Disaccharides: Study with Video Lessons, Practice Problems & Examples

Disaccharides are sugars formed from two monosaccharides, with key examples including maltose, cellobiose, lactose, and sucrose. Maltose, linked by an alpha-1,4 glycosidic bond, is digestible and found in starch. Cellobiose, with a beta-1,4 linkage, is not digestible by most mammals. Lactose, composed of glucose and galactose with a beta-1,4 bond, is generally digestible, though some individuals are lactose intolerant. Sucrose, made of glucose and fructose with a beta-1,2 linkage, is also digestible. Understanding these configurations is crucial for grasping their digestibility and biological roles.

Disaccharides

In this video, we're going to introduce some specific disaccharides. Recall that disaccharides are defined as sugars that consist of exactly 2 monosaccharides; 'di' means 2 and 'saccharides' means sugars. There are many disaccharides that exist in nature, so you're definitely not expected to memorize all of them. However, the most abundant disaccharides are the following four: maltose, cellobiose, lactose, and sucrose.

Notice in our image, we're just giving you information about these four disaccharides. The disaccharide consists of a sugar 1 plus a sugar 2, a glycosidic linkage that is linking those two sugars. In this table, we have sugar 1 in the first row, sugar 2 in the second, and the glycosidic linkage in the third. We then put the name of the disaccharide in this column, provide a bit of information about the disaccharide, and let you know if the disaccharide is digestible or not with check boxes. Also, notice that the table is color-coordinated to the structures of the disaccharides shown below; the yellow corresponds with the yellow, green with the green, white with the white, and blue with the blue.

Let's get started with maltose. Notice that maltose has a glycosidic linkage used to link together these two D-glucose residues, as identified here in this image, which is an alpha 1-4 glycosidic linkage. Maltose, found in the polysaccharide starch, is digestible in many plant-based foods including potatoes.

The next disaccharide is cellobiose. It consists of the same sugar residues as maltose, the two D-glucose residues. Instead of having an alpha 1-4 glycosidic linkage like maltose, cellobiose has a beta 1-4 glycosidic linkage. Cellobiose is found in cellulose, a polysaccharide that we will discuss later in our course. Despite their similarity, the configuration of the anomeric carbon in the glycosidic linkage makes cellobiose indigestible by most mammals, highlighting the fact that configurations matter significantly for digestibility.

Moving to lactose, this disaccharide swaps the D-glucose residue for a D-galactose residue. However, the second sugar remains a D-glucose, like the others we have covered. The glycosidic linkage in lactose is a beta 1-4 glycosidic linkage. Lactose is commonly found in milk and is digestible by most mammals, but not all due to lactose intolerance, which will be discussed later in our course.

Finally, we discuss sucrose, which retains the D-glucose residue for its first sugar but swaps the second sugar for a D-fructose molecule. The glycosidic linkage in sucrose involves two anomeric carbons, the first being an alpha, and the second, on the same side as its highest numbered carbon, is a beta 1-2 glycosidic linkage. Sucrose, commonly found in processed foods, is digestible by most mammals.

Most of the sugars in our list are D-glucose residues except for D-galactose in lactose and D-fructose in sucrose. In the glycosidic linkages, all except sucrose's are 1-4 linkages. Maltose features an alpha linkage, whereas cellobiose and lactose have beta linkages, and sucrose features both alpha and beta. The only indigestible disaccharide in our list is cellobiose. Spending time with this table and committing these structures and components to memory will help you master these concepts quickly. Moving forward, we'll get some practice applying these concepts. See you in our next video.

Which of the following contains galactose as one of the sugar subunits?

• A. Glucose.
• B. Ribose.
• C. Maltose.
• D. Lactose.
• E. Cellobiose.

Which disaccharide forms a 1,1-glycosidic linkage?

• A. Lactose.
• B. Trehalose.
• C. Maltose.
• D. Sucrose.

What is the identity of the disaccharide below?

• A. Cellulose.
• B. Lactose.
• C. Chitin.
• D. Maltose.
• E. Sucrose.
• F. Cellobiose.
• G. Glycogen.

The structure of a disaccharide is shown below. Which statement applies?

• A. Both sugar rings A and B are in equilibrium with their linear chain forms.
• B. Only sugar ring A is in equilibrium with its linear chain form.
• C. Only sugar ring B is in equilibrium with its linear chain form.
• D. Neither sugar ring is in equilibrium with their linear chain form.
• E. None of the above statements are correct.

Name each monosaccharide unit & the glycosidic linkage in the following disaccharide (gentiobiulose).

Alright. This practice problem wants us to name each monosaccharide unit and the glycosidic linkage in the following disaccharide below, which is genteobiolose. All we need to do is name this sugar unit right here and name this sugar unit over here, and then name the glycosidic linkage that's linking together these 2 sugar units. Of course, there are multiple different approaches you can take to this problem. Once you start getting very good at recognizing these Haworth projections, you'll realize that this sugar unit over here is galactose. Now, this first part here is complete, so we can move on to naming the glycosidic linkage here in the middle. We're going to look at the anomeric carbon here, which is carbon number 1, and then all these other carbons we can sequentially number. Notice that the glycosidic bond, the O-glycosidic bond, is going in an upwards direction in the same direction as the highest numbered carbon, which means that it's going to be in the beta configuration. Up above here, we can write in beta configuration. Of course, the first carbon here is being linked. We need to figure out which carbon this is, what number it is. For fructose over here, its anomeric carbon is over here, so this will be carbon 2. So this will be carbon 1. So we have 1, 2. This will be carbon 3. This will be carbon 4. This will be carbon 5. This will be carbon 6. What you'll see is that this is a beta 1-6 Glycosidic Linkage. This is the quick answer to this practice problem. If you are clear on this practice problem, you can end the video here, but if you're looking for a way that you could've recognized this as galactose and recognized this as fructose, then stick around for a bit longer. Notice that, if we want to convert cyclic sugars into linear Fischer projections, all we need to remember are the two words "up, lefting, and down right". We can highlight chemical groups that are going up in green and highlight chemical groups going down in blue. We could do the same here for fructose. The anomeric carbons, we are not going to be concerned about their positions because recall the anomeric carbons are the reactive carbons, and we only care about the chiral carbons that are not going to be reactive. We can see that this sugar has 6 carbons on it. Draw our Fischer projection, making sure there are a total of 6 carbons. This anomeric carbon is part of a hemiacetal group and has 2 OR groups. The structure of this sugar, the linear form, was derived using "up, lefting, and down right." Notice that this looks similar to glucose, except the C4 position has a different configuration from glucose. This is the C4 epimer of glucose. Recalling that the C4 epimer of glucose is galactose, helps us to identify this as galactose. We could have done a similar process over here with D-fructose. Notice that fructose has a total of 6 carbon atoms. Notice the anomeric carbon here on fructose is part of a hemiketal group, indicating that it has one OR group. Derived from ketones, this ketone is going to be on the C2 carbon. From C3 all the way down, this looks a lot like glucose, and the only difference is the top part up here. This is the ketose form of glucose, which is fructose. This is another way we could have identified this sugar unit as being fructose. This here again concludes our practice problem. The answers are galactose, fructose, and a beta 1-6 glycosidic linkage.

Given the following information, draw a Haworth projection for the disaccharide gentibiose:

1. Gentibiose is a dimer of glucopyranoses.

2. The glycosidic linkage is β(1 → 6).

3. The anomeric carbon not involved in the glycosidic linkage is in the α configuration.

Alright. So this practice problem says, given the following information listed below, draw a Haworth projection for the disaccharide, GENTI Bios. And so, notice that the first piece of information that we are given says, GENTI Bios is a dimer of glucopyranosis. And dimer is just telling us that, again, it's going to be a disaccharide consisting of 2 identical sugar units. And those sugar units are going to be glucopyranosis. And so, we know that pyranoses are cyclic sugars with a 6 membered ring, and the glucose tells us that this sugar is going to be derived from glucose. And so, we need to recall the structure of glucose and recall that the structure of glucose looks like this. And so, if we want to convert the linear form of d-glucose into the cyclic pyranose form, glucopyranose, then we're going to need to remember the 2 words, "uplifting" and "downright." And so we can go ahead and start to draw the base of our pyranose, our 6-membered rings. So we know that the 6-membered ring is going to have an oxygen atom in it. And so, we're just drawing our hexagon here. So, there it is, right there, and of course, we know that the carbon atom furthest to the right, we're going to want to be our anomeric carbon. So this will be our carbon number 1. And then we can sequentially number all of these other carbons in order here. So this is carbon number 5. And of course, we know that glucose has a total of 6 carbon atoms. So we need to add our CH2OH group here. So we can go ahead and add that right here. Our CH2OH and that will be our 6th carbon atom and then, of course, looking at all of these chiral carbons here, we want to draw in all of these hydroxyl groups. So notice that the C2 hydroxyl is going to the right and so we think about the word "downright," and so in the Haworth projection, we expect this to be going down, so the hydroxyl will be going down. And then notice that the C3 hydroxyl is going to the left. So, we think about the word "uplifting" and so here we can draw the hydroxyl group going up. And then again, the C4 hydroxyl is going to the right, so we think about "downright," and the hydroxyl will be going down. And so now, we've got the base of our glucopyranose. And so, looking at, the second piece of information that we're given, it tells us that the glycosidic linkage that's going to be linking together the 2 glucopyranoses is going to be a β-1→6 glycosidic linkage. And so, of course, the beta is referring to the configuration of the anomeric carbon and the two bumps of the beta being on the same side tells us that the glycosidic linkage here on the anomeric carbon has to be going to the same side here as the highest numbered carbon. And so it's got to be going in an upwards, fashion here. So we can go ahead and draw that in. And, we know that disaccharides are linked by O-glycosidic linkages. So we know that this is going to be an oxygen atom here. And so, of course, it tells us that, in the third piece of information here that the anomeric carbon that's not involved in the glycosidic linkage is going to be in the alpha configuration. And so this is referring to the anomeric carbon of the second glucopyranose. So we have our first glucopyranose drawn. Now we just need to draw our second one. So essentially, what we can do is take this structure right here, and we can redraw it. And, again, we're told that this is going to be a β-1→6 glycosidic linkage. So here's, the one. It's already in beta configuration. So we now just need this to be linked to our 6th carbon. And so we want to make sure that we redraw our second glucopyranose here, so that it can be linked here like so. And so, essentially what's going to happen is we're going to get rid of this and we will have a connection right here. And so, we'll have our 6th carbon and, we'll still have, the 2 hydrogens here. And so, ultimately, what you can see is that we've got one glucopyranose, our second glucopyranose, and they are linked by a β-1→6 glycosidic linkage. So we can actually write beta here to remind us this is already in the beta configuration. And then, again, this last piece of information is telling us that the anomeric carbon not involved in the glycosidic linkage, which would be this anomeric carbon over here, is going to have the alpha configuration. And alpha, of course, means that it's going to be going the opposite direction of its highest numbered carbon, which means that the hydroxyl group here is going to be going down, and this is the alpha configuration here. And so, really this is it. This is the answer to our practice problem and this is the structure of Genti Bios. And even though we hadn't discussed GENTIBIOS before in our previous videos, notice how we were able to use these clues to draw the structure of GENTI BIOS. And so we have developed a lot of critical skills here, and, again, this is the answer to our practice problem and I will see you guys in our next video.

Here’s what students ask on this topic:

What are the main types of disaccharides and their components?

The main types of disaccharides are maltose, cellobiose, lactose, and sucrose. Maltose consists of two D-glucose molecules linked by an alpha-1,4 glycosidic bond. Cellobiose also consists of two D-glucose molecules but is linked by a beta-1,4 glycosidic bond. Lactose is composed of one D-galactose and one D-glucose molecule, connected by a beta-1,4 glycosidic bond. Sucrose is made up of one D-glucose and one D-fructose molecule, linked by a beta-1,2 glycosidic bond. Understanding these components and linkages is crucial for grasping their digestibility and biological roles.

Why is cellobiose not digestible by most mammals?

Cellobiose is not digestible by most mammals because it has a beta-1,4 glycosidic linkage between its two D-glucose molecules. Most mammals lack the enzyme necessary to break this specific type of glycosidic bond. In contrast, maltose, which has an alpha-1,4 glycosidic linkage, is digestible because mammals possess the enzyme to hydrolyze this bond. The configuration of the glycosidic bond plays a crucial role in determining whether a disaccharide is digestible.

What is the difference between maltose and cellobiose?

Maltose and cellobiose both consist of two D-glucose molecules, but they differ in their glycosidic linkages. Maltose has an alpha-1,4 glycosidic bond, making it digestible by most mammals. Cellobiose, on the other hand, has a beta-1,4 glycosidic bond, which is not digestible by most mammals. This difference in the configuration of the glycosidic bond significantly affects their digestibility and biological roles.

What is the significance of glycosidic linkages in disaccharides?

Glycosidic linkages in disaccharides are crucial because they determine the structure, digestibility, and biological function of the sugar. For example, maltose has an alpha-1,4 glycosidic bond, making it digestible, while cellobiose has a beta-1,4 bond, making it indigestible by most mammals. Lactose has a beta-1,4 bond but is generally digestible, except for those who are lactose intolerant. Sucrose has a beta-1,2 linkage and is digestible. The type of glycosidic bond affects how enzymes interact with the disaccharide, influencing its breakdown and absorption in the body.

Why are some people lactose intolerant?

Lactose intolerance occurs when individuals lack sufficient levels of lactase, the enzyme required to break down lactose into glucose and galactose. Lactose is a disaccharide found in milk, composed of one D-galactose and one D-glucose molecule linked by a beta-1,4 glycosidic bond. Without enough lactase, lactose remains undigested in the intestine, leading to symptoms like bloating, gas, and diarrhea. This condition varies in severity and is more common in certain populations.