Carbohydrates are abbreviated using a three-letter abbreviation followed by their glycosidic bond type. For example, maltose and sucrose can be written respectively as

Glcα (1→4) Glc Glcα (1→2) ßFru
Maltose Sucrose

Provide the structure for the O-type blood carbohydrate set given the following abbreviation:

L-Fucα (1→2) Galß (1→4) GlcNAc

Question

Carbohydrates are abbreviated using a three-letter abbreviation followed by their glycosidic bond type. For example, maltose and sucrose can be written respectively as

Glcα (1→4) Glc Glcα (1→2) ßFru

Maltose Sucrose

Provide the structure for the O-type blood carbohydrate set given the following abbreviation:

L-Fucα (1→2) Galß (1→4) GlcNAc

Accepted Answer

All right, hello everyone. So this question says that the three letter abbreviation with the Glyco cytic bond type is a useful technique for representing carbohydrates. For example, sucrose can be denoted as GLC alpha 12 beta fru using the same concept, draw the structure of the sugar with an abbreviation of G A beta 16 GLC alpha. So first and foremost, what information is being given from the information presented in the abbreviation. So first let's analyze it right here. We can see that the abbreviation is telling us that the sugar is a di saccharide. It's made up of two monosaccharide units. So the first one gal is going to be galactose, right? And the second one GLC is going to be glucose. So these are our two monosaccharide units. But in addition to the monosaccharide units that were discussing for this question, we also have information about their configuration. For example, we know that galactose is in the beta configuration whereas glucose is in the alpha configuration. So what does that mean? Right. Recall that alpha versus beta configuration refers to the orientation of the hydroxy group on the anomic carbon. So it's the carbon derived from the carbonel of the straight chain form of the given monosaccharide. So we'll discuss the difference between alpha and beta later on in this video when drawing out the monosaccharide. For now, let's focus on another piece of information given to us from the abbreviation. And that is these numbers in parentheses, this one arrow six. This if you recall indicates where the glyco cytic bond between the two sugars is. So what this means to say is that the glyco cytic bond of this particular disaccharide is between carbon one of galactose and carbon six of glucose. So now that we have some more context in mind, we can go about drawing out this molecule, drawing out this disaccharide. So first thing I'm going to do is scroll up here or rather scroll down and just bring up these these straight chain structures that I went ahead and drew previously just to save a little bit of time for this video. So here, as you can see, I have the structure of the straight chain structure of D galactose on the left and D glucose on the right. So first, we're going to talk about how to translate these structures into the cyclic form of each. So let's start with D galactose. Now, the first thing I'm going to do is draw out the ring structure for D galactose. So it's going to be a six member ring where oxygen is part of the ring itself. So we called it by convention, right, the anomic carbon that is carbon number one is depicted on this far right side. So just keep that in mind just shading in this first half here just to show three dimensionality. And there we go. Now, what I'm also going to do actually is copy this six member ring structure because we are going to need it again when drawing out glucose in its cyclic form. So let me just do that right now. All right. So now let's focus on galactose specifically. The first thing I want to address is that anomic carbon because we discussed earlier, the difference between alpha and beta configuration gallitos is in the beta configuration. So what that means is that the hydroxy group of carbon number one in this cyclic form should be pointing upwards. So this is in contrast, for example, to glucose, which is in the alpha configuration because glucose is in the alpha configuration, its hydroxy group of the anomic carbon that is position one should be pointing downwards instead. And so from here, we can reference the straight chain forms. There's straight chain structures here, these fisher projections. So recall it when it comes to translating a Fisher projection into a cyclic structure into a Hayworth projection, the direction of hydroxy groups matters. So for example, in the structure of D glucose, in the fissure projection of D glucose, the hydroxy group of carbon number two points to the right. What that means is that in the Hayworth projection, that same hydroxy group should be pointing downwards. And by contrast, the hydroxy group of carbon number three is porting to the left in the fisher projection, which means that it should point upwards in the Hayworth projection, oops in the cyclic structure. So same principle here, the carbon number four's hydroxy group is pointing to the left, which means it should point upwards in our cyclic structure. And the hydroxy group of carbon number five is the oxygen of the ring itself. So we're not concerned with its configuration. However, when it comes to carbon number six, because this is ad sugar. Carbon number six should be pointing above the ring with its hydroxy group. So that's D galactose or de galacto pyin nose because it is the ring structure. So now let's do our glucose unit, our D glu copy nose. So as mentioned before, the anomic carbon has its hydroxy group pointing downwards because glucose is in the alpha configuration. So on carbon number two, we have a hydroxy group pointing to the right in the fission projection, which means it points downwards in the Hayworth projection in our ring. For carbon. Number three, it's to the left. So that's upwards in the ring structure for number four, it's pointing to the right once again. So it should be pointing downwards here once again. Carbon number five, specifically, the hydroxy group of carbon number five is what's making up the oxygen of the ring. So carbon number six should be pointing above the ring because this is ad sugar. OK. So now the question is how do we connect these two monosaccharide now that we've drawn them individually. So first let me go ahead and scroll down and open up just a little bit more space between these two. So first and foremost, where was the glyco cytic bond supposed to be? Well, the glyco cytic bond was supposed to be in between the anomic carbon of galactose and carbon number six of glucose. Now, to be more precise, the hydroxy group of carbon number one in galactose should be connecting with carbon number six of glucose. So now how do you represent that Glyco cytic bond? What modifications do you have to make? Well, first and foremost, let me start by erasing the hydrogen of our anomic hydroxy group in galactose. And what I'm also going to do is remove the hydroxy group attaching to carbon number six of glucose. This is because during the formation of the glyco cytic bond, what's being removed should no longer be present. And so now what I'm going to do is just draw a line representing the bond between these two atoms and there you have it here is our disaccharide. And so with that being said, if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful

Verified Solution

Key Concepts

Glycosidic Bonds

Carbohydrate Abbreviations

Blood Group Carbohydrates