Question

suppose that a tissue engineer wishes to reconstitute an organ that is a cube of side 5 cm. he/she does back of the envelope calculations to get an idea of cell numbers and length scales. using standard estimates of cell densities and length scales of human tissues and cells, calculate the upper and lower limits of the total number of cells that should go into this organ. for each limit, name one organ that would have that approximate cell density

Answer 1

The spherical cell with a diameter of 5 µm would exchange nutrients and wastes more efficiently compared to the cubed-shaped cell with a side length of 7 μm. This is because the spherical cell has a higher surface area-to-volume ratio.

Step-by-step explanation:

The cell with a spherical shape and a diameter of 5 µm would likely exchange nutrients and wastes with its environment more efficiently compared to the cubed-shaped cell with a side length of 7 μm. This is because the surface area-to-volume ratio of a spherical cell is higher than that of a cubed-shaped cell, which allows for more efficient diffusion of substances across the cell membrane.

To quantitatively justify this, we can calculate the surface area-to-volume ratios for both cells:

For the spherical cell with a diameter of 5 µm:

Surface Area = 4π(2.5 µm)^2 ≈ 78.54 µm^2

Volume = (4/3)π(2.5 µm)^3 ≈ 65.45 µm^3

Surface Area-to-Volume Ratio = 78.54 µm^2 / 65.45 µm^3 ≈ 1.20 µm^-1

For the cubed-shaped cell with a side length of 7 μm:

Surface Area = 6(7 μm)^2 = 294 μm^2

Volume = (7 μm)^3 = 343 μm^3

Surface Area-to-Volume Ratio = 294 μm^2 / 343 μm^3 ≈ 0.86 μm^-1

Therefore, the spherical cell with a diameter of 5 µm has a higher surface area-to-volume ratio (1.20 μm^-1) compared to the cubed-shaped cell with a side length of 7 μm (0.86 μm^-1), indicating that it would exchange nutrients and wastes more efficiently.