Help please

This is a biology related doubt.

When we put a seed inside a solution with less concentration of solute, i.e. hypotonic solution, then after endosmosis, what solution would the large vacuole have. Most of them say hypotonic because that is the original solution. But I think that it should be isotonic (equal concentration of solute on both sides) because osmosis would stop only when concentration of solute equalizes on both sides.

Please put across your views. For helping me with this question, you need not be good in biology, this is just simple common sense. Thanks in advance. :-)

Note by Ashish Siva
3 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

It would be isotonic because the water would keep flowing out until the two were balanced. The only cause in which the solution would not become isotonic is where the membrane/wall was only h20 permiable and that way you would have osmosis and the concentration in the cell becomes hypertonic.

Amol Garg - 3 years, 5 months ago

Log in to reply

First of all, would not water keeping flowing into the cell when it is kept in hypotonic solution? And, why do you think that the membrane should be H2OH_2O permeable? Isotonic means that concentration of solute in solution in both sides are equal. And it is not necessarily H2OH_2O permeable membrane that would make this possible. Any other reason? :D Thanks for your reply btw.

Ashish Siva - 3 years, 5 months ago

Log in to reply

Haha oops yes you are right the water would be flowing into the membrane.

As per your second question the membrane of the cell is water permiable because it simply has to be. All cells need water to keep homeostasis and therefore to keep living, the cell would have to be h20 permiable.

As per your third statement/questuon, here is the deal. You are partially correct in that the concentrations would not become equal if only h20 could move. Concentration can be found out by dividing solute by water. You can lower concentration of something by adding water or taking away solute. If the hypotonic solution kept losing water then it's concentration would increase and that water going into the cell would decrease the concentration. Eventually the cell and the solution would meet half way. There is one case where this would not happen, however, if the outside solution was only water (no solute). This would mean the no matter how much water went into the cell the cell wouldn't match concentration of outside (the outside has 0 solute and cell can never match that). This is a special case however where only h20 permiability would prevent solutions from reading isotonicity.

Hope this helps! Whew that was long :)

Amol Garg - 3 years, 5 months ago

Log in to reply

@Amol Garg Great explanation. : +1 : :+ 1: :+1: ;) But then the outer solution would never be hypertonic with respect to the cell membrane, right? :3

Ashish Siva - 3 years, 5 months ago

Log in to reply

@Ashish Siva No it should never be hypertonic in respect to the cell membrane ever if it starts off as hypotonic in cases only involving passive transport.

Amol Garg - 3 years, 5 months ago

Log in to reply

@Amol Garg Sorry, I meant hypotonic yeah you are correct :+1: Thanks for clearing my doubt. So, my answer is valid for all cases except the spaciel case as you mentioned right?

Ashish Siva - 3 years, 5 months ago

Log in to reply

@Ashish Siva Yup :)

Amol Garg - 3 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...