Question

A cell with volume 10^-12 cm^3 contains 60 molecules of Lac repressor. the binding affinity of repressr is 10^-8 M (with lactose) and 10^-9 M (without lactose). the molar conc of repressor in cell is
A. 0.1nM B.1nM C. 10 nM D.100 nM . pl help

Answer 1

how is "the binding affinity of repressr is 10^-8 M (with lactose) and 10^-9 M (without lactose)" relevant to the question?

Molar concentration = moles of solute/L of solution

Vcell=10^-12 mL to L -> Vcell= 10^-15 L

60 molecules of Lac repressor

molariity = [60/(6.022*10^(23))]/(10^-15 L )
=0.0000000996346728661574227831285287279973= 0.99 nM = 1 nM