Biomolecules Class 11 Biology: Proteins, Carbohydrates, Lipids, Nucleic Acids and Enzymes

What Biomolecules is about

Weightage: This chapter is part of Unit III, Cell: Structure and Function, which carries 15 of the 70 theory marks in the school annual exam together with Cell: The Unit of Life and Cell Cycle and Cell Division.Source: CBSE Biology (044) syllabus 2026-27, Class XI course structure, Unit III Cell: Structure and Function

Key concepts

Chemical analysis of living tissue: acid-soluble and acid-insoluble pools

NCERT section 9.1 How to Analyse Chemical Composition?; 9.3 Biomacromolecules

Finding the organic compounds:
1. Grind living tissue (a leaf, a piece of liver) in trichloroacetic acid (Cl₃CCOOH) with a mortar and pestle to get a thick slurry.
2. Strain it through cheesecloth or cotton. Two fractions result:

• Filtrate = acid-soluble pool: small molecules.
• Retentate = acid-insoluble fraction: large molecules.

3. Separate and purify each compound and use analytical techniques to find its formula and probable structure.

All carbon compounds obtained from living tissue are biomolecules.

Finding the inorganic part (a destructive experiment):

• Weigh fresh tissue (wet weight), dry it so all water evaporates (dry weight), then burn it fully. Carbon compounds are oxidised to CO₂ and water vapour and escape. What remains is ash, containing inorganic elements such as calcium and magnesium. Inorganic compounds such as sulphate and phosphate show up in the acid-soluble fraction.

Living vs non-living matter: both contain the same elements, but carbon, hydrogen and oxygen are relatively far more abundant in living things. For example, carbon is about 18.5% of human body weight but only 0.03% of the Earth's crust; silicon is about 27.7% of the crust but negligible in the body.

Micromolecules and macromolecules:

• Acid-soluble pool: molecular weights of about 18 to 800 daltons (Da): micromolecules (simply 'biomolecules'). It roughly represents the composition of the cytoplasm.
• Acid-insoluble pool: only four types: proteins, nucleic acids, polysaccharides and lipids. Except lipids, they have molecular weights of 10,000 Da and above: macromolecules (biomacromolecules), which are polymers.
• Why are lipids in the insoluble pool? Their molecular weight does not exceed about 800 Da, but they are arranged into membranes. Grinding breaks membranes into water-insoluble vesicles that separate with the insoluble pool. So lipids are not strictly macromolecules.

Average composition of a cell (% of total mass): water 70-90, proteins 10-15, nucleic acids 5-7, carbohydrates 3, lipids 2, ions 1. Water is the most abundant chemical in living organisms.

Worked example

Lipids have molecular weights below 800 Da, yet they appear in the acid-insoluble fraction. Explain why, and name the other three compounds of this fraction. (3 marks)

Try it first, then show the solutionSolution

1. Lipids in cells are arranged into the plasma membrane and other membranes. (1)
2. On grinding, these membranes break into vesicles that do not dissolve in water, so they separate along with the acid-insoluble pool. (1)
3. The other compounds are proteins, nucleic acids and polysaccharides (true macromolecules, 10,000 Da and above). (1)

Answer: Membrane fragments form insoluble vesicles; the others are proteins, nucleic acids and polysaccharides.

Small biomolecules: amino acids, lipids, sugars, nucleosides and nucleotides

NCERT section 9.1 How to Analyse Chemical Composition?; 9.2 Primary and Secondary Metabolites

Amino acids

• An amino group (–NH₂) and an acidic carboxyl group (–COOH) are both attached to the same carbon, the α-carbon, so they are α-amino acids.
• They are substituted methanes: the α-carbon's four valencies carry hydrogen, carboxyl, amino and a variable R group.
• Many amino acids exist, but only 20 types occur in proteins. R group examples: H → glycine, methyl → alanine, hydroxymethyl → serine.
• By the number of amino and carboxyl groups: acidic (glutamic acid), basic (lysine), neutral (valine). Aromatic amino acids: tyrosine, phenylalanine, tryptophan.
• The –NH₂ and –COOH groups ionise, so an amino acid's structure changes with pH. The form carrying both a positive and a negative charge is the zwitterion.

Lipids (generally water-insoluble)

• Fatty acid: a carboxyl group attached to an R group (1 to 19 carbons). Palmitic acid has 16 carbons and arachidonic acid 20 carbons, counting the carboxyl carbon.
• Saturated fatty acids have no C=C double bond; unsaturated ones have one or more.
• Glycerol is trihydroxy propane. Fatty acids esterified with glycerol give mono-, di- and triglycerides, called fats or oils depending on melting point. Oils (e.g. gingelly oil) have a lower melting point and stay liquid in winter.
• Phospholipids contain phosphorus and a phosphorylated organic compound; they are found in cell membranes, e.g. lecithin. Neural tissue has lipids with more complex structures.

Nitrogen bases, nucleosides and nucleotides

• Heterocyclic nitrogen bases: adenine, guanine, cytosine, uracil, thymine.
• Base + sugar = nucleoside: adenosine, guanosine, thymidine, uridine, cytidine.
• Base + sugar + phosphate = nucleotide: adenylic, guanylic, thymidylic, uridylic and cytidylic acid.
• DNA and RNA are made only of nucleotides and act as genetic material.

Sugars: e.g. glucose (C₆H₁₂O₆) and ribose (C₅H₁₀O₅).

Primary and secondary metabolites

• Primary metabolites (amino acids, sugars, etc.) occur in all tissues and have known roles in normal physiology.
• Secondary metabolites are found mainly in plants, fungi and microbes; their roles in the organism are not all understood, but many help humans. Examples: pigments (carotenoids, anthocyanins), alkaloids (morphine, codeine), terpenoids, essential oils (lemon grass oil), toxins (abrin, ricin), lectins (concanavalin A), drugs (vinblastin, curcumin), polymeric substances (rubber, gums, cellulose).

Worked example

Distinguish between a nucleoside and a nucleotide with one example of each. Classify valine, lysine and glutamic acid as acidic, basic or neutral. (3 marks)

Try it first, then show the solutionSolution

1. Nucleoside: nitrogen base + sugar, e.g. adenosine. (1)
2. Nucleotide: nitrogen base + sugar + phosphate, e.g. adenylic acid. (1)
3. Valine neutral, lysine basic, glutamic acid acidic. (1)

Answer: Nucleoside lacks phosphate (adenosine); nucleotide has it (adenylic acid). Valine neutral, lysine basic, glutamic acid acidic.

Proteins: heteropolymers and their functions

NCERT section 9.4 Proteins

Proteins are polypeptides: linear chains of amino acids joined by peptide bonds.

• Since proteins contain up to 20 different amino acids, a protein is a heteropolymer. A homopolymer has only one type of monomer repeated *n* times.
• Essential amino acids cannot be made by our body and must come from dietary protein; non-essential amino acids can be made by the body.

Functions (proteins do almost every job in the cell):

• Structural: collagen forms the ground substance between cells.
• Catalytic: trypsin is a digestive enzyme.
• Hormonal: insulin controls blood sugar.
• Defence: antibodies attack infectious agents.
• Sensing: receptor proteins detect smells, tastes and hormones.
• Transport: GLUT-4 carries glucose into cells.

• Collagen is the most abundant protein in the animal world.
• RuBisCO (Ribulose bisphosphate carboxylase-oxygenase) is the most abundant protein in the whole biosphere (it is the main photosynthetic enzyme in every green leaf).

Worked example

Why is a protein called a heteropolymer while cellulose is a homopolymer? Name the most abundant protein in the animal world and in the biosphere. (3 marks)

Try it first, then show the solutionSolution

1. A protein is built from up to 20 different amino acids, so its monomers are of many types: heteropolymer. (1)
2. Cellulose is built from only one monomer, glucose, repeated many times: homopolymer. (1)
3. Collagen (animal world); RuBisCO (biosphere). (1)

Answer: Many kinds of monomers vs one kind; collagen and RuBisCO.

Polysaccharides and nucleic acids

NCERT section 9.5 Polysaccharides; 9.6 Nucleic Acids

Polysaccharides: long chains of sugars (monosaccharides) found in the acid-insoluble pellet.

• Cellulose: a polymer of glucose only, so a homopolymer. Plant cell walls, paper made from plant pulp and cotton fibre are cellulose.
• Starch: a variant of glucose polymer that stores energy in plants.
• Glycogen: the storage variant in animals; it is branched.
• Inulin: a polymer of fructose.
• In a chain such as glycogen, the right end is the reducing end and the left end the non-reducing end.
• Iodine test: starch forms helical secondary structures that can hold I₂ molecules, giving a blue colour. Cellulose has no complex helices, so it cannot hold I₂ and gives no blue colour.
• Complex polysaccharides contain amino-sugars and modified sugars (glucosamine, N-acetyl galactosamine). Example: chitin in the exoskeleton of arthropods. These are mostly homopolymers.

Nucleic acids: polynucleotides. Together with polysaccharides and polypeptides they make up the true macromolecular fraction.

• Building block: the nucleotide, with three parts: (1) a heterocyclic nitrogenous base, (2) a pentose sugar, (3) phosphoric acid / phosphate.
• Bases: adenine and guanine are substituted purines; cytosine, uracil and thymine are substituted pyrimidines.
• Sugar: ribose in RNA (ribonucleic acid); 2′-deoxyribose in DNA (deoxyribonucleic acid).

Worked example

Complete hydrolysis of a nucleic acid sample gives a pentose sugar, phosphoric acid, adenine, guanine, cytosine and uracil. (a) Is the sample DNA or RNA? Give two reasons. (b) Sort the four bases into purines and pyrimidines. (3 marks)

Try it first, then show the solutionSolution

1. Uracil is present and thymine is absent; uracil is found in RNA. (1)
2. So the pentose must be ribose (RNA), not 2′-deoxyribose (DNA): the sample is RNA. (1)
3. Purines: adenine and guanine; pyrimidines: cytosine and uracil. (1)

Answer: RNA (uracil instead of thymine, ribose sugar); purines A and G, pyrimidines C and U.

Structure of proteins: four levels

NCERT section 9.7 Structure of Proteins

Biologists describe protein structure at four levels:

1. Primary structure: the sequence of amino acids, i.e. which amino acid is first, second and so on. A protein is drawn as a line: the left end is the first amino acid (N-terminal) and the right end the last (C-terminal).
2. Secondary structure: the thread is not a rigid rod. Parts of it coil into a helix like a revolving staircase (α-helix; in proteins only right-handed helices occur), and other parts fold in other ways such as the β-pleated sheet.
3. Tertiary structure: the long chain folds on itself like a hollow woollen ball, giving a 3-dimensional shape. This is absolutely necessary for the biological activity of many proteins (an enzyme's active site is a pocket formed by this folding).
4. Quaternary structure: in proteins made of more than one polypeptide (subunit), the way the folded subunits are arranged relative to each other (like a string of spheres, a cube or a plate).

Example: adult human haemoglobin has 4 subunits: two α and two β subunits (2α + 2β), so it has quaternary structure.

Bonds that hold the folded structure include hydrogen bonds and disulphide bonds.

Flow chart: amino acid sequence (1°) → local helix/sheet (2°) → whole chain folded into a ball (3°) → several chains assembled (4°).

Worked example

Haemoglobin has four polypeptides, two of type α and two of type β. Which levels of structure does it show? Why would heating that destroys its tertiary structure stop it from working? (3 marks)

Try it first, then show the solutionSolution

1. Each chain has a primary sequence, helices (secondary) and is folded (tertiary); the arrangement of four subunits is quaternary, so it shows all four levels. (1½)
2. Function depends on the precise 3-D shape produced by tertiary (and quaternary) folding. (1)
3. If heat unfolds the chain, the shape is lost and the protein cannot do its job, even though the primary sequence is unchanged. (½)

Answer: All four levels; activity needs the folded 3-D shape, which heat destroys.

Enzymes: active site, activation energy and the catalytic cycle

NCERT section 9.8 Enzymes (9.8.1 Chemical Reactions; 9.8.2 How do Enzymes bring about such High Rates of Chemical Conversions?; 9.8.3 Nature of Enzyme Action)

Nature: almost all enzymes are proteins; some nucleic acids act as enzymes and are called ribozymes. Folding of the chain (tertiary structure) makes crevices or pockets; one of these is the active site, into which the substrate fits.

Enzymes vs inorganic catalysts: inorganic catalysts work well at high temperature and pressure; enzymes are damaged at high temperatures (say above 40 °C). Exception: enzymes from thermophilic organisms in hot vents and sulphur springs stay active up to 80-90 °C (thermal stability).

Rate of a reaction: amount of product formed per unit time, rate = δP/δt. As a rough rule, the rate doubles or halves for every 10 °C rise or fall in temperature.

How dramatic is catalysis? CO₂ + H₂O → H₂CO₃: without enzyme about 200 molecules per hour; with carbonic anhydrase about 600,000 molecules per second, roughly 10 million times faster.

Metabolic pathway: a multi-step reaction where each step is catalysed by an enzyme. Example: glucose → 2 pyruvic acid through ten enzyme-catalysed steps. The same pathway with one or two extra steps gives lactic acid in skeletal muscle under anaerobic conditions and ethanol in yeast during fermentation.

How enzymes speed up reactions:

• The substrate must bind the active site, forming a short-lived enzyme-substrate (ES) complex.
• While bound, the substrate passes through an unstable, high-energy transition state.
• Activation energy = energy difference between the substrate and the transition state. Enzymes lower this barrier.
• On an energy graph: y-axis potential energy, x-axis progress of reaction. If product P is lower than substrate S, the reaction is exothermic; even then S must climb to the transition state.

Catalytic cycle:
E + S ⇌ ES → EP → E + P
1. Substrate binds the active site.
2. Binding makes the enzyme change shape and fit more tightly around the substrate (induced fit).
3. The active site breaks bonds in the substrate, forming an enzyme-product (EP) complex.
4. The product is released; the free, unchanged enzyme binds another substrate molecule.

Worked example

Define activation energy. Explain, using the ES complex, how an enzyme speeds up a reaction without being used up. (3 marks)

Try it first, then show the solutionSolution

1. Activation energy: the difference in average energy between the substrate and the transition state that must be crossed to form product. (1)
2. The enzyme binds the substrate at its active site (ES complex) and changes shape to fit tightly, so the transition state is reached with less energy: the barrier is lowered. (1)
3. Bonds are broken/made, product is released (EP → E + P) and the enzyme comes out unchanged, ready for another cycle. (1)

Answer: Enzymes lower activation energy via the ES complex and are released unchanged.

Factors affecting enzyme activity, inhibition, classes and co-factors

NCERT section 9.8.4 Factors Affecting Enzyme Activity; 9.8.5 Classification and Nomenclature of Enzymes; 9.8.6 Co-factors

Anything that changes the enzyme's tertiary structure changes its activity.

Temperature and pH: each enzyme works within a narrow range and is most active at its optimum temperature and optimum pH. Activity falls on either side, giving a bell-shaped graph.

• Low temperature keeps the enzyme temporarily inactive (it recovers on warming).
• High temperature denatures the protein and destroys activity permanently.

Substrate concentration: as [S] rises, velocity rises at first, then levels off at a maximum velocity (Vmax). Reason: enzyme molecules are fewer than substrate molecules; once all active sites are saturated, extra substrate has no free enzyme to bind. (The graph also marks Km, the substrate concentration at Vmax/2.)

Inhibition: a chemical that binds the enzyme and shuts off activity is an inhibitor.

• Competitive inhibitor: closely resembles the substrate, competes for the substrate-binding site, so the substrate cannot bind.
• Example: malonate inhibits succinic dehydrogenase because it resembles the substrate succinate.
• Competitive inhibitors are often used to control bacterial pathogens.

Six classes of enzymes (each with 4-13 subclasses, named by a four-digit number):
1. Oxidoreductases / dehydrogenases: oxidation-reduction between two substrates (S reduced + S′ oxidised → S oxidised + S′ reduced).
2. Transferases: transfer a group G (not hydrogen) from one substrate to another (S–G + S′ → S + S′–G).
3. Hydrolases: split bonds by adding water; the bonds may be peptide, glycosidic, ester, ether, C–C, C–halide or P–N.
4. Lyases: remove groups by mechanisms other than hydrolysis, leaving double bonds.
5. Isomerases: inter-convert optical, geometric or positional isomers.
6. Ligases: join two compounds (C–O, C–S, C–N, P–O bonds).

Co-factors: non-protein parts that make some enzymes active. The protein part is then the apoenzyme. Three kinds:

• Prosthetic groups: organic, tightly bound. Example: haem in peroxidase and catalase (which break hydrogen peroxide into water and oxygen); haem is part of the active site.
• Co-enzymes: organic, bound only transiently during catalysis; one co-enzyme can serve many enzymes. Many contain vitamins: NAD and NADP contain niacin.
• Metal ions: form coordination bonds with the active site and the substrate. Example: zinc for the protein-digesting enzyme carboxypeptidase.

Removing the co-factor destroys catalytic activity, proving its importance.

Worked example

Why does an enzyme kept in a fridge recover its activity when warmed, while one that is boiled does not? Name the enzyme class that joins two molecules. (3 marks)

Try it first, then show the solutionSolution

1. Low temperature only slows molecular movement; the enzyme stays temporarily inactive with its shape intact, so it works again on warming. (1)
2. High temperature denatures the protein: the tertiary structure and active site are destroyed permanently, so activity is not recovered. (1)
3. Ligases join two compounds. (1)

Answer: Cold = reversible inactivity; heat = irreversible denaturation; ligases.

Biomolecules in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Enzyme Action, Inhibition and Classes
• Protein Structure: Four Levels

Biomolecules: key facts

Enzyme action

$\text{E}+\text{S}\rightleftharpoons \text{ES}\to \text{EP}\to \text{E}+\text{P}$

ES complex formation is essential; enzyme released unchanged.

Rate of reaction

$\text{rate}=\frac{\delta P}{\delta t}$

Roughly doubles (or halves) for every 10 °C rise (or fall).

Nucleoside and nucleotide

$\text{nucleoside}=\text{base}+\text{sugar};\text{nucleotide}=\text{base}+\text{sugar}+\text{phosphate}$

DNA has 2'-deoxyribose; RNA has ribose.

Haemoglobin subunits

$\text{Hb}=2\alpha +2\beta$

Four subunits: quaternary structure.

Holoenzyme

$\text{active enzyme}=\text{apoenzyme (protein)}+\text{co-factor}$

Co-factor: prosthetic group, co-enzyme or metal ion.

Cell composition (% of mass)

$\text{water }70\text{-}90;\text{proteins }10\text{-}15;\text{nucleic acids }5\text{-}7;\text{carbohydrates }3;\text{lipids }2;\text{ions }1$

Water is the most abundant chemical.

Biomolecules: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Biomolecules

The question patterns to expect, and the method that works for each.

Tricks to remember Biomolecules

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Six enzyme classes: OTHLIL

  Oxidoreductases, Transferases, Hydrolases, Lyases, Isomerases, Ligases: 'Only The Hungry Lions Ignore Lunch'.

  Use it when: Order of the six classes.

• Pure As Gold

  PURines = Adenine and Guanine; the rest (C, U, T) are pyrimidines ('CUT the pie').

  Use it when: Purine/pyrimidine classification.

• Nucleo-T-ide has the phosphaTe

  The 't' in nucleotide reminds you of the phosphate; nucleoside is just base + sugar.

  Use it when: Nucleoside vs nucleotide.

• G-A-S amino acids

  Glycine (H), Alanine (methyl), Serine (hydroxymethyl): R group grows from H to CH₃ to CH₂OH.

  Use it when: Recalling R groups.

• Lyase leaves a double bond

  Lyases 'lyse' a group off without water and leave a double bond; hydrolases need water.

  Use it when: Telling lyases from hydrolases.

• Ni-NAD

  NIacin is in NAD and NADP; zinc is in carboxypeptidase; haem is in catalase.

  Use it when: Co-factor examples.

Biomolecules: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Biomolecules

The slips examiners see most often, and what to write instead.

• ✗Mistake: Calling lipids true macromolecules

  They are found in the acid-insoluble pool.

  ✓Do this: Lipids are small (under about 800 Da); they appear in the insoluble pool only because membrane fragments form vesicles.

  Usually costs: 1 mark.

• ✗Mistake: Writing that high and low temperatures both destroy enzymes

  Both reduce activity on the graph.

  ✓Do this: Low temperature makes enzymes temporarily inactive; high temperature denatures them permanently.

  Usually costs: 1 mark.

• ✗Mistake: Drawing the substrate-concentration graph as a straight line or a bell curve

  Confused with the pH and temperature graphs.

  ✓Do this: It rises and then flattens at Vmax because active sites are saturated.

  Usually costs: 1-2 marks.

• ✗Mistake: Calling adenosine a nucleotide

  The two names look alike.

  ✓Do this: Adenosine = nucleoside (no phosphate); adenylic acid = nucleotide.

  Usually costs: ½-1 mark.

• ✗Mistake: Stating that collagen is the most abundant protein on Earth

  Two 'most abundant' facts are mixed up.

  ✓Do this: Collagen: most abundant in animals; RuBisCO: most abundant in the biosphere.

  Usually costs: ½ mark.

• ✗Mistake: Saying a competitive inhibitor binds somewhere else on the enzyme

  Not reading the definition carefully.

  ✓Do this: A competitive inhibitor resembles the substrate and competes for the same substrate-binding (active) site.

  Usually costs: 1 mark.

NCERT exercise guide: Biomolecules

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Biomolecules

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   Which of these is a secondary metabolite?
   (a) Glucose (b) Alanine (c) Morphine (d) Palmitic acid

   Hint

   Alkaloid.

   Answer

   (c) Morphine

2. Q2 · 1 mark · MCQExam level

   Adult human haemoglobin consists of:
   (a) 2 α and 2 β subunits (b) 4 α subunits (c) 1 α and 3 β subunits (d) 2 subunits only

   Hint

   Two pairs.

   Answer

   (a) 2 α and 2 β subunits

3. Q3 · 2 marks · Very short answerExam level

   Name the purines and pyrimidines found in nucleic acids. Which sugar distinguishes DNA from RNA? (2 marks)

   Hint

   Pure as gold.

   Answer

   Purines: adenine and guanine; pyrimidines: cytosine, uracil, thymine (1). DNA has 2′-deoxyribose; RNA has ribose (1).

4. Q4 · 2 marks · Very short answerHarder exam level

   Why are proteins called heteropolymers? Distinguish between essential and non-essential amino acids. (2 marks)

   Hint

   20 monomers; diet.

   Answer

   Proteins contain up to 20 different amino acids as monomers (1). Essential amino acids must come from the diet; non-essential ones are made by the body (1).

5. Q5 · 2 marks · Very short answerHarder exam level

   Give the difference between fats and oils and between saturated and unsaturated fatty acids. (2 marks)

   Hint

   Melting point; double bonds.

   Answer

   Fats have higher melting points and are solid at room temperature; oils have lower melting points and stay liquid (1). Saturated fatty acids have no C=C double bonds; unsaturated ones have one or more (1).

6. Q6 · 2 marks · Very short answerHarder exam level

   Explain why the rate of an enzyme reaction levels off at high substrate concentration. What is Km? (2 marks)

   Hint

   Saturation.

   Answer

   Enzyme molecules are fewer than substrate molecules; once all active sites are occupied, extra substrate cannot bind, so velocity stays at Vmax (1). Km is the substrate concentration at which velocity is half of Vmax (1).

7. Q7 · 3 marks · Short answerHarder exam level

   Describe the three kinds of co-factors with an example of each. (3 marks)

   Hint

   Tight, transient, metal.

   Answer

   Prosthetic group: organic, tightly bound, e.g. haem in catalase (1). Co-enzyme: organic, transiently bound, often a vitamin derivative, e.g. NAD with niacin (1). Metal ion: forms coordination bonds with active site and substrate, e.g. zinc in carboxypeptidase (1).

8. Q8 · 3 marks · Short answerHarder exam level

   Compare cellulose, starch and glycogen in terms of monomer, where found and reaction with iodine. (3 marks)

   Hint

   All glucose polymers.

   Answer

   All three are glucose polymers (homopolymers) (1). Cellulose: plant cell walls, cotton, paper; no blue with iodine (no helix). Starch: energy store in plants; blue with iodine (helix). Glycogen: branched energy store in animals (2).

9. Q9 · 3 marks · Short answerHarder exam level

   Explain the steps of the catalytic cycle of an enzyme. (3 marks)

   Hint

   Four steps.

   Answer

   Substrate binds the active site (½). Binding induces a shape change so the enzyme fits more tightly (1). Bonds of the substrate are broken and an enzyme-product complex forms (1). Product is released and the free enzyme binds another substrate (½).

10. Q10 · 5 marks · Long answerHarder exam level

    (a) Explain how living tissue is analysed for organic and inorganic compounds. (b) Why is the acid-soluble pool said to represent the cytoplasm? (c) Give the average composition of a cell. (5 marks)

    Hint

    Grind, filter, burn.

    Answer

    (a) Grind tissue in trichloroacetic acid, strain: filtrate is the acid-soluble pool, retentate the acid-insoluble pool; separate and purify compounds and use analytical techniques for formula and structure (1½). For inorganic matter, dry the tissue (wet weight → dry weight) and burn it; ash holds elements like Ca and Mg (1). (b) It holds the small molecules (18-800 Da) that are dissolved in the cytoplasm; macromolecules of cytoplasm and organelles form the insoluble pool (1). (c) Water 70-90%, proteins 10-15%, nucleic acids 5-7%, carbohydrates 3%, lipids 2%, ions 1% (1½).

11. Q11 · 3 marks · Short answerHOTS / competency

    Suppose a vitamin deficiency lowers the amount of niacin in the body. Predict how this could affect many different enzyme-catalysed reactions at once, and explain why removing a co-factor stops an enzyme. (3 marks)

    Hint

    One co-enzyme, many enzymes.

    Answer

    Niacin is a component of the co-enzymes NAD and NADP (1). A single co-enzyme serves many different enzymes, so all reactions needing NAD/NADP (e.g. oxidation-reduction steps) slow down together (1). The co-factor is part of the catalytic machinery; without it the apoenzyme alone is inactive (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): Boiling an enzyme destroys its activity permanently even though its primary structure is unchanged.
    Reason (R): Enzyme activity depends on the tertiary structure, which forms the active site.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    Shape decides function.

    Answer

    (a)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Which acid is used to grind tissue for chemical analysis?Show answer

   Trichloroacetic acid (Cl₃CCOOH).

2. 2.What is left when a dried tissue is burnt completely?Show answer

   Ash, containing inorganic elements such as calcium and magnesium.

3. 3.What is the R group of glycine?Show answer

   A hydrogen atom.

4. 4.How many carbons does palmitic acid have?Show answer

   16, including the carboxyl carbon.

5. 5.Give one example of a phospholipid.Show answer

   Lecithin.

6. 6.Name an alkaloid and a toxin that are secondary metabolites.Show answer

   Morphine (or codeine); ricin (or abrin).

7. 7.Inulin is a polymer of which sugar?Show answer

   Fructose.

8. 8.Which polysaccharide forms the exoskeleton of arthropods?Show answer

   Chitin.

9. 9.What are ribozymes?Show answer

   Nucleic acids (RNA) that act as enzymes.

10. 10.Which compound competitively inhibits succinic dehydrogenase?Show answer

    Malonate.

11. 11.Which vitamin is part of NAD?Show answer

    Niacin.

12. 12.What is the protein part of an enzyme that needs a co-factor called?Show answer

    Apoenzyme.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why lipids land in the acid-insoluble fraction?
• ☐Can you name an acidic, a basic, a neutral and an aromatic amino acid?
• ☐Can you distinguish nucleoside and nucleotide with examples?
• ☐Can you describe the four levels of protein structure?
• ☐Can you explain why starch, but not cellulose, gives a blue colour with iodine?
• ☐Can you draw and explain the activation-energy graph?
• ☐Can you sketch the effect of pH, temperature and substrate concentration on enzyme activity?
• ☐Can you name the six enzyme classes and the three types of co-factor with an example each?

Biomolecules: frequently asked questions