清华大学2005年生物化学期末试卷
清华大学生物化学试卷2005_final_exam_A

B.6
C.6.5
D.7.5
E.No ATP would form under these conditions.
Answer(s): D
3.Indicate which of the following events can occur during the processing of eukaryotic mRNA transcripts.
清华大学本科生考试试题专用纸(A)
考试课程Biochemistry II 2005年1月15日
说明:1.第1-40题为选择题,每题2分,答案可能是一个或多个,请将
正确答案填在适当的横线处;
2.第41-44题为问答题,每题的分数标在了题目后面,请将答案写
在答题纸上;
3.本卷满分为100。
1. Which compound links glycolysis, nucleotide synthesis, and glycogen synthesis?
Answer(s):D
8.Consider the mRNA sequence: (5’) AAUGCAGCUUUAGCA(3’). The sequence ofthe coding strand of DNA is:
A.(5’) ACGATTTCGACGTAA(3’)
B.(3’) TTACGTCGAAATCGT(5’)
E.The value ofG’ois also negative.
Answer(s): AD
13.Which of the following isnota physiological role of nucleotides?
A.allosteric regulators
生物化学期末考试试题及答案 别人给我的复习资料

生物化学?期末考试题A1、蛋白质溶液稳定的要紧因素是蛋白质分子外表形成水化膜,并在偏离等电点时带有相同电荷2、糖类化合物都具有复原性()3、动物脂肪的熔点高在室温时为固体,是因为它含有的不饱和脂肪酸比植物油多。
()4、维持蛋白质二级结构的要紧副键是二硫键。
()5、ATP含有3个高能磷酸键。
()6、非竞争性抑制作用时,抑制剂与酶结合那么碍事底物与酶的结合。
()7、儿童经常晒太阳可促进维生素D的汲取,预防佝偻病。
()8、氰化物对人体的毒害作用是由于它具有解偶联作用。
()9、血糖全然来源靠食物提供。
()10、脂肪酸氧化称β-氧化。
()11、肝细胞中合成尿素的部位是线粒体。
()12、构成RNA的碱基有A、U、G、T。
()13、胆红素经肝脏与葡萄糖醛酸结合后水溶性增强。
()14、胆汁酸过多可反响抑制7α-羟化酶。
()15、脂溶性较强的一类激素是通过与胞液或胞核中受体的结合将激素信号传递发扬其生物()1、以下哪个化合物是糖单位间以α-1,4糖苷键相连:()A、麦芽糖B、蔗糖C、乳糖D、纤维素E、香菇多糖2、以下何物是体内贮能的要紧形式()A、硬酯酸B、胆固醇C、胆酸D、醛固酮E、脂酰甘油3、蛋白质的全然结构单位是以下哪个:()A、多肽B、二肽C、L-α氨基酸D、L-β-氨基酸E、以上都不是4、酶与一般催化剂相比所具有的特点是()A、能加速化学反响速度B、能缩短反响到达平衡所需的时刻C、具有高度的专一性D、反响前后质和量无改E、对正、逆反响都有催化作用5、通过翻译过程生成的产物是:()A、tRNAB、mRNAC、rRNAD、多肽链E、DNA6、物质脱下的氢经NADH呼吸链氧化为水时,每消耗1/2分子氧可生产ATP分子数量() A、1B、2 C、3D、4.E、57、糖原分子中由一个葡萄糖经糖酵解氧化分解可净生成多少分子ATP?()A、1B、2C、3D、4E、58、以下哪个过程要紧在线粒体进行()A、脂肪酸合成B、胆固醇合成C、磷脂合成D、甘油分解E、脂肪酸β-氧化9、酮体生成的限速酶是()A、HMG-CoA复原酶B、HMG-CoA裂解酶C、HMG-CoA合成酶D、磷解酶E、β-羟丁酸脱氢酶10、有关G-蛋白的概念错误的选项是()A、能结合GDP和GTPB、由α、β、γ三亚基组成C、亚基聚合时具有活性D、可被激素受体复合物激活E、有潜在的GTP活性11、鸟氨酸循环中,合成尿素的第二个氮原子来自()A、氨基甲酰磷酸B、NH3C、天冬氨酸D、天冬酰胺E、谷氨酰胺12、以下哪步反响障碍可致苯丙酮酸尿症()A、多巴→黑色素B、苯丙氨酸→酪氨酸C、苯丙氨酸→苯丙酮酸D、色氨酸→5羟色胺E、酪氨酸→尿黑酸13、胆固醇合成限速酶是:()A、HMG-CoA合成酶B、HMG-CoA复原酶C、HMG-CoA裂解酶D、甲基戊烯激酶E、鲨烯环氧酶14、关于糖、脂肪、蛋白质互变错误是:()A、葡萄糖可转变为脂肪B、蛋白质可转变为糖C、脂肪中的甘油可转变为糖D、脂肪可转变为蛋白质E、葡萄糖可转变为非必需氨基酸的碳架局部15、竞争性抑制作用的强弱取决于:()A、抑制剂与酶的结合部位B、抑制剂与酶结合的牢固程度C、抑制剂与酶结构的相似程度D、酶的结合基团E、底物与抑制剂浓度的相比照例16、红细胞中复原型谷胱苷肽缺乏,易引起溶血是缺乏()A、果糖激酶B、6-磷酸葡萄糖脱氢酶C、葡萄糖激酶D、葡萄糖6-磷酸酶E、己糖二磷酸酶17、三酰甘油的碘价愈高表示以下何情况()A、其分子中所含脂肪酸的不饱和程度愈高B、其分子中所含脂肪酸的不饱和程度愈C、其分子中所含脂肪酸的碳链愈长D、其分子中所含脂肪酸的饱和程度愈高E、三酰甘油的分子量愈大18、真核基因调控中最重要的环节是()A、基因重排B、基因转录C、DNA的甲基化与往甲基化D、mRNA的衰减E、翻译速度19、关于酶原激活方式正确是:()A、分子内肽键一处或多处断裂构象改变,形成活性中心B、通过变构调节C、通过化学修饰D、分子内部次级键断裂所引起的构象改变E、酶蛋白与辅助因子结合20、呼吸链中氰化物抑制的部位是:()A、Cytaa3→O2B、NADH→O2C、CoQ→CytbD、Cyt→CytC1E、Cytc→Cytaa31、基因诊断的特点是:()A、针对性强特异性高B、检测灵敏度和精确性高C、有用性强诊断范围广D、针对性强特异性低E、有用性差诊断范围窄2、以下哪些是维系DNA双螺旋的要紧因素()A、盐键B、磷酸二酯键C、疏水键D、氢键E、碱基堆砌3、核酸变性可瞧瞧到以下何现象()A、粘度增加B、粘度落低C、紫外汲取值增加D、紫外汲取值落低E、磷酸二酯键断裂4、服用雷米封应适当补充哪种维生素()A、维生素B2B、V—PPC、维生素B6D、维生素B12E、维生素C5、关于呼吸链的表达以下何者正确?()A、存在于线粒体B、参与呼吸链中氧化复原酶属不需氧脱氢酶C、NAD+是递氢体D、NAD+是递电子体E、细胞色素是递电子体6、糖异生途径的要害酶是()A、丙酮酸羧化酶B、果糖二磷酸酶C、磷酸果糖激酶D、葡萄糖—6—磷酸酶E、已糖激酶7、甘油代谢有哪几条途径()A、生成乳酸B、生成CO2、H2O、能量C、转变为葡萄糖或糖原D、合成脂肪的原料E、合成脂肪酸的原料8、未结合胆红素的其他名称是()A、直截了当胆红素B、间接胆红素C、游离胆红素D、肝胆红素E、血胆红素9、在分子克隆中,目的基因可来自()基因组文库B、cDNA文库C、PCR扩增D、人工合成E、DNA结合蛋白10关于DNA与RNA合成的讲法哪项正确:()A、在生物体内转录时只能以DNA有意义链为模板B、均需要DNA为模板C、复制时两条DNA链可做模板D、复制时需要引物参加转录时不需要引物参加E、复制与转录需要的酶不同四、填空题〔每空0.5分,共15分〕1、胞液中产生的N A DH经和穿梭作用进进线粒体。
生物化学期末考试试题及答案精编

《生物化学》期末考试题 A 一、判断题(15个小题,每题1分,共15分)1、蛋白质溶液稳定的主要因素是蛋白质分子表面形成水化膜,并在偏离等电点时带有相同电荷2、糖类化合物都具有还原性( )3、动物脂肪的熔点高在室温时为固体,是因为它含有的不饱和脂肪酸比植物油多。
( )4、维持蛋白质二级结构的主要副键是二硫键。
( )5、ATP含有3个高能磷酸键。
( )6、非竞争性抑制作用时,抑制剂与酶结合则影响底物与酶的结合。
( )7、儿童经常晒太阳可促进维生素D的吸收,预防佝偻病。
( )8、氰化物对人体的毒害作用是由于它具有解偶联作用。
( )9、血糖基本来源靠食物提供。
( )10、脂肪酸氧化称β-氧化。
( )11、肝细胞中合成尿素的部位是线粒体。
( )12、构成RNA的碱基有A、U、G、T。
( )13、胆红素经肝脏与葡萄糖醛酸结合后水溶性增强。
( )14、胆汁酸过多可反馈抑制7α-羟化酶。
( )15、脂溶性较强的一类激素是通过与胞液或胞核中受体的结合将激素信号传递发挥其生物()1、下列哪个化合物是糖单位间以α-1,4糖苷键相连:()A、麦芽糖B、蔗糖C、乳糖D、纤维素E、香菇多糖2、下列何物是体内贮能的主要形式( )A、硬酯酸B、胆固醇C、胆酸D、醛固酮E、脂酰甘油3、蛋白质的基本结构单位是下列哪个:( )A、多肽B、二肽C、L-α氨基酸D、L-β-氨基酸E、以上都不是4、酶与一般催化剂相比所具有的特点是( )A、能加速化学反应速度B、能缩短反应达到平衡所需的时间C、具有高度的专一性D、反应前后质和量无改E、对正、逆反应都有催化作用5、通过翻译过程生成的产物是:( )A、tRNA B、mRNA C、rRNA D、多肽链E、DNA6、物质脱下的氢经NADH呼吸链氧化为水时,每消耗1/2分子氧可生产ATP分子数量( )A、1B、2 C、3 D、4. E、57、糖原分子中由一个葡萄糖经糖酵解氧化分解可净生成多少分子ATP?( )A、1B、2C、3D、4E、58、下列哪个过程主要在线粒体进行( )A、脂肪酸合成B、胆固醇合成C、磷脂合成D、甘油分解E、脂肪酸β-氧化9、酮体生成的限速酶是( )A、HMG-CoA还原酶B、HMG-CoA裂解酶C、HMG-CoA合成酶D、磷解酶E、β-羟丁酸脱氢酶10、有关G-蛋白的概念错误的是( )A、能结合GDP和GTPB、由α、β、γ三亚基组成C、亚基聚合时具有活性D、可被激素受体复合物激活E、有潜在的GTP活性11、鸟氨酸循环中,合成尿素的第二个氮原子来自( )C、天冬氨酸A、氨基甲酰磷酸B、NH3D、天冬酰胺E、谷氨酰胺12、下列哪步反应障碍可致苯丙酮酸尿症( )A、多巴→黑色素B、苯丙氨酸→酪氨酸C、苯丙氨酸→苯丙酮酸D、色氨酸→5羟色胺E、酪氨酸→尿黑酸13、胆固醇合成限速酶是:( )A、HMG-CoA合成酶B、HMG-CoA还原酶C、HMG-CoA裂解酶D、甲基戊烯激酶E、鲨烯环氧酶14、关于糖、脂肪、蛋白质互变错误是:( )A、葡萄糖可转变为脂肪B、蛋白质可转变为糖C、脂肪中的甘油可转变为糖D、脂肪可转变为蛋白质E、葡萄糖可转变为非必需氨基酸的碳架部分15、竞争性抑制作用的强弱取决于:( )A、抑制剂与酶的结合部位B、抑制剂与酶结合的牢固程度C、抑制剂与酶结构的相似程度D、酶的结合基团E、底物与抑制剂浓度的相对比例16、红细胞中还原型谷胱苷肽不足,易引起溶血是缺乏( )A、果糖激酶B、6-磷酸葡萄糖脱氢酶C、葡萄糖激酶D、葡萄糖6-磷酸酶E、己糖二磷酸酶17、三酰甘油的碘价愈高表示下列何情况( )A、其分子中所含脂肪酸的不饱和程度愈高B、其分子中所含脂肪酸的不饱和程度愈C、其分子中所含脂肪酸的碳链愈长D、其分子中所含脂肪酸的饱和程度愈高E、三酰甘油的分子量愈大18、真核基因调控中最重要的环节是( )A、基因重排B、基因转录C、DNA的甲基化与去甲基化D、mRNA的衰减E、翻译速度19、关于酶原激活方式正确是:( )A、分子内肽键一处或多处断裂构象改变,形成活性中心B、通过变构调节C、通过化学修饰D、分子内部次级键断裂所引起的构象改变E、酶蛋白与辅助因子结合20、呼吸链中氰化物抑制的部位是:( )A、Cytaa3→O2B、NADH→O2C、CoQ→CytbD、Cyt→CytC1E、Cytc→Cytaa31、基因诊断的特点是:( ) A、针对性强特异性高 B、检测灵敏度和精确性高 C、实用性强诊断范围广D、针对性强特异性低E、实用性差诊断范围窄2、下列哪些是维系DNA双螺旋的主要因素( )A、盐键B、磷酸二酯键C、疏水键D、氢键E、碱基堆砌3、核酸变性可观察到下列何现象( )A、粘度增加B、粘度降低C、紫外吸收值增加D、紫外吸收值降低E、磷酸二酯键断裂4、服用雷米封应适当补充哪种维生素( )A、维生素B2 B、V—PP C、维生素B6D、维生素B12E、维生素C5、关于呼吸链的叙述下列何者正确?( )A、存在于线粒体B、参与呼吸链中氧化还原酶属不需氧脱氢酶C、NAD+是递氢体D、NAD+是递电子体E、细胞色素是递电子体6、糖异生途径的关键酶是( )A、丙酮酸羧化酶B、果糖二磷酸酶C、磷酸果糖激酶D、葡萄糖—6—磷酸酶E、已糖激酶7、甘油代谢有哪几条途径( )A、生成乳酸B、生成CO2、H2O、能量 C、转变为葡萄糖或糖原D、合成脂肪的原料E、合成脂肪酸的原料8、未结合胆红素的其他名称是( )A、直接胆红素B、间接胆红素C、游离胆红素D、肝胆红素E、血胆红素9、在分子克隆中,目的基因可来自( )基因组文库 B、cDNA文库 C、PCR扩增 D、人工合成 E、DNA结合蛋白10关于DNA与RNA合成的说法哪项正确:( )A、在生物体内转录时只能以DNA有意义链为模板B、均需要DNA为模板C、复制时两条DNA链可做模板D、复制时需要引物参加转录时不需要引物参加E、复制与转录需要的酶不同四、填空题(每空0.5分,共15分)1、胞液中产生的NADH经和穿梭作用进入线粒体。
清华大学XX年生物化学1本科期末考试试题

清华大学XX年生物化学1本科期末考试试题考试科目:生物化学考试时刻:考试类型:期末试题I.1: how many carbons does Arachidic acid have (20 carbons)2: how many double bonds does Arachidonic acid have (4 double bonds)3: list two advantages that fats have over sugars as stored fuels (more energy gram for gram; no hydration needed)4: where inside the cells are most of the phospholipids degraded (lysosomes)5: oligosaccharide head groups determine the blood type of an individual. How are they attached to the plasma membrane (glycosphingolipids or lipids and surface proteins)6: list at least one genetic disease that could result from abnormal accumulation of membrane lipids (Tay-Sachs, Sandhoff’s, Fabry’s, Gaucher’s, or Niemann-Pick diseases)7: list the three main eicosanoids that produced from arachidonic acid (prostaglandins; thromboxanes; and leukotrienes).8: list one NASID you know (aspirin, ibuprofen, or acetaminophen or meclofenamate)9: list two fat-soluble vitamins (A, D, E, K)10: which vitamin can be derived from beta-carotene (A).11: which year was the fluid mosaic model proposed (1972)12: why the thickness of most biological membranes is thicker than 3nm, the standard thickness of lipid bilayer ( due to association of proteins to the membrane and carbohydrates on the membrane)13: Please define the transition temperature of the lipid bilayer (the temperature above which the paracrystalline solid changes to fluid)14: If a membrane protein has its N-terminus exposed to the outside of the cell while itsC-terminus resides in the cytosolic compartment, is it a type I transmembrane protein (yes).15: How can you predict if a protein has a transmembrane domain (hydropathy or hydropathy index, or hydropathy plot).16: Name the two cell surface receptors that HIV use to enter cells (CCR5 and CD4).17: For the Na+ K+ ATPase, how many Na+ and K+ can it move across the membrane for the hydrolysis of one ATP (2 K+ in, 3 Na+ out).18: what drives F-type ATPases to synthesize ATP (proton or proton gradients)19: The acetylcholine receptor is a _____-gated channel (Ligand)20: The neuronal Na+ channel is a _____-gated channel (voltage)II.D and H(3 points) The antiparallel orientation of complementary strands in duplex DNA was elegantly determined in 1960 by Arthur Kornberg by nearest-neighbor analysis. In this technique, DNA is synthesized by DNA polymerase I from one (alpha-32P)-labelled and three unlabelled deoxynucleoside triphosphates. The resulting product is then hydrolyzed by a Dnase that cleaves phosphodiester bonds on the 3’ s ides of all deoxynucleotides. For example, in the labeled dATP reaction,ppp*A + pppC + pppG +pppT --ô€ƒ† …pCpTp*ApCpCp*ApGp*Ap*ApTp… -ô€ƒ†â€¦+Cp+Tp*+Ap+Cp+Cp*+Ap+Gp*+Ap*+Ap+TpT…If in the dATP labeled reaction the relative radioactivity of Tp*, Gp*, Cp* and Ap* is 0.2, 0.3, 0.4 and 0.1.In the labeled dGTP reaction, Tp* radioactivity will beA) 0.1B) 0.2C) 0.3D) 0.4E) 0In the labeled dCTP reaction, Tp* radioactivity will beF) 0.1G) 0.2H) 0.3J) 0(B)(1 point) A molecule of amylopectin consists of 1000 glucose residues and is branched every 25 residues. How many reducing ends does it haveA) 0B) 1C) 25D) 40E) 41F) 1000(1 point) (D)A mirror image ofB form DNA isA) Z-DNAB) A-DNAC) B-DNAD) does not exist in nature(1 point) (A)Which of the following statements is not trueA) all natural occurring DNA is in B formB) A-DNA probably does not exist in vivoC) Both major groove and minor grove of B-DNA are deepD) B-DNA is right handed, while Z-DNA is left handed.(1 point) (A)Lung fish has 100,000,000 kb (kb=1000 base pairs) for its haploid genome. The total length of this DNA isA) 34 mB) 3.4 mC) 1 mD) 17 mF) 34 mm(3 points) (1.5 point) A, D, EBoth tubes X and Y contain the same DNA in 0.1 M NaCl solution. In tube X more NaCl is added and in tube Y some ethanol is added. Which of the following is (are) correctA) tube X will have higher Tm than YB) tube Y will higher Tm than XC) tube X and Y will have the same TmD) tube X will have higher Tm than originalE) tube Y will have lower Tm than original(1 point) (D)Which of the following is NOT correctA) Glycosaminoglycan chains are linearB) Glycosaminoglycan chains are acidicC) Glycosaminoglycans are composed of repeating disaccharide unitsD) Different glycosaminoglycans have the same disaccharide units(1 point) (C)The so-called table sugar or cane sugar is a disaccharide ofA) galactose and glucoseB) 2 glucose unitsC) glucose and fructoseD) galactose and fructose(1 point) (A)Lactose is a disaccharide ofA) galactose and glucoseB) 2 glucose unitsC) glucose and fructoseD) galactose and fructose(1 point) (B)Choose the INCORRECT statement. In dideoxy sequencing or Sanger’s method,A) the newly synthesized DNA is labeled while the template is notB) the template has to be labeledC) a primer has to be usedD) ddNTPs are used to terminate elongation(1 point) (B)Choose the INCORRECT statement. In Maxam-Gilbert sequencing or chemical method,A) the template is labeledB) a primer is neededC) enzyme is not neededD) nucleotide analogues are not used(1 point) (D)uv radiation in the solar light directly cause which kind of the following DNA damageA) deaminationB) methylationC) DNA breakageD) pyrimidine dimer formationE) depurination(1 point) (C)RNA is sensitive to alkaline hydrolysis becauseA) RNA has uridine instead of thymidine in its sequenceB) RNA has hydroxyl group at 4th and 5th positions of its riboseC) RNA has hydroxyl group at 2nd and 3rd positions of its riboseD) RNA is mostly single strandedE) RNA forms three dimensional structure(1 point) (D)Which of the following is NOT a carbohydrateA) glucoseB) maltoseC) glycogenD) proteglycanE) glycosaminoglycan(3 points)(1.5 point) C, E, FWhich of the following is (are) NOT branchedA) glycogenB) starchC) celluloseD) amylopectinE) amyloseF) chitin(1 point) (A)Shown on the right isA) AB) GC) TD) CE) U(1 point) (C)Shown on the right isA) AB) GC) TD) CE) UIII.1) The follows are the important features of signal transduction systems EXCEPT :A. AmplificationB. Desensitization/turn-offC. SpecificityD. SimplicityE. Integration2) Relative to the outside of a cell, calcium concentration inside the cell is:A. HigherB. LowerC. Approximately equal3) Hydrolysis of GTP to GDP is essential for the normal function of all the GTP-binding proteinsA. TrueB. False4) cAMP-dept kinase (PKA) regulates only sugar metabolismA. TrueB. False5) The follows are among the intracellular second messengers EXCEPT:A. cAMPB. cGMPC. Ca2+D. IP3E. ATP6) How does IP3 work as a second messengerAï¼ژBy activating phaspholipase CBï¼ژBy activating protein kinase CC. By activating Ca2+ channelsD. By activating Gخ±S7) The insulin receptor has tyrosine kinase activity which phsophorylates several different proteins including itselfA. TrueB. False8) The insulin signal is amplified via the MAP kinase cascadeA. TrueB. False9) All signaling pathways are very unique and very specific, and they don’t talk to each otherA. True10) Which of the followings is UNIQUE to TGFخ²receptorsA. transmembrane proteinsB. Ser/The receptor kinasesC. Ligand binding activity present in the extracellular domainD. Become activated upon ligand binding11) The activity of CDK proteins is tightly regulated by the following events EXCEPT :A. PhosphorylationB. DephosphorylationC. Cyclin bindingD. UbiquitinationE. CDK inhibitors12) Caspases play essential roles in apoptosis. The following are among their effects EXCEPT:A. Activation of DNaseB. Activation of other caspasesC. Inducing cell shrinkageD. Stimulating gene expression13) Cancer can be a result of the following events EXCEPTA. Activation of mitogenic signalsB. Cell cycle arrestC. Inactivation of negative regulators for cell growthD. Blockage of cell apoptosis14) In 2002, the Nobel Price for Physiology or Medicine was awarded for the research onA. PhosphorylationB. Cell cycleC. ApoptosisD. G proteinE. OncogenesAnswers for Biosignaling:1) D3) A4) B5) E6) C7) A8) A9) B10) B11) D12) D13) B14) CIV.1. Which of the following statement about 2,3-bisphosphoglycerate (BPG) is wrong: (d)(a) BPG binds at a site distant from the oxygen-binding site of hemoglobin.(b) BPG regulates the oxygen-binding affinity of hemoglobin in relation to the pO2 in the lungs.(c) BPG concentration in normal human blood at sea level is lower than that at high altitudes.(d) BPG greatly increases the affinity of hemoglobin for oxygen.2. Antibodies of the IgG class : (d)(a) consist of four subunits.(b) have noncovalent bonds and disulfide crosslinks.(c) are abundant in the blood.(d) All three choices are correct.3. ELISA allows for rapid screening and quantification of the presense of an antigen in a sample. Which of the following steps of ELISA is wrong: (a)(a) Proteins in a sample are adsorbed to an inert surface, and the surface is washed with a solution of specific protein similar to the protein of interest, to block proteins in subsequence steps from also adsorbing to these surfaces.(b) The sample was treated with a solution containing antibodies against the protein of interest. Unbound antibody is washed away, and the sample is treated with with a solution containing antibodies against the primary antibody.(c) These secondary antibodies have been linked to an enzyme that catalyzes a reaction that forms a colored product.(d) After unbound secondary antibody is washed away, the substrate of the antibody-linked enzyme is added and the product formation is proportional to the concentration of the protein of interest in the sample.4. The major proteins of muscle are: (c)(a) Myosin and hemoglobin(b) Actin and troponin(c) Myosin and actin(d) Myoglobin and actin5. Positive cooperative binding can be identified by (c)(a) a hyperbolic binding curve.(b) a Hill plot with a slope less than one.(c) a Hill plot with a slope greater than one.(d) Choices a) and b) are both correct.6. Which pair of amino acids absorbs the most UV light at 280 nm (b)(a) Thr & His.(b) Trp & Tyr.(c) Phe & Pro.(d) Phe & Pro.7. The strong conclusion from Anfinsen's work on RNaseA was that: (b)(a) 100% enzyme activity corresponds to the native conformation.(b) the sequence of a protein determines its structure.(c) Cys-SH groups are not found in vivo.(d) disulfide bonds (S-S) in proteins can be reduced in vitro.(e) irreversible denaturation of proteins violates the "Thermodynamic Hypothesis".8. Which of the following statement about protein folding is wrong: (c)(a) Some proteins undergo assisted folding by chaperons.(b) Polypeptides fold rapidly by a stepwise process.(c) Misfolding may cause misfunctioning, but does not cause death.(d) A loss of 3-d protein structure sufficient to cause loss of function is called denaturation.9. Hydrogen bonds in a-helices are (d)(a) more numerous than Van der Waals interactions.(b) not present at Phe residues.(c) analogous to the steps in a spiral staircase.(d) roughly parallel to the helix axis.(e) about 5 أ… in length.10. What is the appoximate molecular weight of a protein with 200 amino acid residues in a single polypeptide chain: (b)(a) 11000(b) 22000(c) 44000(d) 66000V.1. What are the two most striking characteristics of enzymes(1 pt, 0.5 pt x 2)Answer:The two most striking characteristics of enzymes are their 1) catalytic power, 2) specificity. 2ï¼ژHow does an enzyme gain its catalytic power (2 pt, 1 pt x 2)Answer:1) Enzymes bring substrates together in an optimal orientation;2) They catalyze reactions by stabilizing transition states;3. Is the structure of the binding site of an enzyme complementary to its substrate or to transition state Why (1 pt, 0.5 pt x 2)Answer: An enzyme is in complementary in structure to the transition state of the substrate because the activation barrier is lowered during such a binding.4ï¼ژHow many general ways of regulations on enzyme Activity (2 pts, 0.5 pt x 4)Answer:1) Feed-back Inhibition2) Regulatory Proteins3) Covalent Modification4) Proteolytic Activation5. List the common features of enzymes (2.5 pts, 0.5 pt x 5)Answer:1) The active site take up small volume compare with the entire protein volume;2) The active sites are composed of amino acid residues located in different positions on a linear sequence;3) An enzyme binds to its substrate via multiple weak interactions;4) Active sites are clefts or crevices;5) The specificity of binding depends on precisely defined arrangement of atoms in an active site.6. How many types of weak interactions involved in the binding of a substrate by an enzyme (2 pts, 0.5 x 4)Answer:1) Electrostatic interactions; 2) Hydrogen bonds; 3) van der Waals forces; 4) Hydrophobic interactions.7. Write down the Michaelis-Menten equation (0.5 pt), what is the meaning of Km (0.5 pt) Under which condition the Km is a measure of the affinity of the enzyme for the substrate (1 pt), what is the meaning of Vmax (0.5 pt) What is the physical limit of the value of k3/km (1 pt)Answer:Michaelis-Menten equation: V = Vmax S/(S + Km)Km is equal to the substrate concentration at which the reaction rate is half its maximal rate value.Km is a measure of the enzyme’s affinity only when K2 >> K3 (i.e., the dissociation rate constant of ES is much greater than the catalytic rate constant)Vmax is reached when all the catalytic sites on an enzyme are saturated with substrate.The physical limit of the value of k3/km is that the enzyme-catalyzed reaction rate is rate of diffusion of both the substrate and the enzyme.VI.The following statements are true (T) or false (F)1. The biological system is an open system; living organisms are at equilibrium with their surroundings. 1.5False: The biological system is indeed an open system, able to absorb and exchange materials with the environment. However the living organisms use energy to, for example, keep their own concentrations of ions inside the cells, which are not necessarily of the same concentrations in the environment.2. Activation energy for a chemical reaction is the energy required for a chemical reaction to convert the reactant to transition state, but does not measure the free energy change between the reactants and products. 1.5True3. The mitochondrion is an organelle that functions as an energy regeneration powerhouse, but does not participate in the regulation of cell survival. 1.5False: Mitochondria indeed play the most important role in energy production; however, they are also a key organelle in the regulation of cell death. In fact, most critical apoptotic or antiapoptotic factors regulate cell death through functional interaction with mitochondria.4. The Buchners’ discovery of fermentation supported the view, as asserted by Pasteur, that fermentation can take place only in living cells. 1.55False: The Buchners’ finding that fermentation can take place in yeast cell extracts demonstrated that cell-free extracts possess all the elements for fermentation from sucrose to alcohol. Pasteur’s view that fermentation is inextricably tied to living cells was wrong. This finding has been regarded as one of the earliest biochemical experiments.5. Endoplasmic reticulum is a place for posttranscriptional RNA processing, translation, and protein modification.1.5False: The ER is a place for protein synthesis (translation), protein modification such as glycosylation and signal peptide cleavage, but not for RNA processing.6. Carbohydrates function as structural components in nucleic acids, amino acids, and protein modification. 1.5E. False: Everything but amino acids7. The present day biochemistry is the interweaving product of historical traditions of biochemistry, cell biology, molecular biology and genetics. 1.5True。
清华大学生物化学2科期末考试试题

清华大学生物化学2科期末考试试题清华大学生物化学本科期末考试试题考试课程Biochemistry II考试时间:------考试类型:期末测试说明:1. 请选择正确答案,填在适当的横线处(1-30题)或写在答题纸上(31-33题);2. 答案可能是一个或多个;3. 每题的分数在标在了题目后面;4. 本卷满分为100.1. If the ΔG` of the reaction A B is –12 kJ/mol, which of the following statements are correct? (Note the prime symbol means that a thermodynamic parameter is measured at pH.7.0) (2 points)A. The reaction will proceed spontaneously from left to right at the given conditions.B. The reaction will proceed spontaneously from right to left at standard conditions.C. The equilibrium constant favors the formation of B over the formation of A.D. The equilibrium constant could be calculated if the initial concentrations of A and B were known.E. The value of ΔG`o is also negative.Answer(s): A, D2. Which of the following statements about ATP and its roles in cells are true? (2 points)A. The ATP molecule is kinetically unstable and is thus consumed within about one minute following its formation in cells.B. ATP provides free energy to a thermodynamically unfavorable reactions by group transfer, always donating a Pi to form a covalent intermediate.C. ATP can be regenerated by coupling with a reaction that releases more free energy than does ATP hydrolysis.D. A transmembrane proton-motive force can drive ATP synthesis.E. The active form of ATP is usually in a complex with Mg2+.Answer(s): C, D, E3. A common moiety for NADP, NAD, FMN, FAD, and coenzyme A is: (2 points)A. A pyrimidine ring;B. A three ring structure;C. An ADP;D. A pyranose ring;E. A triphosphate group.Answer(s): C4. If the C-1 carbon of glucose were labeled with 14C, which of the carbon atoms in pyruvate would be labeled after glycolysis? (2 points)A. The carboxylate carbon;B. The carbonyl carbon;C. The methyl carbon.Answer(s): C5. Which of the following are metabolic products of pyruvate in higher organisms?(2 points)A. GlycerolB. Lactic acidC. AcetoneD. Acetyl-CoAE. EthanolAnswer(s): B, D6. Indicate whether each of the following statements about the pentose phosphate pathway is true (T) or false (F). (5 points)A. It generates NADH for reductive biosyntheses. FB. The reactions occur in the cytosol. TC. Transketolase and transaldolase link this pathway to gluconeogenesis. FD. It is more active in muscle cells than in fat-storage cells. FE. It interconverts trioses, tetroses, pentoses, hexoses, and heptoses. T7. Which of the following statements are correct? The citric acid cycle (2 points)A. does not exist as such in plants and bacteria, because its functions are performed by the glyoxylate cycle.B. oxidizes acetyl CoA derived from fatty acid degradation.C. produces most of the CO2 in anaerobic organisms.D. provides succinyl CoA for the synthesis of carbohydrates.E. provides precursors for the synthesis of glutamic and aspartic acids.Answer(s): B, E8. Match the cofactors of the pyruvate dehydrogenase complex with their corresponding enzyme components and with their roles in the enzymatic steps that are listed. (5 points)A. Coenzyme A: 3,7B. NAD+: 2, 9C. Thiamine pyrophosphate (TPP): 1, 5D. FAD: 2, 6E. Lipoamide: 3, 4,8(1). Pyruvate dehydrogenase component(2). Dihydrolipoyl dehydrogenase(3). Dihydrolipoyl transacetylase(4). Oxidizes the hydroxylethyl group(5). Decarboxylates pyruvate(6). Oxidizes dihydrolipoamide(7). Accepts the acetyl group from acetyllipoamide(8). Provides a long, flexible arm that coveys intermediates to different enzyme component.(9). Oxidizes FADH2.9. Matching the role in fatty acid oxidation and/or mobilization to the appropriate component listed below. (5 points)A. Bile salt: 2B. Serum albumin:5C. ApoC-II:4D. Apolipoprotein:3E. Carnitine:1(1). Acts as a “carrier” of fatty acids across the inner mitochondrial membrane.(2). Acts as a biological detergent, disrupting fat globules into small mixed micelles.(3). Binds and transports triacylglycerols, phospholipids, and cholesterol between organs.(4). Activates lipoprotein lipase, which cleaves triacylglycerols into their components.(5). Binds some fatty acids molibized from adipocytes and transports them in the blood to heart and skeletal muscle.10. Which of the following answers complete the sentence correctly? Surplus dietary amino acids may be converted into (2 points)A. proteins.B. Fats.C. ketone bodies.D. glucose.E. a variety of biomolecules for which they are precursors.Answer(s): A, B, C, D,E11. Which of the following compounds serves as an acceptor for the amino groups of many amino acids during metabolism? (2 points)A. GlutamineB. Asparagine.C. α-ketoglutarate.D. OxaloacetateE. GlutamateAnswer(s):C12. Match the functions for the coenzymes that are involved in amino acid metabolism. (4 points)A. Pyridoxal phosphate: (3)B. Coenzyme B12: (2)C. Tetrahydrobiopterin: (1)D. NAD+: (1)E. Biotin (4)(1). Carries electrons(2). Provides free radicals(3). Carries amino groups(4). Carries CO2.13. Which of the following experimental observations would not support the chemiosmotic model of oxidative phosphorylation? (3 points)A. If mitochondrial membranes are ruptured, oxidative phosphorylation cannot occur.B. Raising the pH of the fluid in the intermembrane space results in ATP synthesis in the matrix.C. Transfer of electrons through the respiratory chain results in formation of a proton gradient across the inner mitochondrial membrane.D. The orientation of the enzyme complexes of the electron transfer chain results in a unidirectional flow of H+.E. Radioactively labeled inorganic phosphate is incorporated into cytosolic ATP only in the presence of an H+ gradient across the inner mitochondrial membrane. Answer(s): B14. Some photosynthetic prokaryotes use H2S, hydrogen sulfide, instead of water as their photosynthetic hydrogen donor. How does this change the ultimate products of photosynthesis? (2 points)A. Carbohydrate (CH2O) is not produced.B. H2O is not produced.C. Oxygen is not produced.D. ATP is not produced.E. The products do not change.Answer(s): C15. Which of the following are constituents of chlorophylls? (2 points)A. Substituted tetrapyrrole.B. Plastoquinone.C. Mg2+.D. Fe2+.E. Phytol.F. Iron porphyrin.Answer(s): A, C, E16. The observation that the incubation of photosynthetic algae with 14CO2 in the light for a very brief time (5s) led to the formation of 14C-labeled3-phosphoglycerate suggested that the 14CO2 was condensing with sometwo-carbon acceptor. That acceptor was in fact which of the following? (2 points)A. AcetateB. Acetyl CoAC. Acetyl phosphateD. AcetaldehdydeE. Glycerol phosphateF. None of the aboveAnswer(s): F17. Which of the following are common features of the syntheses of mevalonate (an intermediate of cholesterol biosynthesis) and ketone bodies? (2 points)A. Both involve 3-hydroxyl-3-methylglutaryl CoA (HMG-CoA).B. Both require NADPH.C. Both require the HMG-CoA cleavage enzyme.D. Both occur in the mitochondria.E. Both occur in liver cells.Answer(s): A18. S-adenosylmethionine is involved directly in which of the following reactions. (3 points)A. Methyl transfer to phosphatidyl ethanolamine.B. Synthesis of glycine from serine.C. Synthesis of polyamines.D. Conversion of homocysteine to methionine.E. Generation of the 5` cap of the eukaryotic mRNAs.Answer(s): A, C, E19.Which of the following does not provide a carbon skeleton for the synthesis of amino acids? (2 points)A. succinate.B. α-ketoglutarateC. Pyruvate.D. Oxaloacetate.E. Ribose-5-phosphate.Answer(s): A20. Which of the following compounds directly provide atoms to form the purine ring? (3 points)A. Aspartate.B. Carbamoyl phosphate.C. Glutamine.D. Glycine.E. CO2.F. N5,N10-methylenetetrahydrofolate.G. N10-formyltetrahydrofolate.H. NH4+.Answer(s): A, C, D, E, G21. Which of the following statements about ribonucleotide reductase are true? (2 points)A. It converts ribonucleoside diphosphates into 2`-deoxyribonucleoside diphosphates in humans.B. It contains coenzyme B12, which generates free radicals needed for the catalysis.C. It accepts electrons directly from FADH2.D. It receives electrons directly from either thioredoxin or glutaredoxin.E. It contains two kinds of allosteric regulatory sites: one for controlling the overall activity and the other for controlling the substrate specificity.Answer(s): A, D, E22. Biosynthetic pathways that require NADPH include which of the following? (2 points)A. Gluconeogenesis.B. Fatty acid biosynthesis.C. Ketone body formation.D. Cholesterol biosynthesis.E. Tyrosine biosynthesis.Answer(s): B,D,E23. Homologous recombination in E. coli is likely to require which of the following?(3 points)A. DnaB protein.B. RecA protein.C. RecBCD complex.D. ATP.E. NAD+.F. Single-strand DNA binding protein.G. DNA-dependent RNA polymerase.H. DNA polymerase I.I. DNA ligase.J. dATP.Answer(s): B, C, D, E, F, H, I, J24. Which of the following statements about E. coli promoters are correct? (2 points)A. They may exhibit different transcription efficiencies.B. For most genes they include variants of consensus sequences.C. They specify the start sites for transcription on the DNA template.D. They have identical and defining sequences.E. They are activated when C or G residues are substituted into their –10 regions by mutation.F. Those that have sequences that correspond closely to the consensus sequences and are separated by 17 base pairs are very efficient.Answer(s): A, B, C, F25. The AAUAAA sequence on a RNA molecule marks (2 points)A. The site where ribosomes bind to initiate polypeptide synthesis.B. The site where transcription stops.C. The site near which the primary transcript is cleaved and a poly (A) sequence is added.D. The site where the release factor will bind to end polypeptide synthesis.E. The site where polyribonucleotide phosphorylase will add a stretch of random sequences.Answer(s): C26. The σ70 subunit of the E. coli RNA polymerase: (2 points)A. acts as the catalytic site for polymerization.B. Recognize promoters.C. Has a proofreading function.D. Increases the processivity of the enzyme.E. Recognizes termination signals.Answer(s): B27. The discontinuity of eukaryotic genes were first revealed by: (2 points)A. Using footprinting techniques.B. DNA sequence comparison.C. RNA sequence analysis.D. Electron microscopic analysis of RNA-DNA hybrid molecules.E. Plant genetic studies.Answer(s): D28. Which of the following mRNA codons can be recognized by the tRNA anticodon ICG. (2 points)A. UGC.B. CGA.C. UGA.D. CGU.E. CGC.Answer(s): B, D, E29. A new compound, vivekine, was recently discovered by a clever undergraduate student. It was isolated from bacteria found in deep sea-dwelling organisms. Vivekine inhibits protein synthesis in eukaryotes: Protein synthesis can initiate, but only dipeptides are formed and these remain bound to the ribosome. This toxin affects eukaryotic protein synthesis by blocking the: (2 points)A. binding of formylmethionyl-tRNA to ribosomes.B. activity of elongation factors.C. activation of amino acids.D. recognition of stop signals.E. formation of peptide bonds.Answer(s): B30. Indicate whether each of the following statements about prokaryotic translation is true (T) or false (F). (7 points)A. An aminoacyl-tRNA synthetase catalyzes formation of an ester bond. ( T )B. An mRNA molecule cannot be used to direct protein synthesis until it has been completely transcribed. ( F )C. The positioning of fMet-tRNA on the A site defines the reading frame. ( F )D. Incoming aminoacyl-tRNA are first bound to the A site. ( T )E. Formation of the 70S initiation complex requires an input of energy. ( T )F. The carboxyl group of the amino acid on the aminoacyl-tRNA is transferred to the amino group of a peptidyl-tRNA.( F )G. Release factors cause the peptidyl transferase activity of the ribosome to useH2O as a substrate. ( T )31. In an attempt to determine whether a given RNA was catalytically active in the cleavage of a synthetic oligonucleotide, the following experimental results were obtained. When the RNA and the oligonucleotide were incubated together, cleavage of the oligonucleotide occurred. When either the RNA or the oligonucleotide was incubated alone, there was no cleavage. When the RNA was incubated with higher concentrations of the oligonucleotide, saturation kinetics of the Michaelis-Menten type were observed. Do these results demonstrate that theRNA has catalytic activity? Explain. (8 points)Answer: These results alone do not establish that the RNA has catalytic activity. A catalyst must be regenerated. It is entirely possible that the results observed could be accounted for by a stoichiometric, as opposed to a catalytic, interaction between RNA and the oligonucleotide. In which the RNA may “commit suicide” as the oligonucleotide is cleaved. In such an interaction, a portion of the RNA would also be cleaved itself as a part of the reaction. Four reaction products would accumulate, two resulting from the cleavage of RNA and two from the cleavage of the oligonucleotide. To show that this particular RNA was catalytic, it would be necessary to demonstrate that it turns over and is regenerated in the course of the reaction.32.Translation involves conversion of the language of nucleotides to that of proteins. In the chain of events leading from a nucleotide sequence on DNA to the production of protein by ribosomes, where precisely does the process of translation occur? Explain. (6 points)Answer: Translation involves conversion of the language of nucleotides to that of proteins. The agent of translation is the appropriate aminoacyl-tRNA synthetase, which must recognize a particular amino acid and link it to a tRNA containing an anticodon for that amino acid.33.Suppose that a bacterial mutant is found to replicate its DNA at a very low rate. Upon analysis, it is found to have entirely normal activity of DNA polymerases I and III, DNA gyrase, and DNA ligase. It also makes normal amounts and kinds of dnaA,dnaB, dnaC, and SSB proteins. The oriC region of its chromosome is found to be entirely normal with respect to nucleotide sequence. What defect might account for the abnormally low rate of DNA replication in this mutant? Explain. (6 points) Answer: A decrease in the activity of primase would account for the low rate of DNA replication. DNA replication requires the prior synthesis of RNA primers. Decreased rates of dNTP synthesis would also slow replication.。
生物化学期末考试试题及答案

生物化学期末考试试题及答案《生物化学》期末考试题一、判断题(15个小题,每题1分,共15分)1、蛋白质溶液牢固的主要因素是蛋白质分子表面形成水化膜 ,并在偏离等电点时带有相同电荷( )2 、糖类化合物都具有还原性( )3、动物脂肪的熔点高在室温时为固体,是由于它含有的不饱和脂肪酸比植物油多。
( )4 、维持蛋白质二级结构的主要副键是二硫键。
( )5 、ATP 含有 3 个高能磷酸键。
( )6、非竞争性控制作用时,控制剂与酶结合则影响底物与酶的结合。
( )7、儿童经常晒太阳可促使维生素 D 的吸取,预防佝偻病。
( )8、氰化物对人体的伤害作用是由于它拥有解偶联作用。
( )9 、血糖基本来源靠食物提供。
( )10 、脂肪酸氧化称β- 氧化。
( )11 、肝细胞中合成尿素的部位是线粒体。
( )12 、构成RN A的碱基有A、U、G、T 。
( )13、胆红素经肝脏与葡萄糖醛酸结合后水溶性增强。
( )14 、胆汁酸过多可反馈抑制7 α- 羟化酶。
( )15、脂溶性较强的一类激素是经过与胞液或胞核中受体的结合将激素信号传达发挥其生物()二、单项选择题(每题1分,共20分)1、以下哪个化合物是糖单位间以α-1,4糖苷键相连:(A)A、麦芽糖B、蔗糖C、乳糖D 、纤维素E、香菇多糖、以下何物是体内贮能的主要形式(E)A、硬酯酸B 、胆固醇C、胆酸D、醛固酮E、脂酰甘油、蛋白质的基本结构单位是以下哪个:)A、多肽B、二肽C、L-α氨基酸D、L-β-氨基酸E、以上都不是、酶与一般催化剂对照所拥有的特点是)A、能加速化学反应速度B、能缩短反应达到平衡所需的时间C、拥有高度的专一性、反应前后质和量无改E、对正、逆反应都有催化作用、通过翻译过程生成的产物是:(D )A、tRNA B、mRNA C、rRNA D 、多肽链E、DNA6、物质脱下的氢经NADH呼吸链氧化为水时,每耗资1/2分子氧可生产ATP分子数量(C)A、1B、2 C 、3 D、4. E 、57、糖原分子中由一个葡萄糖经糖酵解氧化分解可净生成多少分子ATP?(B )A、1B、2C、3D、4 E 、58 、下列哪个过程主要在线粒体进行(E )9A、脂肪酸合成D、甘油分解、酮B 、胆固醇合成E 、脂肪酸β-氧化体生成的C、磷脂合成限速酶是( C )A、HMG-CoA还原酶 B 、HMG-CoA裂解酶C 、HMG-CoA合成酶D、磷解酶 E 、β-羟丁酸脱氢酶10 、有关G- 蛋白的概念错误的是(C )A、能结合GDP和GTP B 、由α、β、γ三亚基组成C、亚基聚合时拥有活性 D 、可被激素受体复合物激活、有潜藏的GTP活性11 、鸟氨酸循环中,合成尿素的第二个氮原子来自( C )A、氨基甲酰磷酸 B 、NH3 C 、天冬氨酸D、天冬酰胺E、谷氨酰胺12 、下列哪步反应障碍可致苯丙酮酸尿症( B )A、多巴→黑色素 B 、苯丙氨酸→酪氨酸C、苯丙氨酸→苯丙酮酸 D 、色氨酸→5羟色胺、酪氨酸→尿黑酸13 、胆固醇合成限速酶是:( B )A、HMG-CoA合成酶 B 、HMG-CoA还原酶C、HMG-CoA 裂解酶D、甲基戊烯激酶 E 、鲨烯环氧酶14 、关于糖、脂肪、蛋白质互变错误是:( D )A、葡萄糖可转变成脂肪B、蛋白质可转变成糖C、脂肪中的甘油可转变成糖 D 、脂肪可转变成蛋白质E、葡萄糖可转变成非必需氨基酸的碳架部分15 、竞争性抑制作用的强弱取决于:(E )A、控制剂与酶的结合部位 B 、控制剂与酶结合的牢固程度C、控制剂与酶结构的相似程度 D 、酶的结合基团E、底物与控制剂浓度的相比较例16、红细胞中还原型谷胱苷肽不足,易引起溶血是缺乏(B )A、果糖激酶B、6-磷酸葡萄糖脱氢酶C、葡萄糖激酶D、葡萄糖6-磷酸酶E、己糖二磷酸酶17、三酰甘油的碘价愈高表示以下何情况(A )A、其分子中所含脂肪酸的不饱和程度愈高B、其分子中所含脂肪酸的不饱和程度愈、其分子中所含脂肪酸的碳链愈长D、其分子中所含脂肪酸的饱和程度愈高、三酰甘油的分子量愈大18、真核基因调控中最重要的环节是( B )A、基因重排 B 、基因转录 C 、DNA的甲基化与去甲基化D、mRNA的衰减 E 、翻译速度19、关于酶原激活方式正确是:( D)A、分子内肽键一处或多处断裂构象改变,形成活性中心B、经过变构调治 C 、经过化学修饰D、分子内部次级键断裂所引起的构象改变、酶蛋白与辅助因子结合20、呼吸链中氰化物控制的部位是:(A )A、Cytaa3→O2 B 、NADH→O2 C 、CoQ→CytbD、Cyt→CytC1 E 、Cytc→Cytaa3三、多项选择题(10个小题,每题1分,共10分)1、基因诊断的特点是:(ABC )A、针对性强特异性高B、检测矫捷度和精确性高C、合用性强诊断范围广D、针对性强特异性低E、合用性差诊断范围窄2、以下哪些是维系DNA双螺旋的主要因素(DE )A、盐键B 、磷酸二酯键C、疏水键D、氢键 E、碱基堆砌3、核酸变性可观察到以下何现象( BC )A、粘度增加 B 、粘度降低 C 、紫外吸取值增加D 、紫外吸取值降低E 、磷酸二酯键断裂4、服用雷米封应合适补充哪一种维生素( BC )A 、维生素B2B 、V —PPC 、维生素B6D 、维生素B12E 、维生素C5、关于呼吸链的表达以下何者正确?( )A 、存在于线粒体B 、参加呼吸链中氧化还原酶属不需氧脱氢酶++E 、细胞色素是递电子体C 、NAD 是递氢体D 、NAD 是递电子体6、糖异生路子的要点酶是( )A 、丙酮酸羧化酶B 、果糖二磷酸酶C 、磷酸果糖激酶D 、葡萄糖—6—磷酸酶E 、已糖激酶7、甘油代谢有哪几条路子( )A 、生成乳酸 B、生成CO2、H2O 、能量 C 、转变成葡萄糖或糖原D、合成脂肪的原料 E 、合成脂肪酸的原料8、未结合胆红素的其他名称是( )A、直接胆红素 B 、间接胆红素 C 、游离胆红素D、肝胆红素E、血胆红素9、在分子克隆中,目的基因可来自( )基因组文库B、cDNA文库C、PCR扩增 D 、人工合成 E 、DNA结合蛋白10关于DNA与RNA合成的说法哪项正确:( )A、在生物体内转录时只能以DNA有意义链为模板B、均需要DNA为模板C 、复制时两条DNA链可做模板D、复制时需要引物参加转录时不需要引物参加E、复制与转录需要的酶不相同四、填空题(每空分,共15分)1、胞液中产生的NADH经和穿越作用进入线粒体。
《生物化学》期末考卷卷

《生物化学》期末考卷卷1 / 71 / 7《生物化学》期末考卷(A卷)~学年——学期班级 __________ 姓名 ___________ 座号 _________成绩 ___________一 二 三 目简答题阐述题选择题分卷人 核人一、选择题1、血清白蛋白( PI 为 4.7)在以下哪一种 PH 值溶液中带正电荷A 、 PH4.0B 、 PH5.0C 、PH6.0D 、PH7.0E 、PH8.02、蛋白质溶液的稳固要素是 A 、蛋白质溶液有分子扩散现象 B 、蛋白质在溶液中有“布朗运动”C 、蛋白质分子表面带有水化膜和同种电荷D 、蛋白质溶液的粘度大E 、蛋白质分子带有电荷 3、蛋白质变性是因为 A 、氨基酸摆列次序的改变 B 、氨基酸构成的改变C 、肽键的断裂D 、蛋白质空间构象的损坏E 、蛋白质 的水解4、蛋白质的一级构造是指下边的哪一种情 况A 、氨基酸种类的数目B 、分子中的各种化学键C 、多肽链的形态和大小D 、氨基酸残基的摆列次序E 、分子中的共价键5、保持 DNA 双螺旋横向稳固性的力是A 、碱基聚积力B 、碱基对之间的氢键C 、螺旋内侧疏水力D 、二硫键E 、磷酸二酯键6、核酸变性后会出现以下哪一种现象A 、减色效应B 、添色效应C 、浮力密度降落D 、粘度增添E 、最大汲取峰发生改变 7、竞争性作用的强弱取决于A 、部位B 、联合的坚固C 、与酶 总分构造的相像程度 D 、酶的联合基团E 、底物与克制剂浓度的相对照例 8、磺胺药的抑菌作用属于 A 、不行逆克制 B 、竞争性克制 C 、非竞争性克制D 、反竞争性克制E 、克制强弱不取决于底物与克制剂浓度的相对照例9、同工酶的正确描绘为A 、催化功能不一样, 理化、免疫学性质同样B 、催化功能、理化性质同样C 、同一种属一种酶的同工酶Km 值不一样D 、同工酶无器官特异性E 、同工酶是由同样基因编码的多肽链 10、对于酶原及其激活的正确表达为 A 、酶原无活性是因为酶蛋白肽链合成不完 成B 、酶原无活性是因为缺少辅酶或辅基C 、体内的酶初泌时都以酶原的形式存在D 、酶原激活过程是酶活性中心形成与裸露的过程 E 、全部酶原都有自己激活功能11、三羧酸循环第一步的反响产物是A 、柠檬酸B 、草酰乙酸C 、乙酰CoAD 、 CO 2E 、 NADH+H +12、三羧酸循环中有底物水平磷酸化的反响是A 、柠檬酸→ α-酮戊二酸B 、 α-酮戊二酸→琥珀酸 CoA2 / 72 / 7C 、琥珀酰 CoA →琥珀酸D 、延胡索酸21、以下哪一种脂蛋白被以为拥有抗动脉粥→苹果酸E 、苹果酸→草酰乙酸样硬化作用13、脂肪酸、胆固醇合成时的供氢体是A 、CMB 、VLDLC 、IDLD 、A 、NADH+H +B 、 NADPH+H +C 、LDLE 、 HDLFADH 2D 、FMNH 2E 、 CoQH 222、呼吸链中细胞色素传达电子的次序是、糖原合成时,葡萄糖的载体是3→b →c 1→ c → 1/2O 2B 、→ 3→14A 、aab aaA 、 ADPB 、GDPC 、CDPD 、1→ c →1/2O 2、 → → 3→b →1/2O 2cC c1 caaTDPE 、UDPD 、c 1 →c → b → aa 3→ 1/2O 2E、 → 1→ cbc15、肌糖原不可以分解为葡萄糖,因为肌肉 → aa 3→1/2O 2中不含有23、氧化磷酸化解偶联作用可使A 、果糖二磷酸酶B 、葡萄糖激酶A 、呼吸链氧化过程停止B 、ATP 合酶C 、磷酸葡萄糖变位酶克制C 、三羧酸循环停止D 、葡萄糖 -6-磷酸酶E 、磷酸已糖异构D 、线粒体能利用氧,但不可以生成 ATP酶E 、线粒体不可以利用氧,但能产生 ATP 16、在糖酵解和糖异生中都起催化作用的 24、以下哪一种激素在调理氧化磷酸化中起 是重要作用A 、丙酮酸激酶B 、丙酮酸羧化酶 A 、肾上腺皮质素 B 、甲状腺素C 、C 、果糖二磷酸酶D 、已糖激酶肾上腺素 D 、生长素 E 、胰岛素E 、 3-磷酸甘没醛脱氢酶25、NADH 氧化呼吸链的偶联部位有17、糖异生门路中催化 1,6-二磷酸果糖转A 、1 个B 、2 个C 、3 个D 、4变为 6-磷酸果糖的酶是个E 、5 个A 、丙酮酸羧化酶B 、磷酸烯醇式丙酮26、氨基酸脱氨基的主要方式为酸羧激酶 C 、磷酸葡萄糖变位酶A 、氧化脱氨基B 、复原脱氨基C 、D 、果糖二磷酸酶E 、葡萄糖 -6- 磷酸酶 水解脱氨基D 、转氨基E 、联合脱18、以下相关酮体的表达中错误的选项是氢基 A 、酮体是脂肪酸在肝中氧化的中间产物 27、ALT 可催化以下哪组底物发生氨基移B 、糖尿病时可惹起血酮体增高换反响C 、饥饿时酮体生成减少D 、酮体能够A 、谷氨酸与丙酮酸B 、丙氨酸与谷氨从尿中排出酸C 、天冬氨酸与 α-酮戊二酸E 、酮体包含丙酮、乙酰乙酸和 β-羟丁酸D 、草酰乙酸与谷氨酸E 、草酰乙酸与19、体内合成胆固醇的原料是α -酮戊二酸A 、乙酰乙酰 CoAB 、乙酰 CoAC 、 28、能直接进行氧化脱氨基作用的氨基酸 苹果酸D 、丙酮酸E 、α磷酸甘油是20、以下各样脂蛋白中,参加运输内源性 A 、甘氨酸 B 、天冬氨酸 C 、谷氨TG 的是酸D 、丝氨酸E 、丙氨酸A 、CMB 、 VLDLC 、IDLD 、 29、肾脏中产生的氨主要来自LDL E、HDL A、氨基酸的联合脱氨基作用B、谷氨3 / 73 / 7酰胺的水解C、氨基酸的氧化脱氨作用D、尿素的水解E、嘌呤核苷酸循环30、体内氨的主要去路是A 、随尿排出体外B 、合成谷氨酰胺C、合成营养非必要氨基酸D、合成丙氨酸E、合成尿素31、体内运输一碳单位的载体是A 、蛋氨酸B、生物素C、维生素B12D、叶酸E、四氢叶酸32、嘌呤核酸的代谢终产物是A、尿素B、黄嘌呤C、尿酸D、次黄嘌呤E、乳清酸33、合成 DNA 的原料是A 、dAMP 、 dAMP 、dTMP 、dCMP B、dATP 、dGTP、 dTTP、dCTPC、dADP、dGDP、dTDP、dCDPD、AMP 、GMP、TMP 、CMPE、 ATP、 GTP、TTP、CTP34、以下对于 DNA 复制的表达,哪一项为哪一项错误A 、半保存复制B、两便条链均连续合成C、合成的方向 5′→ 3′D、以四种dNTP 为原料E、有 DNA连结酶参加35、DNA复制时,模板序列5′-TAGA-3 ′,将合成子链的序列是以下哪一种A 、5′-TCTA-3 ′B、5′-ATCA-3 ′C、5′-UCUA-3 ′D、5′-GCGA-3 ′E、3′-TCTA-5 ′36、DNA复制需要① DNA聚合酶②引物酶③解螺旋酶④DNA拓扑异构酶⑤ DNA连结酶,其作用的次序是A 、①→②→③→④→⑤B、③→④→①→②→⑤C、④→③→②→①→⑤D、②→③→④→①→⑤E、③→②→④→①→⑤37、逆转录过程中需要的酶是A、DNA 指导的 DNA 聚合酶B、RNA指导的 DNA 聚合聚C、RNA 指导的 RNA 聚合酶D、核酸酶E、DNA 指导 RNA 的聚合酶38、参加转录的酶是A、依靠 DNA 的 RNA 聚合酶B、依靠 DNA 的 DNA 聚合酶C、依靠 RNA 的 DNA 聚合酶D、依靠RNA 的 RNA 聚合酶E、端粒酶39、模板 DNA 的碱基序列是 3′-TG-CAGT-5 ′,其转录出 RNA 碱基序列是A 、 5 ′ -AGGUCA-3 ′B、5′-ACGUCA-3 ′C、 5′-UCGUCU-3 ′D 、 5 ′ -ACGTCA-3 ′E、5′-ACGUGT-3 ′40、构成 mRNA 四种核苷酸每相邻三个核苷酸一组,总合能构成多少种密码子A、16B、46C、64D、32E、6141、与 mRNA 中密码子 5′-ACG-3 ′相应的反密码子( 5′→ 3′)是A、CGUB、CGAC、UCGD、UGCE、GCU42、人体内不一样细胞能合成不一样蛋白质,是因为A、各样细胞的基因不一样B、各样细胞的基因同样,而表达基因不一样C、各样细胞的蛋白酶活性不一样D、各种细胞的蛋白激酶活性不一样E、各样细胞的氨基酸不一样43、形成镰刀状红细胞贫血的原由是A、缺少维生素 B12B、缺少叶酸C、血红蛋白的β链 N 尾端缬氨酸变为了谷氨4 / 74 / 7酸D 、血红蛋白β链基因中的CTT 变为了CAT E、血红蛋白β链基因中的CAT 变为了 CTT44、肝脏生物转变作用第一相反响中最重要的酶是微粒体中的A 、加单氧酶B、加双氧酶C、胺氧化酶D、水解酶E、复原酶45、以下哪一种物质不是生物转变中联合物的供体A 、 UDPGAB 、 PAPSC 、 SAMD、乙酰 COAE、UDPG46、血中哪一种胆红素增添可能出此刻尿中A 、联合胆红素B 、未联合胆红素C、血胆红素D、间接胆红素E、胆红素 -Y 蛋白47、血中胆红素的主要运输方式是A 、胆红素 -清蛋白B、胆红素 -Y 蛋白C、胆红素 -Z 蛋白D、胆红素 -球蛋白E、胆红素 -脂蛋白48、正常状况下,人粪便的主要色素是A 、血红素B、胆绿素C、胆红素D、胆素原E、胆素49、维生素 D 的活性形式是A、维生素 D3B、维生素 D2C、24- 羟维生素 D3D、5-羟维生素 D3E、 1, 25-二羟维生素 D350、哪一种维生素既是氨基酸转氨酶的辅酶又是氨基酸脱羧酶的辅酶成分A 、生物素B、硫酸楚C、维生素B6D、维生素 PP E、维生素 B12二、简答题(每题 5 分,共 20 分)51、构成蛋白质的元素有哪几种?哪一种为蛋白质分子中的特色成分?测其含量有何用途?52、简述 RNA 的主要类型与功能。
生物化学期末考试试题及答案

生物化学期末考试试题及答案《生物化学》期末考试题 A一、判断题(15个小题,每题1分,共15分)( )1、蛋白质溶液稳定的主要因素是蛋白质分子表面形成水化膜,并在偏离等电点时带有相同电荷2、糖类化合物都具有还原性( )3、动物脂肪的熔点高在室温时为固体,是因为它含有的不饱和脂肪酸比植物油多。
( )4、维持蛋白质二级结构的主要副键是二硫键。
( )5、ATP含有3个高能磷酸键。
( )6、非竞争性抑制作用时,抑制剂与酶结合则影响底物与酶的结合。
( )7、儿童经常晒太阳可促进维生素D的吸收,预防佝偻病。
( )8、氰化物对人体的毒害作用是由于它具有解偶联作用。
( )9、血糖基本来源靠食物提供。
( )10、脂肪酸氧化称β-氧化。
( )11、肝细胞中合成尿素的部位是线粒体。
( )12、构成RNA的碱基有A、U、G、T。
( )13、胆红素经肝脏与葡萄糖醛酸结合后水溶性增强。
( )14、胆汁酸过多可反馈抑制7α-羟化酶。
( )15、脂溶性较强的一类激素是通过与胞液或胞核中受体的结二、单选题(每小题1分,共20分)1、下列哪个化合物是糖单位间以α-1,4糖苷键相连:()A、麦芽糖B、蔗糖C、乳糖D、纤维素E、香菇多糖2、下列何物是体内贮能的主要形式( )A、硬酯酸B、胆固醇C、胆酸D、醛固酮E、脂酰甘油3、蛋白质的基本结构单位是下列哪个:( )A、多肽B、二肽C、L-α氨基酸D、L-β-氨基酸 E、以上都不是4、酶与一般催化剂相比所具有的特点是( )A、能加速化学反应速度B、能缩短反应达到平衡所需的时间C、具有高度的专一性D、反应前后质和量无改E、对正、逆反应都有催化作用5、通过翻译过程生成的产物是:( )A、tRNA B、mRNA C、rRNA D、多肽链E、DNA6、物质脱下的氢经NADH呼吸链氧化为水时,每消耗1/2分子氧可生产ATP分子数量( )A、1B、2 C、3 D、4. E、57、糖原分子中由一个葡萄糖经糖酵解氧化分解可净生成多少分子ATP? ( )A、1B、2C、3D、4E、58、下列哪个过程主要在线粒体进行( )A、脂肪酸合成B、胆固醇合成C、磷脂合成D、甘油分解E、脂肪酸β-氧化9、酮体生成的限速酶是( )A、HMG-CoA还原酶B、HMG-CoA裂解酶C、HMG-CoA合成酶D、磷解酶E、β-羟丁酸脱氢酶10、有关G-蛋白的概念错误的是( )A、能结合GDP和GTPB、由α、β、γ三亚基组成C、亚基聚合时具有活性D、可被激素受体复合物激活E、有潜在的GTP活性11、鸟氨酸循环中,合成尿素的第二个氮原子来自( )A、氨基甲酰磷酸B、NH3C、天冬氨酸D、天冬酰胺E、谷氨酰胺12、下列哪步反应障碍可致苯丙酮酸尿症( )A、多巴→黑色素B、苯丙氨酸→酪氨酸C、苯丙氨酸→苯丙酮酸D、色氨酸→5羟色胺E、酪氨酸→尿黑酸13、胆固醇合成限速酶是:( )A、HMG-CoA合成酶B、HMG-CoA还原酶C、HMG-CoA裂解酶D、甲基戊烯激酶E、鲨烯环氧酶14、关于糖、脂肪、蛋白质互变错误是:( )A、葡萄糖可转变为脂肪B、蛋白质可转变为糖C、脂肪中的甘油可转变为糖D、脂肪可转变为蛋白质E、葡萄糖可转变为非必需氨基酸的碳架部分15、竞争性抑制作用的强弱取决于:( )A、抑制剂与酶的结合部位B、抑制剂与酶结合的牢固程度C、抑制剂与酶结构的相似程度D、酶的结合基团E、底物与抑制剂浓度的相对比例16、红细胞中还原型谷胱苷肽不足,易引起溶血是缺乏( )A、果糖激酶B、6-磷酸葡萄糖脱氢酶C、葡萄糖激酶D、葡萄糖6-磷酸酶E、己糖二磷酸酶17、三酰甘油的碘价愈高表示下列何情况( )A、其分子中所含脂肪酸的不饱和程度愈高B、其分子中所含脂肪酸的不饱和程度愈C、其分子中所含脂肪酸的碳链愈长D、其分子中所含脂肪酸的饱和程度愈高E、三酰甘油的分子量愈大18、真核基因调控中最重要的环节是( )A、基因重排B、基因转录C、DNA的甲基化与去甲基化D、mRNA的衰减E、翻译速度19、关于酶原激活方式正确是:( )A、分子内肽键一处或多处断裂构象改变,形成活性中心B、通过变构调节C、通过化学修饰D、分子内部次级键断裂所引起的构象改变E、酶蛋白与辅助因子结合20、呼吸链中氰化物抑制的部位是:( )A、Cytaa3→O2B、NADH→O2C、CoQ→CytbD、Cyt→CytC1E、Cytc→Cytaa31 2 3 4 5 6 7 8 9 101、基因诊断的特点是:( ) A、针对性强特异性高 B、检测灵敏度和精确性高 C、实用性强诊断范围广D、针对性强特异性低E、实用性差诊断范围窄2、下列哪些是维系DNA双螺旋的主要因素( )A、盐键B、磷酸二酯键C、疏水键D、氢键E、碱基堆砌3、核酸变性可观察到下列何现象( )A、粘度增加B、粘度降低C、紫外吸收值增加D、紫外吸收值降低E、磷酸二酯键断裂4、服用雷米封应适当补充哪种维生素( )A、维生素B2B、V—PPC、维生素B6D、维生素B12E、维生素C5、关于呼吸链的叙述下列何者正确?( )A、存在于线粒体B、参与呼吸链中氧化还原酶属不需氧脱氢酶C、NAD+是递氢体D、NAD+是递电子体E、细胞色素是递电子体6、糖异生途径的关键酶是( )A、丙酮酸羧化酶B、果糖二磷酸酶C、磷酸果糖激酶D、葡萄糖—6—磷酸酶E、已糖激酶7、甘油代谢有哪几条途径( )A、生成乳酸B、生成CO2、H2O、能量C、转变为葡萄糖或糖原D、合成脂肪的原料E、合成脂肪酸的原料8、未结合胆红素的其他名称是( )A、直接胆红素B、间接胆红素C、游离胆红素D、肝胆红素E、血胆红素9、在分子克隆中,目的基因可来自( )基因组文库 B、cDNA文库 C、PCR扩增 D、人工合成 E、DNA结合蛋白10关于DNA与RNA合成的说法哪项正确:( )A、在生物体内转录时只能以DNA有意义链为模板B、均需要DNA为模板C、复制时两条DNA链可做模板D、复制时需要引物参加转录时不需要引物参加E、复制与转录需要的酶不同四、填空题(每空0.5分,共15分)1、胞液中产生的N A DH经和穿梭作用进入线粒体。
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B.It is regulated by malonyl-CoA.
C.The cytosolic and matrix pools of CoA are distinct and separate.
A.substitution
B.insertion
C.deletion
D.silent
E.B and C
Answer(s): E
5.Which of the followingcorrectlydescribes the nitrogen cycle?
A.Fixation ofatmospheric nitrogen by nitrogen-fixing bacteria yields bioavailable nitrate.
清华大学本科生考试试题专用纸(A)
考试课程Biochemistry II 2005年1月15日
说明:1.第1-40题为选择题,每题2分,答案可能是一个或多个,请将
正确答案填在适当的横线处;
2.第41-44题为问答题,每题的分数标在了题目后面,请将答案写
在答题纸上;
3.本卷满分为100。
1. Which compound links glycolysis, nucleotide synthesis, and glycogen synthesis?
B.are important in the pair of damaged DNA.
C.involve RecA protein.
D.generategenetic diversity.
E.The two types of recombination have nothing in common.
C.(5’) AATGCAGCTTTAGCA(3’)
D.(5’) AAUGCAGCUUUAGCA(3’)
E.(3’) AATGCAGCTTTAGCA(5’)
Answer(s):C
9.The synthesis of palmitate requires:
A.8acetyl-CoA.
B.14 NADH.
D.Once fatty acyl groups have entered the matrix, they are committed to oxidation to acetyl-CoA.
E.All of the above are true.
Answer(s):E
17.In nucleotide metabolism, all of the following are trueexcept:
D.They are catalyzed by the same enzyme.
E.They require the cofactorS-adenosylmethionine.
Answer(s):B
12.If theG’of the reaction A B is –12 kJ/mol, which of the following statements arecorrect? (Note the prime symbol means that a thermodynamic parameter is measured at pH 7.0)
A.hexokinase; ATP
B.phosphofructokinase-1; AMP
C.pyruvate kinase; ATP
D.pyruvate dehydrogenase: NADH
E.pyruvate carboxylase; ADP
Answer(s):B
16.Which of the following describes fatty acid transportinto the mitochondrial matrix?
A.5
B.6
C.6.5
D.7.5
E.No ATP would form under these conditions.
Answer(s): D
3.Indicate which of the following events can occur during the processing of eukaryotic mRNA transcripts.
D.A and B
E.A, B, and C
Answer(s):C
6. Assuming the 5’3’ connection of writing nucleotide sequence, indicate which of the following mRNA codons can be recognized by the tRNA anticodon ICG.
E.The value ofG’ois also negative.
Answer(s): AD
13.Which of the following isnota physiological role of nucleotides?
A.allosteric regulators
B.intermediates for biosynthetic processes
A.The reaction will proceed spontaneously from left to right at the given conditions.
B.The reaction will proceed spontaneously from right to left at standard conditions.
C.Efficient binding of RNA polymerase to DNA occurs only in the presence of CRP and cAMP.
D.cAMP acts as an inducer by binding to the Lac repressor.
E.In the absence of lactose, the repressor binds the operator.
A.acetyl-CoA
B.oxaloacetate
C.citrate
D.glucose 6-phosphate
E.glycerol 3-phosphate
Answer(s): D
2.You have discovered acompound that inhibits fumarase. How many moles of ATP would you expect to be generated from one mole of acetyl-CoA in the presence of this inhibitor?
A.UGC
B.CGA
C.UGA
D.CGU
E.CGC
Answer(s):BDE
7.Both general recombination and transposon-type recombination:
A.occurbetween homologous regions on chromosomes.
E.Splicing together of exons.
F.Differential cutting and splicing to produce two different proteins.
Answer(s):CE
4. AlthoughDNA replication has very high fidelity, mutations do occur. Which of the following types of single base-pair mutations would be most likely to be a lethal mutation?
ponents of many proteins
ponents of the coenzymes NAD, FAD, and CoA
E.intracellular signaling molecules
Answer(s):C
14.Indicate which of the following statements about thelacoperon istrue.
C.7 ATP.
D.A and C.
E.A, B, and C.
Answer(s):D
10.Enoyl-CoA isomerase is needed for the complete-oxidation of
A.unsaturated fatty acids with cis double bonds.
A.All seven genes are transcribed into mRNA, which is then translated into five different proteins.
B.The repressor binds to the structural genes, preventing their transcription.
A.They have a large, negativeG’o.
B.The amino group is transferred to an-keto acid (such as-ketoglutarate) to form the corresponding amino acid.
C.The amino group is transferred from an ammonia molecule.
B.Nitrate is reduced to ammonia in a process known as denitrification.