The activation energy for the acid catalysed hydrolysis of sucrose is 6.22 kJ mol$^{-1}$, while the activation energy is only 2.15 kJ mol$^{-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain.

The activation energy for the acid catalysed hydrolysis of sucrose is 6.22 kJ mol⁻¹, while the activation ener

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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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The enzyme sucrase lowers the activation energy barrier for sucrose hydrolysis from 6.22 kJ mol⁻¹ to 2.15 kJ mol⁻¹ by providing an alternative reaction pathway through an enzyme–substrate complex, making the reaction dramatically faster at biological temperatures.

The question asks you to explain why the activation energy ( $E_a$ ) for sucrose hydrolysis drops so sharply when the catalyst changes from an acid (H⁺) to the enzyme sucrase. The numbers themselves are striking: 6.22 kJ mol⁻¹ vs 2.15 kJ mol⁻¹ — a reduction of about 65%. That is not a small tweak; it is a fundamental change in how the reaction proceeds.

Let’s first recall what activation energy means. It is the minimum energy that reactant molecules must possess for a collision to result in a reaction. A lower $E_a$ means a much larger fraction of molecules have enough energy at a given temperature, so the rate constant $k$ increases exponentially (Arrhenius equation: $k = A e^{-E_a/RT}$ ). So the enzyme makes the reaction happen millions of times faster than the acid-catalysed version under the same conditions.

Now, why does this happen?

Acid catalysis works by brute force. In the acid-catalysed hydrolysis, H⁺ ions (from a strong acid like HCl) protonate the glycosidic oxygen in sucrose. This makes the C–O bond more polar and easier to break. But the transition state — the highest-energy, most unstable arrangement along the reaction coordinate — still involves a highly strained, high-energy species. The acid simply lowers the barrier a bit by stabilising the protonated intermediate, but the pathway remains essentially the same as the uncatalysed reaction, just with a slightly lower hill to climb.
Enzyme catalysis creates a completely new path. Sucrase (also called invertase) is a protein with an active site that binds the sucrose molecule in a precise orientation. The binding itself is not just a passive grip — it involves multiple non-covalent interactions (hydrogen bonds, van der Waals forces, hydrophobic effects) that distort the sucrose molecule toward its transition-state geometry. This is called transition-state stabilisation. The enzyme’s active site is complementary to the transition state, not to the reactant sucrose. So as the substrate binds, the enzyme strains bonds and positions catalytic groups (often acidic/basic amino acid side chains) to directly participate in bond breaking and forming.
The enzyme provides a lower-energy alternative route. Instead of going over the same high hill as the acid-catalysed reaction, the enzyme–substrate complex (ES) proceeds through a series of intermediates (e.g., a covalent glycosyl-enzyme intermediate) that are each much lower in energy than the acid-catalysed transition state. The activation energy of 2.15 kJ mol⁻¹ corresponds to the highest barrier along this new, multi-step pathway — which is far lower than the single-step barrier in the acid-catalysed case.

Tip

A useful mental picture: acid catalysis is like climbing a steep hill with a walking stick; enzyme catalysis is like finding a tunnel through the hill. Both get you to the other side, but the tunnel is far easier.

The entropy factor also matters. In the acid-catalysed reaction, two separate species (sucrose and water) must come together in the correct orientation for hydrolysis — this costs entropy. In the enzyme-catalysed reaction, the substrate is held in the active site with water already positioned nearby, so the entropic cost is largely prepaid by the binding step. This further lowers the free energy of activation ( $\Delta G^\ddagger$ ), which is what the experimental $E_a$ reflects when measured from the temperature dependence of the rate. …

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Common Mistakes & How to Avoid Them

Mistake 1: Thinking "lower activation energy means slower reaction"

The error: Students sometimes confuse the relationship — they see a smaller number (2.15 kJ mol⁻¹) and assume it means the reaction is "weaker" or "slower."

Why it's wrong:

Activation energy ( $E_a$ ) is the energy barrier that reactant molecules must overcome.

Lower $E_a$ → more molecules have enough energy → faster reaction
Higher $E_a$ → fewer molecules can cross the barrier → slower reaction

How to avoid:

Always remember the Arrhenius equation:

k = A e^{-E_a / RT}

$k$ = rate constant
$E_a$ appears in the negative exponent
Smaller $E_a$ → larger $k$ → faster reaction

So the enzyme (sucrase) with $E_a = 2.15$ kJ mol⁻¹ makes the reaction much faster than the acid-catalysed route with $E_a = 6.22$ kJ mol⁻¹.

Mistake 2: Saying "enzymes change the equilibrium constant"

The error: Students claim that because the enzyme lowers activation energy, it also shifts the equilibrium position or changes $\Delta G$ .

Why it's wrong:

Activation energy ( $E_a$ ) affects how fast equilibrium is reached, not the equilibrium constant ( $K_{eq}$ )
The Gibbs free energy change ( $\Delta G$ ) for the overall reaction remains the same with or without the enzyme
Enzymes are catalysts — they lower $E_a$ for both forward and reverse reactions equally

How to avoid:

Draw a reaction energy diagram:

The starting point (reactants) and ending point (products) are at the same energy levels
Only the peak (transition state) is lowered by the enzyme
The net $\Delta G$ is unchanged

Mistake 3: Confusing "catalyst" with "reactant" or "cofactor"

The error: Students write that the enzyme "provides energy" or "gets used up" in the reaction.

Why it's wrong:

Enzymes are biological catalysts — they are not consumed
They work by providing an alternative reaction pathway with a lower activation energy
The enzyme binds to the substrate (sucrose) at its active site, stabilising the transition state

How to avoid:

Remember the lock-and-key model (or induced fit model):

Enzyme = lock
Substrate = key
The enzyme temporarily binds the substrate, lowers the energy barrier, and then releases the product unchanged

Mistake 4: Ignoring the "why" — not explaining the mechanism

The error: Students simply state "enzyme lowers activation energy" without explaining how.

Why it's wrong:

The question asks "Explain." — you must give the reason for the difference.

Correct explanation:

Acid catalysis: H⁺ ions protonate the glycosidic oxygen, making it a better leaving group — but this still requires significant energy
Enzyme catalysis (sucrase): The enzyme's active site binds sucrose and stabilises the transition state through:
- Hydrogen bonding
- Electrostatic interactions
- Proper orientation of the substrate This lowers the energy barrier much more effectively than acid alone

How to avoid:

Always include two parts in your answer:

What happens (lower $E_a$ ) …

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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.The activation energy for the acid catalysed hydrolysis of sucrose is 6.22 kJ mol $^{-1}$ , while the activation energy is only 2.15 kJ mol $^{-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain.

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

Concept: Activation Energy and Catalysis

Method: Comparison of Activation Energies

Steps:

Common Mistakes & How to Avoid Them

Mistake 1: Thinking "lower activation energy means slower reaction"

Mistake 2: Saying "enzymes change the equilibrium constant"

Mistake 3: Confusing "catalyst" with "reactant" or "cofactor"

Mistake 4: Ignoring the "why" — not explaining the mechanism

Concept and Intuition

Step-by-Step Explanation

Q.The activation energy for the acid catalysed hydrolysis of sucrose is 6.22 kJ mol−1^{-1}−1, while the activation energy is only 2.15 kJ mol−1^{-1}−1 when hydrolysis is catalysed by the enzyme sucrase. Explain.

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Q.The activation energy for the acid catalysed hydrolysis of sucrose is 6.22 kJ mol $^{-1}$ , while the activation energy is only 2.15 kJ mol $^{-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain.

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