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Lecture Comments (5)

1 answer

Last reply by: Professor Hovasapian
Mon May 9, 2016 3:27 AM

Post by imene hacene on May 8 at 07:31:11 AM

How to Draw the structure of lysine that predominates at PH =5.5 and PH= 12.7 ?

0 answers

Post by Torrey Poon on January 28, 2014

Thank you Prof. Hovasapian, that explanation helped!

1 answer

Last reply by: Professor Hovasapian
Mon Jan 27, 2014 3:30 PM

Post by Torrey Poon on January 27, 2014

How did you get the value 27,166 g/mol on Example 2, part b?

More Examples with Amino Acids & Peptides

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Example 1 0:22
    • Data
    • Part A: What is the pI of Serine & Draw the Correct Structure
    • Part B: How Many mL of NaOH Solution Have Been Added at This Point (pI)?
    • Part C: At What pH is the Average Charge on Serine
    • Part D: Draw the Titration Curve for This Situation
    • Part E: The 10 mL of NaOH Added to the Solution at the pI is How Many Equivalents?
    • Part F: Serine Buffer Solution
  • Example 2 23:04
    • Data
    • Part A: Calculate the Minimum Molar Mass of the Protein
    • Part B: How Many Tyr Residues in this Protein?
  • Example 3 30:08
    • Question
    • Solution
  • Example 4 48:46
    • Question
    • Solution

Transcription: More Examples with Amino Acids & Peptides

Hello and welcome back to Educator.com, and welcome back to Biochemistry.0000

We just finished discussing amino acids and peptides, and what I thought we would do is just do some more example problems with amino acids and peptides just to make sure we have a complete understanding, give us a little bit of review of the things that we have done before, and just get familiar with it.0004

Let's jump on in.0020

The first example, example 1.0025

OK. Given the following data, answer the questions that follow.0033

Let me do this in blue actually.0059

We have 100mL of a 0.20M serine solution at a pH of 1.60, is titrated with a 2.0M sodium hydroxide solution.0062

We don't care about the sodium; it is the hydroxide that we are concerned about.0098

OK. The pK1 = 2.21.0100

We have a pK2, which is the pKa of the amino, that is equal to 9.15; and then I'll go ahead and give you the molar mass.0108

I'm just going to write m/m for molar mass; m/m and this happens to be 105g/mol for serine.0120

Now, the first question that we want to ask is "What is the pI of serine?", and draw the correct structure.0131

In other words, draw the correct structure at that particular pH, which is the pI.0155

Well, you have got 2 ionizable groups on serine.0162

You have the carboxyl group, and you have the amino group.0165

So, the pI, that means...let me just draw the general...so you've got N, C, C, that; and you have that.0168

Starting off at a pH of 1.6, these are both going to be protonated.0180

Well, so this serine...sorry, I'll go ahead and just do R there; we are not concerned with this serine yet.0184

This is carrying a +1 charge.0192

A pI is the isoelectric point.0194

It is the point where the net charge on the molecule is going to be 0.0196

So, the pI is going to be where the carboxyl group is full.0201

In this particular case, since you have 2 ionizable groups, it is where it is completely ionized.0206

There is no H left on there at all because then the carboxyl group is going to carry a -1.0210

The NH3+ is going to end up carrying a +1 for a net charge of 0.0217

That is the isoelectric point.0224

Well, with 2 ionizable groups, the isoelectric point takes place at the arithmetic mean of the 2 pKs.0225

So, you just add the 2.21, 9.15 and divide by 2.0232

We have pI...oops, not a capital P; this is a small P.0238

The pI, the isoelectric point is equal to pK1 plus - I'm always, always, always doing capitals - pK2 divided by 2.0245

2.21 + 9.15 / 2 gives us a pI of 5.68.0260

There we go.0270

When the pH is 5.68, that means this serine molecule has a net charge of 0.0271

The carboxyl is ionized; the amino is not ionized.0278

OK.0281

Now, let's go ahead and draw the structure of serine.0283

We have got the N; we have C, and we have C.0285

We have our ionized carboxyl group; we have this one which is not ionized.0292

Let me make my carbonyl carbon a little bit better here.0297

And then we go ahead and we have the H, and then, of course, we have CH2 and OH.0300

This is our structure at 5.68.0306

That is it. OK.0310

Let's see, part B.0316

Excuse me.0320

There we go.0324

Our pen didn't work there for a second.0327

Alright.0328

How many milliliters, specifically, of the sodium hydroxide solution, of the NaOH solution, have been added at this point, in other words, at the pI?0329

OK.0355

So, we started off with a pH of 1.6, and now, we are at a pH of 5.68- that is the isoelectric point.0357

We want to know how many milliliters of this 2M sodium hydroxide solution we've actually added to bring it to that point.0363

OK. Let's think about this.0371

First of all, we want to know how many moles of serine there is in there because it is going to depend on how many moles of ionizable groups that we have to titrate because that is what the hydroxide is doing.0374

The hydroxide is going in there, and it is pulling off the hydrogen from the carboxyl group; and when that is done, that is going to be the isoelectric point, and then it is going to go and start pulling off the hydrogens from the amino group until it's done.0386

That is the second plateau on the titration curve, which we will draw in just a minute.0402

Let's find out how many moles of serine we have.0405

Well, we have 100mL of a 0.2M solution.0409

Let me do this one in red.0415

We have 0.100L x 0.20mol/L.0418

This will give us 0.02mol of serine.0426

Well, serine has 2 ionizable groups, but 0.02mol of serine contains 2 moles of ionizable groups, right?0432

You have the carboxyl, and you have the amino.0454

With every amino acid, there is a minimum of 2 ionizable groups per mole of serine, which means that we have 0.04mol of serine cancels, 0.04mol of ionizable groups.0456

Well, the isoelectric point - in this particular case, you have 2 ionizable groups - is when one of them is completely ionized.0478

So, if you have 0.04mol of ionizable groups, when half of them have been ionized, that is your isoelectric point.0484

That means - let me write that out - one of these groups is fully ionized.0496

That is what the pI means- isoelectric point.0513

You have 0.04mol of total ionizable groups.0515

Half of them have been ionized, which mean 0.02mol.0520

Well, since we have 0.02mol of ionizable group that have been ionized, that means, and it's a 1 to 1 ratio, 1 hydroxide per 1 ionizable group, right, is pulling off 1 hydrogen, that means there are 0.02mol of the hydroxide that I have to use.0533

Let me actually, specifically write that.0557

0.02mol of OH- times, and it is 2mol/L, 2mol of hydroxide per liter.0560

What we end up with is 0.01L or 10mL of sodium hydroxide solution.0576

There we go; that's our answer.0588

I hope that made sense.0590

At the pI, we know that 1 group is fully ionized.0592

Well, I need to know how many moles there are, but each mole of serine actually brings 2 ionizable groups, so I have a total of 4 moles that can actually be ionized.0597

Half of them have been ionized at the pI, at the 5.68, so of the 0.04, 0.02mol have been ionized.0608

Well, the 0.02mol that have been ionized comes from, they have been ionized by 0.02mol of hydroxide because it is a 1 to 1 ratio, OH + H.0616

That is what forms the water.0626

This is a titration, the acid base titration.0627

That is where this comes from.0630

I take the 0.02mol of hydroxide; I multiply by the reciprocal of its molarity, and I get its volume, so, 10mL of NaOH are added.0631

That tells me that in order to fully ionize the serine completely, I would just add another 10mL, so I would need 20mL of sodium hydroxide to completely ionize it.0639

OK. Let's see, part C.0650

I hope you don't mind that I'm jumping around from color to color.0656

OK.0658

Actually, you know what, I'm going to go ahead and do this one in...I'm going to go back to black- part C.0663

Excuse me.0673

At what pH is the average charge on serine -½.0674

OK, we want to know the pH where the average charge on the whole molecule is -½.0690

OK. We have 2 ionizable groups.0696

Let me go ahead and just draw this out really quickly.0699

We have N, C, C, O-.0702

I'm just going to put an R1 here.0707

Actually, let me go ahead and put the H+.0712

OK. When both of these are protonated, the molecule is carrying a charge of +1.0716

At the pK1, right, the pK1, that is when half of the carboxyl groups have been ionized.0723

The other half is still protonated, so half protonated, half deprotonated.0730

At that point, this side is carrying a charge of -½, right?0733

Because if it were fully ionized, it would carry a charge of -1, but at the pK1 - remember we said that's when half of the groups are ionized - the protonated and deprotonated form are in equal concentration.0740

So, instead of -1 total, just a solid -1, you get a -½.0754

Well, -½ + 1 gives you a +½, and then when this is fully protonated, that is the pI, then you are going to have -1 and +1, that is going to be 0.0758

Now, at the pK2, that means half of these are going to be ionized.0773

This side of the molecule is going to carry a charge of +½, and this side is already fully ionized because we’ve passed the pK1 mark, that is a -1 charge.0781

So, -1 + ½, that gives us the negative half.0791

The pH at which the total serine molecules carrying a charge of -½ is the pK2- the 9.15.0795

Now, let me go ahead and write that out.0804

At a pH = 1.60, that implies that the charge on this molecule is +1 because nothing has been deprotonated.0808

OK. At the pK1, that implies the -½ + 1 = +½.0817

That is the charge at the pK1.0828

Well, at the pI...I didn't even write it.0832

Actually, you know what, it doesn’t matter; I'll just go ahead and do that here.0839

That is going to be -½ + 1 = +½ and then at the pI, we have, of course, -1 + 1, we have a charge of 0, and now, the pK2.0843

We have the -1 from the deprotonated carboxyl group, and then we have +½ from the half-deprotonated amino group, and that gives us a +½.0857

At a pH that is equal to the pK2; that is equal to 9.15.0869

At that pH, the serine carries a +½ charge.0875

That is it; I'm sorry, -½ charge.0880

-½- that is the one we are looking for.0882

OK, part D.0885

Draw the titration curve for this situation.0892

OK.0907

Well, we know how to do this, not a problem.0908

So, we are going to do this; we'll do something like that.0910

OK. Let's mark off some, let's do 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, I'll just go ahead and put 11, and then, so let's see, this is 1.0916

We said we started at about a 1.6, 2.12.0930

So, this 2.12, that is the pK1, and we said that the pI is 5.68.0937

I'll do 1, 2, 3, 4, 5, 5.68, so the pI is somewhere around there.0947

This is the pI.0953

You know what, let me write what these are here.0955

This is pK1; this is 2.12, 5.68, and then 9.15, 6, 7, 8, 9, 9.15.0959

So, this puts our pK2 right there, and this is 9.15.0972

Here we go.0978

We are going to start right about here.0980

We are going to plateau out.0982

We are going to plateau out, and we are going to go up, and then we are going to plateau again, and then we are going to go up.0987

So, it looks something like that.0993

That is our pK1 mark; this is our pI, the isoelectric point.0997

This is our second dissociation, the pK2, and that is it.1002

It is at this point that we've added the 10mL of sodium hydroxide.1007

We have fully ionized the carboxyl group, and then, of course, somewhere over here, we have added the 20mL.1014

This is going to be mL of OH-, and that is it.1022

The titration curve, that's what it looks like, 2 ionizable groups; you have 2 plateaus.1024

If you have 3 ionizable groups, you'll have 3 plateaus.1029

That is it, nice and simple.1032

This point, this point and this point, those are the important points- pK1, the pK2 and the pI, the isoelectric point.1034

You should be able to go back and forth.1043

If you are given a titration curve with numbers on it, you should be able to say this happens here, this happens here, this happens here; or if you are given a numerical data, you should be able to produce the titration curve.1044

OK.1056

Let's see, E.1058

The 10mL of NaOH added to the solution at the pI, so the 10mL to bring it to the pI is how many equivalents?1063

OK.1088

In general chemistry, we mentioned this thing called an equivalent.1090

We haven't really talked about it very much.1096

In biochemistry, they tend to use it a little bit more.1098

It is not really that big of a deal.1101

You are not going to see it all that often, but equivalents just means how many.1103

So, 1mol of serine, we said, has 2 ionizable groups, right?1110

It has 2mol of ionizable groups.1115

It has 2 equivalents of ionizable groups.1119

Equivalent just means how much OH do I have to add for each H that is going to be deprotonated.1123

In this particular case, in order to get to the pI, I have ionized 1 of the ionizable groups, so I've added 1 equivalent.1130

If I ionized the whole thing, I've added 2 equivalents.1140

That is it.1144

It's just, instead of talking about specific volumes like 10mL, 5.6mL, we just speak of equivalents.1147

We are speaking more globally.1153

A particular amino acid might have 3 ionizable groups.1156

Therefore, in order to fully ionize that amino acid, I have to add 3 equivalents of hydroxide: 1 for the carboxyl, 1 for the alpha-amino, and 1 for the R-group.1160

That is all that's going on.1174

So, in this particular case, 1 equivalent has been added to bring it to the pI.1175

A second equivalent would be added to bring it to full ionization- not the pK2.1181

Remember, the pK1 and the pK2 represent half deprotonation.1187

These are the buffer regions.1194

It is fully ionized here, and then the second one is fully ionized here.1197

That is the second equivalent.1201

This is really, really easy; this is just 1 equivalent, and they will often refer to it that way, qualitatively instead of quantitatively.1203

Quantitative, 10mL- they are giving you a number.1214

1 equivalent is just, they are speaking about how many ionizable groups.1216

That is all they are doing.1221

OK, F.1223

Now, would a serine buffer solution, so let's say I went ahead and had a serine solution, and I created a buffer solution out of it.1228

With a serine buffer solution, the appropriate for an experiment requiring for an experiment that needs to be maintained at a pH equal to 8.7.1242

If I am running an experiment and I need to maintain the pH at about 8.7, would it be appropriate for me to use a serine buffer solution?1280

In other words, I create a serine solution; I add enough acid or base to bring it up to this particularly.1288

Is this a good buffer solution?1294

Is serine a good buffer solution?1297

Well, the answer is yes, and the reason is, well, take a look at the pKa.1298

Where does this 8.7 fall?1303

Well, I need it to be in the buffer region, so it either needs to be here, 2.2 + or - 1, so 1.12 to 3.12- that is the range of a good buffer, or in this case 8.7.1305

Here, the pK2 is 9.15, which means our particular buffer region - let me go ahead and - this right here is going to be from 8.15 all the way to 10.15.1320

So, as long as the pH falls, if I need the pH to be in that range, this serine is actually a good buffer solution to use.1334

That is it.1342

You are just looking at the pKa.1343

That is all you are doing.1345

Note the pI; this is not the buffer region.1347

The buffer region is the horizontal region.1349

The buffer region is actually the part that resists changes in pH the more you add.1351

Notice, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, you keep adding a hydroxide; the pH doesn't go up all that much.1356

That is why.1366

So, in this particular case, experiment 8.7, yes, it falls within the range of the second pK for serine, so serine would be a perfectly good buffer solution to use in this particular experiment.1367

OK.1380

Let's go on to example no. 2.1385

OK.1394

Quantitative analysis reveals that a certain protein contains 0.60% tyrosine by mass.1396

I am going to use the...oh, that's fine; I'll just go ahead and write by mass.1427

Remember m/m, sometimes as w/w, they often say a tyrosine by weight instead of by mass- it is the same thing.1434

OK.1444

Further analysis estimates the molar mass to be about 135,800g/mol, and the molar mass of tyrosine is equal to 181g/mol.1445

This is the information that is given to you.1482

OK.1485

Quantitative analysis reveals that a certain protein, it contains, by mass, 0.60% tyrosine.1486

In other words, all of the tyrosine in there accounts for 0.6% of the total mass of protein.1493

Well, they do a little further analysis, and they are able to estimate the molar mass that is somewhere in the range of about 135,800, give or take.1498

The molar mass is 181g/mol of tyrosine.1507

So, the first question we want to ask you is...and this is the process that you will use.1511

Assuming the protein contains only 1 tyrosine residue, calculate the minimum molar mass of the protein.1520

What we are going to be doing by assuming that we actually have just 1 tyrosine residue, we have the percentage, if we assume just 1 tyrosine residue, we can place a lower limit on the molar mass; and then from there, given this other bit of information, the 135,800, then we can go ahead and find out a little bit more about it.1557

Let's go ahead and do the first part.1580

Well, let's see.1582

Let me do this one in red.1586

You remember, when we speak about a residue, it is the amino acid minus the elements of water.1591

It is minus the elements of water and the elements of water are H2O.1598

What you get is, what you end up with is 181g/mol minus the 18g/mol which is water, so a residue of tyrosine actually is 163g/mol.1607

OK.1626

Because, when we form a peptide bond, when we are forming the protein, it is a condensation reaction.1627

We are removing the elements of water, right, reverse of hydrolysis.1633

A tyrosine residue in a protein is actually missing an OH and an H from the original structure.1637

When we give the mass of the amino acid, it is a181g/mol, but when that amino acid is tied up in a peptide, in a protein, it is actually missing an oxygen and 2 hydrogens.1644

It is missing 18g/mol.1655

So, in this case, 163g/mol is the mass of the residue.1658

Well, we take the 163g/mol, divided by the total protein mass, times 100, and they already told us that this is equal to 0.60, because it’s 0.60%m/m tyrosine, part over the whole, tyrosine over the whole; and when I solve for the mass, I end up with 27,166g/mol.1663

This is the minimum per mole.1692

This is the min. molar mass, and it is based on just using 1 tyrosine residue.1696

Well, interestingly enough, I already have this, an estimation of the total molar mass, the 135,800.1702

Well, if I know that I have a minimum mass of 27,166, now, my second question to you is the following.1709

How many tyrosine residues are there in this protein?1718

Well, I have the minimum for 1 tyrosine residue- that is the relationship.1733

I have the total, so I'll just take the total, divided by the minimum, and that will give me, hopefully, somewhere near a whole number.1740

That is how many residues I have.1748

I hope that makes sense.1750

I’m going to do that on the next page, running out of room here.1751

Actually, you know what, I should be able to do it here; that’s fine.1755

I've got 135,800; that is the total estimated mass of the protein.1759

Based on this 0.6% for 1 residue, I've got 27,166g/mol, and when I divide that, I get about 4.99; so you are looking at 5 tyrosine residues.1768

There you go.1783

You use the basis of one to find the minimum, and then you work up; you divide it by the total.1785

This is not altogether different than the empirical formula, molecular formula calculation, that you do in general chemistry.1790

OK.1798

Let's go ahead and do another example here.1801

Let's go back to blue; I like blue a lot.1804

OK.1810

This is example no. 3.1811

Now, a peptide called methionine enkephalin was subjected to the following procedures with the following results.1815

OK.1853

We are going to ask you to take these results, take a look at them, and deduce the structure of the methionine enkephalin.1854

OK.1876

The first procedure that it was subjected to was 6M hydrochloric acid hydrolysis.1877

Oops...hydrolysis, there we go.1884

We basically just completely broke every single base, released every single free amino acid.1886

Let's see.1895

6M hydrochloric acid hydrolysis and we ended up with the following results: methionine to glycine to phenylalanine to tyrosine, the molar ratios were 1:2 to 1:1.1896

OK.1915

We were not able to count how many of each, but we were able to get the ratios of the amounts of the amino acids in this particular analysis, so 1:2 to 1:1, Met to Gly to Phe to Tyr.1916

OK.1928

The other procedure we subjected it to was FDNB and hydrolysis.1930

I should write FDNB then hydrolysis, then we broke it up - fluorodinitrobenzene, that is the Sanger reagent - and we ended up with the following.1942

We ended up with DNB-Tyr, this molecule's derivative of tyrosine was detected, and also, there were no free tyrosines detected.1959

OK.1985

Well, now, our third procedure, we had fragmentation.1987

We did fragmentation by pepsin, and pepsin breaks up the aromatics on the amino side.1995

OK.2013

And what we ended up was with the following.2016

We ended up with a dipeptide containing Phe and Met, and we ended up with some Tripeptide.2018

We did not sequence this; we did not know the sequence.2037

We just know that this dipeptide contains Phe and met; we don't know which one actually comes first.2039

And a tripeptide containing tyrosine and glycine in a ratio of 1:2.2044

OK.2059

We need to deduce the structure, in other words, what's the amino acid sequence.2060

OK.2066

Well, let's see what we've got.2067

Let me go ahead and go to the next page here.2070

Let me see.2074

Our glycine is the one that's 2, so I'm going to write the glycine first.2075

Glycine to methionine to phenylalanine to tyrosine- that is going to be the 2:1 to 1:1.2086

OK, so I've got this.2095

Well, they said that the FDNB gave me a DNB-Tyr.2097

OK.2108

This tells me that the tyrosine is the N-terminal residue.2109

OK.2122

So, we know that it is tyrosine.2123

Well, let me see.2124

Here is something else they said.2125

They said that there was no free tyrosine after we actually hydrolyzed it, once we reacted it with the FDNB.2128

Remember, the FDNB attaches to the N-terminal, and then once you break it up, everything comes apart, we detect this.2134

We have labeled it.2142

We detect that.2143

The others, we can also check to see what's in there, but there was no free tyrosine.2145

OK.2151

And, since there was no free tyrosine, that means there was only 1 tyrosine residue, and it happened to be the one on the end.2153

So, remember, we labeled it, the FDNB can only react with what's on the end.2180

Once that reaction is done, that's when we break up the protein, and now, you have a bunch of free acid.2188

Well, if you can have a bunch of tyrosines that are also free in addition to the one on the left, but there is no FDNB in there, so it is not going to react with those.2193

That is the whole idea.2201

It reacts with the last one first, and then if you have any other free acids, then that tells you how many that you have, but in this case, there was only 1 DNB-Tyr that was detected, but there were no free tyrosines, which means there is only 1 tyrosine, so this 2:1 to 1:1 is not just a ratio, it is exactly how much we have.2202

We have 2 glycines; we have 1 methionine, 1 phenylalanine, and we have 1 tyrosine.2221

This is going to be a pentapeptide.2228

In other words, we have 5 amino acids that make this up- really, really nice.2232

We know what the far left one is; it's going to be the tyrosine.2235

OK.2239

Now, let's go ahead and take a look at our relationship.2240

Now, pepsin cleaves, like we said, pepsin cleaves the phenylalanine, the tyrosine and tryptophan, but we don't have to worry about tryptophan on the left.2245

When it cleaves it on the left, that means, in other words, if you have, OK, if this is either phenylalanine or tyrosine, it is going to break it right there.2260

So, one of your fragments is going to have a phenylalanine or a tyrosine on the left.2277

One of the dipeptides, they said, contains phenylalanine and methionine.2282

Well, we know that that dipeptide has to have a phenylalanine on the left.2286

So, we know that we are looking at Phe and Met- that is our dipeptide.2291

Well, we also know that tyrosine...there is also a tyrosine and a glycine on a 1:2 ratio.2297

Well, I have already accounted for the phenylalanine and the methionine.2307

I know that tyrosine is on the left, so that I know that I'm looking at this- Gly and Gly.2310

I know there is nothing to the left of the tyrosine because that is the N-terminal amino, therefore, all of this information points to the following: Tyr, Gly, Gly, Phe, Met- tyrosine, glycine, glycine, phenylalanine and methionine.2318

This is the structure or the sequence of methionine enkephalin.2343

There you go; I hope that made sense.2350

You are just, sort of, putting pieces of the puzzle together.2353

There is no one way to do this; you just have to use your intuition, use the things that you know one piece at a time, put it together.2358

This one was reasonably simple because we are dealing with a pentapeptide, not going to be so simple all like that.2367

Other times, you are probably going to have to use a couple of cleaving procedures to see where you have overlaps; and, in fact, that is what we are going to do next.2373

OK.2382

Let's go ahead and take a look at another example of a sequencing of a peptide, but this one a little bit more complicated, a little bit longer.2384

Let's see what we've got.2394

OK.2397

This one I wrote out because there is a lot more analysis going on.2398

Let's take a look.2402

Glucagon is a peptide hormone that is secreted by the pancreas in response to low levels of glucose in the blood.2404

It induces the liver to convert glycogen to glucose - glycogen is a carbohydrate, glucose is a carbohydrate, glycogen is made of - and release it into the bloodstream.2411

Glucagon was subjected to several analytical procedures with the following results; use these results to deduce the - I'm sorry - amino acid sequence of glucagon.2421

So, glucagon just does the opposite of what insulin does.2431

If the blood sugar gets too high, insulin is released.2435

If the blood sugar gets too low, glucagon is released.2439

It's a way of maintaining the blood sugar level at some stable level, hopefully.2441

OK.2448

In this particular case, we did a 6M hydrochloric acid hydrolysis of the whole thing, and an amino acid separation, so we were actually able to count the number of amino acids.2449

Here are the results of the hydrolysis and the counting.2460

Histidine, serine, Gln, Gly, Try, Asn, Phe, Asp, Tyr - these are all the numbers that we have.2467

OK.2473

Now, let's see what other analyses.2475

So, we did an FDNB, and we ended up with the DNB-His.2480

OK, that's good.2485

We have that; we know that histidine, and we noticed that we have the one histidine.2486

In this particular case, that 1 histidine happens to be the N-terminal, so that's some good information.2493

We did a couple of fragmentations on this using 2 different enzymes.2498

We used the Asp-N protease, and what it does is that fragments that cleaves the Asp, the cleaves of the protein to the left of the Asp, and that is what the N means, the amino side.2503

So, when we fragment it, the fragment is going to start with an Asp.2515

Let's see here.2524

Fragmentation and Edman sequencing gave the following fragments.2526

OK, so, in this particular case, we not only fragmented but we also sequenced it; and we came up with these fragments.2529

Here, we are just using the single letter designations for the proteins, and don't worry about them if you still haven't memorized them.2535

I mean, at some point, you are going to have to; but no big deal for right now.2540

This is the first fragment; that is the second fragment, the third and the fourth.2545

So, this A1, A refers to the aspen protease, then we go ahead and take an intact protein; and we subjected it to a second fragmentation procedure with trypsin.2549

Now, trypsin, tends to break lysine and arginine.2560

It cuts them at the lysine and arginine residues, and it cuts the carboxyl sides, so to the right of it.2567

In any particular fragment, you are going to end up having an arginine or a lysine at the end.2573

OK, so let's see what we've got here.2580

And thus, fragments gave us this, and this, and this, and this.2582

Excuse me.2588

Well, OK, so, let's see if we can put this together; and let's see if we can find some overlap between these fragments and these fragments.2590

Let's see what we've got.2600

Let's see what we can do.2603

Alright.2605

Well, we know that this is what we know.2606

We know that the DNB-histidine means that histidine is the N-terminal residue.2609

Yes, so that is nice.2619

It means that His is the N-terminus of this particular protein, which is really, really great.2621

And, we have a couple of the fragments, so let's see.2630

A1 and if I take a look at T3, I notice that T3 is the one that actually has the H on the left, so we know that that fragment goes first.2634

That is fantastic.2645

We have already taken cared of about ¼ of this.2646

Let me go ahead and list this as...so, I've got H, S, E, G, T, F, T, S, D, Y, S.2650

This is the T3 fragment.2668

The T3 fragment goes there.2671

Now, let me go to blue.2673

OK.2676

I take a look at some of the fragments on the A side, and I notice that this thing, this T3 overlaps A4.2677

I'm going to go ahead and put the overlap A4 right underneath.2690

So, it is going to be D, Y, S, and then K, Y, L.2693

OK.2700

Now, let me switch colors again.2701

Now, I go back to my T, and see if there is an overlap there, turns out there is.2704

This actually overlaps the T4 fragment.2710

And again, you can switch back and forth and see that this is absolutely correct.2715

In this particular case, I write the overlap this way.2719

I just tend to do it on a stair step fashion, and then just read it off at the end.2722

You can do it anyway you like, however you want to put the pieces of the puzzle together.2725

D, S, R, OK, now, let me see.2731

I take a look at my A fragments, and yes, there is an overlap here.2735

Let me go back to red.2738

This overlaps A3, so I go ahead and write A3 underneath.2742

That is going to be D, S, R, R, A, E.2751

Let me go back to blue.2758

OK.2760

This one overlaps T1.2761

I go ahead and write T1 underneath.2767

I've got A, E, D, F, V, E, W, L, M, N, and T; and this one, it overlaps A2, and that will be my final sequence here, A2.2770

And, of course, that is the...let me do this in black.2797

This is going to be the D, F, V, E, W, L, M, N, and T; and there you go.2802

Now, I just sort of read it off, and just make sure to skip the overlap part.2811

There and there, and there, and here, and I just read it off.2817

I'm going to do this final one in...I guess I'll do it in red, how's that?2827

So, our final sequence is H, S, E, G, T, F, T, S, T, F, T, S, D, Y, S.2831

I have that overlap, so it is going to be K, Y, L, K, Y, L.2842

I have the Y, L, so it is going to be D, S, R, D, S, R, R, A, E; and then, the rest, D, F, V, E, W, L, M, N, and T.2847

There you go.2872

This is the final amino acid sequence of glucagon.2874

Again, hydrolysis- to count2878

Sanger reagent- to find out where the N-terminal is.2881

Let me go ahead and make sure that there is this little mark here.2885

Fragmentation with one enzyme or chemical procedure fragmentation with another enzyme, check for some overlaps.2890

You are just sort of putting it together.2897

Again, I like to put it together in a stair step fashion like this, and then just read it off.2899

There you go, pretty straightforward.2903

There is nothing difficult going on here.2907

It is just a question of putting the pieces together.2909

That's it.2914

OK.2915

Let's see what we've got here.2918

OK.2921

Let's close this off with one final example.2923

Excuse me.2926

Let's go ahead and do this in blue, and hopefully we can fit this in one page, so example 5.2927

I want to create as much room as possible, so I'm going to say...oh, that's fine.2941

Give a schematic for the Merrifield synthesis of the following tripeptide: Gly, Phe, Leu, so glycine, phenylalanine, and leucine; and use only the 3-letter designations.2952

You don't actually have to write out the structure, not a problem.2986

OK, so let's go ahead and do it.2990

Remember, the Merrifield synthesis, we actually synthesize from right to left, from the carboxyl end toward the amino end.2992

So, we are actually going to be starting with- oops, let's do this in red.3000

We are actually going to be starting with leucine, then phenylalanine, then glycine.3005

It is the opposite of how nature does it.3009

Nature goes amino to carboxyl; Merrifield goes carboxy to amino, because we attaching the first one to an insoluble bead.3010

OK.3019

Let's start off over here.3020

Hopefully I can do it, so let's start off with our leucine, and I'm going to react it with our Fmoc-chloride, and that is going to give me Fmoc-leucine; and then, I'm going to react this with a bead, those polymer beads.3024

What I end up with is Fmoc-leucine, and then the bead; and then I'm going to subject this to trifluoroacetic acid or mild base to deprotect that leucine, get rid of the Fmoc.3050

So, I end up with leucine and the bead.3067

OK.3072

Now, over here, I'm going to go ahead and do this in black.3073

Now, I'm going to go, and I'm going to work with my phenylalanine.3078

I'm going to take phenylalanine, and I'm going to react it with Fmoc-chloride.3082

I'm going to protect its amino group, so I get Fmoc-phenylalanine; then I'm going to react it with the DCC, dicyclohexylcarbodiimide, and I get Fmoc.3087

I get the phenylalanine, and I get my DCC, and this is what I'm actually going to take and react, goes in here; DCC comes out, and what I'm left with is Fmoc.3101

I'm left with Phe; I'm left with Leu.3122

And now, it is still attached to the bead.3125

OK.3128

I subject this to trifluoroacetic acid to release this Fmoc.3129

So, I end up with phenylalanine, leucine; and I end up with a bead.3134

Now, I'll go ahead and prepare my second amino acid; that is going to be glycine.3140

I'm going to react glycine with Fmoc-chloride.3146

I'm going to get Fmoc-Gly.3152

Notice that the Fmoc is on the N-side.3154

OK.3156

And then, I'm going to react this with the DCC to activate the carboxyl, so it'll actually react.3157

I end up with Fmoc-Gly, DCC.3164

I take this, and I react it with that; and DCC comes out, and what I'm left with here is Fmoc-Gly, Phe, Leu, and a bead.3171

OK.3193

You know what, I can keep this on 1 page; it's not a problem.3194

Well, nah, that's fine.3197

Let me go ahead and do the next page.3200

We have got Fmoc; we've got Gly.3203

We have got Phe; we've got Leu, and we've got a bead, and we want to go ahead and subject that to trifluoroacetic acid, and we end up with glycine, phenylalanine, leucine, still attached to a bead, and then we wash this with some hydrofluoric acid.3210

We end up releasing that; we end up breaking this bond, the leucine and bead bond, and we end up with our final glycine, phenylalanine and leucine.3236

There you go.3254

That is a Merrifield synthesis of that particular tripeptide.3255

OK.3261

Well, that takes care of the examples for the amino acids and peptides.3262

Thank you so much for joining us here at Educator.com and Biochemistry.3267

We'll see you next time, bye-bye.3270