A cell has a surface area of $600 \mu m^2$ and a volume of $1000 \mu m^3$. If the cell grows such that its volume doubles while its shape remains the same, what is the new surface area-to-volume ratio?

The original surface area-to-volume ratio is $600/1000 = 0.6$. When volume doubles to $2000\,\mu m^3$, since surface area scales as volume$^{2/3}$, the new surface area $= 600 \times (2000/1000)^{2/3} = 600 \times 2^{2/3} = 600 \times 1.587 \approx 952\,\mu m^2$. The new ratio $= 952/2000 \approx 0.476 \approx 0.48:1$. The closest correct answer reflecting this calculation is approximately $0.48:1$, but selecting $0.38$ would be an error; the correct answer is B ($0.48:1$).