What are reducing sugars?

Exercises · 10.2

Q. What are reducing sugars?

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Concept understanding — Lactose Hydrolysis Products

Lactose Hydrolysis Products – From Intuition to Precision

Imagine you have a glass of milk. That slightly sweet taste comes from a sugar called lactose. But lactose is a disaccharide – it's actually two smaller sugar units joined together. If you could "unstick" those two units, you'd get two simpler sugars. That unsticking process is hydrolysis (water + breaking), and the two simpler sugars you get are the hydrolysis products.

The Intuition: Breaking a Sugar Chain

Think of lactose as a train with exactly two carriages. The coupling between them is a chemical bond. When you add water and the right conditions (like an enzyme called lactase, or an acid), that bond snaps. The train splits into two separate carriages. Each carriage is now a free, smaller sugar molecule.

So the hydrolysis products are simply the two individual sugar units that were originally linked to form lactose.

The Precise Statement

Lactose ( $\text{C}_{12}\text{H}_{22}\text{O}_{11}$ ) is a disaccharide composed of one molecule of D-galactose and one molecule of D-glucose linked by a $\beta(1 \rightarrow 4)$ glycosidic bond. Upon hydrolysis (reaction with water), this bond is cleaved, yielding the two monosaccharides:

\text{Lactose} + \text{H}_2\text{O} \xrightarrow{\text{lactase or acid}} \text{D-Galactose} + \text{D-Glucose}

The two hydrolysis products are:

D-Galactose – a monosaccharide (aldohexose, C6H12O6)
D-Glucose – a monosaccharide (aldohexose, C6H12O6)

Both are reducing sugars, and both have the same molecular formula ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) but differ in the arrangement of the hydroxyl group on carbon 4 (they are C-4 epimers).

Note

In the body, the enzyme lactase (present in the small intestine) performs this hydrolysis so that the resulting glucose and galactose can be absorbed into the bloodstream. Lactose intolerance occurs when lactase activity is low, leaving lactose undigested.

Why This Matters for Exams

Always name both products: galactose and glucose. Never just "sugars" or "monosaccharides."
Know the bond: $\beta(1 \rightarrow 4)$ glycosidic linkage. Hydrolysis breaks this specific bond.
Recognize the reaction: It's a classic example of disaccharide hydrolysis – water adds across the bond, one OH goes to one sugar, one H to the other.
Remember the formula: Lactose + H₂O → Galactose + Glucose. The water molecule is consumed, so the total number of carbon atoms stays the same (12), but you now have two separate C6 units.

Tip

A quick memory aid: Lactose → Lact (milk) + ose (sugar). Its products are Galactose and Glucose – both start with G, but galactose is the one that sounds like "galactic" (less common), while glucose is the body's main fuel.

Lactose hydrolysis into glucose and galactose is drawn directly from the carbohydrates section of the NCERT Class 12 Chemistry chapter on biomolecules, a frequent source of short-answer and important questions in CBSE board exams. Searches for "lactose hydrolysis products class 12 chemistry" or "disaccharide hydrolysis NCERT" will find this beta-1,4-glycosidic-bond explanation matches the syllabus treatment.

Why this formula?

Lactose Hydrolysis Products — Understanding the Why

Lactose is a disaccharide composed of two monosaccharides linked by a glycosidic bond. When it undergoes hydrolysis, the bond is broken, yielding specific products. Let's build the reasoning step by step.

1. What is lactose chemically?

Lactose = galactose $\beta(1 \to 4)$ glucose
The bond is between:
- Carbon-1 of galactose (in $\beta$ configuration)
- Carbon-4 of glucose

So the structural formula is:

\text{Galactose} - O - \text{Glucose}

2. What does hydrolysis do?

Hydrolysis means "splitting with water." The reaction is:

\text{Lactose} + \text{H}_2\text{O} \xrightarrow{\text{lactase or acid}} \text{Galactose} + \text{Glucose}

The water molecule adds across the glycosidic bond:

The H from water attaches to the oxygen of the galactose (forming a free –OH on galactose)
The OH from water attaches to the carbon-1 of glucose (forming a free –OH on glucose)

3. Why are the products exactly galactose and glucose?

Because the glycosidic bond is between specific carbons:

Galactose contributes its anomeric carbon (C1)
Glucose contributes its C4

When the bond breaks, each sugar regains its free anomeric carbon (in the case of galactose) or free hydroxyl at C4 (in the case of glucose). No rearrangement occurs — the monosaccharides are released as they were originally linked.

4. Key formula — the hydrolysis equation

The balanced chemical equation:

\ce{C12H22O11 + H2O -> C6H12O6 + C6H12O6}

Lactose: $\ce{C12H22O11}$
Water: $\ce{H2O}$
Products: two molecules of $\ce{C6H12O6}$ (one galactose, one glucose)

Why the same molecular formula?

Both galactose and glucose are aldohexoses — they have the same molecular formula $\ce{C6H12O6}$ but differ in the arrangement of –OH groups (epimers at C4).

5. The "why" behind the formula

Mass conservation: The total number of C, H, O atoms before and after must match.
- Left: 12 C, 24 H, 12 O
- Right: 6+6 = 12 C, 12+12 = 24 H, 6+6 = 12 O ✓
Bond energy: The glycosidic bond is an acetal linkage. Water provides the –H and –OH needed to convert it into two hemiacetal (free sugar) forms.
Biological significance: Lactase enzyme in the small intestine catalyzes this specific hydrolysis because the active site is shaped to fit the $\beta(1\to4)$ bond — not $\alpha$ bonds or other linkages.

6. Exam tip — what to remember

Component	Formula	Type
Lactose	$\ce{C12H22O11}$	Disaccharide
Galactose	$\ce{C6H12O6}$	Aldohexose
Glucose	$\ce{C6H12O6}$	Aldohexose

Key takeaway: Hydrolysis of lactose yields one molecule each of D-galactose and D-glucose — not two glucoses, not two galactoses. The bond specificity determines the product identity.

Reducing sugars are carbohydrates that can act as reducing agents because they have a free aldehyde group or a free ketone group that can tautomerize to an aldehyde. The key test is their ability to reduce Cu²⁺ (in Benedict’s or Fehling’s solution) to Cu⁺, forming a brick-red precipitate of Cu₂O.

The Core Idea: Why “Reducing”?

The term “reducing sugar” comes from a simple chemical property: the sugar itself gets oxidized while it reduces another substance (usually a metal ion like Cu²⁺ or Ag⁺). For a sugar to do this, it must have a free or potentially free aldehyde group ( $-CHO$ ) or an $\alpha$ -hydroxy ketone group that can isomerize into an aldehyde under basic conditions.

Think of it this way: the aldehyde group is like a chemical “handle” that can easily lose electrons (get oxidized to a carboxylic acid). If that handle is tied up — for example, in a glycosidic bond — the sugar can’t act as a reducing agent anymore.

Step-by-Step Breakdown

1. The Structural Requirement

A reducing sugar must contain either:

A free aldehyde group ( $-CHO$ ), or
A free ketone group ( $>C=O$ ) adjacent to a free $-$ OH group (an $\alpha$ -hydroxy ketone).

The aldehyde is directly oxidizable. The ketone isn’t — but under the alkaline conditions of common tests (like Benedict’s), it undergoes tautomerization to form an aldehyde. This is why fructose, a ketose, is still a reducing sugar.

Watch out

A common mistake is to think that all ketoses are reducing. They are, but only because of the base-catalyzed isomerization. In neutral or acidic conditions, a ketose does not reduce Cu²⁺ directly — the test requires a basic medium.

2. The Mechanism in Benedict’s Test

Take Benedict’s reagent (Cu²⁺ in alkaline citrate solution). When a reducing sugar is heated with it:

The aldehyde group of the sugar is oxidized to a carboxylate ion (the sugar becomes an aldonic acid).
Cu²⁺ is reduced to Cu⁺, which precipitates as brick-red Cu₂O.

The colour change — from blue (Cu²⁺) to green, yellow, orange, and finally brick-red — tells you how much reducing sugar is present.

\text{R-CHO} + \text{H}_2\text{O} \rightarrow \text{R-COOH} + 2\text{H}^+ + 2e^-

3. Which Sugars Are Reducing? A Quick Table

Sugar	Type	Free aldehyde/ketone?	Reducing?
Glucose	Aldohexose	Yes (free $-CHO$ )	Yes
Fructose	Ketohexose	No free $-CHO$ , but tautomerizes	Yes
Maltose	Disaccharide (Glc $\alpha$ 1 $\rightarrow$ 4 Glc)	One free anomeric carbon	Yes
Lactose	Disaccharide (Gal $\beta$ 1 $\rightarrow$ 4 Glc)	One free anomeric carbon	Yes
Sucrose	Disaccharide (Glc $\alpha$ 1 $\rightarrow$ 2 Fru)	Both anomeric carbons bonded	No
Starch	Polysaccharide	All anomeric carbons in glycosidic bonds	No

Tip

The quickest way to check if a disaccharide is reducing: look at the anomeric carbon of each monosaccharide unit. If both are involved in the glycosidic bond (like in sucrose, where C1 of glucose and C2 of fructose are linked), the sugar is non-reducing. If at least one anomeric carbon is free, it’s reducing.

4. The Special Case of Sucrose

Sucrose is the classic non-reducing disaccharide. Glucose and fructose are joined by their anomeric carbons (C1 of glucose and C2 of fructose). This locks both rings in the cyclic form — neither can open to give a free aldehyde or ketone. So sucrose does not reduce Cu²⁺ or Ag⁺.

But if you hydrolyse sucrose (with acid or the enzyme invertase), you get glucose and fructose — both reducing. That’s why honey (which contains invert sugar) gives a positive Benedict’s test.

5. Why Does This Matter in Exams?

Questions on reducing sugars test your understanding of:

Structure-function relationships: Can you identify a free anomeric carbon?
Reactivity under basic conditions: Why does fructose reduce Cu²⁺ even though it’s a ketone?
Hydrolysis products: Sucrose → glucose + fructose (both reducing). Lactose → glucose + galactose (both reducing). Maltose → two glucose (both reducing).

Important

All monosaccharides are reducing sugars. Not all disaccharides are — only those with at least one free anomeric carbon are reducing.

Final Answer

✓Final answer

Reducing sugars are carbohydrates with a free or potentially free aldehyde group that can reduce Cu²⁺ to Cu⁺ in alkaline solution; examples include glucose, fructose, maltose, and lactose, while sucrose is a common non-reducing sugar.

Here are the common mistakes students make when answering "What are reducing sugars?" — and how to avoid each.

✗ Mistake 1: Confusing "reducing" with "sweet" or "easily digested"

What students do wrong:

They think a reducing sugar is one that is sweet, simple, or quickly absorbed. That is not the definition.

How to avoid:

Use the chemical definition: reducing sugars are carbohydrates that reduce Fehling's solution and Tollens' reagent. The property comes from a free aldehydic or ketonic group (a free anomeric carbon in the cyclic form) — nothing to do with taste or digestion.

✗ Mistake 2: Thinking ketoses cannot be reducing sugars

What students do wrong:

They reason "only aldehydes reduce Tollens' reagent, so fructose (a ketose) must be non-reducing."

How to avoid:

Remember the book's own statement: all monosaccharides, whether aldose or ketose, are reducing sugars. Under the alkaline conditions of these tests, a ketose like fructose isomerises to the aldose form and reduces the reagent.

✗ Mistake 3: Getting the disaccharides backwards

What students do wrong:

They label maltose or lactose non-reducing, or call sucrose reducing.

How to avoid:

Check what the glycosidic bond consumes:

Sucrose — the linkage joins C1 of glucose to C2 of fructose, so both reducing groups are tied up → non-reducing.
Maltose and lactose — one anomeric carbon remains free and can open to the aldehyde form in solution → reducing.

✗ Mistake 4: Thinking "has a glycosidic bond" automatically means non-reducing

What students do wrong:

They assume any sugar containing a glycosidic linkage cannot be a reducing sugar.

How to avoid:

Maltose and lactose both contain glycosidic bonds, yet both are reducing. What matters is whether a free aldehydic/ketonic group (free anomeric carbon) remains after the bond forms — not whether a bond exists.

✗ Mistake 5: Omitting the reagents from the answer

What students do wrong:

They write "sugars that act as reducing agents" without naming what gets reduced — losing the easiest marks.

How to avoid:

Always name Fehling's solution (→ brick-red $Cu_2O$ precipitate) and Tollens' reagent (→ silver mirror) as the reagents reduced by these sugars.

✓ Quick Revision Table

Sugar	Free reducing group?	Reducing?
Glucose (aldose)	Yes	✓
Fructose (ketose)	Yes (isomerises)	✓
Maltose	One free anomeric C	✓
Lactose	One free anomeric C	✓
Sucrose	Both tied in the linkage	✗

Final takeaway:

Reducing sugars are carbohydrates that reduce Fehling's solution and Tollens' reagent. All monosaccharides (aldose or ketose) are reducing; maltose and lactose are reducing disaccharides, while sucrose is the classic non-reducing sugar.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
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›Reveal solutionSolution
Maltose is a disaccharide formed from two $\alpha$ -D-glucose units, linked by an $\alpha$ -1,4-glycosidic bond between C-1 of one glucose and C-4 of the other. The correct option is (A).

Concept and Intuition

Carbohydrates are essential biomolecules, and disaccharides are a class of carbohydrates formed by the condensation of two monosaccharide units. This linkage between two monosaccharide units is called a glycosidic linkage. It's essentially an ether linkage ( $-\text{O}-$ ) formed when a hydroxyl group from the anomeric carbon (C-1) of one monosaccharide reacts with a hydroxyl group from another carbon (often C-4 or C-6) of a second monosaccharide, with the elimination of a water molecule.

Maltose, commonly known as malt sugar, is a disaccharide. Understanding its structure requires knowing its constituent monosaccharides and the specific carbons involved in the glycosidic bond. Maltose is formed by two units of $\alpha$ -D-glucose. The " $\alpha$ " designation indicates the configuration of the hydroxyl group at the anomeric carbon (C-1) relative to the $-\text{CH}_2\text{OH}$ group at C-5 in the Haworth projection. For $\alpha$ -D-glucose, the $-\text{OH}$ at C-1 is on the opposite side of the ring from the $-\text{CH}_2\text{OH}$ at C-5 (typically drawn below the plane of the ring).

Step-by-Step Explanation

Identify the constituent monosaccharides of maltose:

Maltose is a disaccharide composed of two units of $\alpha$ -D-glucose. This is a fundamental fact about maltose's structure.

Understand the nature of the glycosidic linkage:

A glycosidic linkage is an ether bond formed between the anomeric carbon (C-1) of one monosaccharide and a hydroxyl group on another carbon of a second monosaccharide. During this reaction, a molecule of water is eliminated. The configuration of the anomeric carbon involved in the linkage (whether $\alpha$ or $\beta$ ) determines the type of glycosidic bond.

Determine the specific carbons involved in the linkage in maltose:

In maltose, the glycosidic linkage is formed between the C-1 of one $\alpha$ -D-glucose unit and the C-4 of the other $\alpha$ -D-glucose unit. This specific connection is crucial for its structure and properties.

Important

The anomeric carbon (C-1) of the first glucose unit is involved in the linkage, and its configuration is $\alpha$ . Therefore, the linkage is an $\alpha$ -glycosidic linkage. Since it connects to C-4 of the second glucose unit, it is specifically an $\alpha$ -1,4-glycosidic linkage.

Evaluate the given options:

(A) C-1 of $\alpha$ -D-glucose and C-4 of $\alpha$ -D-glucose: This option correctly identifies both the constituent monosaccharide units ( $\alpha$ -D-glucose) and the specific carbons involved in the linkage (C-1 and C-4). This matches the known structure of maltose.

(B) C-1 of $\alpha$ -D-glucose and C-4 of $\beta$ -D-galactose: This option is incorrect because maltose is made of glucose units, not galactose.

(C) C-1 of $\beta$ -D-glucose and C-4 of $\alpha$ -D-glucose: This option is incorrect because both glucose units in maltose are in the $\alpha$ -configuration at their C-1 positions (at least the one forming the linkage).

(D) C-1 of $\beta$ -D-glucose and C-4 of $\beta$ -D-glucose: This option is incorrect for the same reason as (C); maltose involves $\alpha$ -D-glucose units.

Therefore, option (A) accurately describes the glycosidic linkage in maltose.

✓Final answer
$\alpha$
TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$Q. A carbohydrate (A), when treated with dilute HCl in alcoholic solution gives two isomers (B) and (C). B on reaction with bromine water gives a monocarboxylic acid 'Z' and 'C' is a ketohexose. What is A? (A) Starch (B) Maltose (C) Sucrose (D) Lactose$
›Reveal solutionSolution
The key is that dilute acid hydrolysis of a carbohydrate gives two isomers, one of which is a ketohexose and the other yields a monocarboxylic acid with bromine water — this points to sucrose, which splits into glucose (aldose → acid) and fructose (ketohexose).

Concept & Intuition

The problem describes a carbohydrate (A) that, under mild acidic conditions in alcohol, breaks into two isomeric sugars (B and C). Isomers here means they have the same molecular formula but different structures — typical of a disaccharide splitting into its two monosaccharide units. One of these (C) is explicitly a ketohexose (a six-carbon sugar with a ketone group, like fructose). The other (B) reacts with bromine water to give a monocarboxylic acid — bromine water oxidizes only aldoses (sugars with an aldehyde group) to their corresponding aldonic acids. So B must be an aldose. Therefore, A is a disaccharide composed of one aldose and one ketohexose. Among the options, only sucrose fits: it is made of glucose (an aldose) and fructose (a ketohexose). Starch is a polysaccharide, maltose is two glucose units (both aldoses), and lactose is glucose + galactose (both aldoses). Let’s verify step by step.

Step-by-step reasoning

Identify the reaction type

Dilute HCl in alcoholic solution is a classic condition for hydrolyzing glycosidic bonds in disaccharides. The products are the constituent monosaccharides. So A is a disaccharide that yields two monosaccharides (B and C).

Characterize product C

The problem states C is a ketohexose. Ketohexoses have a ketone group (e.g., fructose). This immediately rules out disaccharides that yield only aldoses (like maltose and lactose, which give two aldoses each).

Characterize product B

B reacts with bromine water to give a monocarboxylic acid. Bromine water is a mild oxidizing agent that specifically oxidizes the aldehyde group (–CHO) of an aldose to a carboxylic acid (–COOH), producing an aldonic acid. This reaction does not occur with ketoses under these conditions. Therefore, B must be an aldose.

Combine the clues

A must be a disaccharide that, upon hydrolysis, gives one aldose (B) and one ketohexose (C). Let’s check the options:

(A) Starch: A polysaccharide, not a disaccharide; hydrolysis gives many glucose units, not just two isomers. Incorrect.

(B) Maltose: A disaccharide of two glucose units (both aldoses). Hydrolysis gives two aldoses, no ketohexose. Incorrect.

(C) Sucrose: A disaccharide of glucose (aldose) and fructose (ketohexose). Hydrolysis gives exactly one aldose (glucose) and one ketohexose (fructose). Perfect match.

(D) Lactose: A disaccharide of glucose (aldose) and galactose (aldose). Both are aldoses; no ketohexose. Incorrect.

Confirm the bromine water step

Glucose (B) from sucrose reacts with bromine water to form gluconic acid (a monocarboxylic acid). Fructose (C) does not react under these mild conditions. This matches the description exactly.

Watch out

A common mistake is to think that lactose or maltose could yield a ketohexose — but both are composed entirely of aldoses. Only sucrose contains fructose, the common ketohexose.

Tip

Remember: Bromine water is a quick test for aldoses vs. ketoses. Aldoses turn it colorless (as they get oxidized), while ketoses do not react under the same conditions.

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
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›Reveal solutionSolution
Maltose is a disaccharide of two D-glucose units linked α-1,4; both units are α-D-glucose, both are reducing sugars, and the linkage is 1,4-glycosidic. The incorrect statement is the one that introduces fructose — option (B).

Concept & Intuition

Maltose is produced by the partial hydrolysis of starch. Its structure is well‑known: two D‑glucose molecules joined by an α‑1,4‑glycosidic bond. Because the anomeric carbon of the second glucose is free (not involved in the linkage), maltose is a reducing sugar. Any statement that contradicts this structure — especially one that swaps a glucose for a fructose — must be false.

Identify the monosaccharides from maltose hydrolysis Maltose (C₁₂H₂₂O₁₁) hydrolyses in the presence of dilute acid or the enzyme maltase to give two molecules of D‑glucose.

$\text{Maltose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} 2\,\text{D-glucose}$
Both units are glucose; no fructose is produced.

Check the anomeric form of each glucose

The glycosidic bond in maltose is formed between the α‑anomeric carbon (C1) of the first glucose and the C4 hydroxyl of the second glucose. The first glucose is locked in the α‑configuration at C1. The second glucose retains a free anomeric carbon, which can mutarotate between α and β forms, but in the intact disaccharide the second unit is also α‑D‑glucose (the bond does not alter its ring form). Thus both monosaccharide units are α‑D‑glucose.

Evaluate each option

(A) “Both are α‑D‑glucose units only” — True, as explained.

(B) “One is α‑D‑glucose and second one is β‑D‑fructose” — False. Fructose is a ketohexose; maltose contains only aldohexoses (glucose). This statement is incorrect.

(C) “Both are reducing sugars” — True. Each glucose unit has a free anomeric carbon (the second glucose’s anomeric carbon is free; the first glucose’s anomeric carbon is tied up in the bond, but the monosaccharides after hydrolysis are both reducing). The statement refers to the monosaccharides formed, so after hydrolysis both are reducing.

(D) “In maltose, they are joined through 1,4‑glycosidic linkage” — True. The bond is α‑1,4.

Identify the incorrect statement

Only option (B) contradicts the known chemistry of maltose.

Watch out

A common mistake is to confuse maltose with sucrose. Sucrose is glucose + fructose, but maltose is glucose + glucose. Option (B) describes sucrose’s composition, not maltose’s.

Tip

Remember: “Maltose = malt sugar = two glucoses.” If you see fructose mentioned for maltose, it’s automatically wrong.

✓Final answer
$The correct option is (B). ANSWER: B$
TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$Q. Dehydration occurs during the formation of the following type of bonds I. Peptide bond II. Hydrogen bond III. Glycosidic bond IV. Ester bond (A) I and II (B) II and III (C) II and IV (D) I, III and IV$
›Reveal solutionSolution
Dehydration (loss of water) is a defining feature of condensation reactions that form covalent bonds like peptide, glycosidic, and ester bonds, but not hydrogen bonds, which form by electrostatic attraction without water release. The correct answer is (D).

Concept & Intuition

Dehydration synthesis (also called condensation) is a chemical reaction where two molecules join by removing a water molecule (H₂O). This happens when a hydroxyl group (–OH) from one molecule and a hydrogen atom (–H) from another combine to form water, leaving a covalent bond between the two residues. In biology, this is how many key polymers are built. Hydrogen bonds, however, are weak electrostatic attractions between a partially positive hydrogen and a partially negative atom (like oxygen or nitrogen); they form spontaneously without any water molecule being released. So the question asks: which bond types are formed by a dehydration reaction?

Step-by-step reasoning

Peptide bond (I) – A peptide bond links amino acids in proteins. It forms when the carboxyl group (–COOH) of one amino acid reacts with the amino group (–NH₂) of another, releasing a water molecule. This is a classic dehydration reaction.

→ Yes, dehydration occurs.

Hydrogen bond (II) – A hydrogen bond is an intermolecular force, not a covalent bond. It arises from the attraction between a hydrogen atom (bonded to an electronegative atom like O or N) and another electronegative atom. No atoms are removed or added; no water is produced.

→ No dehydration occurs.

Glycosidic bond (III) – This bond joins monosaccharides to form disaccharides or polysaccharides (e.g., maltose, sucrose). It forms when the hydroxyl group of one sugar reacts with the anomeric carbon of another, eliminating a water molecule.

→ Yes, dehydration occurs.

Ester bond (IV) – An ester bond forms between a carboxylic acid and an alcohol (e.g., in lipids, where glycerol and fatty acids join). The –OH from the acid and –H from the alcohol combine to release water.

→ Yes, dehydration occurs.

Thus, dehydration occurs during the formation of peptide, glycosidic, and ester bonds, but not hydrogen bonds.

Watch out

A common mistake is to think hydrogen bonds form by dehydration because they are often mentioned alongside water in biology. But hydrogen bonds are non-covalent and involve no chemical reaction — they are simply electrostatic attractions.

Tip

Remember the mnemonic: PEG — Peptide, Ester, Glycosidic — all form by dehydration. Hydrogen bonds are the odd one out.

✓Final answer
$The correct option is (D) . ANSWER: D$
TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
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›Reveal solutionSolution
Maltose is a disaccharide of two α-D-glucose units linked by a 1,4-glycosidic bond and is a reducing sugar; the statement involving β-D-galactose is incorrect.

Maltose is a common disaccharide, and its structure is a classic example in carbohydrate chemistry. The key to this question is knowing exactly which monosaccharides make up maltose and how they are linked. Many students confuse maltose with lactose (milk sugar), which does contain galactose. That mix-up is the trap here.

Let’s examine each statement:

Statement (A): "It is a reducing sugar"

A reducing sugar has a free anomeric carbon (the carbonyl carbon of the open-chain form) that can reduce Cu²⁺ or Ag⁺ ions. In maltose, the glycosidic bond is formed between the anomeric carbon of one glucose unit and the C-4 hydroxyl of the other. This leaves the anomeric carbon of the second glucose unit free (it can open to an aldehyde). Therefore, maltose is a reducing sugar. This statement is correct.

Statement (B): "It is composed of two α-D-glucose units"

Maltose is formed from two molecules of α-D-glucose joined by an α(1→4) glycosidic linkage. This is a standard fact. This statement is correct.

Statement (C): "It is composed of one β-D-glucose and one β-D-galactose unit"

This describes lactose, not maltose. Lactose is galactose β(1→4) glucose. Maltose contains only glucose, and both units are in the α-configuration. This statement is incorrect.

Statement (D): "It has 1,4 – glycosidic linkage"

The bond in maltose is indeed an α(1→4) glycosidic linkage (carbon 1 of one glucose to carbon 4 of the other). This statement is correct.

Watch out

The most common mistake is mixing up maltose (glucose + glucose) with lactose (galactose + glucose). Always check the monosaccharide names carefully.

Tip

Remember: Maltose = Malt (grain sugar) = glucose + glucose; Lactose = Lactation (milk sugar) = galactose + glucose.

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$Q. Cori cycle occurs between (A) Liver and kidney (B) Kidney and muscle (C) Liver and muscle (D) Muscle and pancreas$
›Reveal solutionSolution
The Cori cycle is the metabolic shuttle that recycles lactate from muscle to l liver for gluconeogenesis, so the correct answer is (C) Liver and muscle.

The Cori cycle is a beautiful example of metabolic cooperation between tissues. When your muscles work hard — say, during a sprint — they rely on glycolysis for quick energy, even when oxygen is limited. That process produces lactate as a byproduct. But lactate isn't just waste; it's a valuable fuel that gets shipped to the liver, which converts it back into glucose. That glucose then returns to the muscle, completing the cycle.

The key insight is that the liver has the enzymes for gluconeogenesis (making new glucose from lactate), while muscle does not. Muscle can only produce lactate; it cannot recycle it. So the cycle must involve both tissues — one that generates lactate and one that consumes it to remake glucose.

In muscle (during intense exercise): Glycolysis breaks down glucose to pyruvate, which is then reduced to lactate (by lactate dehydrogenase) to regenerate NAD⁺. This lactate is released into the bloodstream.

In the liver: The liver takes up lactate from the blood and converts it back to glucose via gluconeogenesis — a process that requires ATP. The newly made glucose is then released back into circulation.

Back to muscle: The muscle takes up that glucose and can use it again for energy, restarting the cycle.

Watch out

A common mistake is to think the kidney plays a major role here. While the kidney can do some gluconeogenesis, the Cori cycle specifically refers to the liver-muscle shuttle. The kidney is not the primary partner in this classic cycle.

The pancreas is involved in hormone secretion (insulin, glucagon) but not directly in the lactate-glucose shuttle. The kidney can perform gluconeogenesis, but the Cori cycle is defined as the cooperation between liver and muscle.

✓Final answer
$The correct option is (C) Liver and muscle .$
TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$Q. The amount of sucrose needed to produce 1 mole of glucose using acid hydrolysis is (A) 360 g (B) 180 g (C) 342 g (D) 171 g$
›Reveal solutionSolution
Acid hydrolysis of sucrose yields one mole of glucose from one mole of sucrose. Since sucrose has molar mass 342 g/mol, the mass needed is 342 g.

The key here is the stoichiometry of the reaction. Sucrose is a disaccharide made of one glucose unit and one fructose unit linked together. When you hydrolyse it in the presence of an acid, that bond breaks, and you get one molecule of glucose and one molecule of fructose.

So the reaction is simply:
$\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 (\text{glucose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{fructose})$
Notice that one mole of sucrose gives exactly one mole of glucose. That’s the whole story — no coefficients to balance, no side products. The water is in excess, so it doesn’t limit anything.

Now, to find the mass of sucrose needed for 1 mole of glucose, you just need the molar mass of sucrose.

Molar mass of sucrose ( $\text{C}_{12}\text{H}_{22}\text{O}_{11}$ ):

Carbon: $12 \times 12 = 144$ g/mol

Hydrogen: $22 \times 1 = 22$ g/mol

Oxygen: $11 \times 16 = 176$ g/mol

Total: $144 + 22 + 176 = 342$ g/mol

Since the mole ratio is 1:1, to get 1 mole of glucose you need exactly 1 mole of sucrose.

Therefore, the mass required is $342$ g.

Watch out

A common mistake is to think that because sucrose contains two monosaccharide units, you need only half a mole of sucrose to get one mole of glucose. That would be true if the two units were identical and both were glucose — but they aren’t. One is glucose, the other is fructose. So one sucrose molecule gives only one glucose molecule.

✓Final answer
$\boxed{342\ \text{g}}$
TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$Q. Match the following lists. List-I A) Phosphoenol pyruvate B) Pyruvic acid C) Triose phosphate D) Glucose-6-phosphate List-II I) Hexokinase II) Enolase III) Pyruvic kinase IV) Aldolase The correct match is: (A) II IV III I (B) III IV I II (C) II III IV I (D) III IV II I$
›Reveal solutionSolution
This question tests your knowledge of enzyme–substrate pairs in glycolysis. The correct matching is A–II (Enolase), B–III (Pyruvic kinase), C–IV (Aldolase), D–I (Hexokinase), which corresponds to option (C).

The key is to recall the specific step in glycolysis where each enzyme acts and which molecule is its substrate. Instead of memorizing blindly, think of the metabolic pathway as a story: each enzyme recognizes a particular molecule and transforms it into the next.

Phosphoenol pyruvate (PEP) → Enolase (II)

Enolase catalyzes the dehydration of 2-phosphoglycerate to PEP. But wait — PEP is the product of enolase, not its substrate. However, the question asks for the enzyme that acts on the given molecule. In the reverse direction (gluconeogenesis) or in the forward direction, enolase is the enzyme that interconverts 2-phosphoglycerate and PEP. Since PEP is directly linked to enolase’s action, the match is A–II.

Pyruvic acid → Pyruvic kinase (III)

Pyruvic kinase catalyzes the final step of glycolysis: transfer of a phosphate from PEP to ADP, producing pyruvic acid (pyruvate) and ATP. So pyruvic acid is the product of pyruvic kinase. Again, the enzyme is named for its role in forming pyruvate. Thus B–III.

Triose phosphate → Aldolase (IV)

Aldolase splits fructose-1,6-bisphosphate into two triose phosphates: dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P). So triose phosphates are the products of aldolase. Hence C–IV.

Glucose-6-phosphate → Hexokinase (I)

Hexokinase phosphorylates glucose to glucose-6-phosphate (using ATP). So glucose-6-phosphate is the product of hexokinase. Thus D–I.

Tip

Notice that in each case, the enzyme is named after the product it creates (e.g., pyruvic kinase makes pyruvate, hexokinase makes glucose-6-phosphate). This pattern helps avoid confusion.

Watch out

A common mistake is to match pyruvic acid with enolase because both sound “pyruvate-like.” But enolase works on 2-phosphoglycerate, not pyruvate. Always trace the step in the pathway.

Thus the correct sequence is A–II, B–III, C–IV, D–I, which matches option (C).

✓Final answer
$The correct option is (C) . ANSWER: C$

←Previous question10.1 — What are monosaccharides?→Next question10.3 — Write two main functions of carbohydrates in plants.

Q.What are reducing sugars?

Concept understanding — Lactose Hydrolysis Products

Lactose Hydrolysis Products – From Intuition to Precision

The Intuition: Breaking a Sugar Chain

The Precise Statement

Why This Matters for Exams

Why this formula?

Lactose Hydrolysis Products — Understanding the Why

1. What is lactose chemically?

2. What does hydrolysis do?

3. Why are the products exactly galactose and glucose?

4. Key formula — the hydrolysis equation

5. The "why" behind the formula

6. Exam tip — what to remember

The Core Idea: Why “Reducing”?

Step-by-Step Breakdown

1. The Structural Requirement

2. The Mechanism in Benedict’s Test

3. Which Sugars Are Reducing? A Quick Table

4. The Special Case of Sucrose

5. Why Does This Matter in Exams?

Final Answer

Method: Classifying a Sugar as Reducing or Non-Reducing

Step 1: Recall the defining behaviour

Step 2: Find the structural basis

Step 3: Apply to each class of sugar

Step 4: Confirm with the test

Final Answer

✗ Mistake 1: Confusing "reducing" with "sweet" or "easily digested"

✗ Mistake 2: Thinking ketoses cannot be reducing sugars

✗ Mistake 3: Getting the disaccharides backwards

✗ Mistake 4: Thinking "has a glycosidic bond" automatically means non-reducing

✗ Mistake 5: Omitting the reagents from the answer

✓ Quick Revision Table

Concept and Intuition

Step-by-Step Explanation

Q.What are reducing sugars?

The Core Idea: Why “Reducing”?

Step-by-Step Breakdown

1. The Structural Requirement

2. The Mechanism in Benedict’s Test

3. Which Sugars Are Reducing? A Quick Table

4. The Special Case of Sucrose

5. Why Does This Matter in Exams?

Final Answer

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