# Calculating Exact Charge & pI of Amino acids,

# Peptides and other Molecules

## 1. Introduction

Amino acids are classified into nonpolar (hydrophobic), polar-uncharged and polar-charged depending on the polarity and charge on their side chain at the neutral pH. They are also classified into neutral, acidic and basic amino acids depending on the acid-base behavior of their R-groups. The side chain of a neutral amino acid is unionizable at all pH. The side chain of an acidic amino acid generally acts as a weak acid and may donate an H^{+} to the solution. The side chain of a basic amino acid generally acts as a weak base and may accept an H^{+} from the solution. Though the side chains of tyrosine and cysteine are mostly uncharged at the neutral pH, they may bear non-zero charge at a pH near to or above their respective pKa values. For example, the side chain of cysteine (pKa = 8.14) bears -0.19 and -0.99-unit charge at pH 7.50 and 10.0, respectively. The side chain of tyrosine (pKa = 10.10) bears -0.44 and -0.99-unit charge at pH 10.0 and 12.0, respectively. At pH values close to or above their respective pKa, ignoring the charges on the side chain of these residues would lead to significant errors in charge and pI values. So, we will treat these two amino acids as acidic amino acids. These two amino acids have acidic side chains because they can donate a proton under suitable conditions. Though their R-groups are around 10^{4} to 10^{6} times weaker acid than those of carboxylic acid groups of amino acids, they act as the catalytic residue at the catalytic center of some enzymes. In this context, the amino acids are categorized depending on the ionizability of their R-groups as follow-

All the standard amino acids have at least two ionizable groups- I. a -NH_{2} group (-NH or imine or secondary amine group in proline) at N-terminal, and II. a -COOH at C-terminal. The acidic, as well as the basic amino acids, have an additional ionizable group- their acidic or basic R-groups, respectively. An ionizable group may act as either a weak acid or a weak base depending on its ability to __donate__ or __accept__ a proton (H^{+}), respectively. The pKa table of amino acids lists a maximum of three pKa values, namely pKa_{1}, pKa_{2}, and pKa_{3}. pKa_{1}, pKa_{2}, and pKa_{3} are the pKa value of the C-terminal -COOH group (a-carboxyl group), N-terminal -NH_{2} group (a-ammonium group) and the R-group (if any) of the specified amino acid. The pKa values may differ among the reference sources, thus may lead to variation in the result for the same amino acid. It is advised to always use the pKa values mentioned in the question or your textbook itself but not using any outside reference value.

The generalized reaction of weakly acidic and basic groups of amino acids can be written as-

Where -COOH is the acidic group at C-terminal of the amino acids, or in R-group of the acidic amino acids

-COO^{–} is the conjugate base of -COOH, and negatively charged.

-NH_{2} is the basic group at N-terminal of the amino acids, or in R-group of the basic amino acids

-NH_{3}^{+} is the conjugate acid of -NH_{2} and is positively charged.

At any pH, an acidic group can exist in either the uncharged, protonated (having an acidic proton, H^{+}) form -COOH, or negatively charged, deprotonated (has lost its acidic H^{+}) form -COO^{–}, but never in the positively charged form. Since the charge on a weakly acidic group is always fractional with a negative sign (will be discussed in coming sections), its value always lie in the range of -1 to 0. At any pH, a basic group can exist in either the uncharged, deprotonated (lacks an acidic proton, H^{+}) form -NH_{2}, or positively charged, protonated (has gained an acidic H^{+}) form -NH_{3}^{+}, but never in the negatively charged form. Since the charge on a weakly basic group is always fractional with a positive sign (will be discussed in coming sections), its value always lie in the range of 0 to +1.

The modified Henderson-Hasselbalch equations derived in the next sections are used to calculate the exact charge on any weakly acidic or basic group. It is also an excellent tool to mathematically determine the isoelectric pH (pI) of amino acids, peptides, drugs, pH indicators or any other molecule having at least one acidic and basic group. The use of Excel or similar data processing software may not be required for amino acids and simple molecules. However, the use of Excel would greatly simplify our work while dealing with complex molecules with multiple acidic and basic groups. It also increases productivity by minimizing the chances of calculation errors and typo errors. Besides calculating charge and pI, it is also the principal tool to correlated pH-dependent characteristics of molecules with one or more weakly acidic and basic groups. For example, percentage and fraction charged or neutral forms of a drug can be determined using this tool, which further gives its organ-specific absorption characteristic at different pH. It simplifies the calculation of pH-dependent fractional charged, uncharged and zwitterion forms of amino acids and its subsequent graphing. The pH-dependent net charge on molecules can also be calculated and graphed using the tool. Few of the applications are mentioned in this paragraph, more uses and applications of this tool may be explored as needed.

## 2. Henderson-Hasselbalch Equation for an Acidic Group

Let us begin with the classical Henderson-Hasselbalch Equation for weak acids-

Equation 2 can be also be re-arranged into equations 5 and 6 without taking the fraction approach as shown below. Let us begin with equation 2 once again.

Note that equations 9 and 10 are equivalent to equations 5 and 6, respectively.

Equations 5 and 9 only give the fraction of the weak acid in its conjugate base form A^{–}– the only charged species of the weak acid. And, the conjugate base A^{–} of a weak acid is negatively (-) charged. To calculate the negative charge on the weak acid (or, more precisely, its conjugate base), multiply the fraction ionized form [A^{–}] with (-1). That is, the charge on a weak acid species is numerically equal to the fraction of its ionized form (conjugate base), and the sign of the charge is negative (the sign of the charge on the conjugate base).

So, the fractional negative charge on the weak acid, C_{WA} is given by-

To distinguish from the classical Henderson-Hasselbalch equation, the derived equations 5 and 9 would now onwards be referred to as the modified Henderson-Hasselbalch equations for weak acids (for calculating charge).

**Example 1:** I. Calculate the protonated to deprotonated form ratio of the C-terminal of aspartic acid pH 3.0. Also, calculate II. its % and III. fractional charged or ionized form. Use pKa = 1.95.

Ans. Note that the C-terminal (-COOH) is a weak base. So, use the Henderson-Hasselbalch equation for a weak acid. In the case of a weak acid, only the conjugate base or deprotonated form -COO^{–} (= A^{–}) is the (negatively) charged species. So, we will calculate the % and fraction of A^{–} for the given acid.

Hence, fractional ionized form of the C-terminal = 0.91817

__Interpretation:__

I. (Protonated: Deprotonated) forms= 1/11.22. There are 11.22 charged acidic groups A^{–} per uncharged group AH (ignoring the whole number counts for molecules).

II. % Charge on C-terminal= 91.817%. Out of the total amount of the chemical species (both AH and A^{–}), 91.817% are in A^{–} or negatively charged form. Or, on a 100 molecules scale, 91.817 molecules (ignoring whole number counts for molecules) are in deprotonated state A^{–} and negatively charged. The rest 8.183% is in their protonated form AH and uncharged.

III. Fractional charge on C-terminal= 0.91817. Each acidic group has a fractional charge of 0.91817, that is, each acidic group bears a partial negative charge of approximately 0.91817 unit (but NOT a complete 1.0000 unit). It also indicates that, on a fraction scale where the sum of the ionized and unionized form of the basic group is taken unity, 0.91817 factions of unity molecules are in the ionized state.

**Example 2:** Aspirin or acetylsalicylic acid (pKa is 3.50) is one of the most commonly used analgesic drugs to treat pain, fever, and inflammations. Only the ionized form of this drug is readily absorbable (as suggested by a recent study on swiss albino mice). If a person takes a 1.000 g tablet with 500.0 mg of active drug, calculate- I. Amount of drug absorbed in the stomach (pH = 2.00), II. Amount and % of drug absorbed in the small intestine (pH = 7.20). Also, determine III. which of the two organs act as the preferred absorption site of the drug?

Step 2: A small quantity of aspirin is first absorbed in the stomach. The unabsorbed drug in the stomach further moves to the small intestine where it will be reabsorbed again. So, first, calculate the amount of unabsorbed drug in the stomach.

Amount of unabsorbed drug in stomach =

The initial mass of drug – Amount absorbed in the stomach

= 500 mg – 15.30 mg = 484.70 mg

Only 484.70 mg of aspiring will reach the small intestine for further absorption.

Now, the amount of drug absorbed in small intestine =

% absorption x Amount of unabsorbed drug in the stomach

= 99.98 % x 484.70 mg = 484.60 mg

Step 3: % drug absorbed = (Amount absorbed / Initial mass of drug) x 100

= (484.60 mg / 500 mg) x 100 = 96.92%

III. Determine the preferred absorption site of aspirin

Most of the aspirin is absorbed in the small intestine (96.92%, 484.60 mg). So, the small intestine acts as the preferred absorption site of aspirin.

2A. Determining pH for the Given Charge on an Acidic Group

As mentioned in the previous section, the modified Henderson-Hasselbalch equations 5 and 9 only give the fraction of the weak acid in its negatively charged conjugate base form- the only charged form of the weak acid. The charge on a weak acid species is numerically equal to the fraction of its ionized form (conjugate base), and the sign of the charge is negative.

To calculate the required pH for a specified charge on the weak acid, we need to calculate the pH for that fraction ionized form or the absolute charge (i.e. the magnitude of charge ignoring its negative sign). Let us begin with the modified Henderson-Hasselbalch equation for calculating the fraction ionization (or, absolute charge) of an acidic group.

Also, note that equation 12 is the same as the classical Henderson-Hasselbalch equation for a weak acid (equation 1), but in a different form where [AH] = 1 – [A^{–}]. Both the equations give fraction charge on an acidic group. However, unlike most other cases, equation 12 does not require the concentration of the weak acid, [AH].

**Example 3:** Calculate the pH at which the -COOH group of alanine bears a net charge of (-)0.50 units. Use, pK_{a} = 2.33.

Ans. Step 1: Given, charge on the acidic group, [A^{–}] = -0.50

Note that the -ve sign of the charge on the conjugate base of weak acid simply indicates the negative charge. The absolute value of the fraction of [A^{–}] is equal to the absolute value of charge, that is, 0.50 in this case.

So, log {[A^{–}] – (1 – [A^{–}])} = log (0.50 / (1- 0.50)} = log 1.00 = 0

Step 2: Putting the value of log {[A^{–}] – (1 – [A^{–}])} in equation 12-

pH = pKa + 0 = 2.33 + 0 = 2.33

Hence, the required pH = 2.33

**Example 4:** Calculate the pH at which the R-group of aspartic acid bears a net charge of (-)0.367 units. Use, pK_{a} = 3.71

Ans. **Step 1:** Given, charge on the acidic group, [A^{–}] = -0.367

Note that the -ve sign of the charge on the conjugate base of weak acid simply indicates the negative charge. The absolute value of the fraction of [A^{–}] is equal to the absolute value of charge, that is, 0.367 in this case.

So, log {[A^{–}] – (1 – [A^{–}])} = log (0.367 / (1- 0.367)} = log 0.580 = -0.237

**Step 2:** Putting the value of log {[A^{–}] – (1 – [A^{–}])} in equation 12-

pH = pKa + (-0.237) = 3.71 -0.237 = 3.473

Hence, the required pH = 3.47

**Example 5:** Determine the pH at which acetic acid molecule bears a net charge of (-)0.77 units. Use, pK_{a} = 4.75.

Ans. Step 1: Given, charge on the acidic group, [A^{–}] = -0.77

Note that the -ve sign of the charge on the conjugate base of weak acid simply indicates the negative charge. The absolute value of the fraction of [A^{–}] is equal to the absolute value of charge, that is, 0.77 in this case.

So, log {[A^{–}] – (1 – [A^{–}])} = log (0.77 / (1- 0.77)} = log 3.348 = 0.52

**Step 2:** Putting the value of log {[A^{–}] – (1 – [A^{–}])} in equation 12-

pH = pKa + 0.52 = 4.75 +0.52 = 5.26

Hence, the required pH = 5.26

**Example 6:** First prove that the sum of the fraction of the ionized forms of the two acidic groups is equal to 1 at the average of the pKa of -COOH (pK_{aC}) and R-COOH groups (pK_{aR}). Then, using the equation derived in the first part, determine the pH at which the sum of charges on the acidic groups (-COOH and R-group) of aspartic is (-) 1.00. Use, pK_{a} of -COOH = 1.93, and pKa of R-group = 3.71.

Ans. Step 1: To prove that the sum of the fraction of the ionized forms of the two acidic groups is equal to 1 at the average of the pKa of -COOH (pK_{aC}) and R-COOH groups (pK_{aR}).

The sum of fraction ionized forms of the two acidic groups can be written as-

Equation 13 proves that the sum of fractional ionized forms of the two acidic groups is equal to 1 at their average pKa value. Since the ionized forms of the weak acids are negatively charged, the fractional ionized forms’ summed value of 1 also indicates a summed charged -1.0 on these two acidic groups. That is, the sum of fractional charges on the two acidic groups is equal to -1.00 at their average pKa value

The pKa of the N-terminal -NH_{2} group is 9.04 or above for the acidic amino acids. The difference between the basic pKa_{2} and the average acidic pKa would in turn always greater than 2.9 units. At this difference of 2.9 or more pH units, the basic group bears a charge equal to or very close to +1.00. That is, at the average of the two acidic pKa values, the total negative charge on the acidic groups is -1.00, and that of the basic group is +1.00, thus the net charge becomes zero. Recalling, pI is the pH at which the net charge on a chemical species is zero, this pH in equation 13 is also equal to the pI of the acidic amino acid.

Also, note that equation 13 is the basis of the statement in section 4B. “I. At the average of two __acidic pKa__ values, the sum of total charge on the __two__ __acidic__ groups is equal to -1.000.”

Step 2: Coming to the question-

The required pH for -1.00 unit summed charge on both the acidic groups of aspartic acid is given by-

pH = (pKa_{C} + pKa_{R}) / 2 = (1.93 + 3.71) / 2 = 2.82

**Example 7:** Determine the pH at which the side chain of cysteine is 36.37% ionized. Use, pK_{a} of the side chain = 8.14

Ans. Given,

% ionization of cysteine = 36.37%

So, fraction ionization, [A^{–}] = % ionization / 100 = 36.37/100 = 0.3637

Putting the values in equation 12-

log {[A^{–}] – (1 – [A^{–}])} = log (0.3637 / (1- 0.3637)} = log 0.5716 = -0.243

And, putting the values in above expression-

pH = 8.14 + (-0.243) = 8.14 – 0.243 = 7.897 = 7.90

## 3. Henderson-Hasselbalch Equation for a Basic Group

Let us begin with Henderson-Hasselbalch Equation for weak bases-

Equation 16 can be re-arranged into equations 19 and 20 without taking the fraction approach. Let us begin with equation 16 once again-

Note that equations 19 and 20 are equivalent to equations 23 and 24, respectively.

Equations 19 and 23 only give the fraction of the weak base in its conjugate acid form- the only charged form of the weak base. And, the conjugate acid BH^{+} of a weak base is positively (+) charged. That is, the charge on a weak base species is numerically equal to the fraction of its ionized form (conjugate acid), and the sign of the charge is positive. Since a numerical value represented without any sign is understood to be positive, the fractional ionization of a weak base can be taken equal to the fraction positive charge on it.

So, the fractional positive charge on the weak base, C_{WB} is given by-

**Example 8:** Calculate the protonated to deprotonated form ratio of the N-terminal (-NH_{2}) of lysine at pH 7. Also, calculate its % and fraction charge. Use pKa = 9.16.

Ans. Note that the N-terminal (-NH_{2}) is a weak base. So, use the Henderson-Hasselbalch equation for a weak base. In the case of a weak base, only the conjugate acid or protonated form -NH_{3}^{+} (= BH^{+}) is the (positively) charged species. So, calculate the % and fraction of BH^{+} for the given base.

**# Interpreting the Results:**

I. (Protonated: Deprotonated) forms= 144.51/1. There are 144.51 (positively) charged basic groups BH^{+} for every uncharged group B (ignoring the whole number counts for molecules).

II. % Charge on N-terminal= 99.313%. Out of the total amount of the chemical species (both BH^{+} and B), 99.313% are in BH^{+} or positively charged form. Or, on a 100 molecules scale, 99.313 molecules (ignoring whole number counts for molecules) are in protonated state BH^{+} and positively charged. The rest 0.687% is in its deprotonated form B and uncharged.

III. Fraction charge on N-terminal= 0.99313. Each basic group bears a fraction charge of 0.99313. That is, each basic group bears a partial positive charge of approximately 0.99313 unit (but NOT a complete 1.000 unit). It also indicates that, on a fraction scale where the sum of the ionized and unionized form of the basic group is taken unity, 0.99313 factions of unity molecules are in the ionized state.

**Example 9:** Amphetamine, a weak base with pKb of 4.10, is prescribed for the treatment of attention deficit hyperactivity disorder (ADHD). Only the unionized form of this drug is readily absorbable. A person orally ingests 40.0 mg of this drug. Calculate- I. Amount of drug absorbed in the stomach (pH = 2.00), II. Amount of drug absorbed in the blood (pH = 7.40). Also, determine III. which of the two sites act as the preferred absorption site of the drug?

Ans.

3A. Determining the pH for the given charge on a basic group

Let us proceed with the modified Henderson-Hasselbalch equation for a basic group.

Equation 26 with pH as the single unknown, gives the required value of pH for the given charge on a basic group. Also, note that equation 26 is the same as the classical Henderson-Hasselbalch equation for a weak base (equation 15), but in a different form where [B] = 1 – [BH^{+}]. Both the equations give fraction charge on a basic group. However, unlike most other cases, equation 26 does not require the concentration of the weak base, [B].

**Example 10:** Calculate the pH at which the R-group of Arginine bears a net charge of 0.50 units. Use, pK_{a} = 12.10

Ans. **Step 1:** Given, charge on the basic group, [BH^{+}] = 0.50

So, (1 – [BH^{+}]) / [BH^{+}] = (1.00 – 0.50) / 0.50 = 1.00

Now, log (1 – [BH^{+}]) / [BH^{+}] = log 1.00 = 0

**Step 2:** Putting the value of log (1 – [BH^{+}]) / [BH^{+}] in equation 26-

pH = pKa + 0 = 12.10 + 0 = 12.10

Hence, the required pH = 12.10

**Example 11:** Calculate the pH at which the R-group of Arginine bears a net charge of 0.367 units. Use, pK_{a} = 12.10

Ans. **Step 1:** Given, charge on the basic group, [BH^{+}] = 0.367

So, (1 – [BH^{+}]) / [BH^{+}] = (1.00 – 0.367) / 0.367 = 1.725

Now, log (1 – [BH^{+}]) / [BH^{+}] = log 1.725 = 0.237

**Step 2:** Putting the value of log (1 – [BH^{+}]) / [BH^{+}] in equation 26-

pH = pKa + 0.237 = 12.10 + 0.237 = 12.337

Hence, the required pH = 12.337 = 12.24

**Example 12:** Determine the pH at which methylamine a net charge of 0.77 units. Use, pK_{b} = 3.35.

Ans. **Step 1:** Given, pK_{b} = 3.35

So, pK_{a} = 14.00 – pKb = 14.00 – 3.35 = 10.65

Also, given, charge on the basic group, [BH^{+}] = 0.77

So, (1 – [BH^{+}]) / [BH^{+}] = (1.00 – 0.77) / 0.77 = 0.299

Now, log (1 – [BH^{+}]) / [BH^{+}] = log 0.299 = -0.52

**Step 2:** Putting the value of log (1 – [BH^{+}]) / [BH^{+}] in equation 26-

pH = pKa + (-0.52) = 10.65 -0.52 = 10.13

Hence, the required pH = 10.13

**Example 13:** First prove that the sum of the fraction of the ionized forms of the two basic groups is equal to 1 at the average of the pKa of -NH_{2} (pK_{aN}) and R-NH_{2} groups (pK_{aR}). Then, using the equation derived in the first part, determine the pH at which the sum of charges on the basic groups (-NH_{2} and R-group) of arginine is 1.00 units. Use, pK_{a} of -NH_{2} = 9.00, and pKa of R-group = 12.10.

Ans. Step 1: To prove that the sum of the fraction of the ionized forms of the two basic groups is equal to 1 at the average of the pKa of -NH_{2} (pK_{aN}) and R-NH_{2} groups (pK_{aR}).

The sum of fractional ionized forms of the two basic groups can be written as-

Equation 27 proves that the sum of fractional ionized forms of the two basic groups is equal to 1 at their average pKa value. Since the ionized forms of the weak bases are positively charged, the summed fractional ionized forms of 1 also indicate a summed charged +1.0 for the two basic groups. That is, the sum of fractional charges on the two basic groups is equal to +1.0 at their average pKa value.

The pKa of the C-terminal -COOH group is 2.15 or below for the basic amino acids. The difference between the acidic pKa_{1} and the average basic pKa would in turn always greater than 5.5 units. At this difference of 5.5 or more pH units, the acidic group bears a charge equal to or very close to -1.00. As shown in equation 27 above, the sum of charges on the two basic groups at their average basic pKa is +1.00. That is, at the average of the two basic pKa values, the total positive charge on them is +1.00, whereas that on the acidic group is -1.00, thus the net charge becomes zero. Recalling, pI is the pH at which the net charge on a chemical species is zero, this pH in equation 27 is also equal to the pI of the basic amino acid.

Note that equation 27 is the basis of the statement in section 4C. “I. At the average of two __basic pKa__ values, the sum of total charge on these __two__ __basic__ groups is equal to +1.0.”

Step 2: Coming to the question-

The required pH for 1.00 unit summed charge on both the basic groups of arginine is given by-

pH = (pKa_{N} + pKa_{R}) / 2 = (9.00 + 12.10) / 2 = 10.55

**Example 14:** Determine the pH at which methylamine is 40% ionized. Use, pK_{a} = 10.65.

Ans. Given,

% ionization of methylamine = 40%

So, fraction ionization, [BH^{+}] = % ionization / 100 = 40/100 = 0.40

Using equation 26-

## 4. Charge and pI of Amino acids and Equivalent Peptides & Molecule

The net charge on an amino acid at a specified pH is the sum of charges on all the ionizable groups at that pH. Examples 1 and 10 in the previous section explain the calculation of the fractional ionized form, and subsequently, the charge on the C-terminal and N-terminal of amino acids using the classical as well as the modified Henderson-Hasselbalch equation approaches. The classical approach is a self-explanatory, direct method and shall be preferred for calculating the charge on molecules bearing a single ionizable group like acetic acid and methylamine, etc.

Amino acids have at least two ionizable groups- the C-terminal COOH, and the N-terminal -NH_{2} group. Adopting the classical approach for calculating the charge on the two ionizable groups would often be cumbersome and time-consuming. The task gradually grows tiresome with an increase in the number of ionizable groups in the molecule. The modified Henderson-Hasselbalch equation approach handles the issue smoothly. When combined with Excel or similar data processing software, this tool can generate the charge and pI of a molecule with any number of ionizable groups. This textbook would now focus the calculation of charge and pI of amino acids, peptides, drugs, pH indicators or any other molecule with acidic and basic groups using the modified Henderson-Hasselbalch equation and Excel.

For the simplicity of generalizing some pI approximation tools, this textbook categorizes the molecules into the following groups-

**4A.** Molecules with one acidic and one basic group (neutral amino acids)

**4B.** Molecules with two acidic and one basic group (acidic amino acids)

**4C.** Molecules with one acidic and two basic groups (basic amino acids)

**4D.** Molecules with an equal number of acidic and basic groups

**4E.** Molecules with an excess of one acid and basic group

**4F.** All other molecules.

The pI of all amino acids can exactly be approximated using the formulae in the groups A, B and C. The mathematical justifications for approximating pI of amino acids are also provided in equations 14, 28 and 30. Groups D and E also present some approximation tools for calculating the pI of peptides. However, care must be taken to account pKa overlaps and decide when to accept or reject the approximation method. Group F includes all other molecules for which a simple approximation tool for pI calculation is practically infeasible.

## 4A. Molecules with one acidic and one basic group (Neutral Amino Acids)

This first group of molecules the neutral amino acids including the nonpolar amino acids (**alanine, glycine, isoleucine, leucine, methionine, phenylalanine, proline, tryptophan, and valine**), and polar, uncharged amino acids (**asparagine, glutamine, serine, and threonine**). The side chain of these amino acids remains uncharged irrespective of the pH. The calculation of charge on the molecule at a specified pH is relatively much easier and straightforward.

pI is the pH at which the amino acid, peptide, drugs or any other molecule bears a net charge of zero or is electrically neutral. There is only one theoretical pH at which the net charge on a specified molecule becomes zero- the degree of precision is higher with accounting greater number of decimal places in the apparent charge on each residue as well as the net charge.

Before we proceed to any conclusion, note the following points-

I. pH = pI when the net charge of the molecule is zero.

Equation 5 and 9 give the fraction of the conjugate base A^{–} or COO^{–} of a weak acid. Since the conjugate base is NEGATIVELY charged, they also give the magnitude of negative charge on that group. To get the magnitude of negative charge on an acidic group, simply multiply the fraction charge with (-1).

III. Equation 19 and 23 give the fraction of the conjugate acid BH^{+} or -NH_{3}^{+} of a weak base. Since the conjugate acid is POSITIVELY charged, they also give the magnitude of positive charge on that group.

- When D(pH-pKa) is numerically equal (ignoring the + or – sign) for an acidic and basic group, their respective HH equations yield the same numerical value. For example, D(pH-pKa) of 3.69 yields an absolute value of 0.99980 unit charge on both the acidic and basic groups in equations 9 and 19, respectively.
- So, if a molecule has one acidic and one basic group, (neutral amino acid, peptides like CAT, etc.) the average value of the two pKa can be treated at the pI. It is because the charge on the acidic and basic group is equal in magnitude but opposite in sign at their average pKa where D(pH-pKa) is the same for both. It eventually a net charge zero charge on the molecule- the characteristic of a molecule in its zwitterion form. See example 18 for the mathematical proof.

** Conclusion:** For the neutral amino acids, peptides and any other equivalent molecule with only one acidic and one basic group, the pI value is equal to the average of acidic and basic pKa values.

pI of molecule with one basic and one acidic group, **pI = (pKa _{1} + pKa_{2}) / 2**

Or, pI of neutral amino acids,** pI = (pKa _{1} + pKa_{2}) / 2**

**Example 15:** Calculate I. the net charge on alanine at pH 7.00. Use pKa_{1} = 2.33, and pKa_{2} = 9.71. Also calculate II. its pI.

Ans. I. Calculate net Charge: Set up an Excel sheet for charge calculation on the basic and acidic group separately using the modified Henderson-Hasselbalch equation as shown below. The sum of charges on the basic group and the acidic group gives the net charge on the molecule at a specified pH. So, the net charge on alanine at pH 7.00 is -0.00192.

II. Calculate pI:

pI of nonpolar amino acids, pI = (pKa_{1} + pKa_{2}) / 2

Or, pI = (9.71 + 2.33) / 2 = 6.02

Let us see if the pH of 6.02 returns a net charge of zero in the net charge table below.

The average of one acidic and one basic pKa for such molecules return an exact zero charge on them. So, pI is equal to the average of two pKa values.

**Example 16:** Determine the pH at which alanine bears a net charge of +0.50. Use pKa_{1} = 2.33, and pKa_{2} = 9.71.

**Ans.** Set up an Excel sheet for charge calculation on the basic and acidic group separately using the modified Henderson-Hasselbalch equation as shown below. Adjust the pH to get a net charge of +0.50. The required pH is 2.33.

Do you notice any similarity between the required pH for the +0.50 unit charge and any of the pKa values? It is equal to the pKa of the acidic group.

At pH = pKa, a group is half-ionized and has half the charge of its completely ionized form. In this case, the acidic group bears a -0.50 unit charge. Since the completely ionized basic group has a +1.00 charge at pKa of the acidic group, the net charge on alanine becomes +1.00 + (-0.50) = +0.50.

**Example 17:** Determine the pH at which alanine bears a net charge of -0.50. Use pKa_{1} = 2.33, and pKa_{2} = 9.71.

**Ans.** Set up an Excel sheet for charge calculation on the basic and acidic group separately using the modified Henderson-Hasselbalch equation as shown below. Adjust the pH to get a net charge of +0.50. The required pH is 9.71.

Do you notice any similarity between the required pH for -0.50 unit charge and any of the pKa values? It is equal to the pKa of the basic group.

At pH = pKa, a group is half-ionized and has half the charge of its completely ionized form. In this case, the basic group bears a +0.50 unit charge. Since the completely ionized acidic group has -1.00 unit charge at pKa of the basic group, the net charge on alanine becomes 0.50 + (-1.00) = -0.50.

**Example 18:** Prove that the pI of a neutral amino acid is equal to the average of the pKa of -NH_{2} (pK_{aN}) and -COOH groups (pK_{aC}).

Ans. The modified Henderson-Hasselbalch for the acidic and basic group just gives the fraction of their respective ionized forms. Further, the fraction ionization of the acidic group is multiplied by (-)1 to yield the negative charge on its conjugate base. So, if the negative charge is ignored, the fraction of ionized forms of both the acidic and the basic groups must be equal at the pI to yield a net charge of zero.

That is,

*f*_{[BH+]} = *f*_{[A-]}

Using the respective modified Henderson-Hasselbalch equations-

That is, at the average of pK_{aN} and pK_{aC}, the fraction of ionized forms of the basic and acidic groups are equal. Since the two groups are oppositely charged, the net charge on the molecule becomes zero when their fraction ionized forms are equal. Recalling, pI is the pH at which the net charge on a chemical species is zero, this pH in equation 29 is also equal to the pI of the neutral amino acid.

## 4B. Molecules with two acidic and one basic group (Acidic Amino acids)

This group of molecules includes all the acidic amino acids (**aspartic acid**, **glutamic acid**, **cysteine**, and **tyrosine**) and all other molecules including peptides (example- PEN) with two acidic and one basic group. Note that though the side chains of cysteine and tyrosine are generally considered to be neutral at pH 7.00, these are acidic groups and ionizable. So, these two amino acids shall also be considered as acidic amino acids wherever present.

Acidic amino acids have the N-terminal -NH_{2} group as the single basic group while two acidic groups- the C-terminal -COOH groups and the acidic R-group. Note the following points used as strategies to calculate pI of acidic amino acids and equivalent peptides-

I. At the average of two __acidic pKa__ values, the sum of total charge on these __two__ __acidic__ groups is equal to **-1.0**. *See equation 13, Example 6 under section 2A for its derivation.*

II. The pKa of the N-terminal -NH_{2} group is 9.04 or above for the acidic amino acids. The difference between the basic pKa_{2} and the average acidic pKa would in turn always greater than 2.9 units. At this difference of 2.9 or more pH units, the basic group bears a charge equal to or very close to +1.0.

III. At a pH equal to the average of the two acidic pKa values, the sum of negative charges on these two acidic groups is -1.0 whereas that on the basic group is +1.0. When summed up (+1.0 – 1.0 = 0), the net charge of the amino acid turns to be zero. So, the average acidic pKa is taken as the pI of acidic amino acids and equivalent molecules. *See equation 14, Example 6 under section 2A for its mathematical proof.*

** Conclusion:** For molecules with two acidic groups and one basic group, including acidic amino acids and equivalent molecules, the pI value is equal to the average of the two acidic pKa values.

pI of acidic amino acids, **pI = (pKa _{1} + pKa_{3}) / 2**

** Or, pI = **(pKa of C-ter + pKa of acidic R-group)/ 2

**Example 19:** Calculate I. the net charge on aspartic acid at pH 7.00. Use pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71. Also calculate II. its pI.

Ans. I. Calculate net charge

Set up an Excel sheet for charge calculation on the basic and acidic groups separately using the modified Henderson-Hasselbalch equation as shown below. The sum of charges on the basic and the acidic groups gives the net charge on the molecule at the specified pH. The net charge of -1.00166 can be approximated to -1.000 (three decimal places).

II. Calculate pI:

pI of acidic amino acids, pI = average of acidic pKa values

Or, pI = (1.93 + 3.71) / 2 = 2.82

Let us see if the pH of 2.82 returns a net charge of zero in the net charge table shown below. The average of the two acidic pKa for such molecules returns a net charge of zero. So, pI is equal to the average of two acidic pKa values.

**Example 20:** Determine the pH at which aspartic acid bears a net charge of +0.50. Use pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71.

Ans. Set up an Excel sheet for charge calculation on the basic and acidic group separately using the modified Henderson-Hasselbalch equation as shown here. Adjust the pH to get a net charge of +0.50. The required pH is 1.9033.

**Example 21:** Determine the pH at which aspartic acid bears a net charge of -0.75. Use pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71.

Ans. Set up an Excel sheet for charge calculation on the basic and acidic group separately using the modified Henderson-Hasselbalch equation as shown here. Adjust the pH to get a net charge of -0.75. The required pH is 4.1996.

## 4C. Molecules with one acidic and two basic groups (Basic Amino Acids)

This group of molecules includes all the basic amino acids (**arginine**, **histidine**, and **lysine**) and all other molecules including peptides (example- HIS) with one acidic and two basic groups.

Basic amino acids have the C-terminal -COOH group as the single acidic group while having two basic groups- the N-terminal -NH_{2} groups and the basic R-group. Note the following points used as strategies to calculate pI of basic amino acids and equivalent peptides-

I. At the average of two __basic pKa__ values, the sum of total charge on these __two__ __basic__ groups is equal to **+1.0**. *See equation 24, example 13 under section 3A for its derivation*.

II. The pKa of the C-terminal -COOH group is 2.15 or below for the basic amino acids. The difference between the acidic pKa_{1} and the average basic pKa would in turn always greater than 5.5 units. At this difference of 5.5 or more pH units, the acidic group bears a charge equal to or very close to -1.0 unit.

III. At a pH equal to the average of the two basic pKa values, the sum of positive charges on these basic groups is +1.0 whereas that on the acidic group is -1.0. When summed up (+1.0 – 1.0 = 0), the net charge on the molecule becomes zero. So, the average basic pKa is taken as the pI of the basic amino acids. *See equation 25, example 13 under section 3A for the mathematical proof.*

** Conclusion:** For the basic amino acids and equivalent peptides with one acidic and two groups, the pI value is equal to the average of the two basic pKa values.

pI of basic amino acids, **pI = (pKa _{2} + pKa_{3}) / 2**

**Or, pI = **(pKa of N-ter + pKa of basic R-group)/ 2

**Or, pI = **Average of the two basic pKa

** **

**Example 22:** Calculate I. the net charge on Arginine at pH 7.00. Use pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10. Also, calculate II. its pI.

Ans. I. Calculate net charge

Set up an Excel sheet for charge calculation on the basic and acidic groups separately using the modified Henderson-Hasselbalch equation as shown below. The sum of charges on the basic group and the acidic group gives the net charge on the molecule at the specified pH.

The net charge of 0.9901 can be approximated to +1.00.

II. Calculate pI:

pI of basic amino acids, pI = average of the basic pKa values

Or, pI = (9.00 + 12.10) / 2 = 10.550

Let us see if the pH of 10.550 returns a net charge of zero in the net charge table shown below. The average of the two basic pKa for such molecules returns a net charge of zero. So, pI is equal to the average of two basic pKa values.

**Example 23:** Determine the pH at which arginine bears a net charge of +1.00. Use pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10.

Ans. Set up an Excel sheet for charge calculation on the basic and acidic group separately using the modified Henderson-Hasselbalch equation as shown here. Adjust the pH to get a net charge of +1.00. The required pH is 5.50.

**Example 24:** Determine the pH at which arginine bears a net charge of +0.47. Use pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10.

Ans. Set up an Excel sheet for charge calculation on the basic and acidic group separately using the modified Henderson-Hasselbalch equation as shown here. Adjust the pH to get a net charge of +0.47. The required pH is 9.0506.

## 4D. Charge and pI of molecules with an equal number of acidic & basic groups (and, the Concept of pKa Overlap)

Based on their location in a peptide chain, the amino acids (residues) can be categorized into-

I. N-terminal amino acid: By convention, a peptide chain is written from its N-terminal to C-terminal. So, the first amino acid of the peptide is its N-terminal amino acid or residue. Its -NH_{2} and R-group (if ionizable) are free to take the ionized forms. However, its -COOH group cannot take any ionized form because of being involved in peptide bond formation with the next residue in the peptide.

II. C-terminal amino acid: By convention, a peptide chain is written from its N-terminal to C-terminal. So, the last amino acid of the peptide is its C-terminal amino acid or residue. Its -COOH and R-group (if ionizable) are free to take the ionized forms. However, its -NH_{2} group cannot take any ionized form because of being involved in peptide bond formation with the preceding residue.

III. The intermediate amino acids: All the amino acids between the N- and C-terminal of a peptide chain are referred to as the intermediate amino acids or residues. Both the -NH_{2} and -COOH groups of the intermediate residues are involved in peptide bond formation with their respective adjacent residues, so they cannot take any ionized form. The ionizability of the R-groups remains unaffected by the peptide bond formation.

A schematic representation of peptide bond formation is shown below.

This group of molecules would generally consist of peptides with an equal number of acidic and basic groups- say two acidic and two basic groups; or three acidic and three basic groups, and so on. We have already discussed the scenario of one acidic and one basic group in 4A (neutral amino acid). As usual, the calculation of charge on the molecule at a specified pH is relatively much easier and straightforward.

All peptides have an N-terminal -NH_{2} (or -NH in case of histidine) and C-terminal -COOH groups. The pKa of the N-terminal amine group is generally above 8.50 and that of the C-terminal -COOH group is below 3.0. These groups do not interfere with pI calculation because these would generally be completely ionized at the isoelectric pH and bear equal and opposite charge on them. However, the approximated pI substantially deviates from actual pI when the number of R-groups (of tyrosine, cysteine, and histidine) with overlapping pKa values increase in the peptide. For example, the pKa of acidic R-group of tyrosine (pKa = 10.10) and cysteine (pKa = 8.14) overlaps into the basic pH region (pH above 7.00). Similarly, the pKa of the basic R-group of histidine (pKa = 6.04) overlaps into the acidic pH region (pH below 7.00).

For a small peptide chain where the number of acidic groups is equal to those of the basic groups, the pI is approximately equal to the average of highest acidic pKa and lowest basic pKa value.

That is,

**pI = (highest acidic pKa + lowest basic pKa value) / 2**

This approximation method may be used as a handy tool to predict the pI of a molecule in cases where the use of Excel is not feasible or permitted (like examination hall, or if the question mentions so).

**Example 25:** Calculate the I. approximated pI, II. Actual pI, and III. Tabulate and summarize the approximated and actual pI values of following peptides-

Case A: AKID or Ala-Lys-Ile-Asp

Case B: AHID or Ala-His-Ile-Asp

Case C: AYKI or Ala-Tyr-Lys-Ile

Case D: AYHI or Ala-Tyr-His-Ile

Use the following pKa values-

Ala, A: pKa_{1} = 2.33, pKa_{2} = 9.71

Lys, K: pKa_{1} = 2.15, pKa_{2} = 9.16 and pKa_{3} = 10.67

Ile, I: pKa_{1} = 1.26, pKa_{2} = 9.60

Asp, D: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

His, H: pKa_{1} = 1.70, pKa_{2} = 9.09 and pKa_{3} = 6.04

Tyr, Y: pKa_{1} = 2.24, pKa_{2} = 9.04 and pKa_{3} = 10.10

Ans. I. Calculate approximated pI values:

Step 1: List and enumerate all the acidic and basic groups in all the four cases

__Case A:__ AKID or Ala-Lys-Ile-Asp

**Basic:** N-ter of Ala, R-group of Lys ; **Acidic:** C-ter at Asp, R-group of Asp

__Case B:__ AHID or Ala-His-Ile-Asp

**Basic:** N-ter of Ala, R-group of His ; **Acidic:** C-ter at Asp, R-group of Asp

__Case C:__ AYKI or Ala-Tyr-Lys-Ile

**Basic:** N-ter of Ala, R-group of Lys ; **Acidic:** C-ter at Ile, R-group of Tyr

__Case D:__ AYHI or Ala-Tyr-His-Ile

**Basic:** N-ter of Ala, R-group of His ; **Acidic:** C-ter at Ile, R-group of Tyr

Step 2: Tabulate the pKa values in an Excel sheet as shown below.

Use “pI = (highest acidic pKa + lowest basic pKa value) / 2” to show the pI, charge on individual groups and the net charge on the molecules.

II. Calculate actual pI values: Set up the data in the Excel sheet as above. Adjust the pH value to get a net charge of zero in each case independently as shown below-

III. Summarize the result for approximated and actual pI values

As shown to the right, the approximated pI is in good accordance with the actual pI value to one decimal place in all the cases except case D.

### The Concept of pKa Overlap

When the pKa of a basic group falls in acidic pH zone, it is to be called the __basic pKa overlap__ of that __basic__ group. To avoid any confusion over the use of “acidic” or “basic” pKa overlap, we would strictly adhere to the prefix “acidic” or “basic” depending on the nature of the group exhibiting overlap into the other pH zone. For example, if the pKa of a __basic____ group__ exhibits overlapping in the other (acidic) pH zone, this overlap shall be called the __basic____ pKa overlap__. Similarly, if the pKa of an __acidic____ group__ exhibits overlapping in the other (basic) pH zone, this overlap shall be called the __acidic____ pKa overlap__.

The pKa of the basic side chain of histidine is 6.04- it falls in the acidic pH zone. So, the __basic__ side chain of histidine exhibits __basic__ pKa overlap. The pKa of acidic side chains of cysteine and tyrosine are 8.14 and 10.10, respectively. Because the pKa of these __acidic groups__ overlap in the pH zone, these two groups exhibit __acidic pKa overlap__.

**Determining pKa overlap:** By taking the neutral pH 7.00 as the reference point, the absolute difference between the overlapping pKa and 7.00 equals the pKa overlap for that group.

pKa overlap = ½7.00 – Overlapping pKa½

For example, the pKa overlap for histidine, cysteine and tyrosine side chains are as follow-

Basic pKa overlap for histidine R-group = ½7.00 – 6.04½= **0.96**

Acidic pKa overlap for cysteine R-group = ½7.00 – 8.14½= **1.14**

Acidic pKa overlap for tyrosine R-group = ½7.00 – 10.10½= **3.10**

The sum of all the basic and acidic pKa overlaps is called the total pKa overlap for the molecule. For example, in case D, total overlap = 0.96 for His + 3.10 for Tyr = 4.06

**The Usefulness of pKa Overlap?** Approximating pI of peptides may yield unexpected results due to several reasons including relative abundance of acidic and basic groups, their pKa values as well the relative abundance of groups with pKa overlaps. Quantifying pKa overlap provides us a tool to predict the deviation from the actual pI on a relative scale.

As summarized in example 25, the approximated pI values deviate from their respective actual pI values to a varying extent in the presence of groups that exhibit acidic and basic pKa overlaps. We will try to seek any relation between the deviation from actual pI (DpI) and pKa overlaps.

ΔpI = I Actual pI – Approximated pKa I

Note the characteristics of the graph plotted by taking the independent variable (pKa overlap) on X-axis and dependent variables on Y-axis (ΔpI) for cases B, C and D. Note the general trend that the deviation of approximated pI from actual pI increases with increase in the total pKa overlap for the peptide. Case B with the least pKa overlap of 0.96 exhibits the least deviation in pI. Case C exhibits an intermediate deviation in pI because of having an intermediate value of pKa overlap. Case D with maximum pKa overlap value exhibits the highest deviation of the approximated pI from actual pI.

Now, it is noteworthy to keep in mind that the extent of deviation of approximated pI from the actual pI may gradually increase with –

I. Increase in the number of R-groups with or without overlapping pKa.

II. Presence of overlapping pKa than in their absence when the number of acidic and basic groups kept constant.

III. Increase in total pKa overlap value for the molecule

IV. pKa deviation in both acidic and basic regions than in any one region.

## 4E. Charge and pI of molecules where the difference

## in the number of acidic & basic groups is one

As usual, the calculation of charge on the molecule at a specified pH is straightforward and does not require any special consideration. The strategy for calculating pI is equivalent to the previous sections 4B (acidic amino acids) and 4C (basic amino acids) as follow-

I. When the number of acidic groups is one more than those of the basic groups, the approximated pI of the peptide (or any other molecule equivalent to an acidic amino acid) is equal to the average of two highest acidic pKa values. This approach yields the sum of -1.0 unit charge for those two accounted acidic groups.

Furthermore, the total charge on “n” basic groups would be numerically equal to those on “n+1” acidic groups.

II. When the number of basic groups is one more than those of the acidic groups, the approximated pI of the peptide (or any other molecule equivalent to a basic amino acid) is equal to the average of the two lowest basic pKa values. This approach yields the sum of a +1.0 unit charge for those two basic groups under consideration.

Furthermore, the total charge on “n+1” basic groups would be numerically equal to those on “n” acidic groups.

III. Care must be taken for the overlapping pKa values- greater is the sum and zones of pKa overlap, greater would be the deviation of approximated pI from actual pI values as discussed above in section 4D.

**Approximating pI:**

**Case I:** When there is an excess of an acidic group (equivalent to an acidic amino acid)

**pI = average of two highest acidic pKa**

**Case II:** When there is one excess of the basic group (equivalent to a basic amino acid)

**pI = average of two lowest basic pKa**

**Example 26:** Calculate I. net charge on the peptide **DRKEEDRAIN** at pH 7.0, and II. its approximated and actual pI. Use the following pKa values-

Asp, D: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

Arg, R: pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10

Lys, K: pKa_{1} = 2.15, pKa_{2} = 9.16 and pKa_{3} = 10.67

Glu, E: pKa_{1} = 2.16, pKa_{2} = 9.58 and pKa_{3} = 4.15

Ala, A: pKa_{1} = 2.33, pKa_{2} = 9.71

Ile, I: pKa_{1} = 1.26, pKa_{2} = 9.60

Asn, N: pKa_{1} = 2.16, pKa_{2} = 8.73

Ans. I. Calculate net charge at pH 7.0

Step 1: List and enumerate all the basic and acidic groups separately-** DRKEEDRAIN**

Basic groups: N-ter at D, R-group of R, K, and R ; [Total number = **4**]

Acidic groups: C-ter at N, R-group of D, E, E, D ; [Total number = **5**]

Step 2: Setup an Excel table for 4 basic groups and 5 acidic groups to calculate the net charge as shown below-

Hence, net charge on the peptide at pH 7.00 = **-0.99855 = -1.00** (two decimal place)

- Calculate approximated and actual pI:

Approximated pI = The peptide has one excess of acidic group.

So, approximated pI = average of two highest acidic pKa

= (4.15 + 4.15) / 2 = **4.15**

# Predicting the validity of approximated pI:

There is no overlapping pKa in this peptide chain. So, the approximated pI is most likely to be acceptably close to the actual pI value.

Actual pI: Set the pH in the above data table to get the net charge of zero. This pH would be equal to the pI value. So, the **actual pI = 4.4383 = 4.44 **(two decimal places)

As predicted in the above paragraph, the approximated pI (=4.15) is satisfactorily close to the actual pI (4.44) because there is not any overlapping pKa in this peptide chain. The approximated pI value can be accepted when there is no tool for calculation provided or instructions are laid to use the approximation method only.

**Example 27:** Calculate I. net charge on the peptide **DRKEYCHAIN** at pH 7.0, and II. its approximated and actual pI. Use the following pKa values-

Asp, D: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

Arg, R: pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10

Lys, K: pKa_{1} = 2.15, pKa_{2} = 9.16 and pKa_{3} = 10.67

Glu, E: pKa_{1} = 2.16, pKa_{2} = 9.58 and pKa_{3} = 4.15

Tyr, Y: pKa_{1} = 2.24, pKa_{2} = 9.04 and pKa_{3} = 10.10

Cys, C: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

His, H: pKa_{1} = 1.70, pKa_{2} = 9.09 and pKa_{3} = 6.04

Ala, A: pKa_{1} = 2.33, pKa_{2} = 9.71

Ile, I: pKa_{1} = 1.26, pKa_{2} = 9.60

Asn, N: pKa_{1} = 2.16, pKa_{2} = 8.73

Ans. I. Calculate net charge at pH 7.0

Step 1: List all the basic and acidic groups separately-** DRKEYCHAIN**

Basic groups: N-ter at D, R-group of R, K, H ; [Total number = **4**]

Acidic groups: C-ter at N, R-group of D, E, Y, C ; [Total number = **5**]

Note that, as discussed earlier, the side chain of tyrosine and cysteine are also acidic.

Leave un-ionizable or neutral groups unaccounted.

Step 2: Setup an Excel table for 4 basic groups and 5 acidic groups to calculate the net charge as shown below-

Hence, net charge on the peptide at pH 7.00 = **0.03000 = 0.03 **(two decimal places)

II. Calculate approximated and actual pI:

Approximated pI = The peptide has one excess of the acidic group.

So, approximated pI = average of two highest acidic pKa

= (10.10 + 8.14) / 2 = **9.12**

# Predicting the validity of approximated pI:

Total pKa overlap for the peptide =

I 7.00 – 6.04 I for His + I 7.00 – 10.10 I for Tyr + I 7.00 – 8.14 ) for Cys

= 0.96 + 3.10 + 1.14 = **5.20**

The total pKa overlap for this peptide is 5.20, far greater than 1.0. So, the approximated pI is most likely to exhibit greater deviation from the actual pI. And, the approximated pI may not be acceptable anymore.

Actual pI: Set the pH in the above data table to get the net charge of zero. This pH would be equal to the pI value. So, the **actual pI = 7.0834= 7.08 **(two decimal places)

As predicted in the above paragraph, the approximated pI (= 9.12) exhibits greater deviation from the actual pI (= 7.08) because the peptide has a very high value of total pKa overlap. The approximated pI value can NOT be accepted because if off from the actual pH by more than 0.5 units.

**Example 28:** Calculate I. net charge on the peptide **DRKKEDRAIN** at pH 7.0, and II. its approximated and actual pI. Use the following pKa values-

Asp, D: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

Arg, R: pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10

Lys, K: pKa_{1} = 2.15, pKa_{2} = 9.16 and pKa_{3} = 10.67

Glu, E: pKa_{1} = 2.16, pKa_{2} = 9.58 and pKa_{3} = 4.15

Ala, A: pKa_{1} = 2.33, pKa_{2} = 9.71

Ile, I: pKa_{1} = 1.26, pKa_{2} = 9.60

Asn, N: pKa_{1} = 2.16, pKa_{2} = 8.73

Ans. I. Calculate net charge at pH 7.0

Step 1: List and enumerate all the basic and acidic groups separately-** DRKKEDRAIN**

Basic groups: N-ter at D, R-group of R, K, K, R ; [Total number = **5**]

Acidic groups: C-ter at N, R-group of D, E, D ; [Total number = **4**]

Step 2: Setup an Excel table for 5 basic groups and 4 acidic groups to calculate the net charge as shown below-

Hence, net charge on the peptide at pH 7.00 = **+0.99982 = +1.00** (two decimal places)

- Calculate approximated and actual pI:

Approximated pI = The peptide has one excess of basic group.

So, approximated pI = average of two lowest basic pKa

= (9.66 + 10.67) / 2 = **10.165**

# Predicting the validity of approximated pI:

There is no overlapping pKa in this peptide chain. So, the approximated pI is most likely to be acceptably close to the actual pI value.

Actual pI: Set the pH in the above data table to get the net charge of zero. This pH would be equal to the pI value. So, the **actual pI = 9.9541 = 9.95 **(two decimal places)

As predicted in the above paragraph, the approximated pI (= 10.165) is satisfactorily close to the actual pI (9.95) because there is not any overlapping pKa in this peptide chain. The approximated pI value can be accepted when there is no tool for calculation provided or instructions are laid to use the approximation method only.

**Example 29:** Calculate I. net charge on the peptide **DRKEHCHAIN** at pH 7.0, and II. its approximated and actual pI. Use the following pKa values-

Asp, D: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

Arg, R: pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10

Lys, K: pKa_{1} = 2.15, pKa_{2} = 9.16 and pKa_{3} = 10.67

Glu, E: pKa_{1} = 2.16, pKa_{2} = 9.58 and pKa_{3} = 4.15

Cys, C: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

His, H: pKa_{1} = 1.70, pKa_{2} = 9.09 and pKa_{3} = 6.04

Ala, A: pKa_{1} = 2.33, pKa_{2} = 9.71

Ile, I: pKa_{1} = 1.26, pKa_{2} = 9.60

Asn, N: pKa_{1} = 2.16, pKa_{2} = 8.73

Ans. I. Calculate net charge at pH 7.0

Step 1: List and enumerate and enumerate all the basic and acidic groups separately-** DRKEHCHAIN**.

Basic groups: N-ter at D, R-group of R, K, H, H ; [Total number = **5**]

Acidic groups: C-ter at N, R-group of D, E, C ; [Total number = **4**]

Step 2: Setup an Excel table for 5 basic groups and 4 acidic groups to calculate the net charge as shown below-

So, net charge on the peptide at pH 7.00 = **+0.12961 = 0.13 **(two decimal places).

II. Calculate approximated and actual pI:

Approximated pI = The peptide has one excess of the basic group.

So, approximated pI = average of two lowest basic pKa

= (6.04 + 6.04) / 2 = **6.04**

# Predicting the validity of approximated pI:

Total pKa overlap for the peptide =

I 7.00 – 6.04 I for His + I 7.00 – 6.04 I for His + I 7.00 – 8.14 I for Cys

= 0.96 + 0.96+ 1.14 = **3.06**

The total pKa overlap for this peptide is 3.06, far greater than 1.0. So, the approximated pI is most likely to exhibit greater deviation from the actual pI. And, the approximated pI may not be acceptable anymore.

Actual pI: Set the pH in the above data table to get the net charge of zero. This pH would be equal to the pI value. So, the **actual pI = 7.2477= 7.25 **(two decimal places).

As predicted in the above paragraph, the approximated pI (= 6.04) exhibits greater deviation (more than 0.5 units) from the actual pI (= 7.25) because the peptide has a very high value of total pKa overlap. The approximated pI value can NOT be accepted because if off from the actual pH by more than 0.5 units.

## 4F. All other Molecules

Proteins are the most diverse class of molecules. Predicting the pI of a peptide or protein molecule often gets complicated and may exhibit unexpected deviations from the actual value depending on the relative abundance of acidic and basic groups as well as overlapping pKa values.

It would be practically infeasible to generalize the approximation method for all types of peptides (say, for a pentapeptide with 1 acidic and 4 basic groups, with 2 acidic and 3 basic groups, 3 acidic and 2 basic groups, and 4 acidic and 1 basic group for instance and many more). The diversity of approximation methods would keep on growing with an increase in the number of ionizable groups, and would even turn unpredictable with the inclusion of groups with overlapping pKa.

So, the best approach is to use the modified Henderson-Hasselbalch equation in the combination of Excel, and set the pH to get a net charge of zero on the molecule- that pH is exactly equal to the pI value.

**Example 30:** Calculate I. net charge on the peptide **KINDPHYSICALCHEMISTRY **at pH 7.0, and II. its actual pI. Use following pKa values-

Lys, K: pKa_{1} = 2.15, pKa_{2} = 9.16 and pKa_{3} = 10.67

Ile, I: pKa_{1} = 1.26, pKa_{2} = 9.60

Asn, N: pKa_{1} = 2.16, pKa_{2} = 8.73

Asp, D: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

Pro, P: pKa_{1} = 1.95, pKa_{2} = 10.47

His, H: pKa_{1} = 1.70, pKa_{2} = 9.09 and pKa_{3} = 6.04

Tyr, Y: pKa_{1} = 2.24, pKa_{2} = 9.04 and pKa_{3} = 10.10

Ser, S: pKa_{1} = 2.13, pKa_{2} = 9.05

Cys, C: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

Ala, A: pKa_{1} = 2.33, pKa_{2} = 9.71

Leu, L: pKa_{1} = 2.32, pKa_{2} = 9.58

Glu, E: pKa_{1} = 2.16, pKa_{2} = 9.58 and pKa_{3} = 4.15

Met, M: pKa_{1} = 2.16, pKa_{2} = 9.08

Thr, T: pKa_{1} = 2.20, pKa_{2} = 8.96

Arg, R: pKa_{1} = 2.03, pKa_{2} = 9.00 and pKa_{3} = 12.10

Ans. I. Calculate net charge at pH 7.0

Step 1: List and enumerate and enumerate all the basic and acidic groups separately for the peptide **KINDPHYSICALCHEMISTRY**.

Basic groups: N-ter at K,

R-group of K, H, H, R ; [Total number = 5]

Acidic groups: C-ter at Y,

R-group of D, Y, C, C, E, Y ; [Total number = 7]

Step 2: Setup an Excel table for 5 basic groups and 7 acidic groups to calculate the net charge as shown below-

So, net charge on the peptide at pH 7.00 = **-0.37638 = 0.38 **(two decimal places).

- Calculate actual pI:

Set the pH in the above data table to get the net charge of zero. This pH would be equal to the pI value. So, the **actual pI = 6.4941= 6.49 **(two decimal places).

## 5. Fraction, percentage and net charge on amino acids as a function of pH

**Example 31:** Plot fraction, percentage and the net charge on alanine as a function of pH.

Ans. Step 1: Calculate the fractional charge on -NH_{2} and -COOH group using the respective modified Henderson-Hasselbalch equation for the acidic and basic groups.

Step 2: Calculate the fractional charge from pH 0 – 14 (or, as required). At least these three pH values must also be included- I. pKa of -COOH group, II. pI of amino acid, and III. pKa of -NH_{2} group.

Calculate % charge or ionized form and the net charge as follow-

% Charge or ionized form = fraction charge or ionized form x 100

Net charge = Total positive charge – Total negative charge

The data are tabulated below-

Step 3: Plot the graph by putting the pH scale on the X-axis and the fractional charge of the two ionizable groups on the Y-axis. Label the regions as shown below or as required. Also, plot net charge as a function of pH in a separate graph.

**Example 32:** Plot fraction, percentage and the net charge on glutamic acid as a function of pH.

Ans. Step 1: Calculate the fractional charge on -NH_{2} (N-ter), -COOH (C-ter) and R-COOH (R-group) group using the respective modified Henderson-Hasselbalch equation for acidic and basic groups.

Step 2: Calculate the fractional charge from pH 0 – 14 (or, as required). Also include these four pH values- I. pKa of -COOH group, II. pKa of R-COOH group, III. pI, and IV. pKa of -NH_{2} group.

Calculate % ionized form and the net charge as follow-

% Charge = fraction charge x 100

Net charge = Total positive charge – Total negative charge

Step 3: Plot the graph by putting the pH scale on the X-axis and fractional charge on the three ionizable groups on the Y-axis. Also, plot net charge as a function of pH in a separate graph.

**Example 33:** Plot the fraction, percentage and the net charge on arginine as a function of pH.

Ans. Step 1: Calculate the fractional charge on -NH_{2}, -COOH (C-ter), R-NH_{2} (R-group) and -NH_{2} (N-ter) group using the respective modified Henderson-Hasselbalch equation for acidic and basic groups.

Step 2: Calculate the fractional charge from pH 0 – 14 (or, as required). At least these four pH values must also be incorporated- I. pKa of -COOH group, II. pKa of R-NH_{2} group, III. pI, and IV. pKa of -NH_{2} group.

Calculate % charge or ionized form and net charge as follow-

% Charge or ionized form = fraction charge or ionized form x 100

Net charge = Total positive charge – Total negative charge

Step 3: Plot the graph by putting the pH scale on X-axis and fractional charge on the three ionizable groups on Y-axis. Also, plot net charge as a function of pH in a separate graph.

The first step in labeling the graph is to draw a 0.50 or 50.0% (half-ionized) line from the Y-axis parallel to the X-axis. The X-intercept from half-ionized line gives the pKa for respective curves (groups). Also, draw the pI line on X-axis.

**Example 34:** I. Calculate the theoretical pI of the peptide DRFRANKENSTEIN. II. Plot its net charge as a function of pH.

Ans. I. Calculate the theoretical pI

Step 1: List and enumerate all the basic and acidic groups separately-** DRFRANKENSTEIN**

Basic groups: N-ter at D, R-group of R, R, K ; [Total number = **4**]

Acidic groups: C-ter at N, R-group of D, E, E ; [Total number = **4**]

The required pH to yield a net charge of zero on the peptide is 7.0710. Hence, its pI is 7.0710.

Step 2: II. Plot its net charge as a function of pH.

Calculate the net charge on the peptide at different pH using the Excel sheet shown above. Also, include the pI as one of the pHs. The net charge on the peptide as a function of pH is summarized below. Plot the graph by putting the pH scale on X-axis and the net charge on the Y-axis. Intercept the zero charge on the Y-axis to the respective pH on the X-axis.

## 6. Predicting Intra- and Intermolecular Interaction in a Peptide

**Example 35:** Draw the structure of the peptide DTLH showing the backbone and side chain atoms in the ionized states as favored at pH 7.0. Show possible intermolecular interactions including H-bonds, ionic interactions and hydrophobic interactions. Use the following pKa values-

Asp, D: pKa_{1} = 1.95, pKa_{2} = 9.66 and pKa_{3} = 3.71

Thr, T: pKa_{1} = 2.20, pKa_{2} = 8.96

Leu, L: pKa_{1} = 2.32, pKa_{2} = 9.58

His, H: pKa_{1} = 1.70, pKa_{2} = 9.09 and pKa_{3} = 6.04

Ans.

Step 1: Determine the charge on ionizable groups.

Enlist and enumerate the weakly acidic and basic groups in the given peptide-

Basic groups: N-ter at D, R-groups of H

Acidic groups: C-ter at H, R-group of D

Calculate fraction charge on each of the ionizable groups using modified Henderson-Hasselbalch equation and Excel as shown to the right.

Step 2: Draw the peptide with predominant ionized forms of the ionizable groups

Among the four ionizable groups in the peptide shown above, only the side chain of histidine bears a charge of +0.10, the others are completely ionized. That is, histidine does not predominantly occur in its ionized form. So, the side chain of histidine shall be left in its unionized form.

Step 3:** Hydrogen Bonds: **The weak attractive interaction between a highly electronegative atom (F, O, N) and an H-atom covalently linked to any such highly electronegative atoms is called **hydrogen bond**. So, there are two criteria for H-bond formation-

- A highly electronegative atom (F, O, N) acting as a hydrogen bond acceptor.
- An H-atom covalently linked to (F, O, N) that acts as a hydrogen bond donor.

An atom or group that accepts H-atom for hydrogen bonding is called an **H-bond acceptor**. Note that “accepting” just implies that the group attracts a suitable H-atom but does not snatches (does not takes up) the H-atom from the H-bond donor.

An atom/group that donates H-atom for hydrogen bonding is called an **H-bond donor**.

** Intramolecular hydrogen bonds:** The hydrogen bond forming between two atoms or groups within a molecule is called an intramolecular hydrogen bond. In a peptide, it can form among -OH, -NH

_{2}, -NH, -NH

_{3}

^{+}, -CO, -COOH, and -COO

^{–}groups. There is no need for showing charges on any group because the formation of H-bonds does not require charged species at all. However, showing the groups in their respective ionized forms may alter the number of O- and N-linked H-atoms, thus affecting the possible number of H-bonds in turn.

Few of the possible intramolecular H-bonds in the given peptide are shown below.

** Intermolecular hydrogen bonds:** The hydrogen bond forming between two different molecules is called an intermolecular hydrogen bond. For example, an H-bond forming between a peptide and water molecule is an intermolecular H-bond. it can form between the H-atom of a water molecule and O-atoms of -OH, -CO, -COOH and -COO_ groups as well as N-atom of -NH, -NH

_{2}, and -NH

_{3}

^{+}groups. There is no need for showing charges on any group because the formation of H-bonds does not require charged species at all. However, showing the groups in their respective ionized forms may alter the number of O- and N-linked H-atoms, thus affecting the possible number of H-bonds in turn.

A few of the possible intramolecular H-bonds forming between water molecules and different groups of the peptide are shown below.

Step 4:** Ionic Interactions or Salt Bridges:** The attraction between oppositely charged side groups is called ionic interaction or a salt bridge. For example, the positively charged side chains exhibit ionic interactions with the negatively charged side chains. Similarly, the negatively charged side chains interact with positively charged side chains. Note that the ionic interaction or a salt bridge is not an ionic bond.

To exert ionic interactions, the adjacent groups must be in their completely ionized forms. For the given peptide, the adjacently placed positively charged N-terminal and the negatively charged side chain of aspartic acid exert ionic interactions. Since the side chain of histidine predominantly exists in its neutral form, it cannot form a salt bridge with the adjacent, negatively charged C-terminal.

One possible ionic interaction or salt bridge forming in the peptide is shown below.

Step 5:** Hydrophobic Interactions:** The attractive interaction between two nonpolar or hydrophobic groups is called hydrophobic interactions. For the given peptide, the side chains of leucine and -CH3 group in the side chain on threonine can exhibit hydrophobic interactions between them. One possible hydrophobic interaction in the peptide is shown below.

## 7. Quantifying Ionized, Unionized and Zwitterion forms of Amino Acids

Recall that the modified Henderson-Hasselbalch equation only gives the fractional ionized form of a weak acid or base. That fraction is either assigned a +ve or -ve sign depending on the type of charge on their respective conjugate chemical species to get its charge. For example, the fractional charge on the weak base is assigned a +ve sign because its conjugate acid, BH^{+} is positively charged. Similarly, the fractional charge on the weak acid is assigned a -ve sign because its conjugate base, A^{–} is negatively charged- it is done by simply multiplying the fraction with (-1). Subtracting the charged fraction from 1 gives the uncharged fraction of the respective chemical species.

## 7A. Quantifying Ionized, Unionized and Zwitterion forms of Neutral Amino Acids

The modified Henderson-Hasselbalch equations give the fraction of an acidic or a basic group in its ionized form. The fraction of the ionized form of an ionizable group also represents its probability of occurrence in the solution. For example, consider the fraction ionized form of the -COOH group at a pH equalling its pKa. Its fractional ionization is 0.50, or the group -COOH is said to be 50.0% ionized. Alternatively, it also means that the probability of finding the -COOH group in its ionized form is 0.50 (on a unity scale) or 50% (on a percentage scale).

To be in its neutral form, both the N-terminal and C-terminal groups must be simultaneously in their unionized forms. And, the probability of finding both the -NH_{2} and -COOH groups in their respective unionized form is equal to the product of their respective probabilities. That is, the probability of finding an alanine molecule with both of its -NH_{2} and -COOH groups in the unionized form (P_{U-NH2, U-COOH}) simultaneously is given by-

P_{U-NH2, U-COOH} = P_{U-NH2 }x P_{U-COOH} – Equation 31

Where,

P_{U-NH2, U-COOH} = Probability of simultaneous occurrence of both the groups in their unionized forms.

Or, Probability of the unionized form of a neutral amino acid

P_{U}_{-NH2 }= Fraction or probability of the **u**nionized form of -NH_{2}

P_{U}_{-COOH }= Fraction or probability of the **u**nionized form of -COOH

And,

% unionized form = P_{U-NH2, U-COOH} x 100 – Equation 32

**Quantifying Zwitterion Form: **

To be in the zwitterion form, both the -NH_{2} and -COOH groups must simultaneously be in their completely ionized forms. The probability of finding a zwitterion form is given by the product of the probability of the ionized forms of -NH_{2} group (P_{I-NH2}) and ionized -COOH group (P_{I-COOH}).

That is, the probability of the zwitterion form (P_{I-NH2, I-COOH}) is given by-

P_{I-NH2, I-COOH} = P_{I-NH2} x P_{I-COOH} – Equation 33

Where,

P_{I-NH2, I-COOH} = Probability of the zwitterion form of a neutral amino acid

P_{I}_{-NH2 }= Fraction or probability of the **i**onized form of -NH_{2}

P_{I}_{-COOH }= Fraction or probability of the **i**onized form of -COOH

And,

% zwitterion forms = P_{I-NH2, I-COOH} x 100 – Equation 34

A completely ionized form of a neutral amino acid is always a zwitterion form, too. Hence, for a neutral amino acid, the probability of its zwitterion form is equivalent to the probability of its completely ionized form because both the forms represent the same molecule. Also, the % zwitterion form is equivalent to the % completely ionized form.

**Example 36:** For a solution of alanine at pH 7.00, calculate the % of its neutral or unionized form.

Ans. Step 1: Calculate the ionized and unionized fractions of the acidic and basic groups of alanine using modified Henderson-Hasselbalch equation as shown below-

As tabulated above, 0.0019461 fractions of the N-terminal and 0.0000214 fractions of the C-terminal groups are in unionized form. The same is pictorially depicted below. where pool N represents the fraction of alanine molecules with unionized -NH_{2} group irrespective of the ionization of the -COOH group. Similarly, pool C represents the fraction of alanine molecules with unionized -COOH group irrespective of the ionization of the -NH_{2} group.

Step 2: Suppose we pick up all the pool N alanine molecules by using some hypothetical molecular tweezers. By all means, we would be able able to pick up only 0.0019461 fractions of the total alanine molecules. It is noteworthy to remember that the fraction of an ionized and unionized form also represents its respective probability of occurrence in the pool of molecules. So alternatively, the probability of finding the pool N molecules with unionized -NH_{2} group, P_{U-NH2} is 0.0019461 (on the unity scale). In this pool N, some of the -COOH groups would be ionized (-COO^{–}, *f*_{I-COOH}) while the rest be unionized (-COOH, *f*_{U-COOH}) as shown below-

The occurrence of -COOH and -COO^{–} groups on the pool N molecules is totally a random incidence. That is, which molecule of the pool N bears which form of the -COOH group can not be predicted. However, in all cases, the probability of finding the unionized form of -COOH group is always equal to the fraction of its unionized form. So, the probability of finding the unionized form of -COOH group in the pool N molecules, P_{U-COOH} is 0.0000214.

To be in its neutral form, both the N-terminal and C-terminal groups must be simultaneously in their unionized forms. And, the probability of finding both the -NH_{2} and -COOH groups in their respective unionized form is equal to the product of their respective probabilities as mentioned in equation 31.

Now, Putting the values in equation 31-

Probability of the unionized form = P_{U-NH2 }x P_{U-COOH} – Equation 31

Or,

P_{U-NH2, U-COOH} = 0.0019461 x 0.0000241 = 0.00000004690101 = 4.69 x 10^{-8}

And, using equation 32-

The % of unionized alanine molecules = 0.00000004690101 x 100

= 0.000004690101% = 4.69 x 10^{-6} %

Hence, % unionized form of alanine at pH 7 = 4.69 x 10^{-6} %

**Example 37:** For a solution of alanine at pH 7.00, calculate the % of zwitterion form.

Ans. From the data summarized in the table above (example 36)-

*f*_{I-NH2}= P_{I-NH2}= 0.9980539*f*_{I-COOH}= P_{I-COOH}= 0.9999786

To be in the zwitterion form, both the -NH_{2} and -COOH groups must simultaneously be in their completely ionized forms. This probability of finding a zwitterion form is given by the product of the probability of finding the ionized -NH_{2} group (P_{I-NH2}) and ionized -COOH group (P_{I-COOH}) as summarized below-

Using equation 33, the probability of the zwitterion form (P_{I-NH2, I-COOH}) is given by-

P_{I-NH2, I-COOH} = P_{I-NH2} x P_{I-COOH} – Equation 33

So,

P_{I-NH2, I-COOH} = 0.9980539 x 0.9999786 = 0.9980325

And, using equation 34-

% of zwitterion forms = P_{I-NH2, I-COOH} x 100 – Equation 34

= 0.9980325 x 100 = 99.80 %

Hence, % zwitterion form of alanine at pH 7 = 99.80 %

**Example 38:** Plot % zwitterion form of alanine as a function of pH.

Ans.

Step 1: Calculate and tabulate the % zwitterion forms of alanine at different pH using equation 9 for the acidic group, equation 23 for the basic group and equation 33 for calculating % zwitterion form as shown below. Also, include all the pKa values as well as the pI values. Note the following information from the above data table-

I. % zwitterion form is maximum when pH = pI because both the groups are completely ionized to their maximal extent simultaneously.

II. At pH = pKa of the -COOH group, 50% of the acidic group is in its ionized form; whereas all (100%) of the basic -NH_{2} group is completely ionized. So, % ionized form at the pKa of -COOH group = 100% (ionized -NH_{2}) x 50% (half-ionized -COOH) = 50%.

III. Similarly, at pH = pKa of the -NH_{2} group, 50% of the basic group is in its ionized form; whereas all (100%) of the acidic -COOH group is completely ionized. So, % ionized form at the pKa of -NH_{2} group = 100% (ionized -COOH) x 50% (half-ionized -NH_{2}) = 50%.

IV. At the extreme acidic pH, most of the acidic -COOH groups are in the unionized form whereas the basic -NH_{2} group is completely ionized. In this case, the % zwitterion form is equal to 0 = 100% (ionized -NH_{2}) x 0% (almost none ionized -COOH) = 0%.

V. Similarly, at the extreme basic pH, most of the basic -NH_{2} group are in the unionized form whereas the acidic -COOH groups are completely ionized. In this case, the % zwitterion form is equal to 0 = 100% (ionized -COOH) x 0% (almost none ionized -NH_{2}) = 0%.

**Step 2:** Plot the % zwitterion form of alanine as a function of pH by putting the percent zwitterion form on the Y-axis, and pH on the X-axis.

Label the axes along with a suitable chart title.

## 7B. Quantifying Ionized, Unionized and Zwitterion forms of Acidic Amino Acids

To be in its neutral form, all three ionizable groups- the NH_{2}, and COOH and the R-group must be simultaneously in their unionized forms. And, the probability of finding these groups in their respective unionized form simultaneously is equal to the product of their respective probabilities. That is, the probability of finding an acidic amino acid molecule with all its three ionizable groups in their unionized form (P_{U-NH2, U-COOH, U-RCOOH}) simultaneously is given by-

P_{U-NH2, U-COOH, U-RCOOH} = P_{U-NH2 }x P_{U-COOH x }P_{U-RCOOH} – Equation 35

And,

% unionized form of an acidic AA = P_{U-NH2, U-COOH, U-RCOOH} x 100 – Equation 36

**Quantifying Zwitterion forms:**

To be in the zwitterion form, a molecule must have two or more ionized groups, and the net charge must be zero, too. An acidic amino acid can have three zwitterion forms as shown below-

Unlike the neutral amino acids, an acidic amino acid has three ionizable groups. One of these groups, -NH_{2} is positively charged in its ionized form, whereas the two others, -COOH and R-COOH are negatively charged in their ionized forms. When all the three groups are completely ionized, the assumption on which our calculations are based, an acidic amino acid bears a net charge of -1 but not zero; hence, it is not a zwitterion form. The product of the fraction or probabilities of the ionized forms of the three ionizable groups gives the probability of the completely ionized form, but no information of the zwitterion form. Unfortunately, this method does not provide any means of calculating the amount of zwitterion III. This zwitterion form is only possible when pH = pI of the amino acid, where we can assume that all (100%) of the molecules exist in the zwitterion form III because the net charge on the molecule is zero.

**Zwitterion I:** Both the N-terminal -NH_{2} and the acidic R-groups are completely ionized whereas the C-terminal -COOH group remains in its completely unionized form. So, the net charge on this molecule turns to be zero.

N-terminal -NH_{2} group: Completely ionized, the charge is +1.0

Acidic R-group: Completely ionized, the charge is -1.0

C-terminal -COOH group: Completely unionized, the charge is 0

Thus, net charge = +1.0 + (-1.0) = 0

The probability of finding the zwitterion I form is equal to the products of the probability of finding the -NH_{2} and acidic R-groups in completely ionized form and the -COOH group in its unionized form.

That is,

P_{zwitterion I} = P_{I-NH2} x P_{I-RCOOH} x P_{U-COOH} – Equation 37

And,

% zwitterion I = P_{zwitterion I} x 100 – Equation 38

**Zwitterion II:** Both the N-terminal -NH_{2} and -COOH groups are completely ionized whereas the acidic R group (= RCOOH) remains in its completely unionized form. So, the net charge on this molecule turns to be zero.

N-terminal -NH_{2} group: Completely ionized, the charge is +1.0

R-group: Completely unionized, the charge is 0

C-terminal -COOH group: Completely ionized, the charge is -1.0

Thus, net charge = +1.0 + (-1.0) = 0

The probability of finding the zwitterion II form is equal to the products of the probability of finding the -NH_{2} and -COOH groups in completely ionized form and the RCOOH group in its unionized form.

That is,

P_{zwitterion II} = P_{I-NH2} x P_{I-COOH} x P_{U-RCOOH} – Equation 39

And,

% zwitterion II = P_{zwitterion II} x 100 – Equation 40

**Zwitterion III:** The zwitterion III form has all its three ionizable groups in the ionized forms, and the net charge is zero. This form only exists when pH = pI. At pH = pI, all the molecules bear a net charge of zero. Thus, it can be assumed that the zwitterion III form exists only at the pI of the molecule with 100% abundance. At all other pHs other than the pI, the % abundance of this form can be assumed to be zero.

**Example 39:** For a solution of aspartic acid at pH 7.00, calculate the % of aspartic acid molecules in unionized form.

Ans. ** **Calculate the ionized and unionized fractions of the acidic and basic groups of aspartic acid using the modified Henderson-Hasselbalch equation as shown below-

It is noteworthy to remember that the fraction of an ionized and unionized form also represents its respective probability of occurrence in the pool of molecules.

To be in its neutral form, all three ionizable groups- the NH_{2}, and COOH and the R-group must be simultaneously in their unionized forms. And, the probability of finding these groups in their respective unionized form simultaneously is equal to the product of their respective probabilities.

Using equation 35, The probability of finding an aspartic acid molecule with all its three groups in the unionized form (P_{U-NH2, U-COOH, U-RCOOH}) simultaneously is given by-

P_{U-NH2, U-COOH, U-RCOOH} = P_{U-NH2 }x P_{U-COOH x }P_{U-RCOOH} – Equation 35

So,

P_{U-NH2, U-COOH, U-RCOOH} = 0.0021830 x 0.0000089 x 0.0005126 = 9.97 x 10^{-12}

And, using equation 36-

% unionized aspartic acid molecules = P_{U-NH2, U-COOH, U-RCOOH} x 100 – Equation 36

= 9.97 x 10^{-12 }x 100 = 9.97 x 10^{-10 }%

Hence, % unionized form of aspartic acid at pH 7 = 9.97 x 10^{-10 }%

**Example 40:** For a solution of aspartic acid at pH 7.00, calculate the % zwitterion forms of aspartic acid.

Ans.

**Zwitterion I:** From the above data table (see example 39), we have-

*f*_{I-NH2} = P_{I-NH2} = 0.9978170

*f*_{I-RCOOH} = P_{I-RCOOH} = 0.9994874

*f*_{U-COOH} = P_{U-COOH} = 0.0000089

Now, putting the values in equation 37-

P_{zwitterion I} = P_{I-NH2} x P_{I-RCOOH} x P_{U-COOH} – Equation 37

P_{zwitterion I} = 0.9978170 x 0.9994874 x 0.0000089 = 8.88 x 10^{-6}

And, using equation 38-

% zwitterion I = P_{zwitterion I} x 100 – Equation 38

% zwitterion I = 8.88 x 10^{-6} x 100 = 8.88 x 10^{-4} %

**Zwitterion II:** From the above data table (see example 39), we have-

*f*_{I-NH2} = P_{I-NH2} = 0.9978170

*f*_{I-COOH} = P_{I-COOH} = 0.9999911

*f*_{U-RCOOH} = P_{U-RCOOH} = 0.0005126

Now, putting the values in equation 39-

P_{zwitterion II} = P_{I-NH2} x P_{I-COOH} x P_{U-RCOOH} – Equation 39

P_{zwitterion I} = 0.9978170 x 0.9999911 x 0.0005126 = 5.11 x 10^{-4}

And, using equation 40-

% zwitterion II = P_{zwitterion II} x 100 – Equation 40

% zwitterion II = 5.11 x 10^{-4} x 100 = 5.11 x 10^{-2} % = 0.0511%

**Zwitterion III:** The zwitterion III form has all its three ionizable groups in the ionized forms, and the net charge is zero. This form only exists when pH = pI. At pH = pI, all the molecules bear a net charge of zero. Thus, it can be assumed that the zwitterion III form exists only at the pI of the molecule with 100% abundance. At all other pHs other than the pI, the % abundance of this form can be assumed to be zero.

**Example 41:** Plot % zwitterion forms of aspartic acid as a function of pH.

Ans. Step 1: Calculate and tabulate the % zwitterion forms I and II of aspartic acid at different pH as shown below. Also, include all the pKa values as well as the pI values.

Use the following equations-

P_{zwitterion I} = P_{I-NH2} x P_{I-RCOOH} x P_{U-COOH} = P_{I-NH2} x P_{I-RCOOH} x (1 – P_{I-COOH})

And, % Zwitterion I = P_{I-NH2} x P_{I-RCOOH} x (1 – P_{I-COOH}) x 100

P_{zwitterion II} = P_{I-NH2} x P_{I-COOH} x P_{U-RCOOH} = _{} P_{I-NH2} x P_{I-COOH} x (1 – P_{I-RCOOH})

And, % Zwitterion II = P_{I-NH2} x P_{I-COOH} x (1 – P_{I-RCOOH}) x 100

Where,

The fraction of the unionized form of -COOH, *f*_{U-COOH} = (1 – P_{I-COOH})

The fraction of the unionized form of -RCOOH, *f*_{RCOOH} = (1 – P_{I-RCOOH})

The fraction of the unionized forms of -COOH and -RCOOH can also be shown in separate rows in the Excel sheet. However, it would require more space in the Excel sheet which would subsequently distort its fitting on the page. So, the fraction of the unionized forms is not shown in the table below.

Step 2: Plot a graph using the data from the above table by putting the % zwitterion forms on the Y-axis, and pH values on the X-axis. Include all the pKa values. Note that the % zwitterion forms are always the highest at the pI of the amino acid. (cells shaded in green; the peak of zwitterion forms).

**Example 42:** For a solution of aspartic acid at pH 7.00, calculate the % completely ionized form of aspartic acid.

Ans. From the above data table (see example 39), we have-

*f*_{I-NH2} = P_{I-NH2} = 0.9978170

*f*_{I-COOH} = P_{I-COOH} = 0.9999911

*f*_{I-RCOOH} = P_{I-RCOOH} = 0.9994874

Now, the probability of the completely ionized form is equal to the product of the probability of the ionized form of each ionizable group.

So,

The probability of the completely ionized form = P_{I-NH2} x P_{I-COOH} x P_{I-RCOOH}

= 0.9978170 x 0.9999911 x 0.9994874 = 0.9973

And,

% of the completely ionized form = P_{I-NH2} x P_{I-COOH} x P_{I-COOH} x 100

= 0.9973 x 100 = 99.73%

## 7C. Quantifying Ionized, Unionized and Zwitterion forms of Basic Amino Acids

To be in its neutral or unionized form, all three ionizable groups- the NH_{2}, and COOH and the basic R-group of a basic amino acid must be simultaneously in their unionized forms. And, the probability of finding these groups in their respective unionized form is equal to the product of their respective probabilities.

That is, the probability of finding a basic amino acid molecule with all its three groups in the unionized form (P_{U-NH2, U-COOH, U-RNH2}) simultaneously is given by-

P_{U-NH2, U-COOH, U-RNH2} = P_{U-NH2 }x P_{U-COOH x }P_{U-RNH2} – Equation 41

And,

% unionized form of a basic AA = P_{U-NH2, U-COOH, U-RNH2} x 100 – Equation 42

**Quantifying Zwitterion Forms:**

To be in the zwitterion form, a molecule must have two or more ionized groups, and the net charge must be zero, too. An arginine molecule can have three zwitterion forms as shown below-

Unlike the neutral amino acids, a basic amino acid has three ionizable groups. Two of these groups, -NH_{2} and R-NH_{2 }are positively charged in their ionized form, whereas the third group, -COOH is negatively charged. When all the three groups are completely ionized, the assumption on which our calculations are based, an arginine molecule bears a net charge of +1 but not zero- thus, it is not a zwitterion form. The product of the fraction or probabilities of the ionized forms of the three ionizable groups gives the probability of the completely ionized form, but no information of the zwitterion form. Unfortunately, this method does not provide any means of calculating the amount of zwitterion III. This zwitterion form is only possible when pH = pI of the amino acid, where we can assume that all (100%) of the molecules exist in the zwitterion form III because the net charge on the molecule is zero.

**Zwitterion I:** Both the N-terminal -NH_{2} and -COOH groups are completely ionized whereas the basic R-group remains in its completely unionized form. So, the net charge on this molecule turns to be zero.

N-terminal -NH_{2} group: Completely ionized, the charge is +1.0

C-terminal -COOH group: Completely ionized, the charge is -1.0

R-group: Completely unionized, the charge is 0

Thus, net charge = +1.0 + (-1.0) = 0

The probability of finding the zwitterion I form is equal to the products of the probability of finding the -NH_{2} and -COOH groups in completely ionized form and the basic R-group (-RNH_{2}) in its unionized form.

That is,

P_{zwitterion I} of a basic AA = P_{I-NH2} x P_{I-COOH} x P_{U-RNH2} – Equation 43

And,

% zwitterion I of a basic AA = P_{zwitterion I} x 100 – Equation 44

**Zwitterion II:** Both the basic R-group (-RNH_{2}) and -COOH groups are completely ionized whereas the N-terminal -NH_{2} group remains in its completely unionized form. So, the net charge on this molecule turns to be zero.

N-terminal -NH_{2} group: Completely unionized, the charge is 0

C-terminal -COOH group: Completely ionized, the charge is -1.0

Basic R-group: Completely ionized, the charge is +1.0

Thus, net charge = +1.0 + (-1.0) = 0

The probability of finding the zwitterion II form is equal to the products of the probability of finding the -RNH_{2} and -COOH groups in completely ionized form and the N-terminal -NH_{2} group in its unionized form.

That is,

P_{zwitterion II} of a basic AA = P_{I-RNH2} x P_{I-COOH} x P_{U-NH2} – Equation 45

And,

% zwitterion II of a basic AA = P_{zwitterion II} x 100 – Equation 46

**Zwitterion III:** The zwitterion III form has all its three ionizable groups in the ionized forms, and the net charge is zero. This form only exists when pH = pI. At pH = pI, all the molecules bear a net charge of zero. Thus, it can be assumed that the zwitterion III form exists only at the pI of the molecule with 100% abundance. At all other pHs other than the pI, the % abundance of this form can be assumed to be zero.

**Example 43:** For a solution of arginine at pH 7.00, calculate the % of its unionized form.

Ans. Calculate the ionized and unionized fractions of the acidic and basic groups of arginine using the modified Henderson-Hasselbalch equation as shown below-

From the data summarized in the table above –

I. *f*_{U-NH2} = P_{U-NH2} = 0.0099010

II. *f*_{U-COOH} = P_{U-COOH} = 0.0000079

III. *f*_{U-RNH2} = P_{U-RNH2} = 0.0000107

The fraction or probability of the unionized form of a basic amino acid is given by equation 41-

P_{U-NH2, U-COOH, U-RNH2} = P_{U-NH2 }x P_{U-COOH x }P_{U-RNH2} – Equation 41

Putting the values in above expression-

P_{U-NH2, U-COOH, U-RNH2} = 0.0099010 x 0.0000079 x 0.0000107 = 8.43 x 10^{-13}

And, using equation 42-

% unionized form of a basic AA = P_{U-NH2, U-COOH, U-RNH2} x 100

= 8.43 x 10^{-13 }x 100 = 8.43 x 10^{-11} %

**Example 44:** For a solution of arginine at pH 7.00, calculate its % zwitterion forms.

Ans.

**Zwitterion I:** Both the N-terminal -NH_{2} and -COOH groups are completely ionized whereas the basic R-group remains in its completely unionized form. So, the net charge on this molecule turns to be zero.

From the data summarized in the table above (example 43)-

*f*_{I-NH2}= P_{I-NH2}= 0.9900990*f*_{I-COOH}= P_{I-COOH}= 0.9999893

III. *f*_{U-RNH2} = P_{U-RNH2} = 0.0000079

Now, putting the values in equation 43-

P_{zwitterion I} of a basic AA = P_{I-NH2} x P_{I-COOH} x P_{U-RNH2} – Equation 43

So, P_{zwitterion I} of arginine = 0.9900990 x 0.9999893 x 0.0000079 = 7.86 x 10^{-6}

And, using equation 44-

% zwitterion form of a basic AA = P_{zwitterion I} x 100 – Equation 44

So, % zwitterion I form of arginine = 7.86 x 10^{-6} x 100 = 7.86 x 10^{-4} %

**Zwitterion II:** Both the basic R-group (-RNH_{2}) and -COOH groups are completely ionized whereas the N-terminal -NH_{2} group remains in its completely unionized form. So, the net charge on this molecule turns to be zero.

From the data summarized in the table above (example 43)-

I. *f*_{U-NH2} = P_{U-NH2} = 0.0099010

II. *f*_{I-COOH} = P_{I-COOH} = 0.9999893

III. *f*_{I-RNH2} = P_{I-RNH2} = 0.9999921

Now, putting the values in equation 45-

P_{zwitterion II} of a basic AA = P_{I-RNH2} x P_{I-COOH} x P_{U-NH2} – Equation 45

So, P_{zwitterion II} of arginine = 0.0099010 x 0.9999893 x 0.9999921 = 9.90 x 10^{-3}

And, using equation 46-

% zwitterion II of a basic AA = P_{zwitterion II} x 100 – Equation 46

So, % zwitterion I form of arginine = 9.90 x 10^{-3} x 100 = 0.99%

**Example 45:** For a solution of arginine at pH 7.00, calculate the % completely ionized form of arginine.

Ans. From the above data table (see example 43), we have-

* f*_{I-NH2} = P_{I-NH2} = 0.9900990

*f*_{I-COOH} = P_{I-COOH} = 0.9999893

*f*_{I-RNH2} = P_{I-RCOOH} = 0.9999921

Now, the probability of the completely ionized form is equal to the product of the probability of the ionized form of each ionizable group.

So,

The probability of the completely ionized form = P_{I-NH2} x P_{I-COOH} x P_{I-RNH2}

= 0.9900990 x 0.9999893 x 0.9999921 = 0.9901

And,

% of the completely ionized form = P_{I-NH2} x P_{I-COOH} x P_{I-RNH2 }x 100

= 0.9901 x 100 = 99.01%

** Example 46:** Plot % zwitterion forms of arginine as a function of pH.

Ans. Step 1: Calculate and tabulate the % zwitterion forms I and II of arginine at different pH as shown below. Also, include all the pKa values as well as the pI values.

Use the following equations-

P_{zwitterion I} = P_{I-NH2} x P_{I-COOH} x P_{U-RNH2}= P_{I-NH2} x P_{I-COOH} x (1 – P_{I-RNH2})

And, % Zwitterion I = P_{I-NH2} x P_{I-COOH} x (1 – P_{I-RNH2}) x 100

P_{zwitterion II} = P_{I-RNH2} x P_{I-COOH} x P_{U-NH2 }= _{} P_{I-RNH2} x P_{I-COOH} x (1 – P_{I-NH2})

And, % Zwitterion II = P_{I-RNH2} x P_{I-COOH} x (1 – P_{I-NH2}) x 100

Where,

The fraction of the unionized form of -NH_{2}, *f*_{U-NH2} = (1 – P_{I-NH2})

The fraction of the unionized form of -RNH_{2}, *f*_{RNH2} = (1 – P_{I-RNH2})

The fraction of the unionized forms of -NH_{2} and -RNH_{2} can also be shown in separate rows in the Excel sheet. However, it would require more space in the Excel sheet which would subsequently distort its fitting on the page. So, the fraction of the unionized forms is not shown in the table below.

Step 2: Plot a graph using the data from the above table by putting the % zwitterion forms on the Y-axis, and pH values on the X-axis. Include all the pKa values. Note that the % zwitterion form is always the highest at the pI of the amino acid (cells shaded in green; the peak of zwitterion forms; the peak of zwitterion I is not visible in the graph).

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