Compute the fraction of inhibited proteins when the concentration is (D)_ =10, m

Question

Compute the fraction of inhibited proteins when the concentration is (D)_ =10, mu M and K_d = 3.0 times 10^-7 Y = [PD]_eq/C_p = % The concentration of folded proteins during a temperature jump relaxation experiment is described by the relation: [F] (t) = [F]_eq + [F] (0) -[F]_eq | e^-(k+k_1)t where tau_relax = 1/(k_ + k_f) = 3.00 s [F] (0) = 2.00 mu M is the initial concentration of folded proteins and [F]_eq = 3.00 mu M is the equilibrium concentration of folded proteins. (a) Compute the concentration of proteins at t = 2.00 s and (b) compute the equilibrium constant for folding if the total protein concentrate proteins at t 3.10 M.

Explanation / Answer

8) Disassociation constant kd is given by the formula,

kd = concentration of products / concentration of reactants

3 x 10-7 = 1 x 10-6 M / concentration of reactants

Therefore the concentration of reactants or total proteins,

= 1 x 10-6 / 3 x 10-7

= 3.3 M

So, the concentration of inhibited proteins = 3.3 M - 1 x 10-6 M

= 3.29 M

The fraction of inhibited proteins = 3.29/3.3

= 0.997

= 99.7%

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