Answer to Question #254239 in General Chemistry for atenz

Question

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

free+P⇌k1,k2 bound, where k1=7.6×10^6M^−1s^−1 and k2=1s^−1. What is the ratio of free to bound receptors at equilibrium in a 3 μM solution of protein? Give the result to 3 

significant figures.

Answer 1

free+P⇌k1,k2

P= $7.6×10^6×100=7.6×10^8$

P=$1×100=1.0×10^2$