Draw the structure of Leu and determine its pI. pK _a1 = 9.6. pK_a2 = 2.4.
a. 7.59                                          
b. 6.0
c. 3.91
d. 5.5

• A. 7.59
• B. 6.0
• C. 3.91
• D. 5.5

Answer: (B)

Verified step by step guidance

1. Identify the structure of leucine (Leu), which is an amino acid with the chemical formula C6H13NO2. It has a central carbon (alpha carbon) bonded to an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and an isobutyl side chain (CH2-CH-(CH3)2).
2. Understand that the isoelectric point (pI) of an amino acid is the pH at which the molecule carries no net electric charge. For amino acids without ionizable side chains, the pI is the average of the pKa values of the amino and carboxyl groups.
3. Given the pKa values: pKa1 (carboxyl group) = 2.4 and pKa2 (amino group) = 9.6, calculate the pI by averaging these two pKa values. The formula for pI is: pI = (pKa1 + pKa2) / 2.
4. Substitute the given pKa values into the formula: pI = (2.4 + 9.6) / 2.
5. Calculate the average to find the pI, which will give you the isoelectric point of leucine.