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Flashcards in Lab 8 Deck (6)
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1

Did you observe any difference in the isosmotic threshold between the NaCl solution and the sucrose solution? Discuss your answer

There was a clear threshold between NaCl and the sucrose solution. The isometric threshold for the sucrose was between 0.4-0.5M and the threshold for NaCl was between 0.2-0.1M. This makes sense as salt normally has half the isosmotic threshold that sucrose does. This is because the salt is a stronger hypertonic solution. It works at a lower concentration to cause plasmolysis in the cell so you don’t need as much to induce the same effect.

2

If we were to use a solution of CaCl2, how would the isosmotic threshold of this solution compare to NaCl and sucrose? Explain your answer.

Using CaCl2, the isometric threshold would go down as you need a smaller concentration of the salt solution to have plasmolysis. This happens because CaCL2 will split into its constituent ions Ca2+ +2Cl-) and therefore cause an increase in the number of solute particles by three fold (instead of just doubling the number of particle like NACL does)

3

Define the following terms: (1 mark each)
i) Hypotonic
ii) Hypertonic
iii) Deplasmolysis
iv) Plasmolysis

i) hypotonic: a solution with lower solute concentration then the solution it’s being compared to.
ii) hypertonic: a solution with a higher solute concentration then the solution it’s being compared to.
iii) deplasmolysis: reversal of plasmolysis. When a living cell swells up due to inward flow of water into the cell. (or after time where slots from the surrounding hypertonic solution will begin to diffuse across the memb)
iv) plasmolysis: shrinking of the cytoplasm away from the wall of a living cell due to outward osmotic flow of water

4

Explain how you could determine more precisely the isosmotic threshold for red onion vacuoles

After establishing the concentration for the isosmotic threshold, you could prepare dilutions that are within that range. For example if the isosmotic threshold was between 0.4-0.5M, create dilutions at 0.41, 0.42,….0.49 and see where the precise point the change occurred.

5

Polyethylene glycol (PEG) is a compound of very high molecular weight and does not enter the plasma membrane of the cells. You are asked by your boss
to carry out an experiment in which pine cultured cells must experience water stress during a period of 50 days. On your lab bench you have a bottle of sucrose and a bottle of PEG. Which compound would you choose to add into the medium for your experiment? Defend your answer.

I would choose to add PEG to the experiment because it does not enter the plasma membrane of the cells. This however does not mean that water cannot exit the cell. By choosing PEG, the cells will experience water stress due to the water leaving the cell by diffusion and traveling to the PEG solution which is lower in water. This works well because the experiment is run for a lengthly time of 50 days and if sucrose was chosen, it might defuse into the cell and alter the cell beyond just water stress, which is not the objective of the experiment.

6

Describe, in a few sentences, one mechanism adopted by cells to help battle water loss if exposed to hypertonic solutions.

A way for the cell to fight against water loss due to hypertonic solutions is to wait for deplasmolysis to occur. By diffusing some of the ions into the cell, the cell itself becomes a hypertonic solution with a low concentration of water. This allows the water to flow back to the cell by diffusion and the cell will regain its turgid sate and gain back the water it initially lost.