Unit 2: Structural Biochemistry-0826

2.1: Protein structure

2.1.1:

Draw the structure of amino acids.

Describe and rank the interatomic forces that give proteins and amino acids their shapes.

Predict whether functional groups on an amino acid will be protonated or deprotonated at physiological pH.

Predict the conformational 构象diversity of the R groups of different amino acids.

Classify amino acids based upon the properties of their R groups (e.g. by size, polarity, charge, or chemically unique properties).

A chart depicting the genetically-encoded amino acids is shown below. Amino acid single letter codes are highlighted, and the name appears below the amino acid. Note that these are all L-amino acids.

2.1.2 protein structure:

Label the atoms and bonds in a polypeptide.
Recognize the unique properties of each of the bonds in a polypeptide chain.
Describe a Ramachandran plot and how it is generated.
Identify the three secondary structures of proteins.
Predict the secondary structure of a polypeptide based on its phi and psi angles.
Explain how hydrogen bonding contributes to the formation of secondary structure.

2.1.3 protein structure:

Explain the difference between a motif, fold and domain.

Motifs: are commonly repeating secondary structure elements. Many motifs together could make up a protein fold
fold:the specific arrangement of secondary structure elements in a protein.
domain:an independently folding unit of a protein

Describe why there is a limited diversity of different protein folds.

2.2: Protein folding

2.2.1

Describe the main forces that determine the folding of proteins.
Summarize how these forces balance in magnitude to promote protein folding.
Learning objectives for video

2.2.2

Summarize the thermodynamic hypothesis.
Explain how Anfinsen's experiment supports the thermodynamic hypothesis.
Explain how folding funnels are used to visualize protein folding thermodynamics.

2.2.3

Explain the role of cooperativity in protein folding.
Summarize the theoretical basis of Levinthal's paradox.
Describe the diffusion collision and nucleation condensation models of protein folding.

2.2.4

Explain why there is a risk of aggregate formation团粒形成 for proteins folding in the cytosol胞液.
Summarize how different chaperones分子伴侣 function to prevent protein aggregation.
Explain why heat shock causes protein aggregation.
Compare the chaperone function伴侣蛋白功能 of HSP70 and HSP60.

2.0 MyMQL

2.3 enzyme catalysis

2.3.1:

Define reaction equilibrium平衡.
Determine if a reaction is spontaneous based on the free energy of the reaction coordinates反应坐标.
Explain how a catalyst 催化剂reduces the activation energy to speed the rate of a reaction.
Understand that an enzyme stabilizes稳定 a transition state to decrease the activation energy活化能 of a reaction.

酶之所以使得反应速度加快,是因为降低了活化能

2.3.2

  • 酶的三大优势:

Transient covalent bonds瞬态共价键 between subtrate and enzyme:激活底物并降低活化能,在酶的活性位点中

Weak non-covalent interactions 弱的非共价键between enzyme and subtrate:形成会释放binding energy,从而会减少activation energy

Exquisite specificity精准的特异性

energy

A represents the activation energy in an uncatalyzed reaction.
B represents the activation energy in an enzyme-catalyzed reaction.
C represents the change in net energy of the whole reaction, from substrate to product.
D represents the difference in transition state free energy between the catalyzed and un-catalyzed reactions, and is the 'binding energy' of the enzyme and its substrate.

  • Desolvation is the process of replacing substrate-environmentbonds with enzyme-substrate bonds。在此过程中会释放binding energy

2.3.3

  • Reasons for the tertiary structure of enzymes:
    Positioning catalytic residues
    Optimizing catalysis by induced fit
    Enzyme regulation
    Enzyme complexes

2.3.4.1: protein processing

  • zymogen酶原:An inactive precursor of an enzyme that can be hydrolyzed into 水解成an active enzyme.
insulin processing pathway

In the first step depicted, the N-terminal signal sequence that directs this preproinsulin to the secretion pathway is cleaved off.

In the second step depicted, proinsulin accumulates inside the lumen of the ER, and intramolecular disulfide bonds link together the N-terminal and C-terminal regions of proinsulin.

In the third step depicted, proteolytic cleavage of insulin occurs inside the secretory vesicles, producing active insulin formed by the N- and C-terminal regions that are connected by disulfide bonds. The region connecting the N- and C-terminal fragments, and produced by proteolysis, is a single polypeptide chain called a C-peptide.

In the fourth and final step depicte, both mature insulin and the C-peptide are secreted by pancreatic cells.

2.3.4.2 - 凝血机制

Provide a detailed explanation of the blood clotting process凝血过程 and the role of proteolytic activation蛋白水解的激活 of proteins implicated in blood coagulation血液凝固 and hemostasis止血.

凝血与血栓

2.3.5 - thrombus detection血栓监测

2.4 Enzyme kinetics酶动力学

2.4.1 Enzyme kinetics

How do enzymes achieve their maximum velocity?
当底物足够的时候,酶已经饱和了,不能催化得更快了。

2.4.2: Enzyme kinetics公式推导

The Michaelis-Menten Model的推导
书籍:https://www.ncbi.nlm.nih.gov/books/NBK22430/
动画展示:http://www.wiley.com/college/pratt/0471393878/student/animations/enzyme_kinetics/index.html

图1 在初始阶段,产物P很少,所以K-2可以忽略
图2 V1和V2分别代表生成和降解ES的速率,当达到稳定状态时两者相等
图3 从而推得Km,当Km高时,说明生成的ES混合物少,即有低的底物亲和性;当Km低时,说明ES生成的多,即有高的底物亲和性
图4 因起始阶段产物较少,K2可以忽略不计,所以Km计算公式可以进一步简化
图5 所有的酶Et=游离的酶E+ES混合物,将其代入Km计算公式以简化运算
图6 左图代表产物P随时间变化的曲线,其中的曲线的切线代表生成产物的速率。Vo代表最开始反应的速率,即ES向P转换的速率。Vmax与Vo的关系参见图7,当底物充足,所有的酶与底物结合的时候,即E全部转换为ES的时候,即Et=ES的时候,Vo达到最大速度Vmax
图7 初始反应速度随底物S的变化曲线,当底物充足,所有的酶与底物结合的时候,即E全部转换为ES的时候,Vo达到最大速度Vmax
图8 由此Km可以进一步简化
图9 从而获得Michaelis Menten equation
图10 当底物S量较少,可以忽略时,方程进一步变化
图11 当底物S量较多,可以忽略Km时,方程进一步变化,Vo=Vmax
图12 当Km=S时,Vo为Vmax的一半。所以Km又可以定义为当Vo为Vmax的一半时的底物浓度S

2.4.3: 怎样利用Lineweaver-Burk计算Vmax and Km

https://www.ncbi.nlm.nih.gov/books/NBK22557/

图1 Lineweaver-Burk变换
图2 横轴是1/s, 纵轴是1/Vo,纵截距是1/Vmax, 斜率是Km/Vmax
图3 怎样求Vmax
图4 怎样求Km

2.4.4.1 Competitive Inhibitor

图1 Competitive Inhibitor When inhibitor is bound, substrate cannot bind。

从宏观上预测竞争性抑制剂的作用:由于I与E结合,导致S与E的亲和性降低了,因此Km会提高,而Vmax=K2*Et,由于K2和Et都没有变化,所以Vmax不会变化。接下来从公式上推导Km和Vmax的变化。

图2 抑制剂对应于Km的参数Ki
图3 Km与Ki比值
图4 替换EI
图5 替换圆圈圈出的内容
图6 整理成线性方程
图7 有抑制剂和无抑制剂的比较:竞争性抑制剂并不影响Vmax
图8 竞争性抑制剂增大了Km,但Vmax不变

竞争性抑制剂的斜率和横截距变大,纵截距不变

2.4.4.2 Uncompetitive&Mixed&Noncompetitive inhibitor

Uncompetitive inhibitor
图1 Uncompetitive inhibitor仅与ES结合,由于消耗了ES,使得第一步反应的平衡态右移
图 2 Uncompetitive inhibitor同时减小了Vmax和Km

图中红色的线是Uncompetitive inhibitor的线,斜率与原先相同,但是纵截距高于原先,横截距低于原先。即Vmaxapp的倒数>Vmax的倒数,即Vmaxapp<Vmax.
横截距也偏左,越偏左说明数值越小。

Mixed inhibitor
图3 同时结合E和ES
图4 Mixed inhibitor使得Vmax减少

I与ES结合消耗了ES。而Vmax=K2*Et,由于ES少了,所以生成P的量也少了,从而K2少了,但Et不变,所以Vmax变小。

图5 Mixed inhibitor可能增大、减小或不变Km

因为I与E结合会增大Km,而I与ES结合会减小Km。所以要视两者强弱而定。

图6 Mixed inhibitor的方程

Mixed inhibitor纵截距高,横截距和斜率或高或低。

图7 Noncompetitive inhibitor

Noncompetitive inhibitor纵截距高,横截距同,斜率高。

图8 所有抑制剂作用的总结

competitive inhibitor斜率和横截距变大,纵截距不变。不影响Vmax,但增大Km到aKm。

Uncompetitive inhibitor斜率不变,纵截距变大,横截距变小。减小Vmax和Km

Mixed inhibitor纵截距变大,横截距和斜率或高或低。减小Vmax,可能增大、减小或不变Km

Noncompetitive inhibitor横截距同,纵截距和斜率高。

2.4.5 - ISM (Iterative Saturation Mutagenesis)迭代饱和突变

  • 用于筛选产生特定功能的突变酶,用的是遗传学的知识。
  • Semi-rational Design综合了Rational Design和Directed Evolution两种方法的优点,最常见的Semi-rational Design是ISM。
  • Directed Evolution适用于酶结构未知的情况
  • 因为 在Rational Design we need to have almost full structural knowledge about the enzyme, so we are able to design a ligand that can modify the properties and activity of the enzyme. And also because of high specificity of enzymes. On the other hand, in Semi-rational Design we only need to identify the active site not the whole protein because we are only applying modifications to the active site. Perhaps that's why it is call "Semi-rational".

2.5 Lipids structure and membrane assembly

2.5.1 动画讲解Lipids structure

  • Lipids all have a large hydrophobic region疏水区

2.5.2

TAG在人体中被转换为糖原,因为糖原供给能量更高效

2.5.3

Understand the structure of a lipid droplet.
Understand how sequestration of lipases脂肪酶 regulates triacylglycerol 三酰甘油代谢metabolism.
Understand hormonal激素调节 regulation of lipases脂肪酶 in triacylglycerol metabolism.
Recall the fate of glycerol 甘油 and fatty acid during lipolysis脂类分解.

2.5.4 Lipid structure → function

Membrane curvature膜曲率
Bilayer asymmetry双分子层不对称
Membrane fluidity膜流动性

2.5.5

Describe how fluorescence recovery after photo-bleaching荧光漂白恢复 (FRAP) is used to study lipid diffusion in a membrane.

Understand that the protein composition of a membrane supports its function.

Distinguish between integral and peripheral membrane proteins.

Understand how lipids and proteins control plasma membrane细胞膜 permeability渗透性.

2.5.6 lipodystrophy脂肪代谢障碍

Understand the causes of lipodystrophy at the cellular and molecular/biochemical levels.
Describe the factors that affect the balance of lipid storage and mobilization.

2.5.7 lipid droplet pathologies脂滴病态

Understand the mechanisms that lead to neutral lipid storage disease中性脂质沉积病.
Understand the mechanisms that lead to metabolic syndrome代谢综合征.

2.6 carbohydrates糖结构

2.6.1 carbohydrates

Identify chemical and structural features of monosaccharides.
State the difference between monosaccharides and polysaccharides.
Predict the location of the carbonyl group in an aldose ("-ose") versus ketose ("-ulose") sugars.
Determine which carbons within a sugar constitute chiral centers.
Contrast the polarized light and Fisher projection nomenclatures for describing monosaccharide stereochemistry.
Recognize a monosaccharide as D or L based on its Fisher projection.

2.6.2

Recall the evidence indicating that hexose and pentose sugars exist mostly in their cyclic form.
Illustrate the spontaneous cyclization reaction for glucose - in which the C5 hydroxyl attacks the C1 aldehyde.
Identify the anomeric carbon of a sugar.
Contrast the formation of alpha- versus beta-D-glucose.
Classify a sugar as pyranose or furanose based on its ring structure.

  • D-glucose转换为α-D-glucose和β-D-glucose的过程
D-glucose葡萄糖弯曲
进而转换为β-D-glucose和α-D-glucose两种结构
D-glucose通过Synthesis method 1转换为α-D-glucose;通过Synthesis method 2转换为β-D-glucose

2.6.3

Understand the reaction of monosaccharides to form disaccharides that contain a glycosidic linkage.
Categorize sugars as 'reducing' or 'non-reducing'.
Utilize carbon numbering to name glycosidic linkages.
Recognize the role of glycosidic linkages in the linear and branching structures of glycogen.
Explain the roles of glycogen synthase, glycogen branching enzyme, and glycogenin in glycogen synthesis.

2.6.4

Explain how to experimentally distinguish between reducing and non-reducing sugars.
Describe the reduction and oxidation of methylene blue.

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