Can someone run through how to do this? I'm reading TBR and it says "pI is found by summing the number of basic amino acids plus one, and taking the pKa with that numerical subscript and averaging it with the next pKa value sequentially up." I don't really understand the "summing basic amino acids plus one" part
pI calculation
- Thread starter JDbb
- Start date
- Joined
- Jul 23, 2011
- Messages
- 850
- Reaction score
- 20
are you sure this is an MCAT topic...sound like overkills...
so...here how you calculate your pI:
you know animo acid have different functional group that have differnt pKa. The OH group have a pkA of 2 and the NH3 have pkA of 9. So, at pH 7, the OH lose the H and become O-, while the NH3+ still retain its H, the overall charge on the amino acid is (-1) + 1 = 0.
When you try to find the pI, you start from the state in which the amino acid charge is 0 (a Zwitterion). next, you find the state where it has a (+1) overall charge. For a simple nonpolar uncharged aminoacid, the (+1) state is when ph < 2 (OH group still intact). Then you find the (-1) state, which in the same case, is when NH3+ had become NH2. You observed, that the 2 functional group participating in change in overcharged is the OH group and the NH3 group, which respectively have pkA of 2 and 9. the pI is 2+9/2 = 5.5 This is also the pI for all nonpolar uncharged amino acids.
ofc, when you are dealing with charged or some polar uncharged with side chains having pkA or even a peptides, things are going to be more complicated. But, the process is still the same, find the -1 and +1 state, and indicate which functional groups or side chains is causing the change in charge; then average their pkA.
I doubt this is going to be on the MCAT, to be able to do this, you need to memorize the structure of all amino acids as well as their pkA
so...here how you calculate your pI:
you know animo acid have different functional group that have differnt pKa. The OH group have a pkA of 2 and the NH3 have pkA of 9. So, at pH 7, the OH lose the H and become O-, while the NH3+ still retain its H, the overall charge on the amino acid is (-1) + 1 = 0.
When you try to find the pI, you start from the state in which the amino acid charge is 0 (a Zwitterion). next, you find the state where it has a (+1) overall charge. For a simple nonpolar uncharged aminoacid, the (+1) state is when ph < 2 (OH group still intact). Then you find the (-1) state, which in the same case, is when NH3+ had become NH2. You observed, that the 2 functional group participating in change in overcharged is the OH group and the NH3 group, which respectively have pkA of 2 and 9. the pI is 2+9/2 = 5.5 This is also the pI for all nonpolar uncharged amino acids.
ofc, when you are dealing with charged or some polar uncharged with side chains having pkA or even a peptides, things are going to be more complicated. But, the process is still the same, find the -1 and +1 state, and indicate which functional groups or side chains is causing the change in charge; then average their pkA.
I doubt this is going to be on the MCAT, to be able to do this, you need to memorize the structure of all amino acids as well as their pkA
This is the shortcut method:
1.) Assume the peptide chain is at a pH of 1.
2.) At a pH of 1, every acidic and basic side chain is protonated, as well as the amino and carboxylic acid terminus. Only basic aminoacids (Histidine, Lysine, Arginine) and the amino terminus, when protonated have a +1 charge.
3.) Add up the total charge. So if there were 3 basic aminos acids in your polypeptide, 3 (basic aminoacids) + 1 (aminoterminal) = a total of +4.
4.) This means, the pI is between pKa4 and pKa5. Add both and divide by 2. This is what the pI is equal to. The idea here is that:
@ Above pKa1 --> -COOH gets deprotonated to COO- (neg charge)
Total Charge: -1 and +4 = +3
@ Above pKa2 --> The first basic amino acid gets deprotonated (neutral).
Total Charge: -1 and +3 = +2
@ Above pKa3 --> The second basic amino acid gets deprotonated (neutral).
Total Charge: -1 and +2 = +1
@Above pKa4 --> The third basic amino acid gets deprotonated (neutral).
Total Charge: -1 and +1 = 0 (ZWITTER ION exits between pKa4 and pKa5)
@Above pKa5 --> The aminoterminus gets deprotonated (neutral)
Total Charge: -1 and 0 = -1
Note: Each Basic aminoacids that gets deprotonated go from +1 to 0 (neutral charge): ie. NH4+ to NH3
The pI is an average of the two pKa's that lead to a zwitteron (neutral species) -- the pKa that leads TO the zwitteron and the pKa that leads AWAY from the zwitteron. In this case, the pI is found by averaging pKa4 and pKa5 (pKa4 +pKa5/2).
For a simple aminoacid: Add the two groups with the closest pKa's:
If it's an acidic R group, add it's pKa to -COOH's pKa and divide by 2.
If it's a basic R group, add it's pKa to -NH4+'s pKa and divide by 2.
If it's a neutral (polar or nonpolar aa), average the amino and carboxylic acid pKa's.
1.) Assume the peptide chain is at a pH of 1.
2.) At a pH of 1, every acidic and basic side chain is protonated, as well as the amino and carboxylic acid terminus. Only basic aminoacids (Histidine, Lysine, Arginine) and the amino terminus, when protonated have a +1 charge.
3.) Add up the total charge. So if there were 3 basic aminos acids in your polypeptide, 3 (basic aminoacids) + 1 (aminoterminal) = a total of +4.
4.) This means, the pI is between pKa4 and pKa5. Add both and divide by 2. This is what the pI is equal to. The idea here is that:
@ Above pKa1 --> -COOH gets deprotonated to COO- (neg charge)
Total Charge: -1 and +4 = +3
@ Above pKa2 --> The first basic amino acid gets deprotonated (neutral).
Total Charge: -1 and +3 = +2
@ Above pKa3 --> The second basic amino acid gets deprotonated (neutral).
Total Charge: -1 and +2 = +1
@Above pKa4 --> The third basic amino acid gets deprotonated (neutral).
Total Charge: -1 and +1 = 0 (ZWITTER ION exits between pKa4 and pKa5)
@Above pKa5 --> The aminoterminus gets deprotonated (neutral)
Total Charge: -1 and 0 = -1
Note: Each Basic aminoacids that gets deprotonated go from +1 to 0 (neutral charge): ie. NH4+ to NH3
The pI is an average of the two pKa's that lead to a zwitteron (neutral species) -- the pKa that leads TO the zwitteron and the pKa that leads AWAY from the zwitteron. In this case, the pI is found by averaging pKa4 and pKa5 (pKa4 +pKa5/2).
For a simple aminoacid: Add the two groups with the closest pKa's:
If it's an acidic R group, add it's pKa to -COOH's pKa and divide by 2.
If it's a basic R group, add it's pKa to -NH4+'s pKa and divide by 2.
If it's a neutral (polar or nonpolar aa), average the amino and carboxylic acid pKa's.
So how do I number the pKas? From lowest to highest?
And basically, the total number of charges is equal to which pKa I choose? (if charge=+7, then I average pKa7 and pKa8)
@tn4596: I'm not sure, it was just in TBR so that's why I'm asking
And basically, the total number of charges is equal to which pKa I choose? (if charge=+7, then I average pKa7 and pKa8)
@tn4596: I'm not sure, it was just in TBR so that's why I'm asking
You just need to realize that:
The lowest pKa is always going to be -COOH
Every pKa after that is either from an Acidic or Basic amino acid, or the NH4+ terminus.
If they gave you a table of pKa's, pKa1 corresponds to the lowest pKa (COOH), the next lowest = pKa2, the third lowest = pKa3, and so on.
And yes to your second question.
- Joined
- Jul 23, 2011
- Messages
- 850
- Reaction score
- 20
I think if you are asking this question, you really need to review how pkA affect charges in amino acid. Majik explanation is right but it is hard to explain this concept without structure...try to google some powerpoint about amino acid, it should makes more sense.
Similar threads
- Question
- Question
