Chapter 9: Biomolecules Application Questions | biology Class 11 CBSE

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Chapter 9: Biomolecules Application Questions | biology Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A biochemist grinds liver tissue in trichloroacetic acid and separates the filtrate from the retentate. When the retentate is analysed, she finds proteins, nucleic acids, polysaccharides, and lipids. With reference to biomacromolecules, explain why each of these is found in the acid-insoluble fraction.

Proteins, nucleic acids, and polysaccharides are true polymeric macromolecules with molecular weights of 10,000 daltons or more; they are too large to pass through the filtration step and thus remain in the acid-insoluble retentate as the macromolecular fraction.
Lipids are small molecular weight compounds (below 800 Da) and would normally pass into the filtrate; however, in the intact cell they are organised into membrane structures (cell membrane, organelle membranes) which, upon grinding, break into water-insoluble vesicles.
These lipid-containing membrane vesicles are not water-soluble and thus sediment with the acid-insoluble fraction along with the true macromolecules, even though lipids are not strictly macromolecules.
Together, the acid-insoluble fraction (macromolecules from cytoplasm and organelles) and the acid-soluble pool (small molecules representing cytoplasmic composition) account for the entire chemical composition of the living tissue.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A student adds iodine solution to two test tubes — one containing a starch solution and another containing a cellulose suspension. Only the starch tube turns blue. With reference to the structural features of these polysaccharides, explain this observation.

Starch forms helical secondary structures due to the specific glycosidic linkages between its glucose units; the interior of these helices is hydrophobic and creates a cavity of the right dimensions to physically trap iodine (I2) molecules.
The inclusion of iodine within the helical coils of starch produces the characteristic blue-black colour observed; this is a physical interaction (inclusion complex) rather than a covalent chemical reaction.
Cellulose, although also a homopolymer of glucose, has different glycosidic linkages (beta-1,4 linkages) that result in a linear, unbranched structure without complex helical coiling; the absence of helices means there are no cavities to trap iodine molecules.
Since cellulose cannot incorporate I2 into any helical cavity, it gives no blue colour with iodine, confirming that the ability to hold iodine is a consequence of specific secondary structure (helix formation) and not merely glucose composition.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

In an experiment, the enzyme carbonic anhydrase converts CO2 and water to carbonic acid (H2CO3). Without enzyme: ~200 molecules of H2CO3 form per hour. With enzyme: ~600,000 molecules form per second. Calculate the approximate fold-increase and explain what this demonstrates about enzymes.

Without enzyme: 200 molecules per hour = 200/3600 per second ≈ 0.056 molecules per second. With enzyme: 600,000 molecules per second. Fold increase = 600,000 / 0.056 ≈ 10,700,000, which is approximately 10 million times — as stated in the chapter.
This enormous rate enhancement demonstrates the remarkable catalytic power of enzymes: they do not merely speed up reactions marginally but can increase rates by millions of times, making physiologically critical reactions occur at speeds compatible with life.
The mechanism behind this acceleration is the lowering of activation energy — enzymes stabilise the transition state of the reaction, reducing the energy barrier that must be overcome, so that many more substrate molecules can cross the energy barrier per unit time.
This example also illustrates that enzyme-catalysed reactions are essential for life: without carbonic anhydrase, the conversion of respiratory CO2 to carbonic acid (needed for CO2 transport in blood) would be far too slow to sustain aerobic life.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

A pharmaceutical company is developing an antibiotic that targets a bacterial enzyme. They design a molecule that structurally resembles the enzyme's natural substrate. With reference to competitive inhibition, explain how this drug would work.

The drug molecule is a competitive inhibitor: because it closely resembles the natural substrate in its molecular structure, it can bind to the active site of the bacterial enzyme, fitting into the same substrate-binding pocket.
The competitive inhibitor occupies the active site, preventing the natural substrate from binding; since the substrate cannot enter the active site, the enzyme cannot catalyse its normal reaction, disrupting a metabolic pathway essential for bacterial survival.
An example from the chapter is malonate, which inhibits succinic dehydrogenase by resembling the substrate succinate; the antibiotic would work by the same principle — structural mimicry allows competition for the active site.
The chapter explicitly states that competitive inhibitors are often used in the control of bacterial pathogens, validating this approach; higher concentrations of the natural substrate could potentially overcome the inhibition, so the drug's efficacy would depend on maintaining sufficient inhibitor concentration relative to substrate.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

During intense exercise, your skeletal muscle runs short of oxygen. Using your knowledge of metabolic pathways from this chapter, explain what happens to the glucose being metabolised and what product accumulates.

During normal aerobic conditions, the glycolytic pathway converts glucose to pyruvic acid through ten enzyme-catalysed steps: C6H12O6 + O2 → 2C3H4O3 + 2H2O; pyruvic acid is then further metabolised via aerobic respiration.
When oxygen is insufficient (anaerobic conditions), as during intense exercise, the metabolic pathway is redirected: instead of entering the aerobic pathway, pyruvic acid is reduced to lactic acid in the skeletal muscle.
This anaerobic conversion of pyruvic acid to lactic acid regenerates NAD+ (needed to keep glycolysis running without oxygen) but results in the accumulation of lactic acid in the muscle tissue.
The accumulation of lactic acid in muscles is responsible for the sensation of fatigue and muscle cramps during intense exercise; this illustrates the chapter's point that the same metabolic pathway can yield different end products depending on the physiological conditions.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

A cook stores one bottle of enzyme-based detergent in the refrigerator and another near the stove at high temperature. After a week, which bottle will be more effective for removing stains, and why? Relate your answer to enzyme properties.

The bottle stored in the refrigerator will be more effective, because low temperatures preserve enzymes in a temporarily inactive but structurally intact state; upon use at room temperature, the enzyme can regain activity as the low temperature only reversibly inhibits it.
The bottle stored near the stove at high temperature will be less or ineffective: high temperatures (above ~40°C) denature the enzyme by disrupting its tertiary structure — specifically the precise three-dimensional folding that creates the active site.
Once the tertiary structure is disrupted by heat, the active site loses its specific shape and the enzyme permanently loses its ability to catalyse reactions; this heat-induced denaturation of the proteinaceous enzyme is irreversible under normal conditions.
This scenario illustrates two key properties of enzymes described in the chapter: (1) low temperature preserves enzymatic activity in an inactive but reversible state, and (2) high temperature destroys enzymatic activity permanently because proteins are denatured by heat.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A biologist discovers bacteria living in a deep-sea hydrothermal vent where the temperature exceeds 80°C. She isolates an enzyme from these bacteria and finds it is active at 85°C. How does this relate to enzyme properties described in the chapter?

The chapter states that enzymes isolated from thermophilic organisms — those living under extremely high temperatures such as hot vents and sulphur springs — are stable and retain their catalytic power even at high temperatures up to 80°–90°C.
Normal enzymes from mesophilic organisms are denatured above approximately 40°C because heat disrupts the tertiary structure; but in thermophilic organisms, the enzyme's amino acid sequence and tertiary structure have evolved to be stable at high temperatures.
The thermal stability of this vent enzyme is described in the chapter as an important quality; such thermostable enzymes maintain their active site geometry even at high temperatures because of additional or stronger stabilising interactions within their tertiary structure.
This observation highlights that the optimum temperature for enzyme activity is not fixed at 37°C for all organisms but varies depending on the organism's natural habitat, and that enzymes can be adapted by evolution to function under extreme conditions.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

With reference to the chapter, explain why eating a protein-rich diet is essential for human health, specifically focusing on the distinction between essential and non-essential amino acids.

Proteins are heteropolymers made from 20 types of amino acids; these amino acids fall into two categories — essential amino acids (which the human body cannot synthesise and must obtain through diet) and non-essential amino acids (which the body can manufacture internally).
Dietary proteins are the primary source of essential amino acids; consuming a protein-rich diet ensures a sufficient supply of all essential amino acids needed for building and maintaining body proteins, including enzymes, hormones, antibodies, and structural proteins like collagen.
A deficiency of essential amino acids would impair the synthesis of critical proteins: for example, without sufficient amino acids, the body cannot produce haemoglobin (transport), insulin (hormone), or antibodies (immunity), leading to serious physiological disorders.
The chapter emphasises that this information will be important in future nutrition lessons, establishing that dietary protein quality (the balance of essential amino acids) rather than just quantity determines nutritional adequacy, making protein-rich foods like legumes, dairy, and meat critical for health.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A patient suffers from haemophilia, a disorder in which blood does not clot properly. Given that blood clotting involves proteins, and with reference to the structural levels of proteins, suggest why a single amino acid change could cause this disease.

The primary structure of a protein is the sequence of amino acids linked by peptide bonds; this sequence directly determines all higher-order structural levels (secondary, tertiary, and quaternary) because the chemical nature of each amino acid influences how the chain folds.
A single amino acid change (point mutation) alters the primary structure at one position; if this substitution changes the chemistry of a critical region — for example, replacing a charged amino acid with a neutral one — it can disrupt the secondary and tertiary folding of the protein.
A disrupted tertiary structure means the protein's active site or binding surface can no longer adopt the correct shape; in the case of clotting factors (proteins), the inability to form the right three-dimensional structure prevents proper substrate binding or protein-protein interactions needed for the clotting cascade.
This illustrates the principle that tertiary structure is absolutely necessary for the biological activity of proteins; even a single amino acid substitution in the primary sequence can propagate to alter the entire three-dimensional architecture and abolish function, explaining how a single genetic mutation causes a clinical disease like haemophilia.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

Insulin is a protein hormone that regulates blood glucose levels. A diabetic patient is prescribed insulin injections rather than oral insulin tablets. Using your knowledge of proteins and digestion from this chapter, explain why oral insulin would be ineffective.

Insulin is a protein — a polypeptide chain of specific amino acid sequence with a precise three-dimensional structure that allows it to bind to receptors on cell surfaces and trigger glucose uptake.
If taken orally, insulin would encounter protein-digesting enzymes (proteases/hydrolases like trypsin) in the digestive system; the chapter states that hydrolases catalyse the hydrolysis of peptide bonds, and these enzymes would break down insulin into its constituent amino acids.
Once hydrolysed into individual amino acids, insulin loses its specific three-dimensional (tertiary and quaternary) structure and therefore its biological activity; the amino acids absorbed from the gut cannot reassemble into functional insulin in the bloodstream.
By contrast, insulin injected subcutaneously bypasses the digestive tract entirely, entering the bloodstream intact and maintaining its tertiary structure; it can then bind to its target receptors and carry out its hormonal function — illustrating the critical importance of protein structure for biological activity.

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Chapter sections

Chapter 9: Biomolecules | Application Questions

A student adds iodine solution to two test tubes — one containing a starch solution and another containing a cellulose suspension. Only the starch tube turns blue. With reference to the structural features of these polysaccharides, explain this observation.

In an experiment, the enzyme carbonic anhydrase converts CO2 and water to carbonic acid (H2CO3). Without enzyme: ~200 molecules of H2CO3 form per hour. With enzyme: ~600,000 molecules form per second. Calculate the approximate fold-increase and explain what this demonstrates about enzymes.

A pharmaceutical company is developing an antibiotic that targets a bacterial enzyme. They design a molecule that structurally resembles the enzyme's natural substrate. With reference to competitive inhibition, explain how this drug would work.

During intense exercise, your skeletal muscle runs short of oxygen. Using your knowledge of metabolic pathways from this chapter, explain what happens to the glucose being metabolised and what product accumulates.

A cook stores one bottle of enzyme-based detergent in the refrigerator and another near the stove at high temperature. After a week, which bottle will be more effective for removing stains, and why? Relate your answer to enzyme properties.

A biologist discovers bacteria living in a deep-sea hydrothermal vent where the temperature exceeds 80°C. She isolates an enzyme from these bacteria and finds it is active at 85°C. How does this relate to enzyme properties described in the chapter?

With reference to the chapter, explain why eating a protein-rich diet is essential for human health, specifically focusing on the distinction between essential and non-essential amino acids.

A patient suffers from haemophilia, a disorder in which blood does not clot properly. Given that blood clotting involves proteins, and with reference to the structural levels of proteins, suggest why a single amino acid change could cause this disease.

Insulin is a protein hormone that regulates blood glucose levels. A diabetic patient is prescribed insulin injections rather than oral insulin tablets. Using your knowledge of proteins and digestion from this chapter, explain why oral insulin would be ineffective.

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