哺乳动物相关的技术有（）①人工授精②胚胎移植③细胞培养④核移植．A. ①③④B. ①②③④C. ②③④D. ①②③【答案】C【解析】①“多利”羊是通过细胞核移植获得的未进行人工授精①错误②“多利”羊形成过程中用到了胚胎移植②正确③核移植细胞需通过细胞培养③正确④“多利”羊形成过程中主要应用了核移植技术④正确11.利用细胞工程方法以H7N9病毒颗粒外膜表面糖蛋白为抗原制备出单克隆抗体下列相关叙述错误的是（）A. 用纯化的糖蛋白反复注射到小鼠体内提取的血清抗体具有异质性B. 经多次抗体检验得到的阳性杂交瘤细胞可注射到小鼠腹水中继续培养C. 利用该单克隆抗体通过抗原—抗体杂交技术可诊断出H7N9病毒感染者D. 经灭活的仙台病毒诱导融合并筛选得到的杂交瘤细胞具有二倍体核型【答案】D【解析】解 A. 将纯化的糖蛋白作为抗原反复注射到小鼠的体内一段时间后从小鼠体内提取出血清抗体该血清抗体只针对H7N9病毒发挥作用所以具有异质性 A正确B. 多次检验后得到的阳性杂交瘤细胞可以冻存起来既可以在体外进行培养也可以在体内（小鼠腹腔）进行培养 B正确C.该单克隆抗体是针对H7N9病毒的具有特异性可与抗原发生特异性反应通过抗原—抗体杂交技术诊断患者 C正确D.杂交瘤细胞是在制备单克隆抗体过程中用骨髓瘤细胞和B淋巴细胞融合而成的两个二倍体细胞融合后产生的杂交细胞是四倍体 D错误故选 D12.有关单克隆抗体的制备下列叙述错误的是（）A. 将抗原注入小鼠体内从骨髓中分离出已免疫的B淋巴细胞B. 用灭活的病毒可促进己免疫的B淋巴细胞与小鼠骨髓瘤细胞融合C. 用抗原﹣﹣抗体杂交法可以筛选出能产生特异性抗体的杂交瘤细胞D. 杂交瘤细胞株的增殖方式和产物分别是有丝分裂和单克隆抗体【答案】A【解析】A、将抗原注入小鼠体内从脾脏中分离出已免疫的B淋巴细胞 A错误B、动物细胞融合过程中可以利用灭活的仙台病毒作诱导剂促进己免疫的B淋巴细胞与小鼠骨髓瘤细胞融合 B正确C、用抗原﹣﹣抗体杂交法可以筛选出能产生特异性抗体的杂交瘤细胞 C正确D、杂交瘤细胞株的增殖方式和产物分别是有丝分裂和单克隆抗体 D正确13.在单克隆抗体制备中关于杂交瘤细胞制备的叙述正确的是（）A. 可利用电激等物理方法直接诱导两个细胞的融合B. 可利用细胞中染色体的数目和形态筛选出杂交瘤细胞C. 利用了细胞膜的流动性和基因重组等原理D. 用选择培养基筛选得到的杂交瘤细胞可直接用于生产单克隆抗体【答案】A【解析】A、单克隆抗体制备中可利用电激等物理方法直接诱导两个细胞的融合 A正确B、因为观察细胞染色体需要对细胞解离等处理细胞已死亡所以不能通过观察染色体的数目和形态筛选杂交瘤细胞 B错误C、利用了细胞膜的流动性和细胞增殖等原理 C错误D、用选择培养基筛选得到的杂交瘤细胞需要再专一抗体检测和克隆化培养后才能用于生产单克隆抗体 D错误14.中科院动物研究所与福州大熊猫研究中心合作通过将大熊猫细胞核植入去核后的兔子卵细胞中在世界上最早克隆出一批大熊猫早期胚胎表明我国大熊猫人工繁殖研究再次走在世界前列．下列描述与克隆无关的是（）A. “复制”B. 无性繁殖C. 细胞培养D. 基因工程【答案】D【解析】A、“复制”属于分子的克隆 A正确B、无性繁殖属于个体水平上的克隆 B正确C、细胞培养属于细胞水平上的克隆 C正确D、基因工程与克隆无关 D错误15.如图是制备乙肝病毒表面抗体的流程下列叙述正确的是（）A. 细胞Ⅰ表示多种T淋巴细胞B. 原生质体融合与过程①均可用灭活病毒处理C. 两次筛选均需进行专一抗体检验D. 图中对小鼠的两次注射目的不同【答案】D【解析】A、图中细胞Ⅰ是从小鼠脾脏中获取已经免疫的B淋巴细胞 A错误B、原生质体融合不能用灭活病毒处理 B错误C、第一次筛选是筛选得到杂交瘤细胞不需要进行专一抗体检验 C错误D、第一次注射的是抗原第二次注射的是产生专一抗体的杂交瘤细胞两次注射目的不同 D正确16.下图表示植物组织培养的简略过程．下列有关叙述不正确的是（）①\xrightarrow\,脱分化②\xrightarrow\,再分化③\rightarrow ④A. 若①是根尖细胞④的叶片为绿色B. 若①是花粉则④是单倍体C. 若提取药物紫草素可从②阶段提取D. 若想制作人工种子应该选用④【答案】D【解析】解 A、若①是根尖细胞则培养的④的叶片的颜色是绿色这说明与叶绿体形成有关的基因表达了 A正确B、若①是花药则④是单倍体植株由于单倍体植株高度不育对其使用秋水仙素处理可获得纯合的正常植株该过程称为单倍体育种 B正确C 、提取药物紫草素可从②愈伤组织阶段提取 C正确D、人工种子是指通过植物组织培养得到的胚状体、不定芽、顶芽和腋芽等为材料通过人工薄膜包装得到的种子所以获得人工种子需将①培养成③ D错误故选 D．17.下列各图表示单克隆抗体制备过程的各阶段图解请根据图分析下列四个选项其中说法不正确的一项是（）A. 单克隆抗体制备过程的顺序是B. H阶段的细胞分泌的抗体能与特定抗原结合C. 单克隆抗体具有能准确识别抗原的细微差异、与特定抗原发生特异性结合并可以大量制备的优点所以可以制成“抗体—药物偶联物”D. 由图中可看出此过程运用的技术手段有动物细胞融合和动物组织培养【答案】D【解析】解 A.单克隆抗体制备过程包括获取免疫能力的效应B细胞和骨髓瘤细胞、诱导细胞融合和筛选出杂交瘤细胞、克隆化培养和抗体检测并筛选出能分泌所需抗体的杂交瘤细胞、体外或体内培养杂交瘤细胞、获取大量单克隆抗体因此按单克隆制备过程的顺序是 A正确B.H阶段的细胞为筛选完的能分泌特定抗体的杂交瘤细胞故分泌的抗体能与特定抗原结合 B正确C.单克隆抗体具有特异性强、灵敏度高并可以大量制备的优点与抗癌药物结合可制成可以制成“生物导弹”（“抗体一药物偶联物”） C正确D.此过程运用的技术手段有动物细胞融合和动物细胞培养 D错误故选 D二、解答题（本大题共计8小题每题10分共计80分）18.（1）根据上述资料分析需要向经过促融处理后的各种细胞混合培养物中加入________ 可使________不能分裂从而筛选出唯一能增殖的________18.（2）上述筛选到的细胞要置于多孔细胞培养板的每一个孔中进行随后的培养转移前通常要将培养细胞稀释到7～10个/mL 每孔中加入0.1mL．这样做的目的是________ 必要时可重复上述过程目的是________18.（3）将获得的上述细胞在多孔培养板上培养用抗原检测从中选择________的杂交瘤细胞在________中培养一段时间后提取单克隆抗体18.（4）若利用生产的单克隆抗体治疗相关病毒（抗原）的感染性疾病请提出相关的治疗思路 ________【答案】氨基蝶呤, 骨髓瘤细胞、骨髓瘤细胞自身融合细胞, 杂交瘤细胞【解析】根据上述资料分析需要向经过促融处理后的各种细胞混合培养物中加入氨基蝶呤可使骨髓瘤细胞、骨髓瘤细胞自身融合细胞不能分裂从而筛选出唯一能增殖的杂交瘤细胞【答案】让每个培养孔中只有一个杂交瘤细胞, 可保证每个培养孔的杂交瘤细胞来自于同一个杂交瘤细胞的克隆【解析】上述筛选到的细胞要置于多孔细胞培养板的每一个孔中进行随后的培养转移前通常要将培养细胞稀释到7～10个/mL 每孔中加入0.1mL．这样做的目的是让每个培养孔中只有一个杂交瘤细胞必要时可重复上述过程目的是可保证每个培养孔的杂交瘤细胞来自于同一个杂交瘤细胞的克隆【答案】能产生特定抗体, 体外或小鼠腹腔【解析】将获得的上述细胞在多孔培养板上培养用抗原检测从中选择能产生特定抗体的杂交瘤细胞在体外或小鼠腹腔中培养一段时间后提取单克隆抗体【答案】利用该单克隆抗体作为药物给患者注射（或将相关抗病毒药物与单克隆抗体结合制成靶向药物给患者注射）【解析】利用生产的单克隆抗体治疗相关病毒（抗原）的感染性疾病的治疗思路利用该单克隆抗体作为药物给患者注射（或将相关抗病毒药物与单克隆抗体结合制成靶向药物给患者注射）19.（1）X、Y、Z细胞的名称分别是_________、_________、_________19.（2）①过程类似于植物组织培养技术中的__________过程与纤维母细胞相比诱导干细胞的全能性较_________ ②过程的实质是__________19.（3）③过程中需用到的生物诱导剂是__________19.（4）④处需要筛选其目的是__________19.（5）用含^32 \mathrmP标记的核苷酸的培养基培养Z细胞细胞质中能测到^32\mathrmP的细胞器有_________和_________【答案】（1）骨髓瘤细胞, 抗原刺激过的B细胞杂交瘤细胞（或浆细胞）, 杂交瘤细胞【解析】解（1）将H1N1病毒外壳蛋白注入乙小鼠获得的是能产生抗体的浆细胞浆细胞与X骨髓瘤细胞融合 Z为杂交瘤细胞【答案】（2）脱分化, 高, 基因的选择性表达【解析】（2）通过①过程已经分化的纤维母细胞失去分化状态变为干细胞类似植物组织培养过程中的脱分化诱导干细胞的分化程度低于纤维母细胞因此其全能性高于后者②过程干细胞基因选择性表达产生不同的组织器官【答案】（3）灭活的病毒【解析】（3）诱导动物细胞融合常用的生物诱导剂是灭活的病毒【答案】（4）筛选出能产生特异性抗体的杂交廇细胞【解析】（4）④是筛选出能产生特异性抗体的杂交瘤细胞【答案】（5）核糖体, 线粒体【解析】（5）核苷酸包括脱氧核苷酸和核糖核苷酸因此能利用^32 \mathrmP标记的核苷酸的细胞器是核糖体和线粒体20.（1）将小鼠骨髓瘤细胞与经抗原免疫过的B细胞融合形成杂交瘤细胞动物细胞融合的诱导因素________、________、电激等对融合的细胞需要进行筛选、克隆培养和________ 进而获得可以产生单克隆抗体的细胞群体20.（2）临床发现利用上述方法产生的抗体具有外源性会被人体免疫系统清除科学家通过分析抗体的结构发现抗体由“C区”和“V区”构成引起人体免疫反应的主要是“C区” 科学家通过“人鼠嵌合抗体”成功解决了这个问题制备“人鼠嵌合抗体”时先从小鼠杂交瘤细胞中提取RNA 在________酶的作用下形成DNA 扩增筛选得到编码抗体________区的基因再与人的________基因通过DNA连接酶连接后插入适当的表达载体导人宿主细胞表达与鼠源单克隆抗体相比“人鼠嵌合抗体”的优点是________【答案】聚乙二醇（PEG）, 灭活病毒, 专一抗体检测【解析】诱导动物细胞融合的方法有聚乙二醇（PEG）、灭活病毒、电激等将小鼠骨髓瘤细胞与经抗原免疫过的B细胞融合形成杂交瘤细胞对融合的细胞需要进行筛选、克隆培养和专一抗体检测而获得可以产生单克隆抗体的细胞群体【答案】逆转录, V, 抗体“C区”, 降低人体对鼠源单克隆抗体的免疫排斥反应提高治疗效果【解析】临床发现利用上述方法产生的抗体具有外源性会被人体免疫系统清除科学家通过分析抗体的结构发现抗体由“C区”和“V区”构成引起人体免疫反应的主要是“C区” 那么我们可以将鼠源抗体的C区换为人的C区操作思路是制备“人鼠嵌合抗体”时先从小鼠杂交瘤细胞中提取RNA 在逆转录酶的作用下形成DNA 扩增筛选得到编码抗体V区的基因再与人的抗体“C区”基因通过DNA连接酶连接后插入适当的表达载体导人宿主细胞表达与鼠源单克隆抗体相比“人鼠嵌合抗体”的优点是降低人体对鼠源单克隆抗体的免疫排斥反应提高治疗效果21.（1）转基因动物通过分泌的乳汁来生产所需要的药品称为乳腺生物反应器将药用蛋白基因与乳腺蛋白基因的_________________________等调控组件重组在一起通过________的方法导入哺乳动物的________中然后使其生长发育成转基因动物21.（2）上述实验前必须给小鼠甲注射H-Y抗原该处理的目的是___________________________21.（3）实践中常用________作为促融剂以提高细胞融合的成功率21.（4）写出以小鼠甲的脾脏为材料制备单细胞悬液的主要实验步骤___________________________________________________________________________ _____21.（5）研究发现在小鼠细胞内DNA的合成有两条途径主要途径（简称D途径）是在细胞内由氨基酸和其他小分子化合物合成核苷酸进而合成DNA的过程该过程中叶酸必不可少氨基蝶呤是叶酸的拮抗剂可以阻断此途径另一辅助途径（简称S途径）是在次黄嘌呤和胸腺嘧啶核苷存在的情况下经酶的催化作用合成DNA 骨髓瘤细胞缺乏催化次黄嘌呤和胸腺嘧啶核苷合成的酶①图中筛选1利用DNA合成途径不同的特点配制的HAT培养基含有多种成分其中添加的______________________具有筛选杂交瘤细胞的作用骨髓瘤细胞及其互相融合细胞不能在HAT培养基上增殖其原因是______________________②图中筛选2含多次筛选筛选所依据的基本原理是____________________ 筛选的目的是_____________________________________________________21.（6）通常选用囊胚期的早期胚胎做雌雄鉴别鉴别后的_____________________胚胎符合要求可做胚胎移植【答案】（1）启动子, 显微注射, 受精卵【解析】解（1）用转基因动物生产药物通常将药用蛋白基因与乳腺蛋白基因的启动子等调控组件重组在一起通过显微注射法等方法导入哺乳动物的受精卵中最终培育的转基因动物进入泌乳期后可以通过分泌的乳汁生产所需要的药物【答案】（2）诱导小鼠甲产生能够分泌H-Y抗体的B淋巴细胞【解析】（2）实验前给小鼠注射H-Y抗原是为了诱导小鼠产生能够分泌抗H-Y抗体的B淋巴细胞【答案】（3）聚乙二醇【解析】（3）为了将B淋巴细胞与骨髓瘤细胞融合实践中常用聚乙二醇（或灭活的病毒）作为促融剂以提高细胞融合的成功率【答案】（4）取小鼠甲脾脏剪碎用胰蛋白酶处理使其分散成单个细胞加入培养液制成单细胞悬液【解析】（4）以小鼠甲的脾脏为材料制备单细胞悬液的主要实验步骤取小鼠甲脾脏剪碎要想动物细胞分散需要用胰蛋白酶处理根据酶的专一性破坏细胞之间的联系使其分散成单个细胞【答案】（5）①氨基蝶呤, 缺乏催化次黄嘌呤和胸腺嘧啶核苷合成的酶无法合成次黄嘌呤和胸腺嘧啶核苷, ②抗体与抗原特异性结合, 获得能产生专一抗体的杂交瘤细胞【解析】（5）①由题干可知氨基嘌呤可以阻断在细胞内由氨基酸和其他小分子化合物合成核苷酸进而合成DNA的过程而骨髓瘤细胞的DNA合成没有其他辅助途径（骨髓瘤细胞缺乏催化次黄嘌呤和胸腺嘧啶核苷合成的酶无法合成次黄嘌呤和胸腺嘧啶核苷）因此配制的HAT培养基中添加氨基嘌呤的成分骨髓瘤细胞及其互相融合细胞不能增殖从而具有筛选杂交瘤细胞的作用②图中筛选2含多次筛选目的是选育出能产生专一抗体的杂交瘤细胞原理是抗原和抗体的特异性结合【答案】（6）雌性【解析】（6）雌性可以产生乳汁通常选用桑椹胚或囊胚期的胚胎做雌雄鉴别鉴别后的雌性胎可做胚胎移植22.（1）简要归纳“离体的细胞”进行植物组织培养获得胚状体的过程离体的细胞\xrightarrow（）愈伤组织\xrightarrow（）胚状体22.（2）“人工种子”大体由胚状体、包埋胚状体的胶质以及人造种皮等三部分组成包埋胚状体的胶质中富含营养物质这是因为________ 为了满足胚状体代谢的需求“人造种皮”则必须具有良好的________性能22.（3）由“人工种子”萌发出的植株是否可育？________22.（4）利用人工种子繁殖后代属于________生殖若利用此项技术制造治疗烫伤、割伤的药物——紫草素培养将进行到________【答案】（1）脱分化, 再分化【解析】解（1）“离体的细胞”进行植物组织培养获得胚状体的过程离体的细胞、组织或器官\xrightarrow脱分化愈伤组织\xrightarrow再分化胚状体\rightarrow 植株【答案】（2）胚状体在萌发初期不能进行光合作用, 透水、透气【解析】（2）胚状体在萌发初期不能进行光合作用因此包埋胚状体的胶质中富含营养。
2023-2024学年高中生物浙科版选修3第2章 克隆技术单元测试(含答案解析)
2023-2024学年浙科版高中生物单元测试班级 __________ 姓名 __________ 考号 __________一、选择题（本大题共计15小题每题3分共计45分）1.阿霉素是一种抗肿瘤抗生素可抑制RNA和DNA的合成阿霉素没有特异性在杀伤肿瘤细胞的同时还会对机体其它细胞造成伤害科研人员将阿霉素与能特异性识别肿瘤抗原的单克隆抗体结合制成抗体–药物偶联物（ADC） ADC通常由抗体、接头和药物三部分组成作用机制如下图所示下列有关叙述错误的是（）A. 阿霉素对肿瘤细胞和已经分化的组织细胞都有杀伤作用B. 将肿瘤抗原反复注射到小鼠体内从小鼠的血清中分离出的抗体为单克隆抗体C. ADC的制成实现了对肿瘤细胞的选择性杀伤D. 利用同位素标记的单克隆抗体在特定组织中的成像技术可定位诊断肿瘤【答案】B【解析】解 A.阿霉素是一种抗肿瘤抗生素可抑制RNA和DNA的合成导致蛋白质也无法合成因此阿霉素对肿瘤细胞和已经分化的组织细胞都有杀伤作用 A正确B.将肿瘤抗原反复注射到小鼠体内从小鼠的血清中分离出的抗体并不是单克隆抗体 B 错误C.将阿霉素与能特异性识别肿瘤抗原的单克隆抗体结合制成抗体–药物偶联物（ADC）实现了对肿瘤细胞的选择性杀伤 C正确D.利用同位素标记的单克隆抗体在特定组织中的成像技术可定位诊断肿瘤 D正确故选 B2.如图为制备人工种子部分流程示意图下列叙述正确的是（）A. 胚状体是外植体在培养基上脱分化形成的一团愈伤组织B. 该过程以海藻酸钠作为营养成分以\ CaCl_2溶液作为凝固剂C. 可在海藻酸钠溶液中添加蔗糖为胚状体提供碳源D. 包埋胚状体的凝胶珠能够隔绝空气有利于人工种子的储藏【答案】C【解析】A、外植体在培养基上脱分化形成的愈伤组织再分化才能形成胚状体 A错误B、该过程以海藻酸钠作为包埋材料以CaCl_2溶液作为凝固剂 B错误C、可在海藻酸钠溶液中添加蔗糖为胚状体提供碳源和能量 C正确D、包埋胚状体的凝胶珠能够保护胚状体有利于人工种子的储藏 D错误3.以H1N1病毒核衣壳蛋白为抗原采用细胞工程制备单克隆抗体下列叙述正确的是（）A. 用纯化的核衣壳蛋白反复注射到小鼠体内产生的血清为单克隆抗体B. 体外培养单个B淋巴细胞可以获得大量针对H1N1病毒的单克隆抗体C. 将等量B淋巴细胞和骨髓瘤细胞混合经诱导融合后的细胞均为杂交瘤细胞D. 利用该单克隆抗体与H1N1病毒核衣壳蛋白特异性结合的方法可以诊断该病毒感染体【答案】D【解析】4.利用细胞工程方法以H7N9病毒颗粒外膜表面糖蛋白为抗原制备出单克隆抗体下列相关叙述错误的是（）A. 用纯化的糖蛋白反复注射到小鼠体内提取的血清抗体具有异质性B. 经多次抗体检验得到的阳性杂交瘤细胞可注射到小鼠腹水中继续培养C. 利用该单克隆抗体通过抗原—抗体杂交技术可诊断出H7N9病毒感染者D. 经灭活的仙台病毒诱导融合并筛选得到的杂交瘤细胞具有二倍体核型【答案】D【解析】解 A. 将纯化的糖蛋白作为抗原反复注射到小鼠的体内一段时间后从小鼠体内提取出血清抗体该血清抗体只针对H7N9病毒发挥作用所以具有异质性 A正确B. 多次检验后得到的阳性杂交瘤细胞可以冻存起来既可以在体外进行培养也可以在体内（小鼠腹腔）进行培养 B正确C.该单克隆抗体是针对H7N9病毒的具有特异性可与抗原发生特异性反应通过抗原—抗体杂交技术诊断患者 C正确D.杂交瘤细胞是在制备单克隆抗体过程中用骨髓瘤细胞和B淋巴细胞融合而成的两个二倍体细胞融合后产生的杂交细胞是四倍体 D错误故选 D5.图为制备人工种子部分流程示意图下列有关叙述正确的是（）A. 外植体需要经过脱分化、再分化等才能形成胚状体B. 在海藻酸钠溶液中可加入适量的无机盐、有机碳源以及农药、抗生素等C. 在人工种皮中加入植物生长调节剂就能促进胚状体的生长发育D. 包埋胚状体的凝胶珠会隔绝空气有利于人工种子的储藏【答案】B【解析】 A 、外植体需要经过脱分化、再分化等才能形成胚状体 A错误B、在海藻酸钠溶液中可加入适量的无机盐、有机碳源以及农药、抗生素等 B正确C 、在人工种皮中加入植物生长调节剂能调节胚状体的生长发育 C错误D、凝胶珠不能隔绝空气 D错误6.HER-2（一种跨膜糖蛋白）高表达通常只出现于胎儿期正常人体组织中其表达量极低在乳腺细胞中高表达可加速细胞分裂使其增殖、分化过程失衡最终转变为乳腺癌抗体–药物偶联物T-DM1由曲妥珠单抗和药物DM1连接而成对HER-2有很强的靶向性下列相关分析错误的是（）A. 胎儿期HER-2高表达可能与胚胎发育有关成人HER-2含量高可为乳腺癌诊断提供参考B. 因为获取的B淋巴细胞可能有多种所以曲妥珠单抗的制备过程需进行专一抗体阳性检测C. DM1与HER-2特异性结合是T-DM1对癌细胞进行选择性杀伤的关键D. 利用同位素标记的曲妥珠单抗在乳腺组织中成像的技术可定位诊断肿瘤的位置【答案】C【解析】解 A.胎儿期HER-2高表达可能与胚胎发育有关成人含量HER-2高可为乳腺癌诊断提供参考 A正确B.获取的B淋巴细胞可能有多种所以曲妥珠单抗的制备过程需进行专一抗体阳性检测B正确C.曲妥珠单抗与HER-2特异性结合是T-DM1对癌细胞进行选择性杀伤的关键 C错误D.利用同位素标记的曲妥珠单抗在乳腺组织中成像的技术可定位诊断肿瘤的位置 D正确故选 C7.某兴趣小组拟用植物组织培养技术繁殖一种名贵花卉其技术路线为“取材→消毒→愈伤组织培养→出芽→生根→移栽” 下列有关叙述错误的是（）A. 消毒的原则是既杀死材料表面的微生物又减少消毒剂对细胞的伤害B. 在愈伤组织培养中加入细胞融合的诱导剂可获得染色体加倍的细胞C. 出芽是细胞再分化的结果受基因选择性表达的调控D. 生根时培养基通常应含α-萘乙酸等生长素类调节剂【答案】B【解析】解 A在植物组织培养中消毒的原则是既杀死材料表面的微生物又减少消毒剂对细胞的伤害要选择适当的消毒剂及浓度 A正确B.愈伤组织细胞含有细胞壁加入细胞融合诱导剂不会诱导细胞融合因此不会得到染色体加倍的细胞 B错误C.愈伤组织再分化形成胚状体后一般先形成芽再形成根而分化的实质是基因的选择性表达 C正确D.α-萘乙酸是生产中常用的植物生长素类似物具有促进植物生根的功能 D正确故选 B8.组培苗在移栽前一般需要炼苗目的在于提高组培苗对外界环境条件的适应性下图是不同炼苗方式组合对白兰地红枫组培苗移栽存活率的影响以下叙述正确的是（）闭盖炼苗 T_1=3d, T_2=6d, T_3=12d开盖炼苗 t_1=1d, t_2=2d, t_3=3d不同炼苗方式组合\left( T+t\right)对白兰地红枫移栽存活率的影响A. 炼苗初期应创设与室外环境相同的条件促使组培苗适应环境B. 闭盖炼苗期间应控制的环境条件为无菌、无光并保持一定的湿度C. 白兰地红枫组培苗的最佳炼苗方式是闭盖锻炼6d后再开盖炼苗3dD. 白兰地红枫组培苗的移栽存活率随着闭盖锻炼时间的增加先升高后降低【答案】D【解析】解 A.炼苗开始数天内应和培养时的环境条件相似炼苗后期则要与预计的栽培条件相似从而达到逐步适应的目的 A错误B.组织培养期间需要无菌操作长成幼苗后不需要无菌的环境条件了 B错误C.分析题图可知白兰地红枫组培苗的最佳炼苗方式是闭盖锻炼6d后再开盖炼苗2d C 错误D.分析题图白兰地红枫组培苗的移栽存活率随着闭盖锻炼时间的增加先升高后降低 D 正确故选 D9.单克隆抗体制备过程中要用特定的培养基筛选出杂交瘤细胞在这种培养基上不能存活、增殖的细胞有（）①B淋巴细胞②小鼠骨髓瘤细胞③B淋巴细胞自身融合细胞④小鼠的骨髓瘤细胞自身融合的细胞⑤杂交瘤细胞A. ①②③④B. ①②③C. ②④D. ②③【答案】A【解析】特定培养基培养细胞的目的是筛选出既能大量增殖、又能产生特异性抗体的杂交瘤细胞因此其他的细胞都不能在此培养基上生存所以在这种培养基上不能存活和繁殖的细胞有①B淋巴细胞、②小鼠的骨髓瘤细胞、③B淋巴细胞自身融合的细胞、④小鼠的骨髓瘤细胞自身融合的细胞故选A10.下列关于现代生物技术应用的叙述正确的是（）A. 基因治疗可以把缺陷基因诱变成正常基因B. “生物导弹”治疗癌症是利用单克隆抗体特异性对抗癌细胞C. ES细胞可以治疗人类的阑尾炎、过敏性反应等D. 利用植物细胞工程技术可以生产无花叶病毒的甘蔗等作物新品种【答案】D【解析】解 A、基因治疗是把正常基因导入病人体内有缺陷的细胞中使该基因的表达产物发挥功能而不是把缺陷基因诱变成正常基因 A错误B、“生物导弹”治疗癌症是借助了单克隆抗体的导向作用 B错误C 、人类的阑尾炎只需手术切除即可不需要使用ES细胞来治疗 C错误D、利用植物组织培养技术可以生产无病毒的甘蔗等作物新品种 D正确．故选 D．11.人绒毛膜促性腺激素（HCG）是女性怀孕后胎盘滋养层细胞的分泌的一种糖蛋白制备抗HCG单克隆抗体可用于早孕的诊断如图是抗HCG单克隆抗体制备流程示意图相关叙述正确的是（）A. ①过程所用的促融方法与植物体细胞杂交完全相同B. ③过程以抗HCG单克隆抗体为抗原进行检测C. 应给小鼠多次注射HCG 以获得较多的记忆细胞D. ④过程需要添加抗生素等物质以防止病毒污染【答案】B【解析】A项①过程为诱导动物细胞融合所用的促融方法与植物体细胞杂交不同B项③过程的目的是筛选出能产生抗HCG抗体的杂交瘤细胞需以抗HCG单克隆抗体为抗原进行检测C项给小鼠注射HCG的目的是获得免疫的B淋巴细胞D项④过程为采用动物细胞培养技术在体外生产抗HCG单克隆抗体需要添加抗生素等物质以防止杂菌污染故选B12.下图是利用基因工程培育抗虫植物的示意图以下相关叙述不正确的是（）A. ①→②利用两种不同限制酶处理能避免含抗虫基因的DNA片段自身环化B. ②→③可用氯化钙处理农杆菌有助于促进重组Ti质粒转化到农杆菌细胞中C. ③→④用农杆菌侵染植物细胞重组Ti质粒整合到植物细胞的染色体上D. ④→⑤用植物组织培养技术培养利用了植物细胞的全能性【答案】C【解析】解A.①→②利用两种不同限制酶处理能避免含抗虫基因的DNA片段自身环化 A正确B.②→③可用氯化钙处理农杆菌使之成为易于吸收周围环境中DNA分子的感受态这样有助于促进重组Ti质粒转化到农杆菌细胞中 B正确C.③→④用农杆菌侵染植物细胞重组Ti质粒的T-DNA片段整合到植物细胞的染色体上 C错误D.④→⑤用植物组织培养技术培养原理是植物细胞的全能性 D正确故选 C13.如图表示抗人体胃癌的单克隆抗体的制备过程有关叙述不正确的是（）A. 图中实验小鼠注射的甲是能与抗人体胃癌抗体特异性结合的抗原B. 利用聚乙二醇、灭活的病毒和电激等方法均可诱导细胞融合获得乙C. 用特定的选择培养基对乙筛选融合细胞均能生长未融合细胞均不能生长D. 丙需要进行克隆化培养和抗体检测经多次筛选后可获得大量能分泌所需抗体的丁【答案】C【解析】A．图中从注射了甲的小鼠中获得的B淋巴细胞可以产生特异性抗体所以甲为相应的抗原 A正确B．利用聚乙二醇（化学法）、灭活的病毒（生物法）和电激（物理法）等方法均可诱导细胞融合获得杂种细胞 B正确C．用特定的选择培养基对乙筛选同种细胞的融合细胞和未融合细胞均不能生长只有融合的杂种细胞（杂交瘤细胞）能生长 C错误D．杂交瘤细胞经过选择培养基的筛选并经单一抗体检测呈阳性后培养能产生特定抗体的杂交瘤细胞即可制备单克隆抗体 D正确故选C14.下列生物技术的应用实例中运用的原理与其他各项不同的是（）A. 利用花药离体培养获得单倍体植株B. 通过植物组织培养获得人工种子并发育成植株C. 将转入贮存蛋白基因的向日葵细胞培育成植株D. 利用生物反应器悬浮培养人参细胞提取人参皂甙【答案】D【解析】A项利用花药离体培养获得单倍体植株利用了植物组织培养技术原理为植物细胞的全能性B项通过植物组织培养获得人工种子并发育成植株利用了植物组织培养技术原理为植物细胞的全能性C项将转入贮存蛋白基因的向日葵细胞培育成植株利用了植物组织培养技术原理为植物细胞的全能性D项利用生物反应器悬浮培养人参细胞提取人参皂甙利用的是细胞增殖的原理所以D 项运用的原理与其他各项不同故选D15.单克隆抗体技术在疾病诊断和治疗以及生命科学研究中具有广泛的应用下列关于单克隆抗体的叙述错误的是（）A. 特异性强灵敏度高B. 与抗癌药物结合可制成“生物导弹”C. 体外培养B淋巴细胞可大量分泌单克隆抗体D. 由浆细胞与骨髓瘤细胞融合而成的杂交瘤细胞分泌【答案】C【解析】A．单克隆抗体的优点是特异性强、灵敏度高、可大量制备 A正确B．制备的单克隆抗体与抗癌药物结合可制成“生物导弹” 直接作用于癌细胞 B正确C、D．经过免疫的B淋巴细胞（浆细胞）与骨髓瘤细胞融合而成的杂交瘤细胞能在体外培养的条件下大量增殖并分泌单克隆抗体 C错误、D正确故选C二、解答题（本大题共计6小题每题10分共计60分）16.（1）题中“荧光RT—PCR技术”所用的试剂盒中应该含有检测者的mRNA、_________、________、逆转录酶、引物、四种脱氧核苷酸、缓冲体系16.（2）如果要同时扩增两种基因则试剂盒中的引物应该有________种引物的作用是______________________________________________________________________________16.（3）理论上在检测过程中有荧光标记的“杂交双链”出现则说明检测结果呈________（填“阴”或“阳”）性但为了保证检测结果的准确性一般要达到或超过阈值时才确诊现有甲乙两个待检样本检测时都出现了上述形态的曲线但甲的a点比乙的a点明显左移请给这种结果做出科学合理的解释（试剂盒合格且正富操作过程规范且准确）______________________________________________________________________________16.（4）除核酸检测外研究人员还研发了利用生产单克隆抗体的技术生产出能与新冠病毒结合的特异性抗体当杂交瘤细胞在体外条件下做大规模培养时为防止细胞代谢产物积累对细胞自身造成危害应采取的措施是_____________________________【答案】（1）耐高温的DNA聚合酶, 荧光标记的新冠病毒核酸探针【解析】解（1）根据题目信息可知荧光PCR法基本原理是将新冠病毒mRNA逆转录为DNA 通过采用多重荧光RT-PCR技术监测因此荧光PCR法所用的试剂盒中通常都应该含有耐高温的DNA聚合酶、荧光标记的新冠病毒核酸探针、逆转录酶、引物、四种脱氧核苷酸、缓冲体系【答案】（2）4, 使DNA聚合酶能够从引物的3´端开始连接脱氧核苷酸【解析】（2）根据PCR技术的原理在扩增DNA分子时每一条链均需与相应的引物结合才能进行扩增因此如果要成功将上述两种基因同时扩增出来则在试剂盒中的引物应该有4种引物是一小段单链DNA或RNA 作为DNA复制的起始点作用是使热稳定DNA聚合酶能够从引物的3´端开始连接脱氧核苷酸【答案】（3）阳, 甲样本中的新冠病毒含量更高达到阈值所需的循环数更少【解析】（3）理论上在检测过程中有荧光标记的“杂交双链”出现则说明检测结果呈阳性为了保证检测结果的准确性一般要达到或超过阈值时才确诊检测结果甲的a点比乙的a点明显左移说明甲样本中的新冠病毒含量更高达到阈值所需的循环数更少【答案】（4）定期更换培养液【解析】（4）当杂交瘤细胞在体外条件下做大规模培养时为防止细胞代谢产物积累对细胞自身造成危害应采取的措施是定期更换培养液最后所得抗体是单克隆抗体优点是特异性强、灵敏度高、可大量制备17.（1）生产单克隆抗体时一般不直接利用效应B淋巴细胞主要原因是________．17.（2）①过程中常用的与植物细胞融合相同的诱导剂是________．17.（3）②过程为________培养培养液中还需通入气体物质其中通入CO_2的目的是________．17.（4）④表示杂交瘤细胞的扩大培养其既可在体外培养也可注射到小鼠________内培养．体内培养与体外培养相比其优点是________．17.（5）为了从培养液中分离出杂交瘤细胞需在③过程向培养液中加入________ 理由是杂交瘤细胞可利用________途径合成DNA 从而才能在该培养液中增殖．【答案】B淋巴细胞不能无限增殖【解析】B淋巴细胞不能无限增殖故生产单克隆抗体时一般不直接利用一个B淋巴细胞．【答案】聚乙二醇【解析】）①诱导动物细胞融合常用的与植物细胞融合相同的诱导剂是聚乙二醇．【答案】细胞, 维持培养液的pH【解析】②过程为细胞培养培养液中还需通入气体其中通入CO_2的目的主要是维持培养液的pH．【答案】腹腔, 无需再提供营养物质以及无菌环境操作简便．【解析】④表示杂交瘤细胞的扩大培养其既可在体外培养也可注射到小鼠腹腔内培养．体内培养与体外培养相比其优点是无需再提供营养物质以及无菌环境操作简便．【答案】氨基嘌呤, 淋巴细胞中的S【解析】在培养液中加入氨基嘌呤培养后收集增殖细胞．加入氨基嘌呤后使D合成途径阻断仅有D全合成途径的骨髓细胞及其彼此融合的细胞就不能增殖淋巴细胞一般不分裂增殖．因此培养液中加入该物质后不能生长的细胞有骨髓瘤细胞、融合的淋巴细胞、融合的骨髓瘤细胞、淋巴细胞．人淋巴细胞和骨髓癌细胞融合后的杂种细胞中可以利用淋巴细胞中的S途径合成DNA而增殖．18.（1）克隆动物是_______________（填“有性”或“无性”）生殖的产物18.（2）克隆产生的动物表现型与体细胞供体动物接近但不是对体细胞供体动物100%的复制理由是_____________________18.（3）为了提高动物细胞克隆成功率可以选择适宜的培养基添加___________等天然成分以滋养细胞支持其生长克隆动物除了用到动物细胞工程的技术手段外还必须用到______________、_______________等胚胎工程技术手段18.（4）若要鉴别XY型性别决定的动物胚胎的性别可以获取该种动物囊胚的_____________细胞并提取其DNA 用Y染色体上特异性DNA做探针检测这种技术称为_____________________【答案】（1）无性【解析】解（1）克隆动物的培育不需要两性生殖细胞的结合属于无性生殖【答案】（2）克隆动物绝大部分DNA来自于供体细胞核但其核外还有少量的DNA （线粒体中的 DNA）是来自于受体卵母细胞此外核供体动物生活的环境与克隆动物所生活的环境不会完全相同【解析】（2）克隆产生的动物表现型与体细胞供体动物接近但不是对体细胞供体动物100%的复制原因是克隆动物绝大部分DNA来自于供体细胞核但其核外还有少量的DNA （线粒体中的DNA）是来自于受体卵母细胞此外核供体动物生活的环境与克隆动物所生活的环境不会完全相同【答案】（3）动物血清, 早期胚胎培养, 胚胎移植技术【解析】（3）为了提高动物细胞克隆成功率可以选择适宜的培养基添加动物血清等天然成分以滋养细胞支持其生长克隆动物除了用到动物细胞工程的技术手段外还必须用到早期胚胎培养、胚胎移植技术等胚胎工程技术手段【答案】（4）滋养层, DNA分子杂交技术【解析】（4）若要鉴别XY型性别决定的动物胚胎的性别可以采用DNA分子杂交技术获取该种动物囊胚的滋养层细胞并提取其DNA 用Y染色体上特异性DNA做探针检测19.（1）图中cDNA文库________基因组文库（大于、等于、小于）19.（2）为在短时间内大量获得目的基因可用的技术是________ 其原理________ 19.（3）目的基因获取之后需要进行________ 此步骤是基因工程的核心目的是19.（4）将该目的基因导入某双子叶植物细胞常采用的方法是农杆菌转化法其能否在此植物体内稳定遗传的关键是________【答案】【解析】【答案】【解析】【答案】【解析】【答案】【解析】20.（1）制备特异性的单克隆抗体首先要对小鼠注射特定________20.（2）从单克隆抗体制备过程可以看出它运用了________（至少两种）等细胞工程技术20.（3）若只考虑细胞的两两融合理论上融合的细胞有________种类型符合要求的细胞的特点是________20.（4）体外培养杂交瘤细胞的培养基与用于植物组织培养的培养基在物理性质上的主要区别是前者要用________ 从培养基的成分来看前者培养基中除了添加必要的营养物质外还需要加入含天然成分的________20.（5）特异性抗体也可以通过常规方法来制备向动物反复注射某种抗原使动物产生抗体然后从动物的血清中分离所需抗体与这种常规方法相比单克隆抗体具有________等优点（答出三点即可）【答案】（1）抗原【解析】（1）在制备单克隆抗体时首先要对小鼠注射特定的抗原刺上激小鼠产生相应的B淋巴细胞（浆细胞）【答案】（2）动物细胞培养, 动物细胞融合【解析】（2）单克隆抗体制备过程运用了动物细胞培养和动物细胞融合等动物细胞工程技术【答案】（3）3, 能产生特异性抗体和无限增殖【解析】（3）动物细胞两两融合融合的细胞一共有3种其中只有杂交瘤细胞符合要求其具有无限增殖和产生特异性抗体的特点【答案】（4）液体培养基, 动物血清, 血浆【解析】（4）体外培养杂了交瘤细胞使用的是液体培养基该培养基中需要加入含天然成分的上动物血清、血浆【答案】（5）特异性强, 灵敏度高, 能大量制备【解析】（5）单克隆抗体具有特异性强、灵敏度高、能大量制备等优点21.（1）长期以来人们获得抗体的方法是向动物体内反复注射某种抗原然后从动物的________中分离出抗体用这种方法制得抗体的缺点是________21.（2）本次基因敲除技术的受体细胞不是小鼠的受精卵而是________细胞再将处理过的该细胞重新植入小鼠脾脏中增殖21.（3）动物细胞融合技术常用的诱导因素有聚乙二醇、电激和________ 培养液中共有________种类型的融合细胞（只考虑细胞两两融合的情况）可通过________培养基获得杂交瘤细胞21.（4）利用上述技术生产的单克隆抗体可制作成诊断盒用于准确、快速诊断H7N9病毒感染者这种诊断运用了________杂交技术。
人教版高中生物选择性必修3第2章细胞工程章末整合提升练含答案
章末整合提升发展路径1认同细胞工程的发展是一代又一代科学家经过长时间探索的结果,其中我国科学家发挥了重要作用。
问题情境1.据报道,我国科学家以一只雌性猕猴胎儿成纤维细胞为核供体成功克隆出两只猕猴,成为全球首例体细胞克隆灵长类动物。
下列相关叙述错误的是()A.这项技术证明高度分化的动物体细胞具有全能性B.研究中需要使用MⅡ期卵母细胞作为核移植的受体C.培育出的两只雌性猕猴的基因组成几乎完全相同D.克隆猕猴的成功为大量培育灵长类遗传病模型动物提供了可能解析:克隆的两只猕猴是以一只雌性猕猴胎儿成纤维细胞为核供体发育而来的,该过程经过了细胞核移植,只能说明高度分化的动物体细胞的细胞核具有全能性。
卵母细胞含有丰富的营养物质,且体积比较大,容易操作;同时卵母细胞的细胞质中存在促进细胞核全能性表达的物质,因此研究中需要使用培养至MⅡ期的卵母细胞作为核移植的受体;同一个体所有细胞细胞核的基因组成是相同的,因此培育出的两只雌性猕猴的基因组成几乎完全相同;克隆猕猴的成功为大量培育灵长类遗传病模型动物提供了可能。
答案:A素养阐释1.在20世纪70年代,我国科学家就运用核移植技术,将鲤鱼囊胚细胞的细胞核移植到已去核的鲫鱼未受精卵中,培育出了鲤鲫移核鱼,该种鱼兼有鲫鱼和鲤鱼的特点;本题所述的体细胞克隆猴的问世,是我国科学家多年努力的结果,是全球首例体细胞克隆灵长类动物。
由此可见,在核移植技术研究方面,我国科学家发挥了重要作用。
2.不仅在核移植技术方面,在细胞工程的其他方面,我国科学家也发挥了举足轻重的作用。
细胞工程的发展是一代又一代科学家长时间探索的结果。
发展路径2认同细胞工程在农业生产、医疗卫生等方面发挥了重要作用。
问题情境2.下列关于细胞工程技术在生产实践中的应用的叙述,错误的是()A.愈伤组织细胞分裂能力强,可作为体细胞诱变育种的材料B.植物花粉粒细胞容易获取,可作为培育单倍体植株的理想材料C.自体皮肤生发层细胞分裂能力强且不会排斥,可作为烧伤患者皮肤移植母细胞D.动物胚胎细胞分化程度低、易恢复全能性,可优先作为动物体细胞核移植的受体细胞解析:愈伤组织细胞分裂能力强,可作为体细胞诱变育种的材料;植物花粉粒细胞容易获取,可作为培育单倍体植株的理想材料;自体皮肤生发层细胞分裂能力强且不会排斥,可作为烧伤患者皮肤移植母细胞;动物体细胞核移植的受体细胞一般是去核的卵母细胞。
CCNA第一学期各章习题和参考答案.doc
CCNA第一学期各章习题和参考答案第二章网络通信1、TCP/IP网络接入层有何作用？A路径确定和数据包交换B数据表示、编码和控制C可靠性、流量控制和错误检测E将数据段划分为数据包2、下列哪些陈述正确指出了中间设备在网络中的作用？（选择三项）B发起数据通信D发送数据流F数据流最后的终止点3、下列哪三项陈述是对局域网(LAN) 最准确的描述？（选择三项）C LAN 中的不同网段之间一般通过租用连接的方式连接。
D此类网络的安全和访问控制由服务提供商控制。
F此类网络的每个终端通常都连接到电信服务提供商(TSP)。
004 什么是PDU？A传输期间的帧损坏B在目的设备上重组的数据C因通信丢失而重新传输的数据包005 OSI 模型哪两层的功能与TCP/IP 模型的网络接入层相同？（选择两项）018哪个应用层协议通常用于支持客户端与服务器之间的文件传输？A HTMLB HTTP D Telnet019哪个应用层协议中规定了Microsoft 网络中用于文件共享的服务？A DHCPB DNS D SMTP E Telnet020服务器上的应用层通常如何处理多客户端服务请求？A终止与服务的所有连接B拒绝与单一守护程序的多个连接C暂停当前连接，建立新连接第四章OSI传输层001下列哪两项是用户数据报协议(UDP) 的功能？（选择两项）A流量控制 D 面向连接E 序列和确认002请参见图示。
此Wireshark 捕获输出的第7 行中执行的是哪一项TCP 操作？A会话创建B 数据段重传C 数据传输D 会话断开003数据段的TCP 报头中为什么包含端口号？A指示转发数据段时应使用的正确路由器接口B 标识接收或转发数据段时应使用的交换机端口C确定封装数据时应使用的第3 层协议让接收主机转发数据到适当的应用程序E让接收主机以正确的顺序组装数据包004OSI 模型哪一层负责规范信息从源设备到目的设备准确可靠地流动？A应用层 B 表示层 C 会话层传输层 E 网络层005请参见图示。
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The switch will issue an ARP request to confirm that the source exists.
The switch will map the source MAC address to the port on which it was received.
The command will cause the messageAuthorized personnel Onlyto display before a user logs in.
The command will generate the error message% Ambiguous command: "banner motd" ”to be displayed.
Layer 2 switches can send traffic based on the destination MAC address.
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12
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Which two statements are true about EXEC mode passwords? (Choose two.)
1
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Refer to the exhibit. How many collision domains are depicted in the network?
1
2
4
6
7
8
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2
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Refer to the exhibit. The switch and workstation are administratively configured for full-duplex operation. Which statement accurately reflects the operation of this link?
Enable CDP on the switch.
Change passwords regularly.
Turn off unnecessary services.
Enable the HTTP server on the switch.
Use the enable password rather than the enable secret password.
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14
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Refer to the exhibit. The exhibit shows partial output of theshow running-configcommand. The enable password on this switch is "cisco." What can be determined from the output shown?
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10
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Which command line interface (CLI) mode allows users to configure switch parameters, such as the hostname and password?
user EXEC mode
privileged EXEC mode
The hosts extend their delay period to allow for rapid transmission.
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9
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Where is the startup configuration stored?
DRAMห้องสมุดไป่ตู้
NVRAM
ROM
startup-config.text
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7
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Which statement is true about the commandbanner login "Authorized personnel Only"issued on a switch?
The command is entered in privileged EXEC mode.
incorrect vty lines configured
incorrect default gateway address
incompatible Secure Shell version
missingtransport input sshcommand
vty lines that are configured to allow only Telnet
The switch ends an acknowledgement frame to the source MAC of this incoming frame.
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5
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What are two ways to make a switch less vulnerable to attacks like MAC address flooding, CDP attacks, and Telnet attacks? (Choose two.)
The hosts return to a listen-before-transmit mode.
The hosts creating the collision have priority to send data.
The hosts creating the collision retransmit the last 16 frames.
This line represents most secure privileged EXEC mode password possible.
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15
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Which two statements are true regarding switch port security? (Choose two.)
The command will cause the messageEnd with the character “%”to be displayed after the command is entered into the switch.
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8
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When a collision occurs in a network using CSMA/CD, how do hosts with data to transmit respond after the backoff period has expired?
A username/password combination is no longer needed to establish a secure remote connection to the switch.
The switch requires remote connections via proprietary client software.
The enable secret password command stores the configured password in plain text.
The enable secret password command provides better security than the enable password.
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6
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Refer to the exhibit. What action does SW1 take on a frame sent from PC_A to PC_C if the MAC address table of SW1 is empty?
SW1 drops the frame.
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4
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When a switch receives a frame and thesourceMAC address is not found in the switching table, what action will be taken by the switch to process the incoming frame?
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3
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What happens when thetransport input sshcommand is entered on the switch vty lines?
The SSH client on the switch is enabled.
Communication between the switch and remote users is encrypted.
The enable password is encrypted by default.
An MD5 hashing algorithm was used on all encrypted passwords.
Any configured line mode passwords will be encrypted in this configuration.
Best practices require both the enable password and enable secret password to be configured and used simultaneously.
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13
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Refer to the exhibit. The network administrator has decided to allow only Secure Shell connections to Switch1. After the commands are applied, the administrator is able to connect to Switch1 using both Secure Shell and Telnet. What is most likely the problem?