Match the following enzymes given in Column I with the reactions they catalyse given in Column II. Column I (E

NCERT Exemplar · Q64

Q. Match the following enzymes given in Column I with the reactions they catalyse given in Column II. Column I (Enzymes): (A) Invertase; (B) Maltase; (C) Pepsin; (D) Urease; (E) Zymase. Column II (Reactions):

(1) Decomposition of urea into NH

_3

and CO

_2

;

(2) Conversion of glucose into ethyl alcohol;

(3) Hydrolysis of maltose into glucose;

(4) Hydrolysis of cane sugar;

(5) Hydrolysis of proteins into peptides.

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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. …

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. …

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This is a matching problem linking enzymes to their specific biochemical reactions. The correct matches are: Invertase → hydrolysis of cane sugar (4), Maltase → hydrolysis of maltose into glucose (3), Pepsin → hydrolysis of proteins into peptides (5), Urease → decomposition of urea into NH₃ and CO₂ (1), Zymase → conversion of glucose into ethyl alcohol (2).

The key to solving this lies in understanding what each enzyme actually does — not just memorising names, but knowing the substrate it acts on and the product it yields. Enzymes are highly specific; each one catalyses a particular chemical transformation.

Let’s go through them one by one.

Invertase (A) – This enzyme acts on sucrose (cane sugar). Sucrose is a disaccharide made of glucose and fructose. Invertase hydrolyses the glycosidic bond in sucrose, yielding an equimolar mixture of glucose and fructose, often called "invert sugar." So it matches with (4) Hydrolysis of cane sugar.
Maltase (B) – Maltose is another disaccharide, composed of two glucose units linked by an α-1,4 bond. Maltase specifically hydrolyses this bond to produce two molecules of glucose. Hence it matches with (3) Hydrolysis of maltose into glucose.
Pepsin (C) – This is a proteolytic enzyme (a protease) secreted in the stomach. It breaks down proteins into smaller peptide fragments (not individual amino acids — that comes later). So it matches with (5) Hydrolysis of proteins into peptides.
Urease (D) – This enzyme catalyses the breakdown of urea into ammonia and carbon dioxide. Urea is a waste product from protein metabolism, and urease is found in certain bacteria and plants. The reaction is:

$\text{CO(NH}_2)_2 + \text{H}_2\text{O} \xrightarrow{\text{urease}} 2\text{NH}_3 + \text{CO}_2$

So it matches with (1) Decomposition of urea into NH₃ and CO₂. …

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Concept: Enzyme-Substrate Specificity

Enzymes are biological catalysts that act on specific substrates. The correct matching depends on knowing the substrate each enzyme acts upon and the product formed.

Method: Substrate–Product Mapping

Step 1: Identify the substrate for each enzyme.

Enzyme	Substrate
Invertase	Cane sugar (sucrose)
Maltase	Maltose
Pepsin	Proteins
Urease	Urea
Zymase	Glucose (or fructose)

Step 2: Identify the reaction each enzyme catalyses.

Invertase hydrolyses sucrose → glucose + fructose (inversion of sugar) → matches (4)
Maltase hydrolyses maltose → 2 glucose molecules → matches (3)
Pepsin breaks proteins → peptides (proteolysis) → matches (5)
Urease decomposes urea → $\text{NH}_3 + \text{CO}_2$ → matches (1)
Zymase (yeast enzyme complex) converts glucose → ethanol + $\text{CO}_2$ (fermentation) → matches (2) …

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Here is a breakdown of the common mistakes students make when matching enzymes to their reactions, along with strategies to avoid them.

Mistake 1: Confusing Invertase with Maltase (Sucrose vs. Maltose)

The Mistake:

Students often mix up which sugar is broken down by which enzyme. They might match Invertase (A) with the hydrolysis of maltose (Reaction 3), or Maltase (B) with the hydrolysis of cane sugar (Reaction 4).

Why it happens:

Both are disaccharide-digesting enzymes, and the names sound similar. The key is to remember the source of the sugar name.

Malt (from barley) contains maltose.
Invert sugar is a mixture of glucose and fructose, produced from sucrose (cane sugar).

How to Avoid:

Memorize the specific substrate: Invertase acts on sucrose (cane sugar). Maltase acts on maltose.
Use a mnemonic: "Maltase breaks maltose" (both start with "Malt"). Therefore, Invertase must break the other one (cane sugar/sucrose).

Mistake 2: Confusing Zymase with Invertase (Fermentation vs. Hydrolysis)

The Mistake:

Students match Zymase (E) with the hydrolysis of cane sugar (Reaction 4) or match Invertase (A) with the conversion of glucose to alcohol (Reaction 2).

Why it happens:

Both enzymes are involved in the overall process of making alcohol from sugar cane. Students forget the sequence of reactions.

First, Invertase hydrolyses sucrose into glucose and fructose.
Then, Zymase ferments those simple sugars into ethanol and CO $_2$ .

How to Avoid:

Learn the sequence: In the production of alcohol, Invertase is the preparation step (breaking down the big sugar), and Zymase is the final step (making the alcohol).
Focus on the key word: Zymase is associated with "conversion into ethyl alcohol" (fermentation). Invertase is associated with "hydrolysis" (breaking down).

Mistake 3: Forgetting the Specific Product of Pepsin

The Mistake:

Students match Pepsin (C) with the hydrolysis of proteins into amino acids (a common wrong answer), rather than into peptides (Reaction 5).

Why it happens:

Students know pepsin digests proteins, but they forget that digestion is a stepwise process. Pepsin works in the stomach and only breaks proteins into smaller chains (peptides). The final breakdown into amino acids happens later in the small intestine with other enzymes (trypsin, peptidases).

How to Avoid:

Remember the "P" rule: Pepsin produces Peptides.
Visualize the process: Think of a long protein chain. Pepsin is the "rough cutter" that makes medium-sized pieces (peptides). Other enzymes are the "fine cutters" that make individual amino acids. …

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

AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$\alpha-D-glucose$
›Reveal solutionSolution
Both statements accurately describe the standard disaccharide compositions of sucrose and lactose.

Concept and Intuition

Disaccharides are named by which two monosaccharide units (and in which anomeric form) are joined by a glycosidic bond. Sucrose's non-reducing character comes specifically from the fact that BOTH anomeric carbons (C1 of glucose and C2 of fructose) are involved in the glycosidic bond, locking the ring forms as $\alpha$ -D-glucose and $\beta$ -D-fructose. Lactose, by contrast, is a reducing sugar because glucose's anomeric carbon is left free; the fixed unit is $\beta$ -D-galactose joined via $\beta$ -1,4 linkage to D-glucose, and standard descriptions state it as $\beta$ -D-galactose and $\beta$ -D-glucose.

Step-by-Step Solution

Statement I: Sucrose = $\alpha$ -D-glucopyranose + $\beta$ -D-fructofuranose, linked $C1(glucose) \to C2(fructose)$ . This is the textbook description. Correct. …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$\alpha$
›Reveal solutionSolution
Tests whether you remember lactose's actual monomer composition and why it is still classed as a reducing sugar.

Concept and Intuition

Disaccharides are named by which two monosaccharides are linked and through which carbons. Sucrose (glucose + fructose, both anomeric carbons involved) is the classic non-reducing sugar because neither free anomeric OH survives the glycosidic bond. Lactose and maltose are the classic reducing sugars because one anomeric carbon is left free.

Step-by-Step Solution

Lactose = $\beta$ -D-galactose $+$ D-glucose, joined $\beta(1\to4)$ between galactose C1 and glucose C4.

Statement-I claims lactose is " $\alpha$ -D-glucose $+$ $\beta$ -D-glucose" — this describes maltose's/only-glucose composition, not lactose's actual galactose+glucose composition. False.

Because the glycosidic bond uses galactose's C1 and glucose's C4, glucose's own C1 (anomeric carbon) stays free with a free -OH. …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\alpha$
›Reveal solutionSolution
This tests the exact monosaccharide composition and anomeric forms in sucrose vs lactose. The answer is (C).

Concept and Intuition

Disaccharides are formed by a glycosidic linkage between two monosaccharide units, and the specific anomeric form ( $\alpha$ or $\beta$ ) of each unit is a precise structural fact that must be remembered correctly, since sucrose and lactose have different compositions and linkages.

Step-by-Step Solution

Sucrose (cane sugar): formed by a glycosidic bond between C1 of $\alpha$ -D-glucose and C2 of $\beta$ -D-fructose. Statement-I matches this exactly — correct.

Lactose (milk sugar): formed by a glycosidic bond between C1 of $\beta$ -D-galactose and C4 of $\beta$ -D-glucose (glucose unit here is in $\beta$ form as it provides the free anomeric carbon, though the ring can open to $\alpha$ / $\beta$ equilibrium — the standard textbook description names $\beta$ -D-galactose and glucose, not $\alpha$ -D-glucose). …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$Q. Given below are two statements Assertion (A): Hydrolysis of sucrose results in change in the optical rotation from dextro (+) to laevo (-) Reason (R): Both the products from the hydrolysis are leavorotatory The correct answer is (A) Both A and R are correct and R in the correct explanation of A (B) Both A and R are correct but R is not the correct explanation of A (C) A is correct but R is incorrect (D) A is incorrect but R is correct$
›Reveal solutionSolution
The assertion is correct: sucrose hydrolysis inverts optical rotation from dextro to laevo. The reason is incorrect because one product (glucose) is dextrorotatory, not both laevorotatory. So the correct choice is (C).

Concept & Intuition

Optical rotation measures how a substance rotates plane-polarized light. Sucrose is a disaccharide made of glucose and fructose. When hydrolyzed, it breaks into these two monosaccharides. The key is that sucrose itself is dextrorotatory (rotates light to the right, +), but the mixture of glucose and fructose after hydrolysis is laevorotatory (rotates light to the left, –). This phenomenon is called inversion of sucrose, and the product mixture is called invert sugar. The reason given claims both products are laevorotatory — that’s the trap. In reality, glucose is dextrorotatory, fructose is strongly laevorotatory, and the net effect is laevorotatory because fructose’s leftward rotation outweighs glucose’s rightward rotation.

Step-by-step reasoning

Identify the specific rotations

Sucrose: $[ \alpha ]_D = +66.5^\circ$ (dextrorotatory)

Glucose: $[ \alpha ]_D = +52.7^\circ$ (dextrorotatory)

Fructose: $[ \alpha ]_D = -92.4^\circ$ (laevorotatory)

Hydrolysis reaction

$\text{Sucrose} + \text{H}_2\text{O} \rightarrow \text{Glucose} + \text{Fructose}$
One molecule of sucrose yields one molecule each of glucose and fructose.

Net rotation after hydrolysis The observed rotation of the mixture is the weighted average of the rotations of the products. Since both are produced in equal molar amounts:

$\text{Net rotation} = \frac{(+52.7^\circ) + (-92.4^\circ)}{2} = \frac{-39.7^\circ}{2} = -19.85^\circ$
This is negative (laevorotatory). So the mixture is laevorotatory, even though glucose alone is dextrorotatory.

Evaluate Assertion (A) …
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AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
$\beta$
›Reveal solutionSolution
Cellulose has only $\beta$ -1,4 glycosidic bonds; starch (via amylopectin's branching) has both $\alpha$ -1,4 and $\alpha$ -1,6 bonds.

Concept and Intuition

Both cellulose and starch are glucose polymers, but the way the glucose units are joined determines their structure and digestibility. Cellulose is built entirely of $\beta$ -D-glucose units connected end to end by $\beta$ -1,4-glycosidic bonds. This linkage lets the chains lie flat and hydrogen-bond into rigid, fibrous microfibrils — ideal for cell walls, but not digestible by human enzymes (which only cleave $\alpha$ linkages).

Starch, by contrast, is made of $\alpha$ -D-glucose units. Its two components are amylose (a straight chain held together by $\alpha$ -1,4 bonds) and amylopectin (a branched molecule with an $\alpha$ -1,4 backbone plus $\alpha$ -1,6 bonds at branch points, roughly every 24–30 residues). Because starch as a storage polysaccharide is really this combination, both $\alpha$ -1,4 and $\alpha$ -1,6 bonds are correctly attributed to it.

Step-by-Step Solution

Identify cellulose's bond type: only $\beta$ -1,4-glycosidic bonds (no branching, no $\alpha$ bonds). …
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AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$Q. Hydrolysis of sucrose gives (A) Dextrorotatory glucose & Laevorotatory fructose (B) Dextrorotatory fructose & Laevorotatory glucose (C) Dextrorotatory glucose & Dextrorotatory fructose (D) Laevorotatory glucose & Laevorotatory fructose$
›Reveal solutionSolution
Hydrolysis of sucrose ("inversion") gives dextrorotatory glucose and laevorotatory fructose — the sign flip (net + to net −) is why it's called inversion of sugar.

Concept and Intuition

Sucrose is built from $\alpha$ -D-glucopyranose and $\beta$ -D-fructofuranose joined C1–C2 through their anomeric carbons, which locks both anomeric centres and makes sucrose a non-reducing sugar with no free aldehyde/ketone. Hydrolysing this glycosidic bond liberates both monosaccharides in their free, mutarotating forms, each with its own intrinsic optical rotation.

Step-by-Step Solution

Sucrose itself is dextrorotatory, $[\alpha]_D = +66.5^\circ$ .

Acid hydrolysis (or the enzyme invertase) breaks the glycosidic bond: $\text{Sucrose} + H_2O \rightarrow \text{Glucose} + \text{Fructose}$ .

Free D-glucose is dextrorotatory, $[\alpha]_D = +52.5^\circ$ .

Free D-fructose is strongly laevorotatory, $[\alpha]_D = -92^\circ$ (fructose's rotation is large and negative — it is sometimes called laevulose for this reason). …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$Q. Hydrolysis of which disaccharide in presence of enzyme maltase give glucose only? (A) Sucrose (B) Cellulose (C) Lactose (D) Maltose$
›Reveal solutionSolution
Maltose is a disaccharide built from two glucose units, so its enzymatic hydrolysis by maltase produces glucose exclusively — unlike sucrose or lactose, which each yield two different monosaccharides.

Concept and Intuition

Disaccharides hydrolyse into their two constituent monosaccharides, and the specific enzyme named must match the specific glycosidic bond being cleaved. Maltase is the enzyme that hydrolyses the $\alpha(1\to4)$ bond in maltose; since maltose's two building blocks are both glucose, hydrolysis gives only glucose as product.

Step-by-Step Solution

Maltose = glucose + glucose (joined by an $\alpha(1\to4)$ glycosidic linkage). Enzyme maltase hydrolyses this bond $\Rightarrow$ 2 glucose molecules only.

Sucrose = glucose + fructose, hydrolysed by sucrase/invertase $\Rightarrow$ gives glucose and fructose, not glucose alone.

Lactose = glucose + galactose, hydrolysed by lactase $\Rightarrow$ gives glucose and galactose, not glucose alone. …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$Q. If sucrose is boiled with dilute. HCl in alcoholic solution the ratio in which glucose and fructose are formed is (A) 1:1 (B) 1:2 (C) 2:1 (D) 4:1$
›Reveal solutionSolution
Acid hydrolysis of sucrose (inversion) cleaves its single glycosidic linkage to give exactly one glucose and one fructose molecule per sucrose molecule — a 1:1 ratio, option (A).

Concept and Intuition

Sucrose is a disaccharide formed by the condensation of one molecule of alpha-D-glucose and one molecule of beta-D-fructose, joined through a glycosidic linkage between C1 of glucose and C2 of fructose. Because this glycosidic bond is the only bond joining the two monosaccharide units, hydrolyzing it (by boiling with dilute acid, a reaction historically called 'inversion' because the optical rotation changes sign) breaks sucrose into exactly one glucose unit and one fructose unit — there is no possibility of an unequal split, since each sucrose molecule contains precisely one of each monosaccharide.

Step-by-Step Solution

Recall the structure of sucrose: glucose + fructose joined by one glycosidic bond (1→2 linkage), with the molecular formula $C_{12}H_{22}O_{11}$ .

Acid hydrolysis reaction: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6\,(\text{glucose}) + C_6H_{12}O_6\,(\text{fructose})$ . …
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AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$(C_6H_{10}O_5)_n + nH_2O \xrightarrow[2\text{-}3\ atm]{H^+,\ 393K}$
›Reveal solutionSolution
Complete acid hydrolysis of starch under heat and pressure breaks the glycosidic bonds all the way down to its monosaccharide unit, glucose.

Concept and Intuition

Starch is a polysaccharide made of many glucose units linked by glycosidic bonds ( $\alpha$ -1,4 and $\alpha$ -1,6 linkages in amylose/amylopectin). Acid-catalyzed hydrolysis under heat and elevated pressure cleaves all these glycosidic bonds completely, releasing the individual glucose monomer units — this is the industrial process used to make glucose syrup from starch.

Step-by-Step Solution

Starch's repeating unit formula is $(C_6H_{10}O_5)_n$ .

Complete hydrolysis adds one water molecule per glycosidic bond broken: $(C_6H_{10}O_5)_n + nH_2O \xrightarrow[2\text{-}3\ atm]{H^+, 393K} nC_6H_{12}O_6$ .

The product, $C_6H_{12}O_6$ , is glucose — the single repeating monosaccharide unit of starch. …
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Q.Match the following enzymes given in Column I with the reactions they catalyse given in Column II. Column I (Enzymes): (A) Invertase; (B) Maltase; (C) Pepsin; (D) Urease; (E) Zymase. Column II (Reactions):

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