Titration of Aspartate with Hydroxide
On the left in both the chemical reaction and the titration curve, you should imagine that aspartic acid is in a very acidic solution at a pH of about 0. As you move from left to right accross the page, you are adding hydroxide to the solution. This increases the hydroxide concentration and decreases the hydrogen ion concentration. Note that aspartate loses protons as you move from left to right. At the first pKa, the alpha-carboxyl dissociates. At the second pKa, the R-group carboxyl dissociates, At the third pKa, the alpha-amino dissociates.
Note that at each pKa, the solution is buffered. That is, it resists changes in pH as hydroxide is added. Also note, that the pI occurs where aspartate has no net charge.
The main objective of this section is "Given any amino acid you must be able to predict the isoelectric point." How do I start? How do I figure this out without a graph to look at?
For this course, there are only four categories of isoelectric points that you need to learn:
- amino acids without any charged R-group (alanine, glycine, ...)
- lysine and arginine
- aspartate and glutamate
- histidine
The isoelectric point (pI) is the pH at which an amino acid or protein has no net charge and will not migrate towards the anode or cathode in an electric field. The charges on any amino acid at a given pH are a function of their pKas for dissociation of a proton from the alpha-carboxyl groups, the alpha-amino groups, and the side chains (R-group). The pKa for the alpha-amino groups and the alpha-carboxyl groups are about 2 and 10.
You start by having a very rough idea of the structure of the amino acid. What are the acidic groups and what are their pKas. Next, you try to visualize the amino acid fully associated with hydrogen and what the charge on the molecule would be. Next, you visualize removing hydrogen ions by titrating with hydroxide ions. You will remove hydrogen ions from the group with the lowest pKa first and, then from the next higher pKa. Eventually you reach the pI.
Example: Calculate the pI for Aspartate. Aspartic acid has an alpha-carboxyl, and alpha-amino and an R-Group that is a carboxyl group.
At very acidic pHs, the R-group is in the COOH form, the alpha-amino group is in the –NH3+ form and the alpha-carboxyl group is in the COOH form so aspartate has a net charge of +1
As we titrate with hydroxide ion, we remove hydrogen ions. They combine with hydroxide ions and become water. When we reach pH 2, the protons on half the alpha-carboxyl groups are removed. This is not the pI because half of the alpha-carboxyl have a negative charge but all of the alpha-amino groups –NH3+ have a positive charge and all of the R-Group (COOH ) have no charge. So, the net charge on the aspartate molecules is a positive 0.5.
As we titrate with more hydroxide ions, we reach a point half way between pKa1 and pKa2. At this pH, half of the protons have been removed from the two carboxyl groups and half of the carboxyl groups are not dissociated so the net charge on the carboxyl groups is a -1. The alpha-amino group is fully charged so it has a net charge of +1. The net charge on the aspartate molecules is 0. This is the pI
As we titrate with more hydroxide ions, we reach the pKa2, At this pH, all of the protons have been removed from the alpha-carboxyl group and half of the half of the protons have been removed from the R-group carboxyl groups. The net charge on the carboxyl groups is a -1.5. The alpha-amino group is fully charged so it has a net charge of +1. The net charge on the aspartate molecules is -0.5.
As we titrate with more hydroxide ions, we reach the pKa3, a point where all the carboxyl groups are dissociated and only half of the alpha-amino groups still have a positive charge. The net charge is a negative 1.5. We did not have to go this far to determine the pI but I thought it might be useful.
To review, you started with knowing that aspartate had three dissociable groups and the pKas for those groups. You know that as you titrate, the molecule will change as follows:
COOH, COOH, –NH3+ | at a pH below pKa1. The net charge is about +1. |
COO-, COOH, –NH3+ | at a pH between pKa1 and pKa2. Net charge is 0. |
COO-, COO-, –NH3+ | at pH above pKa2. The net charge is about -1. |
COO-, COO-, –NH2 | at a pH above pKa3, the net charge is about-2. |
The only neutral solution of aspartate must be when the pH is between pKa1 and pKa2. The pI is half way in between.pKa1 and pKa2
ACID-BASE PROBLEMS AND THE HENDERSON-HASSELBALCH EQUATION
A. A pharmaceutical molecule with antifungal properties is only active when deprotonated and negatively charged (A-). The protonated state (HA) is inactive. If the pKa of this drug is 10.0, (a) calculate the ratio of protonated to deprotonated compound at physiological pH (7.4). (b) Is this drug likely to be a useful pharmaceutical agent?
(a) calculate the ratio of protonated to deprotonated compound at physiological pH (7.4).
Since we are given both the pH and pKa of the compound, we can use the Henderson-Hasselbalch equation to solve for the ratio of [HA] to [A-].
pH = pKa - log([HA] / [A-])
log([HA] / [A-]) = pKa – pH
log([HA] / [A-]) = 10.0 – 7.4
log([HA] / [A-]) = 2.6
([HA] / [A-]) = 398.11
([HA] / [A-]) = 400 (correct sig figs)
The ratio of protonated (inactive) compound to deprotonated (active) compound is 400 to 1 at physiological pH.
(b) Is this drug likely to be a useful pharmaceutical agent?
Since the vast majority of the compound is in the inactive form at physiological pH, it is unlikely to be a useful pharmaceutical agent.* Ideally, most of the compound would be active in the body.
*However, if the active compound is highly potent, it is possible that a small fraction of active compound is sufficient for useful antifungal activity.
B.
Absorption of aspirin (acetylsalicylic acid, C9H8O4,) into the bloodstream occurs only when the molecule is in its conjugate base form.
(a) If a patient takes two tablets of aspirin (325 mg each), how many grams of aspirin are available for immediate absorption in the stomach? The pH of the stomach is 1.6, and the pKa of aspirin is 3.5.
Since we are given both the pH of the stomach and the pKa of aspirin, we can use the Henderson-Hasselbalch equation to solve for the ratio of [HA] to [A-].
pH = pKa - log([HA] / [A-])
log([HA] / [A-]) = pKa – pH
log([HA] / [A-]) = 3.5 – 1.6
log([HA] / [A-]) = 1.9
([HA] / [A-]) = 79
The ratio of protonated aspirin to its conjugate base is 79 to 1.
So one-eightieth (1/80) of the total aspirin taken will be in the conjugate base form and available for immediate absorption in the stomach:
2 x 325 mg x (1/80) = 8.75 mg
9 mg
(b) Would you expect more or less aspirin to be absorbed in the small intestine (pH ≈ 7.5) compared to the stomach? Briefly explain your answer (no calculation is required).
More aspirin will be absorbed in the small intestine. The higher pH in the intestine means that more aspirin will be in the conjugate base form and therefore available for absorption.
TEST YOURSELF
1.1 Which of the following is the smallest of all amino acids?
A).Glycine, B).Valine, C).Alanine, D).Serine
1.2 Which of the following amino acids can form a covalent bond called a disulfide bond?
A). C, B). K, C). M, D). Y
1.3 Which of the following amino acids is actually an imino acid?
A). Glutamate, B). Alanine, C). Proline D). Glycine
1.4 Which of the following amino acids is the largest of all amino acids?
A). E, B). Y, C). W, D). H
1.5 Which of the following amino acids is not positively charged?
A). H, B). K, C). D D). R
1.6 The three letter “asn” is for which amino acid?
A). Alanine, B). Aspartic acid, C). Asparagine, D). Arginine
1.7 Which of the following is a basic amino acid with a positive charge?
A). Lys, B). Glu C). Trp, D). Gln
1.8 Which of the following amino acids’ side chain is a single methyl group?
A). I, B). V, C). A, D). L
1.9 Which of the following amino acids can form an ionic bond?
A). E & V, B). D & K, C). R & Q, D). P & G
1.10Which of the following amino acids() are polar?
A). All of these, B). Tyrosine, C). Serine, D). Arginine
1.11Titration of Valine by a strong base reveals two pKa’s. The titration reaction occurring at pKa2(pKa2=9.62) is:
A. –COOH + OH- à -COO- + H2O C. –COO- + -NH2+- à -COOH + -NH2
B. –COOH + -NH2 à -COO- + -NH2+ D. –-NH3+ + OH- à -NH2 + H2O
1.12For amino acids with a neutral R group, at any pH below the pI of the amino acid, these amino acids in solution will have:
A). a net negative charge, B). a net positive charge, C). no charged groups, D). no net charge
1.13The peptide AEGAL has:
A). a S-S bridge, B). 5 peptide bonds, C). 4 peptide bonds, D). no free carboxyl groups.
1.14Charmaine was upset that her boyfriend dumped her over her best friend; so she decided to commit suicide by overdosing with Panado (pKa=3) and drinking it with lots of fresh milk (pH 6.7). She was admitted to the hospital but was later discharged without any serious complication. Drug tests revealed that most of the Panado remained in the stomach rather than being absorbed, can you explain these? (include calculations).
1.15You are working for a MERCK Pharmaceutics and the manager gives you a task of designing a drug to treat breast cancer and produce it in the form of a pill. The drug must have an absorption success of atleast 60% in the stomach (pH=2), which is where it must be in unionized form. What will the pKa of your drug be?
1.16A student titrates acidified Lysine using NaOH to completed all of its titrable groups. Draw the titration curve of Lysine showing the state of these Ionizable groups at each point and also calculate its pI.
1.17Hemoglobin is a protein responsible for transporting oxygen from the lungs to the tissue via the blood circulatory system. For it to be able to perform its function which of the following groups of amino acids will dominate the surface of hemoglobin?
A). A, V, F, L B). H, R, L, E C). Y, G, M, I D). W, D, A, V
1.18Insulin binds to insulin receptor on cell surfaces to be able to facilitate uptake of glucose by the cells. Which types of amino acids will dominate the surface of the Insulin receptor?
A). AVFL B). HRLE C). YGMN D). WDAV
1.19All the 19 amino acids have a chiral carbon center except for A). F B). G C). H D). W.
1.20Suppose you were a bacterium and had a choice of living in the small intestines (pH 8) or the stomach (pH 1). A new antibacterial drug (pKa = 2.5) has just been released into the market to be taken by humans. Which environment would choose to live in? The stomach or small intestine?
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