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.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$\alpha$
›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. …
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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. …
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TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$\alpha$
›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. …
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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. …
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TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$\alpha$
›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" …
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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. …
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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 …
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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) …
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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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