PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Exam II will be on material covered inLecture 8 (Lipids & Membranes)throughLecture 13 (Allostery & Hemoglobin)Tonight 7-9 p.m.(for room assignments, see web site)MCB 450 Review Session for Exam IIExam 2 Sp15 Rev-1~30-33 Multiple choice Qs:~1/3 @ 6 pts, including calculations~2/3 @ 4 ptsExam II FormatThings that have come up• Lipid-related questions: what to know• Gs and what they mean• Enzyme kinetics: Km, kcat,kcat/Km, ki• Enzyme inhibition• Chymotrypsin mechanismburst kineticsExam 2 Sp15 Rev-2Fatty acid chain lengths & degree of unsaturation:Know the C16:0, C18:2(9,12) etc. nomenclature and correspondingn-hexadeanoate, cis,cis-9,12-octadecadienoate nomenclatureTm trends & chain length & degree of unsaturationKnow 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, PSSlides 8-6, 8-7, 8-10Slides 8-13, 8-23, 8-24Slides 8-25, 8-26Slides 8-27, 8-28, 8-29Slide 8-24Exam 2 Sp15 Rev-3Predicting spontaneity of a reaction from GFor reaction A + B C + D G = 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) G ≠ 0: Reaction will spontaneously proceed to a state of lower G(i.e. towards equilibrium)….G < 0: Reaction will proceed spontaneouslyin the forward () directionG > 0: Reaction will proceed spontaneouslyin the reverse () direction∆Greaction = ∆Gproducts - ∆GreactantsExam 2 Sp15 Rev-4∆G = change in free energy, = energy available to do work in a chemical reactionG Gibbs free energy∆G0 = free energy for reaction under standard conditions1M 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) conditionscan be modified by [A] and [B], also temperatureReaction A + B C + D Exam 2 Sp15 Rev-5DON'T ASSUMESTANDARD TEMPCONDITIONS∆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) conditionscan be modified by [A] and [B], also temperature[C][D][A][B]G°' = -RT ln EQUILIBRIUM CONSTANT UNDER STANDARDCONDITIONS DEFINED AS K'eq G°' = -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 Gin living cellsIntroduce a NEW TERM, G',that can be modified by [A] & [B]….“GENERAL” OR “TRUE” GWHEN THE SYSTEM IS AWAY FROM EQUILIBRIUM(AS IT USUALLY IS IN A LIVING CELL)[C][D][A][B] G' = G°' + 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-eqTo make G' < 0,G' = - RT ln Keq + RT ln Knon-eqneeds to be < Keq[REACTANTS] NEEDS TO BE > [PRODUCTS], MAKINGlogs of #s <1 are -vei.e. FAVORABLE[C][D][A][B]< 1How to make G' favorableG°' = -RT ln K'eq [C][D][A][B] G' = G°' + RT ln CELLULAR, NON-EQ. CONDITIONSExam 2 Sp15 Rev-8G' in cells can be made favorable (-ve)by making [reactants, A & B] >> [products C & D]and G' in cells can be -ve even if G°' is +veHow to make G' favorableIN BIOCHEMICAL REACTIONS IN METABOLIC PATHWAYS,PRODUCTS OFTEN REMOVED, &REACTANTS KEEP COMING,SO [REACTANTS] >> [PRODUCTS], KEEPING G -VEExam 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, G0’, 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 G’? 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/molE. - 1.11 kJ/molG' = G°’+ [RT lnKnon-eq]G' = G°’+ RT ln[PEP][2-PGP](0.09 M)(0.068 M)G' = 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/molSample G' calculationExam 2 Sp15 Rev-10Equations to rememberActual G in living cells:[C][D][A][B]G°' = -RT ln EQUILIBRIUM CONSTANT UNDER BIOCHEMICAL STANDARD CONDITIONS DEFINED AS K'eq G°' = -RT ln K'eq [C][D][A][B] G' = G°' + RT ln CELLULAR, NON-EQ. CONDITIONSExam 2 Sp15 Rev-11V0, initial velocity, depends on [S]LINEARRANGE Exam 2 Sp15 Rev-12Reflects formation ofES complex and fillingof all catalytic sites with SAll enzyme moleculesare reacting as fast asthey canEffect of [substrate] on V0 of enzyme-catalyzed reaction Approaches “saturation”rectangularhyperbolaExam 2 Sp15 Rev-13The Michaelis-Menten equationMICHAELIS CONSTANT Km =COMBINED RATE CONSTANT:Exam 2 Sp15 Rev-14Km on the V0 versus [S] plotExam 2 Sp15 Rev-15At [S] >> Km, Km term in denominatorof M-M equation becomesinsignificant, and equationsimplifies to:plateau athigh [S]Michaelis-Menten eq. fits experimental observationsV0 =Vmax [S]Km + [S]At [S] << Km, [S] is an insignificantaddition to Km, so M-Mequation simplifies to:….and V0 has lineardependence on [S]……..as seenexperimentally 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 FOREFFECTIVE CATALYSIS TO OCCURExam 2 Sp15 Rev-18V0 =Vmax [S]Km + [S]Exam 2 Sp15 Rev-19Experimental determinationof Km and Vmax:Determine Vo for reactionat a range of [substrate]Plot 1 Vo [S]1 versus From intercepts,calculate Vmaxand KmDetermination of Km and Vmax from double reciprocal plot YOU MAY BE GIVEN x & yINTERCEPT VALUES INmM OR mM-1. MULTIPLY1/x AXIS INTERCEPT BY -1Exam 2 Sp15 Rev-20OF THE AFFINITY Exam 2Sp15Rev-21kcat : rate constant for the rate-limiting stepin an enzyme-catalyzed reactionTotal amount of enzyme = EtFor, rate constant for limiting step when enzyme is saturated with S is k2Knowing Vmax and
View Full Document
Unlocking...