Minireview
Regulation of the cholinergic gene locus by the repressor
element-1silencing transcription factor/neuron restrictive
silencer factor (REST/NRSF)
Masahito Shimojo,Louis B.Hersh *
Department of Molecular and Cellular Biochemistry,University of Kentucky,
Chandler Medical Center,800Rose Street,Lexington,KY 40536-0298,USA
Received 17July 2003;accepted 21August 2003
Abstract
The cholinergic gene locus is comprised of two genes,the choline acetyltransferase gene and the vesicular acetylcholine transporter gene.The vesicular acetylcholine transporter gene is located within the first intron of the choline acetyltransferase gene.This arrangement permits coordinate regulation of the locus.Protein kinase A regulates expression of the cholinergic gene locus in PC12cells.This regulation was found to be dependent on the presence of a 21-bp DNA sequence known as the repressor element-1(RE-1)/neuron-restrictive silencer element (NRSE).Repressor element-1silencing transcription factor (REST)/neuron-restrictive silencer factor (NRSF),which binds to the RE-1/NRSE,is a zinc finger containing transcriptional repressor that blocks the expression of many neuronal RE-1/NRSE containing genes in nonneuronal cells.However,REST/NRSF expression has also been observed in neurons as well as the PC12cell line used in these studies.REST/NRSF truncated isoforms were expressed in neuronal cells,suggesting they also function in regulating neuronal gene expression.A study of REST4,one of the REST/NRSF isoforms,suggests that it regulates transcription of the cholinergic gene locus by blocking the repressor activity of REST/NRSF.Protein kinase A regulation of the cholinergic gene locus in PC12cells can thus be attributed,at least in part,to increased synthesis of REST4,which in turn derepresses the repressor activity of REST/NRSF.
D 2004Elsevier Inc.All rights reserved.
Keywords:Gene transcription;Gene regulation;Cholinergic gene;Transcriptional repression
0024-3205/$-see front matter D 2004Elsevier Inc.All rights reserved.doi:10.1016/j.lfs.2003.08.045
*Corresponding author.Tel.:+1-859-323-5549;fax:+1-859-323-1727.
E-mail address:lhersh@https://www.doczj.com/doc/4412911093.html, (L.B.Hersh).
https://www.doczj.com/doc/4412911093.html,/locate/lifescie
Life Sciences 74(2004)
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Introduction
The ability of a cell to regulate its phenotype is dependent upon its ability to carefully and precisely control which genes are expressed and which are not.In the case of neurons there are many genes whose expression is necessary to produce a neuronal phenotype,but whose expression outside the central nervous system is not needed and likely to be detrimental.Recent evidence suggests that one mechanism utilized to prevent expression of neuronal genes outside the central nervous system is that of gene repression.This review focuses on the regulation of one neuron specific gene locus,the cholinergic gene locus,by transcriptional repression.In addition to describing the regulation of the cholinergic gene locus,the complexity of this system,its application to other neuronal genes,and the many questions yet to be answered are noted.
Cholinergic neurotransmission in the central nervous system is a key process dependent on the coexpression of proteins involved in the synthesis,storage,and release of the neurotransmitter acetylcholine (ACh).ACh plays an important role in fundamental brain processes such as memory,learning,and sleep (Karczmar,1976;Bartus et al.,1982;Aigner and Mishkin,1986).ACh is used in a cyclic process involving the uptake of choline by a high-affinity uptake system (Jope,1979),the formation of ACh through acetylation of choline catalyzed by the enzyme choline acetyltransferase (ChAT)(EC 2.3.1.6)and the transport of ACh into synaptic vesicles by the vesicular ACh transporter (V AChT).Upon stimulation,synaptic vesicles fuse with the plasma membrane and release ACh,which can then be hydrolyzed by acetylcholinesterase to produce free choline,thereby completing the cycle.The cholinergic gene locus is comprised of both the choline acetyltransferase gene as well as the vesicular acetylcholine transporter gene,the latter being located within the first intron of the ChAT gene.ChAT is primarily if not exclusively a cytosolic protein whereas V AChT has 12transmembrane domains and is integrated in the membrane of synaptic vesicles.In the CNS,ChAT and V AChT are both required for cholinergic neurotransmission.In some neurodegenerative disorders,i.e.Alzheimer’s disease,amyotrophic lateral sclerosis,and schizophrenia,there is a dysfunction of central cholinergic neurons involving a loss of,or abnormality in,basal forebrain ChAT activity (Davies and Maloney,1976;Coyle et al.,1983;Price,1986;Wu and Hersh,1994).
Genomic structure and transcription of the cholinergic gene locus
The genes for mouse (Misawa et al.,1992;Pu et al.,1993),rat (Brice et al.,1989;Ishii et al.,1990),and human ChAT (Kong et al.,1989;Hersh et al.,1993)have been isolated and shown to contain three 5V -noncoding exons;exon 1,referred to as the R exon;exon 2,referred to as the N exon;and exon 3,referred to as the M exon (Misawa et al.,1992;Kengaku et al.,1993),Fig.1.The alternative splicing of these three exons results in multiple 5V -mRNA species,which are transcribed from three different promoters upstream of the R,N,and M exons,Fig.1.In the mammalian gene,the M exon resides f 5kb downstream from the R exon and transcription from its promoter appears to represent the major promoter used for generating ChAT mRNA.The V AChT gene in mammalian species (Bejanin et al.,1994;Erickson et al.,1994)as well as C.elegans (Alfonso et al.,1993)was found to lie uniquely within the first intron of the cholinergic gene locus between the first (R)and second (N)non-coding exons.This gene arrangement is conserved across such diverse species such as the nematode,Drosophila ,and mammals.
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There are also multiple mRNA species found for V AChT that differ in their5V-untranslated sequences (Bejanin et al.,1994;Cervini et al.,1995).These mRNA isoforms appear to be generated from either a transcriptional start site upstream of the R exon,which may also be used for ChAT,or from sites within the V AChT exon(Bejanin et al.,1992;Misawa et al.,1992;Kengaku et al.,1993;Alfonso et al.,1994; Cervini et al.,1995),Fig.1.The human ChAT gene(and thus the cholinergic gene locus)has been mapped to chromosome10,region10q11.2–10q22.2(Cervini et al.,1991).
Coordinate regulation of ChAT and V AChT gene expression
The organization of the cholinergic gene locus suggested the possibility of coordinate regulation of the ChAT and V AChT genes at the transcriptional level(Berrard et al.,1995;Berse and Blusztajn,1995;
Erickson et al.,1994;Shimojo et al.,1998).Prior to the discovery of the gene structure within the cholinergic gene locus,several studies reported that ChAT activity was regulated by growth factors (Hefti et al.,1986;Gould and Butcher,1989;Saadat et al.,1989;Knusel et al.,1991)and cytokines (Kamegai et al.,1990).Using a PC12mutant cell line,in which protein kinase A (PKA)activity is dramatically reduced by the expression of a defective PKA regulatory subunit,Inoue et al.(1995)reported that PKA could regulate expression of the ChAT gene at the transcriptional level.Transient transfection analysis using a human genomic reporter gene revealed that PKA acts through a site on the human ChAT gene upstream of the R exon.Lonnerberg et al.(1996)reported that this region of the rat ChAT gene contains both enhancer and silencer elements.Misawa et al.(1993)reported that dbcAMP increased ChAT mRNA and activity as well as transcription from the R,N,and M exons,suggesting that the cholinergic gene locus might be regulated by a PKA dependent pathway.
Although a number of studies had reported regulation of the ChAT gene (see above),several reports emerged suggesting that the expression of ChAT and V AChT might be regulated coordinately in the CNS (Erickson et al.,1994;Berrard et al.,1995;Berse and Blusztajn,1995;Misawa et al.,1995).Berse and Blusztajn (1995)studied the expression of the ChAT and V AChT genes in a murine septal cell line,SN56,and found that ciliary neurotrophic factor (CNF),retinoic acid,leukemia inhibitory factor (LIF),and dbcAMP increased the mRNA levels for both ChAT and V AChT,although the effects appeared quantitatively different for the two genes.Studies on the regulation of ChAT and V AChT gene transcription in PC12cells using selective inhibitors and two protein kinase A (PKA)mutant PC12cell lines suggested that PKA signaling pathways involving PKAII coordinately regulate both the ChAT and V AChT genes at the transcriptional level (Shimojo et al.,1998).
Transcriptional regulation of cholinergic gene locus by REST/NRSF
Studies from a number of laboratories suggest that the transcriptional regulation of the ChAT and V AChT genes involves multiple cell specific regulatory elements.The region of the cholinergic gene locus upstream of the R-type promoter contains a consensus repressor element-1(RE-1)/neuron-restrictive silencer element (NRSE)sequence and a cholinergic specific enhancer that presumably operate in a cooperative manner to control cholinergic gene expression in the CNS,Fig.1.The RE-1/NRSE is a 21-basepair sequence originally isolated from the promoters of the type II sodium channel (Maue et al.,1990;Mori et al.,1992;Kraner et al.,1992)and SCG10(Mori et al.,1990,1992),and is implicated in silencing the cholinergic gene locus in non-neuronal cells (Lonnerberg et al.,1996).This sequence is found in a large number of neuron-specific genes including the genes for the N-methyl-D-aspartate receptor (NMDAR)(Bai et al.,1998),synapsin I (Schoch et al.,1996),the m4muscarinic acetylcholine receptor (Wood et al.,1996)and BDNF (Timmusk et al.,1999)as well as choline acetyltransferase (Lonnerberg et al.,1996).The transcription factor that binds to the RE-1/NRSE sequence was cloned independently by two groups;one naming it the RE-1silencing transcription factor or REST (Chong et al.,1995),the other naming it the neuron-restrictive silencing factor or NRSF (Schoenherr and Anderson,1995).The RE-1/NRSE sequence,through its binding of REST/NRSF,leads to the repression of the expression of neuron-specific genes in nonneuronal cells (Chong et al.,1995;Schoenherr and Anderson,1995).REST/NRSF is a member of the Gli-Kruppel family of transcription zinc-finger proteins being a 210-kDa glycoprotein containing nine zinc finger domains (Fig.2).
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It was shown that REST/NRSF acts as a silencer of neuron-specific gene expression in nonneuronal tissues and in undifferentiated neuronal progenitor cells(Chong et al.,1995;Schoenherr and Anderson, 1995).The absence of the RE-1/NRSE sequence(Kallunki et al.,1997)or expression of a dominant negative form of REST/NRSF can produce expression of neuronal genes in both nonneuronal tissues and central nervous system progenitors(Chen et al.,1998).A mouse model in which the REST/NRSF gene was deleted,although embryonic lethal,showed aberrant expression of SCG10and tubulin h III in nonneuronal tissues(Chen et al.,1998).Although it was thought that expression of REST/NRSF was limited to non-neuronal cells where neuron-specific genes are repressed,recent evidence suggests that REST/NRSF may control the expression of neuron-specific genes containing an RE-1/NRSE element in neurons(Kallunki et al.,1997;Palm et al.,1998;Timmusk et al.,1999).
To date five REST/NRSF isoforms arising from alternative splicing of the REST/NRSF pre-mRNA have been reported(Palm et al.,1998)and referred to as REST1through REST5.Two of these REST isoforms,REST4and REST5,are expressed specifically in mature neurons of adult brain,albeit at low levels.REST4and REST5contain an insertion of either16nucleotides or28nucleotides respectively,in the region of the gene encoding a spacer between zinc fingers5and6,Fig.2.These insertions in REST4 and REST5,as well as similar insertions in REST2and REST3lead to truncated isoforms that contain only 5of the9zinc finger domains found in REST/NRSF.REST1differs in that it contains only4zinc finger domains,Fig.2.Interestingly,REST4formation is induced by neuronal activity produced by kainate-induced seizure(Palm et al.,1998).REST4exhibits weak binding to the RE-1/NRSE element(Lee et al., 2000),and thus cannot directly regulate neuronal gene expression in RE-1/NRSE containing genes.
As noted above the level of transcription of the cholinergic gene locus is greatly reduced in two PC12 cell lines;A123.7(Correll et al.,1989;Ginty et al.,1991)and A126.1B2(Van Buskirk et al.,1985)that have reduced protein kinase A https://www.doczj.com/doc/4412911093.html,ing reporter gene analysis it was found that the RE-1/NRSE
element was required for repression of the gene in the protein kinase A deficient PC12cells.Introduction of protein kinase A into the PKA deficient cells restored expression from the cholinergic gene locus,that is ChAT and V AChT mRNAs were now detectable.The finding that the concentration of REST/NRSF was essentially the same in both wild type and protein kinase A deficient PC12cells suggests that protein kinase A does not affect the expression levels of REST/NRSF.
REST/NRSF in nuclear extracts from the protein kinase A deficient cells bound to the cholinergic RE-1/NRSE element as determined by electrophoretic mobility shift assays.On the other hand nuclear extracts derived from wild type PC12cells exhibited barely detectable binding to the RE-1/NRSE sequence,despite the fact that both nuclear extracts contained the same levels of REST/NRSF immunoreactive protein.The finding of active REST/NRSF in the protein kinase A deficient PC12cells suggests that transcription of the cholinergic locus was repressed in these cells by REST/NRSF,thus accounting for the absence of ChAT and V AChT mRNAs.However,the reason for the absence of repression of the cholinergic locus by REST/NRSF in wild type PC12cells was unclear.
The possibility that protein kinase A phosphorylated REST/NRSF and generated an inactivate repressor was considered.However,purified protein kinase A catalytic subunit did not phosphorylate recombinant REST/NRSF.Wild type,but not protein kinase A deficient,PC12cells expressed REST4,the neuron-specific isoform of REST/NRSF described above and in Fig.2.It was shown that REST4forms a heterodimer with REST/NRSF.This heterodimer does not bind to the RE-1/NRSE element from the cholinergic gene locus.Thus even though REST/NRSF is expressed in wild type PC12cells,the presence of REST4inhibits the silencing of the cholinergic gene by blocking the repressor activity of REST/NRSF through heterodimer formation.Treatment of PC12cells with forskolin,which increases intracellular cAMP levels by activating adenylate cyclase,induced REST4expression within 4h and ChAT mRNA expression within 6to 12h.These kinetics are consistent with cAMP activating protein kinase A,which in turn leads to an increase in REST4levels,followed by derepression of the cholinergic locus and increased levels of ChAT mRNA.Thus it appears that protein kinase A regulation of the cholinergic gene locus occurs by PKA increasing the splicing of REST/NRSF pre-mRNA to produce REST4.REST4in turn blocks the repressor activity of REST/NRSF leading to increased gene transcription.
Palm et al.(1998)found that a truncated form of REST/NRSF,REST2-5trunc ,also acted as a transcriptional repressor of a REST/NRSF containing reporter gene in Neuro-2A cells,a cell line that does not contain detectable REST/NRSF.On the other hand relatively weak repression of the REST/NRSF containing reporter gene was seen in C6cells,which contain relatively high levels of endogenous REST/NRSF.It was concluded that truncated forms of REST/NRSF produce repression independent of the activity of REST/NRSF and that the repressor activity of the truncated forms resides in an N-terminal repressor domain.It was further suggested that the repressor activity of the REST/NRSF truncated forms could be attributed to the formation of a complex with essential components of the transcriptional machinery.
Regulation of REST/NRSF function by its REST4isoform
The finding that REST4can block REST/NRSF repressor activity,coupled with REST4expression in neuronal cells,can account for the presence of REST/NRSF in neurons without neuronal gene expression being repressed.Palm et al.(1998)reported the presence of REST/NRSF in adult neurons,
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albeit at rather low levels relative to its expression in non-neuronal cells.It was suggested that the inability of REST/NRSF to repress gene expression in neuronal cells could be explained by its low concentration.However,as noted above,it appears that the relative levels of REST/NRSF and REST4 determine the extent of transcriptional repression.Although REST/NRSF levels appear to be constant in PC12cells,the levels of REST4can be regulated by cAMP through protein kinase A activity(Shimojo et al.,1999).In the PC12cell line we studied,the cholinergic gene,although transcriptionally active,was not maximally active since there was an excess of REST/NRSF relative to REST4.Thus modulation of transcriptional repression occurs through modulation of REST4,not REST/NRSF.This mechanism, rather than cAMP acting on a cAMP response element within the cholinergic locus,can explain at least
a Fig.3.Schematic showing the possible mechanisms for the de-repression activity of REST4.A.The formation of heterodimers between REST4and REST/NRSF prevents the binding of REST/NRSF to the cholinergic RE-1/NRSE and leads to de-repression of the cholinergic gene locus.Zinc finger domains are represented by numbered black circles.B.One of two repressor domains is located at the N-terminal domain and binds Sin3and histone deacetylase(HDAC).The other,located at the C-terminal domain is the CoREST binding site.All of the REST truncated forms retain the Sin3/HDAC binding site,but not the CoREST binding site.The truncated REST isoforms could form complexes with Sin3/cofactor complexes,resulting in the supression of available Sin3.C.Regulation of RNA splicing.The various truncated forms of REST shown in Fig.2are produced by RNA splicing as illustrated.Exons are shown as boxes and introns are shown as lines.The ORF of each transcript is indicated as a black box.The neuron-specific exon(N)located between exons V and VI has a stop codon thus when inserted produces truncated proteins.The relative expression levels for REST/NRSF and the splice variant REST4is illustrated for neurons and non-neuronal cells.
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part of the effects of cAMP on cholinergic gene expression in PC12cells.Whether this mechanism can be extended to the cholinergic gene locus in vivo remains to be determined.Supporting this hypothesis is the report of Tabuchi et al.(2002)who found that REST4affects the activity-dependent activation of the BDNF gene promoter I (BDNFp-I)in cultured rat cortical neurons.As we observed in PC12cells,REST4antagonized silencing of the BDNF gene by REST/NRSF,suggesting a role for REST4in preventing the neuron-specific BDNF gene from being transcriptionally repressed by REST/NRSF.In this case REST4activity varied in response to a variety of neuronal stimuli.
Recently Magin et al.(2002)reported that humanized REST4had no discernable effect in NS20Y neuroblastoma cells.REST4did not impair transcriptional repression of the human synapsin I promoter by REST nor did it have a direct effect on the synapsin I promoter.The reason for this discrepancy with our studies and those of Tabuchi et al.(2002)remain unclear,but could be related to the cell type studied.As Tabuchi et al.(2002)suggested,the mechanism of regulation of gene transcription by REST/NRSF isoforms is complex in neurons.There are several mechanisms whereby REST isoforms can function to regulate transcription in the neuron.As shown in Fig.3A ,one mechanism is by heterodimer formation between REST/NRSF and REST4.Since REST4and other REST isoforms retain the N-terminal repressor domain,but do not bind efficiently to DNA,an alternative,or possibly additional mechanism,is that REST isoforms regulate gene expression by competing for binding of the transcriptional repressor complex made up of Sin3and HDAC as illustrated in Fig.3B (Palm et al.,1998;Roopra et al.,2001).
The mRNA levels of REST/NRSF and REST isoforms are determined by the extent of alternative splicing of REST/NRSF pre-mRNA (Fig.3C).Our studies suggest that protein kinase A regulates REST/NRSF pre-mRNA splicing.The ability of a cell signaling pathway to regulate neuronal gene transcription through the REST/NRSF -REST isoform system provides a mechanism for a neuron to regulate its plasticity in response to neuronal activity,neuronal stimuli,or neuronal development.Role of zinc finger domains of REST/NRSF
REST/NRSF contains nine Cys 2-His 2type zinc finger domains,the first of these is near the N-terminus and is followed by a cluster of seven zinc finger domains,with the last zinc finger domain being near the C-terminus of the molecule,Fig.2.REST4,being a truncated form of REST/NRSF contains only zinc finger domains 1–5,Fig.2.It binds weakly to the RE-1/NRSE with the strength of binding dependent on the number of zinc finger domains present (Lee et al.,2000).These findings suggest that zinc finger domains 2–5are required for DNA binding,while zinc finger domains 6–8contribute to the strength of the binding interaction (Shimojo et al.,2001;Chong et al.,1995).In order to study the effect of zinc finger conformation on REST/NRSF activity,mutational analysis was conducted in which zinc finger domains 6–8were studied.Conversion of a Cys to Arg within a zinc finger domain was shown to disrupt the conformation and function of the Cys 2-His 2type zinc finger domain (Tapia-Ramirez et al.,1997).The results of analyzing zinc finger mutations in REST/NRSF suggest that a combination of three zinc finger domains (zinc finger domains 6–8)contribute to DNA binding,however two adjacent zinc fingers,either zinc finger 6plus7or zinc finger 7plus 8,are required to produce maximal DNA binding.
Analysis of the REST4sequence reveals that it lacks the putative nuclear localization signal found in the C-terminal region of REST/NRSF (Chong et al.,1995;Grimes et al.,2000).However,REST4is
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efficiently targeted to the nucleus.Thus the translocation of REST4and REST/NRSF to the nucleus appears to be regulated by different mechanisms.Either REST/NRSF contains an additional N-terminally located nuclear localization signal,or a different nuclear localization signal in REST4is functional.Deletion mutagenesis indicated that the nuclear targeting signal of REST4resides within the C-terminal region containing zinc finger domains2–5(Lee et al.,2000;Shimojo et al.,2001).Fusion protein analysis showed that zinc finger domains2through5,but not2through4,could target h-galactosidase to the nucleus(Shimojo et al.,2001).Mutation of zinc finger domain5of REST4did not prevent its nuclear localization.However,the REST isoform REST1,which is similar to REST4but lacks zinc finger domain5,was not targeted to the nucleus.These data indicate the importance of the region around zinc finger5and suggest that there is a nuclear targeting signal,other than the zinc finger domain,located within the region preceding and going through zinc finger domain5.An alternative explanation is that deletion of the region around zinc finger5changes the structure of REST4such that a critical conformation needed for nuclear targeting is disrupted.
To further study whether the zinc finger domain structure is important for nuclear targeting,REST4 with combinations of Cys to Arg point mutations in the zinc finger domains was fused to GFP.None of the combinations of point mutations altered the localization of REST4from nuclear to cytoplasmic.These data suggest that a functional structure of the zinc finger domains is not required to target REST4to the nucleus.However,certain combinations of zinc finger domain mutations produced abnormal nuclear localization.In these cases,REST4-GFP constructs appeared to localize to the edge of the nucleus,as if they were attached to the nuclear envelope(Shimojo et al.,2001).These data are consistent with previous findings that suggest that zinc finger domains not only function in DNA recognition(Wolfe et al.,2000), but can also be involved in protein-protein interactions(Lee et al.,1993;Milne and Segall,1993). REST/NRSF Repressor Domains
REST/NRSF is a modular protein that represses neuronal gene expression via two distinct repressor domains.Repression through an amino terminal domain is mediated by a Sin3-histone deacetylase (HDAC)complex,which affects chromosomal structure by promoting histone deacetylation(Tapia-Ramirez et al.,1997;Grimes et al.,2000;Huang et al.,1999;Roopra et al.,2001;Naruse et al.,1999). The other repressor domain is found in the carboxyl-terminal domain and binds a novel protein, CoREST and HDAC(Ballas et al.,2001;Andres et al.,1999).In situ hybridization studies of CoREST and Sin3A expression as well as REST/NRSF expression suggest that Sin3is expressed more ubiquitously during early embryogenesis(embryonic day8.5),but at latter stages of embryogenesis (day11.5)CoREST and Sin3A are both expressed fairly ubiquitously throughout the embryo.Evidence has recently been obtained that gene silencing by REST through these domains involves both histone deacetylation and ATP-dependent chromatin remodeling(Battaglioli et al.,2002).
The future for REST
It is evident that the regulation of neuronal gene expression by REST/NRSF and its truncated isoforms is complex and far from understood.As noted above REST mediated gene repression involves two distinct repressor domains,one binding Sin3A and the other binding CoREST.Why two distinct
repressor domains are present in REST/NRSF is puzzling.It is unclear whether both domains are utilized simultaneously or represent different modes of gene regulation that are utilized differentially.The need for a complex system whereby REST4modulates REST/NRSF activity (Fig.3)is also not clear,and whether REST4and the other REST isoforms have additional functions yet to be elucidated is an intriguing question.To date a function has only been attributed to REST4,a function for REST1,REST2,REST3,and REST5remain to be described.
It has been shown that a signaling molecule,cAMP,acting through protein kinase A and REST/NRSF pre-mRNA splicing,can regulate cholinergic gene expression in PC12cells.Is this a general mechanism in all neurons and does cAMP and other signaling pathways regulate transcription of neuronal genes in response to cellular stimuli?On the surface it would appear that protein kinase A,through it increase in REST4,would lead to derepression of all REST/NRSF responsive genes.Whether this is the case or whether there are additional mechanisms that permit only a subset of REST/NRSF responsive genes to be affected by REST4remains to be established.Clearly there are many unanswered questions and much to be done to understand how the REST/NRSF system regulates the cholinergic gene locus as well as a host of other neuron specific genes.
Acknowledgements
This work was supported in part by grants AG46734and AG05893from the National Institutes of Health.
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2015年普通高等学校招生全国统一考试（新课标I卷） 理科综合能力侧试 请注意基础学习 一、选择题 1. 下列叙述错误.的是 A. DNA与ATP中所含元素的种类相同 B. 一个tRNA分子中只有一个反密码子 C. T2噬菌体的核酸由脱氧核糖核苷酸组成 D. 控制细菌性状的基因位于拟核和线粒体中的DNA上 【答案】D 【解析】 扎DXA与A7?中所含■元素的种类都是CH0NP n左一个分子有三个相邻的碱基，对应一个反密码子. C, T：ffi菌体是DMA病畫，橈酸只W DMA,由』兑氧核德核昔酸组咸.D细菌是康核生物，没有线粒体 2. 下列关于植物生长素的叙述，错误的是 A. 植物幼嫩叶片中的色氨酸可转变为生长素 B. 成熟茎韧皮部中的生长素可以进行非极性运输 C. 幼嫩细胞和成熟细胞对生长素的敏感程度相同 D. 豌豆幼苗切段中乙烯的合成受生长素浓度的影响 【答案】C 【解析】幼嫩的细胞对生长素敏感，成熟细胞则比较迟钝。 3. 某同学给健康实验兔静脉滴注0.9%的NaCl溶液（生理盐水）20mL后，会出现的现象是 A. 输入的溶液会从血浆进入组织液 B. 细胞内液和细胞外液分别增加10mL C. 细胞内液Na+的增加远大于细胞外液Na+的增加 D. 输入的Na+中50%进入细胞内液,50%分布在细胞外液 【答案】A
【解析】输入的溶液进入血液，随血液运输，同时从血浆通过毛细血管壁细胞，进入组织液。人体
体液中，细胞内液和细胞外液所占比重不同，注射的20mL的生理盐水到体内的在细胞内液和细胞 外液的分配不同。Na+主要分布在细胞外，细胞内液Na+的增加远小于细胞外液Na+的增加。 4. 下列关于初生演替中草本阶段和灌木阶段的叙述，正确的是 A. 草本阶段与灌木阶段群落的丰富度相同 B. 草本阶段比灌木阶段的群落空间结构复杂 C?草本阶段比灌木阶段的群落自我调节能力强 D.草本阶段为灌木阶段的群落形成创造了适宜环境 【答案】D 【解析】草本阶段与灌木阶段群落相比，草本阶段丰富度低，空间结构简单，自我调节能力差，为灌木阶段群落形成创造了条件。 5. 人或动物PrP基因编码一种蛋白（PrP c）,该蛋白无致病性。PrP c的空间结构改变后成为PrP Bc（朊粒），就具有了致病性。PrP Bc可以诱导更多PrP c的转变为PrP Bc，实现朊粒的增一一可以引起疯牛病. 据此判一一下列叙述正确的是 A. 朊粒侵入机体后可整合到宿主的基因组中 B. 朊粒的增殖方式与肺炎双球菌的增殖方式相同 C?蛋白质空间结构的改变可以使其功能发生变化 D. PrP c转变为PrP Bc的过程属于遗传信息的翻译过程 【答案】C 【解析】 肮粒再蛋白质，其结构与基因差别较大，不能整合到基因组中「肺炎収球菌的増殖方式为二分裂；m注的空间结构改变后成为矶产生了致病性，原因是空间结构改变导致功能改变；P H P U转变为氏PBc的过程为蛋吕质到蛋m质的过程，属于蛋白质结构的改变. 6. 抗维生素D佝偻病为X染色体显性遗传病，短指为常染色体显性遗传病，红绿色盲为X染色体 隐性遗传病，白化病为常染色体隐性遗传病。下列关于这四种遗传病特征的叙述，正确的是
微软公司的历史 微软公司是世界PC机软件开发的先导，比尔·盖茨是它的核心。微软公司1981年为IBM-PC 机开发的操作系统软件MS-DOS曾用在数以亿计的IBM-PC机及其兼容机上。但随着微软公司的日益壮大，Microsoft与IBM已在许多方面成为竞争对手。1991年，IBM公司和苹果公司解除了与微软公司的合作关系，但IBM与微软的合作关系从未间断过，两个公司保持着既竞争又合作的复杂关系。微软公司的产品包括文件系统软件（MS-DOS和Xenix）、操作环境软件（窗口系统Windows系列）、应用软件MS-Office等、多媒体及计算机游戏、有关计算机的书籍以及CDROM产品。1992年，公司买进Fox公司，迈进了数据库软件市场。 1975年，19岁的比尔·盖茨从哈佛大学退学，和他的高中校友保罗·艾伦一起卖BASIC 语言程序编写本。当盖茨还在哈佛大学读书时，他们曾为MITS公司的Altair编制语言。后来，盖茨和艾伦搬到阿尔伯克基，并在当地一家旅馆房间里创建了微软公司。1979年，MITS公司关闭，微软公司以修改BASIC程序为主要业务继续发展。 1977年，微软公司搬到西雅图的贝尔维尤(雷德蒙德)，在那里开发PC机编程软件。1980年，IBM公司选中微软公司为其新PC机编写关键的操作系统软件，这是公司发展中的一个重大转折点。由于时间紧迫，程序复杂，微软公司以5万美元的价格从西雅图的一位程序编制者帕特森手中买下了一个操作系统的使用权，再把它改写为磁盘操作系统软件（MS-DOS）。公司目前在60多个国家设有分支办公室，全世界雇员人数接近44,000人。 IBM-PC机的普及使MS-DOS取得了巨大的成功，因为其他PC制造者都希望与IBM兼容。MS-DOS在很多家公司被特许使用，因此80年代，它成了PC机的标准操作系统。到1984年，微软公司的销售额超过1亿美元。随后，微软公司继续为IBM、苹果公司以及无线电器材公司的计算机开发软件，但在91年后，由于利益的冲突，IBM、苹果公司已经与Microsoft 反目。1983年，保罗·艾伦患霍奇金氏病离开微软公司，后来成立了自己的公司。艾伦拥有微软公司15%的股份，至今仍列席董事会。1986年，公司转为公营。盖茨保留公司45%的股权，这使其成为1987年PC产业中的第一位亿万富翁。1996年，他的个人资产总值已超过180亿美元。1997年，则达到了340亿美元，98年超过了500亿大关，成为理所当然的全球首富。 微软的拳头产品Windows98/NT/2000/Me/XP/Server2003成功地占有了从PC机到商用工作站甚至服务器的广阔市场，为微软公司带来了丰厚的利润：公司在Internet软件方面也是后来居上，抢占了大量的市场份额。在IT软件行业流传着这样一句告戒：“永远不要去做微软想做的事情”。可见，微软的巨大潜力已经渗透到了软件界的方方面面，简直是无孔不入，而且是所向披靡。微软的巨大影响已经对软件同行构成了极大的压力，也把自己推上了反垄断法的被告位置。连多年来可靠的合作伙伴Intel也与之反目，对薄公堂。2001年9月，鉴于经济低迷，美国政府有意重振美国信息产业，拒绝拆分微软。至此，诉微软反垄断法案告一段落。 微软的产品微软生产的软件产品包括了很多的种类： Windows -称为「视窗」的图形操作系统；它有很多版本。目前桌上版最新版本是Windows XP，服务器最新版本是Windows Server 2003。Windows几乎预装在所有的IBM 兼容的个人电脑上。请参看Microsoft Windows的历史获取更多详细资料。 MS-DOS -微软公司的早期产品，它是一个命令行界面。早期的Windows版本要在MS-DOS下运行，但是到了Windows NT以及以后的产品已经可以脱离MS-DOS运行了，但基于用户因软硬件在Windows NT不能正常运作，微软同时间继续推出Windows 95, Windows 98, Windows Me在MS-DOS下运行的过渡产品。 Microsoft Office -它是微软公司的办公软件套件，根据版本不同可能包括Word（文字处理）、Excel（试算表）、Access（桌面数据库）、PowerPoint（幻灯片制作）、Outlook
2014年高考全国1卷理综化学试题 可能用到的相对原子质量：H 1 C 12 N 14 O 16 F 19 Al 27 P 31 S 32 Ca 40 Fe 56 Cu 64 Br 80 Ag 108 一、选择题（每小题6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） 7.下列化合物中同分异构体数目最少的是( ) A.戊烷 B. 戊醇 C. 戊烯 D. 乙酸乙酯 8. 9.2 2 2 2 H2O2 + I?→H2O + IO?慢； H2O2 + IO?→H2O + O2 + I?快；下列有关该反应的说法正确的是( ) A.反应的速率与 I?的浓度有关 B. IO?也是该反应的催化剂 C. 反应活化能等于 98 kJ·mol?1 D. (H2O2)= (H2O)= (O2) 10.W、X、Y、Z 均是短周期元素，X、Y 处于同一周期，X、Z 的最低价离子分别为 X2?和Z－，Y＋和 Z－离子具有相 同的电子层结构。下列说法正确的是( ) A.原子最外层电子数：X＞Y＞Z B. 单质沸点：X＞Y＞Z C. 离子半径：X2?＞Y＋＞Z－ D. 原子序数：X＞Y＞Z 11.溴酸银(AgBrO3)溶解度随温度变化曲线如图所示，下列说法错误的是( ) A.溴酸银的溶解是放热过程 B.温度升高时溴酸银溶解速度加快 C.60℃时溴酸银的K sp 约等于6×10－4 D.若硝酸钾中含有少量溴酸银，可用重结晶方法提纯 12.下列有关仪器的使用方法或实验操作正确的是( ) A.洗净的锥形瓶和容量瓶可以放进烘箱中烘干 B.酸式滴定管装标准液前，必须先用该溶液润洗 C.酸碱滴定实验中，用待滴定溶液润洗锥形瓶以减少实验误差 D.用容量瓶配溶液时，若加水超过刻度线，立即用滴定管吸出多余液体。 13. 利用右图所示装置进行下列实验，能得出相应实验结论的是( ) 浓硫酸具有脱水性、氧化性 与可溶性钡盐均可以生
2015第Ⅰ卷(选择题共126分) 可能用到的相对原子质量：H 1 C 12 N 14 O 16 Cl 35.5 K 39 Cr 52 Fe 56 Cu 64 Br 80 Ag 108 I 127 一、选择题：本题共13小题，每小题6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．下列叙述错误的是 A．DNA与ATP中所含元素的种类相同 B．一个tRNA分子中只有一个反密码子 C．T2噬菌体的核酸由脱氧核糖苷酸组成 D．控制细菌性状的基因位于拟核和线粒体中的DNA上 2．下列关于植物生长素的叙述，错误的是 A．植物幼嫩叶片中的色氨酸可转变为生长素 B．成熟茎韧皮部中的生长素可以进行非极性运输 C．幼嫩细胞和成熟细胞对生长素的敏感程度相同 D．豌豆幼苗切段中乙烯的合成受生长素浓度的影响 3．某同学给健康实验兔静脉滴注0.9％的NaCl溶液(生理盐水)20 mL后，会出现的现象是 A．输入的溶液会从血浆进入组织液 B．细胞液和细胞外液分别增加10 mL C．细胞液Na＋的增加远大于细胞外液Na＋的增加 D．输入的Na＋中50％进入细胞液，50％分布在细胞外液 4．下列关于初生演替中草本阶段和灌木阶段的叙述，正确的是 A．草本阶段与灌木阶段群落的丰富度相同 B．草本阶段比灌木阶段的群落空间结构复杂 C．草本阶段比灌木阶段的群落自我调节能力强 D．草本阶段为灌木阶段的群落形成创造了适宜环境 5．人或动物PrP基因编码一种蛋白(PrP°)，该蛋白无致病性。PrP°的空间结构改变后成为PrP°°(朊粒)，就具有了致病性。PrP°°可以诱导更多的PrP°转变为PrP°°，实现朊粒的增殖，可以引起疯牛病。据此判断，下列叙述正确的是 A．朊粒侵入机体后可整合到宿主的基因组中 B．朊粒的增殖方式与肺炎双球菌的增殖方式相同 C．蛋白质空间结构的改变可以使其功能发生变化 D．PrP°转变为PrP°°的过程属于遗传信息的翻译过程 6．抗维生素D佝偻病为X染色体显性遗传病，短指为常染色体显性遗传病，红绿色盲为X染色体隐性遗传病，白化病为常染色体隐性遗传病。下列关于这四种遗传病遗传特征的叙述，正确的是 A．短指的发病率男性高于女性 B．红绿色盲女性患者的父亲是该病的患者 C．抗维生素D佝偻病的发病率男性高于女性 D．白化病通常会在一个家系的几代人中连续出现 7．我国清代《本草纲目拾遗》中记叙无机药物335种，其中“强水”条目下写道：“性最烈，能蚀五金……其水甚强，五金八石皆能穿滴，惟玻璃可盛。”这里的“强水”是指 A．氨水 B．硝酸 C．醋 D．卤水 8．N A为阿伏加德罗常数的值。下列说确的是 A．18 g D2O和18 g H2O中含有的质子数均为10 N A B．2 L0.5 mol·L－1亚硫酸溶液中含有的H＋离子数为2 N A C．过氧化钠与水反应时，生成0.1 mol氧气转移的电子数为0.2 N A D．密闭容器中2 mol NO与1 mol O2充分反应，产物的分子数为2 N A 9．乌洛托品在合成、医药、染料等工业中有广泛用途，其结构式如图所示。将甲醛水溶液与氨水混合蒸发可制得乌洛托品。若原料完全反应生成乌洛托品，则甲醛与氨的物质的量之比应为
2015年普通高等学校招生全国统一考试 理科综合能力测试（全国2） 可能用到的相对原子质量：H 1 C 12 N 14 O 16 F 19 Na 23 AI 27 P 31 S 32 CL 35.5 Ca 40 Fe 56 Zn 65 Br 80 第I卷 一．选择题：本题共13小题，每小题6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.将三组生理状态相通的某种植物幼根分别培养在含有相同培养液的密闭培养瓶下，一段时间后，测定根吸收某一矿质元素离子的量。培养条件及实验结果见下表： 下列分析正确的是 A.有氧条件有利于该植物幼根对该离子的吸收 B.该植物幼根对该离子的吸收与温度的变化无关 C.氮气环境中该植物幼根细胞吸收该离子不消耗ATP D.与空气相比，氮气环境有利于该植物幼根对该离子的吸收 2.端粒酶由RNA和蛋白质组成，该酶能结合到端粒子上，以自身的RNA为模板合成端粒子DNA 的一条链。下列叙述正确的是 A.大肠杆菌拟核的DNA中含有端粒 B.端粒酶中的蛋白质为RNA聚合酶 C.正常人细胞的每条染色体两端都含有端粒DNA D.正常体细胞的端粒DNA随细胞分裂次数增加而变长 3.下列过程中不属于胞吐作用的是 A.浆细胞分泌抗体到细胞外的作用 B. mRNA从细胞核到细胞质的过程 C.分泌蛋白从胰腺的腺泡细胞到胞外的过程 D.突触小泡中的神经递质释放到突触间隙的过程
4.下列有关生态系统的叙述，错误 ．．的是 A.生态系统的组成成分中含有非生物成分 B.生态系统相对稳定时无能量输入和散失 C.生态系统持续相对稳定离不开信息传递 D.负反馈调节有利于生态系统保持相对稳定 5.下列与病原体有关的叙述，正确的是 A.抗体可以进入细胞消灭寄生在其中的结核杆菌 B.抗体抵抗病毒的机制与溶菌酶杀灭细菌的机制相同 C. Rous肉瘤病毒不是致瘤因子，与人的细胞癌变无关 D.人感染HIV后的症状与体内该病毒浓度和T细胞数量有关 6.下列关于人类猫叫综合征的叙述，正确的是 A.该病是由于特定的染色体片段缺失造成的 B.该病是由于特定染色体的数目增加造成的 C.该病是由于染色体组数目成倍增加选成的 D.该病是由于染色体中增加某一片段引起的 7.食品千操剂应无毒、无味、无腐蚀性及环境友好。下列说法错误 ．．的是A.硅胶可用作食品干操剂 B.P 2O 5 不可用作食品干操剂 C.六水氯化钙可用作食品干燥剂 C.加工后具有吸水性的植物纤维可用作食品干燥剂 8.某羧酸酯的分子式为C 18H 26 O 5 ，1mol该酯完全水解可得到1mol羧酸和2mol乙醇,该羧酸的分 子式为 A.C 14H 18 O 5 B.C 14 H 16 O 4 C.C 16H 22 O 5 D.C 16 H 20 O 5 9.原子序数依次增大的元素a、b、c、d，它们的最外层电子数分别为1、6、7、1。a-的电子层结构与氦相同，b和c的次外层有8个电子，c-和d+的电子层结构相同。下列叙述错误 ．．的是 A.元素的非金属性次序为c>b>a B.a和其他3种元素均能形成共价化合物 C.d和其他3种元素均能形成离子化合物 D.元素a、b、c各自最高和最低化合价的代数和分别为0、4、6
2014年全国高考理综试题 大纲卷 化学部分 一、单选题 1.下列叙述正确的是（） A. 锥形瓶可用作加热的反应器 B. 室温下，不能将浓硫酸盛放在铁桶中 C. 配制溶液定容时，俯视容量瓶刻度会使溶液浓度偏高 D.用蒸馏水润湿的试纸测溶液的 P H, —定会使结果偏低 A . SO 2使溴水褪色与乙烯使 KMnO 4溶液褪色的原理相同 B .制备乙酸乙酯时可用热的 NaOH 溶液收集产物以除去其中的乙酸 C.用饱和食盐水替代水跟电石反应，可以减缓乙炔的产生速率 D. 用AgNO3溶液可以鉴别 KCl 和KI F 图是在航天用高压氢镍电池基础上发展起来的一种金属氢化物镍电池。下列有关说法不正确 的是（） 2. N A 表示阿伏加德罗常数，下列叙述正确的是（ A . lmol Fel 2与足量氯气反应时转移的电子数为 2N A 3. B . C. 2 L0.5 molLL ，硫酸钾溶液中阴离子所带电荷数为 N A 1 mol Na 2O 2固体中含离子总数为 4 N A D.丙烯和环丙烷组成的 42 g 混合气体中氢原子的个数为 6 N A 下列叙述错误的是（） 4.
NiOOH +H 20+e —T NiQHL+OH- 6. 下列离子方程式错误的是( ) A.向 Ba(OH)2 溶液中滴加稀硫酸： Ba 2++2OH +2H ++ SO42"=BaSQ J +2H2O B.酸性介质中 KMnO 4 氧化 H 2O 2 : 2MnO 4"+5H 2O 2+6H 十二 2Mn^+5O 2 +8H 2O C.等物质的量的 MgCl 2、Ba(OH )2 和 HCI 溶液混合：Mg 2++2OH "= Mg(OH )2 J B .电池的电解液可为 KOH 溶液 C.充电时负极反应为: MH +OH —T +H 2O + M +e- D. MH 是一类储氢材料, 其氢密度越大，电池的能量密度越高 选项 被提纯的物质 杂质 除杂试剂 除杂方法 A NmOR 容液、浓H2EO4 洗气 B NKjCXaq) F 眉 NmOR 容液 过痣 C C12(S ) H% 饱和食盐水、浓出$04 洗气 D Ha2CO3Cs) NaHCO3(s) —— 灼烧 5.下列除杂方案错误的是(选项以表格内容为准) 被提纯的物质: A. () CO(g);杂质：CO 2(g);除杂试剂：NaOH 溶液、浓H2SO4除杂方法：洗 B. 被提纯的物质: NH4CI(aq)；杂质：Fe 讥aq);除杂试剂：NaOH 溶液；除杂方法：过滤 C. 被提纯的物质: Cl2(g);杂质：HCI(g);除杂试剂：饱和食盐水、浓 H2SO4除杂方法：洗 D. 被提纯的物质: Na2CO3(s);杂质：NaHCO3(s);除杂试剂：无；除杂方法：灼烧 A .放电时正极反应为:
文科综合能力测试试卷 第1页（共40页） 文科综合能力测试试卷 第2页（共40页） 绝密★启用前 2014普通高等学校招生全国统一考试（全国新课标卷1）理 科综合能力测试 使用地区：陕西、山西、河南、河北、湖南、湖北、江西 本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。满分300分，考试时间150分钟。 注意事项： 1. 答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。 2. 回答第Ⅰ卷时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用橡皮擦干净后，再选涂其它答案标号。写在本试卷上无效。 3. 回答第Ⅱ卷时，将答案写在答题卡上，写在本试卷上无效。 4. 考试结束后，将本试卷和答题卡一并交回。 可能用到的相对原子质量：H —1 C —12 N —14 O —16 F —19 Al —27 P —31 S —32 Ca —40 Fe —56 Cu —64 Br —80 Ag —108 第Ⅰ卷(选择题 共126分) 一、选择题（本题共13小题，每小题6分，共78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. 关于细胞膜结构和功能的叙述，错误的是 （ ） A. 脂质和蛋白质是组成细胞膜的主要物质 B. 当细胞衰老时，其细胞膜的通透性会发生改变 C. 甘油是极性分子，所以不能以自由扩散的方式通过细胞膜 D. 细胞产生的激素与靶细胞膜上相应受体的结合可实现细胞间的信息传递 2. 正常生长的绿藻，照光培养一段时间后，用黑布迅速将培养瓶罩上，此后绿藻细胞的叶绿体内不可能发生的现象是 （ ） A. 2O 的产生停止 B. 2CO 的固定加快 C. ATP/ADP 比值下降 D. NADPH/NADP +比值下降 3. 内环境稳态是维持机体正常生命活动的必要条件，下列叙述错误的是 （ ） A. 内环境保持相对稳定有利于机体适应外界环境的变化 B. 内环境稳态有利于新陈代谢过程中酶促反应的正常进行 C. 维持内环境中Na +、K +浓度的相对稳定有利于维持神经细胞的正常兴奋性 D. 内环境中发生的丙酮酸氧化分解给细胞提供能量，有利于生命活动的进行 4. 下列关于植物细胞质壁分离实验的叙述，错误的是 （ ） A. 与白色花瓣相比，采用红色花瓣有利于实验现象的观察 B. 用黑藻叶片进行实验时，叶绿体的存在会干扰实验现象的观察 C. 用紫色洋葱鳞片叶外表皮不同部位观察到的质壁分离程度可能不同 D. 紫色洋葱鳞片叶外表皮细胞的液泡中有色素，有利于实验现象的观察 5. 如图为某种单基因常染色体隐性遗传病的系谱图（深色代表的个体是该遗传病患者，其余为表现型正常个体）。近亲结婚时该遗传病发病率较高，假定图中第Ⅳ代的两个个体婚配生出一个患该遗传病子代的概率为1/48，那么，得出此概率值需要的限定条件是（ ） A. 2Ⅰ和4Ⅰ必须是纯合子 B. 1Ⅱ、1Ⅲ和4Ⅲ必须是纯合子 C. 2Ⅱ、3Ⅱ、2Ⅲ和3Ⅲ必须是杂合子 D. 4Ⅱ、5Ⅱ、1Ⅳ和2Ⅳ必须是杂合子 6. 某种植物病毒V 是通过稻飞虱吸食水稻汁液在水稻间传播的。稻田中青蛙数量的增加可减少该病毒在水稻间的传播。下列叙述正确的是 （ ） A. 青蛙与稻飞虱是捕食关系 B. 水稻与青蛙是竞争关系 C. 病毒V 与青蛙是寄生关系 C. 水稻和病毒V 是互利共生关系 7. 下列化合物中同分异构体数目最少的是 （ ） A. 戊烷 B. 戊醇 C. 戊烯 D. 乙酸乙酯 9. 已知分解1 mol 22H O 放出热量98 kJ 。在含少量I 的溶液中，22H O 分解的机理为 222I O H O I H O --??→++ 慢 2222IO O O O I H H --??++→+ 快 下列有关该反应的说法正确的是 （ ） ------------- 在 --------------------此 -------------------- 卷--------------------上 -------------------- 答-------------------- 题--------------------无 -------------------- 效---------------- 姓名________________ 准考证号_____________
2014年普通高等学校招生全国统一考试理科综合能力测试（新课标第I卷） 注意事项： 1. 本试卷分第I卷（选择题）和第II卷（非选择题）两部分。答卷前，考试务必将自己的姓名、准考证号填写在答题卡上。 2. 回答第I卷时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用橡皮擦干净后，再选涂其它答案标号，写在本试卷上无效。 3. 回答第II卷时，将答案写在答题卡上，写在本试卷上无效。 4. 考试结束后，将本试卷和答题卡一并交回。 可能用到的相对原子质量：H1 C12 N14 O16 F19 AL27 P31 S32 Ca 40 Fe56 Cu64 Br 80 Ag108 一、选择题：本题共13小题，每小题6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.关于细胞膜结构和功能的叙述，错误的是 A.脂质和蛋白质是组成细胞膜的主要物质 B.当细胞衰老时，其细胞膜的通透性会发生改变 C.甘油是极性分子，所以不能以自由扩散的方式通过细胞膜 D.细胞产生的激素与靶细胞膜上相应受体的结合可实现细胞间的信息传递 2. 正常生长的绿藻，照光培养一段时间后，用黑布迅速将培养瓶罩上，此后绿藻细胞的叶绿体内不可能发生的现象是 A.O2的产生停止 B.CO2的固定加快 C.ATP/ADP比值下降 D.NADPH/NADP+比值下降 3. 内环境稳态是维持机体正常生命活动的必要条件，下列叙述错误的是 A.内环境保持相对稳定有利于机体适应外界环境的变化 B.内环境稳态有利于新陈代谢过程中酶促反应的正常进行 C.维持内环境中Na+、K+浓度的相对稳定有利于维持神经细胞的正常兴奋性 D.内环境中发生的丙酮酸氧化分解给细胞提供能量，有利于生命活动的进行 4. 下列关于植物细胞质壁分离实验的叙述，错误的是 A.与白色花瓣相比，采用红色花瓣有利于实验现象的观察 B.用黑藻叶片进行实验时，叶绿体的存在会干扰实验现象的观察 C.用紫色洋葱鳞片叶外表皮不同部位观察到的质壁分离程度可能不同 D.紫色洋葱鳞片叶外表皮细胞的液泡中有色素，有利于实验现象的观察 5. 下图为某种单基因常染色体隐性遗传病系谱图（深色代表的个体是该遗传病患者，其余为表现型正常个体）。近亲结婚时该遗传病发病率较高，假定图中第Ⅳ代的两个个体婚配生出一个患该遗传病子代的概率为1/48，那么，得出此概率值需要的限定条件是 A.Ⅰ－2和Ⅰ－4必须是纯合子
2015年全国卷1理综 第Ⅰ卷(选择题共126分) 可能用到的相对原子质量：H 1 C 12 N 14 O 16 Cl 35.5 K 39 Cr 52 Fe 56 Cu 64 Br 80 Ag 108 I 127 一、选择题：本题共13小题，每小题6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．下列叙述错误的是 A．DNA与ATP中所含元素的种类相同 B．一个tRNA分子中只有一个反密码子 C．T2噬菌体的核酸由脱氧核糖核苷酸组成 D．控制细菌性状的基因位于拟核和线粒体中的DNA上 2．下列关于植物生长素的叙述，错误的是 A．植物幼嫩叶片中的色氨酸可转变为生长素 B．成熟茎韧皮部中的生长素可以进行非极性运输 C．幼嫩细胞和成熟细胞对生长素的敏感程度相同 D．豌豆幼苗切段中乙烯的合成受生长素浓度的影响 3．某同学给健康实验兔静脉滴注0.9％的NaCl溶液(生理盐水)20 mL后，会出现的现象是A．输入的溶液会从血浆进入组织液 B．细胞内液和细胞外液分别增加10 mL C．细胞内液Na＋的增加远大于细胞外液Na＋的增加 D．输入的Na＋中50％进入细胞内液，50％分布在细胞外液 4．下列关于初生演替中草本阶段和灌木阶段的叙述，正确的是 A．草本阶段与灌木阶段群落的丰富度相同 B．草本阶段比灌木阶段的群落空间结构复杂 C．草本阶段比灌木阶段的群落自我调节能力强 D．草本阶段为灌木阶段的群落形成创造了适宜环境 5．人或动物PrP基因编码一种蛋白(PrP°)，该蛋白无致病性。PrP°的空间结构改变后成为PrP°°(朊粒)，就具有了致病性。PrP°°可以诱导更多的PrP°转变为PrP°°，实现朊粒的增殖，可以引起疯牛病。据此判断，下列叙述正确的是 A．朊粒侵入机体后可整合到宿主的基因组中 B．朊粒的增殖方式与肺炎双球菌的增殖方式相同 C．蛋白质空间结构的改变可以使其功能发生变化 D．PrP°转变为PrP°°的过程属于遗传信息的翻译过程 6．抗维生素D佝偻病为X染色体显性遗传病，短指为常染色体显性遗传病，红绿色盲为X 染色体隐性遗传病，白化病为常染色体隐性遗传病。下列关于这四种遗传病遗传特征的叙述，正确的是 A．短指的发病率男性高于女性
微软模拟火车初始版安装 我有言在先；本版本是给微软模拟火车迷一个方便，主目的是为了喜欢微软模拟火车而不知道安装的人一个方便。它是由[微软模拟火车].train.simulator_CD1.iso，[微软模拟火车].train.simulator_CD2.iso，DTLite4451-0236.exe（虚拟光驱）组成。下面我介绍您介绍微软模拟火车的安装方法。 第一步；安装虚拟光驱，您要打开DTLite4451-0236.exe 这个可执行文件。然后选择中文按确定。 按下一步。然后出现：
单击“我同意”。然后： 单击免 费许可，再单击下一步，然后出现：
如果您 不需要，请把工具栏，主页，搜索的勾去掉（用鼠标在勾上点一下）即可，如果您需要，请点击下一步，然后出现 最好不要安装到C盘，因为安装到C盘电脑很容易卡，所以单击浏览，最好安装到E盘，
就像我这样，（再此之前要在E盘新建文件夹，如： （单击右键，然后有一个新建，在第一个有一个新建文件夹，然后电脑就会在E盘创建一个文件夹，你要编辑它，然后在文件夹字的框框里输入：“虚拟光驱”就行了。下一步单击安装，然后等待一下，最后出现：
单击完成。虚拟光驱就搞定了！——————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————下一步—————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————下一步—————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————
2014年普通高等学校招生全国统一考试 理科综合能力测试 一、选择题：本题共13小题，每小题6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. 关于细胞的叙述，错误的是 A. 植物细胞的细胞连丝具有物资运输的作用 B. 动物细胞间的粘着性细胞膜上的糖蛋白有关 C. A TP水解释放的能量可用于细胞内的吸能反应 D. 哺乳动物的细胞可以合成蔗糖，也可以合成乳糖 2.同一动物个体的神经细胞与肌细胞在功能上是不同的，造成这种差异的主要原因是 A.二者所处的细胞周期不同 B. 二者合成的特定蛋白不同 C. 二者所含有的基因组不同 D. 二者核DNA的复制方式不同 3.关于在正常情况下组织液生成与回流的叙述，错误的是 A.生成与回流的组织液中氧气的含量相等 B. 组织液不断生成与回流，并保持动态平衡 C. 血浆中的有些物质经毛细血管动脉端进入组织液 D. 组织液中的有些物质经毛细血管静脉端进入血液 4.讲某植物花冠切成大小和形状相同的细条，分为a、b、c、 d、e和f组（每组的细条数相等），取上述6组细条数分别 置于不同浓度的蔗糖溶液中，浸泡相同时间后测量各组花冠 细条的长度，结果如图所示。假如蔗糖溶液与花冠细胞之间 只有水分交换则 A.试验后，a组液泡中的溶质浓度比b组的高 B.浸泡导致f组细胞中液泡的失水量小于b组的 C.A组细胞在蔗糖溶液中失水或吸水所耗ATP大于b组 D.使细条在浸泡前后长度不变的蔗糖浓度介于0.4~0.5mol-L-1之间 5.关于核酸的叙述，错误的是 A.细胞核中发生的转录过程有RNA聚合酶的参与 B.植物细胞的线粒体和叶绿素体中均可发生DNA的复制 C.双链DNA分子中一条链上的磷酸和核糖是通过氨键链接的 D.用甲基绿和吡罗红染色可观察DNA和RNA在细胞中的分布 6.关于光合作用和呼吸作用的叙述，错误的是 A.磷酸是光反应中合成A TP所需的反应物
2014年全国高等学校招生考试全国Ⅱ卷理科综合一、选择题：本题共13小题，每小题6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. 关于细胞的叙述，错误的是 A. 植物细胞的细胞连丝具有物资运输的作用 B. 动物细胞间的粘着性细胞膜上的糖蛋白有关 C. ATP水解释放的能量可用于细胞内的吸能反应 D. 哺乳动物的细胞可以合成蔗糖，也可以合成乳糖 2.同一动物个体的神经细胞与肌细胞在功能上是不同的，造 成这种差异的主要原因是 A.二者所处的细胞周期不同 B. 二者合成的特定蛋白不同 C. 二者所含有的基因组不同 D. 二者核DNA的复制方式 不同 3.关于在正常情况下组织液生成与回流的叙述，错误的是 A.生成与回流的组织液中氧气的含量相等 B. 组织液不断生成与回流，并保持动态平衡 C. 血浆中的有些物质经毛细血管动脉端进入组织液 D. 组织液中的有些物质经毛细血管静脉端进入血液 4.讲某植物花冠切成大小和形状相同的细条，分为a、b、c、d、e和f组（每组的细条数相等），取上述6组细条数分别置于不同浓度的蔗糖溶液中，浸泡相同时间后测量各组花冠细条的长度，结果如图所示。假如蔗糖溶液与花冠细胞之间只有水分交换则
A.试验后，a组液泡中的溶质浓度比b组的高 B.浸泡导致f组细胞中液泡的失水量小于b组的 C.A组细胞在蔗糖溶液中失水或吸水所耗ATP大于b组 D.使细条在浸泡前后长度不变的蔗糖浓度介于0.4~0.5mol-L-1之间 5.关于核酸的叙述，错误的是 A.细胞核中发生的转录过程有RNA聚合酶的参与 B.植物细胞的线粒体和叶绿素体中均可发生DNA的复制 C.双链DNA分子中一条链上的磷酸和核糖是通过氨键链接的 D.用甲基绿和吡罗红染色可观察DNA和RNA在细胞中的分布 6.关于光合作用和呼吸作用的叙述，错误的是 A.磷酸是光反应中合成ATP所需的反应物 B.光合作用中叶绿素吸收光能不需要酶的参与 C.人体在剧烈运动时所需要的能量由乳酸分解提供 D.病毒核酸的复制需要宿主细胞的呼吸作用提供能量 7.下列过程没有发生化学反应的是 A.用活性炭去除冰箱中的异味 B.用热碱水清除炊具上残留的污垢 C.用浸泡过高锰酸钾溶液的硅藻土保鲜水果 D.用含硅胶、铁粉的透气小袋与食品一起密封包装 8.四联苯的一氯代物有 A. 3种 B. 4种 C. 5种 D. 6种 9.下列反应中，反应后固体物质增重的是
Trainz模拟火车教程 编写：江南QQ：396037170 各位玩家，相信大部分同学都是最先开始接触msts微软模拟火车才开始转战TRS的，msts的各种局限导致游戏不能满足玩家的与时俱进的对游戏的要求。然而TRS给大家提供了这么一个游戏平台，无论是从游戏的画面还是游戏的可玩性，都是其它任何一款模拟火车游戏不可比拟的。下面让我们开始我们的trainz 火车世界之旅吧。 第一章：游戏下载及安装 1.1 Trainz游戏下载：各种版本 T rs2006：https://www.doczj.com/doc/4412911093.html,/topics/56160/ Trs2009：https://www.doczj.com/doc/4412911093.html,/topics/2727138
Trs2010：https://www.doczj.com/doc/4412911093.html,/topics/2789870/ 本人使用的是TRS2009，感觉比较稳定，建议电脑配置低的考虑实际情况下载适合自己的版本。（各种版本的配置要求是递增的）。 当你费了一天的时间下载好了游戏文件，那么我们就开始安装吧！ 1.2游戏安装 游戏下载好了之后你会得到很多个压缩文件，你把压缩文件其中的一个解压就会得到以下的文件，第一个是种子文件没用的，而第四个就是最重要的安装文件了： 这个时候游戏新手可能会问虚拟光驱是什么，请看下载地址： https://www.doczj.com/doc/4412911093.html,/soft/2345.html 安装好虚拟光驱之后运行，之后操作如图：
注意：提醒安装2010版本的同学在安装时光驱运行到75%左右的时候会提示你插入第二章光盘，这个时候你记得把第二个ISO镜像文件导入进去。 第二章：游戏前准备工作 2.1 辅助软件 游戏安装完成了，先别急，我们先来游戏前的一些准备工作。 首先，请先确认你的电脑中安装了DX软件，这个就是游戏加速软件，必须要安装，否则进不了游戏！ 下载地址： https://www.doczj.com/doc/4412911093.html,/game_tool/000009.rar?key=9186f597e6363862075 48d40d3546d93 2.2 界面介绍
2014新课标卷2—理综—生物部分 一、选择题，每小题6分，只有一个符合题意。 1、关于细胞的叙述，错误的是----（D ） A．植物细胞的胞间连丝具有物质运输的作用 B．动物细胞间的粘着性与细胞膜上的糖蛋白有关 C．ATP水解释放的能量可用于细胞内的吸能反应 D．哺乳动物的细胞可以合成蔗糖，也可以合成乳糖 2.同一动物个体的神经细胞与肌肉细胞在功能上是不同的，造成这种差异的主要原因是A．两者所处的细胞周期不同B．两者合成的特定蛋白不同 C．两者所含有的基因组不同D．两者核DNA复制的方式不同 3.关于在正常情况下组织液的生成与回流的叙述，错误的是 A．生成与回流的组织液中氧气的含量相等 B．组织液不断生成与回流，并保持动态平衡 C．血浆中的有些物质经毛细血管动脉端进入组织液 D．组织液中的有些物质经毛细血管静脉端进入血液 4.将某植物的花冠切成大小和形状相同的细条分为a,b,c,d,e和f组 （每组的细条数相等），取上述细条分别置于不同浓度的蔗糖溶液中， 浸泡相同时间后测量各组花冠细条的长度，结果如下图所示。假如蔗 糖溶液与花冠细胞之间只有水分交换，则 A．实验后，a组液泡中的溶质浓度比b组的高 B．浸泡导致f组细胞中液泡的失水量小于b组的 C．a组细胞在蔗糖溶液中失水或吸水所耗ATP大于b组 D．使细条在浸泡前后长度不变的蔗糖浓度介于0.4～0.5mol·L-1之间 5.关于核酸的叙述，错误的是 A．细胞核中发生的转录过程有RNA聚合酶的参与 B．植物细胞的线粒体和叶绿体中均可发生DNA的复制 C．双链DNA分子中一条链上的磷酸和核糖是通过氢键连接的 D．用甲基绿和吡罗红染色可观察DNA和RNA在细胞中的分布
2015年高考物理全国卷1及答案解析
第Ⅰ卷(选择题)2015年理综全国卷1 物理部分 二、选择题：本题共8小题，每小题6分。在每小题给出 的四个选项中。第l4～18题只有一项符合题目要求。 第l9～21题有多项符合题目要求。全部选对的得6分。 选对但不全的得3分。有选错的得0分。 14．两相邻匀强磁场区域的磁感应强度大小不同、方向平行。一速度方向与磁感应强度方向垂直的带电粒子(不计重力)，从较强磁场区域进入到较弱磁场区域后，粒子的 A．轨道半径减小，角速度增大B．轨道半径减小，角速度减小 C．轨道半径增大，角速度增大D．轨道半径增大，角速度减小 15．如图，直线a、b和c、d是处于匀强电场中的两组平行线，M、N、P、Q是它们的交点，四点处的电势分别为 由M ?、N?、P?、Q?。一电子 M 点分别运动到N点和P点的过程中，电场力所做的负功相等。则 A．直线a位于某一等势面内，M?> ? Q B．直线c位于某一等势面内，M?>N? C．若电子由M点运动到Q点，电场力做正功 D．若电子由P点运动到Q点，电场力做负功 16．一理想变压器的原、副线圈的匝数比
为3：l ，在原、副线圈的回路中分别接有阻值相同的 电阻，原线圈一侧接在电压为220 V 的正弦交流电源 上，如图所示。设副线圈回路中电阻两端的电压为U ， 原、副线圈回路中电阻消耗的功率的比值为k 。则 A ．U=66V ，k=19 B ．U=22V ，k=19 C ．U=66V ，k=13 D ．U=22V ，k=13 17．如图，一半径为R 、粗糙程度处处相同的半圆形轨道 竖直固定放置，直径POQ 水平。一质量为m 的质点 自P 点上方高度R 处由静止开始下落，恰好从P 点进 入轨道。质点滑到轨道最低点N 时，对轨道的压力为 4 mg ，g 为重力加速度的大小。用W 表示质点从P 点 运动到N 点的过程中克服摩擦力所做的功。则 A ．W =12 mgR ，质点恰好可以到达Q 点 B ．W >12 mgR ，质点不能到达Q 点 C ．W =12 mgR ，质点到达Q 点后，继续上升一段距离 D ．W<12 mgR ，质点到达Q 点后，继续上升一段距离 18．一带有乒乓球发射机的乒乓球台如图
MSTS线路制作教程之一地形区域创建1 进入MSTS线路地形生成器（英文简称RGE）后选择文件->新建线路…
2 线路参数的设置 2.1模板不用改， 2.2线路名的第一行填的是玩家看到线路的名字（如：外南铁路）第二行是线路创建的文 件夹名（如WaiNan） 2.3在线路规格中可以选择是电气化铁路还是非电气化铁路，如果是选择电气化铁路，则 需要输入接触网的高度，单位是米 2.4在线路限速中填写线路的警戒区速度和线路最高限速 2.5如果需要KM/H为速度单位时，最后那个框不要勾 以下是我创建线路的参数设置，大家可以根据自己的需要填写
3 图中的地球仪上找到自己想要创建线路的大致位置，然后右键单击所要放大的区域，选择地区缩放或者放大 继续缩放，此时可以看到地图上以红色线条标注的就是铁路，如果是新的线路可能无法找到，就要用Google Earth或者其他的工具，找到你所建立线路的大致经纬度进行缩放，在缩放期间注意用滚动条来找到缩放的中心位置，不要缩放错了
4 把地图一直放大到不能再放大为止，如下图的放大选项已经变成灰色 5此时在程序的左下角的状态栏上显示的是[线路名没有四分树] 经纬度信息等
6选择文件->新建四分树 此时刚才的状态栏变成了[线路名标准地形] 除此之外，地图上没有任何的变化 7用鼠标左键框选你想要的线路区域，然后用单击右键，添加所选择的全部区块，如果要添加多块区域，只要重复此步骤多次即可
8添加完后如下图出现了很多正方形的小黑格最小的那种方格就是我们刚才框选的区域 9此时用鼠标右键单击你所要建设线路的周围小方格，选择标记开关，那个方格就会变成蓝色，如果再对同一个方格选择标记开关一次，就是取消蓝色，变回原来的黑色