Lehninger Principles of Biochemistry 习题答案chapter 2
《无机及分析化学》课程思政元素的挖掘
2021年第4期广东化工第48卷总第438期 · 167 · 《无机及分析化学》课程思政元素的挖掘龚绍峰，张益萍，肖新生，谭英芝，刘芳(湖南科技学院化学与生物工程学院，湖南永州425199)[摘要]无机及分析化学课程中开展思政教育应着重课程思政元素的挖掘以及思政元素与专业知识的融合。
本文从科学家故事引申出科研精神和家国情怀，化学定律和原理中蕴含的思政元素和人生哲理，专业知识升华为伦理道德和职业素养三个方面对无机及分析化学课程中的思政元素进行挖掘和实践，力求将无机及分析化学课程与思政教育无缝融合，实现全员、全程和全方位育人的教学目标。
[关键词]无机及分析化学；课程思政；高等院校；思政元素[中图分类号]G642.0；O65；O61 [文献标识码]A [文章编号]1007-1865(2021)04-0167-02Excavation of Ideological and Political Instruction Elements in Inorganic andAnalytical Chemistry CourseGong Shaofeng, Zhang Yiping, Xiao Xinsheng, Tan Yingzhi, Liu Fang(College of Chemical and Biological Engineering, Hu‘nan University of Science and Engineering, Yongzhou 425199, China) Abstract: The Ideological and Political Instruction in Inorganic and Analytical Chemistry should focus on the excavation of ideological and political elements and the integration of Ideological and political elements and professional knowledge. This paper excavates the Ideological and Political Elements in Inorganic and Analytical Chemistry from three aspects of scientists' stories extending to scientific research spirit and feelings of home and country, ideological and political elements and philosophy of life contained in chemical laws and principles, professional knowledge sublimated to ethics and professional quality, and strive to seamlessly integrate Inorganic and Analytical Chemistry courses with Ideological and Political Education to achieve the teaching goal of all-staff, whole-process and all-around education.Keywords: inorganic and analytical chemistry；ideological and political instruction in professional courses；colleges and universities；ideological and political elements1 引言2019年3月18日，习近平总书记在学校思想政治理论课教师座谈会的重要讲话中指出:“要坚持显性教育和隐性教育相统一，挖掘其他课程和教学方式中蕴含的思想政治教育资源，实现全员、全程、全方位育人”[1]。
生物化学(英文版)biochemistry-chapter1英
Through biochemical research, people can understand the basic processes and mechanisms of organisms, explore the mechanisms of disease occurrence and development, and provide scientific basis for disease prevention, diagnosis, and treatment.
Biochemistry is an interdisciplinary field that aims to reveal the chemical essence behind life phenomena.
Definition of Biochemistry
Biochemistry helps to increase crop yield and improve quality in agriculture, and contributes to the development of nutritious, safe, and reliable new types of food in the food industry.
Large molecules in living organisms, such as proteins, nucleic acids, and polysaccharides, all have carbon as their core.
Carbon atoms can form complex network structures through covalent bonds, which endow biomolecules with high diversity and stability.
principles of biochemistry
principles of biochemistry1. 引言1.1 概述生物化学是一门研究生物体内基本分子和化学反应的科学，它揭示了细胞和生物体的运作机制。
通过细致研究和理解生命过程中所涉及的分子结构、功能和代谢途径，我们可以更好地理解生命现象以及许多与人类健康相关的疾病。
1.2 文章结构本文将按照以下结构进行论述：第二部分将介绍生物化学的基本原理。
我们将探讨生物化学的定义、重要概念，并深入了解不同生物分子在维持生命活动中扮演的角色，还会涵盖不同生化反应和代谢途径的重要性。
第三部分将聚焦于分子生物学，强调DNA、RNA、蛋白质等基因组成分与其功能之间的关系。
我们将探索DNA的结构和功能，以及如何通过转录和翻译过程合成RNA和蛋白质。
第四部分将关注酶与酶动力学。
我们将介绍酶的基本特性和分类，深入探讨酶催化机制以及速率方程式对酶反应的描述。
此外，我们还将探讨酶动力学参数及其调节因子在生物体内的重要性。
最后一部分将介绍代谢途径与能量转换。
我们将详细讨论糖类代谢途径的重要性以及相应的调控机制，同时深入了解脂类代谢和β氧化反应在能量产生和存储中的作用。
此外，我们还将研究氨基酸代谢和尿素循环对蛋白质降解和氮排除过程的贡献。
1.3 目的本文旨在通过深入探索生物化学原理，提供关于生物分子结构、功能以及与之相关的生化反应和代谢途径的全面知识。
通过理解这些基本概念，读者可以更好地理解细胞和生物体内发生的化学过程，并认识到其对于维持生命活动、健康和疾病发展等方面的重要性。
2. 基本原理：生物化学是研究生物体内分子结构、组成和转化过程的科学。
在生物化学中，有一些基本原理需要了解。
2.1 生物化学概念：生物化学研究的主要对象是生物分子，如蛋白质、核酸、碳水化合物和脂类等。
通过研究它们的结构和功能，可以深入了解生命的基本单位——细胞的运作方式。
2.2 生物分子的重要性：生物分子在细胞中扮演着各种重要角色。
例如，蛋白质是细胞内许多反应的催化剂，帮助维持生命活动；核酸则携带遗传信息，并参与蛋白质合成；碳水化合物提供能量和结构支持；脂类则构成细胞膜，并参与信号传导等。
1. 氨基酸
Born july 31, 1800, Escherheim; Germany Died september 23, 1882, Gö ttingen; Germany
目录
（二）“燃烧”学说使“活力论”再次遭遇重创
Justus von Liebig (12 May 1803 – 18 April 1873) was a German chemist who made major contributions to agricultural and biological chemistry, and worked on the organization of organic chemistry. As a professor, he devised the modern laboratory-oriented teaching method, and for such innovations, he is regarded as one of the greatest chemistry teachers of all time. He is known as the "father of the fertilizer industry" for his discovery of nitrogen as an essential plant nutrient, and his formulation of the Law of the Minimum which described the effect of individual nutrients on crops. He also developed a manufacturing process for beef extracts, and founded a company, Liebig Extract of Meat Company, that later trademarked the Oxo brand beef bouillon cube.
国外生物化学优秀教材分析与思考
教育研究课程教育研究84 学法教法研究生物化学是研究生命机体中生物活性分子结构功能及其代谢的一门学科。
生物化学作为医学、农林、综合院校的专业核心课程，其是一门飞速发展的学科同样也是学生最难学透彻的课程之一。
教材的作用不言而喻，其是学生获取知识的重要方式，也是教师教学活动的重要参考依据，教材的合理选用与教学效果有着必然联系。
对比国内外教材可以发现，国外教材平均三四年更新一次版本，更新速度较快，能够及时吸收最新研究成果。
国内教材内容相对陈旧，版本更新周期较长，教材插图比较少，且大多数直接引自国外教材，同时缺乏实验原始数据分析，以黑白印刷的原版彩图效果差强人意。
本文就具有一定国际知名度的《Biochemistry （6th edition ）》、《Lehninger Principles of Biochemistry （7th edition ）》、《Biochemistry （8th edition ）》、《Principles of Biochemistry （5h edition ）》、《Biochemistry （4th edition ）》五本教材进行了初步的分析比较与探讨，旨在全面了解国际上生物化学教材的内容特色、发展趋势和教学方法，希望对国内生物化学教材编写和教学工作有所帮助。
一、教材简介表格一教材基本情况比较二、教材特点介绍1.《Biochemistry （6th edition ）》该教材由弗吉尼亚大学Reginald H.Garrett 、Charles M.Grisham 两位教授主编，他们一位是生物学博士，另外一位是化学博士，他们以各自领域内的独特视角来看待生物化学这个共同的内容，完成了本教材的编写，这也是本书的亮点之一。
本教材具有如下一些特点：（1）除了基础生物化学的内容外，独立编写了第16章马达分子，详细的介绍了马达分子的结构性质等；（2）本版新增如：生化试剂盒的研制、低聚糖与蛋白质折叠、脂筏作用、超级呼吸复合体等；（3）书中穿插了很多实验室技术小box ，介绍了很多实验室常用的研究技术，意在使学生明白实验的重要性；（4）知识覆盖全面，增加了生物化学在治疗疾病、改善公共卫生、改善环境污染，开发更廉价更安全的天然产品等这些与人类联系更加紧密的这些内容；（5）这本书的每一章都围绕着一个这是一个很重要的问题，它邀请学生积极参与到他们的学习中，激发他们对主题的好奇心和想象力；（6）每章末均列出了部分相关推荐阅读文献，同时还有对本章重点内容的概括，全书最后附有章节习题解答。
生物化学之蛋白质三级结构PPT教学课件
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α Helix (Cont’d)
• The structure is stabilized by a hydrogen bond between the hydrogen atom attached to the electronegative nitrogen atom of a peptide linkage and the electronegative carbonyl oxygen atom of the fourth amino acid on the amino-terminal side of that peptide bond.
– For the same reason, many adjacent Lys and/or Arg residues, which have positively charged R groups at pH 7.0, also prevent the formation ofα helix.
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α Helix (α-螺旋)
• The polypeptide backbone is tightly wound around an imaginary axis drawn longitudinally through the middle of the helix.
• Within theα-helix, every peptide bond (except those close to each end of the helix) participates in hydrogen bonding.
生物化学(双语)课程教学大纲
《生物化学（双语）》课程教学大纲课程编号：2130053学时：48学分：3.0授课学院：药物科学与技术学院适用专业：药学教材：Lehninger Principles of Biochemistry, Fourth Edition （2005）.作者： David L. Nelson and Michael M. Cox 出版社：W. H. Freeman一．课程的性质、目的及任务生物化学是一门利用化学的理论和方法作为主要手段来研究生命现象，揭示生命的奥秘的学科。
其主要任务有两个方面：一方面研究构成生物体的基本物质：蛋白质、核酸、糖、脂及对体内的生物化学反应起催化和调节作用的酶、维生素、激素的结构性质和功能，即所谓静态生物化学，另一方面研究构成生物体的基本物质在生命活动过程中进行的化学变化，即新陈代谢过程中的物质和能量转换和调节，即动态生物化学部分。
二．教学基本要求（一）基础理论和基本知识1、蛋白质的结构和功能掌握蛋白质的化学组成、理化性质、分子结构及其与功能的关系，熟悉蛋白质的提取和纯化方法。
2、酶掌握酶的基本概念及影响酶活性的因素，酶结构与功能的关系。
掌握酶反应速率的相关理论。
3、糖化学及其代谢掌握糖的分类及结构，糖酵解、三羧酸循环途径。
熟悉糖原的合成代谢及分解代谢。
4、脂类化学及其代谢掌握脂类的组成及结构，熟悉脂肪的合成代谢及磷脂的合成。
5、核酸化学掌握核苷酸的结构，核酸的结构与功能，熟悉核酸分离纯化的机理。
了解核酸的医药用途。
6、基因信息的贮存和表达掌握DNA、RNA 的分子结构和功能、中心法则、DNA的半保留复制。
熟悉DNA 损伤修复，RNA和蛋白质生物合成、核酸的分子杂交、基因工程及聚合酶链反应。
（二）、基本技能掌握DNA提取及琼脂糖、聚丙烯酰胺凝胶电泳技术，熟悉紫外－可见分光光度计的用法，了解利用HPLC分离鉴定蛋白质的方法。
三．教学内容糖、脂、蛋白质、酶、核酸化学及基因信息的储存和表达。
生物化学之蛋白质三级结构
• Secondary structure refers to the local conformation of some part of a polypeptide. • The most prominent secondary structures are the α helix and β conformations predicated by Caltech Trio in 1951.
α Helix (Cont’d)
• The structure is stabilized by a hydrogen bond between the hydrogen atom attached to the electronegative nitrogen atom of a peptide linkage and the electronegative carbonyl oxygen atom of the fourth amino acid on the amino-terminal side of that peptide bond. • Within theα-helix, every peptide bond (except those close to each end of the helix) participates in hydrogen bonding. • Each successive turn of the helix is held to adjacent turns by three to four hydrogen bonds.
生物化学：Chapter 1-2 Foundations_of_Biochemistry
Sept. 11, 2012
Chapters 1-2
The realm of Biochemistry
Professor Zengyi Chang
(昌增益 教授）
changzy@ Room 204, New Life Science Building
Tel. 6275-8822
two books into Chinese!
The eight classic papers (to be read by all students)
Topic 1 the prebiotic Origin of Life
Stanley L. Miller.(1953) “A Production of Amino Acids Under Possible Primitive Earth Conditions”, Science, 117:528-9.
Books on the history of Biochemistry:
1. 昌增益（译者）《蛋白质、酶和基因：化学与生物 学的交互作用》，清华大学出版社，2005年1月。
Fruton, J. S. (1999). Proteins, Enzymes, Genes: The Interplay of Chemistry and Biology. New Heaven and London: Yale University Press.
What is life?
Life: A process or condition but a definition not yet achieved.
Possessing the following basic features:
– Homeostasis;
Lehninger Principles of Biochemistry 习题答案chapter 19
Electrons generally fall in energy through the chain - from complexes I and II to complex IV O2
NADH (reductant) + H+ + ? O 2 (oxidant) NAD+ + H2O
3
Oxidative Phosphorylation
2
Electron transfer from NADH to O2 involves multisubunit inner membrane complexes I, III, & IV, plus coenzyme Q and cytochrome c. Within each complex, electrons pass sequentially through a series of electron carriers.
4
Structure of mitochondrial cytochrome c
Heme is covalently linked to the protein via S atoms
Iron-sulfur centers (Fe-S)
- electron transfer proteins may contain multiple ironsulfur centers. - transfer only one electron even if they contain two or more iron atoms, because of the close proximity of the iron atoms.
• contains all of TCA cycle enzymes {except, succinate
Principles of Biochemistry 习题答案chapter 6
moles of each enzyme in cell volume of cell in liters 1.6 mm3
pr2h (3.14)(0.50 mm)2(2.0 mm) 1.6 10 12 cm3 1.6 10 12 mL 1.6 10 15 L Amount (in moles) of each enzyme in cell is (0.20)(1.2 g/cm3)(1.6 mm3)(10 12 cm3/mm3) (100,000 g/mol)(1000) 3.8 10
21
mol
Average molar concentration
3.8 10 21 mol 1.6 10 15 L 2.4 10
6
mol/L
2.4
10
6
M
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1. Keeping the Sweet Taste of Corn The sweet taste of freshly picked corn (maize) is due to the high level of sugar in the kernels. Store-bought corn (several days after picking) is not as sweet, because about 50% of the free sugar is converted to starch within one day of picking. To preserve the sweetness of fresh corn, the husked ears can be immersed in boiling water for a few minutes (“blanched”) then cooled in cold water. Corn processed in this way and stored in a freezer maintains its sweetness. What is the biochemical basis for this procedure? Answer After an ear of corn has been removed from the plant, the enzyme-catalyzed conversion of sugar to starch continues. Inactivation of these enzymes slows down the conversion to an imperceptible rate. One of the simplest techniques for inactivating enzymes is heat denaturation. Freezing the corn lowers any remaining enzyme activity to an insignificant level. 2. Intracellular Concentration of Enzymes To approximate the actual concentration of enzymes in a bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000. Assume also that the bacterial cell is a cylinder (diameter 1.0 mm, height 2.0 mm), that the cytosol (specific gravity 1.20) is 20% soluble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell. Answer There are three different ways to approach this problem. (i) The concentration of total protein in the cytosol is (1.2 g/mL)(0.20) 100,000 g/mol 0.24 10
双底物酶促反应动力学机理
Km
K2 K3 K1
K2 K1
K3 K1
Ks
K3 K1
Km是ES分解速度（K2+K3）与形成速度（K1）的比值，它包含ES解 离趋势（K2／K1）和产物形成趋势（K3／K1）。
Ks是底物常数，只反映ES解离趋势（底物亲和力），1/Ks可以准确 表示酶与底物的亲和力大小。
只有当K1 、 K2>>K3时，Km≈Ks，因此，1/Km只能近似地表示底物亲 和力的大小。
E+P E + SP
- dES dt
=
k2 ES
+ k3 ES
当进入稳态时:
d ES dt
=
-
d ES dt
即: k1 E S = k2 ES + k3 ES = ES k2 + k3
稳态理论下米氏方程的推导
又由酶的恒等式: E0 = E + ES
E = E0 - ES
- 则: k1 E0 ES S = ES k2 + k3
E+S
k1 ES k3 E + P
k2
k4
布氏提出以下假设: 1.反应开始时, P = 0 ,不足已产生逆反应.测定酶的反应速度
一般是测定反应的初速度,即产物浓度变化在5%以内的速度.
2. E
S0 即反应中 S =. S0
3.反应开始后,反应立刻进入恒态状态, ES 浓度处于稳定状态 ES 的形成速度等于其分解速度.
三、米氏方程的讨论
Km与天然底物 Km最小的底物称该酶的最适底物或天然底物。因为： Km愈小（达到Vmax一半所需的底物浓度愈小）表示V对△[S] 越灵
敏。
V
1/2Vmax
Chapter 1 & 2 The Foundations of Biochemistry
The major types of biomolecules were revealed
The major types of biomolecules found in ALL types of living organism: proteins, carbohydrates, lipids and nucleic acids. Proteins, carbohydrates, and lipids were all discovered before the 19th century. Nucleic acids were the last of these to be isolated, in 1868, by Johann Friedrich Miescher, a Swiss, twenty-four years old.
Nobel prizes for Biochemical studies
1901-2006
A remarkable number of Nobel prizes have been won by biochemists
The aim of study:
To be able to read what is not written and to hear what is not said!
彻未书之道，悉言外之意
Lectured by 史 锋 shifeng@ 校园移动网：614263 Please feel free to contact me if you need！
The Nobel Prize in Physiology or Medicine 1988
Sir James W. Black
Gertrude B. Elion
生物化学原理(1)
丙氨酸 精氨酸 天冬酰胺 天冬氨酸 半胱氨酸 谷氨酸 谷氨酰胺 甘氨酸 组氨酸 异亮氨酸 亮氨酸 赖氨酸 甲硫氨酸 苯丙氨酸 脯氨酸 丝氨酸 苏氨酸 色氨酸 酪氨酸 缬氨酸
Classification of Amino Acids
Very Small Amino Acid
Lipoamino Acid
•第七章 糖类化合物代谢 •第八章 生物氧化和能量转换 •第九章 脂类物质的合成与分解
•第十章 蛋白质的降解和氨基酸代谢
•第十二章 核酸的生物合成 •第十三章 蛋白质的生物合成 •第十四章 代谢调节
•第十一章 核酸的降解与核苷酸的代谢
生物化学的发展简史
1897年 Buchner 发现酵母细胞质能使糖发酵
组氨酸的咪唑基也能发生烷基化反应。生 成烷基咪唑衍生物，并引起酶活性的降低 或丧失
肽
肽：氨基酸与氨基酸之间通过 -氨基和 羧基反应形成的酰胺键（肽键）共价地结 合在一起形成的产物。
肽链：
氨基末端
羧基末端
Ser H H3N
+
O C N H
Val H C CH
O C N H
Tyr H C CH2
肽链的水解：
酸水解：常用6 mol/L的盐酸或4 mol/L的硫酸在105110℃条件下进行水解，反应时间约20小时。此法的优 点是不容易引起水解产物的消旋化。缺点是色氨酸被 完全破坏；含有羟基的氨基酸如丝氨酸或苏氨酸有一 小部分被分解；门冬酰胺和谷氨酰胺侧链的酰胺基被 水解成了羧基。 碱水解：一般用5 mol/L氢氧化钠煮沸10-20小时。由 于水解过程中许多氨基酸都受到不同程度的破坏，产 率不高。部分的水解产物发生消旋化。该法的优点是 色氨酸在水解中不受破坏。 酶解：应用酶水解多肽不会破坏氨基酸，也不会发生 消旋化。水解的产物为较小的肽段。
Principles of Biochemistry 习题答案chapter 11
Total surface area of one cell (mm2) 98 29.8 99.4
Source: Data from Gorter, E. & Grendel, F. (1925) On bimolecular layers of lipoids on the chromocytes of the blood. J. Exp. Med. 41, 439–443.
Answer The conclusions are justified for dog erythrocytes but not for sheep or human erythrocytes. The table provides the total surface area of the lipid monolayer. To determine the monolayer surface area per cell, first calculate the total number of cells. For example, for dog erythrocytes, the number of cells is 8 106 per mm3 8 109 per cm3 (or per mL). In 40 mL, there is a total of (40 mL)(8 109 cells/mL) 3 1011 cells. From the table, this number of cells yielded a monolayer surface area of 62 m2 6.2 105 cm2. Dividing the surface area by the number of cells gives 6.2 105 cm2 3 1011 cells 2 10
Lehninger Principles of Biochemistry 习题答案chapter 3
11.30 9.60 (IV)
P O
C
NH2
(V)
10
8 5.97
pH
6
(III)
4 2.34 2 (I) 0.5 1.0 1.5 2.0 OH (equivalents) (II)
0
Note: before considering statements (a) through (o), refer to Figure 3–10. The three species involved in the titration of glycine can be considered in terms of a useful physical analogy. Each ionic species can be viewed as a different floor of a building, each with a different net charge:
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Chapter 3 Amino Acids, Peptides, and Proteins
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H3NOCH2OCOOH H3NOCH2OCOO H2NOCH2OCOO
CH3 H3N C H C O O H2N CH3 C H C OH O
Zwitterionic
Uncharged
(a) Why is alanine predominantly zwitterionic rather than completely uncharged at its pI? (b) What fraction of alanine is in the completely uncharged form at its pI? Justify your assumptions. Answer (a) The pI of alanine is well above the pKa of the a-carboxyl group and well below the pKa of the a-amino group. Hence, at pH pI, both groups are present predominantly in their charged (ionized) forms. (b) From Table 3–1, the pI of alanine is 6.01, midway between the two pKa values 2.34 and 9.69. From the Henderson-Hasselbalch equation, pH pKa log ([A ]/[HA]). For the carboxyl group: log [A ] [HA] [HA] [A ] 6.01 10
生物化学(英文版)教学大纲
《生物化学》（英文版）教学大纲课程编号：总学时数：86学时（还有实验教学33学时）学分：5学分一、课程性质、目的和要求生物化学是生物科学和生物技术专业的一门重要的必修专业基础课。
学习和掌握生物化学的大体理论和技术对于其它生物学科的学习及此后的相关工作都具有十分重要的意义。
开展生物化学双语教学的目的是使学生在系统学习和掌握生物化学的理论和和实践知识的同时，更好地掌握和应用专业外语，培育成具有国际合作意识、国际交流与竞争能力的外向型人材。
鉴于生物化学是一门知识涉及面广，复杂程度高的学科，所以，咱们将生物化学双语教学分为不同层次开设，即基础生物化学和高级生物化学的双语教学。
在基础生物化学的双语教学中，着重要求学生学习和掌握生物化学的大体理论、大体知识和大体原理；同时也要求了解一些本学科进展前沿的新理论、新动向，使学生打下生物化学大体理论和知识的坚实基础，适应以后的高级生物化学双语教学及相关学科的学习。
同时，采用原版英文生化教材，有利于学生尽快接触生化进展的新知识，扩大知识面，提高学生阅读英文专业文献的能力，适应生物化学的高速进展。
不仅使学生取得较为广博的生物化学大体理论和基础知识，而且使学生初步树立科学的世界观和人生观，学会用辩证唯物主义的正确观点熟悉生命的现象和本质，从而使学生进一步学习和工作打下坚实的基础。
二、教学内容、要点和课时安排本课程的教学内容共分十三章。
Section A Cell organization 5课时A1 ProkaryotesA2 EukaryotesA3 MicroscopyA4 Cellular fractionationSection B Amino acids and proteins 9课时B1 Amino acidsB2 Acids and basesB3 Protein structureB4 Myoglobin and hemoglobinB5 CollagenB6 Protein purificationB7 Chromatography of proteinsB8 Electrophoresis of proteinsB9 Protein sequencing and peptide synthesisSection C Enzymes 6课时C1 Introduction to enzymesC2 ThermodynamicsC3 Enzyme kineticsC4 Enzyme inhibitionC5 Regulation of enzyme activitySection D Antibodies 6课时D1 The immune systemD2 Antibody structureD3 Polyclonal and monoclonal antibodiesD4 Antibody synthesisD5 Antibodies as toolsSection E Membranes 6课时E1 Membrane lipidsE2 Membrane protein and carbohydrateE3 Membrane transport: small moleculesE4 Membrane transport: macromoleculesE5 Signal transudationSection F DNA structure and replication 6课时F1 DNA structureF2 ChromosomesF3 DNA replication in bacteriaF4 DNA replication in eukaryotesSection G RNA synthesis and processing 8课时G1 RNA structureG2 Transcription in prokaryotesG3 The lac operonG4 The trp operonG5 Transcription in eukaryotes: an overviewG6 Transcription of protein-coding genes in eukaryotesG7 Regulation of transcription by RNA Pol ⅡG8 Processing of eukaryotic Pre-mRNAG9 Ribosomal RNAG10 Transfer RNASection H Protein synthesis 6课时H1 The genetic codeH2 Translation in prokaryotesH3 Translation in eukaryotesH4 Protein targetingH5 Protein glycosylationSection I Recombinant DNA technology 6课时I1 Restriction enzymesI2 Nucleic acid hybridizationI3 DNA cloningI4 VirusesI5 DNA sequencingI6 Polymerase chain reactionSection J Carbohydrate metabolism 10课时J1 Monosaccharides and disaccharidesJ2 Polysaccharides and oligosaccharidesJ3 GlycolysisJ4 GluconeogenesisJ5 Pentose Phosphate PathwayJ6 Glycogen MetabolismJ7 Control of glycogen metabolismSection K Lipid metabolism 6课时K1 Structures and roles of fatty acidsK2 Fatty acid breakdownK3 Fatty acid synthesisK4 TriacylglycerolsK5 CholesterolK6 LipoproteinsSection L Respiration and energy 6课时L1 Citric acid cycleL2 Electron transport and oxidative phosphorylationL3 PhotosynthesisSection M Nitrogen metabolism 6课时M1 Nitrogen fixation and assimilationM2 Amino acid metabolismM3 The urea cycleM4 Hemes and chlorophylls三、教学方式采用英文原版教材，实行双语教学，制作PPT课件，利用多媒体技术和适当的板书辅助教学。
lehninger principles of biochemistry配套课件和教学大纲
lehninger principles of biochemistry配套课件和教学大纲在生物化学领域，Lehninger Principles of Biochemistry是一本经典的教科书，广泛用于大学本科生物化学课程的教学。
为了更好地教授这门课程，配套的课件和教学大纲扮演着重要的角色。
本文将探讨Lehninger Principles of Biochemistry配套课件和教学大纲的重要性以及如何设计和使用它们来促进学生的学习。
I. Lehninger Principles of Biochemistry配套课件的设计与使用配套课件在教学中起着重要的辅助作用。
它们能够帮助学生更好地理解和消化教材中的概念，并提供相关的图表和动画来支持讲解。
以下是设计和使用Lehninger Principles of Biochemistry配套课件的一些关键考虑因素。
A. 内容与结构配套课件的内容应紧密跟随教材的结构，以保持一致性和连贯性。
课件应突出重点内容，并通过简洁而清晰的方式呈现。
同时，还应该充分考虑学生的学习需求和能力水平，避免过于晦涩难懂的语言和概念。
B. 图表与动画图表和动画是生物化学教学中不可或缺的工具。
配套课件应提供清晰而引人入胜的图表和动画，以帮助学生更好地理解复杂的生化过程和分子结构。
这样的可视化呈现可以使抽象的概念更具体和形象化，让学生更容易学习和记忆。
C. 互动与讨论配套课件可以通过添加互动元素和讨论题目，促进学生的主动参与和思考。
例如，在课件中引入问题或案例分析，鼓励学生在课堂上与教师和同学进行讨论和交流。
这样的互动可以提高学生的学习动机和理解深度，培养他们的批判性思维和解决问题的能力。
II. Lehninger Principles of Biochemistry教学大纲的设计与使用教学大纲是指导课程教学的重要工具，它有助于教师组织和安排课程内容，确保教学的连贯性和完整性。
00生物化学导论
讲师：赵丹丹
生物化学导论
3
习题册
余琼,李盛贤,赵丹丹.新编生物化学辅导与习题精选. 北京:化学工业出版社, 2009.
讲师：赵丹丹
生物化学导论
4
主要参考书
沈同, 王镜岩. (第2版), 1990; 王镜岩, 朱圣庚, 徐长法. (第3版) , 2019. 北京: 高等教育出版社
讲师：赵丹丹
生物化学导论
1937 Norman Haworth (carbohydrates and vitamin C), Paul Karrer (carotenoids, flavins and vitamins A and B2).
讲师：赵丹丹
生物化学导论
27
1939 Adolf Butenandt (sex hormones), Leopold Ruzicka (terpenes).
Biochemistry(gravure), 2th ed.. Beijing: Higher Education Press, 2019.
周海梦等. Lehninger生物化学原理(第3版) (中文版).高等教育出版社, 2019.
讲师：赵丹丹
生物化学导论
7
主要内容
一、生物化学的涵义 二、生物化学的研究内容 三、生物化学的分类 四、生物化学的发展简史和研究现状 五、学好生物化学的重要性
讲师：赵丹丹
生物化学导论
26
1929 Arthur Harden, Hans von Euler-Chelpin (fermentation of sugar and fermentative enzymes).
1929 Christiaan Eijkman (antineuritic vitamin ), Sir Frederick Hopkins (growth-stimulating vitamins).
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chapter
Water
2
Answer Ethanol is polar; ethane is not. The ethanol —OH group can hydrogen-bond with water. Answer Using pH log [H ]: 5 (a) log (1.75 10 ) 4.76; (b) (d) log (1.50 10 5) 4.82. log (6.50 10
0.072 M gastric HCl
log [H ] and that HCl is a strong acid, pH log (7.2 10
2
)
1.1
5. Calculation of the pH of a Strong Acid or Base (a) Write out the acid dissociation reaction for hydrochloric acid. (b) Calculate the pH of a solution of 5.0 10 4 M HCl. (c) Write out the acid dissociation reaction for sodium hydroxide. (d) Calculate the pH of a solution of 7.0 10 5 M NaOH.
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Chapter 2 Water
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9. Simulated Vinegar One way to make vinegar (not the preferred way) is to prepare a solution of acetic acid, the sole acid component of vinegar, at the proper pH (see Fig. 2–14) and add appropriate flavoring agents. Acetic acid (Mr 60) is a liquid at 25 C, with a density of 1.049 g/mL. Calculate the volume that must be added to distilled water to make 1 L of simulated vinegar (see Fig. 2–15). Answer From Figure 2–15, the pKa of acetic acid is 4.76. From Figure 2–14, the pH of vinegar is ~3; we will calculate for a solution of pH 3.0. Using the Henderson-Hasselbalch equation pH pKa log [A ] [HA]
and the fact that dissociation of HA gives equimolar [H ] and [A ] (where HA is CH3COOH, and A is CH3COO ), we can write 3.0 1.76 4.76 log ([A ]/[HA]) log ([HA]/[A ])
10
6.52
3.02
10
7
M;
4. Acidity of Gastric HCl In a hospital laboratory, a 10.0 mL sample of gastric juice, obtained several hours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of NaOH was required. The patient’s stomach contained no ingested food or drink, thus assume that no buffers were present. What was the pH of the gastric juice? Answer Multiplying volume (L) by molar concentration (mol/L) gives the number of moles in that volume of solution. If x is the concentration of gastric HCl (mol/L), (0.010 L)x x Given that pH (0.0072 L)(0.1 mol/L)
)
9.19; (c)
log (1.0
10 4)
4.0;
3. Calculation of Hydrogen Ion Concentration from pH What is the H concentration of a solution with pH of (a) 3.82; (b) 6.52; (c) 11.11? Answer Using [H ] (a) [H ] 10 3.82 (c) [H ] 10 11.11 10 pH: 1.51 10 4 M; (b) [H ] 7.76 10 12 M.
O CH3 C O CH2 CH2 CH3 N CH3 CH3
Acetylcholine Choline
H2O
CH3 HO CH2 CH2 N CH3 CH3 C O
Acetate
O
H
CH3
In a typical analysis, 15 mL of an aqueous solution containing an unknown amount of acetylcholine had a pH of 7.65. When incubated with acetylcholinesterase, the pH of the solution decreased to 6.87. Assuming that there was no buffer in the assay mixture, determine the number of moles of acetylcholine in the 15 mL sample. Answer Given that pH the reaction: log [H ], we can calculate [H ] at the beginning and at the end of 7.65 6.87 [H ] [H ]
1.76 3
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Chapter 2 Water
Answer z (a) HCl y H Cl (b) HCl is a strong acid and fully dissociates into H and Cl . Thus, [H ] [Cl ] [HCl]. pH log [H ] log (5.0 10 4 M) 3.3 (two significant figures) z (c) NaOH y Na OH (d) NaOH is a strong base; dissociation in aqueous solution is essentially complete, so [Na ] [OH ] [NaOH]. pH pOH 14 pOH log [OH ] pH 14 log [OH ] 14 log (7.0 10 5) 9.8 (two significant figures) 6. Calculation of pH from Concentration of Strong Acid Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 M HCl to a final volume of 100 mL with H2O. Answer Because HCl is a strong acid, it dissociates completely to H Cl . Therefore, 3.0 mL 2.5 M HCl 7.5 meq of H . In 100 mL of solution, this is 0.075 M H . pH log [H ] log (0.075) ( 1.1) 1.1 (two significant figures) 7. Measurement of Acetylcholine Levels by pH Changes The concentration of acetylcholine (a neurotransmitter) in a sample can be determined from the pH changes that accompany its hydrolysis. When the sample is incubated with the enzyme acetylcholinesterase, acetylcholine is quantitatively converted into choline and acetic acid, which dissociates to yield acetate and a hydrogen ion:
10
1. Solubility of Ethanol in Water Explain why ethanol (CH3CH2OH) is more soluble in water than is ethane (CH3CH3).