Question

An E. coli cell with an internal volume of 2 femtolitres contains 10 molecules of a repressor protein in its cytosol. The concentration of the repressor protein is ............ nM (round off to 1 decimal place)

Hint:

Concentration is calculated by dividing the number of molecules by the volume. Always ensure the units are consistent.

Answer 1

Correct Answer: 8.2

Step 1: Use the formula for concentration.
The concentration \( C \) of a substance is given by: \[ C = \frac{N}{V} \] where \( N \) is the number of molecules and \( V \) is the volume. Step 2: Convert units.
The number of molecules is \( N = 10 \) and the volume is \( V = 2 \, \text{femtolitres} = 2 \times 10^{-15} \, \text{L} \). We need the concentration in nanomolar (nM), where \( 1 \, \text{nM} = 10^{-9} \, \text{mol/L} \). So, the formula becomes: \[ C = \frac{10}{2 \times 10^{-15}} = 5 \times 10^{15} \, \text{mol/L} \] Step 3: Convert to nanomolar.
Now, convert to nanomolar: \[ C = 5.0 \, \text{nM} \] Step 4: Conclusion.
Thus, the concentration of the repressor protein is \( \boxed{5.0} \, \text{nM} \).