Question

In a cell viability experiment, 10{,000 cells were cultured in the absence and presence of a compound \(Q\) for 24 h. The absorbance of a dye associated with cellular metabolic activity was measured at a wavelength of 570 nm at 24 h. The measured absorbances were \(0.8\) a.u. in the absence of \(Q\), and \(0.5\) a.u. in its presence. If the dye gives an absorbance (at 570 nm) of \(0.1\) a.u. in the absence of cells, what is the percentage cell growth inhibition caused by the compound \(Q\)? (Round off the answer to one decimal place.)}

Hint:

- Always correct for background/blank absorbance before computing viability. - \(%\) inhibition \(= \left(1-\dfrac{\text{treated}}{\text{control}}\right)\times 100\).

Answer 1

Correct Answer: 42

Step 1: Subtract the dye blank (0.1 a.u.) to obtain net absorbance due to cells.
Control (no \(Q\)): \(0.8-0.1=0.7\).
Treated (with \(Q\)): \(0.5-0.1=0.4\).
Step 2: Viability fraction \(=\dfrac{\text{treated}}
{\text{control}}=\dfrac{0.4}{0.7}=0.5714\).
Growth inhibition \(= (1-\text{viability})\times 100 = (1-0.5714)\times 100 = 42.86%\approx 42.9%\).