Common Monosaccharides: Study with Video Lessons, Practice Problems & Examples

Monosaccharides, categorized as aldoses or ketoses, are essential in biochemistry. Key examples include D-glucose, D-mannose, and D-galactose, each with distinct configurations. D-glucose, an aldohexose, serves as a reference for memorizing other hexoses. The conversion from linear Fischer projections to cyclic Haworth projections involves understanding the positioning of hydroxyl groups based on their orientation. This knowledge is crucial for grasping carbohydrate structures and their functions in metabolic pathways, including catabolic and anabolic reactions.

Common Monosaccharides

In this video, we're going to begin talking about some of the most common types of monosaccharides. So, you'll notice listed on this entire page, we have some of the most common monosaccharides in nature grouped as being aldoses and ketoses and based on how many carbon atoms they have. And so really this page is just meant to be a reference page to help you guys out as you move forward through our course. And it's important to note that the monosaccharides that your professors might expect you to memorize may vary from course to course. However, if your professor does want you to memorize any monosaccharides, the likelihood is that they're somewhere on this page and so, again, it's just meant to be a reference page to help you out. And what you'll notice for the aldoses, what we have listed are D-glyceraldehyde, D-erythrose, D-ribose, D-arabinose, D-xylose, D-glucose, D-mannose, and D-galactose. And for the ketoses listed below, what we have are dihydroxyacetone or DHA, D-erythrulose, D-ribulose, D-xylulose, and D-fructose. And so again, this page is only meant to be a reference page to help you out as you move along through the course. In our next video, I'll give you a recommendation of which amino acids I would recommend committing to memory. So, I'll see you in that next video.

Common Monosaccharides

So in this video, we're going to talk about some monosaccharide structures that might be worth memorizing. It's important to note that the monosaccharide structures worth memorizing will vary from course to course and from professor to professor. Some professors might not expect you to memorize any monosaccharides at all, whereas others might expect you to memorize these six monosaccharides and some additional ones. However, these six monosaccharides shown down below are definitely some of the common ones. If your professor does want you to memorize any monosaccharide structures, the likelihood is that they're going to include these six monosaccharides. We're going to do our part here at Clutch Prep and help you guys with memorizing these six monosaccharides. It's important to note that these six monosaccharides are all in their linear forms and they all have the D-configuration, which is really no surprise, since we know that life has a preference for the D-configuration of monosaccharides.

Alright, so let's get started with this monosaccharide way over here on the left-hand side. It turns out that this monosaccharide is actually the most common and the most abundant monosaccharide in all of nature. We've actually already talked a lot about this monosaccharide in our previous lesson videos and moving forward, we're going to continue to talk a lot about this monosaccharide. If you haven't already guessed it, this is the monosaccharide D-glucose. D-glucose is an aldohexose, which means that it has an aldehyde group and it has a total of six carbon atoms. If we wanted to number these carbon atoms, of course, we know that the carbon atom that's part of this aldehyde group at the top must be assigned the lowest possible number, which means that this carbon atom is going to be carbon number 1, and then we can sequentially number all of the other carbon atoms from top to bottom. Notice that carbons 2, 3, 4, and 5 are all chiral carbons, and the alcohol groups on these chiral carbons are taking the pattern that we see here. The C2 alcohol group is going to the right with a D-configuration. The C3 alcohol group is going to the left with an L-configuration, and the C4 and C5 alcohol groups are both going to the right with D-configurations. Glucose's alcohol groups end up taking the pattern that we see right here. Now, wait a second. It kind of looks like D-glucose might be trying to tell me something here with this pattern of alcohol groups. It kind of looks like it's trying to give me the finger. Now as much as I don't appreciate that, you've got to recognize this pattern here when you see it. And at the very least, there is quite a resemblance between these two images right here. And so, as unfortunate as it might be, thinking about somebody flicking someone off might actually help you guys with memorizing glucose's structure.

So memorizing glucose's structure is going to be important because we're actually going to use glucose as a reference for memorizing the other hexoses on this page. And so the next monosaccharide that we have here is actually D-mannose. D-Mannose is another aldohexose that's going to be numbered in the same exact way as glucose. What helps me remember D-mannose's structure is that it's actually spelled with two 'n's. This helps me remember that it's actually going to be the C2 epimer of glucose. Referring back to the structure, this means that everything in D-mannose's structure is going to be identical to glucose, from the C3 carbon down. That means its C3 hydroxyl is still pointing to the left, and its C4 and C5 hydroxyls are still pointing to the right. The only thing that's going to differ is that it's the C2 epimer. Its C2 hydroxyl, instead of pointing to the right, like it did with glucose, is going to be pointing to the left. We can go ahead and fill that in, and really, that's it for the structure of mannose.

The next hexose we have here is actually the structure of D-galactose. D-galactose is another aldohexose that's going to be numbered again in the same exact way as glucose and mannose. What helps me remember D-galactose's structure is that it's actually got these four vowels in it. This reminds me that this is going to be the C4 epimer of glucose. Another way to help you remember D-galactose's structure is that D-galactose kind of sounds like a galactic ship, a rocket ship. Those of you that play Call of Duty know that C4s are explosives. You could think that the C4 is the explosive that allows the galactic rocket ship to take off. Hopefully, that'll help you guys remember that galactose is the C4 epimer of glucose. What that means is that everything else is going to be identical to glucose except for the C4. The C5 hydroxyl is still going to the right, the C3 hydroxyl is still going to the left, and the C2 hydroxyl is still going to the right. The only_difference is the C4 epimer. Instead of going to the right hand side as it did for glucose, it's going to be going to the left hand side, and really, that's it for the structure of galactose.

The next monosaccharide that we have here is D-fructose. D-fructose is actually a ketohexose, which means that it has a ketone group instead of an aldehyde group at the top, but it's still going to be numbered from top to bottom here on this page since that's what allows the ketone group to be assigned the lowest possible number. What helps me remember D-fructose is that it's really the only ketone on this entire page. This helps me remember that it's going to be the ketose form

Common Monosaccharides

Alright. So now that we've talked about how to memorize the linear forms of the most common monosaccharides, in this video we're going to talk about how we can use those linear forms of the sugars to derive their cyclic forms. And so notice down below, we have all of the Haworth projections for each of the 6 sugars that we talked about in our last lesson video. And so really the skill of converting a Fischer projection for a linear monosaccharide into the Haworth projection of a cyclic monosaccharide is a skill that we already talked about before in our previous lesson videos. And so recall that we only need to remember the 2 words, "up, lifting," and "down, right." And so recall that all of the chemical groups pointing to the left of a Fischer projection end up pointing upwards in the Haworth projection. And all of the chemical groups pointing to the right of a Fischer projection end up pointing downwards in the Haworth projection. And so this ends up being the key that allows us to use the linear forms of the sugars from our last lesson video to derive their cyclic forms in this lesson video.

Another helpful tool is actually to number all of the carbon atoms so that we can keep track of those carbon atoms easier. Now in our last lesson video, we already numbered all of the carbon atoms of the linear form. So all we need to do is number the carbon atoms of the cyclic forms down below, which means that we want to assign the lowest possible number to the anomeric carbon. And so I'll go ahead and help you guys out For all of the sugars on this page. The anomeric carbon is the carbon atom that's furthest to the right on the Haworth projection. So we're circling all of the anomeric carbons and then we want to assign the lowest possible number. So for this sugar up here, it will be number 1. For this sugar over here, it will be number 1. This one here will also be number 1. But notice that for this sugar over here, we cannot assign it number 1 because that would be skipping out on this carbon atom up above. And so instead, we give this carbon atom up above number 1 and this carbon atom right here number 2, which allows it to be assigned the lowest possible number. And then for this sugar and this sugar down below, both anomeric carbons will be assigned number 1. And then of course, the highest numbered carbon is going to be pointing upwards in all of these sugars that we see here. And, notice that this sugar and this sugar down below only have a total of 5 carbon atoms. So the highest numbered carbon will be assigned number 5.

Now, let's begin derived the first two cyclic sugars using the linear forms of the sugars up above. And then we're going to pause the video and allow you guys to attempt deriving the rest of the sugars before we give you the answers. And so we're starting off with beta-D-glucopyranose over here. The first thing that you see is the beta, which is referring to the configuration of the anomeric carbon. And because the bumps of the beta are on the same side, that reminds us that the highest numbered carbon and the alcohol group of the anomeric are going to be pointing to the same side of the Haworth projection. And so that means that we expect the alcohol group to be pointing upwards here at this position. And then we can go ahead and number all of the other carbon atoms going around here. And so we want to focus on carbons 2, 3, and 4 here. And notice that the glucose prefix here is telling us that this sugar is derived from glucose. And so if we took a look up, at the structure of glucose from our previous lesson video over here, notice that the C-2 and C-4 hydroxyls are going to the right, whereas the C-3 hydroxyl is going to the left. And so in terms of up, lifting, and down, right, this translates to down, down, up, down. For C-2, C-3, and C-4. So, down, is here, and then up is here, and down is over here. And so this is where we expect the hydroxyls to go. So for C-2, we expect the hydroxyl to be going down, and C-4, we expect the hydroxyl to be going down, and then for the C-3, we expect the hydroxyl to be going up. And then all we have to do is fill in the hydrogens at all of these other positions, and really that is it. This is the structure here of beta-D-glucopyranose. And, really, you can see that the key was just using these two words, up, lifting and down right.

Now, let's move on to the next sugar here, alpha-D-mannopyranose. And then the first thing that you see here is the alpha configuration, which is the configuration of the anomeric carbon. And the alpha reminds us that it's going to be pointing down to the ants, and so we know that the alcohol group on the anomeric carbon is going to be going in the opposite direction reaching down for the ants. And so, we can go ahead and put the alcohol group right here. And then going ahead and, numbering all of the carbon atoms here, we can see that we're going to be paying attention to carbons 2, 3, and 4. And because the prefix manno is in here, that tells us that this sugar is derived from mannose. And so if we go up to the structure of mannose from our previous lesson video, right here, you'll notice that the C-2 and C-3 are both pointing to the left, whereas the C-4 hydroxyl is pointing to the right. And so in terms of uplifting and downright, this translates to up, up, down, for C-2, C-3, and C-4. So, up, up, down would be up here, up, here, and then down for C-4 and then all we have to do is fill in the hydroxyls, so we have OH here, OH here, and then an OH going down over here and then all we need to do is fill in all of the hydrogens at these empty positions. And that is the structure of alpha-D-mannopyranose right here. And so at this point, what we're going to do is allow you guys to pause the video and go ahead and give an attempt at deriving these cyclic sugars using the linear forms and using this key right here. So, we'll assume that you guys have paused the video and given it a try. And so, really, the answer to that is right here. And so, what you'll notice is that, all of these have alpha configuration, which means that the anomeric carbon is going to have their hydroxyl groups pointing downwards. And then, of course, all we have to do is use the up, lifting, and down right to derive the positions of the other hydroxyl groups. And so this here concludes our video on how to use the linear forms of the sugars to derive their cyclic forms. And so really, if you're able to commit these 6 sugars from our last lesson video to memory, then you're able to memorize a total of 12 sugars because you'll be able to get the cyclic forms, no problem. And so that concludes this video, and we'll be able to get some practice in our next video. So I'll see you guys there.

The sugar α-D-Mannose is a sweet-tasting sugar. β-D-Mannose, on the other hand, tastes bitter. A pure solution of α-D-mannose loses its sweet taste with time as it is converted into the β anomer. Draw the β anomer:

Alright. So this practice problem says the sugar alpha-D-mannose is a sweet-tasting sugar. Beta-D-mannose, on the other hand, tastes bitter. A pure solution of alpha-D-mannose will lose its sweet taste with time as it's converted into the beta anomer. Draw the beta anomer and so really that's all we need to do in this box is draw beta-D-mannose. Recall from our previous lesson videos that because mannose has 2 ends in it, that reminds us that mannose is going to be the C2 epimer of glucose. So, what we need to do is recall the structure of glucose and then draw the C2 epimer. We know that glucose is going to be an aldohexose. So, it's going to have an aldehyde group at the very top here, and it's a hexose, so we know it's going to have a total of 6 carbon atoms. So, we can go ahead and start to draw those. There's 3, 4, 5, 6 carbon atoms, and of course, the last one is going to be a CH2OH group. We know that the pattern of alcohol groups on glucose kind of looks like it's given us the fingers. The C2 hydroxyl will be going to the right, the C3 hydroxyl will be going to the left, and then, of course, the C4 and C5 hydroxyls are going to the right. Everything else is filled in with hydrogens. This is the structure of glucose, in fact, the structure of D-glucose. If we want to draw the structure of mannose, the two ends tell us that this is going to be the C2 epimer. So, carbon 2 is right here. The only thing to do is redraw the glucose structure and essentially, change this highlighted position. Instead of having the hydroxyl group pointing to the right, we're going to have the hydroxyl group pointing to the left, getting this structure. This is the C2 epimer of glucose, which is mannose or D-mannose. Of course, if it's telling us that this is beta-D-mannose and beta, we know, is the configuration of the anomeric carbon. This is suggesting that this is going to be the cyclic form of beta-D-mannose. What we need to recall at this point is the phrase, "up-lefting" and the other word "down-right." We know from our previous lesson videos that beta-D-mannose is going to form a pyranose. We can draw in the pyranose configuration here, a 6-membered ring with a CH2OH group. Numbering this will be our anomeric carbon, so this will be our carbon 1, 2, 3, 4, 5, and 6, and just for keeping the numbers, keeping track of the carbon atoms, we'll also number these carbons. This tells us that this is the beta configuration, which reveals the configuration of the anomeric carbon. If it's beta, then we know that the bumps are going to be on the same side. That means that the alcohol group on the anomeric carbon must be going up in the same direction as the highest numbered carbon. This will be the position of the alcohol group on the anomeric carbon. We can then look at the C2 carbon and notice that it's going to the left, so we also expect it to be going up on the C2 carbon. The hydroxyl group will be going up here. Notice that the C3 hydroxyl group is also going to the left, which means that it's also going to be going up. We draw the hydroxyl group going up on the C3 carbon. The C4 hydroxyl group is going to the right, so we expect it to be going down on the C4 carbon. This is the beta-D-mannose anomer we were looking for. We don't need to fill in the hydrogens since they are assumed, but again, all carbon atoms have four bonds, so we would expect hydrogens to be coming off. This is the simplest answer we can provide for this question. This is the end of this practice problem. I'll see you guys in our next video.

Draw the α-furanose and β-pyranose forms of D-ribose.

Alright. So this practice problem wants us to draw the alpha furanose and the beta pyranose forms of D-ribose. And so we can draw the alpha furanose form over here on the left-hand side and we can draw the beta pyranose form over here on the right-hand side. And so we need to recall from our previous lesson videos, the structure of ribose. And so we know that D-ribose is an Aldopentose, which means that it has an aldehyde group and it's going to have a total of 5 carbon atoms. So, we have 1 carbon atom here so we can draw the rest. That's 2, 3, 4, and 5 carbon atoms and the last one's going to be a CH2OH group and then, of course, we know that the ribose is going to have all of its R, all of its alcohol groups going to the right. So r for right. And so here we're going to have an alcohol group going to the right, as well as here, and over here as well. And then, of course, we can fill in the hydrogens that are remaining. And so, of course, we know that whenever we want to convert the Fischer projections of a linear monosaccharide into the Haworth projections of cyclic monosaccharides, then, we're going to need to remember the 2 words up, lifting, and down, right. And so notice that we have all of the alcohol groups here pointing to the right-hand side. So, we expect these to be pointing downwards. And so, we can go and number these carbon atoms as well that will also help us keep track of them. So, of course, this will be carbon 1, 2, 3, 4, and carbon 5. And so, we can start by drawing the alpha furanose form. So, recall furanoses are going to be 5-membered sugar rings. And so, we can go ahead and draw that. We know that the oxygen here is going to be part of the ring. And so here is our 5-membered furanose and we want the anomeric carbon to be over here on the right-hand side. So, this is going to be the lowest numbered carbon. 1, 2, this will be 3, and this will be 4, and then of course, we need our 5th carbon, the CH2OH group. So, we can go ahead and add that in up here, CH2OH, and this will be our 5th carbon And then, of course, we know that the alpha here is referring to the configuration of the anomeric carbon and alpha tells us that the alcohol group of the anomeric carbon is going to be going down, towards the ant. So it's going to be going in the opposite direction of the highest numbered carbon here and, it will be alpha configuration. So, this is alpha and then, when we take a look at the C2 and the C3 hydroxyls, notice that they're both going to the right. So, we expect them to be going down on these C2 and C3 carbons. So, we can go ahead and draw these hydroxyls also going down on both of these carbons here going down. Go ahead and redraw this 3 here so you can see it a little bit better. And so, this is going to be the structure of alpha furanose right here and so notice that the hydroxyl group on carbon 4 is the one that reacted with the aldehyde group. So, we don't need to worry about the alcohol group here and this is the final structure of alpha furanose. So, now, we can move on to beta pyranose and recall that pyranoses are 6-membered rings, sugars with 6-membered rings. So, we can go ahead and draw that in. So we know that the oxygen atom is going to be part of that. So it's going to look like this. And of course, we want the anomeric carbon to be the one farthest to the right here. So this will be carbon 1, 2, 3, 4, and carbon 5. And, of course, the beta is telling us that it's going to be on the same side as where the highest numbered carbon would be. So, this means that it's going to be going up and above the ring, the alcohol group. On the anomeric carbon will be going up here and this tells us that this is the beta configuration and then looking at carbons 2, 3, and actually carbons 2, 3, and 4. We'll see that all the alcohol groups are going to the right. So, we expect that they will all be going down in the Haworth projection. So here we expect it to be going down. For here, for carbon 3, we expect it to be going down and for carbon 4 as well. And so this is the final structure of beta pyranose. The final structure of the beta pyranose form of D-ribose. And so, this and this are our final answers for this practice problem. And so, that concludes this practice and I'll see you guys in our next video.

Indicate if the following pairs of sugars are enantiomers, anomers, epimers, or an aldose-ketose pair:

a) α-D-galactopyranose and β-D-galactopyranose. ____________________

b) D-glucose and D-mannose. ____________________

c) D-glucose and D-fructose. ____________________

d) α-D-glucopyranose and β-D-glucopyranose. ____________________

e) D-galactose and D-glucose. ____________________

f) α-D-mannopyranose and α-L-mannopyranose. ____________________

Alright, so this practice problem wants us to indicate if the following pairs of sugars are enantiomers, anomers, epimers, or an aldose-ketose pair. And so, taking a look at our first pair here, we have alpha-D-galactopyranose and beta-D-galactopyranose. Notice the only difference between these two sugars is that one has alpha and the other has beta. This means that they differ at the configuration of the anomeric carbon. So these two are going to be anomers of each other. We can go ahead and write "anomers" here. Now, looking at the next pair, we have D-glucose and D-mannose. Recall from our previous lesson videos, we said that because D-mannose has these two ends in its name, that reminds us that D-mannose is going to be the C2 epimer of D-glucose. Of course, D-glucose and D-mannose are going to be epimers of each other. We can go ahead and write "epimers" here. Now, moving on to the third pair, we have D-glucose and D-fructose. Recall in the list of six monosaccharides that we had talked about in our previous lesson videos that D-fructose was the only ketose on that list, and so that reminds us that D-fructose is the ketose form of D-glucose. Thus, D-glucose and D-fructose are going to be an aldose-ketose pair. We can go ahead and write that in right here. Another way we could have figured out that option C here was an aldose-ketose pair is to recall that D-glucose is an aldohexose and D-fructose is an aldoketose. They are not going to be enantiomers, because that would make them mirror images of each other. They're not going to be anomers, because anomers only relate to cyclic sugars, and these two don't necessarily indicate that they're in their cyclic forms. Epimers only differ in the configuration of just one single chiral carbon. Since they differ because one has an aldehyde and one has a ketone, that's more than just differing by one chiral carbon configuration. So, through process of elimination, we would have been able to figure out that it wasn't any of these three and therefore, it would have only left the aldose-ketose pair, for option C. Now, moving on to the fourth pair, we have alpha-D-glucopyranose and beta-D-glucopyranose. They only differ in the alpha and beta configurations which makes them anomers. Now, moving on to this pair right here, we have D-galactose and D-glucose. Recall from our previous lesson videos that what helps us remember the relationship between these two sugars is that D-galactose has a total of four vowels in it. These four vowels remind us that D-galactose is the C4 epimer of D-glucose. So these two are going to be epimers of each other. The last pair we have is alpha-D-mannopyranose and alpha-L-mannopyranose. Because we have D and L here and the same exact sugar otherwise, this, of course, is telling us that these two must be mirror images of each other. If they're going to be mirror images, then that means they're going to be enantiomers because they have opposite, complete opposite configurations. So, what we can do is write that these are going to be enantiomers. And so, these are the answers to this practice problem and that concludes this practice problem. I'll see you guys in our next video.

Here’s what students ask on this topic:

What are the most common monosaccharides in nature?

The most common monosaccharides in nature include D-glucose, D-mannose, D-galactose, D-fructose, D-ribose, and D-deoxyribose. D-glucose is the most abundant and serves as a reference for memorizing other hexoses. These monosaccharides are categorized as aldoses or ketoses based on their functional groups. Aldoses like D-glucose and D-mannose have an aldehyde group, while ketoses like D-fructose have a ketone group. Understanding these common monosaccharides is crucial for studying carbohydrate structures and their roles in metabolic pathways.

How do you convert a Fischer projection to a Haworth projection?

To convert a Fischer projection to a Haworth projection, follow these steps: 1) Identify the anomeric carbon, which is the carbonyl carbon in the linear form. 2) Number the carbon atoms, ensuring the anomeric carbon gets the lowest possible number. 3) Use the rule 'up, lefting, and down, right' to determine the orientation of hydroxyl groups. Groups on the left in the Fischer projection point upwards in the Haworth projection, while groups on the right point downwards. This method helps visualize the cyclic form of monosaccharides, essential for understanding their biochemical functions.

What is the difference between aldoses and ketoses?

Aldoses and ketoses are types of monosaccharides differentiated by their functional groups. Aldoses contain an aldehyde group (-CHO) at the end of the carbon chain, while ketoses have a ketone group (C=O) typically at the second carbon atom. For example, D-glucose is an aldohexose with an aldehyde group, whereas D-fructose is a ketohexose with a ketone group. This distinction is crucial for understanding their chemical behavior and roles in metabolic pathways.

Why is D-glucose considered the most important monosaccharide?

D-glucose is considered the most important monosaccharide because it is the primary energy source for cells in most organisms. It is an aldohexose, meaning it has six carbon atoms and an aldehyde group. D-glucose is central to metabolic pathways such as glycolysis and the citric acid cycle, where it is broken down to produce ATP, the energy currency of the cell. Additionally, D-glucose serves as a building block for more complex carbohydrates like starch and glycogen.

What are the key differences between D-glucose, D-mannose, and D-galactose?

D-glucose, D-mannose, and D-galactose are all aldohexoses, but they differ in the configuration of their hydroxyl groups. D-glucose has hydroxyl groups on carbons 2, 4, and 5 pointing to the right, and on carbon 3 pointing to the left. D-mannose is the C2 epimer of glucose, with the hydroxyl group on carbon 2 pointing to the left. D-galactose is the C4 epimer of glucose, with the hydroxyl group on carbon 4 pointing to the left. These differences affect their biochemical properties and roles in metabolism.