Solvation layer?

[m25]

The AAMC lists "Solvation layer (entropy)" as a topic to know for the MCAT, and I understand what a solvation later/shell is, but what does entropy have anything to do with this?

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[Labrat07]

From my understanding, Solvation layer "entropy" is in context of protein folding in aquaeous solution. Think "surface area". Protein unfolding= more surface area = more disorder = increase entropy for protein. Protein folding = less surface area = more order = decrease entropy of protein. In term of aquaeous solution (surrounding) , it's opposite. Protein unfolding = increase entropy for protein = decrease water entropy = not favorable for water. Protein folding = decrease entropy for protein = increase water entropy = more favorable.

that's how I come to think of it. I might be a bit confused on water entropy though. @Cawolf @EParker37 Can you clarify?

 

[Cawolf]

You have it down correctly I believe @Labrat07.

It likely is asking how entropy changes when a substance is dissolved in solution. If you put a soluble salt in water, then the entropy change of the salt is positive as it becomes aqueous. In contrast, the entropy change of the water is negative due to the formation of a solvation shell around the ions. This arrangement of water decreases it's entropy.

These competing entropy changes explain why a more concentrated solution eventually becomes saturated. When there are no un-coordinated water molecules, the dissolution of the salt would be so entropically unfavorable due to the high order of water.

 

[EParker37]

As far as proteins this is also why the hydrophobic side chains are interior in a protein fold, if the hydrophobic layers were exterior in say an alpha-helix, it would cause the water molecules to concentrate and decrease entropy which is unfavored, this is basically due to less freedom of movement for the water molecules. Hence why the hydrophilic side chains form on the exterior of the folding to allow for solvation which allows more freedom of movement of the molecules, translating to an increase of entropy oddly enough, hence being favored. But the actual solvation layer is actually a decrease in entropy, but since it pulls from the universe, it does follow the 2nd law of thermodynamics.

 

[Labrat07]

You have it down correctly I believe @Labrat07.

It likely is asking how entropy changes when a substance is dissolved in solution. If you put a soluble salt in water, then the entropy change of the salt is positive as it becomes aqueous. In contrast, the entropy change of the water is negative due to the formation of a solvation shell around the ions. This arrangement of water decreases it's entropy.

These competing entropy changes explain why a more concentrated solution eventually becomes saturated. When there are no un-coordinated water molecules, the dissolution of the salt would be so entropically unfavorable due to the high order of water.

Hmmm, I cannot grasp these two concepts fluidly in my head since it's kinda opposite from protein folding theory. Can I think of them like this??? Salt's disolving into a liquid. So solid>>liquid = entropy increase. While water getting more solute = more solid then before = entropy decrease. So enventuallly when water's become saturated with salts = entropy decrease to a threshold that will be unfavorable.

As far as proteins this is also why the hydrophobic side chains are interior in a protein fold, if the hydrophobic layers were exterior in say an alpha-helix, it would cause the water molecules to concentrate and decrease entropy which is unfavored, this is basically due to less freedom of movement for the water molecules. Hence why the hydrophilic side chains form on the exterior of the folding to allow for solvation which allows more freedom of movement of the molecules, translating to an increase of entropy oddly enough, hence being favored. But the actual solvation layer is actually a decrease in entropy, but since it pulls from the universe, it does follow the 2nd law of thermodynamics.

Another way to look at this. Learn something new. Thanks

 

[EParker37]

Hmmm, I cannot grasp these two concepts fluidly in my head since it's kinda opposite from protein folding theory. Can I think of them like this??? Salt's disolving into a liquid. So solid>>liquid = entropy increase. While water getting more solute = more solid then before = entropy decrease. So enventuallly when water's become saturated with salts = entropy decrease to a threshold that will be unfavorable.

Right, basically the entropy of the solute increases, while the entropy of the solvent decreases. Once you reach a point where there are not enough unsolvated water molecules to causes the solute to come apart, the solution is considered saturated and the solute will just collect on the bottom. Now, we can modify this a bit by using heat, if we heat the solution, we can actually dissolve more solute than we could otherwise, this is a state known as super-saturation.

 

[Labrat07]

@EParker37
Thank you bud.

 

[Borealis]

I find this might be a little easier to grasp, but basically says the same things mentioned in the above, with a little added explanation.

If you are looking at a solute dissolving in a solvent, the nearby solvent molecules form the solvation layer around the solute.

From an enthalpy standpoint, even things like hydrocarbons are going to have more stability in an aqueous solution as opposed to organic solutions (ΔH < 0). Conversely, if an amino acid with a hydrophobic R group is dropped in an aqueous solution, the water molecules in the solvation layer can't hydrogen bond with the side chain. This inability to hydrogen bond forces the water molecules nearby to rearrange themselves into specific patterns to maximize hydrogen bonding. Therefore you have a negative change in entropy , or ΔS. Because you are seeing negative change in entropy (decrease in disorder = increase in order), it's unfavorable. So the overall process is going to be nonspontaneous (ΔG > 0).

However, if there's a hydrophilic residue on the protein's exterior, that will allow these nearby water molecules more freedom in their positioning, which means their entropy will increase (ΔS > 0), and make the overall solvation process spontaneous (ΔG < 0). So a protein is able to achieve maximum stability by pushing hydrophobic groups to the interior of a protein, and keeping hydrophilic groups on the outside.

 

[m25]

I find this might be a little easier to grasp, but basically says the same things mentioned in the above, with a little added explanation.

If you are looking at a solute dissolving in a solvent, the nearby solvent molecules form the solvation layer around the solute.

From an enthalpy standpoint, even things like hydrocarbons are going to have more stability in an aqueous solution as opposed to organic solutions (ΔH < 0). Conversely, if an amino acid with a hydrophobic R group is dropped in an aqueous solution, the water molecules in the solvation layer can't hydrogen bond with the side chain. This inability to hydrogen bond forces the water molecules nearby to rearrange themselves into specific patterns to maximize hydrogen bonding. Therefore you have a negative change in entropy , or ΔS. Because you are seeing negative change in entropy (decrease in disorder = increase in order), it's unfavorable. So the overall process is going to be nonspontaneous (ΔG > 0).

However, if there's a hydrophilic residue on the protein's exterior, that will allow these nearby water molecules more freedom in their positioning, which means their entropy will increase (ΔS > 0), and make the overall solvation process spontaneous (ΔG < 0). So a protein is able to achieve maximum stability by pushing hydrophobic groups to the interior of a protein, and keeping hydrophilic groups on the outside.

So are you saying that entropy of solvent decreases when the solute is hydrophobic and increases when the solute is hydrophilic?

 

[lifetothefullest]

However, if there's a hydrophilic residue on the protein's exterior, that will allow these nearby water molecules more freedom in their positioning, which means their entropy will increase (ΔS > 0), and make the overall solvation process spontaneous (ΔG < 0). So a protein is able to achieve maximum stability by pushing hydrophobic groups to the interior of a protein, and keeping hydrophilic groups on the outside.

I would be careful with how hydrophilic something is. If it is very hydrophilic, what could happen is the entropy of your water will still decrease because they are drawn into a tightly organized solvation shell around the solute. Another way to look at it is to consider the total microstates possible that can be occupied by the water. In just water, the number of microstates that the ensemble of molecules can occupy is quite large. But after solvation, some molecules are "bound" into solvation shells around the solute, so the number of microstates available is smaller. OP, I know you would like a simple rule but most times, that's just not how the world works. You have to think about what's actually happening during these processes.

What determines the spontaneity of a solvation process is not the change in entropy of the water molecules or entropy of the solute, but the total universe change in entropy. Remember that the Second Law only says that the change in entropy of the universe must be positive for spontaneity.

 

[RealityDream]

I find this might be a little easier to grasp, but basically says the same things mentioned in the above, with a little added explanation.

If you are looking at a solute dissolving in a solvent, the nearby solvent molecules form the solvation layer around the solute.

From an enthalpy standpoint, even things like hydrocarbons are going to have more stability in an aqueous solution as opposed to organic solutions (ΔH < 0). Conversely, if an amino acid with a hydrophobic R group is dropped in an aqueous solution, the water molecules in the solvation layer can't hydrogen bond with the side chain. This inability to hydrogen bond forces the water molecules nearby to rearrange themselves into specific patterns to maximize hydrogen bonding. Therefore you have a negative change in entropy , or ΔS. Because you are seeing negative change in entropy (decrease in disorder = increase in order), it's unfavorable. So the overall process is going to be nonspontaneous (ΔG > 0).

However, if there's a hydrophilic residue on the protein's exterior, that will allow these nearby water molecules more freedom in their positioning, which means their entropy will increase (ΔS > 0), and make the overall solvation process spontaneous (ΔG < 0). So a protein is able to achieve maximum stability by pushing hydrophobic groups to the interior of a protein, and keeping hydrophilic groups on the outside.

This explanation is from the Kaplan Biochemistry Review Notes on Solvation Layer.

 

[Mark Kovnikov]

This is my favorite website for chemistry questions
http://chemwiki.ucdavis.edu/Physica...ntermolecular_Forces/Hydrophobic_interactions

 

[scuzum2u]

Question 22 of the Bio/BCM section on Kaplan's FL1 addresses this exact topic actually. I don't want to spoil the FL so I will only post if desired.

 

[m25]

Question 22 of the Bio/BCM section on Kaplan's FL1 addresses this exact topic actually. I don't want to spoil the FL so I will only post if desired.

Please do share!

 

[scuzum2u]

#22. The clustering of non-polar protein groups in water results in a net:

a. Decrease in entropy
b. Increase in entropy
c. Increase in enthalpy
d. Decrease in enthalpy

Explanation: The organization of the solvation layer causes a decrease in entropy, so the clustering of non-polar groups, by virtue of diminishing the layer, causes a favorable increase in entropy. In fact, this increase in entropy is the predominant thermodynamic influence resulting in the clustering of nonpolar groups in polar solvents like water.

 

[m25]

#22. The clustering of non-polar protein groups in water results in a net:

a. Decrease in entropy
b. Increase in entropy
c. Increase in enthalpy
d. Decrease in enthalpy

Explanation: The organization of the solvation layer causes a decrease in entropy, so the clustering of non-polar groups, by virtue of diminishing the layer, causes a favorable increase in entropy. In fact, this increase in entropy is the predominant thermodynamic influence resulting in the clustering of nonpolar groups in polar solvents like water.

So the answer is B? Is it saying that even though the formation of the solvation layer causes a decrease in entropy, the clustering of non-polar groups in protein increases entropy more, resulting in net increase in entropy? How does the clustering of non-polar groups in protein increase entropy though? Wouldn't clustering it decrease the entropy?

 

[kingtal0n]

what the heck is narcan*

it says "if in doubt, narcan"

 

[NETKA755]

So the answer is B? Is it saying that even though the formation of the solvation layer causes a decrease in entropy, the clustering of non-polar groups in protein increases entropy more, resulting in net increase in entropy? How does the clustering of non-polar groups in protein increase entropy though? Wouldn't clustering it decrease the entropy?

It looks like the question is asking for the entropy of water? This way, clustering of a non-polar protein (=increased order and decreased entropy of the protein) would increase the entropy (=disorder) of water, I think..

 

[PharmHoppe]

#22. The clustering of non-polar protein groups in water results in a net:

a. Decrease in entropy
b. Increase in entropy
c. Increase in enthalpy
d. Decrease in enthalpy

Explanation: The organization of the solvation layer causes a decrease in entropy, so the clustering of non-polar groups, by virtue of diminishing the layer, causes a favorable increase in entropy. In fact, this increase in entropy is the predominant thermodynamic influence resulting in the clustering of nonpolar groups in polar solvents like water.

Favorable increase in entropy of the clustered protein?

 

[Mark Kovnikov]

The water forms cages with itself which is highly ordered (h-bonding). The solute breaks the cages causes more disorder which results in positive change in S. G=H-ST
The enthalpy (H) may or may not be positive; however, the large increase in entropy outweighs the unfavorableness of the increase in enthalpy. Anyway, I had the same problem and it's not to be pondered to long because the change in enthalpy is negligible in either case (positive or negative).

 

[Vero Medico]

I find this might be a little easier to grasp, but basically says the same things mentioned in the above, with a little added explanation.

If you are looking at a solute dissolving in a solvent, the nearby solvent molecules form the solvation layer around the solute.

From an enthalpy standpoint, even things like hydrocarbons are going to have more stability in an aqueous solution as opposed to organic solutions (ΔH < 0). Conversely, if an amino acid with a hydrophobic R group is dropped in an aqueous solution, the water molecules in the solvation layer can't hydrogen bond with the side chain. This inability to hydrogen bond forces the water molecules nearby to rearrange themselves into specific patterns to maximize hydrogen bonding. Therefore you have a negative change in entropy , or ΔS. Because you are seeing negative change in entropy (decrease in disorder = increase in order), it's unfavorable. So the overall process is going to be nonspontaneous (ΔG > 0).

However, if there's a hydrophilic residue on the protein's exterior, that will allow these nearby water molecules more freedom in their positioning, which means their entropy will increase (ΔS > 0), and make the overall solvation process spontaneous (ΔG < 0). So a protein is able to achieve maximum stability by pushing hydrophobic groups to the interior of a protein, and keeping hydrophilic groups on the outside.

That totally makes sense until I approached this passage-based Question

 

[TopTenTutoring]

Hopefully an explanation on this passage will help.

The passage says in the second paragraph that the creation of the solvation layer creates an "organized shell surrounding the protein...this organization results in an unfavorable decrease in entropy." So we know the solvation layer decreases the entropy of the water. But the passage also says that "the clustering of hydrophobic protein groups actually causes a decrease in size of the solvation layer." This means that the hydrophobic (non-polar) groups clustering together actually decreases the amount of water needed to form the solvation layer thereby increasing the entropy of the water.

Essentially, we are starting off at a point where the non-polar protein groups are not clustered and more solvation is needed by the water due to the increase in size of the solvation layer (lower entropy) to a state where the non-polar groups are clustered and less solvation is needed by the water due to a decrease in size of the solvation layer (higher entropy state).

Hope this helps!

 

[aldol16]

That totally makes sense until I approached this passage-based Question

Can you clarify which part of it you don't understand? You should always be clear on defining your processes. You aren't comparing free water and free protein. You're comparing non-clustered protein to clustered protein. If they're non-clustered, you have a huge solvation shell and therefore very few degrees of freedom in the water molecules. Imagine taking five spoons and sticking them into a beaker of water far apart from one another. The total surface area exposed to the water is quite large. Now imagine taking the spoons, stacking them on top of one another, and then sticking the whole thing in water. Now, the spoon-spoon surfaces replace what used to be spoon-water surfaces and so the solvation shell decreases in size. This is an important concept you should understand.

Therefore, compared to the non-clustered protein, the clustered protein will cause the water around it to increase in entropy.

 

[Vero Medico]

Can you clarify which part of it you don't understand? You should always be clear on defining your processes. You aren't comparing free water and free protein. You're comparing non-clustered protein to clustered protein. If they're non-clustered, you have a huge solvation shell and therefore very few degrees of freedom in the water molecules. Imagine taking five spoons and sticking them into a beaker of water far apart from one another. The total surface area exposed to the water is quite large. Now imagine taking the spoons, stacking them on top of one another, and then sticking the whole thing in water. Now, the spoon-spoon surfaces replace what used to be spoon-water surfaces and so the solvation shell decreases in size. This is an important concept you should understand.

Therefore, compared to the non-clustered protein, the clustered protein will cause the water around it to increase in entropy.

Thank you so much for the in-depth explanation! I could totally visualize the scenario accurately!
Because of less surface are and less opportunity to contact with water, the solvation layer is decreased and this leads to increased in entropy compared to when the proteins are non-clustered!

 

[Vero Medico]

Hopefully an explanation on this passage will help.

The passage says in the second paragraph that the creation of the solvation layer creates an "organized shell surrounding the protein...this organization results in an unfavorable decrease in entropy." So we know the solvation layer decreases the entropy of the water. But the passage also says that "the clustering of hydrophobic protein groups actually causes a decrease in size of the solvation layer." This means that the hydrophobic (non-polar) groups clustering together actually decreases the amount of water needed to form the solvation layer thereby increasing the entropy of the water.

Essentially, we are starting off at a point where the non-polar protein groups are not clustered and more solvation is needed by the water due to the increase in size of the solvation layer (lower entropy) to a state where the non-polar groups are clustered and less solvation is needed by the water due to a decrease in size of the solvation layer (higher entropy state).

Hope this helps!

Great, Thanks so much for the passage analysis. While I was reading it again, I was thinking, wouldn't it be more accurate to say that clustering of proteins cause the decrease in the number of solvation layer compared to the non-clustered proteins?

 

[Vero Medico]

Hopefully an explanation on this passage will help.

The passage says in the second paragraph that the creation of the solvation layer creates an "organized shell surrounding the protein...this organization results in an unfavorable decrease in entropy." So we know the solvation layer decreases the entropy of the water. But the passage also says that "the clustering of hydrophobic protein groups actually causes a decrease in size of the solvation layer." This means that the hydrophobic (non-polar) groups clustering together actually decreases the amount of water needed to form the solvation layer thereby increasing the entropy of the water.

Essentially, we are starting off at a point where the non-polar protein groups are not clustered and more solvation is needed by the water due to the increase in size of the solvation layer (lower entropy) to a state where the non-polar groups are clustered and less solvation is needed by the water due to a decrease in size of the solvation layer (higher entropy state).

Hope this helps!

Can we say that the reason why folded proteins ( native state) are themodynamically favorable because the SA and the opportunity of contacting with water is less, therefore , there is less solvation layer which results in the increase in entropy.

 

[TopTenTutoring]

Can we say that the reason why folded proteins ( native state) are themodynamically favorable because the SA and the opportunity of contacting with water is less, therefore , there is less solvation layer which results in the increase in entropy.

This is correct for the situation in question 22 because we are strictly talking about the nonpolar groups in the protein. It is really going to depend on the protein, though. If the protein was strictly nonpolar throughout the polypeptide chain, then reducing the surface area would be a strong factor as to why it would be favored as folded.

If you notice, in the passage it also talks a little bit about other situations in the last sentence of paragraph one. It says that an unfolded protein may actually be favored. The passage says that this depends on the groups on the polypeptide chain. For example, if the protein has functional groups that can undergo hydrogen-bonding with the water, then it may be favorable to have those exposed in an unfolded state instead of folding to reduce surface area.

This creates a balancing act when talking about protein folding. The protein wants to have as many water soluble (polar) groups toward the surface as possible, but it also wants to limit the surface area of the nonpolar groups that get solvated.

Hopefully, this makes sense!

 

[TXhopeful]

Sorry to revive this month's old thread.
From EK 9th Edition, Biology 1 (Molecules): "Due to the presence of hydrophobic R-groups on the protein, surrounding molecules assemble into an organized structure known as the solvation layer that forces these hydrophobic groups towards the inner area of the protein. The gathering of hydrophobic R-groups away from the surrounding water is highly favorable because it allows a decrease in size of the highly ordered solvation layer, increasing the entropy of the system."

It all seems so contradicting. When the hydrophobic side chains gather on the inside, wouldn't that be considered more order = decrease entropy? It seems that the protein would be folding in on itself to allow the hydrophobic groups shelter from the aqueous solution, which is a decrease in surface area, which is more favorable.
#haaallppp

 

[aldol16]

It all seems so contradicting. When the hydrophobic side chains gather on the inside, wouldn't that be considered more order = decrease entropy? It seems that the protein would be folding in on itself to allow the hydrophobic groups shelter from the aqueous solution, which is a decrease in surface area, which is more favorable.

You're confusing where the entropy comes from. Having a huge solvation shell raises entropy because of the many water molecules that must form it. Burying hydrophobics lowers the entropy of the protein. In the folded structure, the overall entropy rises because the solvation shell reduction dominates.

 

[TXhopeful]

You're confusing where the entropy comes from. Having a huge solvation shell raises entropy because of the many water molecules that must form it. Burying hydrophobics lowers the entropy of the protein. In the folded structure, the overall entropy rises because the solvation shell reduction dominates.

Think I got it. The salvation layer is organized, which is less disorder = decrease in entropy. Burying hydrophobic groups decreases the size of the solvation layer, which is an increase in disorder and therefore an increase in entropy.