Determine the pI of Glu & mark it on the provided titratio...

Question

Determine the pI of Glu & mark it on the provided titration curve.

a. 6.96

b. 5.93

c. 7.48

d. 3.22

Use the titration curve to draw the predominate structures of Glu at pH = 1.0, pH = 3.0, pH = 7.0, & pH = 11.0.

Glu at pH = 1.0 Glu at pH = 3.0 Glu at pH = 7.0 Glu at pH = 11.0

Accepted Answer

Alright. So here's a 2 part practice problem broken up into part a up above and part b down below. And part a wants us to determine the p I or the isoelectric point of glutamic acid, and then to mark it on the provided titration curve over here on the right. And so we know from our previous lesson videos, the first step to determine the isoelectric point is to just know how the r group ionizes. And so we can remember our mnemonic for the 7 amino acids with ionizable r groups. And that mnemonic is just yucky crazy dragons eat knights riding horses. And of course, we know that glutamic acid's one letter amino acid code is an e, and so it is part of our mnemonic over here. And because dragons eating things is a negative thing, that reminds us that glutamic acid has a negatively charged r group. And so the second step in determining the isoelectric point is to order the pKa values from the smallest value to the largest value, and notice that the pKas are already ordered from smallest to largest in our titration curve. And so I've gone ahead and just rewritten those pKa values right here from smallest to largest. And so we know that alpha carboxyl groups have pKas in the ballpark range of about 2, so this pKa here is for the carboxyl group. Then we know that alpha amino groups have p k a's in the ballpark range of about 9 to 10.5. So this p k a over here is for the amino group, and then of course, the p k r is the p k a of the r group. And so I've gone ahead and labeled those appropriately over here on the left. And so in our 3rd step, all we need to do is use the middle p k a here as a guide and determine the net charges of the predominant structures at any pH between the adjacent pKas. So that's just saying determine the net charge of the predominant structure between these 2 pKas and determine the net charge of the predominant structure between these 2 pKas, if necessary. And so that's gonna correspond with this region here between these 2 p k a's and the titration curve, and this region, between these 2 p k a's on the titration curve. And so I'm gonna start with the region over here on the left because we typically do things from left to right anyways. And so, essentially, all we need to do is pick any pH as long as it falls between these 2 pKas. So I'm gonna pick a pH of 3. And, is compare the pH of the solution that we've chosen, pH of 3, to all of the pKas separately. And so we'll start off with the pKa of the amino group, and the pKa of the amino group, which is 9 point 6 7, is greater than the pH. And when the p k a is greater than the pH, the conjugate acid predominates, and the conjugate acid of the amino group is gonna be a protonated form. So it's gonna have an extra hydrogen, NH 3, and it's gonna be positively charged. Now we can do the same with the carboxyl group next, and notice that the pKa of the carboxyl group is smaller than the pH of the solution which is 3. And so when the pH is greater than the pKa, the conjugate base predominates. And the conjugate base of the carboxyl group is just gonna be a carboxylate anion with a negative charge. Now we know that for the r group of glutamic acid that it's literally because it has the GLUT in it, it's just gonna have an extra CH 2 section or or an extra butt section in comparison to aspartic acid. And we know that aspartic acid is literally just alanine with a carboxyl group. So we can go ahead and draw in glutamic acid structure, so it's gonna be c h 2, another c h 2, and then it's gonna have its carboxyl group down at the bottom. And so this group here, we have to determine if it's going to be ionized or not. So the p k r is 4.25, and that's greater than the pH. And when the p k r is greater than the pH, the conjugate acid predominates, and the conjugate acid of the carboxyl group is gonna be protonated. So it's gonna be Oh with a neutral. And so notice here that we have 2 charges. We've got the positive charge and the negative charge that cancel each other out, and so we found our predominant structure that has a neutral net charge of 0. And so that's exactly what we're looking for when we're trying to, determine the p I. And so all we to do is average the 2 p k a's that are sandwiching the region where the predominant structure has a neutral net charge. So that would be this p k a over here for the carboxyl group and the p k r, since they sandwich this region where the structure has a neutral net charge. So we can go ahead and do our calculation over here. So we have the isoelectric point, it's gonna be the average of those two numbers. So it'll be 4.25 plus 2.19 divided by 2, and of course if you type this in your calculator, it'll come out to about 3 point 22. And so this here is our answer and it matches with option d here, so we can go ahead and give it a check to indicate that d is the correct answer for part a. So now all we need to do is mark this p I on the titration curve. So, p I of 3.22 is gonna show up just above 3. So here would be 3 on, the p h, and so 3.22 would be slightly above, 3. So, it would correspond on this position over here in this yellow region. And so this is where we would expect our p I to be our isoelectric point. So there we go. We've marked it on the titration curve, and part a is completely done. Now we can move on to part b down below. Now in part b, it wants us to use the titration curve to draw the predominant structures of glutamic acid at the indicated p h's. And so we can start with glutamic acid at pH 1. And and so notice up above here, we can see that at pH of 1, that's gonna correspond with this region right here. And this region is a pH that is below the p k a values for all of the other ionizable groups. It's below. And so when the pH is below the p k a's, the conjugate acid form predominates. So we can draw the conjugate acid form or the protonated form for each of the groups here. So it'll be n h three plus for the amino group. It'll be an Oh neutral for the carboxyl groups. And so we only have one charge on here, and it's a plus one charge. So our structure at a pH of 1 is gonna have a positive one charge. Now, at a pH of 3, that's showing up right really close to where our isoelectric point was, which is at 3.22, and it falls into the same yellow region. So what that means is that our structure here is gonna have a neutral net charge in the same yellow region. So we already know it's gonna have a net charge of 0. And so essentially, the only thing that changes is, the carboxyl group is the carboxyl group PKA is now smaller than the pH. And so all of the other groups are gonna stay the same and the carboxyl group's gonna be the only one to change. So we can go ahead and, reflect that here. So the amino group is gonna be the same, the r group is gonna be the same, and the only thing that's gonna change is the carboxyl group which is now gonna be in the conjugate base deprotonated form. So now we can move on to our next region, which is pH of 7, and notice that a pH of 7 falls into this blue region over here. And so, we know that when we cross from one region through a pKa into the next region, that the net charge is gonna go down 1. So we expect the net charge to go down 1 to negative 1 from the previous region. And so now notice that the only thing that's different between the previous structure at pH 3 and the structure at pH 7 is now that the p k r is now smaller than the pH of 7. And so, the only thing that's gonna change between the structure on the left and this structure is the r group. So everything else is gonna remain the same. So the carboxyl group will still be deprotonated as a carboxylate group. This will still be protonated as an NH 3 plus. And the only thing that's gonna change now is we'll have a deprotonated, carboxyl group down here. And so now you can see that these two charges cancel each other out and we only have a negative charge here. So that's why we have a negative one charge. So now for our final region here at a pH of of 11 at a pH of 11, we can go up and check that region. So a pH of 11, notice correspond way up here which is above all of the other pKas. And so when the pH is above the pKa, the conjugate base form predominates. So we'll draw all of the ionizable groups in the conjugate base form. And so this will be an NH 2, which is neutral for the amino group. The carboxyl groups are still gonna be deprotonated in the conjugate base forms as carboxyl anions. And so notice here that we have 2 negative charges, which gives us a net negative charge of negative 2. And so, essentially, that is the end of this problem, and I'll see you guys in our next video.

Struggling with Biochemistry?

Determine the pI of Glu & mark it on the provided titration curve.
a. 6.96
b. 5.93
c. 7.48
d. 3.22

Use the titration curve to draw the predominate structures of Glu at pH = 1.0, pH = 3.0, pH = 7.0, & pH = 11.0.
Glu at pH = 1.0 Glu at pH = 3.0 Glu at pH = 7.0 Glu at pH = 11.0

Watch next

Titrations of Amino Acids with Ionizable R-Groups practice set

Struggling with Biochemistry?

Determine the pI of Glu & mark it on the provided titration curve.a. 6.96 b. 5.93c. 7.48d. 3.22Use the titration curve to draw the predominate structures of Glu at pH = 1.0, pH = 3.0, pH = 7.0, & pH = 11.0.Glu at pH = 1.0 Glu at pH = 3.0 Glu at pH = 7.0 Glu at pH = 11.0

Watch next

Titrations of Amino Acids with Ionizable R-Groups practice set

Determine the pI of Glu & mark it on the provided titration curve.
a. 6.96
b. 5.93
c. 7.48
d. 3.22

Use the titration curve to draw the predominate structures of Glu at pH = 1.0, pH = 3.0, pH = 7.0, & pH = 11.0.
Glu at pH = 1.0 Glu at pH = 3.0 Glu at pH = 7.0 Glu at pH = 11.0