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IUPUI / Chemistry / CHEM 384 / What are the benefits offered by enzymatic catalysis?

What are the benefits offered by enzymatic catalysis?

What are the benefits offered by enzymatic catalysis?

Description

School: Indiana University Purdue University - Indianapolis
Department: Chemistry
Course: Fundamentals of Biochemistry
Professor: Blacklock
Term: Spring 2016
Tags: biochemistry and Biochemistry 1
Cost: 50
Name: Biochemistry Exam 2 Study Guide
Description: Study material for Exam 2 of Biochemistry at IUPUI
Uploaded: 03/02/2018
11 Pages 5 Views 11 Unlocks
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Biochemistry Exam 2 Study:


What are the benefits offered by enzymatic catalysis?



Lecture 8-12 Expectations:

Slide 42 Expectations

1.) Explain benefits offered by enzymatic catalysis

a.) Increased reaction rates compared to uncatalyzed and chemically catalyzed reactions

b.) Milder reaction conditions

c.) Greater reaction specificity with respect to substrates and products

d.) Capacity for regulation

2.) Define…

a.) Apoenzyme: Cofactor that was dissociated

b.) Apoprotein: Helps form a molecule (such as an enzyme) with a prosthetic group c.) Holoenzyme: Enzyme with cofactor

d.) Active site: Region on an enzyme that binds to a protein or other substance during a reaction

e.) Cofactor: Small molecules/metal ions that are required by some enzymes to catalyze reaction, the site at which the chemistry occurs

f.) Coenzyme: Organic cofactors. May associate transiently or be tightly linked. Required as carriers of specific atoms or functional groups


Helps form a molecule with a prosthetic group.



g.) Substrate: The substance on which an enzyme acts

h.) Reaction intermediate: formed from the reactants and reacts further to give products. Detectable

i.) Rate-limiting step: The one with the highest energy barrier (smallest ‘k’) 3.) Differentiate between △G, △G​o​, △G’​o​, △G‡ If you want to learn more check out who is james madison?

a.) △G: Change in free energy

b.) △Go: Standard free energy change. pH of 0, standard conditions

c.) △G’o:Biochemical standard free-energy change. pH of 7.

i.) How likely that a reaction will happen. Thermodynamics

d.) △G‡: Activation energy

i.) How fast it will happen. Kinetics

4.) Explain how enzymes accelerate reactions through their effects on activation energy

a.) Enzymes lower the activation energy (making it faster) by binding energy and binding the transition state

5.) Explain how binding is used to overcome activation energy barrier...and what factors contribute to a high activation energy barrier


What is monosaccharide?



a.) Enzymes bind to the transition state to lower the activation barrier

b.) Reaction intermediates can contribute to high activation energy barrier 6.) Differentiate between general acid-base catalysis and specific acid-base catalysis

a.) General acid-base catalysis: Charge intermediates are not stable and can decay to reactants. Seek to neutralize charges by donating or abstracting protons We also discuss several other topics like what is the absence of government called?

b.) Specific acid-base catalysis: When the donation/abstraction is carried out by water itself

7.) Describe covalent catalysis and metal-ion catalysis

a.) Covalent Catalysis: occurs frequently with general acid-base catalysis i.) Covalent E-S Complex forms

ii.) Results in Different Reaction Pathway than uncatalyzed reaction iii.) As long as steps involved in forming and breaking covalent bonds are faster than uncatalyzed reaction then catalysis will occur If you want to learn more check out What does consumer surplus mean?

b.) Metal-Ion Catalysis: Metals can orient substrates

i.) Metals can contribute binding energy

ii.) Metals can change redox state

8.) Apply your understanding of each of these concepts to new problems

Slide 61 Expectations

1.) Know and be able to use the Michaelis-Menten and double-reciprocal equations

a.) Michaelis Menten: Reaction is zero-order with respect to [S] when [S] >> [E]

i.) Second step is rate-limiting

ii.) Steady state assumption: d[ES]/dt = ~0

iii.) View slide 48 and Michaelis notes from Blacklock

b.) Double-reciprocal: Plots help to easily find Km and Vmax

i.) Slope: Km/Vmax

ii.) See slide 49 If you want to learn more check out What are the Efferent Pathways of the Autonomic Nervous System?
Don't forget about the age old question of what is Ebionites?

2.) Understand the derivation of the M-M equation and assumptions made therein

a.) Slide 48

b.) Notes on Canvas by Blacklock on M-M equation

3.) Understand how the M-M equation is reminiscent of the equation for theta a.) View Blacklock Notes

4.) Be able to find Vmax and Km from Vo vs. [S] and double-reciprocal plots a.) Slides 48, 49

5.) Know the meaning of Km and how it relates to affinity and a KD a.) Km: Concentration of substrate [S] that leads to an initial velocity Vo that is half the maximal velocity (Vmax).

b.) It is a measure of affinity. Not an affinity constant, but gives an idea of affinity Don't forget about the age old question of What are the four P's in marketing mix?

i.) Meaning depends on complexity of enzymatic mechanism ii.) The more complex the mechanism, the more complicated the definition of Km

c.) As a dissociation constant for ES, Km gets smaller, the affinity of the enzyme for its substrate becomes greater

d.) Relation to KD 

i.) Km: Measures how the “affinity” of an enzyme relates to/affects its substrate (or ligand)

ii.) KD: Measures how “tight” an enzyme binds to substrate

e.) Slide 53

6.) Understand what turn-over number, catalytic efficiency, and specificity constants tell us and be able to describe enzymes based on these parameters

a.) Turn-Over number: kcat. # molecules of substrate converted to product by 1 enzyme molecule in 1 second (s^-1)

i.) Rate limiting step at saturation

b.) Catalytic efficiency: 

c.) Specificity constants: Tells how efficiently an enzyme catalyzes the reaction in each encounter between E and S. It is a second-order constant i.) Slide 56

7.) Use your knowledge of catalytic perfection, feedback inhibition, ternary complex and ping-pong mechanisms, and allostery to interpret an enzyme’s activity and its regulation

a.) Catalytic perfection: Achieved when reaction rate is diffusion-controled i.) If Kcat/Km approaches the “diffusion limit”, the enzyme is so efficient that it carries out catalysis every time a collision occurs between E and S

ii.) Slide 57

b.) Feedback Inhibition 

c.) Ternary Complex: Enzyme that has two or more substrates and binds to them in different modes

i.) Random Order: Order does not matter

ii.) Ordered: Order is important

d.) Ping-Pong Mechanisms: No ternary complex is formed

i.) Enzyme binds to Substrate forming product. P1 leaves, S2 enters. (“ping pongs” between substrate and product entering and leaving enzyme)

ii.) Slides 59-60

e.) Allostery 

i.) Binding effector molecules to areas other than the active site

8.) Be able to apply your knowledge to new problems

Slide 104 Expectations

1.) Explain how an enzyme might use acid/base, covalent, preferential transition state binding and metal ion catalytic strategies

2.) Describe the major features of chymotrypsin, including its catalytic triad and role of each residue

3.) Explain the catalytic mechanism of chymotrypsin including describing how it is a two-step mechanism

4.) From a given starting structure, draw steps of chymotrypsin reaction mechanism using curved arrow notation

5.) Understand how pre-steady state kinetics can contribute to what we know about enzyme mechanism

6.) Define and identify burst phase, oxyanion hole, and transition state analog 7.) Differentiate between a tetrahedral intermediate and an acyl-enzyme intermediate

8.) Describe the differences in how HIV protease and chymotrypsin catalyze the exact same reaction (peptide bond hydrolysis for example)

9.) Describe how the HIV protease catalyzes peptide bond hydrolysis 10.) Describe the roles of divalent metal ions in the catalytic mechanism of enolase

11.) Identify the key features of a beta-lactam antibiotic, what makes it effective, and what reaction/enzyme it targets

12.) Describe how bacteria overcome beta-lactam antibiotics

13.) Describe why clavulanic acid is an effective drug

Lecture 118 Expectations

1.) Define homo- and heterotropic allosteric modulator, feedback inhibition, proenzyme zymogen, kinase, phosphatase, motif

2.) Explain how an enzyme can be allosterically regulated

3.) Describe the regulation of aspartate transcarbarmoylase

4.) Rationalize the sigmoidal shape of a rate vs. substrate concentration curve for an allosterically-regulated curve

5.) Describe how covalent modification can be used to regulate the function of an enzyme

6.) Describe the usefulness of phosphorylation in enzymatic regulation and how glycogen phosphorylase is regulated in response to phosphorylation 7.) Describe how cleavage can be used to activate an enzyme

Lecture 13 Expectations:

1.) Define…

a.) Monosaccharide:​ Simple Sugar.

i.) Aldehydes and Ketones

ii.) 3+ carbons

iii.) Single Unit

b.) Disaccharide: ​Two monosaccharide residues

c.) Oligosaccharide: ​Short chains of monosaccharides

i.) Includes Disaccharides

d.) Aldose:​ Carbohydrate that is an aldehyde

e.) Ketose:​ Carbohydrate that is a ketone

2.) Name and Draw boxed linear monosaccharides in D-configurations a.) D-Aldoses

i.) D-Glyceraldehyde

ii.) D-Erythrose

iii.) D-Ribose

iv.) D-Arabinose

v.) D-Xylose

vi.) D-Glucose

vii.) D-Mannose

viii.) D-Galactose

b.) D-Ketoses

i.) Dihydroxyacetone

ii.) D-Erythrulose

iii.) D-Ribulose

iv.) D-Xylulose

v.) D-Fructose

3.) Draw pentoses and hexoses in correct cyclic forms

a.) Furanoses: 5-membered sugar ring

b.) Pyranoses: 6-membered sugar ring

c.) See slides 19,20, and 24 to see how to draw

4.) Identify L vs. D configuration

a.) Draw Fischer with C=O at top

b.) Look at farthest chiral center from C=O

i.) If -OH is on right, sugar is D

ii.) If -OH is on left, sugar is L

5.) Define…

a.) Epimer​: Differ in configuration in exactly one chiral center

b.) Diastereomer​: NOT mirror images.

i.) Differ in configuration in at least one chiral center, but not all! c.) Anomer​: Cyclic epimers

d.) Enantiomer​: Non super-imposable mirror images.

i.) Differ configuration at EVERY chiral center

ii.) Identical chemical and physical properties

6.) How do Diastereomers and Enantiomers differ?

a.) Diastereomers differ at one or more chiral centers (But NOT all) b.) Enantiomers differ at EVERY chiral center

7.) Explain the difference between reducing vs. nonreducing sugars and determine if a given sugar is reducing or non-reducing

a.) Reducing: Can reduce one form to another

i.) Ex. Aldehyde can reduce Ag+ to Ag0 or Cu2+ to Cu+ 

ii.) Can isomerize to aldose via enediol intermediate

iii.) Aldosesand Ketoses are reducing

b.) Non-reducing:

i.) No free anomeric carbon

ii.) Two acetals/ketals

iii.) No hemiacetal to open up into the linear form

8.) Know structures of lactose, maltose, and sucrose

a.) View slide 35

9.) Draw disaccharide when given full formal name, or short-form name (only pentoses or hexoses boxed earlier)

a.) View slide 35-39

10.) Understand why polysaccharides do not have defined molecular weight a.) No template so no start-stop codons

b.) Length determined by enzymatic activity

11.) Describe the structures of glycogen, starch, cellulose, and chitin a.) Glycogen:

i.) Branched homopolysaccharide of glucose

ii.) (α1 → 4) linked chains

iii.) Branched with (α1 → 6) linkers every 8-12 residues

iv.) Main storage polysaccharide in animals

b.) Starch

i.) Mixture of two homopolysaccharide of glucose

ii.) Amylose: Unbranched. (α1 → 4) linked residues

iii.) Amylopectin: branched. Branch-points with (α1 → 6) linkers occur every 24-30 residues

(1) Huge

iv.) Main storage homopolysaccharide in plants

c.) Cellulose

i.) Glucose polymers linked via (β1 → 4)

ii.) Hydrogen bonds

iii.) Great tensile strength

d.) Chitin

i.) Linear, unbranched (β1 → 4) linked chains

ii.) N-acetylglucosamine rather than glucose

iii.) Tough structure

12.) Describe the structural difference between amylose and cellulose a.) Amylose: Unbranched (α1 → 4) linked residues

b.) Cellulose: (β1 → 4)

c.) Rotation about glycosidic bonds makes different conformations possible “phi” and “psi”

d.) View slides 48 and 52 and 54

13.) Define…

a.) Glycosaminoglycan: Anionic polysaccharide chains

i.) Uronic acid + Amino sugar

ii.) Can be extensively sulfate

b.) Proteoglycan: Sulfated glycosaminoglycans + large core protein i.) Mostly carbohydrate by mass (>80-90%)

ii.) Syndecans (proteoglycans) bind to Integrins (receptors for proteoglycans)

c.) Glycolipid: A lipid with covalently bound oligosaccharide

i.) Blood groups

ii.) Lipopolysaccharides

d.) Glycoprotein: Proteins with covalently attached carbohydrates i.) More protein than sugar

14.) Describe two types of glycosylation in a glycoprotein a.) O-linked glycosylation...O-glycosidic bond

i.) Anomeric carbon of sugar + OH of Ser or Thr

ii.) Occurs in Golgi only

iii.) No consensus

b.) N-linked glycosylation...N-glycosyl bond:

i.) Anomeric carbon of sugar + NH2 group of Asn

ii.) Occurs in ER and Golgi

iii.) NX(S/T) consensus

Lecture 14 Expectations

1.) Understand how the first and second laws of thermodynamics apply to living systems

a.) Living Organisms…

i.) Cannot create energy from nothing

ii.) Cannot destroy energy into nothing

iii.) May only transform energy from one form to another

(1) In the process of transforming energy, living organisms must

increase the entropy of the universe by extracting usable 

energy and releasing useless energy (heat) 

2.) Understand the meaning of equilibrium constant K’eq and how it is related to △G’​o 

a.) K’eq = [Product]eq/[Reactants]eq

i.) Lowest energy = more stability

b.) △G’° = -RTln(K’eq)

c.) Equilibrium does not mean Products = Reactants or K’eq = 1

i.) They tell what happens to the reaction when all components start at 1 M

3.) Understand what positive, negative, and 0 values of △G’​o​ imply a.) △G’o relates to the stability of the reactants and products

b.) △G’o >0 means S is more stable than P

c.) △G’o <0 means P is more stable than S

d.) △G’o = 0 means equilibrium is reached.

i.) Products = Reactants and K’eq =1 ONLY at all standard conditions (1 M) 

4.) Understand the difference between △G and △G’°, and how the two are related mathematically…similarly, understand the difference between K’eq and Q

a.) △G = △G’° + RT ln (Q)

i.) Q: “Reaction Quotient”. Products/REactants

(1) Unlike K’eq, it gives the actual conditions in the cell

b.) △G’° tells where you are going. (reactions tend to go towards equilibrium); Equilibrium position

c.) △G = How “Far” from equilibrium you are

d.) △G’° = -RTln(K’eq)

5.) Understand how a reaction can proceed forward even if △G’° > 0

a.) Not all reactions in a cell are exergonic. Some reactions require energy to go forward

b.) To overcome: Need RT ln (Q) to be large and negative

i.) Q << 1

ii.) [Products] << [Reactants]

iii.) One way cells ensure reactions go “forward” is by keeping Q << 1 for reactions that are not thermodynamically favorable by

PRODUCT REMOVAL

c.) Metabolism can “couple” favorable reactions to unfavorable reactions if Q can’t compensate for △G’° > 0

6.) Understand why coupling two chemical reactions together (one thermodynamically favorable, and the other one unfavorable) works a.) Favorable reactions are coupled to unfavorable reactions. Favorable reactions have better solvation, charge separation, and resonance stabilization of products to be able to bind to unfavorable reactions. 7.) Understand why ATP hydrolysis is an exergonic reaction…as is the hydrolysis of other phosphorylated compounds

a.) Electrostatic repulsion within the reactant molecule is relieved b.) Products (phosphate) are stabilized via resonance, or by more favorable solvation

c.) Product (pyruvate) undergoes further tautomerization

d.) Strongly negative △G’°

e.) ATP concentration is usually far above equilibrium concentration 8.) Understand why under cellular conditions, ATP hydrolysis is even more favorable than it would be under standard conditions

a.) Cellular ATP concentration is usually far above the equilibrium concentration

9.) Understand why it’s good that hydrolysis of ATP does not have the highest △G’° a.)

10.) Understand why thioester hydrolysis is even more energetically favorable than oxyester hydrolysis

a.) Thioesters don’t undergo resonance like oxyesters so they start out at a much higher free energy level. Thus, the △G for hydrolysis is larger because you start higher

11.) Describe what is meant by “group transfer” and how this relates to the way ATP is actually used by the cell to drive various processes a.) The pathways from reactants to products don’t matter. The free energy change for that transformation is still the same.

b.) Group Transfer: One or more groups of atoms are transferred from a molecule to another.

i.) Put on a good leaving group

12.) Identify the types of group transfer reactions ATP can participate in a.) Phosphoryl transfer: Oxygen of nucleophile (R--18O) attacks gamma position. Bridge oxygen of product is labeled ADP indicating group transferred from ATP is phosphoryl, not phosphate.

b.) Pyrophosphoryl transfer: Attack on beta position displaces AMP and leads to the transfer of a pyrophosphoryl (not pyrophosphate) group to the nucleophile

c.) Adenylyl transfer: Attack on alpha position displaces PPi and transfers the adenylyl group to nucleophile

d.) See slide 46 in Lecture 14

13.) Define oxidation, reduction, emf, redox couple, reduction potential, endergonic, exergonic

a.) Oxidation: Gaining of bonds to oxygen

b.) Reduction: Gaining of Hydrogen

c.) EMF:Electromotive Force. Proportional to the difference in affinity. d.) Redox couple: Oxidation reactions always occur together with reduction reactions (oxidizing agent accepts electrons from reducing agent) e.) Reduction Potential: A measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Unit: Volts

f.) Endergonic: Heat/Energy absorbed

g.) Exergonic: Heat/Energy released

14.) Identify four types of biological redox reactions

a.) Electrons transferred…

i.) Directly as electrons

ii.) As hydrogen atoms

iii.) As hydride ions

iv.) When something combines directly with oxygen

15.) Understand how standard reduction potentials are determined a.) Determined by seeing which reaction (1 or 2) is more likely

i.) Higher affinity for electrons (reference to test cell): E° >0. Forward reaction.

ii.) Lower affinity for electrons than H+ (test cell to reference cell): E° <0. Reaction proceeds in reverse.

16.) Know the difference between E and E’°

a.) E: E = E’° + RT/nF ln [X]/[X-]

b.) E’°: Biochemical standard conditions. pH 7.0.

17.) Carry out calculations to determine △G, △E, etc (please do assigned sample problems for practice)

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