The structure of a basic amino acid, lysine, is given below:
`(HOOC-underset(.^(o+)NH_(3))underset(|)(C)H-(CH_(2))_(3)-CH_(2)overset(o+)(N)H_(3))(A)`
The `pK_(a_1),pK_(a_2),` and `PK_(a_3)` of `(A)`, respectively, are: `2.18,8.95`, and `10.53`.
What is the `pH` at isoelectric points `(pI)`?

Options

A. `(pK_(a_(1))+pK_(a_(2)))/(2)`
B. `(pK_(a_(1))+pK_(a_(3)))/(2)`
C. `(pK_(a_(2))+pK_(a_(3)))/(2)`
D. `(pK_(a_(1))+pK_(a_(2))+pK_(a_(3)))/(3)`

Answer

To find the pH at the isoelectric point (pI) of the amino acid lysine, we can follow these steps:

### Step 1: Understand the concept of isoelectric point (pI)
The isoelectric point (pI) is the pH at which an amino acid exists predominantly in its zwitterionic form, meaning it has no net charge. For amino acids, the pI is calculated as the average of the pKa values of the ionizable groups that are closest to each other around the zwitterionic form.

### Step 2: Identify the relevant pKa values
For lysine, the given pKa values are:
- pKa1 = 2.18 (carboxylic group)
- pKa2 = 8.95 (α-amino group)
- pKa3 = 10.53 (side chain amino group)

### Step 3: Determine the pKa values to use for calculating pI
Since lysine is a basic amino acid, the zwitterionic form occurs between the pKa values of the α-amino group (pKa2) and the side chain amino group (pKa3). Therefore, we will use pKa2 and pKa3 to calculate the pI.

### Step 4: Calculate the pI
The formula to calculate the isoelectric point (pI) is:
\[ pI = \frac{pKa2 + pKa3}{2} \]

Substituting the values:
\[ pI = \frac{8.95 + 10.53}{2} \]

### Step 5: Perform the calculation
Calculating the sum:
\[ pI = \frac{19.48}{2} = 9.74 \]

### Conclusion
The pH at the isoelectric point (pI) of lysine is **9.74**.

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