For more information, please see full course syllabus of Biochemistry

For more information, please see full course syllabus of Biochemistry

### 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
- Example 1
- 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
- Data
- Part A: Calculate the Minimum Molar Mass of the Protein
- Part B: How Many Tyr Residues in this Protein?
- Example 3
- Example 4

- 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

### Biochemistry Online Course

### 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 pK _{1} = 2.21.*0100

*We have a pK _{2}, which is the pK_{a} 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 NH _{3}^{+} 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 pK _{1} plus - I'm always, always, always doing capitals - pK_{2} 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 R _{1} 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 pK _{1}, right, the pK_{1}, 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 pK _{1} - 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 pK _{2}, 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 pK _{1} 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 pK _{2}- 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 pK _{1}, that implies the -½ + 1 = +½.*0817

*That is the charge at the pK _{1}.*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 pK _{2}.*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 pK _{2}; 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 pK _{1}, 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 pK _{1}; this is 2.12, 5.68, and then 9.15, 6, 7, 8, 9, 9.15.*0959

*So, this puts our pK _{2} 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 pK _{1} mark; this is our pI, the isoelectric point.*0997

*This is our second dissociation, the pK _{2}, 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- pK _{1}, the pK_{2} 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 pK _{2}.*1181

*Remember, the pK _{1} and the pK_{2} 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 pK _{a}.*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 pK _{2} 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 pK _{a}.*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 H _{2}O.*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 count*2878

*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

1 answer

Last reply by: Professor Hovasapian

Mon May 9, 2016 3:27 AM

Post by imene hacene on May 8, 2016

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?