Question

In the figure below ‘A’ shows a cell in the normal state and ‘B’ shows the same cell after leaving it in a certain solution for a few minutes.

(i) Describe the change which has occurred in the cell as seen in B.
(ii) Give the technical term for the condition of the cell as reached in B and as it was in A.
(iii) Define the process which led to this condition.
(iv) What was the solution-isotonic, hypotonic or hypertonic, in which the cell was kept?
(v) How can the cell in B, be brought back to its original condition?
(vi) Name the parts numbered 1 to 3.

Answer 1

(i) Exosmosis (exit of water) has resulted in the shrinkage of the protoplasm of cell B.
(ii) Plasmolysis and deplasmolysis.
(iii) When plant cells are kept in a hypertonic solution, exosmosis takes place. This process is called plasmolysis.
(iv) Hypertonic solution.
(v) The cell in B can be brought back to its original condition by placing it in a drop of distilled water.
(vi) 1. Cell wall, 2. Plasma membrane, 3. Chloroplast.