# tbr cbt 4 # 140 pH in Gel electrophoresis

#### Whiteshoes

##### Member
10+ Year Member
Ions can migrate in an electric field. If the buffered solution used in gel electrophoresis were set at pH 9.0, vasopressin would migrate toward the:

negatively charged cathode.
supplementary information:
Vasopressin. The pKa of the N-terminus is 8.0, the phenolic hydroxyl is 10.0, and the guanidino group is 12.5.

tbr cbt explanation:
First, we must determine the charge on vasopressin when the pH of the surrounding solution is 9.0. A pH of 9.0 is about halfway between a pH of 8.0 (vasopressin has a pKa = 8.0) and a pH of 10.0 (vasopressin has a pKa = 10.0). The charge on vasopressin at pH 9.0 is therefore +1 (given that it carries a +0.5 charge at pH = 8.0 and +1.5 charge at pH = 10.0). This means that vasopressin is a cation at a pH of 9.0. In biology, a positively charged cation is attracted to the cathode, so the cathode must be negatively charged. The best answer is D.
I understand that the cathode is connected to the Negative terminal of the batter so that's why a positive charge would move towards a cathode but I'm having trouble understanding how they figured out that at a ph of 9 vasopressin was positively charged +1. please help.

#### riddler

##### Resident
7+ Year Member
Were the pKas given in a passage?

In any case, since the pKa of the N-Terminus is 8 and the pH of the solution is 9, pKa<pH and the N-terminus will be deprotonated. The phenolic hydroxyl and guanidine group have pKa > pH, and so they will be protonated.

Thus, we have one that is deprotonated (-1), and two that are protonated (+1 and +1). 1+1-1 = +1.

That's how I would think about it...not sure if it's correct reasoning.

#### Whiteshoes

##### Member
10+ Year Member
Were the pKas given in a passage?

In any case, since the pKa of the N-Terminus is 8 and the pH of the solution is 9, pKa<pH and the N-terminus will be deprotonated. The phenolic hydroxyl and guanidine group have pKa > pH, and so they will be protonated.

Thus, we have one that is deprotonated (-1), and two that are protonated (+1 and +1). 1+1-1 = +1.

That's how I would think about it...not sure if it's correct reasoning.
oh so that's how you calculate it? I didn't understand what they meant by .5 and 1.5 in their explanation.. but your explanation makes sense.. thanks!

#### PopeJoja

(apologies for high jacking the thread)

I am having trouble with understanding how you denote what the charge is on the amino acids based on the pH and pka differences.

For instance, on TBR 4 #:111. At pH = 7, the charge on the eleven-amino acid polypeptide would be:

A. &#8211;3. B. 0. B is the best answer. Of the eleven amino acids in the protein, none has an active proton on the side chain. To determine the charge of the entire polypeptide, only the charge on the two terminals must be known. The amino terminal is basic, so at pH = 7, the pH < pKa (RNH3+). Thus, the amino terminal is protonated and carries a positive charge. The carboxyl terminal is acidic, so at pH = 7, the pH > pKa (RCO2H). Thus, the carboxyl terminal is deprotonated and carries a negative charge. The amino terminal is positive and the carboxyl terminal is negative, meaning that the polypeptide is a zwitterion and has a net charge of zero. The best answer is B.

C. +1. D. +4.

Can someone be helpful in explaining how to do this? Thanks a million.

#### kehlsh

##### Medic Commando
7+ Year Member
(apologies for high jacking the thread)

I am having trouble with understanding how you denote what the charge is on the amino acids based on the pH and pka differences.

For instance, on TBR 4 #:111. At pH = 7, the charge on the eleven-amino acid polypeptide would be:

A. 3. B. 0. B is the best answer. Of the eleven amino acids in the protein, none has an active proton on the side chain. To determine the charge of the entire polypeptide, only the charge on the two terminals must be known. The amino terminal is basic, so at pH = 7, the pH < pKa (RNH3+). Thus, the amino terminal is protonated and carries a positive charge. The carboxyl terminal is acidic, so at pH = 7, the pH > pKa (RCO2H). Thus, the carboxyl terminal is deprotonated and carries a negative charge. The amino terminal is positive and the carboxyl terminal is negative, meaning that the polypeptide is a zwitterion and has a net charge of zero. The best answer is B.

C. +1. D. +4.

Can someone be helpful in explaining how to do this? Thanks a million.
look at the side chains, there are no acidic/basic amino acids
so the polypeptide stays as zwitterion over all