Zoology 2022 Paper II 50 marks Draw

Paper II — Q6

(a) Draw the molecular structure of ATP synthase and diagrammatically explain the chemiosmotic concept of ATP synthesis. 20 (b)…

(a)

Draw the molecular structure of ATP synthase and diagrammatically explain the chemiosmotic concept of ATP synthesis. 20 marks

(b)

Illustrate the following :

(i)

A = T and G ≡ C base pairing as a part of DNA double strand

(ii)

Michaelis-Menten kinetic pattern of an enzymatic reaction 15

(c)

Explain the cellular mechanism of action of steroid hormones. 15 marks

हिंदी में प्रश्न पढ़ें
(a)

ए० टी० पी० सिन्थेज की आण्विक संरचना का चित्र बनाइए तथा ए० टी० पी० संश्लेषण की रसोप्रसरणी (केमिऑस्मोटिक) संकल्पना को आरेखीय रूप से समझाइए। 20

(b)

निम्नलिखित को सचित्र समझाइए :

(i)

डी० एन० ए० द्वैलड़ी के एक भाग के रूप में A = T तथा G ≡ C का बेस युग्मन

(ii)

एक प्रकिण्व की अभिक्रिया की माइकेलिस-मेंटेन गतिक प्राक्रता 15

(c)

स्टेरॉयड हार्मोनों की क्रिया की कोशिकीय कार्यप्रणाली की व्याख्या कीजिए। 15

Q6 of the 2022 UPSC Mains Zoology Paper II, as printed
The question as printed in the 2022 Zoology paper

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a) ATP synthase and the chemiosmotic concept of ATP synthesis

ATP synthase (F₁F₀-ATPase, complex V) is a rotary enzyme of the inner mitochondrial membrane. The labelled diagram of its molecular structure should show:

  • F₀ in the membrane: subunit a with two half-channels for H⁺, two b subunits, and a ring of c subunits (eight in vertebrates)
  • F₁ projecting into the matrix: an α₃β₃ hexamer with the catalytic sites on the three β subunits
  • the central stalk (γ, ε) joining the c-ring to the hexamer, and the peripheral stalk (b, OSCP) holding the hexamer still
Diagram: ATP synthase structure

Chemiosmotic concept (Peter Mitchell, 1961; Nobel Prize 1978). The second diagram should show these steps, the intermembrane space H⁺-rich (positive) and the matrix H⁺-poor (negative):

  1. Complexes I, III and IV of the electron transport chain pump H⁺ from the matrix into the intermembrane space (about 10 H⁺ per NADH) as electrons pass to O₂.
  2. The impermeable inner membrane holds the gradient as the proton-motive force, Δp = Δψ − (2.303RT/F)ΔpH (inside minus outside), about 0.2 V in magnitude (Δψ about 0.14 V plus about 60 mV per pH unit).
  3. H⁺ flows back through F₀ down this gradient; protonation and deprotonation of carboxyl groups on the c subunits turn the c-ring and the γ stalk.
  4. The turning γ drives the three β subunits through open, loose and tight states (Boyer's binding-change mechanism): ADP and Pi bind in the loose state, ATP forms in the tight state and is released in the open state. One 360° turn makes 3 ATP.
  5. With a c₈ ring, 8 H⁺ per turn give 2.7 H⁺ per ATP, about 3.7 including H⁺ for Pi and ADP/ATP transport.
Diagram: Chemiosmotic scheme

Evidence: uncouplers such as 2,4-dinitrophenol dissipate the gradient, so electron transport continues but ATP synthesis stops; oligomycin blocks F₀; an artificial pH gradient made ATP in chloroplasts (Jagendorf, 1966) and in vesicles with bacteriorhodopsin and ATP synthase (Racker and Stoeckenius, 1974).

(b)(i) A=T and G≡C base pairing in the DNA double strand

The labelled diagram should show:

  • two antiparallel sugar–phosphate strands, one running 5′→3′ and the other 3′→5′
  • adenine (a purine) joined to thymine (a pyrimidine) by two hydrogen bonds, A=T
  • guanine (a purine) joined to cytosine (a pyrimidine) by three hydrogen bonds, G≡C
  • base pairs stacked 0.34 nm apart, in a helix 2 nm wide with one turn of 3.4 nm per 10 base pairs
Diagram: DNA base pairing

Pairing is complementary (Watson and Crick, 1953). A purine always pairs with a pyrimidine, which keeps the width of the helix uniform, and the hydrogen-bonding groups allow only A with T and G with C, so Chargaff's rule (A = T, G = C) holds. The extra bond makes G≡C pairs stronger, so GC-rich DNA melts at a higher temperature.

(b)(ii) Michaelis–Menten kinetic pattern of an enzymatic reaction

For E + S ⇌ ES → E + P (Michaelis and Menten, 1913; steady-state treatment by Briggs and Haldane, 1925), the initial velocity is v = Vmax[S]/(Km + [S]), where Km = (k₋₁ + k₂)/k₁ and Vmax = k₂[E]total. The labelled diagram should show:

  • a rectangular hyperbola of v against [S]: nearly straight at low [S] (first order), bending over to approach Vmax at high [S] (zero order, enzyme saturated)
  • Vmax as the plateau, and Km as the [S] at which v = Vmax/2
  • a Lineweaver–Burk plot of 1/v against 1/[S]: a straight line with y-intercept 1/Vmax, x-intercept −1/Km and slope Km/Vmax
Diagram: Michaelis Menten and LB

A low Km means high affinity for the substrate.

(c) Cellular mechanism of action of steroid hormones

Steroid hormones (cortisol, aldosterone, oestradiol, progesterone, testosterone) are lipid-soluble derivatives of cholesterol. They travel in the blood bound to carrier proteins and diffuse through the plasma membrane of target cells, so they act on intracellular receptors, unlike peptide hormones.

Genomic action:

  1. The inactive receptor lies in the cytoplasm (glucocorticoid, mineralocorticoid) or already in the nucleus (oestrogen, progesterone), held by heat-shock proteins such as hsp90.
  2. The hormone binds the ligand-binding domain; the receptor changes shape and releases hsp90.
  3. The hormone–receptor complex dimerises and, if cytoplasmic, enters the nucleus through the nuclear pore.
  4. Its DNA-binding domain, with two zinc-containing motifs, binds the hormone response element (for example the glucocorticoid response element) in the promoter or enhancer of target genes.
  5. The complex recruits coactivators or corepressors and chromatin-modifying enzymes, so that RNA polymerase II transcribes or represses the gene.
  6. The mRNA is translated into new proteins (for example gluconeogenic enzymes in the liver under cortisol), which produce the cellular response.

The response is slow (hours) but long-lasting and amplified, because each gene gives many mRNAs and proteins. Some steroid effects occur within seconds to minutes through membrane-associated receptors and second messengers (non-genomic action). The signal ends when the hormone dissociates and is metabolised.

How this answer will be evaluated

Framework: null

What earns full marks: All parts fully answered with labelled diagrams, precise mechanisms, and specific examples

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Draw the molecular structure of ATP synthase and explain the chemiosmotic concept of ATP synthesis with a diagram. 20 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Labelled structure: F₀ (a, b, c-ring) and F₁ (α₃β₃, γ stalk, peripheral stalk)
    • Chemiosmotic sequence: proton pumping by the electron transport chain, proton-motive force, H⁺ return through F₀ turning the rotor, ATP formed in F₁ (Mitchell)
    • Binding-change mechanism (open, loose, tight) converting ADP + Pi to ATP

    Loses marks

    • Unlabelled or missing diagrams
    • Confusing chemiosmosis with substrate-level phosphorylation
    • Placing F₁ on the intermembrane-space side

    Earns more

    • Direction of H⁺ flow and the two sides of the membrane marked
    • Proton numbers: c₈ ring, 2.7 H⁺ per ATP, about 3.7 with transport
    • Experimental evidence: uncouplers, oligomycin, Jagendorf, Racker and Stoeckenius

    Extra mark

    • Components of Δp (Δψ and ΔpH) with magnitude about 0.2 V
  2. (b) Illustrate (i) A=T and G≡C base pairing as part of the DNA double strand and (ii) the Michaelis–Menten kinetic pattern of an enzymatic reaction. 15 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • (i) Antiparallel strands; purine–pyrimidine pairing with two hydrogen bonds for A=T and three for G≡C
    • (ii) Hyperbolic v against [S] curve with Vmax plateau and Km at Vmax/2, and the equation v = Vmax[S]/(Km + [S])

    Loses marks

    • Wrong number of hydrogen bonds (three for A=T, two for G≡C)
    • Curve without Vmax or Km marked
    • Omitting one of the two sub-parts

    Earns more

    • Helix dimensions (2 nm width, 0.34 nm rise, 3.4 nm turn) and Chargaff's rule
    • Lineweaver–Burk plot with intercepts 1/Vmax and −1/Km
    • Meaning of Km as affinity; first-order and zero-order regions

    Extra mark

    • Greater stability of GC-rich DNA
  3. (c) Explain the cellular mechanism of action of steroid hormones. 15 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Lipid-soluble hormone crosses the membrane and binds an intracellular receptor (hsp90 released)
    • Dimerised complex binds the hormone response element and alters transcription through coactivators and RNA polymerase II
    • New mRNA and proteins give a slow, long-lasting response

    Loses marks

    • Describing a cell-surface receptor and cAMP pathway as the main mechanism
    • Omitting the nuclear transcription step

    Earns more

    • Cytoplasmic versus nuclear receptors (glucocorticoid versus oestrogen)
    • Receptor domains: ligand-binding and zinc-containing DNA-binding
    • Rapid non-genomic actions through membrane receptors

    Extra mark

    • Example such as cortisol inducing gluconeogenic enzymes
What "Draw" is asking you to do

Produce the diagram as the answer, not as an ornament to it. Where the question lists several items — circuit, function table, logic symbol, structure — each is separately marked, and the lines of text must refer to the diagram through its own labels.

Structure that answers it

Diagram drawn large and clean → every part, axis and terminal labelled → caption → two or three lines tying it to what was asked

Where marks are lost

Delivering part of the list and leaving the rest, which forfeits those marks directly. In chemistry, a flat sketch where the geometry or stereochemistry was the point; in engineering, unlabelled terminals, missing polarity, or no sign convention stated.

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