Amino Acid Titrations Flashcards
True or False: Amide bonds between constituent amino acids influence the acid-base properties of a protein.
This statement is false. Amide bonds, also known as peptide bonds, do not influence a protein’s acid-base chemistry. They are typically electrically neutral and very stable to changes in pH.
pH at which acid and conjugate base are equal
pKa At an acidic functional group’s pKa, 50% of the acid form of the group will be deprotonated and 50% will remain protonated. Therefore, the solution will contain equal concentrations of the acid form of the functional group and its conjugate base.
True or False: Zwitterions are molecules that contain an equal number of positive and negative charges, which cancel out to give the molecule a net charge of zero.
True
At its second half-equivalence point (pH ~ 9.5), what is the overall net charge of a leucine molecule?
between -1 and 0
Is this amino acid acidic or basic based off of its titration curve?
Acidic
True or false: The isoelectric point of a polyprotic acid is always located at the first equivalence point along its titration curve.
This statement is false. An amino acid’s isoelectric point (pI) corresponds to the pH at which the amino acid contains both positive and negative charges that cancel out to make the overall molecule electrically neutral. Coincidentally, the pI does occur at the first equivalence point for nonpolar and acidic amino acids. However, basic amino acids are isoelectric at the second equivalence point on the curve.
Suppose that 60 mL of 0.1 M NaOH is required to completely neutralize the amine group of glycine in an analyte solution. What quantity of glycine is present in the analyte?
3 mmol
Dimensional analysis is an easy way to perform this calculation. Start by converting all values into scientific notation: 60 mL NaOH = 6 x 10-2 L NaOH. 0.1 M NaOH = 1 x 10-1 mol NaOH / L NaOH. We can simply multiply the volume of NaOH by the molarity to obtain the number of moles added. But because we’re talking about glycine’s amine group, we’re actually dealing with the second equivalence point. At the second equivalence point, 2 moles of NaOH have been added for every 1 mole of glycine present in the analyte. This is because one OH- was required to neutralize the COOH group and a second OH- was needed to neutralize the amine. We need to modify our calculation to account for this. The final setup is: (6 x 10-2 L NaOH) x (10-1 mol NaOH / L NaOH) x (1 mol glycine / 2 mol NaOH) = 3 x 10-3 mol = 3.0 mmol glycine.
If pH is above pi, what should the net charge be on a protein ?
It should be negative
If pH is below pi what should the net charge be on a protein?
It should be positive
If the pi of a protein is 8 and the pH of the surrounding solution is 3, what will the net charge of the protein be?
It should be positive because pH is below pi
