Chapter 9: Biomolecules Case Study Questions | biology Class 11 CBSE

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Case Set 1

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Dr. Meena is a biochemist studying the chemical composition of a freshly isolated sample of human liver tissue. She grinds 5 grams of the tissue in trichloroacetic acid using a mortar and pestle, obtaining a thick slurry. After straining through cheesecloth, she collects two fractions. Fraction A (the filtrate) is a clear liquid, while Fraction B (the retentate) is a viscous residue. She analyses both fractions and finds that Fraction A contains amino acids, glucose, nucleotides, and inorganic ions, whereas Fraction B contains proteins, DNA, RNA, glycogen, and lipid-containing vesicles. She wonders why the lipid vesicles ended up in Fraction B despite lipids being small molecules. This experiment forms the basis of how scientists categorise the chemical constituents of living cells into micromolecules and macromolecules.

Question 1: What are Fraction A and Fraction B technically called, and what molecular weight range characterises the compounds in Fraction A?

Fraction A is called the filtrate or the acid-soluble pool, while Fraction B is called the retentate or the acid-insoluble fraction.
The compounds in Fraction A have molecular weights ranging from 18 to approximately 800 daltons, placing them in the category of micromolecules or simply biomolecules.
Examples of compounds in Fraction A include amino acids, monosaccharides, nucleotides, fatty acids, and inorganic ions such as sodium, potassium, and phosphate.

Question 2: Why are lipids found in Fraction B (the acid-insoluble fraction) even though individual lipid molecules have molecular weights below 800 Da?

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 such as the cell membrane and organelle membranes.
When the liver tissue is ground during analysis, these membranes are disrupted and break into small vesicles that are not water-soluble and therefore cannot pass through the cheesecloth during filtration.
These membrane-derived lipid vesicles sediment with the acid-insoluble fraction along with true macromolecules; the chapter explicitly states that lipids are not strictly macromolecules but appear in the macromolecular fraction due to their membrane-associated organisation.

Question 3: Dr. Meena then burns a dried portion of the liver sample completely and collects the ash. What does the ash represent, and what does the experiment as a whole tell us about the difference between living tissue and earth's crust at the elemental level?

The ash remaining after complete combustion of dried tissue represents the inorganic elements — such as calcium, magnesium, sodium, and potassium — and inorganic compounds like sulphate and phosphate that are present in living tissue; all carbon compounds are oxidised to gaseous CO2 and water vapour during burning.
The experiment as a whole (acid fractionation + combustion) reveals the complete chemical composition of living tissue: organic compounds in both fractions plus inorganic constituents in the ash.
When compared to earth's crust, living tissue contains the same elements but in dramatically different relative proportions: carbon is 18.5% of the human body versus only 0.03% of the earth's crust; hydrogen is 9.5% in the body versus 0.14% in the crust; this difference in relative abundance — not the presence of unique elements — is what distinguishes living matter from non-living matter.

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Case Set 2

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In a plant biochemistry laboratory, a researcher is studying the polysaccharides present in three different plant-derived samples: Sample X is isolated from rice grains, Sample Y is isolated from cotton fibres, and Sample Z is isolated from the exoskeleton of a crab. She performs the iodine test on all three samples. Sample X turns deep blue-black, Sample Y shows no colour change, and Sample Z also shows no colour change. She then performs acid hydrolysis on all three and analyses the hydrolysis products. Samples X and Y both yield glucose, while Sample Z yields a different type of sugar unit. She notes that despite having similar building blocks, these polysaccharides serve very different biological functions in their source organisms.

Question 1: Identify Samples X, Y, and Z based on the experimental observations described, and state the monomer unit of each.

Sample X is starch: it gives a deep blue-black colour with iodine because it forms helical secondary structures that physically trap iodine molecules within the coils; its monomer is glucose.
Sample Y is cellulose: it does not give a colour with iodine because it lacks the complex helical structure needed to trap iodine; it is a linear homopolymer whose monomer is also glucose.
Sample Z is chitin: it does not give a colour with iodine and upon hydrolysis yields a different sugar unit — an amino sugar such as N-acetyl galactosamine or glucosamine — rather than plain glucose; chitin is a complex polysaccharide found in the exoskeletons of arthropods.

Question 2: Explain at the structural level why Sample X gives a blue-black colour with iodine while Sample Y does not, even though both are made of glucose.

Starch (Sample X) forms helical secondary structures due to the specific alpha-glycosidic linkages between its glucose units; the interior of these helices creates a hydrophobic cavity of the right dimensions to physically trap iodine (I2) molecules, forming an inclusion complex that appears blue-black.
Cellulose (Sample Y) has beta-glycosidic linkages between its glucose units, resulting in a linear, unbranched chain that does not coil into helices; without the helical cavity structure, there is no space to trap iodine molecules.
The difference is purely structural — the same glucose monomer arranged with different linkages produces fundamentally different secondary structures, demonstrating how glycosidic bond orientation determines polysaccharide properties.

Question 3: Compare the biological roles of Samples X, Y, and Z in their respective source organisms, and explain how their structural features are suited to these roles.

Starch (rice grains) functions as an energy storage polysaccharide in plant tissues; its helical structure and branching allow relatively compact packing of glucose units that can be readily hydrolysed to release glucose when the plant requires energy for growth and metabolism.
Cellulose (cotton fibres) is a structural polysaccharide forming the cell walls of plants; its linear, unbranched chains form extensive hydrogen bonds with neighbouring chains, creating strong, rigid fibres that provide mechanical support — making plant cells rigid and enabling plants to grow upright; this structural role also explains why paper and cotton fibre are cellulosic.
Chitin (crab exoskeleton) is a structural polysaccharide forming the hard outer covering of arthropods; its amino sugar building blocks (e.g., N-acetyl galactosamine) create a tough, rigid material that protects soft body tissues, resists predation, and provides attachment points for muscles — illustrating how chemical modification of the sugar monomer produces a polysaccharide with unique mechanical properties.

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Case Set 3

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Ramesh, a Class 11 student, is studying enzyme kinetics in his school laboratory. He sets up three test tubes, each containing the same enzyme (amylase) and starch substrate, but under different conditions. Test Tube 1 is incubated at 37°C (body temperature), Test Tube 2 is placed in an ice bath at 4°C, and Test Tube 3 is placed in a boiling water bath at 100°C for 10 minutes before adding substrate. After the reaction period, he tests each tube with iodine solution. Test Tube 1 gives no blue colour (starch fully digested), Test Tube 2 gives an intense blue colour, and Test Tube 3 also gives an intense blue colour. When Test Tube 2 is warmed to 37°C again, starch digestion resumes, but Test Tube 3 never shows any digestion even after prolonged incubation at 37°C.

Question 1: Why did Test Tube 1 show no blue colour with iodine while Test Tubes 2 and 3 both showed intense blue?

Test Tube 1 was incubated at 37°C, which is at or near the optimum temperature for amylase activity; the enzyme was fully active, catalysing the hydrolysis of all the starch into glucose (or maltose), so no starch remained for iodine to bind, giving no blue colour.
Test Tube 2 was at 4°C, well below the optimum temperature for amylase; at low temperature the enzyme is in a temporarily inactive state — the reduced thermal energy means substrate molecules cannot bind effectively or the reaction proceeds too slowly — so starch remained undigested, giving an intense blue with iodine.
Test Tube 3 was exposed to 100°C, which is far above 40°C; this high temperature denatured the amylase by disrupting its tertiary structure and destroying the active site permanently, so no catalysis occurred and starch remained intact, also giving an intense blue with iodine.

Question 2: Explain why Test Tube 2 resumed starch digestion after warming to 37°C, but Test Tube 3 never showed digestion even after prolonged incubation.

In Test Tube 2, the amylase was merely inhibited temporarily by low temperature; the cold did not disrupt the covalent bonds or the tertiary structure of the enzyme, only reducing molecular motion and slowing the reaction rate reversibly.
Upon warming to 37°C, the enzyme molecules regained sufficient thermal energy for substrate binding and catalysis; since the tertiary structure (and therefore the active site) was preserved intact, enzyme activity was fully restored.
In Test Tube 3, the 100°C heat caused irreversible denaturation of the protein: the tertiary structure — the three-dimensional folding stabilised by hydrogen bonds and disulphide bonds — was permanently disrupted; the active site lost its precise geometry and could not be restored by cooling; hence, even at the optimum temperature, the denatured enzyme had no functional active site and could not catalyse starch hydrolysis.

Question 3: If Ramesh had used an enzyme isolated from a thermophilic bacterium found in a volcanic hot spring instead of regular amylase, predict and explain what might have happened in Test Tube 3, and discuss the biological significance of thermostable enzymes.

If a thermostable (thermophilic) amylase were used in Test Tube 3, the enzyme would likely retain its catalytic activity even after exposure to 100°C, because the chapter states that enzymes isolated from thermophilic organisms living in hot vents and sulphur springs are stable and retain catalytic power at temperatures up to 80°–90°C.
Consequently, Test Tube 3 would show starch digestion (no blue colour with iodine), unlike the result with regular amylase; the thermostable enzyme's tertiary structure is stabilised by additional or stronger molecular interactions that resist disruption at high temperatures.
The biological significance of thermostable enzymes is considerable: they allow thermophilic organisms to survive and carry out all metabolic reactions at extreme temperatures that would denature normal enzymes; technologically, thermostable enzymes are of great industrial importance — for example, thermostable DNA polymerase (Taq polymerase from hot-spring bacteria) is essential for PCR, a foundational technique in molecular biology, showing how understanding enzyme properties from the chapter has real-world applications.

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Case Set 4

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A research paper describes the discovery of a new enzyme, ZymX, isolated from a fungal source. ZymX is a protein with a molecular weight of 45,000 Da. When the researchers analysed its tertiary structure, they found a deep cleft on its surface. When substrate molecule 'P' was added, ZymX catalysed the reaction rapidly. However, when molecule 'Q' (structurally very similar to P) was also added to the reaction mixture, the rate of catalysis dropped sharply. When only Q was present (no P), ZymX showed no product formation. The researchers also found that ZymX required a tightly bound haem group to function; removal of haem completely abolished enzyme activity. The enzyme showed maximum activity at pH 7.0 and 37°C.

Question 1: Identify the 'deep cleft' in ZymX's structure and explain its functional significance.

The deep cleft on the surface of ZymX is the active site of the enzyme; it is a crevice or pocket formed when the polypeptide chain of the tertiary structure criss-crosses upon itself, creating a specifically shaped binding region.
The active site is functionally significant because it is the region into which the substrate (molecule P) fits to form the enzyme-substrate (ES) complex — this complex formation is obligatory and essential for catalysis to occur.
The precise three-dimensional shape of the active site determines the enzyme's substrate specificity; only molecules with the correct complementary shape and chemical properties can bind effectively, explaining why ZymX catalyses the reaction with P but shows no product formation with Q alone.

Question 2: Explain the role of molecule Q in inhibiting ZymX activity and classify this type of inhibition.

Molecule Q is a competitive inhibitor: it closely resembles the substrate P in its molecular structure and competes with P for the same substrate-binding site (active site) of ZymX.
Because Q occupies the active site, the actual substrate P cannot bind; with fewer active sites available for P, the rate of ES complex formation decreases and consequently the overall rate of catalysis drops sharply — explaining the observed reduction in reaction rate when both P and Q are present.
This is classified as competitive inhibition; the chapter gives malonate (which competes with succinate for succinic dehydrogenase) as an analogous example; at high enough concentrations of P relative to Q, the inhibition could potentially be overcome because both molecules compete for the same site.

Question 3: The tightly bound haem group in ZymX is a cofactor. Classify this cofactor, explain the role of cofactors in enzyme function, and state what would happen to ZymX's apoenzyme if haem were removed.

The haem group in ZymX is classified as a prosthetic group — an organic cofactor that is tightly and permanently bound to the apoenzyme; the chapter states that in peroxidase and catalase (which also use haem), it is a part of the active site of the enzyme, and the same principle applies to ZymX.
Cofactors in general are non-protein constituents that are bound to certain enzymes to make them catalytically active; without the cofactor, the protein portion alone (apoenzyme) cannot function; cofactors may be prosthetic groups (tightly bound organic compounds), co-enzymes (transiently associated organic compounds, often vitamins), or metal ions (e.g., zinc in carboxypeptidase).
If haem is removed from ZymX, the remaining protein (apoenzyme) would completely lose all catalytic activity — as stated in the chapter, 'catalytic activity is lost when the co-factor is removed from the enzyme'; the apoenzyme may retain its overall protein fold but cannot catalyse the reaction because the prosthetic group is integral to the active site mechanism.

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Case Set 5

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A food technology company is developing a product that uses naturally derived compounds from plants as preservatives, flavourings, and dyes. Their chemist shortlists several plant-derived compounds: curcumin (from turmeric), anthocyanins (from berries), lemon grass oil, morphine (from poppy), rubber (from rubber tree), and vinblastin (from periwinkle). The chemist categorises all of these as secondary metabolites. A colleague asks why these compounds exist in plants and whether all plant cells contain them. The chemist explains using her knowledge of biochemistry that these compounds are found predominantly in plant, fungal, and microbial cells and that they are fundamentally different from the primary metabolites that sustain basic cellular life.

Question 1: Define secondary metabolites and list any four categories of secondary metabolites mentioned in the chapter, with one example from each.

Secondary metabolites are compounds found mainly in plants, fungi, and microbial cells that are distinct from the primary metabolites (amino acids, sugars, nucleotides) present in all living organisms; their roles in the host organism are not always fully understood, though many are useful to human welfare.
Four categories with examples: (1) Alkaloids — morphine, codeine; (2) Pigments — carotenoids, anthocyanins; (3) Essential oils — lemon grass oil; (4) Drugs — vinblastin, curcumin.
Additional categories mentioned include terpenoids (monoterpenes, diterpenes), toxins (abrin, ricin), lectins (concanavalin A), and polymeric substances (rubber, gums, cellulose).

Question 2: How do secondary metabolites differ from primary metabolites in terms of distribution in organisms and known functions? Why do not all plant cells necessarily contain the same secondary metabolites?

Primary metabolites — such as amino acids, sugars, nucleotides, and fatty acids — are found in all living cells (animal, plant, fungal, microbial) and have well-defined, identifiable roles in normal physiological processes such as energy production, growth, and reproduction.
Secondary metabolites are found predominantly in plants, fungi, and microbial cells but not universally in all cells or all species; they are produced by specific organisms or specific tissues within an organism, which is why different plants have different secondary metabolites (e.g., only poppy produces morphine, only turmeric produces curcumin).
The reason not all plant cells contain the same secondary metabolites is that secondary metabolite production is genetically regulated and often tissue-specific; the chapter acknowledges that the functions of many secondary metabolites in the host organism are not fully understood at present, though ecological roles (herbivore deterrence, pathogen resistance, pollinator attraction) and human applications are well-documented.

Question 3: The chemist's colleague suggests that since morphine has no clear function in the poppy plant, it is 'useless' to the plant. Critically evaluate this statement using information from the chapter, and explain the broader significance of secondary metabolites.

The colleague's claim is incorrect or at least incomplete: the chapter explicitly states that 'we do not at the moment understand the role or functions of all secondary metabolites in host organisms' — the absence of a fully understood function does not mean the compound is useless; it may reflect a gap in current scientific knowledge rather than biological irrelevance.
Some secondary metabolites are known to have ecological importance: toxins like abrin and ricin protect plants against herbivores and pathogens; essential oils and scents may attract pollinators; alkaloids like morphine may deter herbivory through their bitter or toxic effects — though the exact ecological function of morphine in poppy requires further research to confirm.
The broader significance of secondary metabolites is enormous: many are of great use to human welfare as drugs (morphine as analgesic, vinblastin as anti-cancer agent, curcumin as anti-inflammatory), spices, scents, pigments, rubber, and gums; therefore, even if a secondary metabolite's function in the host plant is uncertain, its study is vital for medicine, agriculture, and industry, making secondary metabolism one of the most applied areas of natural product biochemistry.

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Chapter 9: Biomolecules | Case Study Questions

Passage

Question 1: What are Fraction A and Fraction B technically called, and what molecular weight range characterises the compounds in Fraction A?

Question 2: Why are lipids found in Fraction B (the acid-insoluble fraction) even though individual lipid molecules have molecular weights below 800 Da?

Question 3: Dr. Meena then burns a dried portion of the liver sample completely and collects the ash. What does the ash represent, and what does the experiment as a whole tell us about the difference between living tissue and earth's crust at the elemental level?

Passage

Question 1: Identify Samples X, Y, and Z based on the experimental observations described, and state the monomer unit of each.

Question 2: Explain at the structural level why Sample X gives a blue-black colour with iodine while Sample Y does not, even though both are made of glucose.

Question 3: Compare the biological roles of Samples X, Y, and Z in their respective source organisms, and explain how their structural features are suited to these roles.

Passage

Question 1: Why did Test Tube 1 show no blue colour with iodine while Test Tubes 2 and 3 both showed intense blue?

Question 2: Explain why Test Tube 2 resumed starch digestion after warming to 37°C, but Test Tube 3 never showed digestion even after prolonged incubation.

Question 3: If Ramesh had used an enzyme isolated from a thermophilic bacterium found in a volcanic hot spring instead of regular amylase, predict and explain what might have happened in Test Tube 3, and discuss the biological significance of thermostable enzymes.

Passage

Question 1: Identify the 'deep cleft' in ZymX's structure and explain its functional significance.

Question 2: Explain the role of molecule Q in inhibiting ZymX activity and classify this type of inhibition.

Question 3: The tightly bound haem group in ZymX is a cofactor. Classify this cofactor, explain the role of cofactors in enzyme function, and state what would happen to ZymX's apoenzyme if haem were removed.

Passage

Question 1: Define secondary metabolites and list any four categories of secondary metabolites mentioned in the chapter, with one example from each.

Question 2: How do secondary metabolites differ from primary metabolites in terms of distribution in organisms and known functions? Why do not all plant cells necessarily contain the same secondary metabolites?

Question 3: The chemist's colleague suggests that since morphine has no clear function in the poppy plant, it is 'useless' to the plant. Critically evaluate this statement using information from the chapter, and explain the broader significance of secondary metabolites.

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