Isoelectric point example (Kaplan)

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CaptainAmazon

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I was doing an isoelectric point calculation with the given information: “Aspartic acid (pKa1 = 1.88, pKa2 = 3.65, pKa3 = 9.60)”
This is an acidic AA, so I used pI = (pKa1 + pKa3) / 2 and got 5.74. But the answer for the concept check says “pI = (1.88 + 3.65) ÷ 2 = 2.77” ... could someone please explain?

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You took the pKa of the wrong group when calculating the PI. With amino acids, you'll learn that there are exceptions to many of the general trends.

PI = pH at which the species net charge is '0'.

Aspartic acid pKas:

1.9 - carboxyl
3.7 - side chain
9.6 - amino group

Think about the species at low, neutral and high pH. When pKa > pH, the molecule wins the proton.

Low pH (~1): everything is protonated
1.9 COOH
3.7 R-COOH
9.6 NH3+
--------------------------------------
+1 charge


Increase the pH so it is between the first and second groups with lowest two pKas e.g.,

pH ~3
1.9 COO-
3.7 R-COOH
9.6 NH3+
--------------------------------------
+0 charge


An easy way to determine which pKas to use in your calculation would be to order them from least to greatest and draw the species' net charge in between each pKa on a pH gradient. For example:

0---------1.9----------3.7--------9.6------------pKa
-----+1----------0---------- -1 ---------- -2 ---net charge
0----------2-----------3----------10-------------pH


(not drawn to scale)

Since neutral charge lies between the groups with pKas of 1.9 and 3.7, you'd use those numbers when calculating PI.

Let me know if I can expand or clarify.
 
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You took the pKa of the wrong group when calculating the PI. With amino acids, you'll learn that there are exceptions to many of the general trends.

PI = pH at which the species net charge is '0'.

Aspartic acid pKas:

1.9 - carboxyl
3.7 - side chain
9.6 - amino group

Think about the species at low, neutral and high pH. When pKa > pH, the molecule wins the proton.

Low pH (~1): everything is protonated
1.9 COOH
3.7 R-COOH
9.6 NH3+
--------------------------------------
+1 charge


Increase the pH in between the first and second groups with lowest pH e.g.,

pH ~3
1.9 COO-
3.7 R-COOH
9.6 NH3+
--------------------------------------
+0 charge


An easy way to determine which pKas to use your your calculation would be to order them from least to greatest and draw the species' net charge in between each pKa on a pH gradient. For example:

0---------1.9----------3.7--------9.6------------pKa
-----+1----------0---------- -1 ---------- -2 ---net charge
0----------2-----------3----------10-------------pH


(not drawn to scale)

Since neutral charge lies between the groups with pKas of 1.9 and 3.7, you'd use those numbers when calculating PI.

Let me know if I can expand or clarify.
Thank you so much! I think this makes sense to me. I'll do some more practice and if I have any trouble I will let you know.
 
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You took the pKa of the wrong group when calculating the PI. With amino acids, you'll learn that there are exceptions to many of the general trends.

PI = pH at which the species net charge is '0'.

Aspartic acid pKas:

1.9 - carboxyl
3.7 - side chain
9.6 - amino group

Think about the species at low, neutral and high pH. When pKa > pH, the molecule wins the proton.

Low pH (~1): everything is protonated
1.9 COOH
3.7 R-COOH
9.6 NH3+
--------------------------------------
+1 charge


Increase the pH so it is between the first and second groups with lowest two pKas e.g.,

pH ~3
1.9 COO-
3.7 R-COOH
9.6 NH3+
--------------------------------------
+0 charge


An easy way to determine which pKas to use your your calculation would be to order them from least to greatest and draw the species' net charge in between each pKa on a pH gradient. For example:

0---------1.9----------3.7--------9.6------------pKa
-----+1----------0---------- -1 ---------- -2 ---net charge
0----------2-----------3----------10-------------pH


(not drawn to scale)

Since neutral charge lies between the groups with pKas of 1.9 and 3.7, you'd use those numbers when calculating PI.

Let me know if I can expand or clarify.

That is an excellent post!!!
 
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