Question

Suppose the cells of a tumor are idealized as spheres each with a radius of \(5 \mu \mathrm{m}\) (micrometers). The number of cells has a doubling time of 35 days. Approximately how long will it take a single cell to grow into a multi- celled spherical tumor with a volume of \(0.5 \mathrm{cm}^{3}(1 \mathrm{cm}=10,000 \mu \mathrm{m}) ?\) Assume that the tumor spheres are tightly packed.

Short answer

Answer: Approximately 1015 days.

Step-by-step solution

Convert the volume of the tumor to micrometers³

First, convert 0.5 cm³ to micrometers³ using the given conversion (1 cm = 10,000 μm). Multiply the volume by the conversion factor cubed. 0.5 cm³ * (10,000 μm/cm)³ = 0.5 * 1,000,000,000,000 μm³ = 500,000,000,000 μm³

Calculate the volume of a single cell

The single cell is in the shape of a sphere with a radius of 5 μm. Use the formula for the volume of a sphere, which is V = (4/3)πr³, where r is the radius. V = (4/3)π(5 μm)³ = (4/3)π(125 μm³) = 500π/3 μm³

Calculate the number of cells needed

Divide the volume of the tumor by the volume of a single cell to find the total number of cells required to form a tumor with the given volume. Number of cells = 500,000,000,000 μm³ / (500π/3 μm³) ≈ 1,000,000,000/π ≈ 318,309,886

Determine the number of doublings needed

Find the number of times the initial cell needs to double to reach the total number of cells required. Use the formula n = log₂(N), where N is the number of cells and n is the number of doublings. In this case, N ≈ 318,309,886. n = log₂(318,309,886) ≈ 28.191 If we round up the value of n, we will get 29, as we cannot have a fraction of a doubling. So we need 29 doublings to reach the required number of cells.

Calculate the time taken for the necessary number of doublings

To find the total time taken for 29 doublings, multiply the number of doublings by the given doubling time, which is 35 days. Time taken = 29 * 35 days ≈ 1015 days Approximately 1015 days are needed for a single cell to grow into a multi-celled spherical tumor with a volume of 0.5 cm³, assuming that the tumor spheres are tightly packed.