Transcript
- Slide 1
- Chapter 6 Protein Function : Enzymes Part 1
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- Enzymes Physiological significance of enzymes Catalytic power of enzymes Chemical mechanisms of catalysis Mechanism of chymotrypsin Description of enzyme kinetics and inhibition Enzyme Learning Goals: to Know
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- Enzymes Mostly Proteins (a few RNAs are capable of catalysis) Active Site: Substrate Binding + Reaction Products Some require Cofactors (metals) or Coenzymes (organic cpds) Some enzymes have other binding sitesinvolved in regulation, we will see later, Part 2 EOC Problem 1 involves the sweetness of corn affected by corn enzymes and Problem 2 calculates the average molar concentration of enzymes in a bacterial cell: you can take it further to find the number of molecules of each enzyme present in a cell.
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- First Cell Free Prep First to Crystallize Urease Weak bonding at active site results in catalysis Enzyme Pioneers
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- Why biocatalysis over inorganic catalysts? Greater reaction specificity: avoids side products Milder reaction conditions: conducive to conditions in cells Higher reaction rates: in a biologically useful timeframe Capacity for regulation: control of biological pathways Metabolites have many potential pathways of decomposition Enzymes make the desired one most favorable EOC Problem 4: Examines the thermal protection of hexokinase that a substrate brings to the table: maintaining conformation under harsh conditions. Later in Part 2 of this chapter we will see X ray data backing this up.
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- Enzymatic Substrate Selectivity: Phenylalanine hydroxylase No binding Binding but no reaction
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- Class Is the First Part of E.C. Number EC 2.7.1.1 = ATP:glucose phosphotransferase or Hexokinase 2 = Transferase 7 = Phosphotransferase 1 = Transferred to a hydroxyl 1 = Glucose is the acceptor
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- Enzyme Search By Class FMNH 2 + O 2 + RCHO FMN + RCOOH + H 2 O + light Bacterial Luciferase Rxn
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- Continuing with the EC Numbers-1
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- Continuing with EC Numbers-2
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- NiceZyme
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- Enzyme with an Active Site Chymotrypsin Active Site
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- Thermodynamics of a Reaction S + E ES E + P
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- Enzyme Catalyzed Reaction E + S ES EP E + P
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- EOC Problem 3: A rate enhancement problem using Urease, the enzyme that converts: Urea CO 2 + 2 NH 3. The calculation demonstrates how long it would take if urease were not present !
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- Dihydrofolate Reductase Substrate Binds in a Fold or Pocket G B = binding energy Folic Acid NADP + +
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- Enzyme Reactions Bind Substrate then Change Shape to Transition State
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- Triose Phosphate Isomerase Terribly Slow rate with Glyceraldehydephosphate important in stabilizing binding.
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- Rate Enhancement Due to Proximity (Entropy Reduction)
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- Acid/Base Catalysis
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- Catalytic Mechanisms acid-base catalysis: give and take protons covalent catalysis: change reaction paths metal ion catalysis: use redox cofactors, pK a shifters electrostatic catalysis: preferential interactions with Transition State.
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- Acid Base Catalysis Involve Proteins R groups
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- Formation of a Covalent Intermediate
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- Michaelis Menten Curve
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- Michaelis Menten Equation: V max [S] K m + [S] v o = L. Michaelis and Miss Maud L. Menten. 1913. "Die Kinetik der Invertinwerkung" Biochemische Zeitschrift Vol. 49. Invertase Reaction: sucrose + H 2 O glucose + fructose
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- Michaelis Menten Experiment Measure Rate (v) at several concentrations of Substrate (S) Here is one tube with one beginning concentration of S Calculate [S]/min or [P]/min. S P E This enzyme, triosephosphate isomerase is a one substrate, one product enzyme.
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- Michaelis Menten Experiment: Real Data At each [S], the concentration of enzyme is exactly the SAME. Calculate [S]/min for each [S] EOC Problem 6 is about using 340nm light to measure dehydrogenase reactionsthe classic lactate dehydrogenase. Do this at more concentrations of S to get a larger data set used for
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- Initial Velocites are the Dashed Line A
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- Michaelis Menten Plot
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- Michaelis Menten Equation is Non-Linear Straightened Out by reciprocalsLineweaver Burke Equation: 1/v o = (K M /V max )(1/[S]) + 1/V max the Equation of a Straight Line y = mx + b Thus, y = 1/v o, x = 1/[S] and m (the slope) = K M /V max Lets Plot this Outnext slide V max [S] K M + [S] v o =
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- Lineweaver-Burke Plot Double Reciprocal Origin is Zero Data Points are in this range
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- There Are Other Equations to Convert the Michaelis Menten Equation to a Straight Line Eadie Hofstie v = -Km(v/S) + V max Hanes Wolf: S/v = (1/V max )(S) + Km/V max all are y = mx + b
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- K M = is an Intrinsic Property of an enzyme What does this mean? Intrinsic vs Extrinsic?
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- V max is an Extrinsic Property of Enzymes At a high [S], varying only the enzyme conc :
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- k cat comes from V max and [Enz] V max is [molar]/sec [Enz] in molar To get an Intrinsic Catalytic Constant from V max k cat = V max / [Enz]
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- k cat /K m
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- Calculation of Km and Vmax The enzyme, Practicase Studentose Productate Studentose, mM velocity, moles/ml/sec 112 220 429 835 1240 Assay volume = 1 ml/tube Whats in the tube: buffer + enzyme, then add substrate at time Zero. EOC Problems 11(dead easy to do by inspection) and 13 to do by Lineweaver Burke plot
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- Calculation of Km and Vmax Studentose, mM 1/[S] Velocity, 1/v moles/ml/sec 1 112 0.083 2 0.520 0.050 4 0.2529 0.034 8 0.12535 0.029 12 0.08340 0.025 Now Plot this on Lineweaver Burk Plot.remember Zero is near the middle of the graph!
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- Lineweaver Burke Plot of Practicase 1 1/
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- Practicase k cat = an Intrinsic Property In the enzyme assay (one ml), each tube had 10 g of practicase. The molecular weight of practicase is 20,000 D. Thus, each tube had 10 g / 20,000 g/mole = 0.0005 mole practicase k cat = V max / [Enz] = (50 mole/sec)/ 0.0005 mole = 1 x 10 5 s -1 Thus one enzyme reaction takes 1/ 1x 10 5 s -1 = 10 -5 sec or 10 sec.
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- What is Wrong with this L-B graph?
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- Things to Know and Do Before Class 1.Principles of catalysis. 2.Enzyme naming concepts. 3.Intrinsic and Extrinsic values of Enzyme kinetics. 4.Be able to do a Michaelis Menten graph. 5.Be able to do a Lineweaver Burke graph. 6.To do EOC problems 1-6, 11, 13.