Question

What will the total charge be for this peptide (T-H-R-I-P ) in a solution at pH 5.52? Please make sure that your final answer and all associated calculations are clear (circle final answer if necessary)! (note that it is OK to consider any group fully (de)protonated when it is >2.0 pH units from its pKa). Don’t forget that you have scratch paper at the back of the test.

Answer 1

Ans. Calculation of Charge on peptide:

Method 1: Use Henderson-Hasselbalch equation

Net Charge (exact value) at pH 5.52 = 1.768 = 1.8

Note: Histidine is a basic residue.

#I. The residues with acidic side chain and C-terminal are written in RED/BOLD. Their corresponding charge is written in columns of “Acidic residues”. Ignore all other charge/values in “black ink”. The sum of charge in “RED ink” is the total –ve charge on the peptide at specified pH.

#II. The residues with basic side chain and N-terminal are written in Blue/BOLD. Their corresponding charge is written in columns of “Basic residues”. Ignore all other charge/values in “black ink”. The sum of charge in “Blue ink” is the total +ve charge on the peptide at specified pH.

#III. The charge on each residue/ group is calculated using Henderson-Hasselbalch equation as shown in respective column.

#IV. Only residues with acidic and basic side chain are shown in excel for calculation.

Method 2: Approximation method:

Net Charge (approx. value) at pH 5.52 = 2.0

If the difference between pKa of a group and pH is equal to or greater than 2, then assign complete charge on the residues. The nature of charge depends of the protonated or deprotonated state of the acidic/ basic group.

At pH > pKa, COO- form predominate.

At pH < pKa, COOH form predominates.

At pH = pKA, the group is neutral because both protonated and deprotonated form exist in equal concentration.

The “acidic” side chain is treated as :-COOH“ group.

At pH = pKa, the NH2 and NH3+ forms of the amine group exist in equilibrium. So, it is neutral.

If pH < pKa, NH₃⁺ form predominates.

If pH > pKa, NH2 form predominates

The “basic” side chain is treated as “–NH2“group.