Draw histidine at pH 5, pH 8, and pH 10
-please show the net charge on each of the structures
-calculate the pI value of histidine
the pKa values for the amino acid histidine are:
pKa=1.82 for the carboxyl group
pKa=6.0 for its imidazole group
pKa=9.17 for its amino group
please show your work, and thank you for your help!
The Correct Answer and Explanation is:
Here is the structure of histidine at pH 5, pH 8, and pH 10, the net charges, and the calculation of the isoelectric point (pI) based on the provided pKa values:
1. Histidine Structures and Net Charges
pKa values:
- Carboxyl group (–COOH): pKa = 1.82
- Imidazole side chain: pKa = 6.00
- Amino group (–NH3⁺): pKa = 9.17
a. pH 5
At pH 5:
- Carboxyl group: deprotonated (–COO⁻), because pH > 1.82
- Amino group: still protonated (–NH3⁺), because pH < 9.17
- Imidazole group: protonated, because pH < 6.0 → positively charged
Net charge = +1 (NH3⁺) −1 (COO⁻) +1 (protonated imidazole) = +1
b. pH 8
At pH 8:
- Carboxyl group: deprotonated (–COO⁻)
- Amino group: protonated (–NH3⁺)
- Imidazole group: deprotonated, because pH > 6.0 → neutral
Net charge = +1 (NH3⁺) −1 (COO⁻) +0 (imidazole) = 0
c. pH 10
At pH 10:
- Carboxyl group: deprotonated (–COO⁻)
- Amino group: deprotonated to –NH2 → neutral
- Imidazole group: deprotonated → neutral
Net charge = 0 (NH2) −1 (COO⁻) +0 (imidazole) = −1
2. Isoelectric Point (pI) Calculation
The pI is the pH where net charge = 0. For histidine, this occurs between the states where:
- One form has +1 charge (pH 5)
- One form has 0 charge (pH 8)
The neutral form is between:
- Protonated imidazole group (pKa = 6.0)
- Protonated amino group (pKa = 9.17)
So, take average of pKa for imidazole and amino: pI=6.00+9.172=15.172=7.59\text{pI} = \frac{6.00 + 9.17}{2} = \frac{15.17}{2} = \boxed{7.59}
Summary Table
| pH | Carboxyl | Amino | Imidazole | Net Charge |
|---|---|---|---|---|
| 5 | −1 | +1 | +1 | +1 |
| 8 | −1 | +1 | 0 | 0 |
| 10 | −1 | 0 | 0 | −1 |
Explanation (≈300 words)
Histidine is a polar amino acid with a unique imidazole side chain that makes its acid–base behavior distinctive. Understanding how histidine behaves at different pH values is crucial for interpreting protein structure, enzyme function, and buffering capacity.
At low pH (pH 5), the environment is acidic, so all ionizable groups that can hold protons do so. The carboxyl group loses its proton first (pKa 1.82), so it exists as –COO⁻. The amino group remains protonated (–NH3⁺), and the imidazole side chain, with a pKa of 6.0, is still protonated and positively charged. Thus, the net charge on histidine is +1.
At neutral to slightly basic pH (pH 8), the imidazole group loses its proton (since pH > pKa 6.0), becoming neutral. The carboxyl is deprotonated (–COO⁻), and the amino group remains protonated (–NH3⁺) since the pH is still below its pKa (9.17). The charges balance out, giving histidine a net charge of 0. This state is called the zwitterion, where the molecule is electrically neutral but has charged groups.
At pH 10, the environment is more basic than all three pKa values. Thus, the carboxyl is deprotonated (–COO⁻), the amino group is also deprotonated to –NH2 (neutral), and the imidazole remains neutral. The molecule carries a net charge of −1.
The isoelectric point (pI) of histidine lies between the pKa values where the molecule transitions between +1 and 0, and 0 and –1. For histidine, this is between the imidazole (6.0) and amino group (9.17). Averaging these gives pI = 7.59, meaning that at this pH, histidine exists predominantly in its zwitterionic, uncharged form.
