ProblemSet2_Solutions_SP09

# ProblemSet2_Solutions_SP09 - MCB 450 Problem Set 2 Answers...

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MCB 450 Problem Set 2 - Answers 1. From the following data on an enzymic reaction, determine (a) the type of inhibition, (b) K M for the substrate, (c) V max , and (d) K i , assuming that the inhibitor is present at a final concentration of 5mM in the reaction mixture. s v v inhib 1/s 1/v 1/v inhib 2 139 88 0.5 0.007194 0.011364 3 179 121 0.333333 0.005587 0.008264 4 213 149 0.25 0.004695 0.006711 10 313 257 0.1 0.003195 0.003891 15 370 313 0.066667 0.002703 0.003195 Using the graphing program at: http://www.changbioscience.com/stat/ec50.html we then make plots of 1/v vs. 1/s for both the uninhibited and inhibited cases, and get the following values: Inhibited: Results: Model: Linear a x + b a = 0.018798 ( +/- 1.015E-4 ) b = 0.001985 ( +/- 3.005E-5 ) Uninhibited: Results: Model: Linear a x + b a = 0.010269 ( +/- 2.154E-4 ) b = 0.002107 ( +/- 6.376E-5 ) Calculation of V max The y-intercept in each case is approximately 0.002. V max = 1/0.002 = 500 μ g/hr This is a reasonable answer if one eyeballs the numbers in the uninhibited case. Inhibition is competitive since plots intersect on the y-axis.

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Calculation of K M Slope of the uninhibited plot = K M /V max 0.010269 = K M /500 K M = 5.13 mM This is a reasonable answer. K M
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## This note was uploaded on 10/20/2010 for the course MCB 01 taught by Professor Mintel during the Summer '09 term at University of Illinois, Urbana Champaign.

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ProblemSet2_Solutions_SP09 - MCB 450 Problem Set 2 Answers...

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