C/P Section Bank Questions

[itsalwayssunny96]

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54. So I see how my answer is wrong as I was a little confused about this passage's mechanism...but how do you know that the answer is glucose over galactose? In the passage it says, "a glutamate residue (Glu-1) in the active site donates a proton to an oxygen atom," but it doesn't specify which oxygen atom...?

57. I guessed for this question and got it correct but I'm confused based on the fact that they said the second part of the serial dilution was a 1->2 dilution. I just don't see it. Is there a simple formula for dilutions?

 

[aldol16]

You might have to draw a mechanism to visualize the first one. You don't need to know which oxygen gets the proton to solve the question (it's the oxygen that's going to become the leaving group). What's important is what's after that, namely the glutamate residue acting as a nucleophile to kick off glucose. That means that the only thing glucose has that's new is the proton. And then a hydroxide ion derived from water nucleophilically attacks the bound galactose to free it up (you actually highlighted this phrase). Since it's this hydroxide that is derived from water, this is where the labeled O is going to be.

For the second, you can use the M1*V1 = M2*V2 formula or you can use understanding of ratios to solve the problem. Let's trace this experiment backwards using the M1*V1 = M2*V2 formula first. Okay, so in the kinetics runs, they use 1 mL of substrate solution and 1 mL of enzyme solution. So we know that the final concentration is E, the final volume is 2 mL (1 mL + 1 mL), and the initial volume was 1 mL (1 mL of the enzyme solution). So let's plug that in. M1*(1 mL) = E*(2 mL). Therefore, M1 must be 2*E. Now, that solution was made by diluting 0.1 mL of the enzyme solution into 25 mL final volume. So let's use the formula above again: M1*V1 = M2*V2. We know that we want to end up with a solution with 2*E concentration. We know that V2 is 25 mL here. And we know that V1 is 0.1 mL. So let's plug it in: M1*(0.1 mL) = 2*E*25 mL. Therefore, M1 = 500*E.

We can also use ratios to solve this. The stock solution was diluted by a factor of 250 because 0.1 mL is 1/250 of 25 mL. Then this diluted solution is further diluted by a factor of 2 to do the kinetics runs. So to reverse the whole thing, you have to concentrate it by the same factor: 2*250 = 500.

 

[itsalwayssunny96]

You might have to draw a mechanism to visualize the first one. You don't need to know which oxygen gets the proton to solve the question (it's the oxygen that's going to become the leaving group). What's important is what's after that, namely the glutamate residue acting as a nucleophile to kick off glucose. That means that the only thing glucose has that's new is the proton. And then a hydroxide ion derived from water nucleophilically attacks the bound galactose to free it up (you actually highlighted this phrase). Since it's this hydroxide that is derived from water, this is where the labeled O is going to be.

For the second, you can use the M1*V1 = M2*V2 formula or you can use understanding of ratios to solve the problem. Let's trace this experiment backwards using the M1*V1 = M2*V2 formula first. Okay, so in the kinetics runs, they use 1 mL of substrate solution and 1 mL of enzyme solution. So we know that the final concentration is E, the final volume is 2 mL (1 mL + 1 mL), and the initial volume was 1 mL (1 mL of the enzyme solution). So let's plug that in. M1*(1 mL) = E*(2 mL). Therefore, M1 must be 2*E. Now, that solution was made by diluting 0.1 mL of the enzyme solution into 25 mL final volume. So let's use the formula above again: M1*V1 = M2*V2. We know that we want to end up with a solution with 2*E concentration. We know that V2 is 25 mL here. And we know that V1 is 0.1 mL. So let's plug it in: M1*(0.1 mL) = 2*E*25 mL. Therefore, M1 = 500*E.

We can also use ratios to solve this. The stock solution was diluted by a factor of 250 because 0.1 mL is 1/250 of 25 mL. Then this diluted solution is further diluted by a factor of 2 to do the kinetics runs. So to reverse the whole thing, you have to concentrate it by the same factor: 2*250 = 500.

Ah, I see now. I feel like sometimes I just overthink these questions. Thank you!!

 

[DaAlienist]

I have a question from the Chemistry Foundations section bank, passage 7, question 53.

I thought hydrolysis of saccharides ends up with the OH going to the sugar on the right of the glycoside link, and an H going to the sugar on the left of the link, as shown below.

The AAMC says that only galactose (the sugar on the left of lactose) will have a tagged O from water.

The explanation is worthless but from what I can gather, from the first paragraph, they state the following:

They say in a 2nd step that water is converted to a hydroxide before attacking the sugar and breaking up the lactose. Reading this + what I know about sugar hydrolysis, I thought the answer would be both monosaccharides. One labeled O goes to the glucose (as expected from typical hydrolysis of sugars) and then another attacks the galactose as specified in the passage.

They say the answer is galactose only. Was I supposed to infer that this mechanism is completely separate and apart from typical hydrolysis of glycoside because the question specifically says "under the action of lactase?" so that normal hydrolysis is NOT occurring? The AAMC explanation mentions none of this.

Thanks for the help!

 

[aldol16]

They say in a 2nd step that water is converted to a hydroxide before attacking the sugar and breaking up the lactose. Reading this + what I know about sugar hydrolysis, I thought the answer would be both monosaccharides. One labeled O goes to the glucose (as expected from typical hydrolysis of sugars) and then another attacks the galactose as specified in the passage.

They say the answer is galactose only. Was I supposed to infer that this mechanism is completely separate and apart from typical hydrolysis of glycoside because the question specifically says "under the action of lactase?" so that normal hydrolysis is NOT occurring? The AAMC explanation mentions none of this.

Why would you expect one oxygen to go to glucose "as expected from typical hydrolysis"? In typical hydrolysis, the glucose here gets no oxygens from water. It just has to pick up a proton since it already has the required oxygen (it gets the one in the ether linkage). Then water comes in to attack the left-over galactose residue that's bound to the enzyme and so the oxygen of the water gets incorporated here. You only get net incorporation of one water.

 

[DaAlienist]

@aldol16

I never said more than one water has to be involved in typical glycoside hydrolysis, I am asking if this mechanism, where the OH from water goes to the sugar on the left of the disaccharide, is specific to lactase-catalyzed hydrolysis. Can you not see the hydrolysis/condensation equilibrium I have posted above? When water hydrolyzes the glycoside link, one of the sugars in the link will receive an OH, and the other will receive the H, while the opposite occurs when the link is formed via condensation. I posted it again below.

Are you telling me, in typical hydrolysis (forget about lactase involvement), the glucose on the right of the disaccharide above would never receive an OH from water as the bond is lysed, the OH will always go to the galactose? I think that is wrong. If it's not, can you tell me why?

When two monosaccharides come together, it is due to nucleophilic attack by OH group on an anomeric carbon from 1 sugar on a ring carbon in the other sugar. Typically isn't this attacking sugar placed on the left when drawing the disaccharide, and this is the monosaccharide which would retain its O as the attacking nucleophile? Thus, the sugar that is drawn on the right hand side of the final disaccharide product, lost an OH group. Thus, the sugar on the right would get the OH when hydrolyzing the disaccharide (the reverse reaction). Is any of that wrong?

 

[aldol16]

I never said more than one water has to be involved in typical glycoside hydrolysis, I am asking if this mechanism, where the OH from water goes to the sugar on the left of the disaccharide, is specific to lactase-catalyzed hydrolysis. Can you not see the hydrolysis/condensation equilibrium I have posted above? When water hydrolyzes the glycoside link, one of the sugars in the link will receive an OH, and the other will receive the H, while the opposite occurs when the link is formed via condensation. I posted it again below.

Yes. Each enzyme will catalyze hydrolysis in a different way and which sugar gets attacked first is determined by the sterics of the active site. Nature can erect physical barriers that prevent one sugar from being attacked, getting almost 100% selectivity for the other one.

Are you telling me, in typical hydrolysis (forget about lactase involvement), the glucose on the right of the disaccharide above would never receive an OH from water as the bond is lysed, the OH will always go to the galactose? I think that is wrong. If it's not, can you tell me why?

In a typical hydrolysis, you would scramble the water since both C-O bond strengths are roughly the same - you are correct. But this is an enzyme-catalyzed hydrolysis so the steric environment of the active site must be taken into account.

When two monosaccharides come together, it is due to nucleophilic attack by OH group on an anomeric carbon from 1 sugar on a ring carbon in the other sugar. Typically isn't this attacking sugar placed on the left when drawing the disaccharide, and this is the monosaccharide which would retain its O as the attacking nucleophile? Thus, the sugar that is drawn on the right hand side of the final disaccharide product, lost an OH group. Thus, the sugar on the right would get the OH when hydrolyzing the disaccharide (the reverse reaction). Is any of that wrong?

This would be right if you're talking about the simple microscopic reversibility of hydrolysis. In other words, you have to follow the exact same path in the forward as the reverse directions at the microscopic level because of the so-called principle of microscopic reversibility. However, we're talking about an enzyme that's doing a hydrolysis here. That's a key difference. The enzyme that made the linkage is not necessarily the same as the one that took it away. So we're not doing a microscopically reversible reaction here.

 

[DaAlienist]

Yes. Each enzyme will catalyze hydrolysis in a different way and which sugar gets attacked first is determined by the sterics of the active site. Nature can erect physical barriers that prevent one sugar from being attacked, getting almost 100% selectivity for the other one.



In a typical hydrolysis, you would scramble the water since both C-O bond strengths are roughly the same - you are correct. But this is an enzyme-catalyzed hydrolysis so the steric environment of the active site must be taken into account.



This would be right if you're talking about the simple microscopic reversibility of hydrolysis. In other words, you have to follow the exact same path in the forward as the reverse directions at the microscopic level because of the so-called principle of microscopic reversibility. However, we're talking about an enzyme that's doing a hydrolysis here. That's a key difference. The enzyme that made the linkage is not necessarily the same as the one that took it away. So we're not doing a microscopically reversible reaction here.

Man, that is so sneaky. Thank you. I assumed it was generic hydrolysis with lactase assisting and adding another step, not an entirely different mechanism. A pox on the AAMC house for this one!

 

[O-Block Uchiha]

You might have to draw a mechanism to visualize the first one. You don't need to know which oxygen gets the proton to solve the question (it's the oxygen that's going to become the leaving group). What's important is what's after that, namely the glutamate residue acting as a nucleophile to kick off glucose. That means that the only thing glucose has that's new is the proton. And then a hydroxide ion derived from water nucleophilically attacks the bound galactose to free it up (you actually highlighted this phrase). Since it's this hydroxide that is derived from water, this is where the labeled O is going to be.

For the second, you can use the M1*V1 = M2*V2 formula or you can use understanding of ratios to solve the problem. Let's trace this experiment backwards using the M1*V1 = M2*V2 formula first. Okay, so in the kinetics runs, they use 1 mL of substrate solution and 1 mL of enzyme solution. So we know that the final concentration is E, the final volume is 2 mL (1 mL + 1 mL), and the initial volume was 1 mL (1 mL of the enzyme solution). So let's plug that in. M1*(1 mL) = E*(2 mL). Therefore, M1 must be 2*E. Now, that solution was made by diluting 0.1 mL of the enzyme solution into 25 mL final volume. So let's use the formula above again: M1*V1 = M2*V2. We know that we want to end up with a solution with 2*E concentration. We know that V2 is 25 mL here. And we know that V1 is 0.1 mL. So let's plug it in: M1*(0.1 mL) = 2*E*25 mL. Therefore, M1 = 500*E.

We can also use ratios to solve this. The stock solution was diluted by a factor of 250 because 0.1 mL is 1/250 of 25 mL. Then this diluted solution is further diluted by a factor of 2 to do the kinetics runs. So to reverse the whole thing, you have to concentrate it by the same factor: 2*250 = 500.

Thanks aldol, you really are all over this forum helping the people lol, i gotta ask when did you take your MCAT

 

[aldol16]

Thanks aldol, you really are all over this forum helping the people lol, i gotta ask when did you take your MCAT

I took it over a year ago now - a year and a half ago.

 

[rlaboss]

Man, that is so sneaky. Thank you. I assumed it was generic hydrolysis with lactase assisting and adding another step, not an entirely different mechanism. A pox on the AAMC house for this one!

Is there any way to tell that the -OH doesn't end up back on the enzyme when galactose is released? Will it be 50/50 where half of the time the labelled O with be part of galactose and the other half on the enzyme?

 

[aldol16]

Is there any way to tell that the -OH doesn't end up back on the enzyme when galactose is released? Will it be 50/50 where half of the time the labelled O with be part of galactose and the other half on the enzyme?

Which -OH?

 

[rlaboss]

Which -OH?

The labelled OH. When the labelled OH hydrolyzes the bond, releasing galactose, how do we know that the galactose (instead of the enzyme) will end up with the labelled OH?

 

[aldol16]

The labelled OH. When the labelled OH hydrolyzes the bond, releasing galactose, how do we know that the galactose (instead of the enzyme) will end up with the labelled OH?

Well, this one is tricky. Because the passage specifically tells you that Glu-1 deprotonates the water and forms a hydroxide, which acts as a nucleophile. If the hydroxide attacked the Glu, it would be attacking a carbonyl to form a tetrahedral intermediate, which would collapse and release the galactose in alkoxide form, which is really unfavorable. For that to happen, you'd need to protonate the galactose first and that is not mentioned in the mechanism. So you would have to assume that the hydroxide attacks at the galactose, kicking out glutamate in the deprotonated carboxylic acid form, which is stable.

 

[neplina]

*bump* @aldol16 overall, in simplistic terms, how do we know for sure the O ends up on galactose and glucose?

the back and forth on this thread kind of confused me

 

[aldol16]

Why would you expect one oxygen to go to glucose "as expected from typical hydrolysis"? In typical hydrolysis, the glucose here gets no oxygens from water. It just has to pick up a proton since it already has the required oxygen (it gets the one in the ether linkage). Then water comes in to attack the left-over galactose residue that's bound to the enzyme and so the oxygen of the water gets incorporated here. You only get net incorporation of one water.

*bump* @aldol16 overall, in simplistic terms, how do we know for sure the O ends up on galactose and glucose?

The passage tells you the mechanism. Draw out the mechanism. In lactose, you have galactose and glucose linked by an ether linkage. The first step is there is proton donation to the ether O and then a glutamate residue on the enzyme attacks the carbon on galactose to liberate the HO-glucose group. Then water comes in to liberate the galactose. Since the labeled O is on water and water is always only added to galactose, the label will be on galactose. Glucose takes the O that was already there in lactose with it so it's unlabeled.

 

[EssentialOil123]

Can someone draw the mechanism for this? The ones on google images are incomplete. I'm confused on the mechanism after protonating the oxygen of the 1,4 alpha linkage, because which carbon on galactose are you attacking? And why is there a need to deprotonate water to liberate the galactose if a water molecule can already act as a nucleophile to release the intermediate chain from galactose?