In a DNA strand the nucleotides are linked together by: (a) glycosidic bonds (b) phosphodiester bonds (c) pept

NCERT Exemplar · Q1

Q. In a DNA strand the nucleotides are linked together by:

(a) glycosidic bonds

(b) phosphodiester bonds

(d) hydrogen bonds

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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.

In a DNA strand, the individual nucleotides are joined end-to-end by phosphodiester bonds — the backbone of each strand is a sugar-phosphate chain held together by these covalent links.

The question asks how nucleotides are linked within a single DNA strand. This is a fundamental point about nucleic acid structure, and the answer is straightforward once you picture the chain.

Each nucleotide has three components: a phosphate group, a sugar (deoxyribose in DNA), and a nitrogenous base. When nucleotides polymerise to form a strand, the phosphate group of one nucleotide forms a covalent bond with the sugar of the next nucleotide. Specifically, the bond is between the 5' carbon of one sugar and the 3' carbon of the next sugar. This is called a phosphodiester bond because it involves a phosphate group that is now linked to two sugar molecules via ester linkages.

Note

The term "phosphodiester" tells you exactly what it is: a phosphate (phospho-) making two ester bonds (-diester) with two different sugar molecules.

So option (B) is correct. Let's quickly see why the other options are wrong, because each of those bonds appears elsewhere in biology and the confusion is common.

Glycosidic bonds link a sugar to another molecule. In DNA, the base is attached to the sugar by an N-glycosidic bond — but that's within a single nucleotide, not between nucleotides.
Peptide bonds are the links between amino acids in proteins. They have nothing to do with nucleic acids.
Hydrogen bonds are crucial in DNA, but they hold the two strands together (between complementary bases), not the nucleotides within a single strand.

Important

A single DNA strand is held together by strong covalent phosphodiester bonds. The two strands are held together by weaker hydrogen bonds between bases. These are two completely different types of bonding serving different structural roles.

✓Final answer

In a DNA strand, the nucleotides are linked together by phosphodiester bonds (option B). These covalent bonds form the sugar-phosphate backbone of each strand.

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

COMEDK 2025Set 2025-M1 markMCQ
$\alpha-D-$
›Reveal solutionSolution
Maltose is a reducing disaccharide from starch hydrolysis, but its hydrolysis yields two glucose units, not glucose and fructose; thus Assertion is correct, Reason is wrong, so option (C) is correct.

Concept & Intuition

Maltose is a disaccharide composed of two glucose molecules linked by an α(1→4) glycosidic bond. Because one of the glucose units retains a free anomeric carbon (the hemiacetal group), maltose can reduce Cu²⁺ ions (e.g., in Benedict’s test) — that’s what makes it a reducing sugar. The assertion correctly states that maltose is obtained by partial hydrolysis of starch using the enzyme diastase. However, the reason claims that hydrolysis of maltose gives one glucose and one fructose — that’s actually the hydrolysis product of sucrose, not maltose. The classic pitfall here is confusing the hydrolysis products of common disaccharides.

Step-by-step reasoning

Check the Assertion (A)

Maltose is indeed a reducing sugar because its structure has a free anomeric carbon on the non-reducing end? Actually, careful: In maltose, the glycosidic bond involves C1 of one glucose and C4 of the other. The glucose unit that provides C1 has its anomeric carbon tied up in the bond, so it cannot open to a free aldehyde. But the other glucose unit (the one with the free C1) retains a hemiacetal group, which can open to an aldehyde form. Thus maltose reduces Tollens’ or Benedict’s reagent.

Partial hydrolysis of starch (a polymer of glucose) with the enzyme diastase (an amylase) does yield maltose as a major product.

Therefore, Assertion (A) is correct.

Check the Reason (R)

Hydrolysis of one mole of maltose (C₁₂H₂₂O₁₁) with water yields two moles of D-glucose.

The reaction:

$\text{Maltose} + \text{H}_2\text{O} \xrightarrow{\text{acid or enzyme}} 2\, \text{D-glucose}$

The reason says one mole each of α-D-glucose and β-D-fructose. Fructose is a ketohexose, not a component of maltose. That description matches sucrose (which gives glucose + fructose).

Therefore, Reason (R) is wrong.

Determine the relationship

Since A is correct and R is incorrect, the correct option is the one that says “A is correct but R is wrong.”

Watch out

A common mistake is to think maltose gives glucose and fructose because “disaccharide” sounds like it always splits into two different monosaccharides. In fact, maltose = glucose + glucose; lactose = glucose + galactose; only sucrose gives glucose + fructose.

Tip

Remember: “Maltose = malt sugar = two glucoses” — the name even hints at “malt” (from barley), which is all about glucose polymers.

✓Final answer
$The correct option is (C). ANSWER: C$
KCET 2024Set B-41 markMCQ
$Q. Stanley Miller simulated the conditions of pre-biotic earth using spark-discharge apparatus. Which organic compounds were observed by him on analysing the end product of his experiment? (A) Pigments (B) Fats (C) Nitrogen bases (D) Amino acids$
›Reveal solutionSolution
Miller's spark-discharge experiment recovered simple amino acids, demonstrating that organic monomers can form abiotically from a reducing atmosphere.

Step 1 — What the experiment set out to test.

Oparin (Russia) and Haldane (England) had proposed that the first form of life arose from pre-existing non-living organic molecules (e.g. RNA, proteins) — chemical evolution preceding biological evolution. The claim needed an experimental test: can organic molecules form from purely inorganic precursors under primitive-Earth conditions?

Step 2 — The apparatus and the conditions simulated.

Stanley L. Miller, a student of Harold Urey, built a closed flask (1953) and created conditions like those of the primitive, reducing (oxygen-free) atmosphere:

Gases: methane ( $\mathrm{CH_4}$ ), ammonia ( $\mathrm{NH_3}$ ) and hydrogen ( $\mathrm{H_2}$ ), in a $2:1:2$ ratio.

Water vapour supplied by boiling water in a connected flask.

Temperature about $800^\circ\text{C}$ .

Electric discharge between electrodes, simulating lightning — the energy source.

The vapour was condensed and collected in a U-trap, and the accumulated liquid analysed after a week.

Step 3 — The result.

He observed the formation of amino acids (such as glycine, alanine and aspartic acid) — the monomers of proteins. Similar experiments by others later yielded sugars, nitrogen bases and pigments, and analyses of meteorite contents independently revealed the same kinds of compounds — corroborating that this chemistry occurs abiotically elsewhere too. But the compound class Miller himself observed and reported was amino acids.

Step 4 — Why the other options are wrong here.

(A) Pigments and (B) Fats — reported from other, later simulation experiments, not from Miller's analysis. ✗

(C) Nitrogen bases — likewise obtained in subsequent experiments (and found in meteorites); they are not the classic result attributed to Miller. ✗

(D) Amino acids — the textbook result of the Miller–Urey spark-discharge experiment. ✓

Significance: the experiment converted the Oparin–Haldane idea from speculation into a testable, supported hypothesis: the building blocks of life can form spontaneously given a reducing atmosphere and an energy source.

✓Final answer
$The correct option is (D) — Amino acids.$

ANSWER: D
KCET 2023Set D-21 markMCQ
$Q. Sucrose is dextrorotatory but after hydrolysis the mixture show laevorotation, this is because of (A) Laevorotation of glucose is more than dextrorotation of fructose. (B) Sucrose is a non-reducing sugar. (C) Racemic mixture is formed. (D) Laevorotation of fructose is more than dextrorotation of glucose.$
›Reveal solutionSolution
Add the specific rotations of the two hydrolysis products: fructose's large negative rotation outweighs glucose's smaller positive one, so the sign of the mixture inverts.

Step 1 — The reaction
$\underset{\text{sucrose}}{\mathrm{C_{12}H_{22}O_{11}}} + \mathrm{H_2O} \xrightarrow{\;\mathrm{H^+/invertase}\;} \underset{\text{glucose}}{\mathrm{C_6H_{12}O_6}} + \underset{\text{fructose}}{\mathrm{C_6H_{12}O_6}}$
Sucrose is a 1→2 glycosidic disaccharide of $\alpha$ -D-glucose and $\beta$ -D-fructose. Hydrolysis cleaves it into equimolar glucose and fructose.

Step 2 — The specific rotations

Species Specific rotation $[\alpha]_D$
Sucrose $+66.5^\circ$ (dextrorotatory)
D-(+)-Glucose $+52.5^\circ$
D-(−)-Fructose $-92.4^\circ$

Step 3 — Net rotation of the product mixture

Optical rotations of components in a mixture are additive. For a 1:1 mixture the net rotation goes as
$[\alpha]_{\text{mix}} \propto \tfrac12\big(+52.5^\circ\big) + \tfrac12\big(-92.4^\circ\big) = \tfrac12\big(-39.9^\circ\big) < 0 .$
The magnitude of fructose's laevorotation ( $92.4^\circ$ ) exceeds glucose's dextrorotation ( $52.5^\circ$ ), so the sum is negative — the mixture rotates plane-polarised light to the left.

Step 4 — Why this is called "inversion"

The sign of rotation has literally inverted from $+$ (sucrose) to $-$ (product mixture). The 1:1 glucose–fructose mixture is therefore called invert sugar, and the enzyme that does it is invertase.

Step 5 — Rejecting the other options

(A) reverses the facts: glucose is dextrorotatory, fructose is laevorotatory.

(B) Sucrose is non-reducing (no free anomeric $-$ OH), but that is irrelevant to the sign of optical rotation.

(C) A racemic mixture would be optically inactive ( $0^\circ$ ), not laevorotatory — and glucose and fructose are not enantiomers anyway.

✓Final answer
$The correct option is (D) — Laevorotation of fructose is more than dextrorotation of glucose.$

ANSWER: D
COMEDK 2022Set 20221 markMCQ
$\alpha$
›Reveal solutionSolution

So the monosaccharide units of maltose are alpha-D-glucose and alpha-D-glucose.

Concept: structure of common disaccharides.

Maltose = two units of alpha-D-GLUCOSE joined by an alpha(1->4) glycosidic linkage (C1 of the first glucose to C4 of the second). It is a reducing sugar since the second glucose retains a free anomeric -OH.

Compare: sucrose = alpha-D-glucose + beta-D-fructose; lactose = beta-D-galactose + beta-D-glucose.

So the monosaccharide units of maltose are alpha-D-glucose and alpha-D-glucose.

✓Final answer
$\alpha$

ANSWER: A
COMEDK 2021Set 20211 markMCQ
$\alpha$
›Reveal solutionSolution
Hence the correct pair is alpha-D-glucopyranose and beta-D-fructofuranose.

Concept: Structure of the disaccharide sucrose.

Sucrose is a non-reducing disaccharide in which the anomeric C1 of glucose and the anomeric C2 of fructose are joined head-to-head through a glycosidic linkage, so no free anomeric -OH (hemiacetal) remains - that is why it is non-reducing.

The two units are:

alpha-D-glucopyranose (six-membered pyranose ring, alpha at C1)

beta-D-fructofuranose (five-membered furanose ring, beta at C2)

linked as alpha-1 -> beta-2 glycosidic bond.

Hence the correct pair is alpha-D-glucopyranose and beta-D-fructofuranose.

✓Final answer
$\alpha$

ANSWER: B

Species	Specific rotation $[\alpha]_D$
Sucrose	$+66.5^\circ$ (dextrorotatory)
D-(+)-Glucose	$+52.5^\circ$
D-(−)-Fructose	$-92.4^\circ$

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