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Unformatted text preview: Metabolic Biochemistry KEY Summer 2011 Homework 2 (40 pts.) 1) (8 pts.) The following data was collected from kinetic analysis performed on new enzyme: 5 0.2 22 0.045 [S] ( µ mol/L) 1/[S] V (( µ mol/L)min-1 ) 1/V 10 0.1 39 0.026 20 0.05 65 0.015 50 0.02 102 0.010 100 0.01 120 0.008 200 0.005 135 0.007 a. Prepare a Lineweaver-Burke plot from the data and estimate V max and K M (show your work) . You can do this either on graph paper or using a graphing program on a computer (MS Excel works fine). You must turn in either your plot on graph paper or a printout. b. If the total enzyme concentration is 1 nmol/L, and enough substrate has been added to achieve the maximum velocity for the reaction (V max ), how many molecules of substrate are converted to product by one molecule of enzyme in one second? (Show your work.) V max = k cat [E t ] turnover number = k cat = V max /[E t ] 162 x10-6 M/min = 1.62 x10 5 molecules/ min. = 2700 molecules/ sec. 1 x10-9 M 1/V 1/[S] 1/V max : Vmax = approx. 162 ( µ mol/L)min-1-1/K M : K M = approx. 32 µ mol/L 2) (8 pts.) Sketch a Michaelis_Menten plot (V vs. [S]—not a reciprocal plot) for an enzyme with and without an irreversible inhibitor. You can assume that the enzyme molecule is destroyed when it reacts with the inhibitor. Clearly label V max...
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This note was uploaded on 03/27/2012 for the course BIBC BIBC 102 taught by Professor Price during the Summer '08 term at UCSD.
- Summer '08