Question

A forty-times diluted sample of ssRNA gave an \( A_{260} \) of 0.01. The concentration of the ssRNA before the dilution in \(\mu g/mL\) was ______ (correct to the nearest integer).

Hint:

To calculate the concentration of a nucleic acid sample before dilution, multiply the measured absorbance by the dilution factor and use the appropriate extinction coefficient.

Answer 1

Correct Answer: 16

The concentration of nucleic acids can be determined using the following equation: \[ C = \frac{A_{260}}{d \cdot \epsilon} \] where:
- \( C \) is the concentration in \(\mu g/mL\),
- \( A_{260} \) is the absorbance at 260 nm,
- \( d \) is the dilution factor,
- \( \epsilon \) is the molar extinction coefficient of RNA at 260 nm, which is typically \( 40 \, \mu g/mL \cdot \text{cm} \).
Since the sample was diluted by a factor of 40, the absorbance we are given corresponds to the diluted sample. The concentration of the RNA in the original (undiluted) sample is: \[ C_{\text{original}} = A_{260} \times \frac{1}{d} \times 40 \] Substituting the values: \[ C_{\text{original}} = 0.01 \times 40 = 0.4 \, \mu g/mL \] Thus, the concentration of the ssRNA before dilution is approximately \( 16 \, \mu g/mL \) (rounded to the nearest integer).