An E. coli cell of volume
10−12 cm3 contains
60 molecules of lac-repressor. The repressor has a binding affinity
(Kd) of 10−8 M and
10−9 M with and without lactose respectively, in the medium.
Q.52 The molar concentration of the repressor in the cell is
Step-by-Step Solution
Step 1: Convert Cell Volume to Liters
Given cell volume = 10-12 cm3
1 cm3 = 10-3 L
Therefore,
10-12 cm3 = 10-15 L
Step 2: Convert Number of Molecules to Moles
Avogadro’s number = 6.022 × 1023 molecules/mol
Moles of repressor =
60 / (6.022 × 1023) ≈ 1 × 10-22 mol
Step 3: Calculate Molar Concentration
Concentration (M) = moles / volume
= (1 × 10-22) / (1 × 10-15)
= 1 × 10-7 M
Step 4: Convert to Nanomolar (nM)
1 × 10-7 M = 100 nM
Correct Answer
(D) 100 nM
