Protein P and ligand L were mixed at 27°C in a cuvette to the final concentrations of 10 μM and 12 μM, respectively. The

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Protein P and ligand L were mixed at 27°C in a cuvette to the final concentrations of 10 μM and 12 μM, respectively. The P and L molecules can form an adduct LP in a reversible reaction P+L PL. Optical spectroscopy was used to monitor the decrease in concentration of P during the course of the reaction as a function of time after the mixing, the data are tabulated as follows (NB time t=0 corresponds to the moment of mixing): Time (t), s [P]. μM 0 10 1 8 2 6.5 3 5.2 4 4.2 5 3.5 6 3 7 2.6 8 9 10 Answer the following questions: (a) Plot a graph of [P] as function of time (t) after the initiation of the reaction. [10%] (b) Considering the obvious relationships between P, L and PL at any time point t, P(t)+PL(t)=Po and L(t)+PL(t)=Lo, find the concentrations of L and PL at the equilibrium (note the equilibrium concentration of P achieved by 8 s is 2.5 μM). [20%] (c) Based on the initial concentrations Po and Lo and final (equilibrium) concentration of P after 8 s, describe how you expect the time dependences of [L] and [PL] should go with respect to plotted [P] on time. You might want to draw a sketch of the time courses for [L] and [PL] on the same graph plotted for [P]. [12%] (d) From the concentrations of P, L and PL at equilibrium, calculate the equilibrium constant for the reaction. [10%] (e) Using the same approach employed in (b), find the concentrations of L AL LAL 222 6555 2.5 2.5 2.5

(d) From the concentrations of P, L and PL at equilibrium, calculate the equilibrium constant for the reaction. [10%] Using the same approach employed in (b), find the concentrations of L and PL at the earliest experimental time point, that is at 1 s. Calculate mass action ratio I for that time point. [20%] Continued...

(f) Using the relationship AG = -2.3 RT log(K/T), calculate the Gibbs free energy difference (AG) that drives compound PL formation at the beginning of the reaction (1 s after mixing). Use the value for the gas constant R = 8.31 J mol-¹1 K-1 [12%] (g) Explain how the reaction is coming to equilibrium in terms of the change of the K/T ratio and AG. [16%]