Final answer:
The net charge of a peptide at pH 7.4 is determined based on the ionizable groups of its amino acid residues. Without the specific sequence, an example with a net charge of +2 was given, indicating that the peptide likely has more basic residues protonated than acidic residues deprotonated at this pH.
Step-by-step explanation:
To determine the net charge of a peptide in a neutral buffer (pH = 7.4), it is important to consider the ionizable groups within the peptide. Amino acids may have side chains that are basic, acidic, or neutral. At a pH of 7.4, the -COOH (carboxyl group) will be deprotonated and carry a negative charge, and the -NH2 (amino group) will be protonated and carry a positive charge. If the side chains of the amino acids have additional acidic or basic groups, these may also gain or lose protons based on their pKa values and the pH of the solution.
Given that the isoelectric pH (pI) is the pH at which an amino acid has no net charge, and considering that physiological pH is 7.4, basic amino acids with side chains that are positively charged at this pH will have a pI higher than 7.4, and acidic amino acids with side chains that are negatively charged at this pH will have a pI lower than 7.4. In the context of a peptide, this analysis must be done for each amino acid residue.
Without the specific sequence of the peptide, we cannot determine the exact net charge. However, if the peptide consisted of basic residues with a combined charge of 6 positive charges, and acidic residues summing to 6 negative charges, the net charge would be 0 at pH 7.4. If there were 8 positive charges and 6 negative ones, the net charge would be +2. Conversely, if there were 6 positive charges and 9 negative ones, the net charge would be -3. Since it was mentioned that "The net charge is +2", the most accurate conclusion, given the information, is that the peptide will have a net charge of +2 at pH 7.4.