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Answers Test 1

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This document contains questions about protein function and enzyme kinetics calculations. It asks the reader to calculate a dissociation rate constant from given association rate and dissociation constant values. It also asks the reader to calculate the antigen concentration needed for 20% binding based on a given dissociation constant. Regarding enzyme kinetics, it asks the reader to calculate the Michaelis constant using the maximum reaction rate, enzyme concentration, substrate concentration, and known catalytic rate constant.

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Protein Function: Question 1. Protein A binds to protein B with an association rate of 8.9 x 103 M-1s-1 and an overall dissociation constant (Kd) of 10 nM. Calculate the dissociation rate 虜d. Ka = 虜a / 虜d, Kd = 1/Ka, so Kd = 虜d / 虜a 10nM = 虜d / 8.9 x 103M-1s-1 d = 8.9 x 10-5s-1 Question 2: An antibody binds to an antigen with a Kd of 5 x 10-8 M. At what concentration of antigen will 慮 be 0.2? = [L] / [L] + Kd ([L] + Kd) = [L] Kd = [L] + [L] (Kd)/ (1- ) = [L] Solving for [L], [L] = 0.2(5 x 108M)/(1-0.2)) = 0.2(6 x 10-8M) = 1.2 x 10-8M
Enzyme Kinetics Question 1: 虜cat for enzyme A is 600 s-1. Experimentally, when [E] = 20nM, and [S] = 40 袖M, V0 = 9.6 袖Ms-1. Calculate Km for this enzyme. Use Equation V0 = Kcat[Et][S] / Km + [S] 9.6 袖Ms-1 = (600s-1)(20袖M)(40袖M) / Km + 40袖M Km + 40 袖M = 480 袖M2s-1 / 9.6 袖Ms-1 Km + 40 袖M = 50 袖M Km = 10 袖M Question 2 we went over in class.

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Answers Test 1

  • 1. Protein Function: Question 1. Protein A binds to protein B with an association rate of 8.9 x 103 M-1s-1 and an overall dissociation constant (Kd) of 10 nM. Calculate the dissociation rate 虜d. Ka = 虜a / 虜d, Kd = 1/Ka, so Kd = 虜d / 虜a 10nM = 虜d / 8.9 x 103M-1s-1 d = 8.9 x 10-5s-1 Question 2: An antibody binds to an antigen with a Kd of 5 x 10-8 M. At what concentration of antigen will 慮 be 0.2? = [L] / [L] + Kd ([L] + Kd) = [L] Kd = [L] + [L] (Kd)/ (1- ) = [L] Solving for [L], [L] = 0.2(5 x 108M)/(1-0.2)) = 0.2(6 x 10-8M) = 1.2 x 10-8M
  • 2. Enzyme Kinetics Question 1: 虜cat for enzyme A is 600 s-1. Experimentally, when [E] = 20nM, and [S] = 40 袖M, V0 = 9.6 袖Ms-1. Calculate Km for this enzyme. Use Equation V0 = Kcat[Et][S] / Km + [S] 9.6 袖Ms-1 = (600s-1)(20袖M)(40袖M) / Km + 40袖M Km + 40 袖M = 480 袖M2s-1 / 9.6 袖Ms-1 Km + 40 袖M = 50 袖M Km = 10 袖M Question 2 we went over in class.