Amino acid and Pka

[Temperature101]

The pKa for the side chain of histidine in aqueous medium is 6.1. What is true of its pKa in a hydrophobic environment?

A. It is the same, because log functions can not be altered.
B. It would be increased, because the proton leaves more readily to make the species neutral.
C. It would be decreased, because the proton leaves more readily to make the species negatively charge.
D. It would be decreased, because the proton leaves more readily to make the species neutral.

I got the right answer by POE but I am not too sure why it is the right answer... Can someone explain why D is the right answer?

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[sdm33]

The pKa for the side chain of histidine in aqueous medium is 6.1. What is true of its pKa in a hydrophobic environment? I got D as well, by POE,It would be decreased, because the proton leaves more readily to make the species neutral.

Histidine is basic , and If the solution is nonpolar there is limited exchange of protons, so the pka of the amino acid will decrease. At pH between 2.2 and 9.4, the predominant form adopted by α-amino acids contains a negative carboxylate and a positive α-ammonium group, so has net zero charge. This molecular state is known as a zwitterion

 

[Postictal Raiden]

The pKa for the side chain of histidine in aqueous medium is 6.1. What is true of its pKa in a hydrophobic environment?

A. It is the same, because log functions can not be altered.
B. It would be increased, because the proton leaves more readily to make the species neutral.
C. It would be decreased, because the proton leaves more readily to make the species negatively charge.
D. It would be decreased, because the proton leaves more readily to make the species neutral.

I got the right answer by POE but I am not too sure why it is the right answer... Can someone explain why D is the right answer?

I also did this by POE, but here is my rationale:

"like dissolves in like", so in a hydrophobic environment, histidine will "try" its best to remain as nonpolar as possible. The best way to do that is to become chargeless. This eliminates A and C. Now, both B and D have an identical second part of the statement, so we know focus on the first part. Would the pKa increase or decrease if something becomes more acidic, aka better proton donor? We know that stronger acids have larger Ka value. We also know that pKa and Ka have inverse log relationship, so if we want the Ka to increase (become more acidic), pKa must decrease.

 

[Postictal Raiden]

The pKa for the side chain of histidine in aqueous medium is 6.1. What is true of its pKa in a hydrophobic environment? I got D as well, by POE,It would be decreased, because the proton leaves more readily to make the species neutral.

Histidine is basic , and If the solution is nonpolar there is limited exchange of protons, so the pka of the amino acid will decrease. At pH between 2.2 and 9.4, the predominant form adopted by α-amino acids contains a negative carboxylate and a positive α-ammonium group, so has net zero charge. This molecular state is known as a zwitterion

This doesn't make sense.

How would the pKa decrease (making Ka increase) if exchange of protons is limited. Ka measures the extent of acid dissociation, so the larger it is the more dissociated the acid becomes. Therefore, pKa would increase if exchange of protons is limited. However, in this scenario, the answer says that protons dissociate more easily, causing the Ka to increase and the pKa to decrease.

 

[sdm33]

you right I guess the word limited is misleading, The pKa, or acid dissociation constant, of an amino acid is strongly tied to the properties of the surrounding solvent. The hydrophobic core of a protein is a distinctly different environment than the water exposed surface of the protein and the pKa in the core is different than the normal, solvent exposed pKa. This is related to the dielectric constant, or the ease at which charge is "felt" over a distance, which is much lower in the hydrophobic core of the protein.

 

[Temperature101]

I also did this by POE, but here is my rationale:

"like dissolves in like", so in a hydrophobic environment, histidine will "try" its best to remain as nonpolar as possible. The best way to do that is to become chargeless. This eliminates A and C. Now, both B and D have an identical second part of the statement, so we know focus on the first part. Would the pKa increase or decrease if something becomes more acidic, aka better proton donor? We know that stronger acids have larger Ka value. We also know that pKa and Ka have inverse log relationship, so if we want the Ka to increase (become more acidic), pKa must decrease.

I used that reasoning to get the answer as well. I should have taken my biochem class more seriously.

 

[BerkReviewTeach]

One thing to keep in mind on this question is that we are only considering the side chain. And the sidechain pK_a is for the protonated, cationic form of the sidechain. In water the charge is stabilized by the hydrogen bonding of the surrounding solvent, so it can remain charged. But in a hydrophobic environment, that charge is not stabilized, so the H⁺ will leave and find a negative charge to offset. This means that the sidechain will lose the H⁺ more readily in a hydrophobic environment than an aqueous environment, so it is a better proton donor in the hydrophobic solvent, making it more acidic. This would mean it has a lower pK_a.

Is this more or less what the BR explanation says?

 

[Temperature101]

One thing to keep in mind on this question is that we are only considering the side chain. And the sidechain pK_a is for the protonated, cationic form of the sidechain. In water the charge is stabilized by the hydrogen bonding of the surrounding solvent, so it can remain charged. But in a hydrophobic environment, that charge is not stabilized, so the H⁺ will leave and find a negative charge to offset. This means that the sidechain will lose the H⁺ more readily in a hydrophobic environment than an aqueous environment, so it is a better proton donor in the hydrophobic solvent, making it more acidic. This would mean it has a lower pK_a.

Is this more or less what the BR explanation says?

Somewhat but they did not get into H-bonding...