Question

Water potential of a flaccid cell will be

• A. ${\Psi }_w={\Psi }_s$
• B. ${\Psi }_s={\Psi }_p$
• C. ${\Psi }_w=0$
• D. ${\Psi }_w={\Psi }_s-{\Psi }_p$

Answer 1

The correct option is A ${\Psi }_w={\Psi }_s$

If plant cell happens to be bathed in hypertonic solution, it loses water through the process of exosmosis. The loss of water is first from cytoplasm and then central vacuole. As a result, the protoplast is reduced in size. This decreases turgor pressure or pressure potential $\left({\Psi }_p\right)$ and corresponding wall pressure. Solute potential becomes slightly more negative due to loss of water. The cell attains a minimum size when turgor pressure is zero/pressure potential $\left({\Psi }_p\right)$ is zero.
Therefore, ${\Psi }_w={\Psi }_s+{\Psi }_p$
or ${\Psi }_w={\Psi }_s$(as ${\Psi }_p=0$)

A cell which is deficient in turgor is called flaccid.

So, the correct answer is '${\Psi }_w={\Psi }_s$'.