Chapter 6 Protein Function : Enzymes Part 1

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

Enzymes Mostly Proteins (a few RNA’s are capable of catalysis) Active Site: Substrate Binding + Reaction  Products Some require Cofactors (metals) or Coenzymes (organic cpds) Some enzymes have other binding sites…involved 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.

First Cell Free Prep First to Crystallize Urease Weak bonding at active site results in catalysis Enzyme Pioneers

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.

Enzymatic Substrate Selectivity: Phenylalanine hydroxylase No binding Binding but no reaction

Class Is the First Part of E.C. Number EC = ATP:glucose phosphotransferase or Hexokinase 2 = Transferase 7 = Phosphotransferase 1 = Transferred to a hydroxyl 1 = Glucose is the acceptor

Enzyme Search By Class FMNH 2 + O 2 + RCHO  FMN + RCOOH + H 2 O + light Bacterial Luciferase Rxn

Continuing with the EC Numbers-1

Continuing with EC Numbers-2

NiceZyme

Enzyme with an Active Site Chymotrypsin Active Site

Thermodynamics of a Reaction S + E  ES  E + P

Enzyme Catalyzed Reaction E + S ↔ ES ↔ EP ↔ E + P

EOC Problem 3: A rate enhancement problem using Urease, the enzyme that converts: Urea  CO NH 3. The calculation demonstrates how long it would take if urease were not present !

Dihydrofolate Reductase Substrate Binds in a Fold or Pocket ΔG B = binding energy Folic Acid NADP +  +

Enzyme Reactions Bind Substrate then Change Shape to Transition State

Triose Phosphate Isomerase Terribly Slow rate with Glyceraldehyde…phosphate important in stabilizing binding.

Rate Enhancement Due to Proximity (Entropy Reduction)

Acid/Base Catalysis

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.

Acid Base Catalysis – Involve Proteins R groups

Formation of a Covalent Intermediate

Michaelis Menten Curve

Michaelis Menten Equation: V max [S] K m + [S] v o = L. Michaelis and Miss Maud L. Menten "Die Kinetik der Invertinwerkung" Biochemische Zeitschrift Vol. 49. Invertase Reaction: sucrose + H 2 O  glucose + fructose

Michaelis Menten Experiment Measure Rate (v) at several concentrations of Substrate (S) Here is one tube with one beginning concentration of S Calculate Δ[S]/min or Δ[P]/min. S  P E This enzyme, triosephosphate isomerase is a one substrate, one product enzyme.

Michaelis Menten Experiment: Real Data At each [S], the concentration of enzyme is exactly the SAME. Calculate Δ[S]/min for each [S] EOC Problem 6 is about using 340nm light to measure dehydrogenase reactions…the classic lactate dehydrogenase. Do this at more concentrations of S to get a larger data set used for 

Initial Velocites are the Dashed Line A

Michaelis Menten Plot

Michaelis Menten Equation is Non-Linear Straightened Out by reciprocals…Lineweaver 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 Out…next slide V max [S] K M + [S] v o =

Lineweaver-Burke Plot Double Reciprocal Origin is Zero Data Points are in this range

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

K M = is an Intrinsic Property of an enzyme What does this mean? Intrinsic vs Extrinsic?

V max is an Extrinsic Property of Enzymes At a high [S], varying only the enzyme conc :

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]

k cat /K m

Calculation of Km and Vmax The enzyme, Practicase Studentose  Productate Studentose, mM velocity, μmoles/ml/sec Assay volume = 1 ml/tube What’s 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

Calculation of Km and Vmax Studentose, mM 1/[S] Velocity, 1/v μmoles/ml/sec Now Plot this on Lineweaver Burk Plot….remember Zero is near the middle of the graph!

Lineweaver Burke Plot of Practicase 1 1/

Practicase k cat = an Intrinsic Property In the enzyme assay (one ml), each tube had 10 μg of practicase. The molecular weight of practicase is 20,000 D. Thus, each tube had 10 μg / 20,000 μg/μmole = μmole practicase k cat = V max / [Enz] = (50 μmole/sec)/ μmole = 1 x 10 5 s -1 Thus one enzyme reaction takes 1/ 1x 10 5 s -1 = sec or 10 μ sec.

What is Wrong with this L-B graph?

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.