lec08.docx

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Department
Biological Sciences
Course
BIOC12H3
Professor
Shelley A.Brunt
Semester
Fall

Description
lec08 mechanisms of enzyme action1 midterm covers lecture 5 to 8speed of different reactions 2 reactions catalyzed by enzymes3 late for this slideK M1 when the rate constant for product formation k is much smaller than either k or k 211a then the k can be neglected 2b K is then equivalent to kk M112 the lower the value of KMa the more tightly the substrate is bound in ES complex 3 K values are sometimes used todistinguish between enzymes that catalyze the same reactionM4 Concepta K is a measure of the stability of the ES complex Mb K is similar to the rate constant for the conversion of ES to EP when substrate is not catlimiting5 Late for this slideMeaning of K and kkcatcatm1 Late for this slide2 Grapha Region Bi Reaction is dependent on both concentration of enzyme E and concentration of substrate S ii Second order reaction 1 ie dependent on two variablesa verifyiii low concentration of substrate b region A i substrate is not limiting1 ie saturated ii enzyme is saturated with substrateability of enzymes to increase the rates of reactions that would be too slow to be useful without enzymes1 catalytic proficiency is the efficiency of the enzyme a divide the rate constants for the presence and absence of enzymesiabsence vs presence of enzyme1 Determines how well the enzyme actually catalyzes the reaction a ie rate of reaction2 came in late for this slide a 1021 am Measurement of Km and Vmax1 Plot a double reciprocal lineweaverburk plot to determine Vmax and Km a This plot represents the linear transformation of the michaelismenten equationi The key parameters are Km and Vmax ii Can calculate the Kcat from Vmax and absolute concentration of enzyme 1 Verify2 The position the linear transformation hits the x and y axis are defined characteristics of particular enzymesa Slope of line becomes KmVmax b This curve is important in calculating Km and Vmax i How c Lineweaverburk equation leads to the linear transformation of the michaelis menten equation3 Somehow began to talk about inhibitors ahow can you tell if it is a competitive or noncompetitive inhibitor i The Xintercept is Km The Y intercept is the Vmax1 Competitive Km is lowered actually raised because substrate competing with inhibitor for enzyme but Vmax isnt If concentration of substrate is high enough inhibitor wont matter Thus X intercept moves Y intercept doesnt 2 Noncompetitive Km doesnt change because the inhibitor doesnt affect how quickly it binds Vmax lowers because inhibitor prevents the inhibitor from being freed up and which lowers Vmax Thus X intercept doesnt move Y intercept does 4 Once you know the initial substrate and velocity of the data a You can determine the Km and Vmax b You can determine the Kcat if you know the substrate of the concentrationi Contradicting information1 Do you need to know the substrate concentration or do you need to know the absolute concentration of enzyme Kinetics of multisubstrate reactions1 More complicated than simple onesubstrate enzyme kinetics2 Can determine Km for each reaction in the presence of saturating amount of each substrate
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