Waste less time on Facebook — follow Brilliant.

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
9 months ago

No vote yet
1 vote


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 · 9 months ago

Log in to reply

@Amol Garg 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 \(H_2O\) permeable? Isotonic means that concentration of solute in solution in both sides are equal. And it is not necessarily \(H_2O\) permeable membrane that would make this possible. Any other reason? :D Thanks for your reply btw. Ashish Siva · 9 months ago

Log in to reply

@Ashish Siva 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 · 9 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 · 9 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 · 9 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 · 9 months ago

Log in to reply

@Ashish Siva Yup :) Amol Garg · 9 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...