Question 1

What are monosaccharides?

Accepted Answer

A monosaccharide is a carbohydrate that cannot be broken down any further — no acid, base, or enzyme can hydrolyse it into a smaller sugar, because it is already the simplest possible sugar unit. Carbohydrates as a class are defined operationally, by what hydrolysis does to them: - A monosaccharide gives only itself back on hydrolysis — there is nothing simpler to split off. - An oligosaccharide (e.g., sucrose) hydrolyses into a small, fixed number (2–10) of monosaccharide units. - A polysaccharide (e.g., starch, cellulose) hydrolyses into a very large number of monosaccharide units. So a monosaccharide sits at the bottom of that ladder. Structurally, every monosaccharide fits one description: A monosaccharide is a polyhydroxy aldehyde or a polyhydroxy ketone — a single chain of carbon atoms carrying one carbonyl group (either an aldehyde, - CHO, or a ketone, C=O ) and two or more hydroxyl ( - OH ) groups on the remaining carbons. This gives two ways of sub-classifying them, and real monosaccharides are named using both together: 1. By the carbonyl group. If the carbonyl is an aldehyde ( - CHO at the end of the chain), the sugar is an aldose. If the carbonyl is a ketone ( C=O within the chain), it is a ketose. 2. By chain length. A 3-carbon sugar is a triose, 4-carbon a tetrose, 5-carbon a pentose, 6-carbon a hexose, and so on. Combining the two: glucose has an aldehyde group and six carbons, so it is an aldohexose; fructose has a ketone group and six carbons, so it is a ketohexose; ribose has an aldehyde group and five carbons, so it is an aldopentose. Writing the two most familiar examples as condensed formulas (naming every symbol as it appears): Glucose, molecular formula C 6H 12O 6, condensed structure CHO - CHOH - CHOH - CHOH - CHOH - CH 2OH Here the leftmost CHO is the aldehyde (carbonyl) carbon, the four middle CHOH units are the hydroxyl-bearing carbons, and the terminal CH 2OH is a primary alcohol carbon. One aldehyde group plus four hydroxyl groups on a six-carbon chain — an aldohexose. Fructose, also C 6H 12O 6, condensed structure CH 2OH - CHOH - CHOH - CHOH - CO - CH 2OH Here the carbonyl ( CO ) sits at carbon 2, flanked by carbons on both sides, so it is a ketone rather than an aldehyde — a ketohexose. Neither of these molecules loses a smaller carbohydrate fragment on hydrolysis: whatever conditions are applied, glucose and fructose come back unchanged, which is the defining test for a monosaccharide. About 20 monosaccharides occur naturally; besides glucose and fructose, common ones include ribose and 2-deoxyribose (both aldopentoses, and components of RNA and DNA respectively) and galactose (an aldohexose, obtained on hydrolysing lactose). Answer: A monosaccharide is a carbohydrate that cannot be hydrolysed further into a simpler sugar — a single polyhydroxy aldehyde (an aldose, e.g., glucose, C 6H 12O 6 ) or polyhydroxy ketone (a ketose, e.g., fructose, C 6H 12O 6 ) carrying one carbonyl group and two or more - OH groups on a single carbon chain.

Question 2

What are reducing sugars?

Accepted Answer

A sugar is called "reducing" if it can hand its carbonyl group to another reagent and get itself oxidized in the process — which means it needs a free aldehyde ( - CHO ) or a free alpha-hydroxy ketone ( - CO - ) somewhere in the molecule that is not tied up in a bond. A reducing sugar is a carbohydrate that reduces Tollens' reagent (ammoniacal AgNO 3, the complex ion [ Ag(NH 3) 2]^+ ) to metallic silver, and reduces Fehling's solution (an alkaline Cu^2+ -tartrate complex) to a brick-red precipitate of Cu 2 O. In both tests the sugar itself is oxidized — its carbonyl carbon is pushed up to a carboxylic acid — while the metal ion is reduced. Whether a given sugar can do this depends entirely on whether it has a free carbonyl (or the free anomeric - OH that can open up into one) left in the molecule. Why every monosaccharide qualifies Glucose, CH 2 OH -( CHOH ) 4- CHO, is an aldohexose: it already carries a free aldehyde group at C 1 in its open-chain form, and in solution the cyclic hemiacetal form sits in equilibrium with this open aldehyde form through the free anomeric - OH. That aldehyde is what Tollens' reagent oxidizes: CH 2 OH -( CHOH ) 4- CHO + 2[ Ag(NH 3) 2]^+ + 3 OH^- → CH 2 OH -( CHOH ) 4- COO^- + 2 Ag + 4 NH 3 + 2 H 2 O The aldehyde carbon of glucose is oxidized to a carboxylate (the product is gluconate, the anion of gluconic acid ), and the silver ion is reduced all the way to silver metal, which deposits as the "silver mirror" on the inside of the test tube. Fehling's solution runs the same oxidation on the aldehyde, with Cu^2+ being reduced to Cu^+, which precipitates as red Cu 2 O. Fructose, CH 2 OH -( CHOH ) 3- CO - CH 2 OH, has no aldehyde at all — it is a ketohexose, with the carbonyl at C2. It still gives a positive Fehling's/Tollens' test, and this is the point people get wrong: under the alkaline conditions of these tests, the alpha-hydroxy ketone at C1–C2 tautomerizes through an enediol intermediate (base pulls a proton off the carbon next to the - OH, the double bond shifts, and re-protonation can put the carbonyl back down at C 1 instead of C2) into an aldose structure, which is then oxidized exactly as glucose is. So the test does not tell you "aldose vs ketose" — it only tells you whether the sugar is reducing or not, and every monosaccharide, aldose or ketose, is reducing because each one has a free anomeric carbon that can open to expose a carbonyl. Why it stops being automatic for disaccharides In a disaccharide, two monosaccharide rings are joined by a glycosidic bond formed between an - OH on one ring's anomeric carbon and an - OH on the other ring. Whether the disaccharide is reducing depends on whether that bond uses up both anomeric carbons or leaves one free: - Maltose (two glucose units joined α -1,4) uses the anomeric carbon (C 1) of only one glucose unit to make the glycosidic bond. The second glucose unit's anomeric carbon (its own C1) is left free, still able to open into a free aldehyde in solution. So maltose reduces Tollens'/Fehling's — it is a reducing sugar. - Lactose (galactose joined to glucose, β -1,4) is built the same way: the glycosidic bond uses galactose's anomeric carbon, but glucose's anomeric carbon (C 1) is left free. Lactose is also a reducing sugar. - Sucrose is different: the glycosidic bond in sucrose is formed between the anomeric carbon of glucose (C 1) and the anomeric carbon of fructose (C 2) — both ends of the bridge are anomeric carbons, so neither monosaccharide unit has a free anomeric carbon left. There is no - OH left that can open into a free aldehyde or free ketone, so sucrose cannot be oxidized by Tollens' reagent or Fehling's solution. Sucrose is the standard example of a non-reducing sugar. So the working rule is: look for a free anomeric carbon (a free hemiacetal/hemiketal - OH, equivalent to a free - CHO or alpha-hydroxy - CO - in the open-chain form). If at least one exists, the sugar is reducing; if the glycosidic bond consumes every anomeric carbon in the molecule, it is non-reducing. Answer: Reducing sugars are carbohydrates that reduce Tollens' reagent to metallic silver and Fehling's solution to red Cu 2 O, because they possess a free aldehyde group or a free (alpha-hydroxy) ketone group — i.e., a free anomeric carbon that can open into a carbonyl in solution. All monosaccharides (e.g., glucose, fructose) are reducing sugars, and among disaccharides those with at least one free anomeric carbon (e.g., maltose, lactose) are reducing, while sucrose, whose glycosidic bond ties up both monosaccharide units' anomeric carbons, is a non-reducing sugar.

Question 3

Write two main functions of carbohydrates in plants.

Accepted Answer

Carbohydrates in a plant are not fuel and scaffold in the abstract — two specific molecules do those two specific jobs: starch stores energy, cellulose builds structure. A plant does not use one carbohydrate for everything. It builds two different polymers, both made of the same glucose unit, C 6 H 12 O 6, but linked differently — and that difference in linkage is what separates a stored fuel from a structural material. Function 1 — Storage of energy (as starch). Plants make glucose by photosynthesis faster than they can use it immediately, so the surplus is stored as starch, a polymer of α -D-glucose units joined by α -1,4-glycosidic linkages (with α -1,6 branch points in the amylopectin fraction). Starch is laid down in seeds, tubers, and roots — the grain in a wheat seed or the tuber of a potato is starch reserve. Because the glycosidic bond is α, the chain coils into a helical shape that packs compactly and is easily hydrolysed back to glucose by amylase enzymes when the plant needs energy — for germination of a seed, for instance. This is the carbohydrate reserve function. Function 2 — Structural support (as cellulose). Cellulose is also a polymer of glucose, but the units are β -D-glucose joined by β -1,4-glycosidic linkages. This single change from α to β linkage changes the shape of the polymer from a coil to a long, straight, unbranched chain. These straight chains lie side by side and hydrogen-bond extensively to one another, bundling into rigid fibres. Cellulose is the chief constituent of the cell wall in plants, giving the cell — and hence the whole plant — its mechanical strength and rigidity. It is this framework that lets a plant stand upright without a skeleton. The reason one plant polysaccharide can be food reserve and the other can be a building material, even though both hydrolyse to the same glucose, is the stereochemistry of the glycosidic bond ( α vs β ): α -links give a compact, digestible coil (starch, for storage); β -links give a rigid, mechanically strong fibre (cellulose, for structure). Answer: The two main functions of carbohydrates in plants are (1) storage of food/energy reserves, in the form of starch, and (2) providing structural support to the cell wall, in the form of cellulose.

Question 4

Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

Accepted Answer

A carbohydrate is a monosaccharide only if it cannot be split into anything simpler by acid hydrolysis — a disaccharide, by contrast, breaks into exactly two monosaccharide units joined by a glycosidic linkage. So the way to classify each of the six names is to ask: on hydrolysis, does this molecule split into two smaller sugar units, or is it already the simplest unit? Going through each one: Ribose — an aldopentose, molecular formula C 5 H 10 O 5, with the open-chain structure CHO-CHOH-CHOH-CHOH-CH 2 OH. It has a single ring/chain of five carbons with an aldehyde at C 1 and a hydroxyl on every other carbon. There is no glycosidic bond joining two smaller units, so it cannot be hydrolysed further. → monosaccharide. 2-Deoxyribose — also an aldopentose, formula C 5 H 10 O 4, structure CHO-CH 2 -CHOH-CHOH-CH 2 OH. It is ribose with the oxygen at C 2 removed (hence "deoxy"), which is why it has one oxygen fewer than ribose. It is still a single five-carbon sugar unit, not a combination of two. → monosaccharide. Galactose — an aldohexose, formula C 6 H 12 O 6, structure CHO-CHOH-CHOH-CHOH-CHOH-CH 2 OH (an epimer of glucose, differing only in the configuration at C4). One six-carbon chain, no glycosidic linkage to hydrolyse. → monosaccharide. Fructose — a ketohexose, formula C 6 H 12 O 6, structure CH 2 OH-CO-CHOH-CHOH-CHOH-CH 2 OH, carrying a ketone at C 2 instead of an aldehyde at C1. Again a single six-carbon unit. → monosaccharide. Maltose — formula C 12 H 22 O 11. On acid or enzymatic (maltase) hydrolysis it splits into two molecules of glucose, joined originally through an α -1,4-glycosidic linkage between C 1 of one glucose unit and C 4 of the other: maltose + H 2 O → glucose + glucose Because hydrolysis gives two monosaccharide units, not one, this is the defining test for a disaccharide. → disaccharide (glucose + glucose). Lactose — also C 12 H 22 O 11, the sugar of milk. On hydrolysis (lactase) it splits into one molecule of galactose and one of glucose, joined by a β -1,4-glycosidic linkage: lactose + H 2 O → galactose + glucose Two monosaccharide units come out, so this too is a disaccharide — note it is the one case here where the two units are different sugars (galactose and glucose), unlike maltose where both units are glucose. → disaccharide (galactose + glucose). Collecting the results: Answer: Monosaccharides — ribose, 2-deoxyribose, galactose, fructose (each a single, non-hydrolysable 5- or 6-carbon sugar unit). Disaccharides — maltose (hydrolyses to glucose + glucose) and lactose (hydrolyses to galactose + glucose), each held together by a glycosidic linkage that acid or enzymatic hydrolysis breaks.

Question 5

What do you understand by the term glycosidic linkage?

Accepted Answer

A glycosidic linkage is the C–O–C bridge that joins the anomeric carbon of one sugar ring to a hydroxyl carbon of a second sugar (or other group), formed by kicking out one molecule of water — it is what stitches monosaccharides into disaccharides and polysaccharides. Start from the cyclic (Haworth) form of a monosaccharide. When an open-chain aldose or ketose closes into its ring, the carbonyl carbon — C 1 in an aldose such as glucose, C 2 in a ketose such as fructose — picks up an –OH from the ring oxygen and becomes a new stereocentre that carries both an –OH group and the ring –O– on the same carbon. This carbon is called the anomeric carbon, and because it still behaves like the carbonyl-derived –OH of a hemiacetal, that particular –OH is far more reactive than the other ring –OH groups. A hemiacetal –OH can react further with an alcohol to give a full acetal, and that is exactly what happens between two sugar molecules: the anomeric –OH of one monosaccharide condenses with an –OH group (often, but not always, on another anomeric carbon) of a second monosaccharide. One molecule of H 2 O is eliminated, and in its place a new bond forms in which an oxygen atom bridges a carbon of the first sugar to a carbon of the second sugar: Sugar 1 –OH + HO–Sugar 2 → Sugar 1 –O–Sugar 2 + H 2 O This C–O–C bridge, built from the anomeric carbon of at least one of the two sugars, is the glycosidic linkage. It is the sugar-chemistry counterpart of the peptide (amide) linkage between amino acids or the phosphodiester linkage between nucleotides — a condensation bond that links repeating units into a bigger molecule. Two worked examples fix the idea: Maltose is built from two units of α -D-glucose. The anomeric carbon (C 1) of the first glucose unit condenses with the –OH on C 4 of the second glucose unit, releasing H 2 O and leaving a C1–O–C4 bridge. Because the first ring's C 1 was in the α configuration, this bond is named an α -1,4-glycosidic linkage, written Glc (α1→4) Glc. The second glucose unit still has a free anomeric –OH, so maltose is a reducing sugar. Sucrose is built from α -D-glucose and β -D-fructose, but here both partners contribute their anomeric carbons: C 1 of glucose bonds through oxygen to C 2 of fructose. Because both anomeric carbons are used up in the same linkage, there is no free hemiacetal –OH left anywhere in the molecule — this is why sucrose cannot reduce Fehling's or Tollens' reagent and is called a non-reducing sugar. This linkage is described as α1↔β2 -glycosidic (a "head-to-head" linkage, unlike maltose's "head-to-tail" one). The same bond, repeated thousands of times, is what builds polysaccharides: amylose (a component of starch) is a long chain of glucose units joined by α -1,4-glycosidic linkages, while cellulose is a long chain of glucose units joined instead by β -1,4-glycosidic linkages — the change from α to β at each glycosidic linkage is the entire chemical difference between a digestible starch and structural cellulose. Answer: A glycosidic linkage is the oxygen bridge (–O–) that joins the anomeric carbon of one monosaccharide to a hydroxyl-bearing carbon of another monosaccharide, formed by loss of one water molecule; it is the bond that links sugar units together in disaccharides (e.g., the α -1,4-glycosidic linkage in maltose, the α1↔β2 -glycosidic linkage in sucrose) and, repeated many times over, in polysaccharides such as starch and cellulose.

Question 6

What is glycogen? How is it different from starch?

Accepted Answer

Glycogen is the glucose-storage polysaccharide of animals -- "animal starch" -- and it differs from plant starch in exactly one structural respect: it is far more heavily branched. What glycogen is Glycogen is a homopolysaccharide built entirely of α -D-glucose units. It is the form in which animals (and fungi) store carbohydrate, deposited mainly in the liver and in skeletal muscle. Whenever blood glucose rises above what the cells need immediately, the excess is polymerised into glycogen; when glucose is needed later, glycogen is broken back down and the glucose is released into the blood. In this role it is the exact animal counterpart of starch in plants -- both are simply a compact, low-osmotic-pressure way to warehouse many glucose units under one molecule. Structurally, glycogen is a branched chain of glucose units: - The main chains are held together by α(1 → 4) glycosidic linkages -- the bond runs from carbon-1 of one glucose ring (in the alpha configuration) to the oxygen on carbon-4 of the next. - Branches are thrown off by α(1 → 6) glycosidic linkages -- carbon-1 of the branch's first glucose bonds to the oxygen on carbon-6 of a glucose in the main chain. - These branch points occur very frequently, roughly every 8 to 10 glucose units along the chain. What starch is, for comparison Starch is also a glucose homopolysaccharide, but it is not a single structure -- it is a mixture of two components: - Amylose (about 15-20% of starch): a long, unbranched chain of glucose units joined only by α(1 → 4) linkages. Because it is unbranched it coils into a helix and is the component that gives the blue colour with iodine. - Amylopectin (about 80-85% of starch): a branched chain, again α(1 → 4) linked along the main chains with α(1 → 6) linkages at the branch points -- the same two bond types as glycogen -- but the branch points here are much sparser, occurring only about every 25 to 30 glucose units. Naming the difference Both glycogen and amylopectin use identical chemistry -- glucose units, α(1 → 4) main-chain bonds, α(1 → 6) branch bonds -- so the distinguishing feature is not the type of bond but how often the branch bond occurs, plus the fact that starch also contains an unbranched component that glycogen has no counterpart to: 1. Occurrence: glycogen is found in animal cells (liver and muscle); starch is found in plant cells (seeds, tubers, roots). 2. Composition: starch is a two-part mixture, amylose (unbranched) plus amylopectin (branched); glycogen has only one component, and it is branched throughout. 3. Degree of branching: glycogen branches roughly every 8-10 glucose units, making it markedly more branched than amylopectin, which branches only every 25-30 units. This is the key structural distinction the question is asking for. 4. Consequence of the branching: the denser branching of glycogen exposes many more chain ends at once, so enzymes (glycogen phosphorylase) can release glucose from many points simultaneously -- appropriate for an animal's fast, on-demand energy release, whereas starch's sparser branching suits the slower mobilisation needs of a plant. Answer: Glycogen is the branched, glucose-only storage polysaccharide of animals (liver and muscle), built from glucose units joined by α(1 → 4) glycosidic bonds along the chains and α(1 → 6) glycosidic bonds at the branch points. It differs from starch in that starch is a mixture of unbranched amylose and branched amylopectin, whereas glycogen is entirely branched and, compared with amylopectin, far more densely so -- a branch point every 8-10 glucose units in glycogen versus every 25-30 units in amylopectin.

Question 7

What are the hydrolysis products of (i) sucrose and (ii) lactose?

Accepted Answer

Hydrolysis breaks the glycosidic bond that joins two monosaccharide units, and each disaccharide gives back the two simple sugars it was built from. (i) Sucrose Sucrose, C 12 H 22 O 11, is built from one molecule of α -D-glucose and one molecule of β -D-fructose. The glycosidic bond connects C-1 of glucose to C-2 of fructose — and both of these are the anomeric carbons of their respective sugars. Because both anomeric carbons are tied up in the linkage, sucrose has no free anomeric carbon and is a non-reducing sugar. On hydrolysis with dilute acid, or with the enzyme invertase, water adds across this glycosidic bond and the ring is opened at the linkage: C 12 H 22 O 11 ( sucrose ) + H 2 O → C 6 H 12 O 6 ( glucose ) + C 6 H 12 O 6 ( fructose ) So the two hydrolysis products are one molecule of glucose (an aldohexose, condensed formula CH 2 OH-(CHOH) 4 -CHO ) and one molecule of fructose (a ketohexose, condensed formula CH 2 OH-(CHOH) 3 -CO-CH 2 OH ). A step people miss: sucrose itself is dextrorotatory, but the glucose + fructose mixture formed is laevorotatory, because fructose's laevorotation is large enough to override glucose's dextrorotation. This reversal of optical rotation on hydrolysis is exactly why the product mixture is called invert sugar, and why the enzyme is named invertase. (ii) Lactose Lactose, C 12 H 22 O 11, is built from one molecule of β -D-galactose and one molecule of β -D-glucose, joined through a β -1,4-glycosidic linkage: the anomeric carbon (C-1) of galactose is bonded to C-4 of glucose. Here the anomeric carbon of the glucose unit is left free, which is why lactose (unlike sucrose) is a reducing sugar. On hydrolysis with dilute acid, or with the enzyme lactase, the β -1,4 linkage is cleaved: C 12 H 22 O 11 ( lactose ) + H 2 O → C 6 H 12 O 6 ( D-galactose ) + C 6 H 12 O 6 ( D-glucose ) So the two hydrolysis products are one molecule of D-galactose and one molecule of D-glucose, both aldohexoses with the condensed formula CH 2 OH-(CHOH) 4 -CHO (they differ only in the spatial arrangement of the –OH at C-4). Answer: Sucrose hydrolyses to one molecule of glucose + one molecule of fructose (the resulting laevorotatory mixture is called invert sugar). Lactose hydrolyses to one molecule of glucose + one molecule of galactose.

Question 8

What is the basic structural difference between starch and cellulose?

Accepted Answer

The difference is not in the building block — both are made entirely of glucose — but in how the glucose units are hooked together: starch uses α -glycosidic linkages and folds into a coiled, partly branched chain; cellulose uses only β -glycosidic linkages and stays a straight, unbranched chain. Both starch and cellulose are polysaccharides: long chains built by joining thousands of glucose ( C 6H 12O 6 ) units end to end through a glycosidic bond, which forms between the C 1 - OH of one glucose ring and the C 4 - OH of the next, with loss of a water molecule at each junction. Where the two polymers part ways is the stereochemistry of that C 1 carbon (the anomeric carbon) and whether the chain ever branches. Starch is built from α -D-glucose, and it is not one single structure but a mixture of two components. - Amylose (about 15–20% of starch) is the unbranched part: a long chain of α -D-glucose units joined only by α -1,4-glycosidic linkages ( C 1 of one ring to C 4 of the next, with the C 1 - OH oriented below the plane of the ring, which is what " α " means here). This α linkage puts a slight kink between successive rings, so the chain does not lie flat — it curls into a helix. - Amylopectin (about 80–85% of starch) is the branched part: the backbone is joined the same way, by α -1,4-glycosidic linkages, but roughly every 20–25 glucose units a side chain is attached through an α -1,6-glycosidic linkage ( C 1 of one chain to C 6 of a glucose unit already in the backbone). That C 6 attachment is what creates the branch points, giving amylopectin a tree-like, branched shape. Cellulose is built from β -D-glucose, and it has only one kind of chain. The glucose units are joined exclusively by β -1,4-glycosidic linkages — the same C 1 -to- C 4 connection as amylose, but now with the C 1 - OH oriented above the plane of the ring (" β "). This β linkage forces each glucose unit to sit rotated 180^° relative to its neighbor, which lets successive units line up flat instead of curling. The result is a straight, completely unbranched chain — there is no C 6 branching linkage anywhere in cellulose. Many such flat chains then pack side by side and are locked together by extensive hydrogen bonding between adjacent chains, which is why cellulose forms rigid, fibrous bundles (the structural material of plant cell walls) rather than the compact, coiled granules that starch forms. So the basic structural difference reduces to two points: (1) the anomeric configuration of the glycosidic linkage — α throughout starch versus β throughout cellulose — and (2) branching — starch (via its amylopectin component) carries α -1,6 branch points, while cellulose is a single, linear, unbranched chain of glucose units held together only by β -1,4 linkages. This is also why the two behave so differently in the body: digestive enzymes such as amylase can hydrolyze the α -linkages in starch, so starch is a usable energy store, while human enzymes cannot cleave the β -linkages in cellulose, so cellulose passes through as dietary fibre instead. Answer: Starch and cellulose are both glucose polymers, but starch (amylose + amylopectin) is held together by α -D-glucose linkages — α -1,4-glycosidic bonds in the chain and α -1,6-glycosidic bonds at branch points — giving a coiled, partly branched structure, whereas cellulose is held together only by β -1,4-glycosidic bonds between β -D-glucose units, giving a straight, completely unbranched chain.

Question 9

What happens when D-glucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO 3

Accepted Answer

Bromine water only oxidises the aldehyde end of glucose, while dilute HNO 3 oxidises both ends of the chain — and it is exactly this difference in how far each reagent reaches that tells you what groups sit at each end of the glucose molecule. Start from the open-chain structure of D-glucose, written out carbon by carbon: CHO - CHOH - CHOH - CHOH - CHOH - CH 2 OH Here C 1 is an aldehyde carbon ( -CHO ), C 2 through C 5 each carry a secondary alcohol ( -CHOH- ), and C 6 is a primary alcohol carbon ( -CH 2OH ). The three reagents in this question are exactly the classical tests used to establish this structure — each one probes a different part of it. (i) HI (excess, prolonged heating) HI, used with red phosphorus on strong heating, is a powerful reducing agent. It removes every oxygen-bearing group in the chain: the -CHO at C 1 and every -OH at C2–C6 are each replaced by -H (the alcohols go -OH → -I → -H; the aldehyde is reduced the same way after first becoming -CH 2OH ). Because no carbon is left bonded to oxygen, and no new carbon–carbon bond is made or broken, the six carbons of glucose come out as a plain saturated hydrocarbon: CH 2 OH -( CHOH ) 4- CHO [Δ] HI (excess) CH 3- CH 2- CH 2- CH 2- CH 2- CH 3 The product is n-hexane — the straight-chain, unbranched isomer of C 6H 14, not 2-methylpentane or any other branched hexane. Since reduction cannot rearrange the carbon skeleton, the only way to get n-hexane out is for all six carbons of glucose to already have been joined in one continuous, unbranched chain before the reaction. This is the evidence that glucose has a straight-chain carbon skeleton. (ii) Bromine water Bromine water is a mild oxidising agent (it acts as a source of hypobromous acid, HOBr ). Under these mild conditions it oxidises an aldehyde group to a carboxylic acid but leaves alcohol groups — primary or secondary — untouched, and it does not oxidise ketones at all. Applied to glucose, only the C 1 end reacts: CH 2 OH -( CHOH ) 4- CHO Br 2/H 2O CH 2 OH -( CHOH ) 4- COOH The product is D-gluconic acid, a mono-carboxylic acid: only C 1 has changed, from -CHO to -COOH; C 6 is still -CH 2OH. That glucose is oxidised at all by this mild reagent is the key point — a ketone would not react with bromine water — so this reaction is the evidence that the carbonyl group in glucose is an aldehyde ( -CHO ) and not a ketone. (iii) Dilute HNO 3 Dilute nitric acid is a stronger oxidising agent than bromine water, and it reaches both ends of the chain: the aldehyde -CHO at C 1 is oxidised to -COOH (as bromine water already showed happens), and in addition the primary alcohol -CH 2OH at C 6 is oxidised to -COOH as well. The four secondary -CHOH- groups at C2–C5 are, as before, left alone. CH 2 OH -( CHOH ) 4- CHO HNO 3 COOH -( CHOH ) 4- COOH The product is saccharic acid (D-glucaric acid), a dicarboxylic acid with both chain ends now -COOH, still with six carbons total and no branching. Only a primary alcohol can be oxidised this far to a carboxylic acid (a secondary -CHOH- would at most be oxidised to a ketone, and the four middle carbons in fact show no such change here); a tertiary carbon could not be oxidised to -COOH at all. So the fact that C 6 converts cleanly to -COOH under this stronger oxidant is the evidence that C 6 in glucose carries a primary alcohol group, -CH 2OH. Answer: (i) HI (prolonged heating) reduces D-glucose to n-hexane, CH 3(CH 2) 4CH 3, showing the six carbons form a straight, unbranched chain. (ii) Bromine water oxidises only the -CHO at C 1 to give D-gluconic acid, CH 2OH(CHOH) 4COOH, showing the carbonyl group in glucose is an aldehyde, not a ketone. (iii) Dilute HNO 3 oxidises both the -CHO at C 1 and the -CH 2OH at C 6 to give saccharic (D-glucaric) acid, COOH(CHOH) 4COOH, showing that C 6 carries a primary alcohol group.

Question 10

Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

Accepted Answer

Three facts about D-glucose refuse to fit a plain, straight-chain aldehyde picture — they only make sense if the molecule closes itself into a ring. Write the open-chain structure of D-glucose first — — so the mismatch is visible: C 1 HO - C 2 HOH - C 3 HOH - C 4 HOH - C 5 HOH - C 6 H 2 OH Here C 1 is drawn as a free aldehyde ( - CHO ), and C2–C5 each carry a secondary - OH, with a primary - OH at C6. If this open-chain formula were the whole truth, glucose should behave, at every moment, exactly like any other aldehyde. It does not. Reaction 1 — Schiff's test fails. An aldehyde restores the pink/magenta colour of Schiff's reagent. D-glucose does not do this, even though it undergoes HCN addition (giving a cyanohydrin) and reacts with hydroxylamine, NH 2 OH, to give an oxime — both of which need a carbonyl carbon. So glucose behaves as if - CHO is present for some reagents and absent for others — a straight open chain cannot give two different answers to the same question. Reaction 2 — no bisulphite addition product. Aldehydes and methyl ketones add sodium hydrogensulphite, NaHSO 3, across the carbonyl to give a crystalline addition compound. Glucose does not form this addition product. Again inconsistent with a permanently free - CHO at C1. Reaction 3 — the pentaacetate will not react with hydroxylamine. Acetic anhydride converts all five - OH groups of glucose into acetate esters, giving glucose pentaacetate. Five acetylated oxygens is exactly what the open-chain formula predicts (four secondary - OH at C2–C5 plus the primary - OH at C6), and that part is fine. But the open-chain formula also predicts a sixth, untouched functional group — the - CHO at C 1 — which acetic anhydride cannot acetylate and which should therefore still be free to react with NH 2 OH to give an oxime. It does not react. That means there is no free - CHO left in the pentaacetate at all — only five oxygens capable of forming esters exist on the whole molecule, not four - OH plus one - CHO. Reaction 4 — two crystalline forms, and mutarotation. Crystallising D-glucose under different conditions gives two distinct solids: α -D-glucose, melting point 419 K, specific rotation [α] D = +111^°, and β -D-glucose, melting point 423 K, specific rotation [α] D = +19.2^°. A single open-chain structure has one arrangement of atoms and can give only one melting point and one specific rotation — it cannot explain two different solids with the same molecular formula, C 6 H 12 O 6. Worse, when either pure form is dissolved in water, its rotation does not stay fixed: it drifts gradually until both solutions settle at the same value, [α] D = +52.7^°. This gradual drift to a common value is called mutarotation, and an open chain — with no extra stereocentre to interconvert around — has no mechanism to produce it. Why the ring explains all four. The resolution is that the - OH on C 5 attacks the carbonyl carbon, C 1, intramolecularly. The C 5 oxygen's lone pair attacks the electrophilic carbonyl carbon of the - CHO; the C=O π bond breaks, its electrons moving onto the carbonyl oxygen; a new C1–O(ring) bond forms, and a proton transfer converts that oxygen into part of a six-membered ring while leaving a new - OH sitting on C1. This closes glucose into a six-membered cyclic hemiacetal, the pyranose form, written as a ring of O–C1–C2–C3–C4–C5 with the - CH 2 OH hanging off C5. C 1 is no longer a carbonyl carbon — it is a hemiacetal carbon, bonded to one ring oxygen, one - OH, one - H, and C2. A hemiacetal carbon does not reduce Schiff's reagent and does not add NaHSO 3 the way a free aldehyde does, which is exactly Reactions 1 and 2 accounted for. In the pentaacetate, this C 1 - OH is simply the fifth - OH that gets acetylated along with the other four — there is no separate sixth group left over for hydroxylamine to attack, which is Reaction 3 accounted for. Because C 1 was not a stereocentre in the open chain (it only had two different substituents, =O and H) but becomes one on ring closure (now bonded to four different groups: ring-O, OH, H, C2), two new arrangements become possible at C 1: the new - OH can end up on the same side as the reference - OH used to assign D/L configuration ( β -D-glucose) or on the opposite side ( α -D-glucose). These are the two crystalline anomers of Reaction 4, differing only at C 1 — which is why they have different melting points and different rotations despite being the "same" molecule everywhere else. In solution, the ring can transiently reopen back through the same hemiacetal step in reverse — the C1–O(ring) bond breaks, regenerating the free - CHO and the C 5 - OH for an instant — and then recloses randomly as either anomer. Repeating this many times drives both pure α and pure β solutions to the same equilibrium mixture (about 64% β, 36% α, with a negligible trace of open chain), which is why both land on the identical rotation, +52.7^° — mutarotation. Answer: The reactions of D-glucose that an open-chain structure cannot explain are (i) the absence of Schiff's test, (ii) the absence of a NaHSO 3 bisulphite addition product, (iii) the failure of glucose pentaacetate to react with hydroxylamine (showing no free - CHO survives once the five - OH groups are acetylated), and (iv) the existence of two crystalline anomers, α -D-glucose (mp 419 K, [α] D=+111^° ) and β -D-glucose (mp 423 K, [α] D=+19.2^° ), whose rotations both drift on dissolving to a common equilibrium value of +52.7^° (mutarotation). All four are explained by D-glucose existing predominantly as a six-membered cyclic hemiacetal (pyranose ring, formed between the C5–OH and the C1–CHO), in which C 1 becomes a new stereocentre bearing - OH and - H instead of a free carbonyl.

Question 11

What are essential and non-essential amino acids? Give two examples of each type.

Accepted Answer

"Essential" and "non-essential" describe where the amino acid comes from, not how important it is to the body -- every one of the twenty is needed; only some of them the body can build for itself. An amino acid is a molecule with an amino group -NH 2 and a carboxyl group -COOH attached to the same carbon (the α -carbon), along with a side chain (R group) that differs from one amino acid to the next. Proteins are built by joining amino acids through peptide bonds, and the human body needs a supply of all twenty standard amino acids to keep making its own proteins. Non-essential amino acids. These are the amino acids the human body can synthesise on its own, inside its cells, starting from simpler precursors (other amino acids, or intermediates of carbohydrate and fat metabolism). Because the body can make as much as it needs, these amino acids do not have to be supplied through food -- a diet completely lacking one of them still does not cause a deficiency, since the body compensates by synthesising it. Two examples of non-essential amino acids: - Glycine, H 2 N-CH 2 -COOH -- the simplest amino acid, with a hydrogen atom as its side chain. - Alanine, CH 3 -CH(NH 2 )-COOH -- with a methyl side chain. (Aspartic acid and glutamic acid are two more valid examples; the textbook accepts any correctly identified pair.) Essential amino acids. These are the amino acids the human body cannot synthesise at all, or cannot synthesise fast enough to meet its needs, because it lacks the enzymes for the necessary biosynthetic steps. They must therefore be obtained ready-made from the diet -- from proteins in food that are digested down to free amino acids and absorbed. Leaving an essential amino acid out of the diet for a sustained period causes a genuine deficiency, because the body has no internal route to replace it. Two examples of essential amino acids: - Valine, (CH 3 ) 2 CH-CH(NH 2 )-COOH -- with an isopropyl side chain. - Leucine, (CH 3 ) 2 CH-CH 2 -CH(NH 2 )-COOH -- with an isobutyl side chain. (Isoleucine, lysine, threonine, tryptophan, phenylalanine, and methionine are the other essential amino acids in the standard list; any correctly identified pair is acceptable.) The practical consequence of this distinction is nutritional: a diet must supply the essential amino acids directly, because no amount of the other nineteen amino acids or of general protein intake lets the body build valine or leucine from scratch -- only non-essential amino acids can be substituted for by the body's own biosynthesis. Answer: Non-essential amino acids (e.g., glycine, alanine) can be synthesised by the human body itself and so need not come from the diet; essential amino acids (e.g., valine, leucine) cannot be synthesised by the body and must be obtained from food.

Question 12

Define the following as related to proteins (i) Peptide linkage (ii) Primary structure (iii) Denaturation.

Accepted Answer

Each of these three terms describes proteins at a different level — the bond that links two amino acids, the sequence that results, and what happens when the folded chain is disturbed. (i) Peptide linkage A peptide linkage is the amide bond, - CO - NH -, formed between the - COOH group of one amino acid and the - NH 2 group of the next, with the elimination of one molecule of water. Take two amino acids, glycine, H 2 N - CH 2- COOH, and alanine, H 2 N - CH ( CH 3)- COOH. The - OH of glycine's carboxyl group and one - H of alanine's amino group leave together as H 2 O, and the carbon of the former - COOH bonds directly to the nitrogen of the former - NH 2: H 2 N - CH 2- COOH + H 2 N - CH ( CH 3)- COOH → H 2 N - CH 2- CO - NH - CH ( CH 3)- COOH + H 2 O The product, H 2 N - CH 2- CO - NH - CH ( CH 3)- COOH, is the dipeptide glycylalanine (Gly-Ala); the - CO - NH - group joining the two residues is the peptide linkage. This is a condensation reaction — the small molecule lost is water — and it is the only kind of bond that links amino acid residues along a protein chain. A protein of n amino acid residues therefore contains (n-1) peptide linkages, since one water molecule is lost at each junction and the two end residues keep one free - NH 2 and one free - COOH group respectively. (ii) Primary structure A protein may be built from one or several polypeptide chains. Within any one chain, the amino acid residues are joined, residue after residue by peptide linkages, in one specific linear sequence — this exact sequence, read from the free-amino (N-terminal) end to the free-carboxyl (C-terminal) end, is the primary structure of the protein. For example, if a chain has the sequence glycine–alanine–valine–…, its primary structure is written Gly-Ala-Val-…, and this order is fixed for a given protein: even one residue out of place is a different protein. The primary structure carries no information about how the chain twists or folds in space — it is only the "letters in order," analogous to the sequence of words in a sentence before any grammar (folding) is applied. Every higher level of protein architecture (the coiled or pleated secondary structure, the 3-D folded tertiary structure, and the multi-chain quaternary structure) is a consequence of this one underlying sequence. (iii) Denaturation Denaturation is the loss of a protein's biological (native) activity that occurs when its secondary and tertiary structures are disrupted by a physical change (such as heating) or a chemical change (such as a change in pH), while the primary structure — the sequence of peptide-linked residues — remains intact. In its native state a protein exists as one of two structural forms: a fibrous form (chains held parallel by hydrogen bonds and cross-links) or a globular form (chains coiled around themselves and folded into a compact, roughly spherical shape). This native, folded shape is what allows a globular protein such as an enzyme to carry out its function. When the protein is heated or treated with acid or base, the hydrogen bonds that hold the coiled/folded (secondary and tertiary) structure in place are broken. The globules unfold and the chains become disorganised — the protein is denatured. A familiar example is boiling an egg: the soft, soluble albumin (egg white protein) unfolds on heating and becomes a solid, insoluble mass. Because only the folding is destroyed and the peptide-bonded sequence of the chain is unaffected, the primary structure of a denatured protein is the same as that of the native protein, but its biological activity is lost. Answer: (i) Peptide linkage — the amide bond - CO - NH - formed between - COOH of one amino acid and - NH 2 of another with loss of H 2 O (e.g., Gly + Ala → Gly-Ala + H 2 O ). (ii) Primary structure — the fixed, linear sequence in which amino acid residues are joined by peptide linkages in a polypeptide chain. (iii) Denaturation — loss of a protein's biological activity when heat, acid, or base disrupts the hydrogen bonds holding its secondary/tertiary (folded) structure, unfolding globules into disorganised chains, without breaking the primary (peptide-bonded) sequence.

Question 13

What are the common types of secondary structure of proteins?

Accepted Answer

Secondary structure is not about which amino acids are present — it is about how the backbone folds, held together only by hydrogen bonds between the peptide -C=O and -N-H groups. There are two common patterns: the alpha-helix and the beta-pleated sheet. Start from what a peptide bond actually offers for hydrogen bonding. Every residue in the chain contributes one C=O (from its carbonyl carbon) and one N-H (from its amide nitrogen). These two groups are strongly polarised — the carbonyl oxygen carries a partial negative charge and the amide hydrogen a partial positive charge — so a C=O on one part of the chain can hydrogen-bond to an N-H elsewhere on the chain (or on a neighbouring chain). Secondary structure is just the geometric pattern this hydrogen bonding forces the backbone into. Alpha-helix. The polypeptide chain twists into a right-handed coil, like a spring. As the chain winds round, the C=O group of one amino acid residue lines up directly below the N-H group of the amino acid four residues further along the chain — that is, residue i hydrogen-bonds to residue i+4 — because the geometry of a 3.6-residues-per-turn helix places these two groups close together and pointing at each other. This C=O···H-N hydrogen bond is repeated at every residue all the way down the chain, so the helix is held rigid by a continuous, regularly repeating "staircase" of hydrogen bonds running parallel to the helix axis, with the amino acid side chains projecting outward from the coil. Beta-pleated sheet. Here the chain is not coiled but stretched out almost fully, running in a zig-zag ("pleated") path. Two or more such extended chains (or two extended stretches of the same chain folded back on itself) are laid side by side, and hydrogen bonds form between the C=O of one chain and the N-H of the adjacent chain, linking the strands together edge-to-edge into a flat, sheet-like arrangement. Depending on whether the neighbouring strands run in the same N-terminus-to-C-terminus direction or in opposite directions, the sheet is called parallel or antiparallel — but in both cases the stabilising interaction is the same inter-chain C=O···H-N hydrogen bonding. So the two forms differ in where the hydrogen-bonding partner comes from: in the alpha-helix it is found further down the same chain (an intra-chain bond, giving a coiled rod), while in the beta-pleated sheet it is found on a neighbouring chain or strand (an inter-chain bond, giving a flat sheet). Answer: The two common secondary structures of proteins are the alpha-helix, in which the polypeptide chain coils into a right-handed helix stabilised by hydrogen bonds between the C=O of one amino acid residue and the N-H of the residue four places ahead in the same chain, and the beta-pleated sheet, in which extended, zig-zag polypeptide chains lie side by side (parallel or antiparallel) and are held together by hydrogen bonds between the C=O and N-H groups of neighbouring chains.

Question 14

What type of bonding helps in stabilising the α-helix structure of proteins?

Accepted Answer

Hydrogen bonding between the backbone C=O and N-H groups of the peptide chain holds the α-helix together — nothing else in the structure is doing the job. A protein's polypeptide backbone is a repeating chain of peptide bonds: - CO - NH - CHR - CO - NH - CHR - Each peptide linkage contributes one carbonyl group, C=O, and one amide group, N-H, to the backbone. In the α-helix, this chain is coiled into a right-handed spiral. The coiling is not random — it is held in place because the backbone twists in exactly the way needed to bring the C=O of one peptide bond close to the N-H of a peptide bond four residues further along the same chain (the n and n+4 residues). The mechanism, step by step: 1. The oxygen of a C=O group carries a partial negative charge (it is the more electronegative atom, and it also bears lone pairs), while the hydrogen of an N-H group carries a partial positive charge (bonded to the electronegative nitrogen). 2. When the chain coils into the helical geometry, the lone pair on the carbonyl oxygen of residue n lines up with the N-H hydrogen of residue n+4. 3. This pulls the two groups together into a hydrogen bond, C=O H-N, running roughly parallel to the axis of the helix. 4. This pattern repeats at every peptide bond along the chain, so every C=O is hydrogen-bonded to an N-H four residues ahead. The result is a continuous ladder of intramolecular hydrogen bonds that runs the whole length of the helix, and it is this repeating set of bonds — not any bond involving the side chain R — that locks the coil into its helical shape and keeps every turn held tightly against the next. Because each individual hydrogen bond is weak, no single one of them could hold the shape on its own; it is the large number of them, all pointing the same way along the helix axis, that makes the α-helix a stable, rigid secondary structure. Answer: Hydrogen bonds — specifically, intramolecular hydrogen bonding between the C=O group of one peptide bond and the N-H group of a peptide bond four amino acid residues further along the same polypeptide chain — stabilise the α-helix structure of proteins.

Question 15

Differentiate between globular and fibrous proteins.

Accepted Answer

Shape is the root cause: a protein either coils tightly into a compact ball, or it lies out straight and packs side by side into a fibre — every other difference follows from that one geometric choice. A protein's final shape (its tertiary/quaternary structure) comes from how the polypeptide chain folds on itself. Two folding patterns exist, and they give two structurally and functionally distinct classes. Globular proteins - The polypeptide chain coils up and folds back on itself so that the molecule becomes a compact, more or less spherical (globe-like) shape. Hydrophobic side chains tuck into the interior, away from water, and polar/charged side chains sit on the surface, facing the water. - Because the surface is studded with polar and charged groups, globular proteins are soluble in water. - Being water-soluble and compact, they are mobile inside the cell and in body fluids, so they carry out the "active" chemical jobs of the cell: catalysis, transport, and defence. - Examples: insulin (a hormone), haemoglobin and myoglobin (oxygen transport/storage), all enzymes (e.g., pepsin, trypsin), and antibodies (immunoglobulins). Fibrous proteins - The polypeptide chains run parallel to one another, extended along one axis, and are held together side by side by hydrogen bonds and disulphide bonds to form long fibres or sheet-like bundles — like strands laid alongside each other in a rope, not folded into a ball. - This side-by-side, hydrogen-bonded arrangement is mechanically strong but leaves few polar groups exposed in an orientation that lets water molecules surround the chain, so fibrous proteins are insoluble in water. - Being insoluble, tough, and fibre-shaped, they cannot diffuse or act catalytically; instead they serve a purely structural, load-bearing role, giving strength and support to tissues. - Examples: keratin (hair, wool, nails), myosin (muscle fibre), and collagen (tendons, connective tissue). Summarised point by point: - Molecular shape: globular = spherical/globe-shaped (chain coiled and folded); fibrous = thread-like, chains lying parallel (extended, not folded into a ball). - Solubility: globular = soluble in water; fibrous = insoluble in water. - Stabilising forces between chains: globular = mainly folded within one chain, interior hydrophobic packing; fibrous = hydrogen bonds and disulphide bonds cross-linking parallel chains. - Biological role: globular = functional (enzymes, hormones, transport, antibodies); fibrous = structural (support and mechanical strength in tissues). - Examples: globular — insulin, haemoglobin, pepsin, immunoglobulins; fibrous — keratin, myosin, collagen. Answer: Globular proteins have their polypeptide chains coiled and folded into a compact, spherical shape, are soluble in water, and perform functional roles (e.g., insulin, haemoglobin, pepsin, antibodies). Fibrous proteins have their polypeptide chains lying parallel to one another, held together by hydrogen and disulphide bonds into long, thread-like fibres, are insoluble in water, and perform structural roles (e.g., keratin, myosin, collagen).

Question 16

How do you explain the amphoteric behaviour of amino acids?

Accepted Answer

An amino acid carries both an acidic group ( -COOH ) and a basic group ( -NH 2 ) on the same molecule, and in solution these two groups react with each other rather than staying neutral — that internal proton transfer is what lets the molecule act as either an acid or a base depending on what it meets. Take glycine, H 2N-CH 2-COOH (systematic name 2-aminoacetic acid), as the working example — the same argument holds for every amino acid, only the side chain R changes. Step 1 — the internal proton transfer (zwitterion formation). The -COOH group is a proton donor and the -NH 2 group is a proton acceptor, and they are close enough on the same molecule for the acid group to hand its proton straight to the amine group: H 2N-CH 2-COOH →^+H 3N-CH 2-COO^- The species on the right, ^+H 3N-CH 2-COO^-, is called a zwitterion ("dipolar ion") — it carries a full positive charge on the nitrogen and a full negative charge on the oxygen at the same time, with no net charge on the molecule overall. This zwitterion, not the neutral-looking H 2N-CH 2-COOH written on paper, is the form amino acids actually exist in — as a solid and in neutral aqueous solution. Step 2 — behaviour in acidic medium (the amino acid acts as a base). When the zwitterion is placed in excess acid (a solution rich in H^+ ), the negatively charged carboxylate oxygen, -COO^-, is the site with a lone pair available to accept a proton. It picks up an H^+ from the medium: ^+H 3N-CH 2-COO^- + H^+ →^+H 3N-CH 2-COOH Here the amino acid has consumed a proton — that is exactly what a base does. The product, ^+H 3N-CH 2-COOH, is a cation (net charge +1 ); this is the form glycine takes in strongly acidic solution. Step 3 — behaviour in basic medium (the amino acid acts as an acid). When the zwitterion is placed in excess base (a solution rich in OH^- ), the site with a proton to give up is the ammonium group, -NH 3^+. The hydroxide ion pulls this proton off: ^+H 3N-CH 2-COO^- + OH^- → H 2N-CH 2-COO^- + H 2O Here the amino acid has donated a proton (to OH^-, forming water) — that is exactly what an acid does. The product, H 2N-CH 2-COO^-, is an anion (net charge -1 ); this is the form glycine takes in strongly basic solution. Why this counts as amphoteric. A substance is amphoteric when the same species can act as a proton donor in one setting and a proton acceptor in another. Steps 2 and 3 show precisely that: the zwitterion ^+H 3N-CH 2-COO^- - accepts a proton (behaves as a base) when acid is added, giving the cation ^+H 3N-CH 2-COOH; - donates a proton (behaves as an acid) when base is added, giving the anion H 2N-CH 2-COO^-. Because it can go either way depending on the pH of the medium, the amino acid is called an amphoteric electrolyte, or ampholyte. (The one detail students often get backwards: it is the carboxylate oxygen, already negative, that grabs the proton in acid; it is the ammonium nitrogen, already holding the extra proton, that gives it up in base — the charges on the zwitterion tell you which group reacts, not the neutral-looking textbook formula H 2N-CH 2-COOH.) Answer: Amino acids exist as zwitterions, ^+H 3N-CH 2-COO^-, formed by an internal proton transfer from the -COOH group to the -NH 2 group. In acidic solution the -COO^- end accepts a proton, giving the cation ^+H 3N-CH 2-COOH (the amino acid acting as a base); in basic solution the -NH 3^+ end donates a proton, giving the anion H 2N-CH 2-COO^- (the amino acid acting as an acid). Because the same molecule can donate or accept a proton depending on the medium, amino acids are amphoteric electrolytes (ampholytes).

Question 17

What are enzymes?

Accepted Answer

Enzymes are biological catalysts — proteins that speed up the specific biochemical reactions of a living cell without themselves being consumed in the reaction. Almost every enzyme is a globular protein (a few catalytic RNA molecules, called ribozymes, are the rare exception, but at the Class 12 level "enzyme" means protein). A protein catalyst folds into a compact three-dimensional shape, and on its surface there is a small pocket or cleft called the active site. The active site is built from the side chains of a handful of amino acid residues that are far apart along the polypeptide chain but are brought close together in space by the folding of the chain — so its shape and its pattern of charges/hydrogen-bond donors and acceptors are unique to that one enzyme. The molecule the enzyme acts on is called the substrate. Because the active site's shape and chemical environment match only one substrate (or a small family of closely related substrates), an enzyme shows very high specificity — the same feature that in ordinary chemistry we call a "lock and key" fit: 1. The substrate diffuses into the active site and binds it, forming an enzyme-substrate complex E + S ES. 2. Binding strains or polarizes particular bonds in the substrate and holds the reacting groups in exactly the orientation needed for reaction, so the transition state is reached with a far lower activation energy than the uncatalysed path. 3. The bound substrate reacts to give product while still on the enzyme, and the product — no longer a good fit for the site — is released, regenerating the free enzyme: ES → E + P. Because the enzyme comes out unchanged at the end of step 3, a very small amount of enzyme can process a very large amount of substrate — enzymes are effective in extremely low concentration. Two features follow directly from this mechanism and are worth naming because they are what make enzymes different from ordinary inorganic catalysts: - High efficiency: a single enzyme molecule can convert a very large number of substrate molecules to product per second, because forming the ES complex lowers the activation energy enormously compared to the same reaction run without a catalyst. - High specificity: an enzyme catalyses only one reaction, or one small class of closely related reactions, because only a molecule whose shape complements the active site can bind there. (Sucrase, for instance, hydrolyses only sucrose and not other disaccharides such as maltose.) A everyday example: the enzyme invertase (sucrase) hydrolyses the disaccharide sucrose into an equal mixture of glucose and fructose, C 12H 22O 11 (sucrose) + H 2O invertase C 6H 12O 6 (glucose) + C 6H 12O 6 (fructose) and the enzyme maltase similarly hydrolyses maltose specifically to two units of glucose. Digestion, respiration, and essentially every metabolic transformation inside a living cell proceed through chains of such enzyme-catalysed steps, each enzyme handling one specific reaction in the sequence. Answer: Enzymes are protein biocatalysts produced by living cells. Each enzyme has an active site whose shape and chemistry match a particular substrate; the substrate binds there to form an enzyme-substrate complex, which lowers the activation energy and converts substrate to product, after which the unchanged enzyme is released to act again. This lock-and-key type binding gives enzymes their two defining properties — very high catalytic efficiency at very low concentration, and very high specificity for a particular reaction/substrate (e.g., invertase converts only sucrose to glucose and fructose; maltase converts only maltose to glucose).

Question 18

What is the effect of denaturation on the structure of proteins?

Accepted Answer

Denaturation strips away every level of folding above the amino-acid sequence itself — the secondary and tertiary structure collapse — while the primary structure, the exact order in which amino acids are joined by peptide bonds, is left completely intact. To see what "collapse" means here, first name the structure that is being destroyed. A protein's shape is built up in stages: - Primary structure: the linear sequence of amino acids, each joined to the next by a peptide bond ( - CO - NH -, formed between the - COOH of one amino acid and the - NH 2 of the next). This is the covalent backbone. - Secondary structure: the backbone coils or folds into a regular local shape — the α -helix or the β -pleated sheet — held together by hydrogen bonds between the C = O of one peptide bond and the N - H of another, a few residues along the chain. - Tertiary structure: the whole helix/sheet then folds further into a compact three-dimensional globule, held in place by a mix of hydrogen bonds, disulphide bridges ( - S - S -, formed between two cysteine side chains), ionic (salt-bridge) interactions between charged side chains, and hydrophobic interactions between non-polar side chains that bury themselves away from water. It is this specific 3-D shape that gives a protein its biological activity — an enzyme's active site, for instance, exists only because tertiary folding brings the right side chains together in space. What denaturation does, step by step: When a protein is subjected to a physical change (heating, e.g. boiling an egg, or agitation) or a chemical change (a change in pH, addition of a strong acid/base or certain salts, or exposure to alcohol), the weak, non-covalent interactions that hold the secondary and tertiary structure together are the first to give way: 1. Heat or a change in pH disrupts the hydrogen bonds between peptide C = O and N - H groups that hold the α -helix/ β -sheet in place, so the regular coiling of the secondary structure is lost. 2. The same disturbance breaks the hydrogen bonds, ionic interactions, and hydrophobic packing that hold the tertiary globule folded, so the compact 3-D globule unfolds and the polypeptide chain uncoils. 3. The peptide bonds of the backbone themselves are not broken — no covalent bond in the primary sequence is touched — so the amino-acid sequence stays exactly as it was. Because the tertiary shape (and with it, the biological activity) is lost while the primary sequence survives, the net effect is described as: the globular protein unfolds and is converted into a shapeless, extended, fibrous form, and it loses its biological (e.g. enzymatic) activity. Everyday examples of this exact process: - Boiling an egg: the globular protein albumin in egg white denatures on heating and coagulates into an insoluble solid mass — it does not turn back into liquid albumin on cooling, showing the change is not a simple physical one like melting. - Curdling of milk: souring of milk lowers the pH (lactic acid produced by bacteria), which denatures the milk protein casein and causes it to coagulate into curd. Answer: Denaturation destroys a protein's secondary and tertiary structure — the hydrogen bonds, disulphide bridges, ionic and hydrophobic interactions that hold the α -helix/ β -sheet and the folded 3-D globule together are broken by heat or a change in pH — causing the globular protein to unfold into a random, fibrous shape and lose its biological activity. The primary structure (the amino-acid sequence held by peptide bonds) is not affected. Coagulation of egg white on boiling and curdling of milk on souring are common examples of protein denaturation.

Question 19

How are vitamins classified? Name the vitamin responsible for the coagulation of blood.

Accepted Answer

Vitamins are grouped by one physical property: whether they dissolve in fat or in water — and that single property also decides how the body handles them. Vitamins are organic compounds the body cannot synthesize (or cannot synthesize in sufficient amount) and must obtain from food in small quantities. They are classified into two groups based on their solubility. 1. Fat-soluble vitamins These dissolve in fats and oils but not in water: vitamins A (retinol), D (calciferol), E (tocopherol), and K (phylloquinone/menaquinone). Because they are lipid-soluble, they are absorbed along with dietary fat, transported in the blood bound to lipoproteins, and — critically — stored in the liver and in adipose (fatty) tissue. This storage means the body keeps a reserve, so a missed day's intake does not immediately cause deficiency; it also means excess intake can accumulate to toxic levels (hypervitaminosis), which is not a concern with the water-soluble group. 2. Water-soluble vitamins These dissolve in water: vitamin C (ascorbic acid) and the members of the vitamin B complex (B1 thiamine, B 2 riboflavin, B 6 pyridoxine, B 12 cyanocobalamin, niacin, folic acid, etc.). Being water-soluble, they are not stored in the body to any significant extent — any excess is simply excreted in urine. This is the step people get wrong: because there is no depot to draw on, these vitamins must be supplied continuously through the diet, and their deficiency symptoms appear much sooner than for the fat-soluble group. Which vitamin controls blood coagulation Vitamin K is the one responsible for the coagulation (clotting) of blood. It functions as a cofactor for an enzyme that carboxylates glutamate residues on several clotting factors (including prothrombin, i.e., clotting factor II), and this carboxylation is essential for those factors to bind calcium ions and become active in the clotting cascade. Without adequate vitamin K, blood clotting is impaired and even minor injuries can bleed excessively — this is why vitamin K deficiency shows up clinically as increased blood-clotting time. Answer: Vitamins are classified as fat-soluble (A, D, E, K — stored in the liver and adipose tissue) and water-soluble (B complex and C — not stored, excreted if in excess). Vitamin K is responsible for the coagulation of blood.

Question 20

Why are vitamin A and vitamin C essential to us? Give their important sources.

Accepted Answer

Vitamins are organic compounds the body cannot manufacture in adequate amounts, so each one that is missing shows up as a specific deficiency disease — the "importance" of a vitamin is best read off what fails when it is absent. Vitamin A (retinol) Vitamin A is a fat-soluble vitamin built on a beta-ionone ring with a long isoprenoid side chain ending in an alcohol group; in the retina it is oxidised to retinal, which combines with the protein opsin to form rhodopsin (visual purple), the pigment that makes vision in dim light possible. - Function: it maintains this visual pigment, so it is essential for good eyesight, especially at low light intensity. - It also keeps epithelial tissue (skin, and the mucous membranes lining the respiratory, digestive and urinary tracts) healthy, and supports normal growth. - Deficiency: night blindness (nyctalopia — inability to see in dim light) as the first sign; prolonged deficiency leads to xerophthalmia, a hardening and clouding of the cornea. - Sources: fish liver oil, milk, butter, and eggs (animal sources, as preformed vitamin A); carrots, spinach, and other yellow and leafy vegetables (as the precursor beta-carotene, which the body converts to vitamin A). Vitamin C (ascorbic acid) Vitamin C is water-soluble, so unlike vitamin A it cannot be stored in the body in significant amounts and must be supplied continuously in the diet. - Function: it is required for the synthesis and maintenance of collagen, the fibrous protein that acts as the intercellular "cementing material" holding cells together in skin, blood vessel walls and gums; this is what keeps tissue integrity and resistance to infection intact. - Deficiency: scurvy — bleeding and swollen gums, loosening of teeth, and slow healing of wounds, because the cementing collagen between cells breaks down. - Sources: citrus fruits (amla/Indian gooseberry, orange, lemon), and fresh green leafy vegetables — being water-soluble, vitamin C is easily destroyed by cooking and boiling, so these are best consumed fresh or lightly cooked. Answer: Vitamin A (retinol) is essential because it forms the visual pigment rhodopsin needed for eyesight and keeps epithelial tissue healthy — its deficiency causes night blindness and xerophthalmia; its sources are fish liver oil, milk, butter, eggs, carrots and leafy vegetables. Vitamin C (ascorbic acid) is essential because it maintains the collagen that cements cells together in skin, blood vessels and gums — its deficiency causes scurvy (bleeding gums, loose teeth); its sources are citrus fruits (amla, orange, lemon) and fresh green vegetables.

Question 21

What are nucleic acids? Mention their two important functions.

Accepted Answer

A nucleic acid is the polymer that stores and reads out hereditary information — a long chain built from repeating nucleotide units, not a single molecule with one fixed formula. Every nucleotide, the repeating unit, is itself built from three pieces joined together: - a nitrogenous base — either a purine (adenine or guanine, each a fused double ring) or a pyrimidine (cytosine, thymine, or uracil, each a single ring), - a pentose sugar — β -D-ribose in RNA, or β -D-2-deoxyribose (ribose with the 2'-OH replaced by H) in DNA, - a phosphate group, PO 4^3-, attached at the sugar's 5' carbon. The base attaches to the sugar's 1' carbon by an N-glycosidic bond to give a base + sugar unit called a nucleoside (for example adenine + ribose = adenosine). Adding the phosphate at the 5'-OH of that nucleoside gives the nucleotide (for example adenosine 5'-monophosphate). Nucleotides then link into a chain: the 5'-phosphate of one nucleotide's sugar forms a phosphodiester bond to the 3'-OH of the next sugar, and this repeats down the chain — sugar-phosphate-sugar-phosphate — with one base hanging off each sugar. That sugar-phosphate backbone with bases attached is the polynucleotide, and a very long polynucleotide is the nucleic acid. Two nucleic acids exist, distinguished by the sugar and by which pyrimidine they carry: - DNA (deoxyribonucleic acid) — sugar is 2-deoxyribose; bases are adenine, guanine, cytosine, and thymine; two polynucleotide strands wind into a double helix, held together by hydrogen bonds between paired bases (adenine with thymine, guanine with cytosine). - RNA (ribonucleic acid) — sugar is ribose; bases are adenine, guanine, cytosine, and uracil (uracil replaces thymine); RNA is ordinarily a single strand. Their two important functions: 1. Storage and transmission of hereditary information. The specific sequence of the four bases along a DNA strand is the genetic code for an organism, and DNA replication — each strand acting as a template for a new complementary strand — copies that sequence exactly so it passes from parent cell to daughter cell, and from one generation to the next. 2. Protein synthesis. Genetic information is not read directly off DNA at the ribosome; it is first transcribed into messenger RNA (mRNA), which carries the coded message to the ribosome, while transfer RNA (tRNA) reads that code three bases (a codon) at a time and delivers the matching amino acid, and ribosomal RNA (rRNA) forms the structural and catalytic core of the ribosome that links the amino acids into a polypeptide. Through this DNA to mRNA to protein flow, nucleic acids control which proteins — and therefore which enzymes and traits — a cell makes. Answer: Nucleic acids are polynucleotides — long chains of nucleotide units (nitrogenous base + pentose sugar + phosphate group) joined by 3'-5' phosphodiester bonds — occurring as DNA (sugar = deoxyribose; bases A, G, C, T; double helix) and RNA (sugar = ribose; bases A, G, C, U; single strand). Their two chief functions are (i) storage and transmission of hereditary information from one generation to the next via DNA replication, and (ii) protein synthesis, in which mRNA, tRNA, and rRNA together translate that stored information into specific polypeptides.

Question 22

What is the difference between a nucleoside and a nucleotide?

Accepted Answer

A nucleoside is only the sugar joined to the base; a nucleotide is the sugar joined to the base AND to a phosphate group. The phosphate group is what turns one into the other, and it is also what lets nucleotides link up into DNA and RNA. Start with the two pieces that both structures share: A pentose sugar — either β -D-ribose (in RNA) or β -D-2-deoxyribose (in DNA, missing the - OH at carbon 2). A nitrogenous base attached at carbon 1 of that sugar — one of the purines adenine or guanine, or one of the pyrimidines cytosine, thymine or uracil. Nucleoside — base + sugar, nothing else. The base is attached to carbon 1' of the sugar through a β -N-glycosidic bond (an N-C bond joining a ring nitrogen of the base to C1' of the sugar). No phosphate is present. For example, adenine joined to ribose this way is the nucleoside adenosine; cytosine joined to deoxyribose is the nucleoside deoxycytidine. In condensed form, a nucleoside is written simply as (base)–(sugar), e.g. adenine–ribose. Nucleotide — base + sugar + phosphate. A phosphate group is esterified onto the sugar, most commonly at the 5' - OH of the sugar, through a phosphoester bond (an O-P bond between the sugar's 5' -oxygen and the phosphorus of PO 4^3- ). So a nucleotide is (base)–(sugar)–(phosphate), e.g. adenine–ribose–phosphate is adenosine 5'-monophosphate (AMP), and adenine–deoxyribose–phosphate is deoxyadenosine 5'-monophosphate (dAMP). Why this distinction matters structurally: the nucleotide's phosphate group carries a second reactive -OH that can esterify onto the 3' - OH of the sugar of the next nucleotide. This is exactly the phosphodiester linkage that repeats down the backbone of DNA and RNA — so nucleotides, not nucleosides, are the actual building blocks of the polynucleotide chain. A nucleoside, lacking the phosphate, has no way to form that second ester bond and cannot be strung into a chain. Summary in one line: nucleoside = base + sugar (joined by an N-glycosidic bond); nucleotide = base + sugar + phosphate (the sugar–phosphate joined by a phosphoester bond), and it is this added phosphate that lets nucleotides polymerise into nucleic acids. Answer: A nucleoside is a base linked to a pentose sugar (by an N-glycosidic bond) with no phosphate — e.g. adenosine (adenine + ribose). A nucleotide is a base linked to a pentose sugar which is in turn esterified to a phosphate group (e.g. AMP, adenine + ribose + phosphate at C5' ) — the extra phosphate is what allows nucleotides to join into the phosphodiester-linked chain of DNA/RNA, which nucleosides alone cannot do.

Question 23

The two strands in DNA are not identical but are complementary. Explain.

Accepted Answer

Complementary means each base on one strand fixes exactly one particular base opposite it on the other strand — the two strands end up carrying different, paired sequences, not the same sequence. DNA is built from two long polynucleotide chains that wind around a common axis to form a double helix. Each nucleotide in a chain is made of a nitrogenous base — adenine (A) or guanine (G), which are purines with two fused rings, or cytosine (C) or thymine (T), which are pyrimidines with a single ring — joined to a deoxyribose sugar, which is joined to a phosphate group. The sugar-phosphate units link up to form the backbone of each strand, and the bases point inward from the backbone, toward the bases of the other strand. The two strands are held together by hydrogen bonds between the bases facing each other at every rung of this ladder-like structure, and this bonding follows one fixed rule, not a random pairing: - Adenine hydrogen-bonds only with thymine, through two hydrogen bonds: A=T. - Guanine hydrogen-bonds only with cytosine, through three hydrogen bonds: G≡C. This rule is forced by geometry. A purine (A or G) is a two-ring base and is large; a pyrimidine (T or C) is a one-ring base and is small. Only a purine-pyrimidine pair spans the fixed, uniform width of the double helix and lines up the correct hydrogen-bond donor and acceptor atoms at the right distance to form 2 (A-T) or 3 (G-C) hydrogen bonds. A purine opposite a purine (A opposite G, say) would be too wide to fit inside the helix; a pyrimidine opposite a pyrimidine (C opposite T) would leave the two bases too far apart to hydrogen-bond at all. So whatever base sits at a given rung on one strand, only one particular base can chemically and geometrically complete that rung on the other strand. Because of this rule, the base sequence of one strand completely determines the base sequence of the other, position by position: wherever strand 1 has A, strand 2 must have T at that position; wherever strand 1 has G, strand 2 must have C; and vice versa. For example, if one strand, read 5' to 3', is 5'-A-T-G-C-A-3' then the other strand, running antiparallel to it, must read 3'-T-A-C-G-T-5' Strand 2 is not a copy of strand 1 — it does not repeat A opposite A or T opposite T — it carries the base that pairs with (complements) each base of strand 1. The two strands are also antiparallel: one runs 5'→3' while the other runs 3'→5' along the same axis, which is another respect in which they are not identical copies of each other. This is exactly why the two strands are called complementary and not identical: their base sequences differ from each other, yet each sequence is completely predictable from the other because of the fixed A-T and G-C pairing rule. This complementarity is also what makes faithful DNA replication possible — each old strand serves as a template on which a new partner strand is built, one base at a time, according to the same A-T / G-C rule, so that the new double helix reproduces the original genetic information exactly. Answer: The two strands of DNA are complementary, not identical, because base pairing between them follows one fixed geometric rule — adenine pairs only with thymine (A=T, 2 hydrogen bonds) and guanine pairs only with cytosine (G≡C, 3 hydrogen bonds), since only a purine-pyrimidine pair fits the constant width of the double helix. This means the base sequence of one strand is always exactly predictable from the other (complementary), but is a different sequence from it, not a copy (not identical); the strands are also antiparallel, running 5'→3' and 3'→5' in opposite directions.

Question 24

Write the important structural and functional differences between DNA and RNA.

Accepted Answer

The single biggest difference is one hydroxyl group on the sugar — everything else about DNA vs RNA follows from what that group does and does not allow. Structural differences 1. The sugar in the backbone. DNA's repeating sugar is 2-deoxyribose, C 5 H 10 O 4 — a five-carbon sugar with no -OH on carbon 2'. RNA's sugar is ribose, C 5 H 10 O 5 — the same skeleton but with an -OH at C-2'. That extra -OH is the reason RNA is chemically fragile: the 2'-oxygen can attack the adjacent phosphodiester bond and cleave the chain, so RNA hydrolyses easily. DNA, lacking that oxygen, cannot do this and is chemically far more stable — which is exactly why the cell trusts DNA, not RNA, to hold its genetic information for a lifetime. 2. The base set. Both nucleic acids use adenine (A), guanine (G) and cytosine (C), but the fourth base differs: DNA uses thymine (T, 5-methyluracil), RNA uses uracil (U, the same ring without the 5-methyl group). So DNA's four bases are A, G, C, T; RNA's are A, G, C, U. 3. Number of strands. DNA is double-stranded: two polynucleotide chains run antiparallel to each other and coil into a double helix, held together by hydrogen bonds between bases on opposite strands — A pairs with T through two H-bonds, G pairs with C through three H-bonds, so the two strands are complementary and either one can be used to reconstruct the other. RNA is ordinarily single-stranded; it can fold back on parts of itself (as in the cloverleaf shape of tRNA) but does not have a second, separate complementary strand running its whole length. 4. Size. A DNA molecule (a chromosome) is enormous — its molecular mass runs into billions of daltons — because it must encode an organism's entire genome in one continuous molecule. RNA molecules are much shorter, since each is transcribed to do one job: one mRNA carries the message of one gene (or a few), one tRNA is about 75–90 nucleotides, one rRNA matches the size of the ribosomal subunit it belongs to. 5. Location in the cell. DNA is confined almost entirely to the nucleus (with a small amount in mitochondria and chloroplasts). RNA is transcribed in the nucleus but does its actual work mostly outside it, in the cytoplasm, at the ribosome. Functional differences 1. DNA stores hereditary information; RNA acts on it. DNA is the genetic material: it carries the coded instructions for building and running the organism, and this information must be passed on essentially unchanged from one generation of cells to the next. 2. DNA can duplicate itself; RNA is made fresh from DNA whenever needed. DNA replicates — each strand is used as a template to build a new complementary strand, so one DNA molecule becomes two identical ones before a cell divides. RNA is transcribed anew from the DNA template only when a particular gene needs to be expressed; RNA does not routinely copy itself. 3. RNA carries out protein synthesis, using three different molecules for three different jobs. Messenger RNA (mRNA) is transcribed from a gene and carries its coded message out of the nucleus to the ribosome. Ribosomal RNA (rRNA) is a structural and catalytic component of the ribosome itself, the site where the message is read. Transfer RNA (tRNA) reads each three-base codon on the mRNA and delivers the matching amino acid, linking the genetic code to the growing protein chain. DNA performs none of these three jobs directly — it only supplies the template from which RNA is transcribed. 4. An exception worth naming, so the rule is not overstated: in retroviruses (e.g. HIV), RNA itself is the genetic material and is copied by reverse transcriptase — so "DNA stores, RNA executes" describes the normal cell, not every biological system. Answer: DNA has 2-deoxyribose sugar, the bases A, G, C, T, and is a stable double helix of two complementary, hydrogen-bonded strands, confined mainly to the nucleus, with an enormous molecular mass; functionally, it stores hereditary information and replicates itself. RNA has ribose sugar (2'-OH present), the bases A, G, C, U, is usually single-stranded, chemically less stable, of much smaller molecular mass, and found in both nucleus and cytoplasm; functionally, it is synthesized from DNA as needed and carries out protein synthesis through its three forms — mRNA (carries the message), tRNA (delivers amino acids), and rRNA (builds the ribosome).

Question 25

What are the different types of RNA found in the cell?

Accepted Answer

A cell needs three different RNAs because protein synthesis has three separate jobs — carrying the code, reading the code, and doing the assembly — and one molecule cannot do all three. Deoxyribonucleic acid (DNA) stores the genetic information, but DNA itself never leaves the nucleus to make a protein. That job is carried out by ribonucleic acid (RNA), and the cell uses three distinct kinds of RNA, each shaped for a different step of translation. 1. Messenger RNA (mRNA) mRNA is transcribed from a gene on the DNA template strand and carries that gene's coded information, as a sequence of codons (groups of three bases), out of the nucleus to the ribosome. It is a long, single-stranded, largely unbranched molecule — essentially a disposable working copy of one gene's instructions. Because it only carries the message and does not fold into a stable, reusable shape, it is also the shortest-lived of the three RNAs, being degraded soon after it has been read. 2. Transfer RNA (tRNA) tRNA is the smallest of the three RNAs, roughly 75–90 nucleotides long. Its single strand folds back on itself and base-pairs internally to give a cloverleaf shape in two dimensions, which then folds further into an L-shaped three-dimensional structure. This shape gives it two functional ends: - one end carries a specific amino acid (the amino-acid-binding site), and - a loop on the opposite side carries a triplet of bases called the anticodon. During translation, the anticodon on the tRNA base-pairs with the complementary codon on the mRNA, which is how the correct amino acid is delivered to the correct position specified by the code. Because there are (with some codon degeneracy) different tRNAs for different amino acids, tRNA is the molecule that physically "translates" a base sequence into an amino-acid sequence. 3. Ribosomal RNA (rRNA) rRNA is the RNA that, together with proteins, makes up the ribosome itself — the structure on which mRNA is read and the polypeptide chain is built. It is the most abundant of the three types of RNA in the cell, since every ribosome (and a cell contains very many ribosomes) is built largely of rRNA. Unlike mRNA and tRNA, rRNA is not a messenger or an adaptor; it forms the physical framework and, in part, the catalytic machinery of the ribosome that carries out peptide-bond formation. How the three work together: DNA is transcribed into mRNA, which travels to a ribosome built of rRNA and protein; there, tRNA molecules — each charged with one specific amino acid — bring their amino acids to the ribosome and pair their anticodon with the matching mRNA codon, so the amino acids are joined in the order the mRNA specifies. Answer: The cell contains three types of RNA — messenger RNA (mRNA), which carries the genetic code from DNA to the ribosome as a sequence of codons; transfer RNA (tRNA), a small cloverleaf-shaped molecule that reads each codon via its anticodon and delivers the corresponding amino acid; and ribosomal RNA (rRNA), the most abundant RNA, which forms the structural and functional core of the ribosome on which the polypeptide chain is assembled.

NCERT Solutions · Class 12 Chemistry Biomolecules

Exercises 10.1–10.10 (part 1 of 3)

Exercise 10.1

Exercise 10.2

Exercise 10.3

Exercise 10.4

Exercise 10.5

Exercise 10.6

Exercise 10.7

Exercise 10.8

Exercise 10.9

Exercise 10.10

Exercises 10.11–10.20 (part 2 of 3)

Exercise 10.11

Exercise 10.12

Exercise 10.13

Exercise 10.14

Exercise 10.15

Exercise 10.16

Exercise 10.17

Exercise 10.18

Exercise 10.19

Exercise 10.20

Exercises 10.21–10.25 (part 3 of 3)

Exercise 10.21

Exercise 10.22

Exercise 10.23

Exercise 10.24

Exercise 10.25