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Lecture Comments (5)

2 answers

Last reply by: David Gonzalez
Sat Aug 9, 2014 3:24 PM

Post by David Gonzalez on August 7, 2014

Hi professor Hovasapian. As always, great lecture!

I have a semi-related question that I hope you can help me with: what's the best way to gain lab experience (in cell biology/biochemistry) if I'm not going to school? Within the next few years, I'm hoping to have learned enough to start a lab (lofty goal, I know) that deals specifically with diseases of old age.

I'm really dead set on this goal, and have been pushing myself through 8-hour study days (with the help of tutors, textbooks, and online resources like for the past year! Do you think that there are universities/colleges that would allow me to "intern" there for the sake of learning, even though I don't have a degree?

I really appreciate you taking the time to read this professor.

0 answers

Post by Professor Hovasapian on October 31, 2013

Hi Marvin.

I hope you're well.

I decided to skip it because, in most places, a separate class (Molecular Biology) deals exclusively with all aspects of Nucleic Acids. However, many people take only a Bochemistry...or perhaps people take this course without being formally enrolled in a course at a Uni anywhere. It would be a shame for these people to be short-changed vital info on Nucleic Acids.

I'm hoping to remedy the situation by recording an entire Unit on Nucleic Acids in the future, and adding it to this Biochem course.

I apologize for the inconvenience.


0 answers

Post by Marvin Alashker on October 31, 2013

is there a reason you skipped the chapter on nucleic acid?

More Example Problems with Carbohydrates

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 1:09
  • Example 2 2:34
  • Example 3 5:12
  • Example 4 16:19
    • Question
    • Solution
  • Example 5 24:18
    • Question
    • Structure of 2,3-Di-O-Methylglucose
    • Part A
    • Part B

Transcription: More Example Problems with Carbohydrates

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

Today we are going to round out our discussion of carbohydrates by just doing some more example problems with them.0004

If you feel comfortable with it, it is not a problem; you can move on to the next lesson, but I thought it would be nice to sort of recap, and maybe towards the end do something a little bit more complex.0010

Let’s get started.0019

Pretty basic stuff, again, we are mostly concerned with structure.0022

The idea behind carbohydrates is you want to be able to recognize the monomers, recognize the connections of the glycosidic bonds and be able to sort of put, not sort of, be able to put them together; or given a structure, be able to say that "this is this connection, this is this connection".0025

That is really the main purpose; we want you to have a good structural understanding of what is happening with the carbohydrates.0043

The rest of the information is information, the extent to which the level of detail that you need regarding the function of a particular family of carbohydrates over a specific carbohydrate, that is going to depend on what your teacher wants, how much he or she wants you to know.0049

It is structure that we are ultimately concerned with.0066

OK, example no. 1.0069

Let’s go to blue; I like blue.0073

So, example no. 1: draw the linear form of D-ribose.0080

OK, linear form; we are not putting it in a ring.0093

Ribose is going to be a 5 carbon sugar, and it isn't aldose.0098

It has an aldehyde on the no. 1 carbon, not the ketose, which would be a ketone on the no. 2 carbon.0103

Let’s just go ahead and draw our carbon chain; this is 1, 2, 3, 4, 5.0109

Let’s go ahead and put our aldehyde group up there.0116

Let’s go ahead and put the H2OH, the non-chiral carbon and as far as ribose is concerned, ribose happens to have all hydroxys on the right- that is it.0119

This is D-ribose, and notice there is no alpha-beta here because it is not in a ring formation.0130

That alpha-beta only refers to when a particular sugar is in a ring formation and whether the hydroxy is pointing below the ring, alpha or above the ring, beta when you are looking at it in a standard projection with the oxygen in the back right or in the case of a pentose ring, with the oxygen straight back- that is it, nice and simple.0135

OK, example no. 2.0155

Oops, go ahead and do this, example 2.0160

Xylose is a c2-epimer or epimer depending on how you want to pronounce it.0167

It is completely up to you; pronunciation is completely irrelevant unless people really just absolutely don not understand what you are saying- epimer of D-ribose.0175

Oh, and again, the D part, this D versus L, that comes from the chiral carbon that is farthest from the aldehyde; that is this one.0189

D, the hydroxy is on the right in linear form like this, in Fischer projection.0200

If it were an L-ribose, the hydroxy would be on the left, so this specifies the D.0205

OK, xylose is a c2-epimer of D-ribose.0212

Draw its linear structure; draw its linear structure.0216

Well, nice and easy.0222

We know that an epimer just means that at that particular carbon, in this case, the no. 2 carbon, the configuration is switched.0225

So, when we want to switch a configuration, we switch to the substituents.0232

In other words, we just move the OH from one side to the other.0236

I mean obviously over here we have hydrogens; we have not put them in here for the...but they are there.0240

In general, I will leave the hydrogens off.0244

OK, let’s go ahead and draw the structure again, 5 carbons, 1, 2, 3, 4 and 5; and we will go ahead and put our aldehyde.0249

Oops, excuse me; let’s make this a little bit better.0260

That is that; that is the hydrogen.0263

That is our aldehyde group; let’s go ahead and put our H2OH, the non-chiral carbon, so c2-epimer.0265

This is 1; this is 2.0271

This is 3; this is 4.0273

This is 5, so 1, 2...wait, I am sorry.0275

Xylose is a c3-epimer, my apologies.0282

I was going to say something is going on here, so 1, 2.0287

The 2 hydroxy stays the same; that is on the right.0289

c3-epimer, this is the one that goes on the left, and this is that, so there you go.0294

That is xylose, the c3-epimer of D-ribose.0300

No. 3 carbon just switched the configuration- that is all.0305

OK, let’s see; let’s draw some rings here.0308

Example no.3: draw the furanose and pyranose forms of D-xylose.0316

We just had the linear form of D-xylose.0338

Now, what we want to do is they want us to draw the ring configuration, but they are not specifying, well, they are telling me that they want both ring configurations.0340

They want the 5-membered ring, the furanose; let me go ahead and do this in red.0349

The furanose is the 5-membered ring, and pyranose is the 6-membered ring.0353

We are going to take a look at the structure to see which hydroxys we are going to attach to the aldehyde carbon to create the 5 and 6-membered ring.0358

And again, sugars, they can form 5-membered rings, 6-memembered rings, any of the hydroxys because sugars have hydroxys in all of the carbons- any of them can react.0369

So, if they can form a stable ring, they will, sometimes 5-membered, sometimes 6-membered.0377

Now, obviously, there are going to be certain rings that are going to predominate, but in this particular case, you can actually have both.0382

Let’s go ahead and draw our linear structure again; let’s go through our systematic procedure, and it is always best to do this.0390

We have 1, 2, 3, 4 and 5.0396

This is our aldehyde, and we had the hydroxy on the right, hydroxy on the left, hydroxy on the right and our achiral carbon, so that is that.0402

Now, let’s go ahead and do the furanose form first; furanose means 5-membered ring.0412

Well this is 1 carbon; Let’s do this in black.0419

This is no. 1; this is no. 2.0424

This is no. 3; this is no. 4.0428

We got an O to react with that, so I am going to go over here.0431

This is going to be no. 5.0434

OK, in our ring, this oxygen is what is going to attack that.0436

We have to know which oxygen is going to happen.0444

When we do the pyranose in a minute, it is actually going to be this oxygen right here, down below.0447

It is going to look completely different, but it is going to be the same sugar but in a completely different form.0452

OK, what we do next is we rotate this to the right, the top; we bring it down to the right.0459

Let’s go ahead and draw that configuration.0465

Let me see, should I do it...I’ll do it over here, a little higher up, so we have plenty of rooms.0470

1, 2, 3, 4 and 5, now, I have my aldehyde here, and this drops down.0473

This hydroxy goes up; this goes down, and here I will just write it as CH2OH.0483

That is this one right here.0490

Now that I have it in a horizontal fashion, I take the left hand side and I pull it.0492

I push it away from me, and I bring it around.0497

Now, let me draw this; when I do that, I have this.0500

Let me see, make sure I have enough room here.0506

C, C, this is my aldehyde; I have a C here.0509

I have a C here, and I have a CH2OH.0515

So, make sure 1, 2, 3, 4, 5- that is correct; and let me see, on the 2, 1, 2, this hydroxy is down.0522

This hydroxy is up, and this hydroxy is down.0530

OK, we said that this hydroxy on the no. 4 carbon, this one right here, that is the one that is going to react.0535

That is the one that is going to attack the carbonyl, so what we want to do is rotate 90° upward just like this.0540

In the back, this group right here in the back is like that.0552

This is the OH; we want to rotate it like this to put the OH horizontal just so we know where this thing ends up.0556

Let’s go ahead and do that; this is going to go, let me go back to black.0565

OK, now, I am going to redraw this structure with this back group rotated 90° up and just to make the OH horizontal, just to be consistent.0572

We are going to have C, C, C, C; actually you know what, let me make it a little bit lower here.0582

I have got C, C, C, C, and now, I have my OH there; and now, I have my CH2OH here.0595

This is the oxygen; let me finish by drawing in my carbonyl.0606

This is the oxygen that is going to actually attack; this is what is going to attack from above or from below in order to get the alpha-beta.0611

Here is what happens, when that happens, and now, I will go ahead and put equilibrium arrows because the ring structure and the linear structure, there is going to be an equilibrium between those two and here is how it works.0623

We have 1, 2, 3, 4, 5; let me confirm, and, of course, a 5 carbon.0635

We put the oxygen in the back, and I decided not to put the carbons in.0640

I hope you don not mind; I will do...actually you know what, maybe I will.0646

This is O; that goes there.0655

That goes there; this goes there.0657

This goes there, something like that.0659

Now, over here, we are going to have the...I am going to not specify the stereochemistry on that.0662

It could be alpha-beta depending on whether its attack from above or attack from below, but here, no. 2 carbon, the hydroxy is down below the ring.0667

The no. 3 carbon, the hydroxy is above the ring, and, of course, here we have our CH2OH group.0676

There you go; this is D-xylofuranose.0683

Take a good look at this because it is going to look nothing like the D-xylopyranose.0691

Unspecified stereochemistry could be alpha or beta, but these are...they are definitely specified.0698

This is below the ring; this is above the ring, and you have that.0703

This is the no. 1 carbon, just a number 1, 2, 3, 4, 5 carbon 1, 2, 3, 4.0708

This is the fifth member of the ring- that is it.0718

We have to know which oxygen we are using, attached to which carbon, is actually going to form it.0721

This is why it is important to go through the systematic procedure- do this, do this, decide which is what.0727

In a minute, we will see we do not actually have to go through the rotation here when we do the pyranose form- this is it.0733

This is D-xylofuranose; This is the 5-membered ring.0739

OK, now, let’s go ahead and do our D-xylopyranose form.0743

OK, let me go back to blue, and let me start with the third structure that I drew.0749

I had the linear; I turned it around, and I brought that back.0759

Let me start with that; I have got, let me see...a little bit lower here.0761

I have got C, C; now, let me put the carbon backbone in first, C and CH2OH.0770

Let me put my aldehyde in; I have got down below.0779

I have got above the ring, and I have got this.0784

Now, I do not have to do anything to this; now, I need a 6-membered ring, the pyranose form.0788

That means that 3, 4, 5, my sixth member of the ring is this oxygen.0794

It is this oxygen that is going to attack this carbon either from above or from below.0806

Now, nothing changes over here; this hydroxy is going to stay down below, and you are going to get a 6-membered ring.0813

Now, for a 6-membered ring, of course - excuse me - the standard position is oxygen in the back right.0819

Let me go ahead and draw that; so, we have this, this, this, that, that and that.0827

Now, let’s go ahead and see what it is that we have; again, we are going to not specify the stereochemistry that can be alpha or beta, depending on attack above or below.0834

Remember, that is the no. 1 carbon; OK, the no.1 carbon.0844

On the 2 carbon - let me go back to red - the hydroxy is down.0847

On the no. 3 carbon, the hydroxy is up; on the no. 4 carbon, the hydroxy is...wait, where am I?0852

1, 2, 3, 4, 1, 2, see, now, I am getting confused, alright.0864

We have got, on the second carbon, that is down; that is up, and this is down.0871

There we go; that is the no. 4 carbon, and over here, there is nothing at all.0880

As you can see, it is easy to lose your way here.0885

So 1, down the 2, 3 and 4, that is...let me see, this is blue.0888

This is 2; this is 3 and this is 4- good, everything is good.0894

On the 5 carbon, you just have 2 hydrogens; I will leave those hydrogens off- that is it.0898

This is the D-xylopyranose; it looks nothing like the D-xylofuranose.0903

It is just a question of keeping track of which hydroxys go where, what the arrangement is, and that is it- just nice, systematic, but as you can see, you have to be really careful because it is easy to lose your way.0912

OK, let’s see.0925

I have anything, and once again, take note of the fact that this no. 5 carbon, because it is a 5 carbon sugar in the pyranose form, it is forming a 6-membered ring.0932

This carbon, the no. 5 carbon, it does not have anything on it.0943

You are used to seeing something on it either a CH2OH, either above or below, because you are so used to seeing galactose.0948

Well, you are used to seeing glucose, galactose, mannose, things like that, but again, different sugars have different things that are attached to them.0955

In this particular case, you have not missed anything; it is just 2 hydrogens that are attached there, so you are good.0962

OK, well, let’s see; let’s move on to example no. 4.0969

OK, this one is a little longer, so instead of writing it out by hand, I thought I would just present it like this.0980

A biochemist wants to synthesize a new branched polysaccharide.0986

It is an amylose chain with branching at the no. 6 carbon.0992

So, you remember, amylose is the glucose monomers; it is a homopolysaccharide.0997

It has a bunch of glucose monomers that are attached by alpha-(1,4) linkages.1002

There is going to be branching at the no. 6 of one of those glucose.1009

OK, now, the first monomer at the branch point will be the furanose form of alpha-D-xylose- we just did that.1013

The rest of the branch alternates between monomers of alpha-D-xylose and N-acetylglucosamine in the following configuration: GlcNAc-alpha-(1,3)-xylose.1021

The Glc, the N-acetylglucosamine, is connected via alpha-1, and it is connected to the no. 3 carbon of xylose in the glycosidic bond.1033

We want you to draw the structure.1044

OK, basically, real quickly, just to get a sense of what is going on, so we are going to have this amylose chain, just these glucose monomers, again I am just doing a quick schematic before, that, that, that.1046

Let me just draw 4 of them; that is going to be the alpha-(1,4) of the glucose- not a problem.1064

And, let me just actually draw one more, and at the no. 6 carbon of one of those, it is going to be something like that; and it is going to be O, and it is going to be connected.1069

The first monomer of the branch point will be the furanose form of alpha-D-xylose, furanose form, so this is going to be a 5-membered ring, here.1081

Let me just go ahead and just do a little square like that.1092

It is going to be a 5-membered ring, and it is going to alternate between monomers of alpha-D-xylose and N-acetylglucosamine.1095

Let me do a little triangle for N-acetylglucosamine, a square for xylose, a triangle for N-acetylglucosamine.1102

This is what is going to happen; we have glucose monomers making up this chain.1110

At a branch point no. 6, we are going to have in alpha-D-xylose; and then we are going to alternate alpha-D-xylose, GlcNAc, Xyl, GlcNAc.1115

That is what we want to draw.1126

OK, hopefully, we either know; we have either memorized the monomer’s structure, or we just open up our books or look on the internet to get the particular monomer’s structure, and then we just construct our molecule and just remembering that it is alpha-(1,3) in the GlcNAc-xyl glycosidic bond.1127

OK, let me do this one in red; let me see here.1148

I think I will do it on - oops - I think I will do it on the next page.1153

OK, OK, so let’s go ahead and draw our alpha-(1,4)-glucose.1161

Let’s draw out 3 of those; let’s go like that.1166

Let me just go ahead and draw, this is alpha-(1,4)- there we go.1174

Now, let me go ahead and put in my CH2OH.1190

Let me go ahead and draw everything in, and then I will go ahead and talk about it.1202

OH, OH, and this is down below already; that is connected on a glycosidic bond, so we have got CH2.1208

OK, now, I am going to go ahead and connect this one, so I will do a CH2 here.1219

This is my no. 6 carbon; 1, 2, 3, 4, 5, 6, my no. 6 carbon.1224

This is the one that is going to be connected, oxygen; and this is what is going to be connected here.1235

Now, I am going to have a 5-membered ring; well, my 5-membered ring, let me make this...that is OK.1243

I do not have to make it too long; my 5-membered ring is going to look like this.1251

Bam, bam, bam, bam, and, of course, here, we have our oxygen; and then this is xylose, so that is OH there.1255

Let me go ahead and put my CH2OH here.1265

Now they said that it is going to be a 1,3 glycosidic bond with the next monomer, which is the N-acetylglucosamine.1270

I’m going to go ahead and draw this way, O; and now, I will draw in the N-acetylglucosamine.1277

This is N; this is COO.1291

This is CH3; this is OH, and, of course, this is going to be O, and this is going to be alternate with that.1295

So, we have another 5-membered ring.1307

This is OH; this is going to be CH2OH, and this is going to be another O connected to something else.1312

Let’s make sure that I have got everything that I am supposed to have here.1322

This right here...I apologize; I lost my colors.1327

This right here is our Glc-alpha-(1,4)-Glc.1334

This is our main; here is a branching point at carbon no. 6.1344

The first monomer is alpha-D-xylose; this is the alpha-1.1350

This is the alpha connection; this is the xylose ring.1355

And, we said that we have a GlcNAc, N-acetylglucosamine in alpha-(1,3) to xylose-alpha-(1,3).1361

This is the no. 3 carbon; OK, this is connected that way.1380

Then, of course, this goes to another; OK, this connection right here, that is going to be xylose-alpha-(1,4)-GlcNAc.1386

So, if you want you can sort of connect this to this one; actually, let me write it right next to it.1405

Because we are alternating monomers, we have to specify at least 3 of them.1410

So, GlcNAc, this is going to be 4; this is going to be alpha-1, and this is going to be Xyl.1415

Xylose-alpha-(1,4)-glucoseNAc, at the other end of the glucose, the reducing end, that is going to be alpha-(1,3) to the xylose and so on, so xyl-GlcNAc, xyl-GlcNAc this way- that is it.1423

That is our structure; everything is taken care of.1437

All the stereochemistry is represented, the connection, 1,6 branch point, 1,3, alpha-(1,4), here- there you go.1441

That is our polysaccharide.1452

OK, good.1456

OK, now, let’s do example 5.1459

OK, example 5 is a little long in terms of just actually writing it out as far as what is going on, but it is not altogether that difficult.1464

This is a great practice in structures, and it is a great practice on actually handling a carbohydrate, how one deals with finding out certain things about it.1473

OK, now, a biochemist wants to determine the extent of branching in a sample of glycogen.1482

So, we remember glycogen molecule, it has a whole bunch of branching.1489

It is just like the starch, except it is more heavily branched and it is more compact, consists of amylose and amylopectin with 1,4 connections and then 1,6 branching at the no. 6 carbon like we just did.1492

OK, a biochemist wants to determine the extent of branching in a sample of glycogen.1509

The branching takes place on those glucose monomers that have their no. 6 carbon and hydrogen attached.1524

A certain number of those glucose monomers have branching at the alpha-(1,6).1531

We want to know what percentage of those, so mind you, we are not saying what percentage by mass.1538

We are saying what percentage, which means we are talking about number.1543

So, of let’s say, 5,000 monomers that make up some glycogen molecule, how many of those 5,000 are actually monomers of glucose that have a 1,6 branch- that is what we are asking.1548

OK, here is what he does.1560

He takes the glycogen and he treats the sample with methyl iodide in order to methylate any free hydroxy groups.1563

That means, I am going to turn the OH groups, the free hydroxy groups in glycogen into OCH3 groups.1569

I am just converting them into, I am just adding a methyl group; I am replacing this hydrogen with a CH3.1578

OK, he then completely hydrolyzes the glycogen to release the free monomers.1584

So, once I convert the free OHs to OCH3, I split up every single glycosidic bond, and I have a bunch of monomers floating around.1589

He measures the amount of (2,3)-Di-O-methyl-glucose recovered, and he makes his computation.1598

OK, the structure of (2,3)-Di-O-methyl-glucose is as follows.1608

We will draw our regular hexose ring, Di-(2,3), (2,3)-Di-O-methyl.1614

Let’s go ahead and specify; let’s just not specify it.1625

We do not have to do that; that is not important.1628

Normally, we have OH, OH, OH and CH2OH, right?1631

Di-O-methyl, (2,3)-Di-O-methyl, well, this is the no. 1 carbon, no. 2, no. 3.1641

All we have done is replace this with CH3 groups.1647

This is the molecule (2,3)-Di-O-methyl-glucose.1654

So, he methylates the glycogen; he completely hydrolyzes it to release all of the free monomers.1657

Now, remember hydrolysis, elements of water, you are splitting it up while you are adding the elements of water, splitting it up the glycosidic bonds.1663

And, what he recovers, he recovers this molecule, and he measures this molecule, the amount that he has of this molecule to determine what percentage are actually branched.1673

OK, now, let’s go ahead and see what we can do, explaining detail, what is happening in this procedure using structures.1688

OK, let’s go ahead and draw a structure, and let’s talk about what it is that is actually going on.1698

OK, let’s draw a linear; let’s draw a part of this molecule.1704

I will draw it a little bit smaller than usual just so I have room on this same page.1716

OK, boom, boom, boom.1724

OK, I will go ahead and do that, and now, let me go ahead and draw my OHs in.1728

So this goes on that way, and this goes on that way; now, there is going to be a branch point.1733

This is going to be CH2, and that is going to be O, and that is going to be like that; and, of course, it is going to go on, it is going to repeat that way.1739

Oops, there is an O there, then that.1759

OK, now, let’s go ahead and draw in what it is that we have got.1763

We have OH, OH, OH, OH, CH2OH, lots of hydroxys.1767

Carbohydrates are just full of hydroxys.1780

OK, OH, OH, CH2OH, and the last one, OH, OH; and, of course, this goes on that way, and, of course, we have a CH2OH.1784

OK, now, this is what we started off with; this is the piece of our glycogen.1801

What we do to this is we methylate it, so every free hydroxy is going to be methylated.1805

Everywhere there is a free hydroxy, you are going to end up with OCH3 instead of OH.1810

Let’s go ahead and see where those are.1816

Those are going to be, well there, there, there, there, there, there, there, there, here, here and here, here, here and here.1820

Notice something here, on the free monomers, the ones that actually do not have a 1,6, that are not branched at the no. 6 carbon, those are going to end up once you actually methylate this and once you break all of these bonds, once you break them up into individual monomers, you are going to end up with something that is methylated at 3 points- 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3.1836

But, the free monomers that end up here, there is no hydroxy attached to this.1869

This no. 6 carbon and the oxygen attached to it is involved in a glycosidic bond.1877

So, when you break that bond, what you end up with is the (2,3)-Di-methyl-glucose, and what you end up with is also down up.1882

This is the (2,3)-Di-O-methyl-glucose.1908

Once you break that, these, the ones that have the branch points, this is not available for methylation.1917

Only 2 of them are available for methylation, so you get (2,3)-Di-O-methyl-glucose.1924

All of the others that are not involved in branching, that have no branching, those are going to end up having 3 methyl groups.1928

What you are going to end up with there is this.1937

You are going to end up with an OCH3 on the no. 2.1941

You are going to end up with an OCH3 on the no. 3.1945

There is an OH here because that is just a glycosidic bond.1948

The glycosidic bonds, they just end up back as hydroxys. but these OHs, you are going to end up CH2, OCH3.1953

You are going to have (2,3,6)-Tri-O-methyl-glucose.1960

That is what is going to happen when you hydrolyze it after you methylate it.1971

When you methylate it and then hydrolyze it, you are going to end up collecting 2 types of monomers: (2,3)-Di-O-methyl-glucose and (2,3,6)-Tri-O-methyl-glucose.1975

Well, if I can measure the amount of this, that will tell me how many of these are actually involved in branching.1985

Well, if I can take the total number of moles of monomers, or if I can take the number of these that are actually involved in branching by counting this derivative of it, divide it by the total number of monomers, I have my percentage of branching.1993

That is what I am doing here; I hope that made sense.2010

OK, now, let’s go ahead and actually run the calculations.2013

OK, let me see, comes from branching one of the hydroxys, yes, branching.2018

OK, 225mg of the glycogen is treated as above.2027

24.5mg of the (2,3)-Di-O-methyl-glucose is recovered.2032

What percentage of the glucose residues in glycogen are involved in branching?2036

Assume a glucose residue is a 162g/mol.2042

The hint here is recall what we mean when we use the word residue.2047

So, just as a recollection, when we talk about a residue, we are not just not talking about an amino acid residue.2053

We can talk about a glucose residue, an amino acid residue.2059

A residue is a general term for a molecule that has the elements of water removed from it, right?2062

Hydrolysis, you take off the OH, you take of the H, that is what you are doing; that is what a residue is.2070

When we talk about an amino acid residue, it means we have taken off an OH from the carboxyl end.2077

We have taken off an H from the amino end, and we have that residue.2083

So, we have actually lost 18g/mol for individual molecule.2086

When we talk about a residue of glucose, that means it is the glucose molecule that we know; but it is missing a hydroxy, and it is missing an H.2091

In other words 18g is missing, per mole.2100

That is what residue means, and that is going to be important in just a minute.2105

OK, what we want to do is the following.2107

The amount of the (2,3)-Di-O-methyl-glucose divided by the total amount of monomer of glucose times a hundred, that is going to give us our percentage.2113

And again, we did that by derivatizing it in such a way, so that when we finally hydrolyze it, you are going to end with 2 types of monomers.2139

The ones that have the branching points only have 2 methyls- the 2,3.2145

All of the other glucose monomers have the 2,3,6- those we do not care about.2150

We care about the 2,3, not the 2,3,6.2154

OK, let’s go ahead and do the math on this one.2157

Once again, we are not doing percent by mass, so we cannot just take the 24.5, and divide it by the - what is it - 225.2164

That is the whole idea; we have to be very, very careful.2179

It did not say percent by mass or percent by volume or something else; it actually said just what is the percentage, so we are talking about numbers.2182

We cannot use the masses directly; we have to go to moles because a mole is a measure of the amount in chemistry.2190

OK, let’s go ahead and calculate what it is that we have got.2197

We have, what did we say, 225?2202

OK, we have 225mg of glycogen, the total molecule; and we said that the average glucose residue, again, glucose is involved in this polymer.2205

That means that it has undergone a condensation reaction, so the elements of water have been removed from each glucose monomer in the actual molecule.2224

When I talk about 225mg of glucose, the 162g/mol, that is the weight of the residue.2233

Now, I am going to go ahead, since it was given to me in milligrams, I am just going to go ahead and write 1mmol is 162mg.2245

You can use these prefixes milli, centi, deci, kilo, as long as you change both of them, so millimole, milligram.2258

Do not change one of them, so 162g/mol is 162mg/mmol, 162kg/kmol.2265

As long as both, then you can just use the numbers as written.2277

You don’t have to write 225mg as 0.225g- that is it.2280

It is a personal thing; I just prefer to work like this by using the numbers that they gave me.2285

You could have written 0.225g x 1mol / 162g.2290

OK, now, I am not exactly sure, but I think my arithmetic was actually wrong on this.2295

So, I am just going to write down what it is that I have on a piece of paper, but I hope that you will verify this for me.2301

I think I used the wrong number here when I did this original division, but I ended up with 1.5mmol, which I’m sure is the wrong number; but again, the number is irrelevant.2306

It is the process that is important.2317

OK, now, let’s go ahead and do our number of moles of the (2,3)-Di-O-methyl-glucose.2320

We said that we recovered 24.5mg of the 2,3 derivative; I will just put 2,3 like that.2327

Now, 1mmol of that, when you calculate the molecular weight of that, it is going to end up being 208mg.2336

So, I end up with a total of 0.1178mmol of that.2347

OK, good.2356

Now, let’s go ahead, and we are done.2360

We have that; we have that.2365

We will take 0.1178mmol divided by 1.59mmol, which I think is not the right number.2368

225 divided by 162, I do not think it is 1.59, but I hope you will double check with me; and if it is not, just use a different number- it is not a problem.2378

Times a hundred, you end up with 7.5%.2387

7.5%, of all of the individual glucose monomers in this glycogen molecule, have a branching at the no. 6 carbon.2392

The amount of branching, the extent of branching in glycogen, this particular case based on this measurement is 7.5%, which makes sense.2401

It is going to be somewhere in the range of about 7-10% branching.2411

There you go; I hope that made sense.2415

Thank you for joining us here at

We will see you next time, bye-bye.2421