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Diversity. How many different oligosaccharides can be made by linking one glucose, one mannose, and one galactose? Assume that each sugar is in its pyranose form. Compare this number with the number of tripeptides that can be made from three different amino acids.

Short Answer

Expert verified
24 different oligosaccharides, 6 different tripeptides.

Step by step solution

01

Understand the Structure of Oligosaccharides

An oligosaccharide is a short chain of sugar (monosaccharide) units linked together. In this problem, we are using three different monosaccharides: glucose, mannose, and galactose, each in its pyranose form. The arrangement of these sugars can vary in terms of the sequence and the type of bonds between them.
02

Calculate the Arrangement Possibilities for Oligosaccharides

The three different sugars can be arranged in different sequences. Since there are 3 sugars and they are all different, the total number of sequences can be calculated using permutations. The formula for permutations of n distinct objects is given by \(n!\). For 3 sugars, the calculation is \(3! = 6\). Thus, there are 6 different orderings of glucose, mannose, and galactose.
03

Consider Number of Linkages

Besides the sequence, each connection between two sugars can be an α or β glycosidic linkage. Since there are two linkages in a sequence of three sugars, each with 2 possibilities, this adds an additional factor of \(2^2 = 4\) possible linkages.
04

Compute Total Different Oligosaccharides

Combine the sequence and linkage possibilities. The total number of different oligosaccharides is the product of the sequence arrangements and the different linkage possibilities: \(6 \times 4 = 24\). Therefore, there can be 24 different oligosaccharides.
05

Understand the Structure of Tripeptides

A tripeptide is made of three linked amino acids. For three different amino acids, their sequence can be varied but the type of linkage remains constant (peptide bonds).
06

Calculate the Arrangement Possibilities for Tripeptides

Similar to oligosaccharides, the three distinct amino acids can also be arranged in \(3! = 6\) different sequences.
07

Compare Oligosaccharides to Tripeptides

Oligosaccharides can form 24 different combinations as there are several glycosidic linkages possible, which significantly increases the diversity. In contrast, tripeptides formed from three different amino acids have only 6 possible combinations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Monosaccharides
Monosaccharides are the simplest form of carbohydrates, often referred to as simple sugars. They are the building blocks for more complex carbohydrates like oligosaccharides and polysaccharides. Each monosaccharide consists of a single sugar unit, which includes a carbon backbone, generally containing 3 to 7 carbon atoms. Common examples of monosaccharides include glucose, mannose, and galactose.
These sugars can exist in different structural forms. In nature, monosaccharides typically adopt a ring-shaped structure, such as the pyranose form. This cyclic structure is crucial because it determines how monosaccharides will connect with each other to form larger carbohydrate molecules. Different isomers—although they have the same molecular formula—can have different physical and chemical properties due to their structure.
Glycosidic Linkages
Glycosidic linkages are the bonds that form when monosaccharides are connected together to create oligosaccharides and polysaccharides. These linkages are crucial in determining the structure and function of the resulting sugar chain. Any connection between two sugar molecules can be either an α or a β glycosidic linkage.
When we look at the structural diversity of carbohydrates, we consider these different linkage types as an additional factor of variation. For each bond location between sugar molecules, there are two possible linkage configurations:
  • α (alpha) glycosidic linkage
  • β (beta) glycosidic linkage
This dual possibility for each bond greatly increases the number of potential oligosaccharides that can be formed from any given set of monosaccharides.
Amino Acids
Amino acids are organic compounds that serve as the building blocks for proteins. Proteins are essential for virtually every biological process. Each amino acid has a central carbon atom bonded to four different groups: an amino group, a carboxyl group, a hydrogen atom, and a variable side chain (R group) that defines the type of amino acid.
There are 20 standard amino acids in nature, each with a unique side chain that affects its chemical characteristics and properties. When amino acids link together, they form peptides or proteins, which are chains distinguished by the sequence of amino acids in them.
  • The sequence of amino acids determines the protein's structure and function.
  • In the context of tripeptides, diversity comes from the different sequences that the amino acids can form.
Permutations
Permutations are fundamental in determining the different ways objects can be ordered or arranged. In the context of our oligosaccharide and tripeptide problem, permutations help us calculate the different sequence arrangements possible with a set of distinct items.
For example, when you have three different monosaccharides or amino acids, they can be arranged in various sequential orders. The number of different sequences can be calculated using the formula for permutations, which is given by:\[ n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \]Where \(n!\) (n factorial) represents the permutations of \(n\) distinct objects. For three distinct objects, such as three different sugars or amino acids, the number of permutations is calculated as \(3! = 6\). This concept applies directly to calculating the arrangements of elements like sugars in oligosaccharides and amino acids in tripeptides.
Tripeptides
Tripeptides are short chains of three amino acids linked together by peptide bonds. This smallest form of peptide illustrates the basics of protein structure and diversity. The peptide bond is a specific type of linkage that joins the carboxyl group of one amino acid to the amino group of another.
Although the peptide bond itself does not vary like glycosidic linkages in oligosaccharides, the sequence of the amino acids can still be varied. This means that if you have three different amino acids, you can arrange them in \(3! = 6\) different ways, similar to how you would arrange three distinct sugars.
The difference with tripeptides, compared to oligosaccharides, is that the linkage type remains constant, offering fewer diversification possibilities. Consequently, this is why tripeptides formed from three different amino acids offer only 6 possible combinations, emphasizing the role of linkages in oligosaccharide diversity.

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Most popular questions from this chapter

Mapping the molecule. Each of the hydroxyl groups of glucose can be methylated with reagents such as dimethylsulfate under basic conditions. Explain how exhaustive methylation followed by the complete digestion of a known amount of glycogen would enable you to determine the number of branch points and reducing ends.

Word origin. Account for the origin of the term carbohydrate.

Mutarotation. The specific rotations of the \(\alpha\) and \(\beta\) anomers of \(\mathrm{D}\) -glucose are +112 degrees and +18.7 degrees, respectively. Specific rotation, \([\alpha]_{\mathrm{D}},\) is defined as the observed rotation of light of wavelength 589 nm (the \(D\) line of a sodium lamp) passing through \(10 \mathrm{cm}\) of a \(1 \mathrm{g} \mathrm{ml}^{-1}\) solution of a sample. When a crystalline sample of \(\alpha\) -D-glucopyranose is dissolved in water, the specific rotation decreases from 112 degrees to an equilibrium value of 52.7 degrees. On the basis of this result, what are the proportions of the \(\alpha\) and \(\beta\) anomers at equilibrium? Assume that the concentration of the open-chain form is negligible.

Stereospecificity. Sucrose, a major product of photosynthesis in green leaves, is synthesized by a battery of enzymes. The substrates for sucrose synthesis, D-glucose and \(D\) -fructose, are a mixture of \(\alpha\) and \(\beta\) anomers as well as acyclic compounds in solution. Nonetheless, sucrose consists of \(\alpha\) -D-glucose linked by its carbon-1 atom to the carbon- 2 atom of \(\beta\) -D-fructose. How can the specificity of sucrose be explained in light of the potential substrates?

A taste of honey. Fructose in its \(\beta\) -D-pyranose form accounts for the powerful sweetness of honey. The \(\beta\) -D-furanose form, although sweet, is not as sweet as the pyranose form. The furanose form is the more stable form. Draw the two forms and explain why it may not always be wise to cook with honey.

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