Q.From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?

The question tests your understanding of the six major enzyme classes defined by the International Union of Biochemistry and Molecular Biology (IUBMB). Each class is named by the type of reaction it catalyzes. The key phrase here is "removal of groups … by mechanism other than hydrolysis leaving double bonds" — that is the precise definition of a lyase.

1. Isomerases (option A) catalyze geometric or structural rearrangements within a single molecule — they don't remove groups at all. So they're out.

2. Dehydrogenases (option B) are a subclass of oxidoreductases; they remove hydrogen atoms (a type of oxidation) but do not necessarily leave a double bond as the primary product. Their mechanism involves electron transfer, not the non-hydrolytic cleavage described.

3. Hydrolases (option C) cleave bonds by adding water — that's hydrolysis. The question explicitly says "by mechanism other than hydrolysis", so hydrolases are eliminated.

4. Lyases (option D) are enzymes that catalyze the cleavage of C–C, C–O, C–N, or other bonds by elimination, leaving double bonds (or adding groups to double bonds in the reverse direction). They do not use hydrolysis or oxidation. Classic examples: fumarase (converts fumarate to malate by adding water — but in the reverse direction, it removes water leaving a double bond), aldolase (cleaves fructose-1,6-bisphosphate into two trioses), and decarboxylases (remove CO₂ leaving a double bond). This matches the description perfectly.

The key to this question is understanding what defines a ligase. Enzyme classification (the EC system) groups enzymes by the type of reaction they catalyze, not by their name or the substrate alone. Class 6, ligases, are enzymes that join two molecules together by forming a new chemical bond (usually C–O, C–S, C–N, or C–C) and they always couple this bond formation to the hydrolysis of a high-energy phosphate compound — almost always ATP (or sometimes GTP, etc.). In short: ligases build bonds by burning ATP.

Now look at each option through that lens.

1. Glutamine synthetase — This enzyme catalyzes the reaction:

   $\text{glutamate} + \text{NH}_3 + \text{ATP} \rightarrow \text{glutamine} + \text{ADP} + \text{P}_i$

   It forms a C–N bond (an amide bond) between glutamate and ammonia, and it uses ATP to drive that otherwise unfavourable reaction. That is the textbook definition of a ligase. In fact, its EC number is 6.3.1.2 — the "6" confirms it belongs to class 6.

2. Hexokinase — This transfers a phosphate group from ATP to glucose:

   $\text{glucose} + \text{ATP} \rightarrow \text{glucose-6-phosphate} + \text{ADP}$

   That is a transferase (class 2), not a ligase. It moves a functional group; it does not join two separate molecules into one.

3. Malate dehydrogenase — This catalyzes:

   $\text{malate} + \text{NAD}^+ \rightleftharpoons \text{oxaloacetate} + \text{NADH} + \text{H}^+$

   It is an oxidoreductase (class 1) — it transfers electrons (hydride ion) between substrates. No ATP is involved, and no new bond is formed between two molecules.

4. Argininosuccinase — This enzyme (also called argininosuccinate lyase) catalyzes:

   $\text{argininosuccinate} \rightarrow \text{arginine} + \text{fumarate}$

   It breaks a bond, not forms one — it is a lyase (class 4). Lyases cleave bonds by means other than hydrolysis or oxidation, often without ATP. So it is the opposite of a ligase.

Evaluating the Assertion

"Enzymes are inactivated at low temperature." — True.

Each enzyme works best at an optimum temperature. Well below this optimum the enzyme is temporarily inactivated: molecular motion is minimal and catalytic activity effectively stops. The enzyme is not permanently damaged and regains activity as the temperature returns towards the optimum, but at low temperature it is inactive. In the NCERT framing this statement is accepted as true.

Evaluating the Reason

"Temperature changes the structure of the substrate." — False.

Temperature acts on the enzyme, not the substrate. At high temperature the enzyme's tertiary (three-dimensional) protein structure and its active site are disrupted (denaturation); at low temperature activity simply falls. The substrate is a small molecule whose structure is not altered by these ordinary temperature changes. So the Reason is incorrect, and it cannot explain the Assertion.

Conclusion

Assertion true, Reason false.

The question asks about the "essential component" of many co-enzymes. To answer this, you need to understand what a co-enzyme actually is and where it comes from.

An enzyme is a protein that speeds up a biochemical reaction. But many enzymes cannot work alone — they need a non-protein partner called a cofactor. Cofactors come in two types: inorganic (like metal ions) and organic (called co-enzymes). So a co-enzyme is specifically the organic type of cofactor.

Now, where do these organic co-enzymes come from? The body cannot synthesise most of them from scratch. Instead, they are derived from vitamins — the water-soluble B vitamins in particular. For example:

• Niacin (vitamin B₃) is used to make NAD⁺ and NADP⁺.
• Riboflavin (vitamin B₂) is used to make FAD and FMN.
• Thiamine (vitamin B₁) is used to make TPP (thiamine pyrophosphate).
• Pantothenic acid (vitamin B₅) is used to make coenzyme A.

So the essential component — the raw material — that the body uses to build co-enzymes is vitamins.

Let’s check the other options to be thorough:

1. Metal ions — These are indeed cofactors, but they are inorganic cofactors, not co-enzymes. Many enzymes need metal ions (e.g., Zn²⁺ in carbonic anhydrase, Mg²⁺ in kinases), but they are not the essential component of co-enzymes. Co-enzymes are organic, so metal ions are a separate category.

2. Hormones — These are chemical messengers (e.g., insulin, adrenaline) that regulate metabolism. They are not structural components of co-enzymes. Hormones and co-enzymes have completely different roles.

3. Lipids — Some lipids (like vitamin A, D, E, K) are fat-soluble vitamins, but they do not typically serve as co-enzymes. The co-enzymes are almost exclusively derived from water-soluble B vitamins. Lipids are not the essential component.

Concept and Intuition

The question refers to a landmark experiment by Avery, MacLeod, and McCarty, which aimed to identify the chemical nature of the "transforming principle" discovered by Frederick Griffith. Griffith's experiment showed that some substance from heat-killed virulent (S) bacteria could transform non-virulent (R) bacteria into virulent (S) bacteria. However, Griffith's work did not identify what this transforming substance was.

Avery, MacLeod, and McCarty took this a step further. Their intuition was that if they could selectively destroy each major type of biochemical molecule (proteins, RNA, and DNA) from the heat-killed S cells and then test if transformation still occurred, they could pinpoint which molecule was responsible for the transformation. If destroying a particular molecule stopped transformation, then that molecule must be the transforming principle.

They used specific enzymes for this purpose:

• Proteases: Enzymes that digest and destroy proteins.
• RNAase (Ribonuclease): An enzyme that digests and destroys RNA.
• DNAase (Deoxyribonuclease): An enzyme that digests and destroys DNA.

By observing the effect of each enzyme on the transformation process, they could deduce the chemical identity of the genetic material.

Step-by-step Explanation

1. Background: Griffith's Experiment and the Transforming Principle:

   Frederick Griffith, in 1928, demonstrated that a "transforming principle" existed. He observed that when live non-virulent R-strain Streptococcus pneumoniae bacteria were mixed with heat-killed virulent S-strain bacteria and injected into mice, the mice died. Live S-strain bacteria were recovered from the dead mice. This indicated that the R-strain bacteria had been transformed into the S-strain by some substance from the heat-killed S-strain.

2. Avery, MacLeod, and McCarty's Approach:

   In the 1940s, Oswald Avery, Colin MacLeod, and Maclyn McCarty sought to identify the chemical nature of this transforming principle. They purified biochemicals (proteins, RNA, and DNA) from heat-killed S-strain bacteria. They then set up experiments where they would add these purified biochemicals to cultures of live R-strain bacteria.

3. Testing with Specific Enzymes:

   To determine which specific molecule was the transforming principle, they treated the purified extract from heat-killed S cells with different enzymes before adding it to the R-strain bacteria:

   • Treatment with Proteases: When the extract was treated with proteases (enzymes that break down proteins), transformation of R cells into S cells still occurred. This indicated that proteins were not the transforming principle.
   • Treatment with RNAase: When the extract was treated with RNAase (an enzyme that breaks down RNA), transformation of R cells into S cells still occurred. This indicated that RNA was not the transforming principle.
   • Treatment with DNAase: When the extract was treated with DNAase (an enzyme that specifically breaks down DNA), the transformation of R cells into S cells was inhibited. The R cells remained R cells and did not acquire the S-strain characteristics.
   Important

   The key observation was that only when DNA was destroyed by DNAase did the transforming ability cease. This directly implicated DNA as the molecule responsible for carrying the genetic information for virulence.

4. Conclusion:

   The results clearly showed that the transforming principle was destroyed only when DNA was digested. This led Avery, MacLeod, and McCarty to conclude that DNA, not protein or RNA, was the genetic material responsible for heredity and transformation.

The question asks about the effect of digesting DNA with DNAase enzymes. Based on the experiment, this digestion led to the inhibition of transformation.

The question tests your knowledge of the three eukaryotic RNA polymerases and their specific products. In prokaryotes, a single RNA polymerase handles all transcription, but eukaryotes have evolved three distinct enzymes, each dedicated to a different set of genes. This specialisation is a classic exam favourite.

The logic is straightforward: you need to match each polymerase to the RNA it synthesises. Let's check each option.

1. Option (A): RNA polymerase I → rRNA

   RNA polymerase I is found in the nucleolus and transcribes the large ribosomal RNA genes (28S, 18S, and 5.8S rRNA in humans). This is correct.

2. Option (B): RNA polymerase II → hnRNA

   RNA polymerase II transcribes protein-coding genes, producing heterogeneous nuclear RNA (hnRNA), which is then processed into mRNA. This is also correct.

3. Option (C): RNA polymerase III → mRNA

   RNA polymerase III transcribes small, stable RNAs: tRNA, 5S rRNA, and some small nuclear RNAs (snRNA). It does not transcribe mRNA. mRNA is produced from hnRNA by RNA polymerase II. So this combination is wrong.

4. Option (D): DNA polymerase → DNA

   DNA polymerase synthesises new DNA strands during replication. This is correct.

Why RNA Viruses Have Higher Mutation Rates

The mutation rate during viral replication depends critically on the fidelity of the replication machinery. Let's understand why RNA viruses are the speed demons of mutation.

The Core Concept: Proofreading Makes the Difference

When genetic material is copied, errors inevitably occur. DNA polymerases have evolved sophisticated proofreading mechanisms (3'→5' exonuclease activity) that can detect and correct mismatched nucleotides immediately after they're incorporated. This is like having a spell-checker that catches typos as you type.

RNA polymerases (including RNA-dependent RNA polymerases used by RNA viruses) generally lack this proofreading ability. They're fast but sloppy copyists, incorporating incorrect nucleotides at rates 1000-10,000 times higher than DNA polymerases.

Analyzing Each Option

Let me walk through why each answer choice does or doesn't explain the mutation rate difference:

1. Option (A): "Presence of complementary strand in RNA"

   • This is misleading. Both RNA and DNA can exist as single-stranded or double-stranded forms depending on the virus type. Many RNA viruses are single-stranded, but some (like reoviruses) are double-stranded. The mutation rate difference isn't fundamentally about strand complementarity.

2. Option (B): "Absence of complementary strand in DNA"

   • This doesn't make sense. DNA viruses typically DO have complementary strands (double-stranded DNA). This option contains a factual error.

3. Option (C): "Presence of 2'–OH group in each nucleotide of DNA"

   • This is backwards! DNA lacks the 2'–OH group (it has just H at the 2' position — hence "deoxy"). RNA has the 2'–OH group. While this chemical difference is important for stability, it's not the primary reason for mutation rate differences.

4. Option (D): "Catalytic function of RNA"

   • This refers to the fact that RNA can act as both genetic material AND as a catalyst (ribozyme activity). RNA-dependent RNA polymerases are themselves proteins, but they evolved from or work with RNA-based catalytic systems. The key insight here is that RNA polymerases, which replicate RNA genomes, lack the sophisticated error-correction mechanisms of DNA polymerases. The "catalytic function" alludes to the enzymatic properties of the replication machinery.

The Biological Significance

The mutation rate for RNA viruses is approximately 10⁻⁴ to 10⁻⁶ substitutions per nucleotide per replication cycle, compared to 10⁻⁸ to 10⁻¹¹ for DNA-based organisms. This creates "quasi-species" populations where RNA viruses exist as swarms of related variants, enabling rapid evolution and immune evasion.

Concept & Intuition

This is a matching problem that tests your grasp of fundamental molecular biology facts:

• Genome sizes of model organisms (prokaryote vs. eukaryote).
• The directionality of nucleic acid synthesis (always $5' \to 3'$).
• The template strand polarity for continuous (leading) replication. The trick is to recall that E. coli has a much smaller genome than a human cell, and that DNA polymerase (and RNA polymerase) always adds nucleotides to the $3'$ end, so the new strand grows $5' \to 3'$. The template strand for continuous replication is read $3' \to 5'$.

Step-by-step reasoning

1. Match A: E. coli genome size

   E. coli is a bacterium with a circular chromosome of about $4.6 \times 10^6$ base pairs (bp). This is a standard fact.

   → A matches with III ($4.6 \times 10^6$ bp).

2. Match B: Diploid human cell genome size

   A human diploid cell has about $6.6 \times 10^9$ bp (the haploid genome is ~$3.3 \times 10^9$ bp, so diploid is double).

   → B matches with I ($6.6 \times 10^9$ bp).

3. Match C: DNA-dependent RNA polymerase catalyzes in which direction?

   RNA polymerase synthesizes RNA in the $5' \to 3'$ direction, just like DNA polymerase. It reads the template strand $3' \to 5'$ but the new RNA chain grows $5' \to 3'$.

   → C matches with IV ($5' \to 3'$ direction).

4. Match D: Continuous replication is with which polarity?

   During DNA replication, the leading strand is synthesized continuously. The template for the leading strand is oriented $3' \to 5'$ relative to the replication fork, so the new strand grows $5' \to 3'$. But the question asks: "The continuous replication is with" — meaning the template strand polarity for continuous synthesis. That template is read $3' \to 5'$.

   → D matches with II ($3' \to 5'$ polarity).

5. Assemble the matches A → III, B → I, C → IV, D → II. This corresponds to the sequence: III, I, IV, II. Looking at the options:
   • (A) III I IV II ✓
   • (B) III IV II I ✗
   • (C) II III IV I ✗
   • (D) I III II IV ✗

The concept first

Four separate ideas are being tested at once; take them one at a time.

• Cofactors come in three flavours: prosthetic groups (tightly, permanently bound — e.g. haem in peroxidase), coenzymes (organic, loosely and transiently bound — e.g. $\mathrm{NAD^+}$, FAD, derived from vitamins), and metal ions (coordinated to the enzyme — e.g. $\mathrm{Zn^{2+}}$ in carboxypeptidase).
• Enzyme classes (IUB): oxidoreductases, transferases, hydrolases, lyases, isomerases, ligases.
• Inhibition: competitive — the inhibitor mimics the substrate and occupies the active site (malonate vs succinate); non-competitive — the inhibitor is structurally unlike the substrate and binds elsewhere, distorting the active site (heavy metals such as $\mathrm{Cu^{2+}}$, $\mathrm{Hg^{2+}}$, $\mathrm{Ag^{+}}$).
• Enzymes vs inorganic catalysts: enzymes are proteins — exquisitely specific, hugely efficient, but destroyed above roughly $40\text{–}50^\circ\mathrm{C}$. Inorganic catalysts (Pt, $\mathrm{V_2O_5}$) are unspecific and happily operate at high temperature.

Step-by-step

1. A) "Coenzymes are organic compounds." Their defining contrast with prosthetic groups is that their association with the apoenzyme is only transient → iv, and it happens during catalysis → II. So A – iv, II.
2. B) Isomerases. By definition they catalyse interconversion of optical and positional isomers → iii. The stock example from the pentose-phosphate pathway is ribose-5-phosphate isomerase → I. So B – iii, I.
3. C) Non-competitive inhibition. Because the inhibitor does not occupy the active site, it needs no structural similarity to the substrate → i. Textbook agents are heavy-metal ions — a metal ion of copper → IV. So C – i, IV.
4. D) Enzyme catalysts. Their whole point in this list is that they differ from inorganic catalysis → ii, and the crispest way to state that difference is thermal: inorganic catalysts operate at high temperature also, while enzymes denature → III. So D – ii, III.
5. Read the option. A–iv,II; B–iii,I; C–i,IV; D–ii,III is printed as option (A). Note option (C) is a decoy differing only in the last two entries (C–i,III and D–ii,IV), which would absurdly say a copper ion is the way enzymes differ from inorganic catalysts.

ANSWER: A