Exam2KEY - 1 Biochemistry I (CH 339K) Fall 2011 Unique...

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1 Biochemistry I (CH 339K) Fall 2011 Unique #51580 Professor Adrian Keatinge-Clay Exam #2 October 5, 2011 Do not begin work on this exam until you are instructed to begin. The core values of The University of Texas at Austin are learning, discovery, freedom, leadership, individual opportunity, and responsibility. Each member of the University is expected to uphold these values through integrity, honesty, trust, fairness, and respect toward peers and community. The work that you hand in will be considered to be your own original work prepared without assistance. I will uphold the honor code on this exam. Signed Name: KEY Printed Name: _________________________ EID: _______________
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2 Equations: Δ G’º = -RTlnK eq = RTlogK d 6 logK d kJ/mol R = 8.31 Jmol -1 K -1 θ = [L]/([L]+K d ) V 0 = V max [S]/([S]+K m ) Show all work. 1. In the absence of ATP, myosin heads bind strongly to actin filaments (K d = 0.1 nM). a) What is the change in free energy ( Δ G’º) for myosin binding to actin in the absence of ATP (assume 37 °C)? (5 points) Δ G’° = 6 log(K d ) kJ/mol [2 pts] 6 log(0.1·10 -9 M) kJ/mol = -60 kJ/mol [3 pts] Alternatively, Δ G’° = RT ln(K d ) = (8.315 kJ·mol -1 ·K -1 )(310 K) ln(0.1·10 -9 M) = -60 kJ/mol b) When ATP γ S (a non-hydrolyzable analog of ATP) is supplied, myosin heads bind weakly to actin filaments (K d = 100 μ M). If the concentration of actin binding sites was 20 μ M, what fraction of myosin heads would be bound to actin in the presence of ATP γ S? (5 points) θ = [L]/([L] + K d ) [2 pts] θ = (20 μ M)/(20 μ M + 100 μ M) = 0.17 or 17% bound [3 pts] 2. Each of three enzymes (A, B, and C) possesses a catalytic efficiency of 10 8 M -1 s -1 . The K m of A for its substrate is 10 mM; the K m of B for its substrate is 100 μ M; the K m of C for its substrate is 1 μ M. a) Which enzyme has the largest turnover number (k cat )? What is the value of that k cat ? (5 points) Catalytic efficiency = k cat /K m = 10 8 M -1 s -1 k cat (A) = k cat /K m · K m = 10 8 M -1 s -1 · 10·10 -3 M = 1·10 6 s -1 k cat (B) = 10 8 M -1 s -1 · 100·10 -6 M = 1·10 4 s -1 k cat (C) = 10
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This note was uploaded on 02/06/2012 for the course BIOCHEMIST 339K taught by Professor Keating-clay during the Spring '11 term at University of Texas at Austin.

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Exam2KEY - 1 Biochemistry I (CH 339K) Fall 2011 Unique...

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