Question

Ks is the dissociation constant

Note: Numbers without units are often meaningless. 1. The maximum possible rate for a reaction between two molecules occurs when every collision results in reaction. In this case the rate is predicted to be k = (472) (D2+D2)r12 with units of Mi's! Here N is Avagadro's number (6.023 x 1023 molecules/mole), r12 is the critical distance for reaction, and D, and D2 are the diffusion coefficients for the two participants. In a typical enzyme catalyzed reaction, r12 is about 10 cm, Denzyme = 5 x 10 cm²/s and Dsubstrate = 5 x 10 cm/s. What does this predict for the value of k? 2. For a Michaelis-Menten reaction of an enzyme with two different substrates having the values of the rate constants shown below, calculate the values of Ks and Km for each substrate. Also calculate the catalytic efficiency of the enzyme with each substrate. Does the enzyme approach catalytic perfection with either substrate? Substrate # ki, M's-1 k.1, 51 k2,5² Ks KM Cat. Eff. 1 7.0 x 10 1.0 x 103 2.0 x 104 3.0 x 10 8.0 x 105 4.0 x 104 Attach calculations.

Answer 1

Question 2.

For substrate 1, Ks = k-1/k1 = 10³/(7*10⁷) = 1.43*10^-5

KM = (k-1 + k2)/k1 = (10³ + 2*10⁴)/(7*10⁷) = 3*10^-4

Cat. Eff (kcat) = k2 = 2*10⁴ s^-1

For substrate 2, Ks = k-1/k1 = 8*10⁵/(3*10⁷) = 0.0267

KM = (k-1 + k2)/k1 = (8*10⁵ + 4*10⁴)/(3*10⁷) = 0.028

Cat. Eff (kcat) = k2 = 4*10⁴ s^-1