See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: The isoelectric point (pI) is the pH at which the amino acid carries no net charge (zwitterionic form predominates). At pH values below the pI, the amino acid is protonated (positively charged); at pH values above the pI, it is deprotonated (negatively charged). Step 1: Identify the pI. The pI of this amino acid is 6.0. Step 2: Compare the solution pH to the pI. The solution pH is 1.0, which is much lower (more acidic) than the pI of 6.0. Step 3: Determine the protonation state. At pH 1.0, the solution is highly acidic. Both ionizable groups—the alpha-amino group (typical pKa ~9) and the alpha-carboxyl group (typical pKa ~2)—will be fully protonated. The amino group exists as -NH3+ and the carboxyl group exists as -CO2H (undissociated acid). Step 4: Identify the predominant species. The fully protonated species has -NH3+ (positively charged amino group) and -CO2H (neutral carboxylic acid), giving a net positive charge. This corresponds to option (a): H3N+(CH)(R)CO2H. Step 5: Eliminate other options. - Option (b) H2N(CH)(R)CO2H: This has a free amine (unprotonated), which would not be favored at low pH. - Option (c) H3N+(CH)(R)CO2-: This is the zwitterion (predominates near pI = 6.0), not at pH 1.0. - Option (d) H2N(CH)(R)CO2-: This has both groups deprotonated, favored only at high pH. At pH 1.0, which is well below the pKa of the carboxyl group (~2) and far below the pKa of the amino group (~9), the predominant species is the fully protonated cation: H3N+(CH)(R)CO2H. Therefore, the correct answer is A.