Chapter 13-14-15 Enzyme part 2 Enzyme kinetics Michaelis-Menten kinetics

合集下载

03_Enzymes

03_Enzymes

下列有关酶的概念哪一项是正确的 A.所有的蛋白质都有酶活性 B.其底物都是有机化合物 C.其催化活性都需要特异的辅助因子 D.对底物都有绝对专一性 E.酶不一定都是蛋白质
下列关于酶蛋白和辅助因子的叙述,哪一点不正确? A.酶蛋白或辅助因子单独存在时均无催化作用
B.一种酶蛋白只与一种辅助因子结合成一种全酶
同工酶的临床意义:
同工酶谱的改
酶 活 性
心肌梗塞酶谱
变有助于对疾
病的诊断。
正常酶谱 肝病酶谱
LDH
1
2
3
4
5
心肌梗塞和肝病病人血清LDH同工酶谱的变化
同工酶是指:
A. 酶的结构相同而存在部位不同
B. 催化相同反应而酶分子的结构与理化性质不同
C. 催化不同反应而理化性质相同 D. 由同一基因编码但催化不同反应 E. 催化相同反应,理化性质亦相同,但分布不同
酶,氨基酸脱羧酶,半胱氨酸脱硫酶等的
辅酶。
Vit B6包括吡哆醇(pyridoxine),吡哆醛(pyridoxal)和 吡哆胺(pyridoxamine)等三种形式。
5. 辅酶A(HSCoA) ——泛酸(遍多酸) CoA中的巯基可与酰基以高能硫酯键结合, 在糖、脂、蛋白质代谢中起传递酰基的 作用,因此CoA是酰化酶的辅酶。
多酶复合体
-酮脂肪 酰还原酶
-酮脂肪 酰合酶-SH
,-烯脂 肪酰水化酶
HS-ACP
脂肪酰 转移酶
,-烯脂 肪酰还原酶
长链脂肪 酰硫解酶
丙二酰单 酰转移酶
多功能酶
一、酶的分子组成中常含有辅助因子
酶可根据其化学组成的不同,分为两类:
单纯酶(全部是蛋白质) (simple enzyme)

生化--英文题

生化--英文题

生化历来都不画重点的,但留学生老是过不了,于是就有了这个重点资料,和我们考的差不多,但是英语的,求高人翻译啊Brief Exercises of BiochemistryChapter 1 The structure and function of proteinExplain the following terms1. peptide bond2. Amino acid residues3. Primary structure of protein4. isoelectric point5. Secondary structure of protein6. Tertiary structure of protein7. Domain8. Protein denaturation9. Quaternary structure of proteinAnswer the following questions briefly1. What is physiological significance of hemoglobin oxygen dissociation curve as S-shaped?2. Please describe physiological functions of proteins.Discuss the following questions (Essay questions)1. Explain the relationship between the primary and spatial structure and the function of protein.Chapter 2 The structure and function of nucleic acidsExplain the following terms1. primary structure of nucleic acids2. DNA denaturation3. Tm4. DNA renaturation5. nucleic acid hybridizationAnswer the following questions briefly1. What is the structural characteristics of an eukaryocyte mature mRNA?2. What is the biological significance of Tm?Discuss the following questions (Essay questions)1. Please compare the two types of nucleic acids (DNA and RNA) in the chemical composition, molecular structure, cell distribution and biological functions.2. Please describe the structural characteristics of the B-DNA.3. Describe the molecular composition, structural features and functions of tRNA.Chapter 3 EnzymeExplain the following terms1. enzyme2. enzyme active center3. enzyme competitive inhibition4. Km5. isoenzyme6. zymogen activationAnswer the following questions briefly1. Explains with examples the competitive inhibition characteristic and the practical significance.2. What is the relationship between the enzyme cofactor and vitamine?3. What is the physiological significance of zymogen?4. What is isoenzyme? What is clinical significance of isoenzyme?5. How many kinds of essential group of enzyme are there? What is the role of each?Chapter 4 Metabolism of carbohydrateExplain the following terms1. glycolysis2. glycolytic pathway3. tricarboxylic acid cycle4. gluconeogenesis5. blood sugarAnswer the following questions briefly1. Describe briefly source and fate of blood sugar2. Describe briefly the physiological significance of gluconeogenesis3. Describe briefly the physiological significance of glycolysis4. Describe briefly the outline of TCA cycle5. Describe briefly the physiological significance of TCA cycle6. Describe briefly the physiological significance of pentose phosphate pathway7. Outline the reasons for the formation of lactic acid cycle and the physiological significance.8. Overview the important role of B vitamins in glucose metabolism.9. Why 6-phosphate glucose dehydrogenase activity will increase after uptake high-carbohydrate diet?Discuss the following questions (Essay questions)1. Explain how is lactate converted into glucose? (Write down the main reactions and key enzymes)2. Explain how is lactate converted into CO2, H2O and releases ATP? (Write down the main reactions and key enzymes)3. Overview the regulation molecular mechanism of adrenaline on the blood sugar level.4. Please explain why a slimmer has to reduce the intake of carbohydrates from the point of view of nutrients metabolism. (Write down the related pathways, cellular localization, main reactions and key enzyme)Chapter 5 Metabolism of lipidsExplain the following terms1. fat mobilization2. ketone body3. plasma lipoprotein4. apolipoprotein5. essential fatty acid6. blood lipidsAnswer the following questions briefly1. What is the function of bile acid at lipids digestion?2. What is the physiological significance of ketone body generation?3. What are materials of fatty acid synthesis?4. What is the physiological significance of cholesterol?5. What are the functions of apolipoprotein?Discuss the following questions (Essay questions)1. Describe the sources, chemical composition characteristics and main physiological functions of plasma lipoprotein.2. Explain how is the stearic acid converted into CO2, H2O and releases ATP?3. Please describe the oxidation catabolism process of glycerol generated from fat mobilization4. Explain how is the glycerol converted into glycogen?5. Describe the source and fate of acetyl-CoA?Chapter 6 Biological oxidationExplain the following terms1. biological oxidation2. respiratory chain3. oxidative phosphorylation4. substrate level phosphorylationDiscuss the following questions (Essay questions)1. Write down the sequence of two respiratory chainChapter 7 Metabolism of amino-acidExplain the following terms1. essential amino acid2. deamination of amino acid3. transamination of amino acid4. one carbon unit5. hyperammonemiaAnswer the following questions briefly1. What is the physiological significance of one carbon units?2. What is meaning of PAPS, GABA, SAM and FH4 each?3. Write down the deamination of amino acids in vivo.4. Outline the source and fate of blood ammonia.Discuss the following questions (Essay questions)1. How does a glutamate be oxidized to supply energy? What is the final product?2. What are functions of vitamins B in the metabolism of amino acids?3. Use the alanine as an example, try to explain the gluconeogenesis process of glucogenic amino acids.Chapter 8 Metabolism of nucleotideExplain the following terms1. de novo synthesis pathway of purine nucleotide2. nucleotide antimetaboliteAnswer the following questions briefly1. Outline the biological function of nucleotide.2. Outline the physiological significance of salvage synthesis of purine nucleotide.Discuss the following questions (Essay questions)1. Use the 6-mercaptopurine as an example, please explain the mechanism of antimetabolite.Chapter 10 Biosynthesis of DNAExplain the following terms1. semi-conservative replication2. reverse transcription3. replication4. excision repairing5. frame-shift mutationAnswer the following questions briefly1. Outline the classification and function of prokaryote DNA polymerase.2. Outline the classification and function of eukaryote DNA polymerase.3. Outline the factors causing DNA damage.4. Outline the repairing of DNA damage.5. Outline the central dogma.Discuss the following questions (Essay questions)1. Describe the materials involved in prokaryote DNA replication and their functions in that process.2. Describe the biological significance of mutation.Chapter 11 Biosynthesis of RNAExplain the following terms1. transcription2. posttranscriptional process3. hnRNA4. promoter5. ribozyme6. structure geneAnswer the following questions briefly1. Outline the eukaryote posttranscriptional process.2. Outline the products of three kinds of eukaryote RNA polymerases.Discuss the following questions (Essay questions)1. Describe the similarity and dissimilarity of replication and transcription.Chapter 12 Biosynthesis of proteinExplain the following terms1. translate2. polyribosomes3. genetic code4. degeneracy of codonAnswer the following questions briefly1. Describe briefly the RNAs involved in the protein synthesis and their functions in that process.2. Outline the main features of the genetic code.3. Describe briefly the dissimilarity of translation initiation complex formation of prokaryotes and eukaryotes.Discuss the following questions (Essay questions)1. Describe the materials involved in protein biosynthesis and their functions in that process.3. Please comparing the process of translation of prokaryotes and eukaryotes.Chapter 13 The regulation of gene expressionExplain the following terms1. gene expression2. cis-acting element3. trans-acting factor4. operon5. general transcription factor6. enhancerAnswer the following questions briefly1. What is biological significance of regulation of gene expression?2. Outline the function of each component of operon.3. What characteristics does eukaryotic genome structure have?Discuss the following questions (Essay questions)1. Explain the regulation mechanism of lactose operon.Chapter 14 Gene recombination and gene engineeringExplain the following terms1. restriction endonuclease2. genomic DNA3. vector4. cDNA. library5. genetic engineering6. DNA cloning7. homologous recombinationAnswer the following questions briefly1. What are the main selection criteria of gene vector?2. What is the significance of restriction endonuclease of bacteria themselves?3. At present, How many ways to get target genes?4. Outline the basic process of DNA cloning.Discuss the following questions (Essay questions)1. Why plasmid can be used as the vector of genetic engineering?2. Explain how to connect the foreign gene and the vector.3. What is α-complementary? Explain how to screening recombinant by it using an example.Chapter 15 Cellular signal transductionExplain the following terms1. signal transduction2. receptor3. ligand4. signal transduction pathway5. protein kinase6. second messenger7. G proteinAnswer the following questions briefly1. Describe briefly which protein kinases are regulated by intracellular second messenger.2. Outline the classification of receptor and its chemical signals.3. Describe briefly the basic mode of G protein-coupled receptor (seven transmembrane receptor)-mediated signal transduction.4. Describe briefly the signal transduction pathway of intracellular receptor of steroid hormone.Discuss the following questions (Essay questions)1. How does intracellular receptor play its function?2. Explain the process of the glycogen metabolism regulated by glucagon.3. Use fat mobilization as an example, explain the process of cAMP-protein kinase pathway. Chapter 16 Blood biochemistryExplain the following terms1. 2, 3-BPG shuntAnswer the following questions briefly1. Outline the function of plasma protein.Chapter 17 Liver biochemistrExplain the following terms1 biotransformation 2. primary bile acid 3. secondary bile acid4. bile pigment5. jaundiceAnswer the following questions briefly1. Describe briefly the physiological significance of biotransformation.2. Outline the main physiological functions of bile acids.3. Describe briefly production and blood transportation of bilirubin.Discuss the following questions (Essay questions)1. Describe the influence factor of biotransformation.2. Explain the dissimilarity of unconjugated and conjugated bilirubin.Chapter 18VitaminsExplain the following terms1. vitamin2. lipid-soluble vitamin3. water-soluble vitaminAnswer the following questions briefly1. Outline the biochemical function of vitamin E.2. Describe briefly the biochemical function of vitamin D and its deficiency disease.Discuss the following questions (Essay questions)1. Explain the relationship between the water-soluble vitamin and the coenzyme. Chapter 20 Oncogenes, tumor suppressor genes and growth factorExplain the following terms1. oncogene2. proto-oncogene3. tumor suppressor geneAnswer the following questions briefly1. Describe characteristics of proto-oncogene.2. Describe briefly wild-type p53 tumor suppressor gene mechanism.Chapter 21 The Principle and Application of Common Used Techniques in Molecular Biology Explain the following terms1. probe2. PCR3. Gene diagnosis4. gene therapyDiscuss the following questions (Essay questions)1. Describe the definition, type and application of the blotting technique.2. Describe the PCR reaction principle and the basic steps.。

chapter 3 enzymes

chapter 3 enzymes

Induced-fit Theory
substrate
Complex of substrate-enzyme enzyme
Mechanisms Proximity effect and orientation arrange: For a biochemical reaction to occur, the substrate must come into close proximity to catalytic functional groups (side chain groups involved in a catalytic mechanism ) within the active site. In addition, the substrate must be precisely, spatially oriented to the catalytic groups. Once the substrate is correctly positioned, a change in the enzyme’s conformation may result in a strained enzymesubstrate complex. This strain helps to bring the enzyme-substrate complex into the transition state.
3
Enzyme kinetics
The rate or velocity of a biochemical reaction is defined as the change in the concentration of a reactant or product per unit time. Plotting initial velocity v versus substrate concentration [S].The rate of the reaction is directly proportional (first order reaction) to substrate concentration only when [S] is low. When [S] becomes sufficiently high that the enzyme is saturated, the rate of the reaction is zero-order with respect to substrate.

酶(part2)

酶(part2)

经整理得:[ES ]=[Et ][S ] K m +[S ]
(1)
由于酶促反应速度由[ES]决定,即 决定, 由于酶促反应速度由 决定
v = k2 [ES ]
,所以
[ES ] =
将(2)代入(1)得: )代入( ) 当[Et]=[ES]时, 时
[Et ][S ] v = k2 K m + [S ]
所以
Vmax 2
Vmax[S] = Km + [S] Km=[S]
∴Km值等于酶促反应速度为最大反应速度一半时 的底物浓度,单位是mol/L。 的底物浓度,单位是 。
小结: 小结:Km的物理意义 的物理意义
Km值
① Km等于酶促反应速度为最大反应速度一半时的底物浓 度。 ② 意义: 意义:
a) Km是酶的特征性常数之一; 是酶的特征性常数之一; b) Km可近似表示酶对底物的亲和力; 可近似表示酶对底物的亲和力; c) 同一酶对于不同底物有不同的 同一酶对于不同底物有不同的Km值。 值
2.酶活力单位和比活力表示方式 (1)酶活力单位
惯用单位 :酶促反应在单位时间内生成一定量的产物或消 耗一定数量的底物所需的酶量 。 国际单位(IU):在特定条件下,每分钟催化1μ mol底物 国际单位(IU):在特定条件下,每分钟催化1μ mol底物 ):在特定条件下 转化为产物所需的酶量为一个国际单位。(1976) 转化为产物所需的酶量为一个国际单位。 1催量(1 kat)是指在特定条件下,每秒钟使 催量( 催量 )是指在特定条件下,每秒钟使1mol底物转化 底物转化 产物所需的酶量。 产物所需的酶量。(1979) Kat与IU的换算:1IU=16.67×10-9Kat, 与 的换算 的换算: = × , 1Kat=6×107IU = ×

13第十三章 RNA的合成

13第十三章 RNA的合成

2. rRNA的合成
rRNA基因位于染色体的特殊区域称核仁组织 者(nucleolar organizer)。每一个转录单位包 括28S,5.8S及18S rRNA。 RNA 聚合酶I识别 位于非转录间隔区上的启动子,其序列大约位 于-40到+10和-150到-110。首先TFI结合到启动 子上,为此,导致RNA 聚合酶I识别启动子。 当RNA聚合酶I达到下一个转录单位的启动子 时,转录便在非转录间隔区终止。
6×105 分子量 5.5×105 分布 核仁 核质 转录产物 5.8S、18S、 mRNA前体 28S rRNA前体 对利福平 不敏感 不敏感 • 敏感性 对鹅膏蕈碱 不敏感 非常敏感 • 的敏感性
• •
二.转录因子
真核生物转录过程还需要一些蛋白质因子参与,这些因子 能结合到DNA的特殊序列并且与RNA聚合酶结合,促进转录, 这些蛋白因子称转录因子(transcription factors)
第一节 参加RNA合成的酶类与蛋白因子
一.DNA指导的RNA聚合酶 (DNA directed RNA polymerase,DDRP)是RNA合成中最主要
• • • • 的酶类, RNA聚合酶催化如下反应: 1. 双链DNA中的一条链作为RNA合成的模板。 2.四种核糖核苷三磷酸(即ATP、GTP、CTP和UTP) 是该酶的底物。 3. 需要二价金属离子,如Mg2+和Mn2+。
DNA转录单位
RNA聚合酶II 外显子1 内含子 剪切及加入3’多聚A
m7Gppp m7Gppp
外显子2
RNA初 级转录 物 AAAn
甲基化修饰 CH3
m7Gppp 细胞核 细胞浆 m7Gppp
转移到胞浆 RNA拼接

Cellulase Assays 纤维素酶活性测定

Cellulase Assays 纤维素酶活性测定

Chapter 14Cellulase AssaysY.H. Percival Zhang, Jiong Hong, and Xinhao YeSummaryCellulose is a heterogeneous polysaccharide, and its enzymatic hydrolysis requires endoglucanase, exoglucanase (cellobiohydrolase), and b-glucosidase to work together. We summarize the most commonly used assays for individual enzymes and cellulase mixture.Key words:b-Glucosidase, Cellobiase, Cellobiohydrolase, Cellulose, Cellulase assay, Endoglucanase, Sugar assay1. IntroductionCellulose, which is the most abundant renewable biologicalresource, is produced mainly by plant photosynthesis. Cellulosebiodegradation mediated by cellulases or cellulolytic microor-ganisms releases organic carbon in plant, animal, and microbialsediments back to the atmosphere as carbon dioxide and methane.Complete enzymatic crystalline cellulose hydrolysis requires threetypes of enzymes (endoglucanase, exoglucanase or cellobiohy-drolase (CBH), and b-glucosidase) to work together (1–4).Physical heterogeneity of the cellulosic materials and the com-plexity of cellulase enzyme systems (synergy and/or competition)on solid enzyme-accessibility-limited substrate surfaces presentsome challenges for cellulase activity assays (5–8). A number ofcellulase activity assays have been summarized (5, 6). In thischapter, we describe the common cellulase activity assays includingthe total cellulase assays, b-glucosidase assays, endoglucanaseassays, and exoglucanase (CBH) assays. This chapter will providesome useful guidance, especially in Subheading4.Jonathan R. Mielenz (ed.), Biofuels: Methods and Protocols, Methods in Molecular Biology, vol. 581DOI 10.1007/978-1-60761-214-8_14, © Humana Press, a part of Springer Science + Business Media, LLC 2009213214Zhang, Hong, and YeDNS (3,5-dinitrosalicylic acid) reagent. Dissolve 10.6 g of DNS and 19.8 g of NaOH in 1,416 ml of distilled water. After com-plete dissolution, add 360 g of Rochelle salts (sodium potassium tartrate), 7.6 ml of melted phenol (at 50°C) (see Note 1), and 8.3 g of sodium metabisulfite, and then mix well. Titrate 3 ml of the DNS reagent using 0.1 M HCl using the phenolphthalein endpoint pH check. It should take 5–6 ml of HCl for a transi-tion from red to colorless. Add NaOH if required (2 g of NaOH added = 1 ml of 0.1 M HCl used for 3 ml of the DNS reagent) (see Note 2).Citrate buffer (1 M, pH 4.5). Dissolve 210 g of citric acid monohydrate in 750 ml of distilled water, and add 50–60 g solid NaOH until pH is 4.3. Dilute the solution to nearly 1,000 ml and check the pH. If necessary, add NaOH to adjust the pH to 4.5.Citrate buffer (50 mM, pH 4.8). Dilute 1 M citrate buffer (pH 4.5) by adding 19 times distilled water.Filter paper strip (50 mg, 1.0 × 6.0 cm). Cut 1.0 × 6.0 cm Whatman No. 1 paper strips with a paper cutter (see Note 3).Glucose standard stock solution – 10 g/l (see Note 4).1. Tris–HCl buffer (0.5 M Tris, pH 7.0, 0.1 M CaCl 2, and optional 1.5% NaN 3). Prepare 0.5 l of 1 M Tris–HCl buffer (pH 7.0), dissolve 11.1 g of CaCl 2 and 15 g NaN 3, and add distilled water to make up to 1 l.2. Dithiothreitol (DTT, 0.5 M). The DTT solution can be stored at 4°C for at least a half year. Less costly cysteine can replace DTT (9).3. Avicel suspension solution (24.4 g/l). Suspend 20 g of completely dry Avicel (FMC 105 or Sigmacell 20) in 820 ml of distilled water with a magnetic stirrer.4. Glucose standard solution – 1 g/l.5. Phenol aqueous solution (5% w/v). Store at 4°C in dark-ness.6. Sulfuric acid ~98% w/w.1. Sodium acetate buffer, 0.1 M, pH 4.8.2. p NPG (5 mM) in acetate buffer. Dissolve 0.1576 g of p NPG in 100 ml acetate buffer.3. Clycine buffer (0.4 M) pH 10.8. Dissolve 60 g of glycerin in 1,500 ml of distilled water, add 50% w/v NaOH until the pH is 10.8, and then dilute to 2 l.4. p -Nitrophenol (pNP; 20 g/l) in acetate buffer (see Note 5).2. Materials2.1. Total Cellulase Assays2.1.1. Filter Paper Activity Assay2.1.2. Anaerobic Cellulase Assay Using Avicel2.2. b -Glucosidase Assays2.2.1. b -Glucosidase Assay Using p-Nitrophenyl- b -D-Glucoside (pNPG)Cellulase Assays 2151. Cellobiose (15 mM) in citrate buffer (freshly made substrate solution).2. Citrate buffer (50 mM, pH 4.8).1. Citrate buffer (50 mM, pH 4.8).2. CMC (2% w/v) in citrate buffer (above).3. DNS reagent (above).4. Glucose standard (2 g/l).1. Citrate buffer (50 mM, pH 4.8).2. CMC solution (0.05% w/v) in the citrate buffer.3. BCA Solution A. Dissolve disodium 2,2¢-bicinchoninate (97.1 mg) in a solution of 2.714 g of Na 2CO 3 and 1.21 g of NaHCO 3 with a final volume of 50 ml. Solution A will remain stable for 4 weeks at 4°C in darkness.4. BCA Solution B. Dissolve CuSO 4.5H 2O (62.4 mg) and l -serine (63.1 mg) in 50 ml of water. Solution B will remain stable for 4 weeks at 4°C in darkness.5. Working BCA reagent. Mix equal volumes of solution A and B. The reagent is to be made immediately before use.6. Glucose standard solution (0.9 g/l, 5 mM).1. Sodium acetate buffer (50 mM, pH 5.0).2. CMC solution (0.5% w/v, medium viscosity, degree of sub-stitution of 0.5–0.7) in acetate buffer.1. Congo red solution (1 g/l) prepared by dissolving 100 mg Congo red in 99 ml water and 1% ethanol.2. NaCl (1 M) solution.3. Sodium phosphate buffer (0.1 M, pH 6.5).1. CMC (1% w/v, low viscosity) in 1.5% agar medium. Dissolve CMC before adding agar and autoclave.1. CMC (1% w/v, low viscosity) in 0.8% agarose. Dissolve CMC completely before adding agarose.1. CMC (1% w/v) in sodium phosphate buffer whose pH is cho-sen depending on the specific cellulase.1. Avicel (FMC PH 101 or PH 105 or Sigmacell 20).2. Sodium acetate buffer (0.1 M, pH 4.8).3. Phenol (5%) solution.4. Sulfuric acid, ~98%.2.2.2. b -Glucosidase Assay Using Cellobiose2.3. Endoglucanase Assays2.3.1. Endoglucanase Assay Using Carboxymeth-ylcellulose (CMC)/DNS 2.3.2. Endoglucanase Assay Using CMC/Bicinchoninic Acid (BCA)2.3.3. EndoglucanaseAssay Using CMC/Viscosity2.3.4. Semiquantitative Endoglucanase Assay Based on Dye ReleaseMicrobe-SecretedEndoglucanase Assay on Agar Medium Endoglucanase Assay on Agarose Gel Endoglucanase Assay on Polyacrylamide Gel2.4. Exoglucanase Assays2.4.1. Exoglucanase Assay Using Avicel216Zhang, Hong, and Ye1. Sodium acetate buffer (1 M, pH 4.5).2. Phenol (5%) solution.3. Sulfuric acid (~98%).4. RAC (1% w/v). RAC preparation is given below.A total cellulase system consists of three enzymatic activities: endoglucanases, exoglucanases, and b -d -glucosidases. Total cellulase activities are always measured using insoluble substrates, including pure cellulosic substrates such as Whatman No. 1 filter paper, cotton linter, microcrystalline cellulose, bacterial cellulose, algal cellulose, as well as cellulose-containing substrates such as dyed cellulose, a -cellulose, and pretreated lignocellulose (2). The two most common assays (filter paper assay and anaerobic cellulase assay) are described here.FPA is the most common total cellulase activity assay recom-mended by the International Union of Pure and Applied Chemistry (IUPAC) (6). IUPAC recommends a filter paper activity (FPA) assay that differs from most enzyme assays based on soluble substrate for initial reaction rates. This assay is based on a fixed degree of conversion of substrate, i.e. a fixed amount (2 mg) of glucose (based on reducing sugars measured by the DNS assay) released from 50 mg of filter paper (i.e., both amorphous and crystalline fractions of the substrate are hydrolyzed) within a fixed time (i.e., 60 min). In part due to the solid heterogeneous substrate, reducing sugar yield during hydrolysis is not a linear function of the quantity of cellulase enzyme in the assay mixture. That is, twice the amount of enzyme does not yield two times the reducing sugar within equal time. Total cellulase activity is described in terms of “filter-paper units” (FPU) per milliliter of original (undiluted) enzyme solution. The strengths of this assay are that (1) the substrate is widely available and (2) the substrate is reasonably susceptible to cellulase activity. However, the FPA has long been recognized for its complexity and susceptibility to operator errors (10).1. Place a rolled filter paper strip into each 13 × 100 test tube.2. Add 1.0 ml of 50 mM citrate buffer (pH 4.8) to the tubes; the paper strip should be submerged in the buffer.3. Prepare the enzyme dilution series, of which at least two dilu-tions must be made of each enzyme sample, with one dilution releasing slightly more than 2.0 mg of glucose (~2.1 mg) and one slightly less than 2.0 mg of glucose (1.9 mg) (see Note 6).2.4.2. Exoglucanase Assay Using Regenerated Amorphous Cellulose (RAC)3. Methods3.1. Total Cellulase Assays3.1.1. Filter Paper Assay (FPA)ProcedureCellulase Assays 2174. Prepare the dilute glucose standards (GSs) as below:GS1: 1.0 ml of glucose standard + 4.0 ml buffer = 2 mg/ml(1.0 mg/0.5 ml).GS2: 1.0 ml of glucose standard + 2.0 ml buffer = 3.3 mg/ml(1.65 mg/0.5 ml).GS3: 1.0 ml of glucose standard + 1.0 ml buffer = 5 mg/ml(2.5 mg/0.5 ml).GS4: 1.0 ml of glucose standard + 0.5 ml buffer = 6.7 mg/ml (3.35 mg/0.5 ml).Add 0.5 ml of GS1–4 solutions to 13 × 100 mm test tubes,and add 1.0 ml of 0.050 M citrate buffer.5. Prepare the blank and controls.Reagent blank (RB): 1.5 ml of 50 mM citrate buffer.Enzyme controls (EC1–5): 1.0 ml of 50 mM citrate buffer +0.5 ml enzyme dilution series whose enzyme concentrations arethe same as those from E1 to E5 (see Note7).Substrate control (SC): 1.5 ml of 50 mM citrate buffer +filter paper strip.6. Prewarm the enzyme solutions, blank, and controls untilequilibrium.7. Add 0.5 ml of the enzyme dilution series to the tubes withfilter paper substrate (E1–5); add 0.5 ml of the enzymedilution series to the tubes without filter paper substrate(EC1–5).8. Incubate the tubes of E1–5, GSs, RB, EC1–5, and SC in a50°C water bath for exactly 60 min.9. Add 3.0 ml of the DNS reagent to stop the reaction, and mixwell.10. Boil all tubes for exactly 5.0 min (see Note8).11. Transfer the tubes to an ice-cold water bath.12. Withdraw ~0.5 ml of the colored solutions into 1.5-ml micro-centrifuge tubes and centrifuge at ~10,000 g for 3 min.13. Add 0.200 ml of the supernatant into 3-ml spectrophotometercuvette tubes, add 2.5 ml of water, and mix well by using apipette or by inversion several times.14. Measure absorbance at 540 nm, where the absorbance of RBis used as the blank.1. Draw a standard sugar curve (sugar along the x-axis vs. Calculationabsorbance at 540 along the y-axis), as shown in Fig. 1.2. Calculate the delta absorbance of dilute enzyme solutions(D E1–4) for E1–5 by subtraction of the sum of the absorb-ance of EC1–5 and SC.3. Calculate the real glucose concentrations released by E1–5according to a standard sugar curve.218 Zhang, Hong, and Ye4. Draw the relationship between the real glucose concentrations and their respective enzyme dilution rates (EDRs) (Fig. 1).5. Link the points less than 2 mg and greater than 2 mg by a line, and identify the EDR by using the point for 2-mg glucose based on the line (Fig. 1).6. Calculate the FPA of the original concentrated enzyme solution in terms of FPU/ml:0.37FPA EDR=where 2 mg glucose = 2 mg/(0.18 mg/m mol) × 0.5 ml × 60 min = 0.37 m mol/min/ml (see Notes 9, 10).Some cellulases or cellulosomes isolated from anaerobic environ-ments need the presence of a reducing agent and some metal ions, such as calcium, to exert maximal hydrolysis ability, for example, the cellulosome from the thermophilic anaerobic bacterium Clostridium thermocellum (11). Johnson et al. (11) established a turbidometric method based on the change of 0.6 g/l Avicel (FMC RC-591), which is a blend of microcrystalline cellulose and sodium carboxymethylcellulose, but the results often suffer from large variations. The anaerobic cellulosome assay was modified on the basis of the soluble sugar release during the initial hydrolysis period (12, 13).3.1.2. Anaerobic Cellulase Assay Using Avicel0.00.20.40.60.8Glucose standard (mg/0.5 mL)A b s o r b a n c e 450 n m−3−2E n z y m e d i l u t i o nFig. 1 The relationship of absorbance at 540 nm for the DNS assay and EDRs in terms of glucose concentration.Cellulase Assays 2191. Add 4.10 ml of the well-suspended Avicel solution into 16 × 125 mm Hungate tubes, and add 0.50 ml of Tris–HCl buffer (each sample needs triplicate tubes).2. Add the rubber stopper and seal the tubes.3. Vacuum and flush the headspace gas by ~5 psi (ultra) pure nitrogen at least three times.4. Add 0.10 ml of 0.5 M DTT solution using a syringe with a 23G needle before enzyme activity assay.5. Prewarm the tubes in a water bath at 60°C.6. Prepare the enzyme solution.7. Add 0.30 ml of the dilute enzyme solution series into the tube using a syringe with a 23G needle. 8. After the first 10 min of adsorption and reaction, withdraw~0.5 ml of well-suspended sample using a syringe with a 21G needle as the starting point for enzymatic hydrolysis. The larger gauge needle is needed for homogeneous sampling of cellulose slurry.9. Shake the tubes continuously or manually mix them every 10–15 min.10. Withdraw another 0.50 ml of well-mixed Avicel suspen-sion every 1 h using a syringe with a 21G needle into the precooled 1.5-ml microcentrifuge tubes or stop the reaction after 1 h by transferring to an ice-cooled water bath.11. Centrifuge the samples in 1.5-ml microtubes at 13,000 g for 3 min. 12. Measure total soluble sugars in the supernatants by the phenol–sulfuric acid assay.13. Calculate the net soluble sugar release during the hydrolysis process by subtraction of the sugar at the starting point.14. Determine enzyme activity at a linear range between sugars released and enzyme concentrations.1. Add 0.7 ml of sugar-containing solution to 13 × 100 mm disposable glass tubes, and mix with 0.7 ml of 5% phenol solution.2. Add3.5 ml of concentrated sulfuric acid with vigorous mixing (see Note 11). 3. Read absorbance at 490 nm after cooling to room tempera-ture (e.g., 20–30 min).b -Glucosidase can cleave b -1,4-glucosidic bonds of soluble substrates, including cellobiose, longer cellodextrins with a DP from 3 to 6, and chromogenic substrates, such as p -nitroph-enyl-b -d -glucoside, p -nitrophenyl b -d -1,4-glucopyranoside,ProcedurePhenol–Sulfuric Acid Assay(A Linear Range from Sugars in the Samples from 0.005 to 0.1 g/l)3.2. b -Glucosidase Assays220Zhang, Hong, and Yeb -naphthyl-b -d -glucopyranoside, 6-bromo-2-naphthyl-b -d - glucopyranoside, and 4-methylumbelliferyl-b -d -glucopyranoside (2). The term “cellobiase” is often misleading because of this key enzyme’s broad substrate specificity.This p NPG method is an initial reaction rate assay (6).1. Add 1.0 ml of p NPG solution and 1.8 ml of acetate buffer into test tubes.2. Equilibrate at 50°C.3. Prepare the enzyme dilution series.4. Add 0.2 ml of diluted enzymes into the tubes containing thesubstrate, and mix well. 5. Enzyme blanks: Add 0.2 ml of diluted enzymes into the tubescontaining 2.8 ml of acetate buffer, and mix well; Substrate blank: Add 1.0 ml of pNPG solution and 2.0 ml of acetate buffer into test tubes. 6. Incubate all tubes at 50°C for 15 or 30 min. 7. Add 4.00 ml of glycine buffer to stop the reaction.8. Measure the absorbance of liberated products of p -nitrophenolat 430 nm based on the substrate blank. 9. Read the net absorbance of the enzyme solutions bysubtracting readings of the enzyme blanks.10. Determine p -nitrophenol release on the basis of the known concentration of p -nitrophenol diluted by glycine at 430 nm.11. Calculate the enzyme activity on the basis of the linear range between absorbance and enzyme concentrations.The b -glucosidase based on cellobiose assay recommended by IUPAC is based on a fixed amount (1 mg) of glucose formation from cellobiose (6). The glucose concentrations in the cellobiose reaction mixture are determined using at least two different enzyme dilutions. One dilution should release slightly more and one slightly less than 1.0 mg (absolute amount) of glucose in the reaction conditions.1. Dilute the enzyme solution by citrate buffer in a series. At least two dilutions must be made of each enzyme sample inves-tigated. One dilution should release slightly more and one slightly less than 1.0 mg (absolute amount) of glucose in the reaction conditions (i.e., 0.5 mg glucose released/ml).2. Add 1 ml of diluted enzyme solution (DES) to the tubes.3. Equilibrate the enzyme solutions and substrate solutions at50°C. 4. Add 1.0 ml of substrate solution into the tubes containingthe enzyme solution.3.2.1. b -Glucosidase Assay Using pNPG Procedure3.2.2. b -Glucosidase Assay Using CellobioseProcedureCellulase Assays 2215. Incubate at 50°C for exactly 30 min.6. Immerse the tubes in boiling water for exactly 5.0 min to stop the reaction.7. Transfer the tubes to a cold water bath.8. Substrate Blank: A mixture of 1.0 ml of cellobiose solutionand 1.0 ml of citrate buffer. Enzyme Blanks: A mixture of 1.0 ml of cellobiose solution and 1.0 ml of DESs. Treat substrate and enzyme blanks identically as the experimental tubes (i.e., equilibrate at 50°C, heat, boil, and cool).9. Determine glucose release using a commercial glucose oxidase kit (GOD) or a glucose hexose kinase and glucose-6 phosphate dehydrogenase kit (HK/G6PDH) or HPLC.10. Measure the absorbance of all solutions based on the substrate blank.1. Calculate the delta absorbance of dilute enzyme solutions by subtracting absorbance of the respective enzyme blanks. 2. Calculate the real glucose concentrations released according to a standard glucose curve by the enzyme kit. 3. Link the points less than 1 mg and greater than 1 mg by a line, and determine the EDR by using the point that is supposed to produce 1 mg glucose. 4. Calculate cellobiase solution activity (IU/ml) (see Note 12):=0.0926Cellobiase .EDRb -Glucosidase activity can be measured on the basis of initial reaction rates of cellobiose by combining the methods of Subheading 3.2.1 and 3.2.2. The hydrolysis product – glucose – can be measured by the glucose HK/G6PDH kit (14).Endo-b -1,4-D-glucanase (EC 3.2.1.4) randomly cleaves accessible intermolecular b -1,4-glucosidic bonds on the surface of cellulose. Because insoluble cellulose has very low accessible fractionation of b -glucosidase bonds to cellulase (3, 8, 15), water-soluble cellulose derivatives such as CMC and hydroxyethylcellulose (HEC) are commonly used for endoglucanase activity assays. The hydrolysis can be determined by measuring the changes in reducing sugars or viscosity or color. Since CMC is an anionic substrate, its properties change with pH. Nonionic substrates such as HEC are recom-mended sometimes.The IUPAC-recommended endoglucanase (CMCase) assay is a fixed conversion method, which requires 0.5 mg of absolute glucose released under the reaction condition (6). The reducing end concentration is measured by the DNS method.Calculation3.2.3. b -Glucosidase Assay Using Cellobiose3.3. Endoglucanase Assays3.3.1. Endoglucanase Assay Using CMC/DNS222Zhang, Hong, and Ye1. Prepare the enzyme dilution series, of which at least two dilutions must be made of each enzyme sample, with one dilution releasing slightly more than 0.5 mg of glucose and one slightly less than 0.5 mg of glucose.2. Add 0.5 ml of the DESs into test tubes with a volume of atleast 25 ml. 3. Equilibrate the enzyme solution and substrate solution at50°C. 4. Add 0.5 ml of the CMC solution to the test tubes and mixwell. 5. Incubate at 50°C for 30 min.6. Add 3.0 ml of DNS solution and mix well.7. Boil for exactly 5.0 min in vigorously boiling water.8. Place the tubes in an ice-cooled water bath to quench thereaction. 9. Add 20 ml of distilled water and seal with parafilm or by asimilar method. Mix by inverting the tubes several times.10. Read the absorbance at 540 nm based on the substrate blank.11. Prepare the substrate blank (0.5 ml of CMC solution + 0.5 ml of citrate buffer) and the enzyme blanks (0.5 ml of CMC solution + 0.5 ml of dilute enzyme solutions). Treat substrate and enzyme blanks identically as the experimental tubes.12. Prepare the glucose standards:GS1 – 0.125 ml of 2 mg/ml glucose + 0.875 ml of buffer.GS2 – 0.250 ml of 2 mg/ml glucose + 0.750 ml of buffer.GS3 – 0.330 ml of 2 mg/ml glucose + 0.670 ml of buffer.GS4 – 0.500 ml of 2 mg/ml glucose + 0.500 ml of buffer.13. Calculate the glucose released by the enzyme solutions with deduction of the enzyme blank absorbance based on the glucose standard curve.14. Draw the relationship between the real glucose concentra-tions and their respective EDRs.15. Link the points less than 0.5 mg and greater than 0.5 mg by a line, and identify the EDR by using the point for 0.5 mg glucose.16. Calculate the CMCase activity of the original concentrated enzyme solution in terms of IU/ml:=0.1CMCase 85EDRThis initial reaction rate enzymatic assay can be conducted at a very low enzyme concentration. The reducing end concentrationProcedure3.3.2. Endoglucanase Assay Using CMC/BCAis measured by the BCA method, in which the color development reaction is conducted at 75°C in order to avoid b -glucosidic bond cleavage during the color-development process (16).1. Dilute the enzyme solution extensively (e.g., 1,000-fold) using the 50 mM citrate buffer and prepare the dilute enzyme solution series.2. Add 1.8 ml of CMC solution into 13 × 100 mm test tubes.3. Equilibrate at 50°C water bath.4. Add 0.2 ml of DES and mix well.5. Incubate at 50°C for 10 min.6. Add 2 ml of working BCA reagents and mix well.7. Incubate at 75°C for 30 min.8. Read absorbance at 560 nm after subtracting the readings forthe enzyme blanks and the substrate blank.9. Calculate the enzyme activity based on a linear range between reducing end concentrations and enzyme concentrations.Substrate blank: 1.8 ml of CMC solution + 0.2 ml of citrate buffer; enzyme blanks: 1.8 ml of CMC solution + 0.2 ml of dilute enzyme solutions. Treat blanks identically as the experimental samples.Glucose standard: 1 ml of 5 mM glucose diluted by 50 mM citrate buffer by 100-fold to 50 m M glucose standard solution; prepare the sugar standards as below:GS1 – 0.4 ml of 50 m M glucose + 1.6 ml of buffer.GS2 – 0.8 ml of 50 m M glucose + 1.2 ml of buffer.GS3 – 1.2 ml of 50 m M glucose + 0.8 ml of buffer.GS4 – 1.6 ml of 50 m M glucose + 0.4 ml of buffer.GS5 – 2.0 ml of 50 m M glucose.This initial reaction rate assay is based on the reduction in specific viscosity of soluble cellulose derivatives such as CMC and HEC (2). Both endoglucanase and exoglucanase can release new reducing sugar ends from soluble substrates. Within a limited degree of hydrolysis, endoglucanase can decrease specific viscosity greatly, and exoglucanase can decrease specific viscosity slowly (7).1. Add 6.0 ml of prewarmed CMC solution in a water bath at 30°C into an Ostwald viscometer (water flow time of 15 s at 30°C) (see Note 13).2. Add 1.0 ml of the prewarmed DES (see Note 14).3. Determine the flow rates every 5 or 10 min.4. Calculate specific viscosity (h sp ):h −=sp 0t t t Procedure3.3.3. Endoglucanase Assay Using CMC/ViscosityProcedurewhere t is the effluent time of the buffer (s) and t 0 is the efflux time of the buffer (s).5. Plot the increasing rate of the reciprocal of the specific viscosity against the enzyme concentration; a linear relation should be obtained.6. Calculate unit of activity arbitrarily from the linear relation-ship between enzyme concentration/rate of increase of reciprocal of the viscosity of the CMC solution (see Note 15).Endoglucanase activity can be detected semiquantitatively on solid supports by staining polysaccharides with various dyes because these dyes are adsorbed only by long chains of polysac-charides. These methods are suitable for monitoring large numbers of samples but differences in enzyme activity levels of less than twofold are difficult to detect by eye. A linear relationship between the halo diameter and the logarithm of endoglucanase activity can be established as D = K × log(A ) + N , where the D is the diameter, A is the enzyme activity, and K and N are parameters determined by the standard curve of the known enzyme activity solutions. The activity of unknown samples can be calculated on the basis of the standard curve. Three procedures are described involving in vivo as well as in vitro endoglucanase detection.1. Inoculate the endoglucanase-secreted microorganisms on the solid CMC medium. The growth time depends on the growth rate of the microorganism and enzyme activity (see Note 16).2. Stain a 9-cm Petri dish by adding 20 ml of Congo red solution at room temperature for 30 min.3. Rinse the residual dye on the dish using distilled water.4. Destain Congo red with ~20 ml of 1 M NaCl for 30 min. If the halos are not clear, destain the dish by another ~20 ml of NaCl solution.5. Detect the clear, weak yellow halos for endoglucanase activity with the red background.6. Option: In order to increase halo contrast, add ~20 ml of 5% acetate acid or 1 M HCl to the plate at room temperature for 10 min, and pour off. The background of the plate will turn blue.1. Pour ~20 ml of the melted CMC agarose solution (~50°C) into a 9-cm Petri dish.2. Drill wells on the solidified agarose gel with a cork borer, and remove the agarose particles in the wells by suction or a pair of tweezers (see Note 17).3. Add 10–20 m l of the enzyme solution into the holes.3.3.4. Semiquantitative Endoglucanase Assay Based on Dye ReleaseMicrobe-Secreted Endoglucanase Assay on Agar Medium ProcedureEndoglucanase Assay on Agarose Gel Procedure4. Put the plate in the incubator (37°C or desired temperature) for several hours or even overnight.5. Wash the plate with distilled water.6. Add 10 ml of the Congo red solution and incubate at room temperature for 30 min.7. Wash the residual dye on the plate by distilled water.8. Destain the dye by using 20 ml of 1 M NaCl solution at room temperature for 30 min, and decant the destained solution (see Note 18).9. Detect the clear yellow halo with the red background.This method can separate mixed protein components by electro-phoresis and then detect endoglucanase activity on polyacrylamide gels (SDS PAGE or native PAGE). If SDS-PAGE is used, cellulase activity must be detected after SDS removal and protein re-naturation.1. Separate the protein mixtures by native or SDS PAGE.2. Rinse the gel in distilled water for 5 min.3. Soak the gel in the sodium phosphate buffer with gentle shaking for 20 min, and repeat the washing procedure three times to remove the SDS.4. Transfer the gel into the CMC/phosphate buffer for 30 min.5. Rinse the gel with distilled water.6. Incubate the gel in 0.1 M sodium phosphate buffer at 40°C for 1 h.7. Stain the gel with the Congo red solution at room temperature for 30 min.8. Wash the gel with distilled water, and destain the gel in 1 M NaCl solution at room temperature for 30 min (see Note 19).Exoglucanase (CBH, EC 3.2.1.91) can release either glucose and/or cellobiose from ends of cellulose chains. Trichoderma reesei CBH1 and CBH2 cleave cellobiose units from the reducing end and the non-reducing end of cellulose chains, respectively. In contrast to endoglucanase and b -d -glucosidase, exoglucanases are difficult to measure due to the lack of specific substrates and interference from other cellulase components. Accordingly, exoglu-canases have to be assayed in the purified form. The activity of purified exoglucanases is often estimated using Avicel. Avicel is a good substrate for exoglucanase activity assay because of its highest ratio of end/accessibility (3, 7). To some extent, Avicelase is regarded as synonymous with exoglucanase or CBH. In addition, amorphous cellulose can be used for determining of exogluca-nase activity.Endoglucanase Assayon Polyacrylamide GelProcedure3.4. Exoglucanase Assays。

生化名词解释(英文)

生化名词解释(英文)

Definition in English:Chapter 1 Structure and Function of Proteins1) peptide unit; 2) motif; 3) domain; 4) protein denaturation; 5) pI of the protein Peptide unit : The partial double-bond character of the peptide bond that links the COOH carbon and the N of an amide makes Cα1, C, O, N, H, Cα2 six atoms coplanar, Cα1 and Cα2 are trans to each other, the semi-rigid plane composed of those six atoms is termed as peptide unit. A linear sequence of amino acids residues linked together by peptide bondsMotifs are grouping of secondary structural elements that fold to near each other in space and have special functions, such as αα motif, βαβ motif, zinc finger motif. Some motifs consist of only a few conserved functionally important AAs rather than super-secondary structures. e.g. RGD (Arg-Gly-Asp ) motifDomain:the tertiary structure of some proteins can be divided into one or more relatively independent compact regions that may be joined by a flexible segment of the chain, these compact units called domains. one protein may contain several domains, the individual domains have separate functions.Protein denaturation: protein spatial structure is especially sensitive to denaturing agents (high T, urea, strong acids or bases, organic solvents, detergents, heavy metal ions) These agents result in unfolding and disorganization of protein spatial structure without change in primary structure, and associate with loss of biological activity.pI of the protein:Proteins are amphoteric molecules, contain dissociating groups of opposite charge, -COOH, -NH2 at end terminus and side chainsthe net charge of these molecules in solution depends on the pH. pI is the pH at which protein molecular becomes electrically neutral, has no net electric charge.Chapter 2 Enzyme1) active site of enzyme; 2) isoenzyme; 3) zymogen and activation of zymogen4)Allosteric enzymes and allosteric regulation of enzymes 5) Covalent modification of enzymesActive site of enzyme is a three-dimensional, local region of the enzyme, the region is composed of several essential groups of AAs, that has special spatial structure which specifically bindssubstrate and catalyzes it to become product. Coenzymes or prosthetic groups can be involved in the active site.Isozymes(isoenzymes) are a group of enzymes, which catalyze the same reaction but have different protein structure, physicochemical and immunological properties.Zymogens:some enzymes are synthesized and secreted as large inactive precursors called zymogens or proenzymes. Zymogens are activated by the irreversible hydrolysis one or more peptide bonds and forming or exposing the active site in the enzyme molecule.Allosteric enzyme: the large enzyme whose activity can be modulated in the presence of allosteric effector at an allosteric site. The allosteric binding site and catalytic site are distinct and separated spatially on enzyme, and the enzyme activity can be regulated either positively or negatively Allosteric regulation of enzyme: small allosteric effectors, which generally have little or no structural similarity to the substrate, binding to allosteric site of the enzyme by non-covalent bonds triggers changes in enzyme conformation that alter the catalytic capacity of the enzyme. An allosteric activator increases the enzyme activity, while an allosteric inhibitor decreases the activity of an enzyme.covalent modification of enzyme: the structure and activity of many enzymes can be altered reversibly through covalent modification by another enzyme.The most common modification is phosphorylation /dephosphorylation, which is reversible addition and removal of a phosphate at HO-group of Ser, Thr and Tyr.Chapter 3 Metabolism of Carbohydrates1) glycolysis; 2) substrate-level phosphorylation; 3) gluconeogenesis 4) Cori cycle Glycolysis: Converts one molecule of glucose into two molecules of lactate; Anerobically; Occur in cytoplasm; Generate two molecules of ATP and NADH is zero.Substrate-level phosphorylation: ATP is formed by the direct transfer of a phosphate group from a high-energy substrate to ADP; Give an example.Gluconeogenesis: The formation of new glucose from noncarbohydrate, including lactate, amino acid and glycerol et al; is necessary for keeping blood sugar level; the main site of gluconeogenesis is liver.Cori cycle: To meet the glucose needs at exercise; Muscles at work produce lactate from glycolysis when oxygen becomes limiting; Lactate is transported from the muscles to the liver via the bloodstream; In the liver, lactate is converted (via gluconeogenesis) back to glucose, where it is dumped back into the bloodstream for transport to muscle.Chapter 4 Metabolism of Lipids1) Mobilization of triacylglycerol;2) β-oxidation of fatty acids; 3) ketone bodies; 4) Essential fatty acidMobilization of triacylglycerol: triacylglycerols stored in adipose tissue are hydrolyzed; the products glycerol and fatty acid are transported to tissues; fatty acids can be oxidized for energy production; the principal enzyme in mobilizing stored fat is triacylglycerol lipase, which is hormone-sensitive.β- Oxidation of fatty acid: Inside of the mitochondrion; fatty acyl-CoAs are oxidized in a series of cycles that each cycle release a two-carbon fragment, in the form of acetyl-CoA; Each cycle involves four reactions-dehydrogenation, hydration, dehydrogenation, and thiolytic cleavage. Ketone bodies: the ketone bodies acetoacetate, acetone, and β-hydroxybutyrate form in the mitochondrial matrix of liver; are intermediates of FA oxidation that are utilized in extrahepatic tissues; overproduced results in ketoacidosis.Essential fatty acid: Essential fatty acids are called essential because we cannot make them naturally and require dietary sources; are necessary for normal functions of the body such as a healthy nervous system (or other example); give an example of essential fatty acid.Chapter 5 Biological Oxidation1) Biological Oxidation;2) Respiratory chain (electron transfer chain); 3) Oxidative Phosphorylation ; 4) P/O ratioBiological Oxidation: The oxidation taking place in organism is termed biological oxidation. It mainly refers to the enzymatic steps in the oxidative degradation of carbohydrates, fats, and amino acids to CO2 , H2O and energy. Some of the energy is used for the generation of ATP from ADP and Pi, and the other is released as heat energy.Respiratory chain (electron transfer chain):The mitochondrial respiratory chain consists of a series of enzymes with prosthetic groups capable of accepting and donating either one or twoelectrons, most of which are membrane-bonded proteins. Each component of the chain can accept electrons from the preceding carrier and transfer them to the following one, in a specific sequence. Oxidative Phosphorylation :The process of the enzymatic phosphorylation of ADP to ATP coupled to electron transfer from a substrate to molecular oxygen is termed Oxidative phosporylation.P/O ratio: The efficiency of oxidative phosphorylation is determined by the P/O ratio, which is a measure of the molecules of ATP made per mol oxygen atom consumed (per pair of electrons carried through the electron transport).Chapter 6 Metabolism of Amino Acids1)essential amino acids;2)putrefaction of protein;3)One-carbon units;4)alanine - glucose cycle essential amino acids:Some of amino acids that can not be synthesized by human , must be obtained in the diet to meet body’s metabolic needs.they are called essential amino acids, including Val, Ile, Leu, Thr, Met, Lys, Phe, Trp.putrefaction of protein: After the absorption of most of protein, intestinal bacterial activity act on the residue passed into the large intestine, that is called putrefaction of protein.Most of the producer of the putrefaction of protein are harmful to human beings, for example: amine, ammonia, phenol, indole and sulfureted hydrogen.One-carbon units:One-carbon units are groups including one carbon atom which are produced by AA catabolism and carried by tetrahydrofolate, for example: methyl group, methylene group, methenyl group, formyl group and formimino group. One-carbon units take part in the composition of purine and thymine.alanine - glucose cycle:the significances of Glucose-alanine cycle:* To transport ammonia in the nontoxic forms of alanine from muscles to liver.* To regulate blood glucose indirectly and to supply available pyruvate for muscles.Chapter 7 Biochemistry In Liver1. Biotransformation:Some non-nutrientsubstances can be converted to more polar metabolites by various chemical reactions such as oxidation, reduction, hydrolyzation, and conjugation, which are then excreted from the body. These processes mainly occur in liver.2. Enterohepatic circulation of bile acids:Bile acids are secreted from the liver through bile ducts to the gallbladder, then into the intestine, where they aid the emulsification of dietary lipids. The bile acids are then reabsorbed (active and passive ) in the lower small intestine, and returned to the liver for reuse through the portal vein.Chapter 8 Blood2,3-BPG shunt pathway:glucose1,3-BPG 3-(15% ~ 20%)2,3-BPG combines with hemoglobin, causing a decrease in affinity for oxygen and a displacement of the oxyhemoglobin dissociation curve to the right. Thus, its presence in the red cells helps oxyhemoglobin to unload oxygen.。

生物化学-酶

生物化学-酶

酶一级结构的差别也决定了催化性质的不同, 如胰蛋白酶、 胰糜蛋白酶和弹性蛋白酶三种蛋白 酶的活性中心Ser残基附近都有一个在立体结构上 的“口袋”状结构。由于三种蛋白酶的口袋”状结 构不同,决定其与不同底物结合即有不同特异性。
酶的特异的三维空间结构是酶催化功能的基础。 酶的二、三级结构是维持酶的活性中心空间构象的 必需结构。
酶的命名包括习惯命名和系统命名,酶可分为六类。 酶与疾病发生、诊断、治疗等密切相关。
➢一、酶的概念 酶是由生物活细胞产生的具有高效催化功能
和高度专一性的一类特殊蛋白质,又叫生物催化 剂•.绝大多数的酶都是蛋白质。
酶的化学本 质是什麽?
酶的概念
• 一、相关概念 • 酶催化的生物化学反应,称为酶促反应。 • 被酶的催化的物质称为底物(S) • 反应的生成物为产物(P) • 酶所具有的催化能力称酶的活性. • 酶失去催化能力称酶的失活.
第四章 酶 (Enzymes)
内容简介
酶是具有高度催化效率及高度特异性的蛋白质。 酶通过多种机制降低反应活化能使反应速率增加。 酶分子一级结构及空间结构是催化功能的基础。 酶促反应速率受到[S]、[E]、pH、T、抑制剂及激活
剂的影响
酶活性可受到别构调节、共价修饰、酶原激活、关键 酶、多酶体系、同工酶等调节
H N C O
COOH CH
R6
氨基酸
氨基酸
消化道中各种蛋白酶的专一性
3.立体异构特异性:一些酶仅能催化一种立体异
构体进行反应,或其催化的结果只产生一种立体异
构体,酶对立体异构物的选择性称为立体异构特异
性(stereospecificity)。
L-乳酸
D-乳酸
H
H
C
OH

生物化学英文课件-酶

生物化学英文课件-酶

immuno- characters.
Lactate dehydrogenase (LDH)
LDH
pyruvate Lactate
H subunit Tetramer LDH (M.W. 130,000)
M subunit Subunit: H (Chr12) M (Chr11)
Cellulose Acetate Membrane Electrophoresis result of LDH isoenzymes _
monomeric enzyme:only contain a
polypeptide chain with tertiary structure. oligomeric enzyme:contain two or more polypeptide chains associated by noncovalent bonds.
3.6 Clinical Applications of Enzymes
3.1 Structure and Function
of Enzymes
3.1.1 structure of enzymes
3.1.1.1 composition of Enzyme Molecules
Simple enzymes——proteins enzymes Conjugated enzymes (holoenzyme) apoenzyme
approximation and orientation of reactants
as bridge between substrate and enzyme
neutralize negative charges to decrease electrostatic repulsion

Enzyme酶

Enzyme酶
研究作出更大的贡献
第一节 酶的分子结构与功能 一、酶的分子组成
酶分子可根据其化学组成的不同,分为两类: 单纯酶
酶→
酶蛋白
结合酶(全酶)→
辅酶
辅助因子→ 辅基
金属离子
由酶蛋白与辅助因子组成的酶称为全酶。
与酶蛋白疏松结合并与酶的催化活性有关 的耐热低分子有机化合物称为辅酶。
与酶蛋白牢固结合并与酶的催化活性有关 的耐热低分子有机化合物称为辅基。
催化反应历程
一般化学反应历程: S →P
酶促反应历程: S+E →
ES → E + P
(二)具有高度的底物特异性 l 一种酶只作用于一种或一类化合物,以促进一定的化 学变化,生成一定的产物,这种现象称为酶作用的特异 性。
一种酶只作用于一类化合物或一定的化学键,以促进一定的化学变化, 并生成一定的产物,这种现象称为酶的特异性或专一性(specificity)。 受酶催化的化合物称为该酶的底物或作用物(substrate)。
三、酶促反应的机制
(一)中间复合物学说
酶催化时,酶活性中心首先与底物结合生成一种 酶-底物复合物(ES),此复合物再分解释放出酶, 并生成产物。
S + E → SE → E + P
(二) 诱导契合学说
当底物与酶接近时,底物分子可以诱导酶活性中 心的构象发生改变,使之成为能与底物分子密切 结合的构象 。
丙二酸对琥珀酸脱氢酶的竞争性抑制
H2N-
-COOH对氨基苯甲酸
H2N-
-SO2NH对氨R 基苯磺酰胺
磺胺类药物对二氢叶酸合成酶的竞争性抑制
某些抑制剂的化学结构与底物相似,因而能与 底物竟争与酶活性中心结合。当抑制剂与活性 中心结合后,底物被排斥在反应中心之外,其 结果是酶促反应被抑制了。

国际酶制剂企业名单

国际酶制剂企业名单

国际酶制剂(酵素)企业名单(中国酶制剂网整理收集 )序号公司名称国家中文名称网页1 Novozymes A/S 丹麦诺维信2 Genencor International(Danisco收购)美国杰能科3 AB Enzymes GmbH (Associated British Foods收购)德国AB酶制剂公司4 Beldem 比利时焙乐道5 Amano Enzyme Incorporated 日本天野6 Danisco A/S 丹麦丹尼斯克7 Agilent 美国安捷伦8 Stratagene(Agilent Technologies收购)美国—9 Maxygen Incorporated 美国—10 BioCatalytics(Maxygen收购)美国生物催化剂11 Codexis(Maxygen收购)美国克迪科思12 Juelich Fine Chemicals(Maxygen收购)德国精细化工13 Enzis(Maxygen收购)荷兰—14 Advanced Enzyme Technologies Limited 印度康帕斯15 Specialty Enzymes and Biochemicals(Advanced Enzyme Technologies收购)美国专业酶和生化药剂16 Associated British Foods plc 英国联合英国食品17 Roal Oy(Associated British Foods 收购)芬兰罗尔18 Floridienne SA NV 比利时—19 Biochem Europe(Floridienne收购)—欧洲生化20 BBI Holdings PLC 英国转乘控股21 Servac Biotech(BBI Holdings收购)加拿大—22 Theratase plc(BBI Holdings plc收购)英国—23 Biocatalysts Limited 英国生物催化24 Henkel KGaA 德国汉高25 Biozym GmbH(Henkel收购) 德国—26 ICC Industries Incorporated 美国—27 Frutarom Industries(ICC Industries收购)荷兰—28 Johnson & Johnson 美国强生29 Janssen-Ortho(Johnson & Johnson收购)加拿大—30 Shire plc 英国—31 Transkaryotic Therapies(Shire收购) 美国—32 Ultra Bio-Logics Incorporated —超生物逻辑33 Nutriteck(Ultra Bio-Logics收购)—超生物逻辑34 Campo Research Pte Limited —坎普研究35 Abbott Laboratories 美国雅培36 Cargill Incorporated 美国嘉吉37 Genentech Incorporated 美国基因泰克38 Genzyme Corporation 美国健赞39 ImaRx Therapeutics Incorporated 美国—40 Invitrogen Corporation 美国英杰41 MP Biomedicals LLC 美国MP生物医学药品42 New England BioLabs Incorporated 美国纽英伦43 PDL BioPharma Incorporated 美国—44 Promega Corporation 美国普洛麦格45 Verenium Corporation(Diversa和Celunol两公司合并组成)美国维莱尼姆46 Dyadic International Incorporated 美国—47 Allergan Incorporated 美国爱力根48 Adisseo France SAS 法国安迪苏49 Biomethodes 法国—50 Chr. Hansen A/S 丹麦克里斯蒂安·汉森51 CSL Limited 澳大利亚—52 ENMEX SA de CV 墨西哥—53 Iogen Corporation 加拿大—54 Biocon Limited 印度百康55 Kao Corporation 日本花王株式会社56 Lonza Group Limited 瑞士龙沙57 Roche Holding Limited 瑞士罗氏58 BASF AG 德国巴斯夫59 RAG AG 德国鲁尔60 Royal DSM NV 荷兰帝斯曼61 Solvay SA 比利时苏威。

tool enzyme

tool enzyme

5′ ′ 3′ ′
3′ ′
5′ ′
Cutting--- ‘cut’ what on DNA?
5' AT GC CG TA GC AT 3.4nm AT CG
3'- 5‘phosphodiester bond
-O
3'
5'
-O
O P O CH2 O
4' 3' 1' 2' 5'
O-
O P O O-
H RE hydrolyze here
平端
5’-CCCGGG-3’ 3’-GGGCCC-5’
SmaI
5’-CCC-OH + 3’-GGG- p
p -GGG-3’ OH-CCC-5’
Sticky ends can bind to any other end with the same overhanging sequence ( or the single-stranded tail) by base pairing
Sticky Ends
Ligase
--joins sugar-phosphate backbones of dsDNA with a 5’phosphate and a 3’-OH in an ATP-dependent reaction.
--forming phosphodiester bond
末端转移 酶
Alkaline phosphatase
碱性磷酸酶
Polynucleotide kinase
核酸 激酶
RNA polymerase
*template *no need of primer *start at promoter site *5’ to 3’ direction

enzyme文档

enzyme文档

enzyme⽂档Enzyme是⼀个⽤于React的JavaScript测试实⽤程序,它使得更容易断⾔,操作和遍历您的React组件的输出,它模拟了jQuery的API,⾮常直观,易于使⽤和学习。

整理相当API为中⽂,所以资料都是官⽅API翻译⽽已,代码块以最新的粘贴过来它提供三种测试⽅法:shallowrendermount先解释⼀个词:wrapper wrapper是enzyme包装好的类,以供api使⽤shallow⽅法shallow在单元测试的过程中,浅渲染将⼀个组件渲染成虚拟DOM对象,并不会渲染其内部的⼦组件,也不是真正完整的React Render,⽆法与⼦组件互动。

以下shallow浅层渲染代码案例JavaScript1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35import React from 'react';import { expect } from 'chai';import { shallow } from 'enzyme';import sinon from 'sinon';import MyComponent from './MyComponent';import Foo from './Foo';describe('<MyComponent />', () => {it('renders three <Foo /> components', () => {const wrapper = shallow(<MyComponent />);expect(wrapper.find(Foo)).to.have.lengthOf(3);});it('renders an `.icon-star`', () => {const wrapper = shallow(<MyComponent />);expect(wrapper.find('.icon-star')).to.have.lengthOf(1);});it('renders children when passed in', () => {const wrapper = shallow((<MyComponent><div className="unique" /></MyComponent>));expect(wrapper.contains(<div className="unique" />)).to.equal(true); });it('simulates click events', () => {const onButtonClick = sinon.spy();const wrapper = shallow(<Foo onButtonClick={onButtonClick} />);wrapper.find('button').simulate('click');expect(onButtonClick).to.have.property('callCount', 1);});});render⽅法介绍render静态渲染API,使⽤enzyme's 的render函数从React树⽣成HTML,并分析⽣成的HTML结构。

期刊缩写

期刊缩写

ANAL BIOANAL CHEM ANAL BIOCHEM ASSAY DRUG DEV TECHN BIOCONJUGATE CHEM BIOINFORMATICS BIOLOGICALS BIOMED CHROMATOGR BIOMOL ENG BIOTECHNIQUES BMC BIOINFORMATICS BRAIN RES PROTOC CHROMATOGRAPHIA COMB CHEM HIGH T SCR CURR OPIN BIOTECH CYTOM PART A CYTOM PART B-CLIN CY CYTOMETRY ELECTROPHORESIS HYBRIDOMA HYBRIDOM J BIOCHEM BIOPH METH J BIOMED OPT J BIOMOL SCREEN J CHROMATOGR A J CHROMATOGR B J CHROMATOGR SCI J COMPUT BIOL J FLUORESC J IMMUNOASS IMMUNOCH J IMMUNOL METHODS J LABELLED COMPD RAD
返回
期刊影响因子 0.360 1.032 0.413 1.693 1.224 0.333 1.826 8.667 4.326 4.175
11 12 13 14 15 16 17 18 19 20
AMINO ACIDS AMYLOID ANAL BIOCHEM ANNU REV BIOCHEM ANNU REV BIOPH BIOM ANTIOXID REDOX SIGN ANTISENSE NUCLEIC A APOPTOSIS APPL BIOCHEM BIOTECH ARCH BIOCHEM BIOPHYS ARCH INSECT BIOCHEM BBA-BIOENERGETICS BBA-BIOMEMBRANES BBA-GEN SUBJECTS BBA-GENE STRUCT EXPR BBA-MOL BASIS DIS BBA-MOL CELL BIOL L BBA-MOL CELL RES BBA-PROTEIN STRUCT M BBA-PROTEINS PROTEOM BBA-REV CANCER BIOCATAL BIOTRANSFOR BIOCHEM BIOPH RES CO BIOCHEM CELL BIOL BIOCHEM GENET BIOCHEM J BIOCHEM MOL BIOL EDU BIOCHEM SOC SYMP BIOCHEM SOC T BIOCHEM SYST ECOL
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Chapter 13-14-15 Enzyme part 2 Enzyme kinetics Michaelis-Menten kinetics1. Enzyme-substrate complexFormation of an enzyme-substrate complex is the first step in enzyme catalysis.中间产物学说.在酶催化的反应中,第一步是酶与底物形成酶-底物中间复合物。

当底物分子在酶作用下发生化学变化后,中间复合物再分解成产物和酶。

.许多实验事实证明了E-S复合物的存在。

E-S复合物形成的速率与酶和底物的性质有关。

.Brown(1902)和Henri(1903)提出“中间产物”假说。

(1) The existence of ES complexes has been shown in a variety of ways. ES complexes have been directly observed by electron microscopy and X-ray crystallography.. The saturation effect: at a constant concentration of an enzyme, the reaction rate increases with increasing substrate concentrations until the V max is reached.(2) Substrates are bound to a specific region of an enzyme called the active site. Much of the catalytic power of enzymes comes from their bringing substrates together in favorable orientations in enzyme-substrate (ES) complexes.. Mutations: affect the on-rate, the off-rate, kcat, …e tc. Most enzymes are highly selective in their interaction of substrates.. The active site is a three-dimensional entity formed by groups that come from different parts of the linear amino acid sequence.. 结合部位Binding site酶分子中与底物结合的部位或区域(酶的专一性). 催化部位catalytic site酶分子中促使底物发生化学变化的部位(酶催化的性质).酶的活性部位或活性中心:酶的结合部位和催化部位的总称.调控部位Regulatory site酶分子中存在着一些可以与其他分子发生某种程度的结合的部位,从而引起酶分子空间构象的变化,对酶起激活或抑制作用。

酶活性中心的通常官能团主要包括:.亲核性基团:丝氨酸的羟基;半胱氨酸的巯基;组氨酸的咪唑基。

.酸碱性基团:天冬氨酸/谷氨酸的羧基;赖氨酸的氨基;酪氨酸的酚羟基;组氨酸的咪唑基;半胱氨酸的巯基。

. The lock-and-key (锁与钥匙). The active sites of enzymes are complementary inshape to those of their substrates (Emil Fisher). The induced fit (诱导结合). the complementary active sites shapes to those of their substrates only after the substrates are bound(Daniel Koshland). Multiple weak (noncovalent) interactions between substrates and enzymes. Nonpolar character of the active sites generally. polar residues usually participate in the catalytic processes (catalytic groups with specific binding)Flexible or rigid structure. Lock-Key model of enzyme action, Emil Fischer, 1890. Induced-fit model of enzyme action, Daniel koshland, modern variation of Lock-Key model Reorientate specific functional groups into the position to catalysis (need to overcome the entropy increase)(3) Binding energy is the major source of free energy used by enzymes to lower the activation energies of reactions. Formation of each weak interaction in the ES complex is accompanied by a small release offree energy: Binding energy (ΔGb). Weak interactions are maximized when the substrate is converted to the transition state. Transition state theory. the enzyme is evolved “designed” to bind the transition state structure.2. some definition of kinetics. Velocity (速率, 反应速率) The change in the concentration of a reactant or product per unit time (单位时间内生成的产物或者底物的消失). Initial velocity (初始反应速率) The rate of the reaction immediately after mixing the enzyme and substrate. assumed that reversible reaction, linearity, …Reaction order. First-order kinetics (1级动力学). The reaction rate depends on the first order of theconcentration of a reactant.. Second-order kinetics (2级).The reaction rate depends on the concentration of tworeactants. A and B are in the rate-determining step.. Pseudo-first-order kinetics (准1级). The second-order reactions involve reactants such as water that are present in great excess. Then the reaction appears to be the first-order.. Zero-order kinetics (零级). The addition of reactant does not alter a reaction rate. The rate is constant because the reactant concentration is high enough to saturate all the catalytic sites on the enzyme molecules.3. The enzyme kineticsEnzyme kineticsThe quantitative study of enzyme catalysisprovides information about reaction rates.. Measure affinity of enzymes for substratesand inhibitors. Provide insight into catalysis or reaction mechanism. Practical application:. comprehension of the forces that regulate metabolic pathways. design of improved therapies4. Michaelis-Menton equation reflects the kinetic behavior of many enzymes. Initial velocity (Vo). measured at the beginning of the enzyme-catalyzed reaction. substrate concentration to be constant ([S] decreasedas the reaction progresses). Saturation effect (Vmax). plotting the initial velocity (Vo) against the substrate concentration ([S]). Affinity measurement (Km). Substrate concentration when velocity reached half of Vmax. Linear relationship between V o and increasing [S]. at relatively low concentrations of substrateThe investigation of enzyme rates was proposed byLeonor Michaelis and Maud Menten in 1913.单底物结合位点的单底物酶促反应Deduction of M-M equations. Quick balance assumptionE和ES之间存在平衡,ES分解生成产物的速度不足以破坏E和S之间的平衡。

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