501. The correctly completed hormone-classification table is:
| Option | Peptide or protein hormone | Steroid hormone | Amino-acid-derived hormone |
|---|
| A | Adrenaline | Insulin | Sex hormone |
| B | Thyroxine | Glucagon | Insulin |
| C | Insulin | Sex hormone | Adrenaline |
| D | Sex hormone | Adrenaline | Glucagon |
ⓐ. Row A
ⓑ. Row B
ⓒ. Row C
ⓓ. Row D
Correct Answer: Row C
Explanation: Insulin is a peptide or protein hormone composed of amino-acid residues. Sex hormones such as testosterone and oestrogen belong to the steroid class. Adrenaline is derived from an amino-acid-related precursor and is classified as an amino-acid-derived hormone. The messenger function does not require all hormones to share the same solubility, size or chemical skeleton. Row C correctly assigns one representative example to each major chemical category.
502. A patient has normal digestive-enzyme production and adequate vitamin intake, but a gland releases too little of a blood-borne messenger that normally binds receptors on distant tissues. Which conclusion is most appropriate?
ⓐ. The disorder is a hormone deficiency, not an enzyme or vitamin deficiency
ⓑ. The disorder is a digestive-enzyme deficiency despite normal enzyme production
ⓒ. The disorder is a vitamin deficiency despite adequate vitamin intake
ⓓ. The disorder is a target-receptor deficiency rather than low messenger release
Correct Answer: The disorder is a hormone deficiency, not an enzyme or vitamin deficiency
Explanation: The described substance is secreted by a gland, transported through blood and recognised by receptors on distant target tissues. Those features identify endocrine hormonal communication. Digestive enzymes act mainly by catalysing reactions involving chemical substrates. Vitamins are dietary micronutrients that often assist metabolism but are not generally glandular long-distance messengers. The evidence separates the three regulatory categories by mechanism rather than merely by their importance to health.
503. A graph plots target-cell response against hormone concentration. The response rises steeply at low concentration and then approaches a plateau at high concentration. The most suitable interpretation is:
ⓐ. all receptors are destroyed as hormone concentration begins to rise
ⓑ. hormone concentration falls even though the administered dose increases
ⓒ. response remains proportional to concentration with no upper limit
ⓓ. receptor occupancy raises response until signalling capacity is saturated
Correct Answer: receptor occupancy raises response until signalling capacity is saturated
Explanation: At low hormone concentration, many receptors are unoccupied and additional hormone increases the number of hormone–receptor complexes. The cellular response therefore rises substantially. At higher concentrations, most available receptors or downstream response components become engaged. Further increases then produce progressively smaller effects, giving a plateau. This resembles saturation behaviour but represents regulatory signalling rather than direct enzyme–substrate catalysis. The graph does not imply that the hormone becomes an enzyme or permanently removes all receptors.
504. Match each category in Column I with the most suitable description in Column II.
| Column I | Column II |
|---|
| P. Hormone | 1. Accelerates a reaction by lowering activation energy |
| Q. Enzyme | 2. Regulatory messenger acting through target-cell receptors |
| R. Vitamin | 3. Micronutrient that may support metabolism or form part of a coenzyme |
| S. Nutrient polymer | 4. May provide stored chemical material or energy after breakdown |
ⓐ. P-1, Q-2, R-4, S-3
ⓑ. P-2, Q-1, R-3, S-4
ⓒ. P-3, Q-4, R-2, S-1
ⓓ. P-4, Q-3, R-1, S-2
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: A hormone is a regulatory messenger recognised by receptors on responsive cells, so P matches 2. An enzyme lowers activation energy and increases reaction rate, linking Q with 1. A vitamin is required in small amounts and may serve as a coenzyme precursor or metabolic helper, so R matches 3. Nutrient polymers can store material or energy that becomes available after hydrolysis or metabolic breakdown, linking S with 4. The comparison separates regulation, catalysis, micronutrient support and bulk nutrient storage.
505. An unknown biological polymer gives only glucose on complete hydrolysis. Its chain contains mainly \(\alpha(1\rightarrow4)\) glycosidic linkages with frequent \(\alpha(1\rightarrow6)\) branch points, and it functions as a rapidly mobilised carbohydrate reserve in animals. The polymer is:
ⓐ. cellulose
ⓑ. glycogen
ⓒ. amylose
ⓓ. a polypeptide
Correct Answer: glycogen
Explanation: Glycogen is an animal storage polysaccharide composed entirely of \(\alpha\)-D-glucose residues. Its main chains contain \(\alpha(1\rightarrow4)\) glycosidic linkages, while \(\alpha(1\rightarrow6)\) linkages occur at branch points. Glycogen is more highly branched than amylopectin, allowing many chain ends to be accessed during rapid glucose mobilisation. Amylose is largely unbranched and contains mainly \(\alpha(1\rightarrow4)\) linkages. Cellulose contains \(\beta\)-linked glucose and performs a structural rather than storage role.
506. The row that correctly relates a biomolecule to its building unit and characteristic linkage is:
| Option | Biomolecule | Building unit | Characteristic linkage |
|---|
| A | Protein | Monosaccharide | Glycosidic linkage |
| B | Cellulose | Amino acid | Peptide linkage |
| C | DNA | Nucleoside | Peptide linkage |
| D | DNA | Nucleotide | Phosphodiester linkage |
ⓐ. Row A
ⓑ. Row B
ⓒ. Row C
ⓓ. Row D
Correct Answer: Row D
Explanation: DNA is a polynucleotide constructed from nucleotide units. Adjacent nucleotide residues are joined through phosphodiester linkages in the sugar–phosphate backbone. A nucleoside lacks phosphate and is therefore not the complete polymer-building unit. Proteins are amino-acid polymers joined by peptide bonds. Cellulose is a glucose polymer joined by \(\beta\)-glycosidic linkages, so the other rows interchange unrelated monomers and bonds.
507. A mixture contains \(0.250\,\mathrm{mol}\) sucrose and \(0.150\,\mathrm{mol}\) maltose. Both disaccharides undergo complete hydrolysis. The correct pair for the total amount of glucose formed and the mass of water consumed is:
ⓐ. \(0.550\,\mathrm{mol}\) glucose and \(7.20\,\mathrm{g}\) water
ⓑ. \(0.400\,\mathrm{mol}\) glucose and \(9.90\,\mathrm{g}\) water
ⓒ. \(0.800\,\mathrm{mol}\) glucose and \(7.20\,\mathrm{g}\) water
ⓓ. \(0.550\,\mathrm{mol}\) glucose and \(3.60\,\mathrm{g}\) water
Correct Answer: \(0.550\,\mathrm{mol}\) glucose and \(7.20\,\mathrm{g}\) water
Explanation: \( \textbf{Hydrolysis of sucrose:} \)
\[
\mathrm{sucrose+H_2O\rightarrow glucose+fructose}
\]
Therefore:
\[
0.250\,\mathrm{mol\ sucrose}\rightarrow0.250\,\mathrm{mol\ glucose}
\]
\( \textbf{Hydrolysis of maltose:} \)
\[
\mathrm{maltose+H_2O\rightarrow2\,glucose}
\]
\[
0.150\,\mathrm{mol\ maltose}\rightarrow0.300\,\mathrm{mol\ glucose}
\]
\( \textbf{Total glucose:} \)
\[
n_{\mathrm{glucose}}=0.250+0.300
\]
\[
n_{\mathrm{glucose}}=0.550\,\mathrm{mol}
\]
Each mole of disaccharide consumes one mole of water.
\[
n_{\mathrm{H_2O}}=0.250+0.150=0.400\,\mathrm{mol}
\]
Using \(M_{\mathrm{H_2O}}=18.0\,\mathrm{g\,mol^{-1}}\):
\[
m_{\mathrm{H_2O}}=0.400\times18.0
\]
\[
m_{\mathrm{H_2O}}=7.20\,\mathrm{g}
\]
Maltose contributes two glucose molecules per molecule, whereas sucrose contributes one glucose and one fructose.
508. Assertion: Starch and cellulose give glucose on complete hydrolysis but show different digestibility in humans.
Reason: Starch contains mainly \(\alpha\)-glycosidic linkages, whereas cellulose contains \(\beta(1\rightarrow4)\) glycosidic linkages that human digestive enzymes cannot efficiently hydrolyse.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Both starch and cellulose are polymers of glucose, so their complete hydrolysis can produce the same monosaccharide. Their glycosidic configurations, however, are different. Human enzymes can hydrolyse the \(\alpha\)-linked structures in starch but lack cellulase for efficient cleavage of cellulose’s \(\beta(1\rightarrow4)\) linkages. The \(\beta\)-linked chains also form strong hydrogen-bonded fibres. The difference in digestibility therefore arises from linkage stereochemistry rather than from monomer identity.
509. Two graphs reach plateaus.
Graph P plots the specific optical rotation of a freshly prepared glucose solution against time.
Graph Q plots initial enzyme rate against substrate concentration at constant enzyme concentration.
The plateaus most appropriately represent:
ⓐ. complete glucose decomposition in P and enzyme denaturation in Q
ⓑ. anomeric equilibrium in P and active-site saturation in Q
ⓒ. constant glucose concentration in P and a changing equilibrium constant in Q
ⓓ. peptide hydrolysis in P and vitamin accumulation in Q
Correct Answer: anomeric equilibrium in P and active-site saturation in Q
Explanation: In the glucose graph, optical rotation changes as \(\alpha\) and \(\beta\) anomers interconvert through the open-chain form. The plateau appears when the anomeric composition reaches dynamic equilibrium and the observed rotation becomes constant. In the enzyme graph, rate rises as additional substrate occupies more active sites. At high substrate concentration, most active sites are occupied for much of the time, so the rate approaches a maximum. The two plateaus arise from different processes: equilibrium composition in one case and finite catalytic capacity in the other.
510. A compound contains one amino group and one carboxyl group attached to the same carbon atom. In aqueous solution it can exist as a zwitterion and can undergo condensation to form a peptide linkage. The compound is most directly classified as:
ⓐ. an amino acid
ⓑ. a monosaccharide
ⓒ. a nucleotide
ⓓ. a steroid hormone
Correct Answer: an amino acid
Explanation: In an \(\alpha\)-amino acid, the amino group and carboxyl group are attached to the \(\alpha\)-carbon. Proton transfer between these groups can produce a dipolar zwitterionic form in aqueous solution. The carboxyl group of one amino acid can condense with the amino group of another to form a peptide bond. Monosaccharides instead contain several hydroxyl groups and a carbonyl-derived framework. Nucleotides contain a base, pentose and phosphate, while steroid hormones possess a fused-ring skeleton.
511. At saturating substrate concentration, an enzyme preparation initially has a maximum rate of \(40.0\,\mathrm{\mu mol\,min^{-1}}\). A treatment denatures \(30\%\) of the enzyme molecules. The remaining preparation is then concentrated so that its enzyme concentration becomes \(1.50\) times the value immediately after denaturation. The new maximum rate is:
ⓐ. \(28.0\,\mathrm{\mu mol\,min^{-1}}\)
ⓑ. \(60.0\,\mathrm{\mu mol\,min^{-1}}\)
ⓒ. \(18.7\,\mathrm{\mu mol\,min^{-1}}\)
ⓓ. \(42.0\,\mathrm{\mu mol\,min^{-1}}\)
Correct Answer: \(42.0\,\mathrm{\mu mol\,min^{-1}}\)
Explanation: \( \textbf{Initial maximum rate:} \)
\[
v_{\max,1}=40.0\,\mathrm{\mu mol\,min^{-1}}
\]
\( \textbf{Active fraction after denaturation:} \)
\[
1-0.30=0.70
\]
If no concentration step occurred, the new maximum rate would be:
\[
v_{\max,\mathrm{denatured}}=0.70\times40.0
\]
\[
v_{\max,\mathrm{denatured}}=28.0\,\mathrm{\mu mol\,min^{-1}}
\]
\( \textbf{Concentration factor:} \)
\[
1.50
\]
Maximum rate is proportional to active enzyme concentration under substrate saturation.
\[
v_{\max,2}=1.50\times28.0
\]
\[
v_{\max,2}=42.0\,\mathrm{\mu mol\,min^{-1}}
\]
Equivalently:
\[
v_{\max,2}=40.0\times0.70\times1.50
\]
\[
v_{\max,2}=42.0\,\mathrm{\mu mol\,min^{-1}}
\]
Concentration increases the amount of surviving active enzyme but cannot reactivate the denatured fraction.
512. Examine the following statements.
Statement I: Protein denaturation may destroy biological activity without hydrolysing the primary peptide sequence.
Statement II: An enzyme changes the equilibrium constant of the reaction it catalyses.
Statement III: Excessive intake of some fat-soluble vitamins may cause accumulation-related toxicity.
Statement IV: A circulating hormone directly affects only cells capable of recognising its signal.
The acceptable statements are:
ⓐ. I and II only
ⓑ. II, III and IV only
ⓒ. I, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, III and IV only
Explanation: Denaturation commonly disrupts secondary, tertiary or quaternary structure while leaving peptide bonds intact, so Statement I is acceptable. Enzymes alter reaction rate but not the equilibrium constant, making Statement II unacceptable. Fat-soluble vitamins may be stored in liver or adipose tissue and can accumulate excessively, so Statement III is acceptable. Hormone action requires suitable receptors and response machinery, making Statement IV acceptable. Catalysis, storage and signalling must be distinguished even though all three involve biologically important molecules.
513. Match each biomolecule class in Column I with the most suitable description in Column II.
| Column I | Column II |
|---|
| P. Protein | 1. Nucleotide polymer carrying genetic information |
| Q. Polysaccharide | 2. Amino-acid polymer with structural or catalytic roles |
| R. Nucleic acid | 3. Monosaccharide polymer used in storage or structure |
| S. Hormone | 4. Regulatory messenger recognised by target cells |
ⓐ. P-2, Q-3, R-1, S-4
ⓑ. P-3, Q-1, R-4, S-2
ⓒ. P-1, Q-4, R-2, S-3
ⓓ. P-4, Q-2, R-3, S-1
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: Proteins are amino-acid polymers that may function as enzymes, transporters or structural materials, linking P with 2. Polysaccharides contain many monosaccharide residues and commonly serve storage or structural roles, linking Q with 3. Nucleic acids are nucleotide polymers involved in genetic information, so R matches 1. Hormones are regulatory messengers whose effects depend on target-cell recognition, linking S with 4. The mapping uses monomer identity and biological function together.
514. Use the structural descriptions below.
Species P is a polyhydroxy aldehyde containing six carbon atoms.
Species Q contains an \(\alpha\)-carbon attached to \(\mathrm{-NH_2}\), \(\mathrm{-COOH}\), hydrogen and a variable side chain.
Species R contains a nitrogenous base, pentose sugar and phosphate.
The correct identification is:
ⓐ. P is a nucleotide, Q is a hexose, and R is an amino acid
ⓑ. P is a protein, Q is a nucleoside, and R is a vitamin
ⓒ. P is an aldohexose, Q is an amino acid, and R is a nucleotide
ⓓ. P is a ketohexose, Q is a fatty acid, and R is a nucleoside
Correct Answer: P is an aldohexose, Q is an amino acid, and R is a nucleotide
Explanation: A six-carbon polyhydroxy aldehyde is classified as an aldohexose, so P fits the structural class of glucose in its open-chain representation. Species Q has the general \(\alpha\)-amino-acid framework. Species R contains all three nucleotide components: base, pentose and phosphate. A nucleoside would lack phosphate. The classifications are determined from functional groups and molecular components rather than from biological role alone.
515. A double-stranded DNA segment contains \(2400\) base pairs. Adenine constitutes \(30\%\) of all nucleotides in the segment. Using \(0.34\,\mathrm{nm}\) as the axial rise per base pair, the correct pair for the total number of hydrogen bonds and the DNA length is:
ⓐ. \(6240\) hydrogen bonds and \(408\,\mathrm{nm}\)
ⓑ. \(5760\) hydrogen bonds and \(816\,\mathrm{nm}\)
ⓒ. \(6240\) hydrogen bonds and \(816\,\mathrm{nm}\)
ⓓ. \(5760\) hydrogen bonds and \(1632\,\mathrm{nm}\)
Correct Answer: \(5760\) hydrogen bonds and \(816\,\mathrm{nm}\)
Explanation: \( \textbf{Total nucleotide count:} \)
\[
N_{\mathrm{nt}}=2\times2400=4800
\]
\( \textbf{Adenine count:} \)
\[
N_{\mathrm{A}}=0.30\times4800=1440
\]
In double-stranded DNA:
\[
N_{\mathrm{T}}=1440
\]
Hence, the number of \(\mathrm{A-T}\) pairs is:
\[
N_{\mathrm{A-T}}=1440
\]
Total guanine and cytosine nucleotides:
\[
4800-(1440+1440)=1920
\]
Therefore:
\[
N_{\mathrm{G}}=N_{\mathrm{C}}=960
\]
The number of \(\mathrm{G-C}\) pairs is \(960\).
\( \textbf{Hydrogen bonds:} \)
\[
N_{\mathrm{H-bonds}}=2(1440)+3(960)
\]
\[
N_{\mathrm{H-bonds}}=2880+2880=5760
\]
\( \textbf{Axial length:} \)
\[
L=2400\times0.34\,\mathrm{nm}
\]
\[
L=816\,\mathrm{nm}
\]
Base percentages apply to nucleotides, while axial length depends on the number of base pairs.
516. Four unknown biological substances give the observations below.
| Substance | Observation |
|---|
| P | Hydrolysis yields amino acids; heating destroys activity without releasing free amino acids |
| Q | Hydrolysis yields ribose, phosphate and nitrogenous bases including uracil |
| R | Hydrolysis yields only glucose; the polymer contains \(\beta(1\rightarrow4)\) linkages |
| S | Required in small amounts; prolonged excessive intake may cause accumulation in adipose tissue |
The correct identification is:
ⓐ. P is cellulose, Q is protein, R is RNA, and S is a water-soluble vitamin
ⓑ. P is protein, Q is RNA, R is cellulose, and S is a fat-soluble vitamin
ⓒ. P is DNA, Q is starch, R is glycogen, and S is an enzyme
ⓓ. P is protein, Q is DNA, R is amylose, and S is a hormone
Correct Answer: P is protein, Q is RNA, R is cellulose, and S is a fat-soluble vitamin
Explanation: Hydrolysis to amino acids identifies P as a protein, while loss of activity without amino-acid release indicates denaturation rather than complete hydrolysis. Ribose and uracil identify Q as RNA. A glucose polymer containing \(\beta(1\rightarrow4)\) linkages is cellulose, identifying R. Storage in adipose tissue and accumulation risk are characteristic of fat-soluble vitamins, identifying S. The combined evidence uses hydrolysis products, linkage type, structural change and physiological handling.
517. A preparation contains \(N\) amino-acid residues distributed among \(c\) separate linear polypeptide chains. If no cyclic chains or interchain peptide bonds are present, the total number of peptide bonds is:
\[
\underline{\hspace{2.0cm}}
\]
ⓐ. \(N+c\)
ⓑ. \(N-1\)
ⓒ. \(c-N\)
ⓓ. \(N-c\)
Correct Answer: \(N-c\)
Explanation: A single linear chain containing \(n\) amino-acid residues has \(n-1\) peptide bonds. For several chains, this relation must be applied to each chain separately. If their lengths are \(n_1,n_2,\ldots,n_c\), then:
\[
N_{\mathrm{peptide}}=(n_1-1)+(n_2-1)+\cdots+(n_c-1)
\]
\[
N_{\mathrm{peptide}}=(n_1+n_2+\cdots+n_c)-c
\]
\[
N_{\mathrm{peptide}}=N-c
\]
The expression \(N-1\) would incorrectly treat all residues as belonging to one continuous chain.
518. A patient has normal enzyme proteins but lacks an organic helper derived from a dietary micronutrient. In another experiment, a blood-borne substance binds a receptor on liver cells and alters their metabolism without catalysing the reaction directly. The two missing or acting substances are most appropriately classified as:
ⓐ. a vitamin and an enzyme
ⓑ. a vitamin-derived coenzyme and a hormone
ⓒ. a hormone and a vitamin-derived coenzyme
ⓓ. an enzyme and a vitamin
Correct Answer: a vitamin-derived coenzyme and a hormone
Explanation: Several vitamins serve as precursors of organic coenzymes required by enzyme proteins. The enzyme may be synthesised normally yet remain inactive when its required coenzyme is unavailable. The second substance acts as a regulatory messenger because it travels through blood and produces a receptor-dependent response in target cells. A hormone does not need to catalyse the metabolic reaction directly. The passage distinguishes catalytic assistance from long-distance cellular regulation.
519. Graph P shows the average net charge of an amino acid decreasing from positive to zero and then negative as \(\mathrm{pH}\) rises. Graph Q shows enzyme activity increasing to a maximum and then decreasing as \(\mathrm{pH}\) rises. The most accurate interpretation is:
ⓐ. The zero crossing in P is the isoelectric point, while the maximum in Q is the enzyme’s optimum \(\mathrm{pH}\)
ⓑ. Both special points represent complete hydrolysis of the molecules
ⓒ. The zero crossing in P shows maximum positive charge, while the maximum in Q shows enzyme saturation by substrate
ⓓ. Both graphs prove that peptide bonds are broken at neutral \(\mathrm{pH}\)
Correct Answer: The zero crossing in P is the isoelectric point, while the maximum in Q is the enzyme’s optimum \(\mathrm{pH}\)
Explanation: At low \(\mathrm{pH}\), an amino acid is generally more protonated and has a more positive average charge. As \(\mathrm{pH}\) increases, deprotonation lowers the net charge until it becomes zero at the isoelectric point. Further increase makes the average charge negative. Enzyme activity has a different \(\mathrm{pH}\) relationship because active-site ionisation and protein conformation determine an optimum region. The two graph features are both condition-dependent boundaries, but they describe charge balance and catalytic performance rather than the same chemical event.
520. A sample contains \(0.0100\,\mathrm{mol}\) of identical linear tripeptide molecules and \(0.00500\,\mathrm{mol}\) of identical linear RNA molecules, each containing \(21\) nucleotide residues. Complete hydrolysis cleaves all peptide bonds and all RNA phosphodiester bonds, but the resulting nucleotides are not further hydrolysed. The correct pair for the total amount of water consumed and the total amount of monomer units formed is:
ⓐ. \(0.115\,\mathrm{mol}\) water and \(0.135\,\mathrm{mol}\) monomer units
ⓑ. \(0.120\,\mathrm{mol}\) water and \(0.105\,\mathrm{mol}\) monomer units
ⓒ. \(0.120\,\mathrm{mol}\) water and \(0.135\,\mathrm{mol}\) monomer units
ⓓ. \(0.135\,\mathrm{mol}\) water and \(0.120\,\mathrm{mol}\) monomer units
Correct Answer: \(0.120\,\mathrm{mol}\) water and \(0.135\,\mathrm{mol}\) monomer units
Explanation: \( \textbf{Peptide-bond count per tripeptide:} \)
\[
3-1=2
\]
\( \textbf{Water consumed by peptide hydrolysis:} \)
\[
n_{\mathrm{H_2O,peptide}}=0.0100\times2
\]
\[
n_{\mathrm{H_2O,peptide}}=0.0200\,\mathrm{mol}
\]
\( \textbf{Amino acids formed:} \)
\[
n_{\mathrm{amino\ acids}}=0.0100\times3
\]
\[
n_{\mathrm{amino\ acids}}=0.0300\,\mathrm{mol}
\]
Each linear RNA molecule with \(21\) residues contains:
\[
21-1=20
\]
phosphodiester bonds.
\( \textbf{Water consumed by RNA-backbone hydrolysis:} \)
\[
n_{\mathrm{H_2O,RNA}}=0.00500\times20
\]
\[
n_{\mathrm{H_2O,RNA}}=0.100\,\mathrm{mol}
\]
\( \textbf{Nucleotides formed:} \)
\[
n_{\mathrm{nucleotides}}=0.00500\times21
\]
\[
n_{\mathrm{nucleotides}}=0.105\,\mathrm{mol}
\]
\( \textbf{Total water consumed:} \)
\[
n_{\mathrm{H_2O,total}}=0.0200+0.100
\]
\[
n_{\mathrm{H_2O,total}}=0.120\,\mathrm{mol}
\]
\( \textbf{Total monomer units formed:} \)
\[
n_{\mathrm{monomers}}=0.0300+0.105
\]
\[
n_{\mathrm{monomers}}=0.135\,\mathrm{mol}
\]
The RNA products are counted as nucleotides because the question stops hydrolysis after backbone cleavage.
