This question involves determining the charge of peptides at pH 7.0 based on their side chains (R groups). At pH 7.0, which is neutral, the charge of a peptide depends on the ionization states of its acidic and basic groups.

The general structure given is a dipeptide with two variable substituents R₁ and R₂. Each peptide also has a free amino group (NH₂) at the N-terminus and a free carboxylic acid group (COOH) at the C-terminus. These terminal groups can ionize:

• The N-terminal amino group can accept a proton and become positively charged (NH₃⁺).
• The C-terminal carboxylic acid group can donate a proton and become negatively charged (COO^-).

Additionally, the side chains R₁ and R₂ may contain ionizable groups that affect the overall charge.

Key ionizable groups and their behavior at pH 7.0:

• Carboxylic acid group (-COOH): pK_a ~4-5. At pH 7.0 (which is above its pK_a), it is deprotonated and carries a negative charge (COO^-).
• Amino group (-NH₂): pK_a ~9-10. At pH 7.0 (which is below its pK_a), it is protonated and carries a positive charge (NH₃⁺).
• Amide group (-CONH₂): Not ionizable; neutral at all pH.
• Hydroxyl group (-OH): Not significantly ionizable at pH 7; neutral.
• Alkyl group (e.g., -CH₃): Neutral.

Step-by-step analysis for each peptide:
For each peptide, we must consider the charges from:

1. N-terminal NH₂ group: +1 charge (protonated at pH 7.0).
2. C-terminal COOH group: -1 charge (deprotonated at pH 7.0).
3. Side chains R₁ and R₂: Additional positive or negative charges based on their ionizable groups.

The net charge is the sum of all these charges. A peptide is positively charged if the net charge > 0.

Detailed peptide analysis:

• Peptide I: R₁ = H, R₂ = H (both neutral). Net charge = +1 (N-term) + (-1) (C-term) = 0. Not positive.
• Peptide II: R₁ = H, R₂ = CH₃ (neutral). Net charge = +1 -1 = 0. Not positive.
• Peptide III: R₁ = CH₂COOH (contains carboxylic acid, so negative charge at pH 7), R₂ = H (neutral). Net charge = +1 (N-term) + (-1) (C-term) + (-1) (R₁) = -1. Not positive.
• Peptide IV: R₁ = CH₂CONH₂ (neutral amide), R₂ = (CH₂)₄NH₂ (contains amino group, so positive charge). Net charge = +1 (N-term) + (-1) (C-term) + (+1) (R₂) = +1. Positively charged.
• Peptide V: R₁ = CH₂CONH₂ (neutral), R₂ = CH₂CONH₂ (neutral). Net charge = +1 -1 = 0. Not positive.
• Peptide VI: R₁ = (CH₂)₄NH₂ (positive), R₂ = (CH₂)₄NH₂ (positive). Net charge = +1 (N-term) + (-1) (C-term) + (+1) (R₁) + (+1) (R₂) = +2. Positively charged.
• Peptide VII: R₁ = CH₂COOH (negative), R₂ = CH₂CONH₂ (neutral). Net charge = +1 -1 -1 = -1. Not positive.
• Peptide VIII: R₁ = CH₂OH (neutral), R₂ = (CH₂)₄NH₂ (positive). Net charge = +1 -1 +1 = +1. Positively charged.
• Peptide IX: R₁ = (CH₂)₄NH₂ (positive), R₂ = CH₃ (neutral). Net charge = +1 -1 +1 = +1. Positively charged.

Final Answer: Peptides IV, VI, VIII, and IX are positively charged. So, the number is 4.

Related Topics and Formulae

Isoelectric Point (pI): The pH at which a molecule carries no net electrical charge. For amino acids and peptides, it is calculated as the average of the pK_a values of the ionizable groups.

Henderson-Hasselbalch Equation: Used to calculate the pH of a solution when the ratio of conjugate base to acid is known: $\mathrm{pH}=\mathrm{pK}a+\log \left[\frac{[A{]}^{-}}{\left[\mathrm{HA}\right]}\right]$

Amino Acid Ionization: Amino acids have at least two ionizable groups: the amino group (pK_a ~9-10) and the carboxyl group (pK_a ~2-3). Some have additional ionizable side chains.