Exam 2 Sp15 Rev 1 MCB 450 Review Session for Exam II Exam II will be on material covered in Lecture 8 Lipids Membranes through Lecture 13 Allostery Hemoglobin Exam II Format 30 33 Multiple choice Qs 1 3 6 pts including calculations 2 3 4 pts Tonight 7 9 p m for room assignments see web site Exam 2 Sp15 Rev 2 Things 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 mechanism burst kinetics Exam 2 Sp15 Rev 3 Fatty acid chain lengths degree of unsaturation Know the C16 0 C18 2 9 12 etc nomenclature and corresponding n hexadeanoate cis cis 9 12 octadecadienoate nomenclature Slides 8 6 8 7 8 10 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 Slide 8 24 Know your headgroups PE PC PS Slides 8 13 8 23 8 24 Slides 8 25 8 26 Slides 8 27 8 28 8 29 Exam 2 Sp15 Rev 4 Predicting spontaneity of a reaction from G For reaction A B C D Greaction Gproducts Greactants 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 spontaneously in the forward direction G 0 Reaction will proceed spontaneously in the reverse direction Exam 2 Sp15 Rev 5 G Gibbs free energy G change in free energy energy available to do work in a chemical reaction 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 6 G0 biochemical standard free energy change at pH 7 298K concentration of water 1 G RT ln i e C D A B EQUILIBRIUM CONSTANT UNDER STANDARD CONDITIONS DEFINED AS K eq G RT ln K eq R Gas constant EXPRESSED IN kJ mol 8 315 J mol 1 K 1 Kelvin 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 DON T ASSUME STANDARD TEMP CONDITIONS Exam 2 Sp15 Rev 7 Actual concentrations of A B can modify G in living cells Introduce a NEW TERM G that can be modified by A B MASTER EQUATION G GENERAL G RT ln AT EQUILIB C D A B Knon eq GENERAL GENERAL OR TRUE G WHEN THE SYSTEM IS AWAY FROM EQUILIBRIUM AS IT USUALLY IS IN A LIVING CELL Exam 2 Sp15 Rev 8 How to make G favorable G G RT ln C D CELLULAR NON EQ CONDITIONS A B G RT ln K eq G RT ln Keq RT ln Knon eq To make G 0 C non eq D non eq A non eq B non eq needs to be Keq i e FAVORABLE REACTANTS NEEDS TO BE PRODUCTS MAKING C D A B 1 logs of s 1 are ve Exam 2 Sp15 Rev 9 How to make G favorable G 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 ve IN BIOCHEMICAL REACTIONS IN METABOLIC PATHWAYS PRODUCTS OFTEN REMOVED REACTANTS KEEP COMING SO REACTANTS PRODUCTS KEEPING G VE Exam 2 Sp15 Rev 10 Sample G calculation The 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 B C D E 5 08 kJ mol 2 49 kJ mol 1 11 kJ mol 2 49 kJ mol 1 11 kJ mol G G RT lnKnon eq G G RT ln PEP 2 PGP G 1 8 kJ mol RT ln 0 09 M 0 068 M 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 Exam 2 Sp15 Rev 11 Equations to remember Actual G in living cells G G RT ln C D A B G RT ln K eq G RT ln C D A B EQUILIBRIUM CONSTANT UNDER BIOCHEMICAL STANDARD CONDITIONS DEFINED AS K eq CELLULAR NON EQ CONDITIONS Exam 2 Sp15 Rev 12 V0 initial velocity depends on S LINEAR RANGE Exam 2 Sp15 Rev 13 Effect of substrate on V0 of enzyme catalyzed reaction rectangular hyperbola Approaches saturation Reflects formation of ES complex and filling of all catalytic sites with S All enzyme molecules are reacting as fast as they can Exam 2 Sp15 Rev 14 The Michaelis Menten equation MICHAELIS CONSTANT Km COMBINED RATE CONSTANT Exam 2 Sp15 Rev 15 Km on the V0 versus S plot Exam 2 Sp15 Rev 16 Michaelis Menten eq fits experimental observations At S Km S is an insignificant addition to Km so M M equation simplifies to V0 Vmax S At S Km Km S Km term in denominator of M M equation becomes insignificant and equation simplifies to and V0 has linear dependence on S as seen experimentally plateau at high S Exam 2 Sp15 Rev 17 Exam 2 Sp15 Rev 18 Small 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 19 V0 Vmax S Km S Exam 2 Sp15 Rev 20 Determination of Km and Vmax from double reciprocal plot Experimental determination of Km and Vmax Determine Vo for reaction at a range of substrate Plot 1 1 versus Vo S From intercepts calculate Vmax and Km YOU MAY BE GIVEN x y INTERCEPT VALUES IN mM OR mM 1 MULTIPLY 1 x AXIS INTERCEPT BY 1 Exam 2 Sp15 Rev 21 OF THE AFFINITY Exam 2 Sp15 Rev 22 kcat 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 So Vmax k2 Et making Vmax dependent on Et 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 Rearranging k2 Vmax Et the TURNOVER NUMBER kcat Limiting rate of any enzyme catalyzed reaction at saturation …
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