Exam II will be on material covered in Lecture 8 (Lipids & Membranes) through Lecture 13 (Allostery & Hemoglobin) Tonight 7-9 p.m. (for room assignments, see web site) MCB 450 Review Session for Exam II Exam 2 Sp15 Rev-1 ~30-33 Multiple choice Qs: ~1/3 @ 6 pts, including calculations ~2/3 @ 4 pts Exam II FormatThings that have come up • Lipid-related questions: what to know • ΔGs and what they mean • Enzyme kinetics: Km, kcat,kcat/Km, ki • Enzyme inhibition • Chymotrypsin mechanism burst kinetics Exam 2 Sp15 Rev-2Fatty acid chain lengths & degree of unsaturation: Know the C16:0, C18:2(Δ9,12) etc. nomenclature and corresponding n-hexadeanoate, cis,cis-Δ9,Δ12-octadecadienoate nomenclature Tm trends & chain length & degree of unsaturation Know and recognize the types of lipid: fatty acyl-containing: di- and triacyglycerol sphingolipids (no ester linkages) sterols (cholesterol & related molecules) Know your headgroups: PE, PC, PS Slides 8-6, 8-7, 8-10 Slides 8-13, 8-23, 8-24 Slides 8-25, 8-26 Slides 8-27, 8-28, 8-29 Slide 8-24 Exam 2 Sp15 Rev-3Predicting spontaneity of a reaction from ΔG For reaction A + B C + D ΔG = 0: Reaction is at equilibrium and there is no net change in the concentrations of reactants and products (no net flow in the forward or reverse directions) ΔG ≠ 0: Reaction will spontaneously proceed to a state of lower G (i.e. towards equilibrium)…. ΔG < 0: Reaction will proceed spontaneously in the forward (→) direction ΔG > 0: Reaction will proceed spontaneously in the reverse (←) direction ∆Greaction = ∆Gproducts - ∆Greactants Exam 2 Sp15 Rev-4∆G = change in free energy, = energy available to do work in a chemical reaction G Gibbs free energy ∆G0 = free energy for reaction under standard conditions 1M concentrations of A, B, C, and D, 298K ∆G0' = biochemical standard free energy change at pH 7, 298K (concentration of water = 1) ∆G' = free energy change at pH 7, under cellular (non-equilibrium) conditions can be modified by [A] and [B], also temperature Reaction A + B C + D Exam 2 Sp15 Rev-5DON'T ASSUME STANDARD TEMP CONDITIONS ∆G0' = biochemical standard free energy change at pH 7, 298K (concentration of water = 1) ∆G' = free energy change at pH 7, under cellular (non-equilibrium) conditions can be modified by [A] and [B], also temperature [C][D] [A][B] ΔG°' = -RT ln EQUILIBRIUM CONSTANT UNDER STANDARD CONDITIONS DEFINED AS K'eq ΔG°' = -RT ln K'eq i.e. EXPRESSED IN kJ/mol R (Gas constant) = 8.315 J mol-1 K-1 (Kelvin) Reaction A + B C + D Exam 2 Sp15 Rev-6Actual concentrations of A & B can modify ΔG in living cells Introduce a NEW TERM, ΔG', that can be modified by [A] & [B]…. “GENERAL” OR “TRUE”ΔG WHEN THE SYSTEM IS AWAY FROM EQUILIBRIUM (AS IT USUALLY IS IN A LIVING CELL) [C][D] [A][B] ΔG' = ΔG°' + RT ln "AT EQUILIB." "GENERAL" "GENERAL" MASTER EQUATION: Knon-eq Exam 2 Sp15 Rev-7[C]non-eq[D]non-eq [A]non-eq[B]non-eq To make ΔG' < 0, ΔG' = - RT ln Keq + RT ln Knon-eq needs to be < Keq [REACTANTS] NEEDS TO BE > [PRODUCTS], MAKING logs of #s <1 are -ve i.e. FAVORABLE [C][D] [A][B] < 1 How to make ΔG' favorable ΔG°' = -RT ln K'eq [C][D] [A][B] ΔG' = ΔG°' + RT ln CELLULAR, NON-EQ. CONDITIONS Exam 2 Sp15 Rev-8$ΔG' in cells can be made favorable (-ve) by making [reactants, A & B] >> [products C & D] and ΔG' in cells can be -ve even if ΔG°' is +ve How to make ΔG' favorable IN BIOCHEMICAL REACTIONS IN METABOLIC PATHWAYS, PRODUCTS OFTEN REMOVED, & REACTANTS KEEP COMING, SO [REACTANTS] >> [PRODUCTS], KEEPING ΔG -VE Exam 2 Sp15 Rev-9The enzyme enolase catalyzes the conversion of 2-phosphoglycerate (2-PG) to phosphoenolpyruvate (PEP) + H2O. The biochemical standard free energy change for this reaction, ΔG0’, is + 1.80 kJ/mol. If the concentration of 2-PG is 68 mM and the concentration of PEP is 90 mM, which of answers A-E is closest to ΔG’? Assume the temperature is 25°C. A. - 5.08 kJ/mol B. - 2.49 kJ/mol C. + 1.11 kJ/mol D. + 2.49 kJ/mol E. - 1.11 kJ/mol ΔG' != ΔG°ʼ+ [RT lnKnon-eq]!ΔG'!= ΔG°ʼ+ RT ln![PEP]![2-PGP]!(0.09 M)!(0.068 M)!ΔG'!= 1.8 kJ/mol + RT ln!= 1.8 kJ/mol + (8.315 J/mol x 298) ln 1.324!= 1.8 kJ/mol + (2.478 kJ x +0.28)!= 1.8 kJ/mol + (0.694 kJ/mol) = +2.494 kJ/mol!Sample ΔG' calculation Exam 2 Sp15 Rev-10Equations to remember Actual ΔG in living cells: [C][D] [A][B] ΔG°' = -RT ln EQUILIBRIUM CONSTANT UNDER BIOCHEMICAL STANDARD CONDITIONS DEFINED AS K'eq ΔG°' = -RT ln K'eq [C][D] [A][B] ΔG' = ΔG°' + RT ln CELLULAR, NON-EQ. CONDITIONS Exam 2 Sp15 Rev-11V0, initial velocity, depends on [S] LINEAR RANGE !Exam 2 Sp15 Rev-12Reflects formation of ES complex and filling of all catalytic sites with S All enzyme molecules are reacting as fast as they can Effect of [substrate] on V0 of enzyme-catalyzed reaction Approaches “saturation” rectangular hyperbola Exam 2 Sp15 Rev-13The Michaelis-Menten equation MICHAELIS CONSTANT Km = COMBINED RATE CONSTANT: Exam 2 Sp15 Rev-14Km on the V0 versus [S] plot Exam 2 Sp15 Rev-15At [S] >> Km, Km term in denominator of M-M equation becomes insignificant, and equation simplifies to: plateau at high [S] Michaelis-Menten eq. fits experimental observations V0 = Vmax [S] Km + [S] At [S] << Km, [S] is an insignificant addition to Km, so M-M equation simplifies to: ….and V0 has linear dependence on [S]… …..as seen experimentally Exam 2 Sp15 Rev-16Exam 2 Sp15 Rev-17Small Km means 1/2 maximal V achieved at relatively low [S] Large Km means 1/2 maximal V achieved at relatively high [S] IS A MEASURE OF THE [S] REQUIRED FOR EFFECTIVE CATALYSIS TO OCCUR Exam 2 Sp15 Rev-18V0 = Vmax [S] Km + [S] Exam 2 Sp15 Rev-19Experimental determination of Km and Vmax: Determine Vo for reaction at a range of [substrate] Plot 1 Vo [S] 1 versus From intercepts, calculate Vmax and Km Determination of Km and Vmax from double reciprocal plot YOU MAY BE GIVEN x & y!INTERCEPT VALUES IN mM OR mM-1. MULTIPLY 1/x AXIS INTERCEPT BY -1 Exam 2 Sp15 Rev-20OF THE AFFINITY Exam 2 Sp15 Rev-21kcat : rate constant for the rate-limiting step in an enzyme-catalyzed reaction Total amount of enzyme = Et For , rate constant for limiting step when enzyme is saturated with S is k2 Knowing Vmax and [Et], we have the TURNOVER NUMBER of the enzyme, = # of substrate molecules an enzyme can convert into product per unit time when the enzyme is saturated with substrate. So,
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