Chemistry of Carbohydrates questions and answer

Q.1. What are carbohydrates? 

Carbohydrates are organic substances containing C, H, and O. Hydrogen atoms are present usually in the ratio of 2:1 as it occurs in a water molecule. Example: glucose, fructose, lactose, starch, etc. There may be exceptions to the above, e.g. C2H4O2 is acetic acid and not a carbohydrate through H and O are in the ratio of 2:1. 

Q.2. Define carbohydrates in chemical terms. 

Carbohydrates are defined chemically as aldehyde or ketone derivatives of the higher polyhydric alcohols or compounds that yield these derivatives on hydrolysis. 

Q.3. How will you classify carbohydrates? 

Carbohydrates are classified into four major groups: 

• Monosaccharide (“simple sugars”): They cannot be hydrolyzed into simpler forms. 

• Disaccharides: They yield two molecules of the same or different monosaccharide units on hydrolysis. 

• Oligosaccharides: They yield three to six molecules of monosaccharides on hydrolysis. 

• Polysaccharides (glycans): They yield more than 6 molecules of monosaccharides on hydrolysis. 

Q.4. How are monosaccharides further classified? 

Monosaccharides are further classified into two groups depending on: • The number of carbon atoms they possess, e.g. trioses, tetroses, pentoses, hexoses, etc. • Whether aldehyde (-CHO) or ketone (-CO) group is present, e.g. aldoses, ketoses.

Q.5. Give an example of an aldohexose and a ketohexose which is of biological importance. 

• Aldhexose: D-glucose 

• Ketohexose: D-fructose 

Q.6. How will you classify polysaccharides? 

Polysaccharides are classified into two main groups: 

• Homopolydisaccharides (hemoglycans): Polymer of same monosaccharide units, e.g. starch, glycogen, insulin, dextrins, cellulose, etc. 

• Heteropolysaccharides (heteroglycans): Polymer of different monosaccharide units or their derivatives, e.g. mucopolysaccharides (MPS). 

Q.7. What is asymmetric carbon? 

A carbon atom to which four different atoms or groups of atoms are attached is said to be an asymmetric carbon. 

Q.8. What are the effects of the presence of asymmetric carbon in a compound? 

The presence of asymmetric carbon atoms in a compound produces the following effects: 

• Gives rise to the formation of stereoisomers of that compound. 

• Also confers optical activity to the compound. 

Q.9. What are stereoisomers? 

The compounds which are identical in composition and differs only in the spatial configuration are called stereoisomers. Two such stereoisomers of glucose, D-glucose and L-glucose is mirror-images of each other. 

Q.10. What are optical isomers? When a beam of plane-polarized light is passed through sugar solution exhibiting optical activity, it will be rotated to the right or left according to the type of the compound. Such compounds are called “optical isomers” (or “enantiomorphs”). 

Q.11. Explain ‘D’ and ‘L’ forms of sugar and d(+) and l(–) sugars. 

• D and L (capital letters) are used to represent the configuration of the sugar molecules. The orientation of H and OH groups around the carbon atom just adjacent to the terminal primary alcohol group -CH2OH, i.e. C atom 5 in glucose determines the series. When the -OH the group is on the right it belongs to D-Series and -OH Group is on the left it is a member of the L-series. 

• d(+) and l(–) denote the nature of the optical activity. When the plane-polarized light is rotated to the right, the compound is called “dextrorotatory” or d(+) and when rotated to left, the compound is called “laevorotatory” or l(–). 

Q.12. What is Vant Hoff’s rule of n? 

According to Vant Hoff’s rule of ‘n’; 2n gives the possible stereoisomers of that compound; where n represents the number of asymmetric carbon atoms in a compound. 

Q.13. How many stereoisomers are formed by glucose? 

Glucose has 4 asymmetric carbon atoms, hence as per Vant Hoff’s rule of ‘n’; glucose will have 2n = 16 stereoisomers; of which 8 belongs to D-series and 8 belongs to L-series. 

Q.14. What are the α and β forms of glucose? When the aldehyde group of C-1, on cyclization, condenses with the alcoholic-OH group on C-5 on the same molecule, two different cyclic forms are produced. When -OH group on C-1 of cyclic form extends to the right, it is called αD-glucose and when -OH group extends to left, it is called β-D-glucose. 

Q.15. What is an anomer and what is an anomeric carbon? 

• After cyclization, the two cyclic compounds α and β have different optical rotation but they are not “mirror images” of each other. Such compounds are called “anomers”. • Carbon 1 after cyclization becomes asymmetric as it has new 4 different groups attached to it and is called “anomeric carbon”. 

Q.16. What do you mean by mutarotation? 

When an aldohexose is first dissolved in water and the solution is put in the optical path so that the plane-polarized light is passed, the initial optical rotation is shown by the sugar gradually changes until a “constant fixed” rotation characteristic of the sugar is reached. This phenomenon of change of rotation is called mutarotation.

Q.17. What is the mechanism of mutarotation? 

Explain. Ordinary crystalline glucose is an α form. The mutarotation represents a conversion of α glucose when in solution to an equilibrium mixture of α and β forms. This involves the opening of the hemiacetal ring to form traces of the “free” aldehyde form which is transitory and recondensation to the cyclic forms again. 

Q.18. What are the optical rotations shown by glucose in solution? 

The glucose solution shows rotation according to its form: α form shows + 112o and β-form shows + 19o. When the solution has an equilibrium mixture of α and β forms, it shows a fixed rotation of + 52.5°. 

Q.19. What is meant by pyranose form? 

The pyranose forms of the sugars are internal hemiacetals formed by the combination of the aldehyde or ketone group of the sugar with the -OH group on the 5th carbon atom from the aldehyde or ketone group. 

Q.20. What is meant by the furanose form of the sugar? 

The furanose forms of the sugars are formed by the reaction between the aldehyde or ketone group with the -OH group on the 4th carbon from the aldehyde or ketone group. 

Q.21. What is an epimer? Give two examples. 

• When two sugars differ from one another only in the configuration of H and OH around a single carbon atom, they are called “epimers”. Examples: Glucose and galactose are epimeric pairs that differ in the orientation of H and OH groups on C-4. • Similarly glucose and mannose are epimers in respect of C-2. 

Q.22. What is epimerization? Give one example. 

• The process by which one epimer is converted to other is called “epimerization” and it requires the enzyme” “epimerase”. • Example: Conversion of UDP-glucose to UDP-galactose in the liver (reversible reaction) by the enzyme “epimerase”.

Q.23. What are osazones? What is its importance? 

Osazones are crystalline derivatives of sugars. They have characteristic features: • Melting points • Crystal structures and • Precipitation times. Thus, they are valuable in the identification of sugars. 

Q.24. Describe the osazones of different sugars. Different sugars form characteristic osazone crystals which can be seen under the microscope. Thus, Osazones of Time taken Appearance • Glucose and fructose 2-5 minutes “Hay like” structure (same osazone) (corn leaf life) • Maltose 10-15 minutes Sunflower like • Lactose 10-12 minutes “Powderpuff” like • Galactose 7 minutes Sunflower but narrow blades 

Q.25. How many molecules of phenylhydrazine are required in the reaction and what are the by-products? 

• Three molecules of phenylhydrazine react with the sugar molecule. • NH3 and aniline are by-products in the reaction. 

Q.26. Why galactose forms different osazone than glucosazone? 

The structure of galactose differs on C-4 and this part of the molecule is unaffected in osazone formation. Hence, galactose forms different osazone from glucose and fructose. 

Q.27. What are the oxidation products of glucose? 

Oxidation products of glucose depend on the type of oxidation: • With mild oxidation with Br2—water, it forms D-gluconic acid (aldonic acid) • With strong oxidizing agent like conc-HNO3, it forms dicarboxylic acid called D-glucaric acid (saccharic/ aldaric acid). • If the -CHO group is protected and the primary -CH2OH the group is oxidized to COOH group, it forms D-glucuronic acid (uronic acid). 

Q.28. What happens when D-galactose is oxidized with hot conc HNO3? 

Galactose on oxidation with hot conc HNO3 acid produces dicarboxylic acid called Mucic acid. The crystals of mucic acid have a characteristic shape, hence, it can be used for the identification of galactose. 

Q.29. What are the reduction products of glucose, mannose, galactose and fructose? 

On reduction, the monosaccharides produce sugar alcohols. Thus, • D-glucose → D-sorbitol • D-galactose → D-dulcitol • D-mannose → D-mannitol • D-fructose → D-mannitol + D-sorbitol 

Q.30. If glucose is dissolved in weak alkali such as Ba(OH)2/ or Ca(OH)2 and kept for some time, one can detect mannose and fructose in the solution.

 Explain. Glucose, fructose, and mannose are interconvertible in solutions of weak alkalinity such as Ba(OH)2 or Ca(OH)2. These interconversions are because all three give same Enediol form, which tautomerizes to all three sugars, the reaction is called Lobry de Bryan Van Ekenstein reaction. 

Q.31. Name one biological fluid that is rich in fructose. What are the source of fructose in this fluid and its importance? 

Seminal fluid is rich in fructose. • Source: It is formed from glucose in the seminiferous tubular epithelial cells. • Importance: Spermatozoa utilizes fructose for energy. 

Q.32. Describe the chemistry of reduction of Benedict’s qualitative reagent by glucose and fructose.

 Glucose/fructose having free -CHO/= CO group respectively undergo enolization in weakly alkaline solution. The “enediol” forms of the sugars are highly reactive. The Cu++/(ic) ions take electrons from the enediols and oxidize them to sugar acids and are in turn reduced to Cu+ (ous) ions. Cu+(ous) ions combine with -OH to form the yellow cuprous hydroxide, which upon heating is converted to red cuprous oxide. 

Q.33. Name the ingredients present in Benedict’s qualitative reagent and mention their functions. 

• Benedict’s Qualitative reagent contains CuSO4, sodium citrate, sodium carbonate. Functions: Sodium citrate in the reagent prevents precipitation of cupric carbonate by forming soluble, slightly dissociable complexes which dissociate sufficiently to provide a supply of readily available Cu++ (ic) ions for oxidation. • Sodium carbonate, a weak alkali analyzes the sugars and thereby causes them to be strong reducing agents. Enolization is better in weak alkali than strong alkali. 

Q.34. What are aminosugars? 

Sugars containing an NH2 group in their structure are called aminosugars. The alcoholic OH group on carbon 2 is usually replaced by -NH2 group. Examples: D-glucosamine, D-galactosamine. 

Q.35. What is the biomedical importance of aminosugars? 

• N-acetyl derivatives of D-glucosamine and D-galactosamine occur as constituents of certain mucopolysaccharides (MPS). • Certain antibiotics such as erythromycin, carbomycin, etc. contain aminosugars which are probably responsible for the antibiotic activity. 

Q.36. What are glycosides? 

Glycosides are compounds containing carbohydrate and a non-carbohydrate residue in the same molecule; Carbon 1 of carbohydrate is attached to the noncarbohydrate residue by an acetal linkage. Q.37. What is aglycone? 

And mention its nature. The non-carbohydrate residue present in the glycoside is called an aglycone. The aglycone can be a simple substance like methyl alcohol, glycerol, phenol or complex substances like sterol, hydroquinone or anthraquinones.

Q.38. Name some glycosides and mention their biomedical importance. 

Some of the important glycosides and their functions are as follows: • Cardiac glycosides: used in cardiac insufficiency. Examples are digitonin, strophanthin, etc. • Ouabain: a sodium pump inhibitor. • Phlorhizin: displaces Na+ from the binding site of “carrier protein” and thus prevents the transport of glucose across the mucosal cells of the small intestine and renal tubular epithelial cells (producing glycosuria). 

Q.39. Name three disaccharides of biological importance. 

They are maltose, lactose, and sucrose. 

Q.40. Mention the hydrolytic products of these three disaccharides. 

Hydrolytic products • Maltose (Malt sugar) → D-glucose + D-glucose. • Lactose (Milk sugar) → One mol of D-glucose + One mol of D-galactose. • Sucrose (Table sugar) → One molecule of D-glucose + one mol of D-fructose. 

Q.41. What will be the oxidative product of lactose after prolonged boiling with conc HNO3 acid? 

After boiling with conc HNO3 acid lactose is hydrolyzed to glucose and galactose. Galactose on oxidation produces mucic acid. 

Q.42. Name the disaccharide which is non-reducing and why it is so? Sucrose is a non-reducing disaccharide. • Reason: In sucrose, both the aldehyde group of glucose and ketone group of fructose are linked together by α1 → 2 linkage. Hence, no free aldehyde or ketone group is available for enolization and reducing action. 

Q.43. What are invert sugars? 

What is meant by inversion? • Sucrose is dextrorotatory (+ 62.5o) but it's hydrolytic products are laevorotatory (– 19.5o), as fructose has a greater specific laevorotation than the dextrorotation of glucose. • As the hydrolytic products invert the rotation from + to –, the resulting mixtures of glucose and fructose, i.e. the hydrolytic products are called “invert sugars” and the process is called “inversion”. 

Q.44. Mention the biomedical importance of maltose. 

• Various baby food preparations contain maltose which is easily digested and thus of nutritional significance. • Dietary maltose is hydrolyzed in the small intestine by the enzyme maltase to two molecules of glucose which are absorbed and utilized for energy by the body. 

Q.45. State the biomedical importance of lactose. 

• Dietary lactose is hydrolyzed in the small intestine by the enzyme lactase to glucose and galactose and absorbed for utilization in the body. • In the lactating mammary gland, lactose is synthesized from glucose by duct epithelium. Lactose, present in breast milk is a good source of energy for the newborn. • Lactose is fermented by E.coli which is usually nonpathogenic and not fermented by the typhoid bacilli which is pathogenic. Thus, lactose fermentation is utilized to differentiate non-pathogenic and pathogenic bacteria. • Many organisms that are found in milk, e.g. E.coli, Str. lactis, A. aerogenes convert lactose of milk to lactic acid (LA) causing “souring of milk”. 

Q.46. What is the effect of sucrose when given parentally? 

Sucrose cannot be hydrolyzed when introduced parentally due to a lack of enzyme sucrase in the blood. The presence of sucrose in the blood changes the osmotic condition and causes a flow of water from the tissues to the blood. Thus, clinicians use it in edema like cerebral edema. Q.47. What is the biomedical importance of oligosaccharides? • Antibodies, blood group substances, and coagulation factors contain oligosaccharides. • The oligosaccharides units of glycoproteins are rich in information and are functionally important.

Q.48. What are sialic acid and its importance? 

• Sialic acid is N-acetyl neuraminic acid (NANA). Neuraminic acid is an amino sugar acid and structurally an aldol condensation product of pyruvic acid (PA) and D-mannosamine. • Neuraminic acid and silica acids (NANA) occur in several mucopolysaccharides (MPS) and in glycolipids like gangliosides. 

Q.49. Name the enzyme which hydrolyses sialic acids. 

“Neuraminidase” is the enzyme that hydrolyzes to split NANA from the compound. 

Q.50. Name some homopolysaccharides (homoglycans). 

Some of the important homopolysaccharides are starch, glycogen, dextrins, dextran, insulin, cellulose, agar, etc. 

Q.51. What are the polymeric units present in the starch granule? 

Two polymeric units of glucose present in a starch granule is: • Amylose and • Amylopectin 

Q.52. Enumerate the essential differences between amylose and amylopectin. 

Essential differences between amylose and amylopectin are: Amylose Amylopectin 1. Amount 15-20% 80-85% 2. Mol Wt Low High (approx 60,000) (approx 500,000) 3. Reaction with Blue color Reddish violet color dilute I2 solution 4. Structure • Unbranched • Highly branched straight-chain • 250-300 • Mainstem and branD-glucose units ches-α 1 → 4 linkages; joined by α 1 → 4 at branch point α 1 → 6 linkages. linkages. • Twists into a • Approx 80 branches, helix, 6 glucose one branch after every unit per turn. 24 to 30 glucose units.