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1．课程简介课程名称：生物化学

课程时间：第 3 学期总课时数：102

具体理论和实验课时如下：科目授课学时数

理论课 72

实验 30

课程简介：生物化学与医学是密切联系的两个学科。生物化学是一门基础医学的必修课

程，是研究生物体内生物大分子以及它们发生的化学反应的科学，主要内容包

括蛋白质、酶、核苷酸和核酸的结构与功能、物质代谢及其调节、基因信息的表达

与调控和基因工程。机体的生理状态依赖于体内物质代谢的平衡，而病理状态反

应了生物大分子以及物质代谢的异常。生物化学为医学各学科从分子水平上研究

疾病发生机制与治疗提供了理论与技术基础。

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Name of Course: Biochemstry

Course Commences: The third semester

Total teaching hours are approximately 102

Subject Teaching hoursLecture 72Practices 30

Course Description:

Biochemstry and medicine are intimately related to each other. The Biochemistry

course introduces the princeples of modern molecular biology and biochemistry as

applied to medicine. It is the science concerned with studying the various molecules

that occur in living cells and organisms as well as with their chemical reactions. It

enhances our understanding of the structures and functions of proteins, enzymes,

nucleotides and nucleic acids and explains how various cellular reactions on

molecular levels. It also explores the expression and regulation of genes and the

principles of recombinant DNA technology. Healthy states depend on the homeostasis

of biochemical reactions occurring in the body while disease states reflect

abnormalities in biomolecules, biochemical reactions or processes arising from

genetic and environmental factors. Biochemical approaches are often fundamental in

illuminating the causes of diseases and in designing appropriate therapies.

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2. 教学大纲

Syllabus of Biochemistry

Department of Biochemistry and Molecular

Biology

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2012.5

Syllabus of Biochemistry

(For International Students)PrefaceThis syllabus is based on the outline of biochemistry teaching for international

medical students. Harper’s Illustrated Biochemistry (the twenty-sixth edition) is

selected as one of the references. The overall objective of this curriculum is to provide

the basic principles of biochemistry and molecular biology. Total credit hours of

Biochemistry are 102. The credit hours of lecture are 72.

Section I. Structures & Functions of Proteins &

Enzymes

Chapter 3 Amino Acids & Peptides

OBJECTIVES1. Mastering: 20 L-α-amino acids, isoelectric pH, peptide bond formation,

characteristics of the peptide bond2. Comprehending: the properties of individual amino acids, properties of peptides3. Understanding: free D-amino acids, methods for determination of primary structure

of peptides4. Focus and difficulty: amino acids may have positive, negative or zero net charge,

primary structure of peptides

COURSE CONTENT1. Introduction and biomedical importance2. Property of amino acids

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(1) 20 L-α-amino acidsStructure of amino acids, L-α-amino acids in proteins, the genetic code, 20 L-α-

amino acids (2) Additional amino acids occur in specific proteins (3) Mammals contain certain free D-amino acids (4) Amino acids may have positive, negative or zero net charge (5) Physical properties of amino acids 3. Classification of amino acids4. Properties of individual amino acids5. Chemical reactions of amino acids

(1) Most important reaction – peptide bond formation(2) Characteristics of the peptide bond(3) What is peptide?

(4) Peptide structural formula (5) Abbreviations of amino acids used in peptide (6) Amino acids sequence determines primary structure (7) Primary structure affects biologic activity (8) Peptides can contain unusual amino acids6. Various techniques of separation amino acids

Chapter 5 Proteins: Higher Orders of Structure

OBJECTIVES1. Mastering: primary structure, secondary structure, tertiary structure, quaternary

structure2. Comprehending: configuration and conformation3. Understanding: various forces of the stability of structures4. Focus and difficulty: the four orders of protein structure

COURSE CONTENT1. Introduction and biomedical importance2. Classification of proteins3. The four orders of protein structure (1) Primary structure (2) Secondary structure Configuration and conformation, various forces, various secondary structure (3) Tertiary structure

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(4) Quaternary structure4. Physical techniques for study higher orders5. Protein folding

Chapter 6 Proteins: Myoglobin and Hemoglobin

OBJECTIVES1. Mastering: structures and functions of myoglobin and hemoglobin2. Comprehending: relation of structure and biologic function3. Understanding: sickle cell anemia, prion diseases4. Focus and difficulty: relation of structure and biologic function

COURSE CONTENT1. Introduction

(1) Relation of the primary structure and function (2) Relation of the higher structure and function

2. Myoglobin (1) Function (2) Structure (3) Linkage of Mb structure and biologic function3. Hemoglobin (1) Function (2) Structure

(3) Linkage of Mb structure and biologic function4. Sickle cell anemia5. Prion diseases

Chapter 7 Enzymes: Mechanism of Action

OBJECTIVES1. Mastering: general properties of enzymes, prosthetic groups, coenzymes, cofactors,

the active site of the enzyme, isozymes2. Comprehending: classification of enzymes, the enzyme name3. Understanding: mechanisms of enzyme-catalyzed reactions, detection of the

catalytic activity of enzymes4. Focus and difficulty: the active site of the enzyme, characteristic of the active site

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COURSE CONTENT1. Introduction and Biomedical Importance

2.General properties of enzymes (1) Highly efficiency(2) Highly specificityAbsolute specificity

Relative specificity Optical specificity3. Classification of enzymes

(1) Six classes of enzymes (2) The nomination of enzymes(3) Additional information may follow in parentheses

(4) A code number4. Many enzymes require coenzymes or prosthetic groups

(1) Prosthetic groups(2) Coenzymes(3) Cofactors

5. The catalytic activity of enzymes(1) What is the active site of the enzyme?(2) Characteristic of the active site

6. Mechanisms of enzyme-catalyzed reactions(1) Enzymes enhance reactant proximity and local concentration (2) Acid-Base catalysis (3) Catalysis by strain(4) Covalent catalysis(5) Substrates induce conformational changes in enzymes

7. Isozymes (1) What are isozymes?

(2) Diagnostic value of isozymes8. The catalytic activity of enzymes (1) What is IU? (2) Detection of the catalytic activity of enzymes

Chapter 8 Enzymes: Kinetics

OBJECTIVES

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1. Mastering: multiple factors affect the rates of enzyme-catalyzed reactions2. Comprehending: the kinetics of enzymatic catalysis3. Understanding: general theory of chemical reaction4. Focus and difficulty: multiple factors affect the rates of enzyme-catalyzed reactions

COURSE CONTENT1. Introduction and Biomedical Importance2. General theory of chemical reaction

(1) G determine the direction and equilibrium state of chemical reaction△(2) Numerous factors affect the reaction rate

3. The kinetics of enzymatic catalysis(1) Enzymes lower the energy barrier

(2) Enzymes provide transition states4. Multiple factors affect the rates of enzyme-catalyzed reactions

(1) Temperature(2) pH(3) Enzyme concentration(4) Substrate concentration

Effect of substrate concentration, the michaelis-menten equation, the

significance of Km, the determination of Km(5) Inhibitors

Chapter 9 Enzymes: Regulation of Activities

OBJECTIVES1. Mastering: active regulation of metabolism2. Comprehending: passive regulation of metabolism3. Understanding: biomedical importance, homeostasis4. Focus and difficulty: allosteric regulation, covalent modification

COURSE CONTENT1. Introduction and Biomedical Importance2. Homeostasis (1) What is the homeostasis?

(2) Regulation of metabolism achieves homeostasis3. Passive regulation of metabolism4. Characteristics of metabolism in the cells

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(1) Rate-limiting enzyme (2) Unidirection (3) Compartmentalization5. Active regulation of metabolism

(1) Regulation of enzyme quantity Enzyme synthesis by inducers and repression, degradation

(2) Regulation of the intrinsic catalytic activities of enzymes Allosteric regulation, covalent modifications

Section II Bioenergetics & the Metabolism of

Carbohydrates & Lipids

Chapter 10 Bioenergetics: The Role of ATP

OBJECTIVES1. Mastering: ATP /ADP cycle2. Comprehending: high-energy phosphates play a central role in energy capture and

transfer3. Understanding: endergonic processes proceed by coupling to exergonic processes4. Focus and difficulty: ATP acts as the "energy currency"

COURSE CONTENT1. Free energy is the useful energy in a system G: (Gibbis change in free energy) the portion of the total energy change in a

system that is available for doing work2. Biologic systmes conform to the general laws of thermodynamics (1) The total energy of a system, including its surroundings, remains constant (2) Chemical energy may be transformed into heart, electrical energy

The total entropy of a system must increase if a process is to occur

spontaneously.3. Endergonic processes proceed by coupling to exergonic processes (1) Vital processes obtain energy by coupling to oxdative reactions (2) Mechanism Synthesizing a high-energy intermediate biology, in the living cell, the main high-

energy compound is ATP4. High-energy phosphates play a central role in energy capture and transfer

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(1) Phosphate is divided into two groups based on the value for hydrolysis of

terminal phosphate of ATP(2) ATP acts as the "energy currency" Three major sources of ATP:

a. Oxidative phosphorylation: the free energy to drive this process comes from

respiratory chain oxidation using molecular O2. b. Glycolysis: catalyzed by phosphoglycerate kinase and pyruvate kinase.

c. The citric acid cycle: one ~P is generated directly in the cycle at the succinyl

thiokinase.d. Another: creatine phosphateATP tranfers of high-energy: a. Act as a phosphate donor to form those compounds of lower free engergy of

hydrolysisb. Activation reactionsc. Allows the coupling of thermodynamically unfavorable reactions to favorable

onesd. Synthesis e. Endergonic processes: muscular contraction, active transport, nervous excitation(3) ATP /ADP cycle

Chapter 11 Biological Oxidation

OBJECTIVES1. Mastering: enzymes involved in oxidation and reduction2. Comprehending: biologic oxidation, characteristics of biologic oxidation3. Understanding: the reactions catalyzed by enzymes involved in oxidation and

reduction 4. Focus and difficulty: the functions of enzymes involved in oxidation and reduction

COURSE CONTENT1. Introduction and Biomedical Importance

(1) What is biologic oxidation?(2) What are the characteristics of biologic oxidation?

2. Enzymes involved in oxidation and reduction are designated (1) Oxidases: catalyze the removal of hydrogen from a substrate using oxygen as a

hydrogen acceptor and form water or hydrogen peroxide. (2) Dehydrogenases: do not use oxygen as a hydrogen acceptor, and transfer of

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hydrogen from one substrate to another (3) Hydroperoxidases: using H2O2 or an organic peroxide as substrate, to protect the

body against harmful peroxides and free radicals(4) Oxygenases catalyze the direct transfer and incorporation of oxygen into a

substrate molecule

Chapter 12 The Respiratory Chain & Oxidative

Phosphorylation

OBJECTIVES1. Mastering: two respiratory chains, oxidative phosphorylation2. Comprehending: many poisons inhibit the respiratory chain3. Understanding: the chemiosmotic theory 4. Focus and difficulty: the pivotal role of the respiratory chain in metabolism

COURSE CONTENT1. Introduction and biomedical importance

(1) The system in mitochondria that couples respiration to the generation of the

high-energy intermediate, ATP(2) Capture a far greater proportion of the available free energy of respiratory

substrates compared with anaerobic organism(3) Drugs and poisons inhibit oxidative phosphorylation

2. The respiratory chain collects and oxidizes reducing equivalents (1) What is the respiratory chain?

(2) Components of the respiratory chain: components of the respiratory chain are

arranged in order of increasing redox potential(3) Two respiratory chains

3. Oxidative phosphorylation4. The respiratory chain provides most of the energy captured in metabolism5. Many poisons inhibit the respiratory chain6. The chemiosmotic theory7. Oxidation of extramitochondrial NADH is mediated by substrate shuttles

Chapter 16 The Citric Acid Cycle: The Catabolism of

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Acetyl-CoA

OBJECTIVES1. Mastering: the reactions of the citric acid cycle, the pivotal role of the citric acid cycle in

metabolism2. Comprehending: biomedical importance3. Understanding: regulation of the citric acid cycle4. Focus and difficulty: the key reactions in the citric acid cycle, ATP formation,

regulation of the citric acid cycle, the pivotal role of the citric acid cycle in

metabolism


(1) What is the citric acid cycle? (2) What is the biomedical importance?

a. Act as the final common pathway for the oxidation of carbohydrate, lipids

and protein b. Has a central role in gluconeogenesis, transamination, deamination and

lipogenesis2. Reactions of the citric acid liberate reducing equivalents and CO2

(1) Reactions of the citric acid(2) The cycle is located in the matrix of mitochondria(3) The reduced coenzymes are oxidized by the respiratory chain linked to

formation of ATP(4) Vitamins play key roles in the citric acid cycle

3. Regulation of the citric acid cycle(1) Respiratory control(2) Depends primarily on a supply of oxidized cofactors(3) Pyruvate dehydrogenase, citrate synthase, isocitrate dehydrogenase, and α-

ketoglutarate dehydrogenase(4) [ATP]/[ADP] and [NADH]/[NAD+] ratios

4. A pivotal role in metabolism (1) It provides the substrates for amino acid synthesis by trans-amination, as well as

for gluconeogenesis and fatty acid synthesis(2) Because it functions in both oxidative and synthetic processes, it is amphibolic.

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Chapter 17 Glycolysis & the Oxidation of Pyruvate

OBJECTIVES1. Mastering: main pathway of glycolysis and pyruvate oxidation to acetyl-CoA, how

many ATP yielded up by oxidation of glucose under aerobic condition and when O2

is absent2. Comprehending: biomedical importance and regulation of glycolysis3. Understanding: lactic acidosis4. Focus and difficulty: the significance of the metabolism of glucose (or glycogen) to

pyruvate and lactate or the metabolism of glucose (or glycogen) to pyruvate and

acetyl-CoA


(1) What is glycolysis?(2) Biomedical importancea. The ability of glycolysis to provide ATP in the absence of oxygen is especially

important because it allows skeletal muscle to perform when oxygen supply is

insufficient.b. Erythrocytes, which lack mitochondria, are completely reliant on glucose as their

metabolic fuel and metabolize it by anaerobic glycolysis2. Glycolysis constitute the main pathway of glucose utilization

(1) Occurs in the cytosol of all cells.(2) Overall equation of glycolysis:

(3) Separated steps only for explanation:3. Glycolysis is regulated at three steps involving nonequilibrium reactions (1) Hexokinase (and glucokinase), phosphofructokinase, and pyruvate kinase, are the

major sites of regulation of glycolysis(2) 2, 3-bisphosphoglycerate pathway in erythrocytes

4. Pyruvate oxidation to acetyl-CoA(1) Pyruvate is oxidized to acetyl-CoA by a multienzyme complex, pyruvate

dehydrogenase (2) The oxidation of pyruvate to acetyl-CoA is the irreversible route(3) Oxidation of glucose yields up to 30 or 32 mol of ATP under aerobic condition

but only 2 mol of ATP when O2 is absent

5. Clinical aspects

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Inhibition of pyruvate metabolism leads to lactic acidosis

Chapter 18 Metabolism of Glycogen

OBJECTIVES1. Mastering: main pathways of glycogenesis and glycogenolysis2. Comprehending: regulation of glycogen metabolism3. Understanding: glycogen storage disease4. Focus and difficulty: regulation of glycogen metabolism is balance in activities of

glycogen synthase and phosphorylase

COURSE CONTENT1. Introduction and Biomedical Importance (1) What is glycogen?(2) Metabolism of glycogen contains glycogenesis and glycogenolysis(3) Biomedical importance

2. Glycogen biosynthesis (glycogenesis) (1) Glycogenesis occurs mainly in muscle and liver(2) The pathway of glycogen biosynthesis

3. Glycogenolysis Glycogenolysis is not the reverse of glycogenesis but is a separate pathway.

4. Regulation of glycogen metabolism (1) It is influenced by a balance of glycogen synthase and phosphorylase

(2) Glycogen synthase and phosphorylase activity are regulated by allosteric

mechanisms and covalent modifications(3) Cyclic AMP integrates the regulation of glycogenolysis and glycogenesis(4) Thus, inhibition of glycogenolysis enhances net glycogenesis, and inhibition of

glycogenesis enhances net glycogenolysis5. Clinical aspects

Glycogen storage disease

Chapter 19 Gluconeogenesis & Control of the Blood Glucose

OBJECTIVES1. Mastering: the pathways of gluconeogenesis, the sources and usage of blood

glucose, regulation of the blood glucose2. Comprehending: glycolysis and glyconeogenesis must be regulated reciprocally

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3. Understanding: clinical aspects4. Focus and difficulty: the sources and usage of blood glucose, regulation the blood

glucose

COURSE CONTENT 1. Introduction and Biomedical Importance2. The pathways of Gluconeogenesis

(1) Gluconeogenesis involves glycolysis, the citric acid cycle, plus some special

reaction(2) Relationship between glycerol and gluconeogenesis

Source: metabolism of adipose tussue Enzyme: glycerol kinase

Reaction:3. Glycolysis and glyconeogenesis must be regulated reciprocally

Three types of mechanism as responsible for regulating the activity of enzymes: (1) Changes in the rate of enzyme synthesis, (2) Covalent modification by reversible phosphorylation, and

(3) Allosteric effectsThe most potent positive allosteric effector of phospho-fructokinase-1 and

inhibitor of fructose-1, 6-bisphosphatase in liver is fructose 2, 6-bisphosphate4. Blood glucose

(1) The concentration of blood glucose is regulated within narrow limits (3.89-

6.11mmol/L)(2) Blood glucose is derived from the diet, gluconeogenesis, and glycogenolysis (3) Blood glucose is used by oxidation, glycogen, the pentose phosphate pathway,

amino acid and adipose tissue(4) Regulation of the blood glucose

Glucokinase is important in regulating blood glucose after a meal Insulin plays a central rose in regulating blood glucoseGlucagon opposes the actions of insulinOther hormones affect Blood glucose

5. Clinical aspects

Chapter 20 The Pentose Phosphate Pathway & Other

Pathways of Hexose Metabolism

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OBJECTIVES1. Mastering: biomedical importance of the pentose phosphate pathway 2. Comprehending: the pentose phosphate pathway generates NADPH and ribose

phosphate3. Understanding: erythrocyte hemolysis4. Focus and difficulty: the pentose phosphate pathway generates NADPH and ribose

phosphate

COURSE CONTENT 1. Introduction and biomedical importance (1) The formation of NADPH for synthesis of fatty acids and steroids (2) The synthesis of ribose for nucleotide and nucleic acid formation

2. The pentose phosphate pathway (1) Reactions of the pentose phosphate pathway occur in the cytosol (2) The pentose phosphate pathway generates NADPH and ribose phosphate

3. Clinical aspectsErythrocyte hemolysis

Chapter 21 Biosynthesis of Fatty Acids

OBJECTIVES1. Mastering: the controlling step, main source of NADPH and the building block in

fatty acid synthesis2. Comprehending: the fatty acid synthase complex and elongation of fatty acid

chains3. Understanding: the regulation of lipogenesis4. Focus and difficulty: How is lipogenesis de-regulated in disease state?

COURSE CONTENTS1. Biomedical Importance2. The main pathway for de novo synthesis of fatty acids (lipogenesis) occurs in the

cytosol(1) Production of malonyl-CoA is the initial and controlling step in fatty acid

synthesis(2) The fatty acid synthase complex is a polypeptide containing seven enzyme

activities(3) The main source of NADPH for lipogenesis is the pentose phosphate

pathway

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(4) Acetyl-CoA is the principal building block of fatty acids(5) Elongation of fatty acid chains occurs in the endoplasmic reticulum

3. The nutritional state regulates lipogenesis4. Short- and long-term mechanisms regulate lipogenesis

(1) Acetyl-CoA carboxylase is the most important enzyme in the regulation of

lipogenesis(2) Pyruvate dehydrogenase is also regulated by Acyl-CoA(3) Insulin also regulates lipogenesis by other mechanisms(4) The fatty acid synthase complex and acetyl-CoA carboxylase are adaptive

enzymes

Chapter 22 Oxidation of Fatty Acids: Ketogenesis

OBJECTIVES1. Mastering: transportion and activation of fatty acids; β-oxidation is a cyclic

reaction2. Comprehending: a large quantity of ATP is produced by oxidation of fatty acids; the

concept and generation of ketone bodies3. Understanding: the regulation of ketogenesis4. Focus and difficulty: high rate of fatty acid oxidation promotes ketogenesis

COURSE CONTENTS1. Biomedical Importance2. Oxidation of fatty acids occurs in mitochondria

(1) Fatty acids are transported in the blood as free fatty acids (FFA)(2) Fatty acids are activated before being catabolized(3) Long-chain fatty acids penetrate the inner mitochondrial membrane as

carnitine derivatives3. β-oxidation of fatty acids involves successive cleavage with release of acetyl-CoA

(1) The cyclic reaction sequence generates FADH2 and NADH(2) Oxidation of a fatty acid with an odd number of carbon atoms yields acetyl-

CoA plus a molecule of propionyl-CoA(3) Oxidation of fatty acids produces a large quantity of ATP(4) Peroxisomes oxidize very long chain fatty acids

4. Oxidation of unsaturated fatty acids occurs by a modified β-oxidation pathway5. Ketogenesis occurs when there is a high rate of fatty acid oxidation in the liver

(1) 3-Hydroxy-3-Methylglutaryl-CoA (HMG-CoA) is an intermediate in the

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pathway of ketogenesis(2) Ketone bodies serve as a fuel for extrahepatic tissues

6. Ketogenesis is regulated at three crucial steps7. Clinical aspects

(1) Impaired oxidation of fatty acids gives rise to diseases often associated with

hypoglycemia(2) Ketoacidosis results from prolonged ketosis

Chapter 25 Lipid Transport and Storage

OBJECTIVES1. Mastering: the concepts of lipoproteins, chylomicrons, VLDL, LDL, HLD and fat

mobilization2. Comprehending: the metabolisms of chylomicrons, VLDL, LDL and HLD and the

central role of liver in lipid transport and metabolism3. Understanding: the regulation of fat mobilization by hormones4. Focus and difficulty: the properties of four groups of lipoproteins and their

metabolism

COURSE CONTENTS1. Biomedical Importance2. Lipids are transported in the plasma as lipoproteins

(1) Four major lipid classes are present in lipoproteins(2) Four major groups of plasma lipoproteins have been identified(3) Lipoproteins consist of a nonpolar core and a single surface layer of

amphipathic lipids(4) The distribution of apolipoproteins characterizes the lipoprotein

3. Free fatty acids are rapidly metabolized4. Triacylglycerol is transported from the intertines in chylomicrons and from the liver

in very low density lipoproteins5. Chylomycrons and very low density lipoproteins are rapidly catabolized

(1) Triacylglyceols of chylomicrons and VLDL are hydrolyzed by lipoprotein

lipase(2) The action of lipoprotein lipase forms remnant lipoproteins(3) The liver is responsible for the uptake of remnant lipoproteins

6. LDL is metabolized via the LDL receptor7. HDL takes part in both lipoprotein triacylglycerol and cholesterol metabolism

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8. The liver plays a central role in lipid transport and metabolism(1) Hepatic VLDL secretion is related to dietary and hormonal status

9. Clinical aspects(1) Imbalance in the rate of triacylglycerol formation and export causes fatty

liver(2) Ethanol also causes fatty liver

10. Adipose tissue is the main store of triacylglycerol in the body(1) The provision of glycerol 3-phosphate regulates esterification: lipolysis is

controlled by hormone-sensitive lipase(2) Increaed glucose metabolism reduces the output of free fatty acids

11. Hormones regulate fat mobilization(1) Insulin reduces the output of free fatty acids(2) Several hormones promote lipolysis(3) A variety of mechanisms have evolved for fine control of adipose tissue

metabolism12. Brown adipose tissue promotes thermogenesis

Chapter 26 Cholesterol Synthesis, Transport, and Excretion

OBJECTIVES1. Mastering: the sources of cholesterol in our bodies; the substrates and key step in

cholesterol synthesis; regulation of cellular cholesterol level2. Comprehending: serum cholesterol is involved in the pathogenesis of

atherosclerosis and coronary heart disease and how to control serum cholesterol

level3. Understanding: conversion and removal of body cholesterol4. Focus and difficulty: cholesterol is derived from both diet and biosynthesis and this

determines how to alter serum cholesterol level effectively

COURSE CONTENTS1. Biomedical importance2. Cholesterol is derived about equally from the diet and from biosynthesis

(1) Acetyl-CoA is the source of all carbon atoms in cholesterol(2) Farnesyl diphosphate gives rise to dolichol and ubiquinone

3. Cholesterol synthesis is controlled by regulation of HMG-CoA reductase4. Many factors influence the cholesterol balance in tissues

(1) The LDL receptor is highly regulated

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5. Cholesterol is transported between tissues in plasma lipoproteins(1) Plasma LCAT is responsible for virtually all plasma cholesteryl ester in

humans(2) Cholesteryl ester transfer protein facilitates transfer of cholesteryl ester from

HDL to other lipoproteins6. Cholesterol is excreted from the body in the bile as cholesterol or bile acids (salts)

(1) Bile acids are formed from cholesterol(2) Most bile acids return to the liver in the enterohepatic circulation(3) Bile acid synthesis is regulated at the 7α-hydroxylase step

7. Clinical aspects(1) The serum cholesterol is correlated with the incidence of atherosclerosis and

coronary heart disease(2) Diet can play an important role in reducing serum cholesterol(3) Lifestyle affects the serum cholesterol level(4) When diet changes fail, hypolipidemic drugs will reduce serum cholesterol

and triacylglycerol(5) Primary disorders of the plasma lipoproteins (Dyslipoproteinemias) are

inherited

Chapter 27 Integration of Metabolism – The Provision of

Metabolic Fuels

OBJECTIVES1. Mastering: metabolic changes in fed and starving states2. Comprehending: metabolic interrelationships between adipose tissue, liver and

extrahepatic tissues3. Understanding: abnormal metabolism results in certain diseases4. Focus and difficulty: there is no net conversion of most fatty acids to glucose, but

most amino acids and glycerol can be used for gluconeogenesis

COURSE CONTENTS1. Biomedical importance2. Many metabolic fuels are interconvertible3. A supply of metabolic fuels is provided in both the fed and starving states

(1) Glucose is always required by the central nervous system and erythrocytes(2) In the fed state, metabolic fuel reserves are laid down

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(3) Metabolic fuel reserves are mobilized in the starving state4. Clinical aspects

Section III Metabolism of Proteins & Amino Acids

Chapter 28 Biosynthesis of the Nutritionally

Nonessential Amino Acids

OBJECTIVES1. Mastering: humans can synthesize 12 of the 20 common amino acids from the

amphibolic intermediates of glycolysis and of the citric acid cycle2. Comprehending: the biosynthetic pathways of glutamate, glutamine, alanine,

aspartate and asparagine3. Understanding: the biosynthetic pathways of hydroxyproline, hydroxylysine,

valine, leucine, and isoleucine4. Focus and difficulty: glutamate dehydrogenase, glutamine synthetase, and

aminotransferases occupy central positions in amino acid biosynthesis

COURSE CONTENTS1. Nutritionally nonessential amino acids have short biosynthetic pathways

(1) Glutamate and glutamine(2) Alanine (3) Aspartate and asparagine

2. Biosynthetic pathways of various amino acids (1) Serine(2) Glycine (3) Proline(4) Tyrosine(5) Hydroxyproline and hydroxylysine(6) Valine, leucine, and isoleucine(7) Seleoncysteine, the 21st amino acid

Chapter 29 Catabolism of Proteins & of

Amino Acid Nitrogen

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OBJECTIVES1. Mastering: biosynthesis of urea2. Comprehending: the degrading pathway of proteins3. Understanding: metabolic disorders of the urea cycle 4. Focus and difficulty: biosynthesis of urea

COURSE CONTENTS1. Proteins are degraded

(1) Proteins are degraded by ATP-dependent (2) Proteins are degraded by ATP-independent pathways(3) Ubiquitin targets many intracellular proteins for degradation

2. Biosynthesis of urea(1) Urea biosythesis occurs in four stages: transamination, oxidative deaminationof

glutamate, ammonia transport, and reactions of the urea cycle(2) L-Glutamate dethdrogenase occupies a central position in nitrogen metabolism (3) Urea is the major end product of nitrogen catabolism in humans(4) Hepatic urea synthesis takes place in part in the mitochondrial matrix and in

part in the cytosol(4) Urea synthesis is a cyclic process(5) The major metabolic role of ornithine, citrulline, and argininosuccinate in

mammals if urea synthesis(6) Changes in enzyme levels and allosteric regulation of carbamoyl phosphate

synthase by N-acetylglutamate regulate urea biosynthesis3. Clinical aspects

Metabolic disorders of the urea cycle

Chapter 30 Catabolism of the Carbon

Skeletons of Amino Acids

OBJECTIVES1. Mastering: Excess amino acids are catabolized to amphibolic intermediates used as

sources of energy or for carbohydrate and lipid biosynthesis2. Comprehending: transamination typically initiates amino acid catabolism3. Understanding: clinical aspects 4. Focus and difficulty: the pathway of amino acids catabolized to substrates for

carbohydrate and lipid biosynthesis

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COURSE CONTENTS1. Transamination typically initiates amino acid catabolism2. Reactions remove any additional nitrogen and restructure the hydrocarbon skeleton

for conversion to oxaloacetate, α- ketoglutarate, pyruvate, and acetyl-CoA(1) Asparagines, aspartate, glutamine, and glutamate (2) Six amino acids form pyruvate(3) Twelve amino acids form acetyl-CoA(4) The initial reactions are common to all three branched-chain amino acids

3. Clinical aspects (1) Phenylketonuria(2) Metabolic disorders of branched-chain amino acid catabolism

Chapter 31 Conversion of Amino

Acids to Specialized Products

OBJECTIVES1. Mastering: the specialized products of main amino acids2. Comprehending: neurotransmitters derived from amino acids3. Understanding: clinical aspects 4. Focus and difficulty: conversion of methionine and tyrosine to specialized products

COURSE CONTENTS1. Glycine participates in additional biosynthetic processes

(1) Glycine participates in the biosynthesis of heme, purines, and creatine (2) Glycine is conjugated to bile acids and to the urinary metabolites of many drugs

2. Serine participates in additional biosynthetic processes(1) Serine participates in phospholipid and sphingosine biosynthesis(2) Serine provides carbons 2 and 8 of purines and the methyl group of thymine

3. Glutamate and ornithine form the neurotransmitter γ-aminobutyrate (GABA)4. S-Adenosylmethionine, the methyl group donor for many biosynthetic processes5. Tyrosine forms both epinephrine and norepinephrine, and its iodination forms

thyroid hormone

Section IV Structure, Function, & Replication of

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Informational Macromolecules

Chapter 33 Nucleotides

OBJECTIVES1. Mastering: structure and function of nucleotides 2. Comprehending: neurotransmitters derived from amino acids3. Understanding: clinical aspects 4. Focus and difficulty: nucleotides chemical composition, structural formula,

numbering systems, linked bonds

COURSE CONTENTS1. Structure and function of nucleotides.

(1) Nucleosides & Nucleotides

Chemistry of purine, pyrimidine, their nucleotides and nucleotides chemical

composition, structural formula, numbering systems, linked bonds

(2) Nucleic Acids Also Contain Additional Bases(3) Major biochemical functions of purine and pyrimidine nucleotides

2. Synthetic nucleotide analogs are used in chemotherapy3. Polynucleotides are directional macromolecules

Chapter 34 Metabolism of Purine &

Pyrimidine Nucleotides

OBJECTIVES1. Mastering: de novo pathway, salvage pathway, catabolism of purine and pyrimidine

nucleotides2. Comprehending: biosynthesis of purine nucleotides and pyrimidine nucleotides3. Understanding: clinical aspects 4. Focus and difficulty: salvage pathway, catabolism of purine and pyrimidine

nucleotides

COURSE CONTENTS1. Biosynthesis of purine nucleotides (1) The sources of the nitrogen and carbon atoms of purine ring (2) The some features of de novo pathway, conversion of IMP to AMP and GMP,

25

regulation of synthesis (3) Salvage pathway (4) Reduction of NDP forms dNDP2. Biosynthesis of pyrimidine nucleotides (1) The sources of the nitrogen and carbon atoms of pyrimidine ring (2) Some features of de novo pathway, conversion of UMP to CTP and (d) TMP,

regulation of synthesis3. Catabolism of purine and pyrimidine nucleotides (1) Purine nucleotide purine ring → uric acid (2) Pyrimidine nucleotide

Chapter 35 Nucleic Acid Structure & Function

OBJECTIVES1. Mastering: the structure of DNA molecule and RNA molecule 2. Comprehending: the functions of DNA molecule and RNA molecule3. Understanding: other RNAs 4. Focus and difficulty: The important features of the Watson-Crike model of DNA,

denaturation, renaturation, hybrid, structures and functions of the three main types

of RNA

COURSE CONTENTS1. The structure of DNA molecule (1) The important features of the Watson-Crike model of DNA

(2) The physical and chemical nature of DNA molecule Denaturation, renaturation, hybrid

(3) DNA exists in relaxed and supercoiled form2. The structural features of RNA molecule (1) The structural features (2) The three main types of RNA (structures and functions) mRNA, tRNA, rRNA

and other RNAs: snRNA et al

Chapter 36 DNA Organization, Replication & Repair

OBJECTIVES1. Mastering: semi-conservative replication, semi-discontinuous replication, the

leading strand, the lagging strand, Okazaki fragment, replication fork, enzymes

26

and protein factors involved in DNA replication, the process of DNA synthesis2. Comprehending: DNA damage and repair3. Understanding: reverse transcription, reverse transcriptase4. Focus and difficulty: semi-discontinuous replication

COURSE CONTENT1. The basic principles of replication

(1) Definition of replication (2) The origin and direction of replication (3) Semi-conservative replication (4) Semi-discontinuous replication2. Enzymes and protein factors involved in DNA replication (1) DNA helicase (2) Topoisomerase (3) Single stranded binding protein (4) Primase (5) DNA polymerase (6) DNA ligase3. The process of DNA synthesis (1) Unwinding of double helix and supercoil (2) Synthesis of primer (3) DNA chain extension4. Reverse transcription5. DNA damage and repair (1) Mismatch repair (2) Base excision-repair

(3) Nucleotide excision-repair(4) Double-strand break repair

Chapter 37 RNA Synthesis, Processing & Modification

OBJECTIVES1. Mastering: template strand, coding strand, asymmetric transcription, promoter,

RNA polymerase, the process of transcription2. Comprehending: RNA processing and modification3. Understanding: ribozyme4. Focus and difficulty: the regulation of transcription

27

COURSE CONTENT1. Basic principles of transcription

(1) Definition of transcription (2) Some important features for transcription (3) Terminology for transcription template strand, coding strand, asymmetric transcription, promoter, transcription

initiation site, primary transcript2. Transcription in prokaryotes (1) E. coli RNA polymerase (2) Promoter and pol binding (3) Three phases of transcription RNA chain initiation, RNA chain elongation, RNA chain termination3. Transcription in eukaryotes (1) Three RNA polymerases in nucleus (2) RNA polymerase II promoters (3) The transcription regions in class II gene (4) Transcription factors in class II gene (5) Formation of the basal transcription complex (6) The assembly of the transcription complex4. RNA processing and modification (1) Pre-mRNA processing and modification Adding cap at the 5’ end, adding poly(A) tail at the 3’ end, RNA splicing, RNA

editing (2) Pre-tRNA processing and modification

(3) Pre-rRNA processing and modification

Chapter 38 Protein Synthesis & the Genetic Code

OBJECTIVES1. Mastering: codon, the features of genetic code, anticodon, the action of three types

of RNA in the protein synthesis, the process of protein synthesis2. Comprehending: polyribosomes, posttranslational processing3. Understanding: mutation4. Focus and difficulty: degeneracy, wobble base pair

COURSE CONTENT1. Basic principles of translation

28

(1) Definition of translation (2) The system of protein synthesis (3) The action of three types of RNA in the protein synthesis mRNA, tRNA, rRNA2. The process of protein synthesis (1) The activation and transfer of amino acid (2) Ribosomal cycle Initiation, elongation (binding of aminoacyl-tRNA to the A site, peptide bond

formation, translocation), termination3. Posttranslational processing4. The comparison of the features of replication, transcription, reverse transcription

and translation5. Mutation (1) Single base-pair substitution (point mutation) (2) Frameshift mutation

Chapter 39 Regulation of Gene Expression

OBJECTIVES1. Mastering: biomedical importance of regulation gene expression, regulation of the

lac operon, DNA cis-active elements, trans-acting protein2. Comprehending: principles of gene regulation3. Understanding: mutation4. Focus and difficulty: regulation of the lac operon, alternative RNA processing


(1) What is gene?(2) What is gene expression?(3) Gene expression can be controlled(4) Biomedical Importance (5) Principles of gene regulation

2. Regulation of prokaryotic gene expression(1) Some features of prokaryotic gene expression are unique(2) Analysis of lactose metabolism in E. col led to the operon hypothesis

3. Special features are involved in regulation of eukaryotic gene transcription(1) The features are involved in regulation of eukaryotic gene

29

(2) Transcription initiation controllingTranscription controls operate at the level of protein-DNA and protein-protein

interactions RNA polymerase DNA cis-active elements Trans-acting protein Interacting of the three aspects

4. Gene regulation in prokaryotes and eukaryotes differs in important respects(1) Alternative RNA processing is another control mechanism(2) Regulation of messenger RNA stability provides another control mechanism

Chapter 40 Molecular Genetics, Recombinant

DNA & Genomic Technology

OBJECTIVES1. Mastering: the general procedures of DNA cloning involved2. Comprehending: cloning vectors, the enzymes involved in DNA cloning3. Understanding: practical applications of recombinant DNA technology4. Focus and difficulty: cloning vectors, the enzymes involved in DNA cloning, the

method of screening recombinants for inserted DNA fragments, the method of gaining

for inserted DNA fragments


(1) What is recombinant DNA technology?(2) Biomedical importance

2. Elucidation of the basic features of DNA led to recombinant DNA technology(1) DNA is a complex biopolymer organized as a double helix(2) Base pairing is a fundamental concept of DNA structure and function

Denaturion, renaturation, hybridization(3) DNA is organized into genes(4) Gene expression and regulation of gene expression

3. Recombinant DNA technology involves isolation and manipulation of DNA to

make chimeric molecules(1) Restriction enzymes cut DNA chains at specific sites(2) Restriction enzymes and DNA ligase are used to prepare chimeric DNA

30

molecules(3) Reverse transcriptase(4) Cloning amplifies DNA

What is clone?Cloning vectors: bacterial plasmids, bacteriophage, cosmidsThe method of screening recombinants for inserted DNA fragmentsA library is a collection of recombinant clonesThe polymerase chain reaction (PCR) amplifies DNA sequences

4. Practical applications of recombinant DNA technology are numerous(1) Gene mapping localizes specific genes to distinct chromosomes(2) Proteins can be produced for research & diagnosis(3) Recombinant DNA technology is used in the molecular analysis of disease

教学方法生物化学教学应以学生掌握生物化学基本内容为目标。1.理论(1)讲授：利用多媒体通过讲授向每个学生最大程度上地传授相关知识和信息(2)讨论：有关疾病与生物化学的关系问题组织学生进行讨论(3)研讨会：安排一些时间，对一些重要的专题，请相关领域的专家与学生进行研讨(4)自主学习：对学生感兴趣的知识，在教师指导下让学生自主学习2.实验(1)实验操作：实验包括验证性实验和综合性实验。全班分为15组，每2人一组①验证性实验：按照操作步骤进行②综合性实验：从蛋白质分离和纯化技术、蛋白质定量测定，到蛋白质功能研究技术一系列关于蛋白质的生化实验技术。

31

(2)实验报告：做完实验后写出报告，并对实验结果做出分析

参考书目1. Albert Lehninger, Davia L Nelson, Michael M Cox. 生物化学原理, 5th Edition,

McGraw-Hill, 2008

2. Philip Kuchel, Simon Easterbrook-Smith, Vanessa Gysbers, J. Mitchell Guss, et al.

生物化学概要, 3th Edition, McGraw-Hill, 2010

3. Jeremy M. Berg, John L, Tymoczko and Lubert Stryer. 生物化学, Sixth Edition,

McGraw-Hill, 2006

4. 生物化学杂志

考试形式和成绩分配总分100

1.理论考试（100分）：(1)期中考试（30分）(2)期末考试（30分）(3)小测验（20分）(4)讨论（5分）(5)研讨会（5分）(6)自主学习（10分）2.实验考试（100分）:

(1)期末考试（50分）(2)实验报告：50分

32

学时分配学时分配：102 （1 个学期）具体理论和实习课时如下：教学内容理论学时实验学时总学时1. 氨基酸和肽 2 9

2. 蛋白质高级结构 3

3. 肌红蛋白和血红蛋白 1

4. 酶作用机制 1

5. 酶促反应动力学 2 3

6. 酶活性调节 2

7. 生物能：ATP 的功能 1

8. 呼吸链和氧化磷酸化 2

9. 三羧酸循环 2

10. 糖酵解和丙酮酸氧化 2 3

11. 糖原代谢 1 3

12. 糖异生和血糖调节 2

13. 磷酸戊糖途径 1

14. 脂肪酸的生物合成 2

15. 脂肪酸的氧化 2

33

16. 脂的运输和存储 2

17. 胆固醇的合成、转运和转化 2

18. 代谢整合 2

19. 营养非必需氨基酸的合成 2

20. 氨基酸分解代谢 2 3

21. 氨基酸碳骨架分解代谢 2

22. 个别氨基酸的代谢 2

23. 嘌呤和嘧啶核苷酸的代谢 2

24. 核苷酸结构域功能 2 3

25. DNA 结构、复制与修复 6

26. RNA 合成、加工与修饰 6

27. 蛋白质合成与遗传密码子 4

28. 基因表达调控 4

29. 重组基因重组与基因工程 4

30. 生化实验概述 3

31. 总复习 3

总学时: 72 30 102

34

TEACHING – LEARNING METHODOLOGY The biochenmistry teaching should be performed with the purpose of making the

students grasp the basic contents of biochenmistry.

1. Theory

(1) Lectures: teach students related knowledge and information to the largest extent

by multimedia

(2) Discussion: organize the students to discuss the questions associated with

35

biochemical reason of the causeof diseases cause of diseases

(3) Seminars: seminars on some important topics will be planned, in which experts in

corresponding fields are invited to discuss with the students

(4) Self-teaching: teach students the interesting knowledge by themselves under the

instruction of teachers, and write reports

2. Practice

(1) Experiment operation: the experiments contain verification experiments and

comprehensive experiments. The whole class will be divided into 15 groups and each

group contains 2 students

① Verification experiments: Do the experiments as the procedures

② Comprehensive experiments: a set of experiments in which separation and

purification technology, quantitative assay, and function research technology of

protein contain.

(2) Experiment reports: Write the experiment reports, and analyze the final results

TEXT-BOOKS RECOMMENDED 1. Albert Lehninger, Davia L Nelson, Michael M Cox. Lehninger Principles of

Biochemistry, 5th Edition, McGraw-Hill, 2008

2. Philip Kuchel, Simon Easterbrook-Smith, Vanessa Gysbers and J. Mitchell

Guss .Schaum's Outline of Biochemistry, 3th Edition, McGraw-Hill, 2010

3. Jeremy M. Berg, John L, Tymoczko and Lubert Stryer.Biochemistry, Sixth Edition,

McGraw-Hill, 2006

4. The Journal of Biological Chemistry [J]

EXAMANATION PATTERN & MARKS DISTRIBUTION

1. Theory Exam: 100 marks：(1) Final examination (30 marks)

(2) Middle examination (30 marks)

(3) Quiz (20 marks),

(4) Discussion (5 marks)

(5) Seminars (5 marks)

36

(6) Overview (10 marks)

2. Practicals Exam: (100 Marks)

(1) Final examination (50 marks)

(2) Experiments reports: (50 marks)

TEACHING HOURS DISTRIBUTIONTotal numbers of teaching hours are approximately 102 (one semester)

Distribution of teaching hours for theory and practicals are as follows

Subject Lecture Practices sum

1. Amino Acids & Peptides 2 92. Higher Orders Structure 3 3. Myoglobin and Hemoglobin 1 4. Mechanism of Enzyme 2

Action 15. Enzymes: Kinetics 2 3 6. Regulation of Enzymes Activities 27. Bioenergetcs: The Role of ATP 1 8. The Respiratory Chain & 2Oxidative Phosphorylation

9 .The Citric Acid Cycle 210. Glycolysis& the Oxidation of Pyruvate 2 311. Metabolism of Glycogen 1 312. GluconeoGenesis and Control of the 2

Blood Glucose 13. The Pentose Phosphate Pathway 114. Biosynthesis of Fatty Acids 215. Oxidation of Fatty Acids 216. Lipid Transport & Storage 217. Cholesterol Synthesis, 2 Transport, & Excretion

18. Integration of Metabolism 219. Biosynthesis of the 2

Nutritionally Nonessential Amino Acids20. Amino Acids Catabolism of 2 3

Proteins & of Amino Acid Nitrogen 21. Catabolism of the 2

37

Carbon Skeletons of Amino Acids22. Conversion of Amino Acids to 2

Specialized Products 23. Nucleotides Metabolism of Purine 4

& Pyrimidine Nucleotides 24. Nucleic Acid Structure & Function 2 325. DNA Organization, Replication, 6

& Repair26. RNA Synthesis, Processing, 6

& Modification27. Protein Synthesis 4

& the Genetic Code28. Regulation of Gene Expression 429. Molecular Genetics, Recombinant DNA, 4

& Genomic Technology30. Introduction to the Biochemistry Laboratory 331. Total Review 3

Total teaching hours: 72 30 102

38

3．教学进度天津医科大学基础医学院教学进度表

学年第学期专业：年级：班级：课程名称：　　周次 Week

日期 Date

讲课内容 Contents

学时 Teaching

hours

节次Time

主讲教师 Teachers

实验内容 Experiment Contents

学时 Teaching hours

实验教师 Experiment teacher

1 　　　　　　　　23456 　　　　　　　7 　　　　　　　　8

39

9

10

11

12

13

14

15

16

17 　　　　　　　　18 　　　　　　　　19 　　　　　　　

总学时理论实验　　　　　上课时间上课地点主任签字：

40

天津医科大学基础医学院理论课教学进度表2012 学年—2013 学年第一学期

专业：留学生 1 班年级：2011

课程名称：生物化学周次 Week

日期 Date

讲课内容 Contents

学时 Teaching

hours

节次Time

主讲教师 Teachers

1 2012-8-27(1)Amino Acids & Peptides(2)Higher Orders Structure

3 1-3 Geng Xin

1 2012-8-29(1)Higher Orders Structure(2)myoglobin and hemoglobin

2 1-2 Geng Xin

2 2012-9-3(1)Mechanism of Enzyme Action(2)Enzymes: Kinetics

3 1-3 Geng Xin

2 2012-9-5 (1)Regulation Of Enzymes Activities 2 1-2 Geng Xin

3 2012-9-10(1)Bioenergetcs: the Role of ATP(2)The Respiratory Chain & Oxidative Phosphorylation

3 1-3 Geng Xin

3 2012-9-12 (1) The Citric Acid Cycle 2 1-2 Geng Xin

4 2012-9-17 (1)Glycolysis& the Oxidation of Pyruvate 3 1-3 Geng Xin

4 2012-9-19(1)Glycolysis& the Oxidation of Pyruvate(2) Metabolism of Glycogen

2 1-2 Geng Xin

5 2012-9-24(1)Gluconeogenesis and Control of the Blood Glucose (2)The Pentose Phosphate Pathway

3 1-3 Geng Xin

5 2012-9-26 (1)Biosynthesis of Fatty Acids 2 1-2 Li Haidong 7 2012-10-8 (1)Oxidation of Fatty Acids: Ketogenesis 3 1-3 Li Haidong

7 2012-10-10 (1)Lipid Transport & Storage 2 1-2 Li Haidong

8 2012-10-15(1)Cholesterol Synthesis, Transport, & Excretion

3 1-3 Li Haidong

8 2012-10-17 (1)Integration of Metabolism 2 1-2 Li Haidong

9 2012-10-22(1)Biosynthesis of the Nutritionally Nonessential Amino Acids

3 1-3 Li Haidong

9 2012-10-24 (1)Amino Acids Catabolism of Proteins & 2 1-2 Li Haidong

42

of Amino Acid Nitrogen (2)Catabolism of the Carbon Skeletons of Amino Acids

10 2012-10-29(1)Conversion of Amino Acids to Specialized Products

3 1-3 Li Haidong

10 2012-10-31(1)Nucleotides Metabolism of Purine & Pyrimidine Nucleotides

2 1-2 Li Haidong

11 2012-11-5 (1)Nucleic Acid Structure & Function 3 1-3 Li Haidong

11 2012-11-7(1)DNA Organization, Replication, & Repair

2 1-2 Yu Gongyuan

12 2012-11-12(1)DNA Organization, Replication, & Repair

3 1-3 Yu Gongyuan

12 2012-11-14(1)RNA Synthesis, Processing, & Modification

2 1-2 Yu Gongyuan

13 2012-11-19(1)RNA Synthesis, Processing, & Modification

3 1-3 Yu Gongyuan

13 2012-11-21 (1)Protein Synthesis & the Genetic Code 2 1-2 Yu Gongyuan

14 2012-11-26(1)Protein Synthesis & the Genetic Code (2)Regulation of Gene Expression

3 1-3 Yu Gongyuan

14 2012-11-28 (1)Regulation of Gene Expression 2 1-2 Yu Gongyuan

15 2012-12-3(1)Molecular Genetics, Recombinant DNA, & Genomic Technology

3 1-3 Yu Gongyuan

15 2012-12-5(1)Molecular Genetics, Recombinant DNA, & Genomic Technology

2 1-2 Yu Gongyuan

16 2012-12-10 (1)Total Review 2 1-2 Yu Gongyuan总学时 102 理论 72 实验 30 上课时间：周一 1-3；周三 1-2上课地点: 西楼 606

主任签字生物化学与分子生物学系

2012年 7 月 1 日

43

天津医科大学基础医学院实验课教学进度表2012 学年—2013 学年第一学期

专业：留学生 1 班年级：2011

课程名称：生物化学实验周次实验内容教室时间1 Introduction to the Biochemistry

Laboratory 刘欣、于林（主楼 302）

杨宇虹、刘敏（主楼 303）

王琨、康英姿（主楼 305）

李宪奎、李艳芸（主楼 306）

周一 5-7 节2 Separation of Hemoglobin and

Protamine by Gel Filtration Chromatography

3 Separation of Serum Proteins by Cellulose Acetate Electrophoresis

4 Isolation and Identification of Nucleic acid

5 The Effects of Temperature, pH, Activator and Inhibitor on the Enzyme Activity

7 The Isolation and Quantitative Analysis of Vitamin C

8 The Folin-Ciocalteu Assay of Protein Concentration

9 The Isolation and Identification of Hepatic Glycogen

10 Separation of LDH ( Lactate Dehydrogenase) Isozymes by Agarose Gel Electrophoresis

11 Preparation for Test

44

主任签字生物化学与分子生物学系

2012年 7 月 1 日

4．教案书写要求

国际学院关于规范留学生教学教案书写要求为了进一步规范从事留学生教学教师的教案，现将该教案（讲稿）的书写要求建议如

下:

一、教案（讲稿）是教师本人编写制定的处理教材与组织课堂教学的课程教学方案。一

门课程的教案（讲稿）应包括课程、章节及一次授课三个层面的教学内容、教学基本要求、

教学手段、教学方法设计，主要解决教什么、怎么教的问题。教案（讲稿）的教学内容要严

格按教学大纲编写，并根据社会发展及对人才培养的新要求及时增加和补充前沿内容。二、就一次课而言，教案（讲稿）的内容原则上应包括本次课的教学目的和教学要求、

教学内容、教材分析、时间安排、作业布置、教学后记等方面。其中，教学目的和教学要求是

课堂教学活动围绕的中心和力求达到的目标；教学内容是教案（讲稿）的主体，要按引入

45

新课、讲授、总结与巩固三方面详细设计；教材分析则要找出本次课的重、难点及关键，并

确立突出重点、克服难点、抓住关键的方案；时间安排要求教师在有效的时间内，根据教学

内容合理安排好教学时间；作业布置一项，要求教师在课毕进行归纳小结，并适当布置预

习和复习作业，为下一轮的讲课做好铺垫工作；教学后记是教案（讲稿）执行情况的经验

总结，目的在于改进和调整教案（讲稿），为下一轮课讲授设计更加良好的教学方案。三、教案（讲稿）设计的详细与否，可因人而异，一般来说，年轻教师的教案（讲

稿）、第一次开课教师的教案（讲稿）必须详细写，同时应该有相应的电子教案（讲稿）。

国际医学院

二 00八年十二月十九日

5．授课教案天津医科大学46

国际学院留学生教学教案

20 ——20 学年第学期

教师姓名：

课程名称：

授课班级：

47

天津医科大学国际学院制

Teaching Objectives ;

Teaching Requirements

Teaching Content

Teaching Focus;

Difficult Problems and Their Solutions

Time Allotment

48

Assignment

Postscript

Memo

Teaching Plan for International Students, TMUTitle of the Course： Chapter:

Teacher’s Name: Prof. Title:

Class： Grade: Department:

Time：： 00 --- ： 00 Date（D/M/Y）：

Lecture Notes：

49

天津医科大学国际学院教案与讲稿Teaching Plan for International Students, TMU

Title of the Course： Biochemistry Chapter: Chapter 7 Enzymes: Mechanism of Action

Teacher’s Name: Wang Kun Prof. Title: Lecturer

Class： Ⅱ Grade: 20 11 Department: Biochemistry and Molecular Biology

Time： 14 ： 00 --- 15 ： 00 Date（D/M/Y）： 9 / 4 / 20 12



1. Mastering: general properties of enzymes, prosthetic groups, coenzymes, cofactors, the active site of enzyme, isozymes

2. Comprehending: classification of enzymes, the enzyme name3. Understanding: mechanism of enzyme-catalyzed reactions, detection of the catalytic

activity of enzymes

Teaching Content

1. Introduction and Biomedical Importance2.General properties of enzymes

(1) Highly efficiency(2) Highly specificity

Absolute specificity Relative specificity Optical specificity3. Classification of enzymes

(1) Six classes of enzymes (2)The enzyme name has two parts

50

(3) Additional information may follow in parentheses(4) A code number

4. Many enzymes require a coenzymes prosthetic groups(1) Prosthetic groups(2) Coenzymes(3) Cofactors

5. The active site of enzymes(1) What is the active site of enzyme(2) Characteristic of the active site

6. Mechanism of enzyme-catalyzed reactions(1) Enzymes enhance reactant proximity and local concentration (2) Acid-Base catalysis (3) Catalysis by strain(4) Covalent catalysis(5) Substrates induce conformational changes in enzymes

7. Isozymes (1) What are isozymes?(2) Diagnostic value of isozymes

8.The catalytic activity of enzymes (1) What is IU? (2) Detection of the catalytic activity of enzymes

Teaching Focus;

Difficult Problems and their Solutions

The active site of enzyme, characteristic of the active siteMechanism of enzyme-catalyzed reactions

Time Allotment

1.General properties of enzymes (5 minutes)2. Classification of enzymes (5minutes)3. Many enzymes require a coenzymes prosthetic groups(10 minutes)4. The active site of enzymes(10 minutes)5. Mechanism of enzyme-catalyzed reactions(10 minutes)6. Isozymes (10 minutes)

Assignment • What are prosthetic groups, cofactors and coenzymes?• What is the the active site of the enzyme? • What are isozymes?

Postscript

51

Memo

Lecture Notes：Chapter 7 Enzymes: Mechanism of Action

Enzymes are biologic polymers that catalyze the chemical reactions. With the exception of a

few catalytic RNA molecules called ribozymes, the vast majority of enzymes are proteins.

Ⅰ Enzymes are effective and highly specific catalysis Enzymes catalyze the conversion of one or more compounds (substrates) into one or more

different compounds (products). They have a very high catalytic capacity which can enhance the

rates of the corresponding noncatalyzed reaction by factors of at least 106. In addition to being highly efficient, enzymes are also extremely selective catalysts. Enzymes

are specific both for the type of reaction catalyzed and for a single substrate or a small set of

closely related substrates.

ⅡEnzymes are classified by reaction type and mechanism The International Union of Biochemists (IUB) developed a system of enzyme nomenclature. In

the IUB system, each enzyme has a unique name and code number that reflect the type of reaction

catalyzed and the substrates involved. Enzymes are grouped into six classes, each with several

subclasses.1. Oxidoreductases catalyze oxidations and reductions (transfer electrons from donors to

acceptors).2. Transferases catalyze transfer of groups such as methyl or glycosyl groups from a donor

molecule to an acceptor molecule.3. Hydrolases catalyze the hydrolytic cleavage of C-C,C-O,C-N,P-O, and certain other

bonds (catalyze cleavage of bonds by the addition of water, producing two products).4. Lyases catalyze cleavage of C-C,C-O,C-N and other bonds by elimination, leaving

double bonds, and also add groups to double bonds.5. Isomerases catalyze geometric or structural changes within a single molecule

(interconvert isomeric forms by transferring groups within the same molecule).

52

6. Ligases catalyze the joining together of two molecules, coupled to the hydrolysis of ATP.

Ⅲ Prosthetic groups, cofactors & coenzymes play important roles in catalysis While some enzymes are composed only of proteins, many other enzymes contain small

nonprotein molecules that participate directly in substrate binding or catalysis. They are termed

prosthetic groups, cofactors, and coenzymes. 1. Prosthetic groups are tightly integrated into an enzyme’s structure

Prosthetic groups are distinguished by their tight, stable incorporation into a protein’s structure by covalent or noncovalent forces. Examples include flavin

mononucleotide (FMN), flavin dinucleotide (FAD). Metals are the most common prosthetic

groups. The enzymes that contain tightly bound metal ions are termed metalloenzymes.

2. Cofactors associate reversibly with enzymes or substratesCofactors serve functions similar to those of prosthetic groups but bind in a transient,

dissociable manner either to the enzyme or to a substrate such as ATP. The most common

cofactors are also metal ions. Enzymes that require a metal ion cofactor are termed metal-activated enzymes.3. Coenzymes serve as substrate shuttles

Coenzymes serve as recyclable shuttles—or group transfer reagents—that transport many substrates from their point of generation to their point of utilization.

Many coenzymes are derived from B vitamins. The water-soluble B vitamins supply

important components of numerous coenzymes. For example, nicotinamide and riboflavin are

components of the redox coenzymes NAD+ and NADP+, and FMN and FAD, respectively. They

participate in the catalytic process but are not metabolites themselves.

Ⅳ Catalysis occurs at the active site The overall structure of an enzyme is designed to fulfill functional requirements such as cellular

localization and binding of substrates and regulatory molecules. The three-dimensional catalytic center that binds and activates the substrates is the active site (binding site + catalytic site) ,

which occupies a relatively small percent of the overall molecule.Characteristic of the active site:• Many aminoacyl residues contribute to the active site• Catalytic residues are highly conserved• The residues that comprise the the active site may be distant one from another in the primary

structure but spatially close in the tertiary structure• Active site often are located in clefts

53

• Active site of multimeric enzymes may reside at sununit interfacesⅤ Mechanism of enzyme-catalyzed reactions(1) Enzymes enhance reactant proximity and local concentration ● When an enzyme binds substrate molecules in its active site, it creates a region of high local

substrate concentration and enhances the rate of reaction. ● This environment also orients the substrate molecules spatially in a position ideal for them to

interact.(2) Acid-Base catalysis The ionizably functional groups aminoacyl side chains and of prosthetic groups contribute to

catalysis by acting as acids or bases (3) Catalysis by strain In the lytic reactions enzymes bind substrates in a conformation slightly unfavorable for bond(4) Covalent catalysis Enzyme and substrate form a transient modified enzyme that is more favorable to reaction(5) Substrates induce conformational changes in enzymes Induced fit model Induced fit model, addressed by Daniel Koshland, states that when substrates approach and bind

to an enzyme, they induce a mutual conformational changes in one another that facilitate substrate

recognition and catalysis, just like a change analogous to placing a hand into a glove. (Fig 7-5)

Ⅵ IsozymesIsozymes are separable forms of a given enzyme present in different tissues, different

cell types or subcellular compartments of the same organism. Isozymes may catalyze the same reaction, but their physical, chemical and immunological properties exhibit many significant differences. Isozymes have a multimetric quaternary structure composed of more than

one type of subunit. Each subunit is produced by its own gene expressed at its own rate in a given

tissue.Lactate dehydrogenase catalyzes the reaction of transfer of two electrons and one H+ from

lactate to NAD+, while lactate turns to pyruvate.

COO- COO-

∣ LDH ∣ HO-C-H + NAD+ —→ C=O + NADH + H+

∣ ∣CH3 CH3

(Lactate) (Pyruvate)

54

LDH isozymes differ at the level of quaternary structure. LDH molecule consists of four

subunits of two types, H and M. Only the tetrameric molecule possesses catalytic activity. These

subunits might be combined in the following five ways. LDH isozyme subunits

I1 HHHH I2 HHHM I3 HHMM I4 HMMM I5 MMMM

Distinct genes whose expression is differentially regulated in various tissues encode the H

and M subunits. Since heart expresses the H subunit almost exclusively, isozyme I1

predominates in this tissue. By contrast, isozyme LDH5 predominates in liver. LDH isozymes

and that their relative proportions changed significantly in pathologic conditions. This fact

has been put to use in the clinical diagnosis of several diseases. The involvement of any organ

in a disease frequently results in death and breakdown of cells from that organ and release of

their components into the blood serum. In normal states LDH exist mainly in cells. But in

pathological conditions, certain types of LDH isozymes will be released into serum. For

example, myocardial infarction is likely to result in the elevated amounts of LDH1 in serum.

On the other hand, liver disease (e.g., infectious hepatitis) is likely to be accompanied by

elevated levels of LDH5 in the serum, since these isozymes predominate in liver cells. Hence,

the determination of the level and isozyme distribution of serum LDH is useful in the

diagnosis of a variety of other hematological, cardiac, and hepatic diseases. (Fig 7-11)

55


Title of the Course： Biochemistry Chapter: Chapter 8 Enzymes: Kinetics

Teacher’s Name: Wang Kun Prof. Title: Lecturer

Class： Ⅱ Grade: 20 11 Department: Biochemistry and Molecular Biology

Time： 15 ： 00 --- 17 ： 00 Date（D/M/Y）： 9 / 4 / 20 12



1. Mastering: multiple factors affect the rates of enzyme-catalyzed reactions

2. Comprehending: the kinetics of enzymatic catalysis

3. Understanding: general theory of chemical reaction

Teaching Content

1. Introduction and Biomedical importance

2. General theory of chemical reaction

(2) G determine the direction and equilibrium state of chemical reaction△(2) Numerous factors affect the reaction rate

3. The kinetics of enzymatic catalysis

(1) Enzymes lower the energy barrier

(2) Enzymes provide transition states

(3) The active site

The active site of enzymes, a rigid catalytic site model, induced fit model, characteristic of

the active site

4. Multiple factors affect the rates of enzyme-catalyzed reactions

(1) Temperature

(2) pH

(3) Enzyme concentration

(4) Substrate concentration

Effect of substrate concentration, the michaelis-menten equation, the significance of

Km, the determination of Km

(5) Inhibitors

Teaching Focus;

Difficult

Multiple factors affect the rates of enzyme-catalyzed reactionsEffect of substrate concentration, the michaelis-menten equation, the significance of Km,

56

Problems and their Solutions

Time Allotment

1. Introduction and Biomedical importance (10min)

2. General theory of chemical reaction (20min)

3. The kinetics of enzymatic catalysis (30min)

4. Multiple factors affect the rates of enzyme-catalyzed reactions (40min)

Assignment 1. What is the the active site of enzyme? Please explicate the property of enzymes catalysis by induced fit model?2. What is the michaelis-menten equation? Please discuss the significance of Km.3. What is the competitive inhibitor? What are the characteristic of it?

Postscript

Memo

Lecture Notes：Chapter 8 Enzymes: Kinetics

Enzyme kinetics is the field of biochemistry concerned with the quantitative measurement

of the rates of enzyme catalyzed reactions and the systematic study of factors that affect these

rates.

Ⅰ Chemical reactions are described using balanced equationsChemical reactions are described using balanced equations. For example, A+B P+Q the

double arrows indicate reversibility, an intrinsic property of all chemical reactions. That is, if A

and B can form P and Q, then P and Q can also form A and B. At equilibrium, the rate of

conversion of substrates to products therefore equals the rate at which products are converted to

substrates. So at equilibrium the overall concentrations of reactants and products remain constant.

Keq is termed the equilibrium constant. The following important properties of a system at

equilibrium must be kept in mind:(1) Enzymes act to increase the rate of reaction, but they do not change the equilibrium of

the reaction. (Enzymes have no effect on equilibrium constant.)

57

(2) At equilibrium, the reaction rates of the forward and back reactions are equal.(3) Equilibrium is a dynamic state. Although there is no net change in the concentration of

substrates or products, individual substrate and product molecules are continually being

interconverted.

ⅡChanges in free energy determine the direction and equilibrium state of chemical reactions The free energy change ΔG equals the sum of the free energies of formation of the products

minus the sum of the free energies of formation of the substrates. If the free energy of the

products is lower than that of the substrates, the sign of ΔG will be negative, indicating that the

reaction is favored in the direction from left to right. Such reactions are referred to as

spontaneous. Enzymes can only catalyze the reaction that is energically favorable (spontaneous reaction). ΔG can provide information only about the direction and equilibrium

state of the reaction. It is independent of the mechanism of the reaction and therefore provides no

information concerning rates of reactions.

Ⅲ The mechanism of enzymes catalyzing reactions1. Reactions proceed via transition states

The concept of the transition state is fundamental to understanding the chemical and

thermodynamic basis of catalysis. As the induced fit model said, when substrate bind to the

active site of enzyme, it causes mutual conformational changes in both substrate and enzyme,

and lead to form an immediate state called the transition state.

The enzyme- catalyzed reaction can be well explained by this model

E+S ES E+P

This model involves a substrate, which binds reversibly with enzyme to form an ES

complex. The ES complex can have either of two outcomes. It can continue the reaction to

form product or it can break down to the enzyme and substrate without reacting. ES complex

is the transition state in the reaction.

The formation of transition state intermediates requires surmounting of energy barriers. The activation energy--the energy required to surmount determines the rates of reactions.

2. Enzymes lower the activation energy barrier for a reactionAll enzymes accelerate reaction rates by lowering the activation energy barrier, for

formation of the transition states. And this increases the probability of product formation.

3. Three important characteristics of enzymes(1) They are not changed by the reaction they catalyze, although they may be

58

temporarily changed during the reaction.(2) They do not change the equilibrium of the reaction, so they can not force a reaction that is not energetically favorable (nonspontaneous).(3) They increase reaction rates by decreasing the activation energy.

Ⅳ Multiple factors affect the rates of enzyme-catalyzed reactions

Assays of enzyme-catalyzed reactions typically measure the initial velocity (vi). Because under

initial rate conditions, only little product accumulate, hence the rate of the reverse reaction is

negligible. Then with the reaction went on, the rate of reaction begins to slow down as substrate

becomes depleted and product builds up.

1. TemperatureRaising the temperature will increase the motion of reacting molecules, thus increases the

rate of reactions. Within a limited range of temperature, the velocity of enzyme catalyzed

reactions increases as temperature rises. On the other hand, since enzymes are proteins, they

will gradually denature or lose their activity completely when the temperature increases

continuously. So in that case, the velocity of an enzyme-catalyzed reaction will decrease.

These two contrary effects exist in same reaction. For each enzyme, there is a temperature

optimum at which the enzyme activity is at its maximum.

2. Hydrogen ion concentration (pH) pH affects enzyme activity by changing the charges on ionizable groups of enzyme, and also

affects the degree of dissociation the substrate. Each enzyme has an optimum pH at which the

rate of the reaction is at its maximum. Most intracellular enzymes exhibit optimal activity at

pH values between 5 and 9.

3. Substrate concentration(1) The shape of the curve that relates reaction velocity to substrate concentration is

hyperbolic (Fig 8-3). For a typical enzyme, as substrate concentration is increased, vi

increases until it reaches a maximum value Vmax. Further increases in substrate

concentration do not further increase vi. At any given instant, only substrate molecules that

are combined with the enzyme as an ES complex can be transformed into product. As shown

in the figure, at points A or B, only a fraction of the enzyme may be present as an ES

complex. So increasing or decreasing [S] will increase or decrease the number of ES

complexes with a corresponding change in vi. At point C, essentially all the enzyme is present

as the ES complex (all enzymes are saturated with substrate). Since no free enzyme remains

available for forming ES, further increases in [S] cannot increase the rate of the reaction

59

anymore. (Fig 8-4)

(2) The Michaelis-Menten equation & the Michaelis constant

The Michaelis-Menten equation illustrates the relationship between initial reaciton velocity

vi and substrate concentration [S]

Vmax is the maximal velocity achieved when enzyme becomes saturated with substrate.

The Michaelis constant Km is the substrate concentration at which vi is half the maximal

velocity (Vmax/2). Km has the following significance:

① When [S] is much less than Km, the term Km+[S] is essentially equal to Km. Replacing

Km+[S] with Km we can get

Since Vmax and Km are both constants, their ratio is a constant. In other words, when [S] is

considerably below Km, the initial reaction velocity is directly proportionate to [S].② When [S] is much greater than Km, the term Km+[S] is essentially equal to [S]. Replacing

Km+[S] with [S] we can get Vi = Vmax. Thus when [S] greatly exceeds Km, the reaction

velocity is maximal and unaffected by further increases in substrate concentration.③ When [S]=Km, Vi=1/2Vmax④ Km measures the affinity of the enzyme for the substrate. A low Km corresponds to a high

affinity and vice versa. (3) A liner form of the Michaelis-Menten equation is used to determine Km & VmaxA liner form of the Michaelis-Menten equation simplifies determination of Km & Vmax.

1/Vi= (Km/Vmax)1/[S] + 1/Vmax Such a plot is called a double reciprocal or Lineweaver-Burk plot. It is the equation for a straight line, y=ax+b, where y=1/Vi, x=1/[S], slope is

Km/Vmax, y intercept is 1/Vmax. Setting the y term equal to zero and the x intercept is –1/Km. So Km is most easily calculated form the x intercept, and Vmax from y intercept.

V=Vmax [S]

Km + [S]

vi = Km

Vmax[S]

60

4. InhibitorsKinetically, we distinguish two classes of inhibitors based upon whether raising the substrate

concentration does or does not overcome the inhibition.(1) Competitive inhibitors.

The structures of most classic competitive inhibitors tend to resemble the structures of a

substrate and thus are termed substrate analogs. It competes with the substrate for binding to the

active site. So the number of free enzyme molecules available to bind substrates is decreased. A competitive inhibitor and substrate exert reciprocal effects on the concentration of the EI and

ES complexes. Since binding substrate removes free enzyme available to combine with inhibitor,

increasing the [S] decreases the concentration of the EI complex and raises the reaction velocity.

So the effect of competitive inhibitors can be overcome by raising the concentration of the

substrate. If the substrate concentration is high enough, the enzyme can become saturated with the

substrate even in the presence of inhibitor and will be at Vmax, but it can raise Km because it

decreases the affinity of enzyme for the substrate. So a competitive inhibitor has no effect on Vmax but raises Km.

(2) Noncompetitive inhibitors Noncompetitive inhibitors bind enzymes at sites distinct from the substrate-binding site and

61

generally bear little or no structural resemblance to the substrate. Therefore, binding of the

inhibitior does not affect binding of substrate. So noncompetitive inhibitors don’t affect Km, however, while the enzyme-inhibitor complex can

still bind substrate, its efficiency at transforming substrate to product, reflected by Vmax, is

decreased. Excess substrate can’t outcompete a noncompetitive inhibitor.

62

天津医科大学国际学院教案与讲稿

Teaching Plan for International Students, TMUTitle of the Course： Biochemistry Chapter: 9

Teacher’s Name: Geng Xin Prof. Title: Associate Professor

Class： 2 Grade: 2011 Department: Biochemistry and Molecular Biology

Time： 10 ： 10 --- 12 ： 00 Date（D/M/Y）： 5/9/2012



Chapter 9: Enzymes: Regulation of Activities1. Mastering: active regulation of metabolism2. Comprehending: passive regulation of metabolism3. Understanding: biomedical importance, homeostasis4. Focus and difficulty: allosteric regulation, covalent modification

Teaching Content

1.Introduction and Biomedical importance (1) what is the homeostasis

(2) regulation of metabolism achieves homeostasis2.Passive regulation of metabolism 3.Characteristics of metabolism in the cells (1) unidirection (2) compartmentalization (3) rate-limiting enzyme4.Active regulation of metabolism

(1) regulation of enzyme quantitycontrol of enzyme synthesis by inducers and repression

control of enzyme degradation(2) regulation of the intrinsic catalytic activities of enzymes

① allosteric regulation ATCase is a model allosteric enzyme

allosteric and catalytic sites are spatially distinctallosteric effects may be on Km or on Vmaxfeedback regulation is not synonymous with feedback inhibitionmany hormones act through allosteric second messengers② covalent modifications

63

Regulation covalent modification can be reversible or irreversibleProteases may be secreted as catalytically inactive proenzymesProenzymes facilitate rapid mobilization of an activity in response to physiologic

demandActivation of prochymotrypsin requires selective proteolysisReversible covalent modification regulates key mammalian enzymesProtein phosphorylation is extremely versatile

Covalent modification regulates metabolite flow

Teaching Focus;


(1) rate-limiting enzyme(2) control of enzyme synthesis by inducers and repression(3) allosteric regulation

allosteric effects may be on Km or on Vmaxmany hormones act through allosteric second messengers

(4)covalent modifications Regulation covalent modification can be reversible or irreversible

Time Allotment

1.Introduction and Biomedical importance (10min)2.Passive regulation of metabolism (10min)3.Characteristics of metabolism in the cells (20min)4.Active regulation of metabolism

(1) regulation of enzyme quantity (10min) (2) regulation of the intrinsic catalytic activities of enzymes

① allosteric regulation (20min)② covalent modifications (30min)

Assignment

1. describe the active regulation of metabolism2. compare the passive regulation and the active regulation of metabolism3. Explain the terms: homeostasis; allosteric regulation, covalent modification; rate-limiting enzyme

PostscriptBefore the beginning of this class, review the chapter 7 and 8Before the end of this class, summarize the chapter 9

Memo

Harper’s Illustrated Biochemistry 26th editionLehninger BiochemistryScriver CR, et al. The metabolic and molecular bases of inherited diseases, 8 th ed. McGraw-Hill, 2000

Lecture Notes：Chapter 9 Enzymes: Regulation of Activities

64

1. Introduction and biomedical importance(1) What is the homeostasis

Walter Cannon subsequently coined the term “homeostasis” to describe the ability of animals to

maintain a constant intracellular environment despite changes in their external environment. We

now know that organisms respond to changes in their external and internal environment by

balanced, coordinated changes in the rates of specific metabolic reactions. Many human diseases,

including cancer, diabetes, cystic fibrosis, and Alzheimer’s disease, are characterized by

regulatory dysfunctions triggered by pathogenic agents or genetic mutations.(2) Regulation of metabolism achieves homeostasis

Knowledge of factors that control the rates of enzyme-catalyzed reactions thus is essential to an

understanding of the molecular basis of disease. This chapter outlines the patterns by which

metabolic processes are controlled and provides illustrative examples. Subsequent chapters

provide additional examples.

2. Passive regulation of metabolismEnzymes that operate at their maximal rate cannot respond to an increase in substrate

concentration, and can respond only to a precipitous decrease in substrate concentration.

For most enzymes, therefore, the

average intracellular concentration of their substrate tends to be close to the Km value, so that

changes in substrate concentration generate corresponding changes in metabolite flux (Figure 9–

1). Responses to changes in substrate level represent an important but passive means for

coordinating metabolite flow and maintaining homeostasis in quiescent cells. However, they offer

limited scope for responding to changes in environmental variables.

3. Characteristics of metabolism in the cells(1) Unidirection

65

Despite the existence of short-term oscillations in metabolite concentrations and enzyme levels,

living cells exist in a dynamic steady state in which the mean concentrations of metabolic

intermediates remain relatively constant over time (Figure 9–2).

(2) Compartmentalization

Segregation of certain metabolic pathways within specialized cell types can provide further

physical compartmentation. Alternatively, possession of one or more unique intermediates can

permit apparently opposing pathways to coexist even in the absence of physical barriers.

(3) Rate-limiting enzymeActive control of homeostasis is achieved by regulation of only a small number of enzymesRate-limiting enzyme is one that the reaction it catalyzed is slow relative to all others in the

pathway

66

4. Active regulation of metabolism (1) Regulation of enzyme quantity

Control of enzyme synthesis by inducers and repressionControl of enzyme degradation

The absolute quantity of an enzyme reflects the net balance between enzyme synthesis and

enzyme degradation, where ks and kdeg represent the rate constants for

the overall processes of synthesis and degradation, respectively.

Changes in both the ks and kdeg of specific enzymes occur in human

subjects.

(2) Regulation of the intrinsic catalytic activities of enzymes① Allosteric regulationFeedback inhibition refers to inhibition of an enzyme in a biosynthetic pathway by an end

product of that pathway. For example, for the biosynthesis of D from A catalyzed by enzymes

Enz1 through Enz3, high concentrations of D inhibit conversion of A to B. Inhibition results not

from the “backing up” of intermediates but from the ability of D to bind to and inhibit Enz 1 .

Typically, D binds at an allosteric site spatially distinct from the catalytic site of the target

enzyme. Feedback inhibitors thus are allosteric effectors and typically bear little or no structural

similarity to the substrates of the enzymes they inhibit. In this example, the feedback inhibitor D

acts as a negative allosteric effector of Enz 1.

67

ATCase is a model allosteric enzyme Aspartate transcarbamoylase (ATCase), the catalyst for the first reaction unique to pyrimidine

biosynthesis (Figure 34–7), is feedback-inhibited by cytidine triphosphate (CTP). Following

treatment with mercurials, ATCase loses its sensitivity to inhibition by CTP but retains its full

activity for synthesis of carbamoyl aspartate. This suggests that CTP is bound at a different

(allosteric) site from either substrate. ATCase consists of multiple catalytic and regulatory

subunits. Each catalytic subunit contains four aspartate (substrate) sites and each regulatory

subunit at least two CTP (regulatory) sites.

Allosteric and catalytic sites are spatially distinctAllosteric effects may be on Km or on VmaxFeedback regulation is not synonymous with feedback inhibitionMany hormones act through allosteric second messengersThe primary or “first” messenger is the hormone molecule or nerve impulse. Second messengers

include 3',5′-cAMP, synthesized from ATP by the enzyme adenylyl cyclase in response to the

hormone epinephrine, and Ca2+, which is stored inside the endoplasmic reticulum of most cells.

② Covalent modifications Regulation covalent modification can be reversible or irreversibleProteases may be secreted as catalytically inactive proenzymesProenzymes facilitate rapid mobilization of an activity in response to physiologic demandActivation of prochymotrypsin requires selective proteolysis

68

Reversible covalent modification regulates key mammalian enzymes

Protein phosphorylation is extremely versatile

Covalent modification regulates metabolite flow

69

Regulation of enzyme activity by phosphorylationdephosphorylation has analogies to regulation

by feedback inhibition. Both provide for short-term, readily reversible regulation of metabolite

flow in response to specific physiologic signals. Both act without altering gene expression. Both

act on early enzymes of a protracted, often biosynthetic metabolic sequence, and both act at

allosteric rather than catalytic sites. Feedback inhibition, however, involves a single protein and

lacks hormonal and neural features. By contrast, regulation of mammalian enzymes by

phosphorylationdephosphorylation involves several proteins and ATP and is under direct neural

and hormonal control.

Summary:1. All biochemical pathways are regulated to maintain the ordered state of living cells.

2. Regulation is accomplished by passive mechanism complemented by active mechanism

3. Active mechanism contains regulation of enzyme quantity and regulation of the intrinsic

catalytic activities of enzymes

4. Compartmentalization is another regulation mode.

Questions:1. What are the characteristics of metabolism in the cells?

2. What kinds of covalent modifications are classified? Please discuss the phosphorylation and

dephosphorylation.

3. What is allosteric regulation? What are the properties of it?

4. To achieve homeostasis, the rates of metabolism must respond to physiologic need. How is this

achieved?


70

Title of the Course： Biochemistry Chapter: Chapter 22 Oxidation of fatty acids: Ketogenesis

Teacher’s Name: Liu Xin Prof. Title: Lecturer

Class： Grade: Department: Biochemistry and Molecular Biology

Time： 14 ： 00 --- 14 ： 50 Date（D/M/Y）： 9/6/2012



1. Mastering: transportion and activation of fatty acids; β-oxidation is a cyclic reaction2. Comprehending: a large quantity of ATP is produced by oxidation of fatty acids; the concept and generation of ketone bodies3. Understanding: the regulation of ketogenesis

Teaching Content

1. Biomedical Importance2. Oxidation of fatty acids occurs in mitochondria(1) Fatty acids are transported in the blood as free fatty acids (FFA)(2) Fatty acids are activated before being catabolized(3) Long-chain fatty acids penetrate the inner mitochondrial membrane as carnitine derivatives3. β-oxidation of fatty acids involves successive cleavage with release of acetyl-CoA(1) The cyclic reaction sequence generates FADH2 and NADH(2) Oxidation of a fatty acid with an odd number of carbon atoms yields acetyl-CoA plus a molecule of propionyl-CoA(3) Oxidation of fatty acids produces a large quantity of ATP(4) Peroxisomes oxidize very long chain fatty acids4. Oxidation of unsaturated fatty acids occurs by a modified β-oxidation pathway5. Ketogenesis occurs when there is a high rate of fatty acid oxidation in the liver(1) 3-Hydroxy-3-Methylglutaryl-CoA (HMG-CoA) is an intermediate in the pathway of ketogenesis(2) Ketone bodies serve as a fuel for extrahepatic tissues6. Ketogenesis is regulated at three crucial steps7. Clinical aspects(1) Impaired oxidation of fatty acids gives rise to diseases often associated with hypoglycemia(2) Ketoacidosis results from prolonged ketosis

Teaching Focus;


1. Focus: (1) Mobilization of stored fat(2) The ATP produced by β-oxidation of fatty acids.(3) Comparison of the synthesis and degradation of even number fatty acids.(4) Ketogenesis and the oxidation of ketone bodies in extrahepatic tissues.(5) Diseases related to β-oxidation of fatty acids and ketone bodies.2. Difficulty: high rate of fatty acid oxidation promotes ketogenesis

Time (1) β-oxidation of fatty acids (15min)

71

Allotment (2) Ketogenesis and the oxidation of ketone bodies in extrahepatic tissues (15min)(3) Diseases related to β-oxidation and ketone bodies (15min)

Assignment (1) Compare the synthesis and degradation of even number fatty acids.(2) Explain the mechanism of diabetes related to ketone bodies.

Postscript

Memo

Lecture Notes：Chapter 22 Oxidation of fatty acids: Ketogenesis

1. Mobilization of stored fat

2. Activation of hormone-sensitive lipase (HSL)One of several hormones (primarily epinephrine) → Receptor→ activate adenylate cyclase→

cAMP →protein kinase A→HSL→ FA + glycerol.Fate of glycerol:

72

Adipose tissue (lack of glycerol kinase) Liver: phosphorylated to glycerol phosphate, DHAP by glycerol phosphate dehydrogenaseFate of fatty acids: transported by albumin in plasma, then enter the cell mitochondria to be

oxidized for energy.3. Conversion of fatty acid to fatty acyl-CoAFatty acid + ATP + CoA → acyl-CoA + PPi +AMP4. Long-chain fatty acids (LCFA) (more than 12 carbons) penetrate the inner mitochondrial

membrane as carnitine derivative.

5. Enzymes: 1) Carnitine palmitoyl transferase -I 2) Carnitine-acylcarnitine translocase3) Carnitine palmitoyltransferase-II6. β-oxidation:β-oxidation of fatty acids involves successive cleavage with release of acetyl-coA ---between αand β carbon atoms. Enzymes: fatty acid oxidase Location: in the mitochondrial matrix or inner membrane Results: β-oxidation coupling with phosphorylation of ADP to ATP.

73

Procedure:(1) Penetration(2) Removal of two hydrogen atoms: FAD(3) Add water to saturate double bond (4) Further dehydrogenation: enzyme: NADH(5) Split of 3-keto acyl-CoA:

Products: acetyl-CoA and acyl-CoA (two carbons shorter than the original acyl-CoA) Oxidation of fatty acids produces a large quantity of ATP

Es. Palmitate (16C) 1) 1 cycle: FADH2, NADH→ 5 ATP2) 7 cycle: 5 X 7=35 ATP3) 8 acetyl-CoA: 8 x 12 = 96 ATP4) Substrate initial activation: -2 ATPTOTAL: 129 ATPTotal energy: 129 x 51.6= 6656 kJ (68%)Free energy of palmitate: 9791 kJ

7. Oxidation of fatty acids with an odd number of carbons Proceeds: two carbons at a time (producing acetyl CoA) until the last three carbons (propionyl

CoA). This compound is converted to methylmalonyl CoA, which is then converted to succinyl

CoA.8. Oxidation of unsaturated fatty acids occurs by a modified β-oxidation pathway:(1) Δ3-cis-acyl-CoA compound → Δ2-trans-acyl compound → to β-oxidation→ hydration→

oxidation.

74

(2)Δ4-cis-acyl-CoA →enter point: Δ2-trans-Δ4-cis-acyl-CoA converted by acyl-CoA

dehydrogenase → then converted to Δ3-trans compound → enzyme attacts Δ3-trans double bond

to produceΔ2-trans-acyl-CoA compound, an intermediate in β-oxidation.9. Comparison of the synthesis and degradation of even number fatty acids.

Genetic Defects in Fatty Acyl–CoA Dehydrogenases Cause Serious Disease Among northern Europeans, the frequency of carriers (individuals with this recessive

mutation on one of the two homologous chromosomes) is about 1 in 40, and about 1 individual in

10,000 has the disease—that is, has two copies of the mutant MCAD allele and is unable to

oxidize fatty acids of 6 to 12 carbons. The disease is characterized by recurring episodes of a syndrome that includes fat

accumulation in the liver, high blood levels of octanoic acid, low blood glucose (hypoglycemia),

sleepiness, vomiting, and coma.10. Ketogenesis occurs when there is a high rate of fatty acid oxidation in the liver Ketone bodies: acetoacetate, 3-hydroxybutyrate, acetone (volatile)Location: liver, mitochondria11. Ketone Bodies, Formed in the Liver, Are Exported to Other Organs as Fuel.

75

12. Overproduction of ketone bodies in uncontrolled diabetes or severely reduced calorie intake

can lead to acidosis or ketosis.Summary1. In the first stage of -oxidation, four reactions remove each acetyl-CoA unit from the carboxyl end

of a saturated fatty acyl-CoA. The shortened fatty acyl-CoA then reenters the sequence.2. In the second stage of fatty acid oxidation, the acetyl-CoA is oxidized to CO2 in the citric acid cycle.

A large fraction of the theoretical yield of free energy from fatty acid oxidation is recovered as ATP by

oxidative phosphorylation, the final stage of the oxidative pathway.3. Malonyl-CoA, an early intermediate of fatty acid synthesis, inhibits carnitine acyltransferase I,

preventing fatty acid entry into mitochondria. This blocks fatty acid breakdown while synthesis is

occurring.4. Genetic defects in the medium-chain acyl-CoA dehydrogenase result in serious human disease, as do

mutations in other components of the -oxidation system.5. The ketone bodies— acetone, acetoacetate, and D--hydroxybutyrate —are formed in the liver. The

latter two compounds serve as fuel molecules in extrahepatic tissues, through oxidation to acetyl-CoA

and entry into the citric acid cycle. 6. Overproduction of ketone bodies in uncontrolled diabetes or severely reduced calorie intake can lead

to acidosis or ketosis.

76


Title of the Course：Biochemistry Chapter: The Citric Acid Cycle Teacher’s Name: Li Haidong Prof. Title: Prof essor

[email protected] 022-83336833(Office Telephone) Department: Biochemistry and Molecular Biology

Class： Ⅱ Grade: 2011

Time：8:00--- 1 0: 0 0 Date（D/M/Y）： 9 /1 2 / 20 1 2



(1) Describe the reactions of the citric acid cycle and the reactions that lead to

the production of reducing equivalents that are oxidized in the mitochondrial

electron transport chain to yield ATP.

(2) Explain why the citric acid cycle is the central hub in the metabolism.

(3) Explain how hyperammonemia can lead to loss of consciousness.

Teaching Content

(1) Introduction: Objectives and Biomedical importance.

(2) How does the citric acid cycle proceed, provide substrates for the respiratory

chain and liberate reducing equivalents and CO2?

(3) How ten ATPs are formed per turn of the citric acid cycle?

(4) Vitamins play key roles in the citric acid cycle.

(5) The citric acid cycle plays a pivotal role in metabolism and is involved in

gluconeogenesis, transamination, deamination and fatty acid synthesis

(6) Regulation of the citric acid cycle depends primarily on a supply of oxidized

cofactors.

Teaching Focus;


(1) the key reactions in the citric acid cycle

(2) ATP formation

(3) Regulation of the citric acid cycle

Time Allotment

(1) Introduction: 10min

(2) Reactions in the citric acid cycle: 40min

(3) ATP generation: 10min

(4) The roles of vitamins: 10min

77

mailto:[email protected]

(5) The central role of the citric acid cycle in metabolism: 20min

(6) Regulation of the citric acid cycle: 10min

Assignment

(1) Describe the key reactions in the citric acid cycle

(2) Explain how ten ATPs are formed

(3) Explain why the citric acid cycle plays a pivotal role in metabolism

PostscriptRobert K. Murray, Daryl K. Granner, Peter A. Mayes and Victor W. Rodwell

(2003) Harper’s Illustrated Biochemistry, 26th Edition, McGraw-Hill Companies.

Memo Krebs HA and Weitzman PDJ (1987) Krebs' citric acid cycle: half a century and

still turning, London: Biochemical Society, ISBN 0-904498-22-0.

Lecture Notes：Chapter 16 The Citric Acid Cycle: The Catabolism of Acetyl-CoA

1. Biomedical Importance• Final common pathway for the oxidation of carbohydrate, lipid, and protein• Central role in gluconeogenesis, lipogenesis, and interconversion of amino acids• Hepatitis, cirrhosis, genetic defects of citric acid cycle enzymes• Hyperammonemia

2. Provides substrates for the respiratory chain• Catalytic role of oxaloacetate

78

3. Reactions in the citric acid cycle

80

4. ATP generation

5. Vitamins play key roles in the citric acid cycle• The role of B vitamins (riboflavin, niacin, thiamin, pantothenic acid)

6. The central role of the citric acid cycle in metabolism• Oxidation of two carbon units• Transamination and deamination (aminotransferase)

81

• Gluconeogenesis (phosphoenolpyruvate carboxykinase)• Fatty acid synthesis (ATP-citrate lyase)

7. Regulation of the citric acid cycle• Respiratory control• Depends primarily on a supply of oxidized cofactors• Pyruvate dehydrogenase, citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate

dehydrogenase• [ATP]/[ADP] and [NADH]/[NAD+] ratios

8. Hyperammonemia• Glutamate dehydrogenase (NH3, NH4

+, [NADH]/[NAD+])• Glutamine synthetase

82

6．备课记录Teaching Preparation

Dept. of Biochemistry and Molecular Biology

Semester: 11-12 ( 1 ) Date:

Topic:

Main Speaker: Lecture / Laboratory

Teachers:

83

7．教学计划变更执行审批表天津医科大学国际医学院教学计划变更执行审批表

班级课程名称教师

教学计划变更方案

及变更原因

审批意见

签字

年月日

84

审批编号

8．调课申请表天津医科大学国际医学院教师调整课程申请表

班级课程名称教师

排定上课时间排定上课地点调整后上课时间调整后上课地点

调课理由教学主管领导

意见

85

9．课堂和实验室纪律规范

天津医科大学国际医学院学生课堂和实验室纪律规范（中英文）课堂是教师向学生传播知识和技能的地方，是学生学习吸收将来神圣职业所需本领的

圣堂。为了使学生能够在一个整洁、和谐、安静、有序的环境充分学到所教的知识和技能，颁

布如下规定，希望学生严格遵守。1、学生应勤奋学习，自觉遵守课堂和实验室规定和学习纪律。按时上课，认真听

讲。2、因病、因特殊原因请假者，应到留学生办公室请假。病假必须有医院证明，并

及时报告老师。凡未请假而缺课者，均按旷课论。3、学生应按照课程安排准时来教室上课，不得旷课和无故迟到；迟到十分钟按

旷课论。4、上课期间不得随便出入教室，不准接打手机、吸烟、吃东西、喝饮料。上课时不

86

准出教室接打手机。5、上课期间不得与他人交谈、睡觉、听音乐，不得影响别人听课。6、上课期间不得请假；中途离开者，按照旷课论处。7、严格遵守实验室的规定和纪律。实验课、临床见习、实习必须穿白大衣；遵照老

师的要求积极动手完成实验、见习和实习课程。实验结束后，主动帮助老师收

拾好实验用具、打扫卫生。8、损坏实验设备者要照价赔偿。

本规定自公布之日起执行。

天津医科大学国际学院

2008-12-23

Discipline Standard in Classroom and Laboratory

Classroom is a sacred place where the teachers spread knowledge and skills; it is also where the students study and absorb all the professional knowledge for their future career. In order to give students a tidy, harmonious, quiet and orderly environment so that they can sufficiently get the knowledge, we hereby issue the following rules, and we hope every student will observe them

87

strictly.

1. You should study hard, and willingly observe the rules and disciplines for classrooms and laboratories. You should attend the class on time, and listen to the teachers carefully.

2. If you have to take a day off and be absent from class, you need to come to the foreign student office and ask for it. If you are sick, you need to present the medical certificate from the doctor, and report to your teacher in time. If you don’t do so, it will be regarded as truancy.

3. You should come to your classroom on time according to your timetable, you can’t play truant or be late for class. If you are more than 10 minutes late for class, it is regarded as truancy.

4. During class time, you can’t go in and out of the classroom. You can’t make or receive phone calls from your mobile phone. Smoking is strictly forbidden, and no food or drinks are allowed inside the classroom. Besides, you can’t go outside the classroom to make or receive phone calls.

5. You can’t talk to other students during class. No sleeping or listening to music. You can’t interfere others listening to the lecture.

6. You can’t ask for a leave during class time; if you leave the classroom half way through the class, this will be regarded as truancy.

7. You should observe the rules and disciplines in the laboratory. You have to wear lab-coat during experiment, clinical observation and internship. You should actively do and finish what you are required by your teacher in a hand-on approach. Upon finishing the experiment, you should voluntarily help the teacher clean the lab and put the things back neatly.

8. If you break anything in the lab, you have to pay for it.

These rules and disciplines take effect from the date when it is issued.

International Medical School, TMU

2008-12-25

10．学生实验报告Gel Filtration Chromatography

[Principle]

88

[Method]

1. Packing the columnPut a whose size is the same as or a little smaller than the internal diameter of the tube

at the end of the column. Then close . Add distilled water until cm above the bottom

and use glass rod to press the nylon membrane gently so as to eliminate the air bubbles. The

column is then filled with and clamped vertically. The suspension is allowed to settle down

under gravity and excess distilled water run off. When the gel particles deposit about , open

the stopcock, wait until .

2. Adding the sampleAdd drops of sample onto the center of the filter paper. Pay attention not to drip the

sample to the internal wall of the tube. Open to let the filter paper emerges again. Add

2-3 drops of distilled water and wait until the filter paper emerges again.

3. ElutionElute with and eluting rate is .

4. Regenerating the columnAfter the last colored band is eluted out of the column, add , which equal to -fold

bed volume of the column. Repeat several times until the gel color returns to its original color.

89

Then put the gel back to the flask.

90

[Results]

[Discussion]

Separation of Serum Proteins by Cellulose Acetate Membrane Electrophoresis

[Principle]

91

[Method]

1. Preparation and application of the sample (1) Membrane size:The cellulose acetate membrane should be touched as little as possible. It is better to use

forceps to handle the membrane at all stages. Cut the membrane into strips.(2) Inspection:The strips should be thoroughly inspected for the such as ridges or spots. Strips with

these faults may produce irregular bands and poor separation, and therefore should be

discarded.(3) Drawing line on the membrane:Examine carefully to distinguish the smooth and the rough sides of the membrane. Draw a using a pencil.

(4) Equilibrating the membrane with buffer:Wet a membrane strip by placing it into the barbital buffer with the rough side facing down.

Immerse the strip completely by gently rocking the dish for at least 10 minutes, and then

92

remove it with forceps. Lightly blot the strip with filter paper so that no excess liquid is seen.

Put both ends of the strip on two glass plates with the .(5) Applying the sample:Three to five microliters of the x-ray film. Press the x-ray film in the previously

marked line on the cellulose acetate membrane quickly and then remove it.2. Electrophoresis

Place the trough and connect it to the buffer compartments with . The

membrane strips should be placed on the two shoulder pads with . Both ends of the strip

are pressed firmly against the pads to ensure proper contact. The sample line should be close to

the cathode. Cover the lid of tank tightly.3. Staining

Immerse the membrane in staining solution for 5 . After the staining is completed, the

membrane is immediately transferred , which contains washing solution. The washing is

continued until the background of the membrane is and the washing solution is colorless.

This process usually takes only . The membrane is then blotted with filter paper to remove

any residue washing solution.

[Results]

[Discussion]

11．监考记录及考场规则93

天津医科大学国际医学院监考记录20 ___---20___学年第____学期期____考试

考试科目：______________ 年级：______________

考试时间：______________ 专业：______________

考试地点：______________ 班级：______________

1.监考老师及考生是否于考前 15 分钟进入考场？是否2.考前是否宣读考试规定？是否3.是否核查了考生的有关证件？是否4.是否准确统计了考生人数？是否

考场人数：__________名，实考__________名，缺考__________名。缺考学生记录：序号姓名学号

5.违纪考生记录：

6.是否按时收回试卷？是否7.监考教师签字：备注

Examination Rules for International Students

94

1. You must enter the examination room 15 MINUTES before the time scheduled for the commencement of

the examination.

2. You must leave all of your study materials and bags in the front of the examination room and put your

Student Card on the upper left corner of your desk for check.

3. You must write your roll number and your name on the examination paper first.

4. During the examination, you are not allowed to bring your mobile phones into the examination room.

5. You must observe absolute silence in the examination room. You must not communicate by word of mouth

or otherwise with other students. You must not borrow any stationery (e.g. pens, rulers, erasers, calculators,

etc.) from others during the examination. You must not peep at other students’ papers or exchange papers

with others.

6. You must use black or dark blue ball-point pens / pens for writing answers. Examination papers completed

by pencils or red pens will be null (examination questions which require drawing pictures excluded).

7. You must observe the prevailing non-smoking rule in the examination room. Food and drink are also not

allowed.

8. You must remain seated at the end of the examination. You must not communicate with other students

while all completed examination papers are being collected by the invigilators

9. You must not talk or discuss outside the examination room after submitting your paper in advance, as this

may disturb other students who may still be sitting for the examination.

International Medical College

12．巡考记录95

200 —— 200 学年第学期（期中、末）考试领导巡视考场记录

时间：地点：记录：

签字：

13．试卷套头Mid/Final term Examination Paper (A/B)

for International Students, Grade 2009

（1st /2nd Semester, 20 ——20 ）Course：(教师填写) Class： Name： Roll No.：

96

-1-

Serial No. I II III IV V VI VII VIII IX X TotalScore

Marking Person

□I.1.2.

□II.1.2.

□III.1.2.

□IV.1.2.

□V.1.2.

□VI.1.2.3.

97

-2-

Course： Class： Name： Roll No.：

装订线

98

-3-

14．试卷样卷及答案 Mid-term Examination of Biochemistry Paper (A)

99


（1st Semester, 2011—2012）Specialty： Class： Name： Roll No.：

-1-

Serial No. I II III IV TotalScore

Marking Person

I .Give the definition accurately for the following specialized terms (four marks for each term, 4X6=24%)

1. Gluconeogenesis

2. Domain

3. The glucose-alanine cycle

.

4. Lipoprotein lipase (LPL)

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-2-

5. Allosteric regulation

6. Nutritional essential fatty acids

Ⅱ. Outline the following questions briefly. (Eight marks for each question, 8X3=24%) 1. What is the secondary structure of protein? Please describe the properties of the two

most common types of secondary structure of protein. 2. What is the Michaelis-Menten equation? Please discuss the significance of Km. 3. Please describe the biochemical processes that cause ketoacidosis and the principles

of clinic treatment.

Ⅲ. Elaborate the following questions. (Twelve marks for each question, 12X2=24%)

1. Can you list the difference between anaerobic and aerobic metabolism of glucose?2. Please describe the source and fate of cholesterol and metabolic related the drugs to

lower plasma cholesterol.

Ⅳ. For each of the questions below, five suggested answers are given. Choose the best one to each of the questions. (One mark for each question, 1X28=28%)

No. 1 2 3 4 5 6 7 8 9 10

Answer

No. 11 12 13 14 15 16 17 18 19 20

Answer

No. 21 22 23 24 25 26 27 28

Answer

101

Specialty： Class： Name： Roll No.：1. Which one of the following types of bonds or interactions is most important in

determining the tertiary structure of proteins?A. hydrogen bonds B. electrostatic bonds C. hydrophobic interactions D. disulfide bonds E. none of above

2. Which of the following statements about the regulation of metabolite is FALSE?A. all regulation of metabolite is passiveB. the Km values for most enzymes are at steady state concentrations of their substrates inside the living cell

C. compartmentalization ensures metabolic efficiency and simplifies regulationD. regulation is the most effective by controlling a single committed stepE. covalent modification can be reversible or irreversible

3. Which of the following amino acid makes the major contribution to the ability of most proteins to absorb light in the region of 280 nm?

A. tryptophanB. lysineC. glycine D. serineE. none of above

4． The zinc finger motif belongs to which of the following structure?A. primary structureB. secondary structureC. supersecondary structureD. tertiary structureE. none of above

5. Which statement is correct about myoglobin and hemoglobin?A. the oxygen dissociation curve of myoglobin is sigmoidal, where as that of

hemoglobin is hyperbolic.B. they both take heme as a prosthetic groupC. they both have quaternary structure.D. hemoglobin has a higher affinity for oxygen than does myoglobin

102

-3-

103

-4-

E. none of above6. Which of the following factors will be the inhibitor of lipogenesis?

A. glucagons B. insulin C. citrate D. well-fed state E. acetyl-CoA

7. How many ATPs will be produced during the β-oxidation if the initial substrate is 18 carbons saturated Stearic acids? A. 113 B. 122 C. 125D. 120 E. 118

8. Which of the following enzymes can catalyze the conversion of glucose 6-phosphate to glucose in liver?A. glucose-6-phosphataseB. glycogen phosphorylaseC. debranching enzymeD. glucokinaseE. hexokinase

9. The patients who lack apoCⅡ have a very increasing component in plasma, this component is A. triacylglycerol B. free cholesterol C. phospholipids D. HDLc E. cholesterol ester

10. Which of the following statements about enzyme is FALSE?A. they are mostly protein catalystsB. most of they are highly specific for their substratesC. they don't change the equilibrium of the reactionD. some of them can have not active siteE. Enzymes do not alter themselves in enzyme-catalyzed reaction.

104

Specialty： Class： Name： Roll No.：11. The similarity of different type of isozymes is

A. they have identical physical propertiesB. they catalyze the same reactionC. they have identical chemical propertiesD. their molecular structures are same.E. none of above all

12. The enzyme that catalyzes the citrate cleavage to acetyl-CoA and oxaloacetate in cytosol is A. acetyl-CoA carboxylase B. ATP-citrate lyase C. citrate synthase D. isocitrate dehydrogenase E. malic enzyme

13. Which of the following statements about coenzyme is FALSE? A. coenzymes may be regarded as second substrates B. many coenzymes are derivatives of B vitamins C. some coenzymes are derivatives of adenosine monophosphate D. all enzymes must require a coenzyme E. coenzymes serve as recyclable shuttles14. Which of the following statements about the citric acid cycle is true?

A. It contains no intermediates for glucogenesisB. It contains intermediates for amino acid synthesisC. It generates fewer molecules of ATP than glycolysis, per mole of glucose

consumedD. It is an anaerobic processE. It is the major anabolic pathway for glucose synthesis

15. The correct order of passage of electrons through the cytochromes of the respiratory chain is A. aa3bcc1 B. aa3bc1c C. bc1caa3

D. aa3cc1b E. cc1b aa3

105

-5-

106

-6-

16. All the following energy-related activities occur in mitochondria EXCEPT A. the citric acid cycleB. fatty acid oxidation C. electron transportD. glycolysisE. oxidative phosphorylation

17. There are three irreversible steps in glycolysis. They are: A. hexokinase, phosphoglycerate kinase, and pyruvate kinaseB. hexokinase, phosphofructokinase, and pyruvate kinaseC. phosphofructokinase, aldolase, and phosphoglyceromutaseD. phosphoglucose isomerase, glyceraldehydes 3-phosphate dehydrogenase, and

enolaseE. triose phosphate isomerase, phosphoglycerate kinase, and enolase

18. Which of the following amino acids includes two carboxyl groups?A. alanine B. phenylalanineC. glutamic acidD. serineE. histidine

19. NADPH is the main source of reducing equivalents for biosynthesis of fatty acids. Which pathway does it formed by?

A. glycolysisB. oxidative phosphorylationC. citrate acid cycleD. pentose phosphate pathwayE. β-oxidation of fatty acids

20. The Cori cycle may be described asA. the interconversion between glycogen and glucose 1-phosphateB. the synthesis of alanine from pyruvate in skeletal muscle and the synthesis of

pyruvate from alanine in liverC. the synthesis of urea in liver and degradation of urea to carbon dioxide and

ammonia by bacteria in the gutD. the production of lactate from glucose in peripheral tissues with the resynthesis

of glucose from lactate in liver

107

Specialty： Class： Name： Roll No.：E. none of above all

21. During starvation, gluconeogenesis increases to maintain the levels of blood glucose, which one of the following will be enhanced? A. liver pyruvate kinase activityB. the secretion of insulin by the pancreasC. muscle phosphoglucomutase activityD. the metabolism of acetyl CoA to pyruvateE. the metabolism of glutamate to glucose 6-phosphate

22. Which of the following substances acts as the “energy currency” of the cell? A. GTPB. ATPC. glucoseD. fatty acidE. creatine phosphate

23. In which fraction of the cell does the biosynthesis of fatty acids occur? A. mitochondria

B. microsomal C. golgi D. cytosol

E. endoplasmic reticulum24. The vitamine which acts as cofactor of acetyl-CoA carboxylase is

A. Vit B1 B. Vit B2 C. Vit PP D. Vit B6 E. Biotin

25. Which of the following lipoproteins can remove cholesterol from the tissues to the liver?

A. CM B. VLDL C. IDL D. LDL E. HDL

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-7-

26. Which of the following statements is true about competitive inhibitors?A. they resemble the substrate B. they can lower the apparent Km of relevant enzymesC. they can affect the Vmax D. they don’t compete with substrates for the same binding sites on the enzymeE. they are not reversed by increasing substrate concentration

27. Which of the following enzymes is the rate-limiting enzyme of biosynthesis of cholesterols?

A. HMG-CoA synthaseB. HMG-CoA reductaseC. HMG-CoA lyase D. Mevalonate kinase E. None of above all

28. The apolipoprotein that can stimulate the lecithin:cholesterol acyltransferase (LCAT) is A. apoB100 B. apoAI C. apoC D. apoD E. apoE

109

-8- Mid-term Examination Paper (A) ANSWER




Marking Person

□I .Give the definition accurately for the following specialized terms (four marks for each term, 4X6=24%)1. Gluconeogenesis

Gluconeogenesis is the process of converting noncarbohydrates to glucose or glycogen. It is of particular importance when carbohydrate is not available from the diet. Significant substrates are glucogenic amino acids, lactate, glycerol, and propionate.

2. Domain Domain is a section of protein structure that folds independently into a stable

conformation. It is sufficient to perform a particular chemical or physical task such as binding of a substrate or other ligand. Different domains work together to provide the complete function of the protein.

3. The glucose-alanine cycleTransports glucose from liver to muscle with formation of pyruvate, followed by transamination to alanine, then transports alanine to the liver, followed by gluconeogenesis back to glucose.

4. Lipoprotein lipase (LPL)LPL located on the walls of blood capillaries and synthesized in extrahepatic tissues. It hydrolyzed triacylglycerols in chylomicrons and very low density lipoproteins into fatty acids and glycerols.

5. Allosteric regulationBinding of metabolites or second messengers to sites distinct from the catalytic site of enzymes triggers conformational changes that alter catalytic activity of the enzymes.

110

6. Nutritional essential fatty acids. Some polyunsaturated fatty acids such as linoleic (18:2: 9,12),linolenic (18:3: 9,12,15)

and arachidonic acid (20:4: 5,8,11,14), which are essential for human , but can’t be synthesized by themselves and must be supplied by diet, are defined as nutritional essential fatty acids.

□Ⅱ. Outline the following questions briefly. (Eight marks for each question, 8X3=24%) 1. What is the secondary structure? Please describe the properties of the two most

common types of secondary structure of protein. (1) Secondary structure is the folding of short contiguous segments of polypeptide

into geometrically ordered units. Side chains are not involved in the secondary structure. (2marks)

(2) The Alpha helices: (3marks)• A complete turn of the helix contains an average of 3.6 aminoacyl residues• The R groups of each aminoacyl residue in an α helix face outward.• The stability of an α helix arises primarily from hydrogen bonds

(3) The Beta sheets (3marks)• The amino acid residues of a -sheet form a zigzag or pleated pattern •The R groups of adjacent residues point in opposite directions• -sheets derive much of their stability from hydrogen bonds.

2. What is the Michaelis-Menten equation? Please discuss the significance of Km.

(1) The Michaelis-Menten equation: (4 marks) V=

Vmax [S]

Km + [S]

(2) The significance of Km: (4 marks)1) When[S]is very much less than Km, Vi is directly proportionate to [S]2) When[S]is very much greater than Km, Vi = Vmax3) When [S] = Km, Vi = 1/2 Vmax4) Km measures the affinity of the enzyme for the substrate. A low Km correspond

to a high affinity and vice versa3. Please describe the biochemical processes that cause ketoacidosis and the

principles of clinic treatment.(1) Biochemical processes (4 points)

111

1) Under some conditions, the body exhausts available carbohydrate as in starvation or it cannot use carbohydrate efficiently to provide energy as in uncontrolled diabetes mellitus

2) There is increased fat mobilization. Ketone bodies are overproduced to a level beyond the capacity of extrahepatic tissues to oxidize them.

3) There are higher than normal quantities of ketone bodies present in the blood and urine. Acetoacetic and 3-hydroxybutyric acids are both moderately strong acids and are buffered when present in blood or other tissues. However, their continual excretion in quantity progressively depletes the alkali reserve, causing ketoacidosis.

(2) Treatment (4 points)1) Restore normal acid-base balance by administration of bases.2) Restore normal carbohydrate metabolism by providing glucose in the case of

starvation or administration (injecting) of insulin in diabetes mellitus..

Ⅲ. Elaborate the following questions. (Twelve marks for each question, 12X2=24%) 1. Can you list the difference between anaerobic and aerobic metabolism of glucose?

(1) The pathway (4 marks)The anaerobic metabolism: the pathway of all mammalian cells for the metabolism of glucose (or glycogen) to pyruvate and lactate. The aerobic metabolism: aerobic conditions, lactate does not accumulate and pyruvate is oxidized to acetyl-CoA by a pyruvate dehydrogenase complex. Then acetyl-CoA is degraded by the citric acid cycle and the respiratory chain to CO2, H2O and ATP

(2) Energy production: (2 marks) The anaerobic metabolism: 2 ATP

The aerobic metabolism: 30 or 32 ATP (3) Location: (2 marks)

The anaerobic metabolism: cytosolThe aerobic metabolism: cytosol and mitochondria

(4) End product: (2marks)The anaerobic metabolism: lactate The aerobic metabolism: CO2, H2O

112

(5) Biomedical importance (2 marks)The anaerobic metabolism: the ability of glycolysis to provide ATP in the absence of oxygen is especially important because it allows skeletal muscle to perform when oxygen supply is insufficient and erythrocytes, which lack mitochondria, are completely reliant on glucose as their metabolic fuel and metabolize it by anaerobic glycolysis.

2. Describe the source and fate of cholesterol and the metabolic related drugs to lower plasma cholesterol.

(1) Cholesterol is derived about equally from the diet and from biosynthesis (4 marks)1) Cholesterol is derived from foods of animal origin such as egg yolk, meat, liver, and

brain2) All tissues containing nucleated cells are capable of cholesterol synthesis3) The liver accounts for approx. 10%, intestines for about 10%4) Cholesterol synthesis occurs in the endoplasmic reticulum and the cytosol

Synthesis: Acetyl-CoA is the source of all carbon atoms in cholesterolit is controlled by regulation of HMG-CoA reductase

(2) Fate (4 marks) Convert to bile acids.

Convert to steroid hormones. Convert to 1, 25-(OH)2 D3.

(3) Hypolipidemic drugs (4 marks)1) Cholestyramine resin, interrupting(blocking) the enterohepatic circulation of bile acids2)The statin drugs such as Mevastatin and lovastatin can inhibit HMG-CoA reductase.


No. 1 2 3 4 5 6 7 8 9 10

Answer C A A C B A D A A D

No. 11 12 13 14 15 16 17 18 19 20

Answer B B D B C D B C D D

No. 21 22 23 24 25 26 27 28

Answer E B D E E A B B

113

Final Examination of Biochemistry



-1-


Marking Person

Ⅰ. Give the definition accurately for the following specialized terms (four marks for each term, 4X6=24%)

1. Nutritional essential amino acids

2. Reading frame

3. Semi-conservative replication

4. Salvage synthesis

114

5. Promoter

6. Vector

Ⅱ. Outline the following questions briefly. (Six marks for each question, 6X4=24%) 1. What is the function of S-adenosyl methionine in the body?2. Please describe the biochemical processes that cause classic phenylketonuria (PKU)?3. Please describe the processing and modification of mRNA precursors in eukaryotes.4. Please describe the process of the activation of amino acid in protein synthesis.

Ⅲ. Elaborate the following questions. (Twelve marks for each question, 12X2=24%) 1. Please describe the structure of lac operon and the function of each component.2. Try to compare the features of replication, transcription and reverse transcription.


No. 1 2 3 4 5 6 7 8 9 10

Answer

No. 11 12 13 14 15 16 17 18 19 20

Answer

No. 21 22 23 24 25 26 27 28

Answer

115

-2-Specialty： Class： Name： Roll No.：

116

1. The precursors of DNA synthesis are A. dAMP, dGMP, dCMP and dTMPB. dADP, dGDP, dCDP and dTDP C. dATP, dGTP, dCTP and dTTPD. ATP, GTP, CTP and UTP

E. ATP, GTP, CTP and TTP2. Reverse transcription

A. catalyzes RNA strand synthesis using DNA template B. catalyzes DNA strand synthesis using DNA templateC. requires NTP as precursorsD. requires dNTP as precursorsE. does not require primer

3. During DNA replication A. each strand of two parental strands acts as templateB. coding strand acts as templateC. leading strand acts as templateD. sense strand acts as template

E. lagging strand acts as template4. Mechanism of DNA repair is

A. mismatch repair B. base excision-repairC. nucleotide excision-repairD. double strand break repairE. all of above are correct

5. The correct description of Okazaki fragment is A. a piece of DNA strand on the template strandB. a piece of DNA strand on the coding strand C. DNA fragment in lagging strandD. DNA fragment in leading strand

E. a short length of RNA formed and catalyzed by primase

-3-

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-4-

6. 5’-ACGTACG-3’ is a newly synthesized DNA strand, which of the following strand is its template strand?

A. 3’-ACGTACG-5’B. 5’-TGCATGC-3’C. 3’-TGCATGC-5’D. 5’-UGCAUGC-3’

E. 3’-UGCAUGC-5’7. Which step is the crucial regulation point in the expression of genes?

A. initiation of transcription B. messenger RNA degradationC. protein synthesis D. posttranslational modification of proteinsE. protein degradation

8. The stop codon is recognized by A. a specific uncharged tRNAB. a specific aminoacyl-tRNAC. a specific mRNAD. a specific protein factorE. a specific ribosomal subunit

9. The sedimentation coefficent of the large subunit of ribosome in eukaryotes isA. 30S B. 80SC. 60SD. 70SE. 40S

10. In E.coli RNA polymerase，core enzyme is A. α2ββ’ωσB. α2ββ’ωC. α2βωD. α2β2ωE. αββ’ω

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Specialty： Class： Name： Roll No.：11. Which subunit can recognize the transcription start site in prokaryotes?

A. σ subunitB. core enzymeC. β subunitD. β’ subunitE. α subunit

12. The exon implies that the sequence A. is not transcribedB. is both transcribed and translatedC. is transcribed but not translatedD. is translated but not transcribedE. is not translated

13. Which of the following vitamins takes part in biosynthesis of thymidine

monophosphate (TMP)?A. vitamin B1 B. vitamin PP C. folic acid D. vitamin B6 E. vitamin B2

14. The coenzyme of transaminase is A. Pyridoxal phosphateB. NADHC. NADPHD. FADE. FMN

15. Which of the following statements of enhancer is FALSE? A. they can exert their influence on transcription even when separated by thousands of base pairs from a promoterB. they work when oriented in either directionC. they are one of the DNA cis-active elementsD. they are one of the trans-acting proteins E. they facilitate or enhance initiation at the promoter

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-5-16. Degeneracy of the genetic code means that

A. a given base triplet can code for more than one amino acidB. there is no punctuation in the code sequenceC. the third base in a codon is not important in coding D. codons are not ambiguousE. a given amino acid can be coded for by more than one base triplet

17. Which of the following dose not belong to one carbon units？A. –CH3

B. –CH2–C. –CH=D. –CHOE. CH4

18. Which of the following substances is not converted by tyrosine? A. epinephrine B. norepinephrine C. creatinine D. triiodothyronine E. dopamine

19. Which of the following enzymes defect can lead to Lesch-Nyhan syndrome?A. hypoxanthine-guanine phosphoribosyl transferase, B. deoxycytidine kinaseC. PRPP transferaseD. adenylosuccinate synthaseE. adenosine kinase

20. Which statement about Tm is right? A. It is connect with the length of the DNA chain onlyB. It is proportion to the content of G-C base pairC. It is not influenced by the salt concentration of the solution D. It has no relationship with the composition of base E. All the Tm in eukaryotic cells is same

21. The amino acid that can produce α-ketoglutarate by deamination isA. tryptophanB. glycineC. phenylalanine

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-6-Specialty： Class： Name： Roll No.：

D. tyrosineE. glutamic acid

22. Which of the following amino acids can form inhibitory neurotransmitter -Aminobutyrate (GABA) by decarboxylation? A. prolineB. glycineC. glutamineD. glutamic acidE. aspartic acid

23. Which of the following statements of bacterial plasmids is FALSE? A. plasmids are small, linear, duplex DNA moleculesB. plasmids exist as single or multiple copies within the bacterium and replicate

independently from the bacterial DNAC. plasmids are smaller than the host chromosome and are therefore easily separatedD. the natural function of plasmids is to confer antibiotic resistance to the host cell to

bacteriaE. plasmids are the cloning vectors in which some chimeric or hybrid DNA molecules can

be constructed. 24. In which fractions of the cell does the urea cycle occur?

A. mitochondria and golgiB. microsomal and golgiC. mitochondria and cytosolD. cytosol and endoplasmic reticulum

E. endoplasmic reticulum and golgi25. Which is the tertiary structure of DNA in eukaryotic cells?

A. nucleosomeB. cyclic DNAC. double-helixD.α-helixE. none of above

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-7-26. What is the reason which cause the denaturation of DNA

A. low temperature is the only reasonB. the break of phosphodiester bondsC. polynucleotide chain disaggregateD. methylation of basesE. the hydrogen bonds between complementary bases break

27. hnRNA is the precursor of A. tRNAB. eukaryotic rRNAC. eukaryotic mRNAD. prokaryotic rRNAE. prokaryotic mRNA

28. Which one is the end product of pyrimidine catabolism in human being? A. UreaB. allantoin C. HypoxanthineD. β-aminoisobutyrateE. Uric acid

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-8-

Final-term Examination Paper (A) Answer




Marking Person

. Give the definition accurately for the following specialized terms (four Ⅰ marks for

each term, 4X6=24%)：1. Nutritional essential amino acids ：The amino acids which cannot be synthesized in the body and must be provided from foods are defined as nutritional amino acids.2. Reading frame: since the sequence of an mRNA molecule is read in groups of three nucleotides(codons) from the 5’ end, It can be read in three possible reading frames, depending on which nucleotide is used as the first codon. The correct reading frame is set in vivo by recognition by the ribosome of the initiation codon AUG 3. Semi-conservative replication: when the two parental strands separate, each acts as a template for making a new complementary strand. Each daughter duplex contains one parental strand and one new strand. In other words, one of the parental strands is “conserved” in each daughter duplex. This is called Semi-conservative replication.

4. Salvage synthesis：the pathway of using purines, pyrimidines or their nucleosides as the starting materials to synthesize nucleotides.5. Promoter: A DNA sequence at which RNA polymerase may bind, leading to initiation of transcription6. Vector: A plasmid or bacteriophage into which foreign DNA can be introduced for the purposes of cloning and they can replicate in a host cell under their own control systems.

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. Outline the following questions briefly. (Six Ⅱ marks for each question, 6X4=24%)

1. What is the function of S-adenosyl methionine in the body?It participates in many methylation reactions to provide methyl group (3 marks)It also participates directly in spermine and spermidine biosynthesis (3 marks)

2. Please describe the biochemical processes that cause classic phenylketonuria (PKU)?(1) Phenylketonuria arises from defects in phenylalanine hydroxylase. (2 marks) (2) In health condition Phenylalanine is first converted to tyrosine catalyzed by

phenylalanine hydroxylase. Subsequent reactions are those of tyrosine(2 marks)

(3) In patients with PKU alternative catabolites are excreted and elevated urinaryphenylpyruvate.L-Phenylalanine → Phenylpyruvate → Phenylacetate and Phenyllactate → Phenylacetylglutamine (2 marks)

3. Please describe the processing and modification of mRNA precursors in eukaryotes.(1) The cap structure is added to the 5′ end of the newly transcribed mRNA precursor

(2marks)(2) Poly (A) tails are added to the 3′ end of mRNA molecules (2marks)(3) Introns are removed and exons are spliced together (1mark)(4) RNA editing (1mark)

4. Please describe the process of the activation of amino acid in protein synthesis.(1) The attachment of an amino acid to a tRNA is catalyzed by an enzyme called

aminoacyl-tRNA synthetase. (2marks)(2) The overall reaction is: (4marks)

Aminoacid + ATP + tRNA→aminoacyl-tRNA + AMP + PPi PPi is driven by the subsequent hydrolysis of pyrophosphate to two inorganic

phosphates.

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. Elaborate the following questions. (Twelve Ⅲ marks for each question, 12X2=24%) 1. Please describe the structure of lac operon and the function of each component (1) lacZ: encodes β-galactosidase, which hydrolyzes lactose into glucose and galactose

(2marks)(2) lacY: encodes a permease, which is responsible for the permeation of galactose into

the cell (1mark)(3) lacA: encodes a thiogalactoside transacetylase, which is also responsible for the

permeation of galactose into the cell with permease. (1mark)(4) lacI: encodes the lac operon repressor protein, which binding to the operator turns

the lac operon OFF (2marks)(5) Promoter: the site where the RNA polymerase binds (2marks)(6) Operator: the site where the repressor protein binds (2marks)(7) CRP binding site: the site where the CRP binds. CRP in conjunction with cAMP

bind to the CRP binding site to accelerates the expression of lac operon. (2marks)

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2. Try to compare the features of replication, transcription and reverse transcription.

Replication Transcription Reverse transcription

Template(2marks)

Double stranded DNA A segment of gene on DNA

RNA

Precursors(2marks)

4 dNTP (A, G, C, T) 4 NTP (A, G, C, U) 4 dNTP (A, G, C, T)

Primer(1mark)

Yes No Yes

Main enzyme(1mark)

DNA-dependent DNA polymerase

DNA-dependent RNA polymerase

RNA-dependent DNA polymerase

Direction of new strand synthesis(1mark)

5’→3’ 5’→3’ 5’→3’

Bond(1mark)

3’,5’-phosphodiester bond



Base pairing(1mark)

A-T, G-C A-T, A-U, G-C A-T, A-U, G-C

Product(1mark)

DNA RNA DNA

Fidelity(1 mark)

Higher Lower Lower

Product processing(1mark)

No Yes No


No. 1 2 3 4 5 6 7 8 9 10

Answer C D A E C C A D C B

No. 11 12 13 14 15 16 17 18 19 20

Answer A B C A D E E C A B

No. 21 22 23 24 25 26 27 28

Answer E D A C A E C D

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Biochemstry Experiment Examination Paper



装订线

Serial No. TotalScore

Marking Person

Answer the following questions (100 marks):

1. What is chromatography? Give out the classification according to the principle.2. Describe the principle of Gel filtration chromatography. What is the order of Separation and elution of Hemoglobin and Protamine? 3. Try to state the effect of trichloroacetic acid, 2, 6-dichlo-rophenol indophenol, nigrosin stain, iodine solution and Benedict’s reagent in experiment.4. Summarize the principle and calculation formula of quantitative assay of vitamin C. 5. Discuss the effects of temperature, pH, activator and inhibitor on the enzyme activity.6. What is electrophoresis and mobility? Discuss the factors affecting electrophoresis. 7. Please illustrate the principle of separation of serum proteins by acetate cellulose film electrophoresis 8. Try to explain the isozyme and its clinical significance.

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Biochemstry Experiment Examination Paper Answer


（1st Semester, 2011—2012）Answer the following questions (100 marks):

1. What is chromatography? Give out the classification according to the principle.Chromatography separates molecules on the basis of different size, shape, mass, quantity of

charge, solubility and adsorption properties. (6 marks)According to the principle, chromatography is generally divided into four types: adsorption

chromatography, partition chromatography, ion-exchange chromatography and gel filtration chromatography. (6 marks)2. Describe the principle of Gel filtration chromatography. What is the order of Separation and elution of Hemoglobin and Protamine?

This method exploits the physical property of molecular size to achieve separation. Molecules of different size can be separated when passing down a column containing swollen particles of a gel. The stationary phase consists of inert particles that contain small pores of a controlled size. A solution containing solutes of various molecular sizes is allowed to pass through the column under the influence of continuous solvent flow. Solute molecules that are larger than the pores cannot enter the interior of the gel beads, so they are limited to the space between the beads. As a result, they are not slowed in their progress through the column and will be rapidly eluted in a single zone. Small molecules that are capable of diffusing in and out of the beads have a much larger volume available to them. Therefore, they will be delayed in their journey through the column bed. Molecules of intermediate size will migrate through the column at a rate somewhere between those for large and small molecules. Therefore, the order of elution of the various solute molecules is directly related to their molecular (8 marks)

The large molecules (hemoglobin) leave the column first and the smaller ones (DNP-protamine) last. (4 marks)

3. Try to state the effect of trichloroacetic acid, 2, 6-dichlo-rophenol indophenol, nigrosin stain, iodine solution and Benedict’s reagent in experiment.

Trichloroacetic acid: protein can be precipitated by trichloroacetic acid. (2 marks)2, 6-dichlo-rophenol indophenol: redox titrant (2 marks)

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Nigrosin stain: protein staining solution (2 marks)Iodine solution: Maltose can’t react with I2 (iodine). But the I2 solution will be changed

from blue, purple, deep brown, light brown to red with the dextrin molecule from big to small. So we can judge the enzyme activity from the color of the hydrolysate with I2 reaction. (4 marks)

Benedict’s reagent: Glucose can reduce the two-valency copper ion which exists in an alkaline solution of a copper salt (Benedict’s reagent) and react to form precipitate of rust-brown cuprous oxide. (4 marks)4. Summarize the principle and calculation formula of quantitative assay of vitamin C. Reduced ascorbic acid in a neutral or a weak acidic solution can act as a donor of reducing equivalents and reduce the dye named 2,6-dichlo-rophenol indophenol (oxidized pattern) into the reduced pattern. At the same time, the reduced ascorbic acid is oxidized into oxidized ascorbic acid. (6 marks) (VA-VB)×0.088 Vit.C mg/100g= –––––––––––––––×100×C DW VA: the average amount of the dye consumed in the three conical flasksVB: the amount of the dye consumed in the contrast solution (which contains no extract)C: the total volume of the solution extracted from the sampleD: the volume of the solution titrated W: the weight of the sample0.088: the coefficient, which indicates how many milligrams Vit.C corresponding to one-milliliter dye consumed (6 marks)5. Discuss the effects of temperature, pH, activator and inhibitor on the enzyme activity.

As general chemical reactions, the velocity of reaction increases as temperature rises and the velocity of reaction decreases as temperature drops. On the other hand, enzymes are proteins. As temperature increases, the velocity of denaturation rises, meanwhile the velocity of enzyme-catalyzed reaction decreases or makes enzyme lose their activity completely. These two contrary effects exist in same reaction. Only in one certain temperature, the velocity of enzyme-catalyzed reaction comes to the maximum. We call this point temperature optimum (6 marks)

129

pH affects enzyme activity by changing the charges on ionizable groups of the enzyme, and affects dissociation degree of the substrate also. Each enzyme has an optimum pH at which the rate of the reaction is at its maximum. (3 marks)

Activators increase the enzyme activity and inhibitors decrease the activity. (3 marks)6. What is electrophoresis and mobility? Discuss the factors affecting electrophoresis. Electrophoresis is the movement of charged particles in an electric field. (4 marks) Mobility is the velocity of the molecule per electric field intensity , also called electrophoretic velocity. (4 marks) Factors affecting electrophoresis: pH of the medium, ionic strength of the buffer, electric field intensity, electro-osmosis (4 marks)7. Please illustrate the principle of separation of serum proteins by acetate cellulose film electrophoresis.

The serum proteins isoelectric points are all less than pH 7.0. Therefore, they all ionized into negative ions in pH 8.6 buffer and they migrate toward anode (4 marks). Since different serum proteins have different isoelectric points, they load different charges at the same pH and they have different molecular weights. Moreover, their migration velocity is

different (5 marks). So we can divide the serum proteins into albumin、α1 -globulin 、α2-

globulin、β- globulin and γ- globulin such five zones (5 marks).8. Try to explain the isozyme and its clinical significance.

Isozymes are the separable forms of a given enzyme present in different tissues, different cell types or subcellular compartments of the same organism. Isozymes may catalyze the same reaction, but their physical and chemical properties exhibit many significant differences. (6 marks)

Medical interest in isozymes was stimulated by the discovery that human sera contained several LDH isozymes and that their relative proportions changed significantly in pathologic conditions. Thus a myocardial infarction is likely to result in the release into the serum of elevated amounts of LDH with the isozyme pattern typical of heart muscle. On the other hand, liver disease (e.g., infectious hepatitis) is likely to be accompanied by elevated levels of the M-type isozymes in the serum, since these isozymes predominate in liver cells. Hence, the determination of the level and isozyme distribution of serum LDH is useful in the diagnosis of a variety of other hematological, cardiac, and hepatic diseases. (8marks)

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15．试卷批阅要求

天津医科大学国际医学院关于规范考试试卷批阅操作的要求为加强留学生批阅试卷的规范化，现将批阅考试试卷操作的有关要求通知如下：

一、批阅试卷统一用红色笔。

二、批阅过程中统一用给分的方法标记每项结果，给分用“+X 分”表示，每题批阅

后统一在题号前写出该题得分。

三、试卷卷首的得分统计表务必填写齐全；阅卷人在试卷首卷上签字。

四、在将各题得分汇总成卷面总成绩时，务必认真细致，保证准确无误。

五、卷面分数如有改动，应由改动人在改动处签名。

各单位务请将以上要求通知到每一位任课教师，并研究具体的检查督促措施。

131

国际医学院

二零零八年十二月十九日

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16．成绩分析记录天津医科大学国际学院 - 学年第学期课程考核试卷分

析记录一、基本情况课程名称　课程代码　主讲教师试卷来源 □1.试题库□2.试卷库□3.校内统一命题□4.校外教师命题□5.任课教师命题　阅卷方式 □1.微机阅卷 □2.流水阅卷 □3.任课教师阅卷阅卷教师考试方式 □1.闭卷 □2.开卷 □3.上机 □4.综述 □5.论文 □6.设计 □7.其它考试方法 □1.笔试 □2.口试 □3.实际操作考试对象年级：专业：应考人数实考人数考试时间：

二、成绩分析统计人数　　　　　　　　　　　　　　　　

30 　　　　　　　　　　　　　　　28 　　　　　　　　　　　　　　　　26 　　　　　　　　　　　　　　　　24 　　　　　　　　　　　　　　　　22 　　　　　　　　　　　　　　　　20 　　　　　　　　　　　　　　　　18 　　　　　　　　　　　　　　　　16 　　　　　　　　　　　　　　　　14 　　　　　　　　　　　　　　　　12 　　　　　　　　　　　　　　　　10 　　　　　　　　　　　　　　　　8 　　　　　　　　　　　　　　　　6 　　　　　　　　　　　　　　　　4 　　　　　　　　　　　　　　　　2 　　　　　　　　　　　　　　　　0 　　　　　　　　　　　　　　　　

30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

133

成绩情况最高分 ; 最低分 ; 平均分百分记分制

成绩分布人数百分比成绩＜ 60 　

60 ≤ 成绩 ≤ 65 　65 ＜成绩 ≤ 70 　70 ＜成绩 ≤ 75 　75 ＜成绩 ≤ 80 　80 ＜成绩 ≤ 85 　85 ＜成绩 ≤ 90 　90 ＜成绩 ≤ 95 　95 ＜成绩 ≤ 100 　

三、综合分析难度评价容易□；较容易□；适中□；偏难□；难□；

试题份量偏多□; 适中□; 偏少□; 三基型(基本知识.基本理论.基本技能)( %)；综合运用型( %)；提高扩展型( %)；实际考试所用时间：多数学生在规定时间内完成□；多数学生完不成□；多数学生只用了(1/3□)(1/4□)(2/3□)时间

卷面质量安排合理:是□,否□; 规范用字:是□,否□;笔迹工整:是□,否□;图表准确清晰:是□,否□;

答题出错较多的主要原因 1.试题较难□；2.学生对基本知识、基本理论掌握不扎实□；3.学生分析应用能力较差□； 4.教学过程中有疏漏□；

试题覆盖面 60%以下□；60-69%□；70-79%□；80-89%□；90-100%□。

试卷综合分析与评价　

四、改进意见

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课程负责人签字：年月日

135

天津医科大学 2011--2012 学年一学期 2010 级留学生生物化学课程考试试卷

试题分析报告===========================试卷份数: 30试卷题数: 40============试卷实得分分段人数统计:满分段,10,9,8,7,6,5,4,3,2,1段,0 分段,平均分数(以下顺序相同) 0 5 10 7 7 1 0 0 0 0 0 0 76.9667 试卷均值：76.9667试卷难易度：.7697试卷自相关区分度：8.5711试卷自相关加权区分度：101.2322

================试题 1题型：　正态分布试题均值：　4试题难易度：　1试题自相关区分度：　0试题自相关加权区分度：　0试题互相关区分度：　0试卷与试题趋势相关系数：　0该题实得分分段人数统计: 30 0 0 0 0 0 0 0 0 0 0 0 4

试题 2题型：　正态分布试题均值：　3.8333试题难易度：　.9583试题自相关区分度：　.2889试题自相关加权区分度：　.2056试题互相关区分度：　.8944试卷与试题趋势相关系数：　.196该题实得分分段人数统计: 26 0 0 3 0 1 0 0 0 0 0 0 3.8333

试题 3题型：　正态分布试题均值：　3.8667

136

试题难易度：　.9667试题自相关区分度：　.2311试题自相关加权区分度：　.1156试题互相关区分度：　.8622试卷与试题趋势相关系数：　.252该题实得分分段人数统计: 26 0 0 4 0 0 0 0 0 0 0 0 3.8667

试题 4题型：　正态分布试题均值：　1.8试题难易度：　.45试题自相关区分度：　1.36试题自相关加权区分度：　2.4267试题互相关区分度：　8.96试卷与试题趋势相关系数：　.5717该题实得分分段人数统计: 7 0 0 3 0 7 0 0 3 0 0 10 1.8

试题 5题型：　正态分布试题均值：　3.3333试题难易度：　.8333试题自相关区分度：　.8889试题自相关加权区分度：　1.2889试题互相关区分度：　6.9778试卷与试题趋势相关系数：　.6109该题实得分分段人数统计: 20 0 0 4 0 4 0 0 0 0 0 2 3.3333



137

试题难易度：　.9778试题自相关区分度：　.24试题自相关加权区分度：　.1822试题互相关区分度：　.6622试卷与试题趋势相关系数：　.1542该题实得分分段人数统计: 27 0 2 0 1 0 0 0 0 0 0 0 5.8667





138

试题难易度：　.8028试题自相关区分度：　3.3622试题自相关加权区分度：　18.0322试题互相关区分度：　27.6878试卷与试题趋势相关系数：　.648该题实得分分段人数统计: 21 0 2 0 0 2 0 1 0 0 0 4 9.6333


试题 13题型：　０－１分布试题均值：　1试题难易度：　1试题自相关区分度：　0试题自相关加权区分度：　0试题互相关区分度：　0试卷与试题趋势相关系数：　0该题实得分分段人数统计: 30 0 0 0 0 0 0 0 0 0 0 0 1

试题 14题型：　０－１分布试题均值：　.9333试题难易度：　.9333试题自相关区分度：　.1245试题自相关加权区分度：　.1245试题互相关区分度：　.2978试卷与试题趋势相关系数：　.0839该题实得分分段人数统计: 28 0 0 0 0 0 0 0 0 0 0 2 .9333

试题 15题型：　０－１分布试题均值：　.9

139

试题难易度：　.9试题自相关区分度：　.18试题自相关加权区分度：　.18试题互相关区分度：　.2967试卷与试题趋势相关系数：　.0695该题实得分分段人数统计: 27 0 0 0 0 0 0 0 0 0 0 3 .9


试题 17题型：　０－１分布试题均值：　.9667试题难易度：　.9667试题自相关区分度：　.0644试题自相关加权区分度：　.0644试题互相关区分度：　-.5344试卷与试题趋势相关系数：　-.2093该题实得分分段人数统计: 29 0 0 0 0 0 0 0 0 0 0 1 .9667


试题 19题型：　０－１分布试题均值：　1

140

试题难易度：　1试题自相关区分度：　0试题自相关加权区分度：　0试题互相关区分度：　0试卷与试题趋势相关系数：　0该题实得分分段人数统计: 30 0 0 0 0 0 0 0 0 0 0 0 1





141

试题难易度：　.9试题自相关区分度：　.18试题自相关加权区分度：　.18试题互相关区分度：　-.17试卷与试题趋势相关系数：　-.0398该题实得分分段人数统计: 27 0 0 0 0 0 0 0 0 0 0 3 .9





142

试题难易度：　.6667试题自相关区分度：　.4444试题自相关加权区分度：　.4444试题互相关区分度：　-.2778试卷与试题趋势相关系数：　-.0414该题实得分分段人数统计: 20 0 0 0 0 0 0 0 0 0 0 10 .6667





143

试题难易度：　.5333试题自相关区分度：　.4978试题自相关加权区分度：　.4978试题互相关区分度：　1.0511试卷与试题趋势相关系数：　.1481该题实得分分段人数统计: 16 0 0 0 0 0 0 0 0 0 0 14 .5333



试题 34题型：　０－１分布试题均值：　.4667试题难易度：　.4667试题自相关区分度：　.4978试题自相关加权区分度：　.4978试题互相关区分度：　2.2156试卷与试题趋势相关系数：　.3121该题实得分分段人数统计: 14 0 0 0 0 0 0 0 0 0 0 16 .4667


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试题难易度：　.5667试题自相关区分度：　.4911试题自相关加权区分度：　.4911试题互相关区分度：　1.5522试卷与试题趋势相关系数：　.2201该题实得分分段人数统计: 17 0 0 0 0 0 0 0 0 0 0 13 .5667





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试题难易度：　.9试题自相关区分度：　.18试题自相关加权区分度：　.18试题互相关区分度：　.33试卷与试题趋势相关系数：　.0773该题实得分分段人数统计: 27 0 0 0 0 0 0 0 0 0 0 3 .9


====================================================试题实得分分段人数统计一览表:满分段,10,9,8,7,6,5,4,3,2,1段,0 分段,平均分数(以下顺序相同)第 1 题: 30 0 0 0 0 0 0 0 0 0 0 0 4 第 2 题: 26 0 0 3 0 1 0 0 0 0 0 0 3.8333 第 3 题: 26 0 0 4 0 0 0 0 0 0 0 0 3.8667 第 4 题: 7 0 0 3 0 7 0 0 3 0 0 10 1.8 第 5 题: 20 0 0 4 0 4 0 0 0 0 0 2 3.3333 第 6 题: 19 0 0 4 0 1 0 0 1 0 0 3 3.4 第 7 题: 27 0 2 0 1 0 0 0 0 0 0 0 5.8667 第 8 题: 19 0 3 0 4 1 0 0 0 0 0 3 4.9333 第 9 题: 16 0 4 0 2 0 0 1 0 0 0 7 4.2 第 10 题: 22 0 0 0 2 0 0 3 0 0 0 2 5.2667 第 11 题: 21 0 2 0 0 2 0 1 0 0 0 4 9.6333 第 12 题: 0 0 5 0 2 7 2 0 1 6 2 5 4.6667 第 13 题: 30 0 0 0 0 0 0 0 0 0 0 0 1 第 14 题: 28 0 0 0 0 0 0 0 0 0 0 2 .9333 第 15 题: 27 0 0 0 0 0 0 0 0 0 0 3 .9 第 16 题: 30 0 0 0 0 0 0 0 0 0 0 0 1 第 17 题: 29 0 0 0 0 0 0 0 0 0 0 1 .9667 第 18 题: 30 0 0 0 0 0 0 0 0 0 0 0 1 第 19 题: 30 0 0 0 0 0 0 0 0 0 0 0 1 第 20 题: 26 0 0 0 0 0 0 0 0 0 0 4 .8667 第 21 题: 27 0 0 0 0 0 0 0 0 0 0 3 .9 第 22 题: 27 0 0 0 0 0 0 0 0 0 0 3 .9

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第 23 题: 27 0 0 0 0 0 0 0 0 0 0 3 .9 第 24 题: 23 0 0 0 0 0 0 0 0 0 0 7 .7667 第 25 题: 22 0 0 0 0 0 0 0 0 0 0 8 .7333 第 26 题: 23 0 0 0 0 0 0 0 0 0 0 7 .7667 第 27 题: 20 0 0 0 0 0 0 0 0 0 0 10 .6667 第 28 题: 20 0 0 0 0 0 0 0 0 0 0 10 .6667 第 29 题: 25 0 0 0 0 0 0 0 0 0 0 5 .8333 第 30 题: 26 0 0 0 0 0 0 0 0 0 0 4 .8667 第 31 题: 16 0 0 0 0 0 0 0 0 0 0 14 .5333 第 32 题: 19 0 0 0 0 0 0 0 0 0 0 11 .6333 第 33 题: 23 0 0 0 0 0 0 0 0 0 0 7 .7667 第 34 题: 14 0 0 0 0 0 0 0 0 0 0 16 .4667 第 35 题: 17 0 0 0 0 0 0 0 0 0 0 13 .5667 第 36 题: 19 0 0 0 0 0 0 0 0 0 0 11 .6333 第 37 题: 16 0 0 0 0 0 0 0 0 0 0 14 .5333 第 38 题: 24 0 0 0 0 0 0 0 0 0 0 6 .8 第 39 题: 27 0 0 0 0 0 0 0 0 0 0 3 .9 第 40 题: 20 0 0 0 0 0 0 0 0 0 0 10 .6667

试卷各题按各卷总分由高至低排列的分数一览表

情况说明1. 命题主要考核点及大纲要求：

本课程要求学生掌握生物化学基本概念，生物大分子结构、功能以及理化性质，物质代谢及调节。在本次考试中，共有四种题型，名词解释、简答题、论述题，和单选题，用于考察学生生物化学基础知识掌握情况、综合分析问题能力情况。2. 命题难易程度、覆盖面：命题难易程度适中。在本次考试中，命题全面覆盖了教学大纲要求的知识面，覆盖面较好。3. 学生掌握情况及存在问题分析：

1）通过阅卷和各题型得分的分析，可知：试卷中选择题得分率较高，试题难易度 P值符合国内标准，说明学生对生物化学基本

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概念普遍掌握较好，知识点较扎实。名称解释的第四题和论述题的第二题试题难易度 P值分别为 0.45、0.3889，低于国内标准和国外标准，主要问题是学生没能准确掌握知识。

2）影响考试成绩的因素及应注意的问题：① 个别学生学习的目的性和自主性较差，受周围环境影响，学习比较浮躁。② 一些学生学习该课程的方法还需改进，在学习或复习时缺乏对相关知识的融会贯通。个别学生不能较好地在整体水平理解物质代谢以及相关联系，导致分数偏低。

4. 任课教师为提高教学质量今后应采取的措施：1）本门课程授课内容多，因此授课教师在课堂上加强学生专业英文词汇的学习，帮助学生尽快进入学习状态。2）课程中重要知识点布置课下练习，督促学生及时完成。严格要求学生，必须独立完成作业。3）在教学活动中，采取多种教学手段来激发学生的学习兴趣，注意对学生学习方法

的正确引导。4）在教学过程中多与学生辅导员沟通，及时反映学生的情况，通过辅导员做好学生

的思想工作，引导他们尽快转换学习方法，增加自我控制的能力。

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17．试卷保存登记表天津医科大学国际医学院试卷保存登记表

课程名称学时学分考试时间适用专业出题教师审定人试卷号

注：该表中的试卷号要同时在电子文档试卷中注明18．专家听课评价表

149

天津医科大学国际医学院专家听课记录表授课教师所在院系：年月日星期第节至第节

课程名称授课教师职称教材版本教室教学班号

授课主要内容

授课主要优点、特点及改进建议

听课教师：备注

150

19．同行听课评价表天津医科大学国际医学院同行听课评价表

开课单位：课程名称： □ 公共基础课 □ 学科基础课 □ 专业课授课教师：年龄层次：□老 □中 □青授课班级：应到人数：实到人数教材版本：课堂讲授内容：听课后总体印象：□优 □良 □中 □差听课后的分项评价：请在下列各评价项目之后的相应评价等级位置填入您的选项，只限单选。选项标准：A 完全同意，B同意，C 一般，D不同意，E 完全不同意

评价项目 A B C D E

1 讲课有热情，精神饱满2 讲课有感染力，能吸引学生的注意力3 对问题的阐述深入浅出，有启发性4 对问题的阐述简练准确，重点突出，思路清晰5 对课程内容娴熟，运用自如6 讲述内容充实，信息量大

7教学内容能反映或联系学科发展的新思路，新概念，新成果

8 能给予学生思考、联想、创新的启迪9 能调动学生情绪，课堂气氛活跃10 能有效地利用各种教学媒体11 学生上课迟到、早退、缺勤等情况12 学生遵守课堂纪律情况

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13 学生听课学习状态对课堂内容或其它方面的具体意见或建议：

天津医科大学国际学院 2008.9 制表听课人签字：听课时间：地点：

20．学生听课反馈表Teaching assessment, Grade 2010 Foreign Students

Biochemstry Specialty: Class

Chpter & teacherdegree of satisfaction (√)

Completely

satisfied

Very

satisfiedcommon dissatisfied

Higher Orders StructureGeng Xin

Regulation Of Enzymes

Geng Xin

Gluconeogenesis and

controlLi Haidong

Lipid Transport & Storage

Li Haidong

RNA Synthesis,

Processing, &

ModificationYong Gongyuan

Regulation of Gene

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Expression

Yong Gongyuan

Overall assessment:

Date：

153

21．学生反馈评价表BIOCHEMSTRY Teaching Feedback

This form is a serious attempt to get feedback from you regarding the quality of instruction you have received. Your honest and constructive opinion will be helpful to improve our teaching level. Please take your time and answer carefully all the questions below, where appropriate, according to the following scale.

1. What was the most impressive thing when you were learning

Biochemstry (about teacher, chapter or others)?

2. What aspects of this course were most beneficial to you?

3. What do you suggest for this course?

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22．教研室备课记录

____________学年第______学期基础医学院教学备课记录

年月日教研室主讲人参加人员

学生年级、专业

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备课内容摘要

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目录 - tianjin medical university · web viewassays of enzyme-catalyzed reactions typically...

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