Answer to Question #253884 in Organic Chemistry for atenz

Question

A protein P binds to receptors on the surface of a cell according to the rule

V\text{free}+\text{P} \underset{k_2}\overset{k_1}\rightleftharpoons}}\text{bound},\]

where \(k_1 = 7.6 \times 10^6 M^{-1}s{-1}\) and \(k_2 = 1 54-1}\). What is the ratio of free

to bound receptors at equilibrium in a 3 \(\mu MV) solution of protein? Give the result to 3

significant figures.

Answer 1

7.6× 10⁶/156

= 4.9×10⁴