SME
Biochemistry
Protein structure
Use this for “Lysine has pKₐ values 2.18, 8.95 and 10.53 (side chain). Find its isoelectric point and its net charge at pH 7.40.”
Diagram
the
amino
acid:
pI
=
8.95 + 10.53
2
=
9.74,
the
mean
of
the
pKₐ
values
either
side
of
the
neutral
form.
At pH 7.40 the net charge is 0.972.
Each group loses its proton half-way through its step, where pH = pKₐ.
This is an example. Type your own values to start a new problem.
R
Side chain
example
pKₐ₁
Carboxyl pKₐ
example
pKₐ₂
Amino pKₐ
example
pKₐR
Side-chain pKₐ
example
pH
pH
example
z
Net charge
worked out (not typed)
pI
Isoelectric point
worked out (not typed)
Formulas
z
=
−
1
1 +
10
pKₐ₁ − pH
+
1
1 +
10
pH − pKₐ₂
+
the
share
of
the
R
side
chain
+0.9719
=
−
1
1 +
10
2.18 − 7.4
+
1
1 +
10
7.4 − 8.95
+
the
share
of
the
basic
side
chain
pI
=
mean
of
the
two
pKₐ
either
side
of
the
neutral
form,
from
pKₐ₁,
pKₐ₂
and
the
R
side
chain
9.74 = mean of the two pKₐ either side of the neutral form, from 2.18, 8.95 and the basic side chain
Assumptions
•
A free amino acid in water. Each group is half ionized at pH = pKₐ, and each pH unit away shifts it tenfold (Henderson–Hasselbalch).
•
The carboxyl group carries −1 once it loses its proton and the amino group +1 while it holds one; an acidic side chain counts −1, a basic one +1.
•
Typical pKₐ₁, pKₐ₂ and pKₐR: glycine 2.34 and 9.60; aspartate 1.88, 9.60 and 3.65; lysine 2.18, 8.95 and 10.53.
•
pI is the pH where the net charge is 0: the mean of the two lowest pKₐ for an acidic side chain, the two highest for a basic one.
Step-by-step
WE KNOW
R: basic, pKₐ₁ = 2.18, pKₐ₂ = 8.95, pKₐR = 10.53, pH = 7.4
FIND
net charge (z), isoelectric point (pI)
Step 1 · Find net charge (z)
Use z = −1 ÷ (1 + 10pKₐ₁ − pH) + 1 ÷ (1 + 10pH − pKₐ₂) + the share of the R side chain
Add each group’s share: −1 times the carboxyl’s ionized share, +1 times the amino group’s protonated share, and the side chain’s.
z
=
−
1
1 +
10
pKₐ₁ − pH
+
1
1 +
10
pH − pKₐ₂
+
1
1 +
10
pH − pKₐR
z
=
−
1
1 +
10
2.18 − 7.4
+
1
1 +
10
7.4 − 8.95
+
1
1 +
10
7.4 − 10.53
z
=
−
1
1 +
10
−5.22
+
1
1 +
10
−1.55
+
1
1 +
10
−3.13
z
=
−
1
1 + (6.026 ×
10
-6
)
+
1
1 + 0.02818
+
1
1 + 0.0007413
z
=
−1
+
1
1.028
+
1
1.001
z
=
−1
+
0.9728
+
0.999
z
=
+0.9719
Step 2 · Find isoelectric point (pI)
Use pI = mean of the two pKₐ either side of the neutral form, from pKₐ₁, pKₐ₂ and the R side chain
A basic side chain: the form with no net charge has lost two protons, so it sits between the two highest pKₐ.
pI
=
pKₐ₂ + pKₐR
2
pI
=
8.95 + 10.53
2
pI
=
19.48
2
pI = 9.74
Check
Put the numbers back in. Both sides should match.
0.9719
=
−
1
1 +
10
2.18 − 7.4
+
1
1 +
10
7.4 − 8.95
+
1
1 +
10
7.4 − 10.53
✓
9.74
=
8.95 + 10.53
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Length of an α-helix or a β-strand
Use this for “How long is a stretch of 18 residues as an α-helix, and how many turns does it make? How long would it be as a β-strand?”
Levels of protein structure
Use this for “Is a disulfide bond between two cysteines far apart in the chain part of the primary, secondary, tertiary or quaternary structure?”