Isoelectric Point of Amino Acids, Part 2 | MCAT Organic Chemistry Prep
HTML-код
- Опубликовано: 6 фев 2025
- Need help preparing for the Organic Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach you Part 2 about isoelectric point of amino acids. Watch this video to get all the MCAT study tips you need to do well on this section of the exam!
Basic Amino Acids
Determining the isoelectric point of basic amino acids requires a slightly different approach from that of acidic and polar or non-polar amino acids. For both basic and acidic amino acids, isoelectric points can be more complicated to calculate, because we have three different acidic groups and three different pKa values to deal with. It is also worth making note of one common convention in the discussion of isoelectric points: The value given for basic functional groups is that of the pKa of the conjugate acid, or the pKb - so for a primary amine, like that of lysine, the pH above which the neutral form of the base, rather than its conjugate acid, starts to predominate. This is significantly different from the more common convention for acids, so keep an eye out!
For the sake of demonstration, let’s take a look at the properties of lysine: Its carboxylic acid group has a pKa value of about 2, its amino group has a pKa of about 9, and the side chain of lysine has a pKa of about 10.7. If all three of these groups are protonated, as shown, the molecule must be in a solution environment of very low pH, one that is significantly lower than the lowest of the functional group pKas. Let’s call this value approximately a pH of 1, enough to ensure that virtually all our functional groups are protonated.
And same as before, we're going to start at a pH of one, and we're going to gradually increase the pH until all four acidic functional groups have been deprotonated. Now, starting at a pH of one, as you increase the pH, the first acidic functional group to be deprotonated is the carboxylic acid. Once we exceed a pH of about two, the pH will be greater than this pKa of two, so this carboxylic acid will be mostly in its deprotonated form, the carboxylate.
As we can continue to increase the pH, the next group that is going to become deprotonated is the amino group on the end terminus of the amino acid. And that is going to happen once we exceed this pKa value of 9 in pH. Now we have the third protonation state: Both the carboxyl group and this amino group is deprotonated.
As the pH of the environment is increased further, to the point that the pH exceeds the pKa value of the final amine group (10.7), the side chain will also become deprotonated. So now all three groups are deprotonated. To make any use of this information, we need to assign net charges to all 4 possible states of lysine. In the beginning, the left-most structure is where all three groups are protonated, we have a +1, +1 and no charge (amino terminus, primary amine of lysine, and carboxyl terminus, respectively). So this molecule has a charge of +2. Next, we have a +1 charge on amino terminus, a +1 on the lysine residue and a -1 on the carboxylate, adding up to +1.On our third molecule, we have charges of +1, 0, and -1, for a total of 0. Since this molecule over here has both positive and negative charges, but the net charge is zero, this species is our zwitterion. And finally, in the maximally deprotonated form, all we have left is a single negative charge for a net charge of -1.
Now that we've identified our zwitterion, we know enough to calculate the pI: Take the pKa values that flanked the zwitterion, 9 and 10.7, and plug them into our previous equation.
You should understand the difference between determining the isoelectric points of polar or non-polar amino acids, acidic ones and basic ones. Note also that while both histidine and arginine have more complex residues (imidazole and guanidino functional groups, respectively), calculations to arrive at their isoelectric points are performed identically to those of other basic amino acids.
MEDSCHOOLCOACH
To watch more MCAT video tutorials like this and have access to study scheduling, progress tracking, flashcard and question bank, download MCAT Prep by MedSchoolCoach
IOS Link: play.google.co...
Apple Link: apps.apple.com...
#medschoolcoach #MCATprep #MCATstudytools
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
The calculated pI is not shown but is said to be "8.85," which really should be 9.95 as it is between the two pKas 9 and 10.7. Great video regardless!
Yup, you're right! Small math mistake :(
9.85 actually
By the way these videos are amazing!! THANK YOU
Is there a way to stop the pop ups at the end of the video? I couldn't see what you wrote for the formula because it was being covered by the stethoscope logo and recommended next video.
You are amazing
Can you explain why at pH 1 the carboxyl group has a charge of zero? I thought all the groups were protonated? Thank you!
The protonated state of the carboxylic acid functional group is COOH, which is neutral. The deprotonate state of the carboxylic acid functional group is the carboxylate, which is COO-. Hope that helps!
why does the amino group have a different pka from the side chain?
please i need an answer asap
It is due to the increased basicity of the side chain amino group relative to the alpha amino group. Basicity is influenced by various factors such as electronegativity, electron donating groups (resonance or inductive) etc... There are likely several factrors at play but it seems like it may be due to the side chain group being near more alkyl groups that are inductively electron donating which increase its basicity (increasing its pKa and decreasing its pKb). Another factor is that the alpha amino group is near an electron with drawing group which will stabilize the lone pair of electrons on the amino group making it less basic/more stable, so it will have a decreased pKa and increase pKb (weaker base). This may not be the full reason why there is a differen b/w the pKas so if anyone has anything to add feel free to do so I am just making a some deductions based on my background knowledge and the structure of the amino acid.
master