Stuck on a water potential question

Hi, in question 3a)iii of this paper, I don't really understand why the answer is this:
different concentration/solute/ water potential of contents; requires different concentration of external salts/water potential, for movement of water/ to burst the cell
https://castleschoolbiology.wikispaces.com/file/view/BY1%20Jan%202011.pdf/217009756/BY1%20Jan%202011.pdf

I thought haemolysis occured when 100% of the cells have bursted... Can anyone explain?
Thank you.

Reply 1

coconut64

OP

16

https://castleschoolbiology.wikispaces.com/file/view/BY1%20June%202012.pdf/363744046/BY1%20June%202012.pdf

This is a different paper, in 6b, how do you know that you should be talking about the second test tube and not the third test tube. The answer is this :
bathing solution {has a lower water potential / is more concentrated /is more negative / hypertonic} than the water potential of beetrootcells / ORA;water leaves / moved {out of / from} cells / into bathing solution;bathing solution became less dense / lighter than original sucrosesolution

The last point mentions the bathing solution being lighter but didn't water move into the bathing solution, so surely it should be heavier?The mark scheme mentions REJECT reference to water moving into or out of the drop. But it is a drop in the third test tube though.. How does that work? I don't get this question, any help would be appreciated. Thanks!

(edited 8 years ago)

Reply 2

zombiejon

9

Original post by coconut64

Hi, in question 3a)iii of this paper, I don't really understand why the answer is this:different concentration/solute/ water potential of contents; requires different concentration of external salts/water potential, for movement of water/ to burst the cellhttps://castleschoolbiology.wikispaces.com/file/view/BY1%20Jan%202011.pdf/217009756/BY1%20Jan%202011.pdfI thought haemolysis occured when 100% of the cells have bursted... Can anyone explain?Thank you.

Haemolysis refers to the mechanism in which a red blood cell died, not the entire population. In this case, it would be lysis/bursting.

The range of 3.3 - 4.7 g/dm^3 probably is because some cells burst, and released their cellular contents into the solution. The additional cellular contents would end up changing the salt concentration of the solution.

Reply 3

coconut64

OP

16

Original post by zombiejon

Haemolysis refers to the mechanism in which a red blood cell died, not the entire population. In this case, it would be lysis/bursting.

The range of 3.3 - 4.7 g/dm^3 probably is because some cells burst, and released their cellular contents into the solution. The additional cellular contents would end up changing the salt concentration of the solution.

Oh okay, so this causes which water potential to change? The water potential in the solution?

different concentration/solute/ water potential of contents; requires different concentration of external salts/water potential, What does this part of the answer actually mean. I thought it is just water molecules moving down the water potenital gradient.. Thanks

Reply 4

zombiejon

9

Original post by coconut64

Oh okay, so this causes which water potential to change? The water potential in the solution?

different concentration/solute/ water potential of contents; requires different concentration of external salts/water potential, What does this part of the answer actually mean. I thought it is just water molecules moving down the water potenital gradient.. Thanks

Normally the water potential would be at a fixed point. However, when a cell lyses, everything inside the cell comes out, and joins the solution. This would then increase the solute concentration, and decrease the water potential.

An analogy would be four rooms arranged around a central room. Let's say...20 people in each room, making it crowded. The central room already has 16 people inside.

     _
   _|1|_
  |2|_|3|
    |4|

We'll also assume everyone has an identical need for a personal space bubble. So when the door to Room 1 (R1) opens, 2 people come flooding out into the central room, allowing everyone an identical bubble for personal space. For this example, it will be a density of 18 people per room.

Now Room 2's door opens. Since Room 2 (R2) is still more crowded (20 ppl per room) than the central room + Room 1 (18 ppl per room), 1 person leaves R2 until there is an equal amount of space for everyone in Central+R1+R2 (density now around 18.7 ppl per room, or 56/3).

R3's door opens. People also come out, but not as many as when R1 or R2 opened. The final density is now 19 ppl per room.

So when R4's door opens, no one comes out because the overall density of people per room is not going to change at all.

Reply 5

zombiejon

9

Original post by coconut64

https://castleschoolbiology.wikispaces.com/file/view/BY1%20June%202012.pdf/363744046/BY1%20June%202012.pdf

This is a different paper, in 6b, how do you know that you should be talking about the second test tube and not the third test tube. The answer is this :
bathing solution {has a lower water potential / is more concentrated /is more negative / hypertonic} than the water potential of beetrootcells / ORA;water leaves / moved {out of / from} cells / into bathing solution;bathing solution became less dense / lighter than original sucrosesolution

The last point mentions the bathing solution being lighter but didn't water move into the bathing solution, so surely it should be heavier?The mark scheme mentions REJECT reference to water moving into or out of the drop. But it is a drop in the third test tube though.. How does that work? I don't get this question, any help would be appreciated. Thanks!

If in doubt, draw out a flow diagram, and what you already know. And it's an application question, based on a mix of chemistry (fluid density/partial pressure) and biology.

You have a sucrose solution at a known concentration (Tube 1).
Some of the solution (2mL) from Tube 1 is then used bathe beetroot discs, in a tube we shall label Tube 2.

Sample from Tube 2 is placed in Tube 1, and the direction/rate of movement is measured.
So we know that bathing solution can be hyper/iso/hypotonic to the beetroot cubes.
If hypertonic, water flows from the beetroot cubes into the solution.
If isotonic, there's no change.
If hypotonic, water goes from the solution into the beetroot cubes.

Overall, when a droplet rises when placed into another solution, it is because it is less dense than the surrounding solution. This is akin to pouring oil on top of water - oil always goes to the top because it is less dense. As such, the converse should be true - a sinking droplet indicates a higher density.

So this tells us one thing - the density/concentration changed. The sucrose concentration of the bathing solution, although 0.6M initially, was no longer at 0.6M. From 6a, we should know that 0.4M is probably isotonic to the beetroot since the droplet did not move. If 0.4M is isotonic, 0.6M has to be hypertonic to the beetroot disc

Since a hypertonic solution results in a net flow of water from the beetroot disc into the surrounding solution, resulting in a decrease of [sucrose], and thus a lighter density.

Reply 6

coconut64

OP

16

Original post by zombiejon

If in doubt, draw out a flow diagram, and what you already know. And it's an application question, based on a mix of chemistry (fluid density/partial pressure) and biology.

You have a sucrose solution at a known concentration (Tube 1).
Some of the solution (2mL) from Tube 1 is then used bathe beetroot discs, in a tube we shall label Tube 2.

Sample from Tube 2 is placed in Tube 1, and the direction/rate of movement is measured.
So we know that bathing solution can be hyper/iso/hypotonic to the beetroot cubes.
If hypertonic, water flows from the beetroot cubes into the solution.
If isotonic, there's no change.
If hypotonic, water goes from the solution into the beetroot cubes.

Overall, when a droplet rises when placed into another solution, it is because it is less dense than the surrounding solution. This is akin to pouring oil on top of water - oil always goes to the top because it is less dense. As such, the converse should be true - a sinking droplet indicates a higher density.

So this tells us one thing - the density/concentration changed. The sucrose concentration of the bathing solution, although 0.6M initially, was no longer at 0.6M. From 6a, we should know that 0.4M is probably isotonic to the beetroot since the droplet did not move. If 0.4M is isotonic, 0.6M has to be hypertonic to the beetroot disc

Since a hypertonic solution results in a net flow of water from the beetroot disc into the surrounding solution, resulting in a decrease of [sucrose], and thus a lighter density.