Q.(A) Define the following term :
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🔒 Start your 14-day free trial to unlock the full solution →Part (a)Concept understanding — Protein Structure Levels
Protein Structure Levels: From a String to a Working Machine
Imagine you have a long string of beads. Each bead is a different colour, and the order of colours is fixed. If you just lay that string on a table, it's a floppy, useless line. But if you could somehow make that string fold itself into a tiny, precise 3D shape — say, a key that fits a specific lock — you'd have something that actually does a job. That's exactly what a protein is.
A protein starts as a long chain of smaller units called amino acids. There are 20 different kinds, each with a unique side chain (the "colour" of the bead). The exact sequence of these amino acids is determined by your DNA. But a protein isn't just a chain — it's a chain that folds into a specific shape, and that shape determines what the protein does. If the shape is wrong, the protein can't work.
The folding happens in stages, and we call these stages the four levels of protein structure.
Level 1: Primary Structure — The Sequence
This is the simplest level: just the linear order of amino acids in the chain, linked by peptide bonds. Think of it as the sentence written in the language of proteins.
Primary structure = the sequence of amino acids from the N-terminus (start) to the C-terminus (end).
Why does this matter? Because the sequence determines everything else. Change one amino acid in a critical spot, and the entire protein can misfold. Example: sickle cell anaemia is caused by a single amino acid swap in haemoglobin — valine replaces glutamic acid at position 6. One bead out of hundreds changes colour, and the whole protein folds wrong.
Level 2: Secondary Structure — Local Folding Patterns
The chain doesn't stay straight. Hydrogen bonds form between the backbone atoms (not the side chains) of nearby amino acids. These bonds cause the chain to twist or fold into regular, repeating patterns.
Two common patterns:
- Alpha helix (α-helix): The chain coils like a spring or a spiral staircase. Hydrogen bonds form between every 4th amino acid, holding the coil tight.
- Beta sheet (β-sheet): The chain folds back and forth like a pleated fan. Hydrogen bonds form between adjacent segments, creating a flat, sheet-like structure.
Secondary structure is stabilised entirely by hydrogen bonds between the carbonyl oxygen of one amino acid and the amide hydrogen of another — both part of the peptide backbone. Side chains stick out and don't participate.
These patterns are local — they happen in short stretches of the chain. A single protein can have multiple α-helices and β-sheets separated by loops.
Level 3: Tertiary Structure — The Global 3D Shape
Now the whole chain folds into its final, compact, three-dimensional shape. This is where the protein becomes functional. The tertiary structure is stabilised by interactions between the side chains of amino acids that may be far apart in the sequence but come close in space.
What holds it together?
- Hydrophobic interactions: Nonpolar side chains cluster together in the protein's interior, away from water.
- Hydrogen bonds: Between polar side chains.
- Ionic bonds: Between positively and negatively charged side chains.
- Disulfide bridges: Covalent bonds between the sulfur atoms of two cysteine amino acids — these are strong and lock parts of the chain together.
- Van der Waals forces: Weak attractions between closely packed atoms.
A common mistake: thinking tertiary structure is just "more secondary structure." It's not. Secondary structure is local folding; tertiary structure is the global arrangement of the entire chain, including how helices and sheets pack together.
Level 4: Quaternary Structure — Multiple Chains Working Together
Some proteins are made of more than one polypeptide chain. Each chain is a separate subunit, and the quaternary structure describes how these subunits assemble into a functional complex. …
Why this formula?
Protein Structure Levels: Understanding the "Why" Behind the Hierarchy
Protein structure is not defined by a single formula, but by a logical hierarchy of organization. Each level builds on the previous one, and the "formulas" here are really principles of molecular interaction that explain why proteins fold the way they do.
Let's break down each level and the reasoning behind its key features.
1. Primary Structure: The Sequence "Formula"
What it is: The linear sequence of amino acids linked by peptide bonds.
Key "formula":
Protein=NH2-[Amino Acid]1-[AA]2-...-[AA]n-COOH
Why this holds:
- Peptide bond formation is a condensation reaction:
-COOH+NH2-→-CO-NH-+H2O
- This bond is rigid and planar due to resonance (partial double-bond character). This restricts rotation, which directly influences higher-order folding.
- The sequence is determined by DNA (genetic code). Every change in sequence can alter the entire structure — this is why a single mutation (e.g., sickle cell anemia: Glu → Val at position 6) can cause disease.
Exam insight: The primary structure is the only level that is covalently determined. All higher levels are non-covalent interactions.
2. Secondary Structure: Local Folding Patterns
Key patterns: α-helix and β-pleated sheet.
Why these form — the hydrogen bond "formula":
The α-helix
- Hydrogen bonds form between the carbonyl oxygen (C=O) of residue n and the amide hydrogen (N-H) of residue n+4.
- Why n+4? This spacing allows the backbone to coil into a right-handed helix with exactly 3.6 amino acids per turn.
- Reasoning: The peptide bond's planar nature forces the backbone into a specific geometry. The n+4 pattern maximizes H-bonding while minimizing steric clashes.
The β-sheet
- Hydrogen bonds form between adjacent strands (either parallel or antiparallel).
- Why not n+4? The backbone is extended (pleated), so H-bonds occur between different segments, not within the same chain.
Key formula (Ramachandran plot):
Only certain backbone dihedral angles (ϕ,ψ) are allowed:
- α-helix: ϕ≈−57∘, ψ≈−47∘
- β-sheet: ϕ≈−130∘, ψ≈+130∘
Why these angles? Steric hindrance — atoms cannot overlap. The Ramachandran plot shows the only regions where no two atoms clash.
3. Tertiary Structure: The 3D Fold
Key "formula": The hydrophobic effect drives folding.
Why this holds:
- Water molecules form a cage-like structure around nonpolar (hydrophobic) side chains. This is entropically unfavorable (water loses freedom).
- To minimize this, hydrophobic side chains cluster together in the protein's core, away from water.
- Result: The protein collapses into a compact globule, with polar/charged residues on the surface.
Supporting interactions (the "glue"):
| Interaction | Why it matters |
|---|---|
| Hydrogen bonds | Between side chains (e.g., Ser–Glu) |
| Ionic bonds | Between charged groups (e.g., Lys–Asp) |
| Van der Waals forces | Close packing of atoms |
| Disulfide bridges | Covalent S–S bonds (only in oxidizing environments) |
Why not just one formula? Tertiary structure is unique to each protein — it's the sum of all these interactions, not a single equation.
4. Quaternary Structure: Multiple Subunits
Key "formula":
Functional protein=∑i=1nSubuniti
Why this holds:
- Some proteins need multiple polypeptide chains to function (e.g., hemoglobin: α2β2). …
Part (b)Concept understanding — Glucose Cyclization
Glucose Cyclization: From a Straight Chain to a Ring
Imagine you have a long, flexible chain with a hook at one end and an eyelet at the other. If you swing that chain around, the hook can snap into the eyelet, forming a loop. That's the core idea behind glucose cyclization.
Glucose in its simplest written form is a straight chain of six carbon atoms with an aldehyde group (−CHO) at one end. But in water (like in your blood), this chain doesn't stay straight. The aldehyde group reacts with the alcohol group on the fifth carbon, forming a stable six-membered ring.
Why does this happen?
The aldehyde carbon is electrophilic (electron-poor), and the oxygen on carbon-5 has lone pairs (nucleophilic). They attack each other, forming a new bond. This creates a hemiacetal — a carbon bonded to both an −OH and an −OR group. The ring is more stable than the open chain in solution.
The open-chain form of glucose exists in equilibrium with the cyclic form, but at any given time, over 99% of glucose molecules are in the cyclic form.
The precise statement
Glucose cyclization is an intramolecular nucleophilic addition where the aldehyde group at C1 reacts with the hydroxyl group at C5, forming a six-membered pyranose ring (named after pyran, a six-membered oxygen heterocycle). This reaction creates a new chiral center at C1, giving two possible stereoisomers called anomers: α and β.
The two anomers
When the ring forms, the oxygen from C5 becomes part of the ring. The carbon that was the aldehyde (now C1) becomes a new chiral center. The −OH group at this new center can point:
- Down (relative to the ring plane) → α-D-glucose
- Up → β-D-glucose
The α and β anomers are diastereomers, not enantiomers. They differ only at the anomeric carbon (C1). In solution, they interconvert through the open-chain form — a process called mutarotation.
How to draw it (Haworth projection)
- Draw a hexagon with an oxygen atom at the top-right corner.
- Number the carbons clockwise from the oxygen: C1 is the carbon to the right of oxygen, C2 next, and so on.
- For α-D-glucose, the −OH at C1 points down (opposite to the CH2OH group at C5).
- For β-D-glucose, the −OH at C1 points up (same side as the CH2OH group). …
Why this formula?
Glucose Cyclization: Why the Ring Forms
Glucose cyclization is a classic example of an intramolecular reaction — a molecule reacting with itself. Let's build the understanding step by step.
1. The Starting Point: Open-Chain Glucose
Glucose (C6H12O6) in its open-chain form has:
- An aldehyde group (−CHO) at carbon 1 (C1)
- A hydroxyl group (−OH) at carbon 5 (C5)
The aldehyde is electrophilic (electron-deficient at the carbonyl carbon), and the hydroxyl is nucleophilic (electron-rich oxygen with lone pairs).
2. Why Cyclization Happens: Thermodynamic & Kinetic Favorability
The Key Insight
The molecule is flexible — it can bend so that the C5 hydroxyl oxygen approaches the C1 aldehyde carbon. This brings two reactive groups into close proximity.
- Entropy is favorable: One molecule becomes one ring — no loss of translational entropy (unlike two separate molecules reacting).
- Ring strain is manageable: A 5- or 6-membered ring (furanose or pyranose) has minimal angle strain (close to tetrahedral angles).
Result: The reaction is reversible but strongly favors the cyclic form — in aqueous solution, >99% of glucose exists as the ring.
3. The Reaction: Hemiacetal Formation
The nucleophilic oxygen of the C5 hydroxyl attacks the electrophilic carbonyl carbon of C1:
R-CHO+R’-OH⇌R-CH(OH)(OR’)
This is a hemiacetal — a carbon bonded to both an −OH and an −OR group.
Mechanism (simplified)
- Protonation of the carbonyl oxygen (acid-catalyzed) makes C1 more electrophilic.
- Nucleophilic attack by the C5 oxygen.
- Deprotonation yields the cyclic hemiacetal.
4. The Key Formula(e): Ring Size & Anomeric Carbon
Ring Size Determination
The ring size depends on which hydroxyl attacks:
- C5 hydroxyl → pyranose (6-membered ring: 5 carbons + 1 oxygen)
- C4 hydroxyl → furanose (5-membered ring: 4 carbons + 1 oxygen)
For D-glucose, the C5 attack is overwhelmingly favored, giving the pyranose form.
The Anomeric Carbon
The new chiral center formed at C1 is called the anomeric carbon. Two stereoisomers arise:
- α-anomer: −OH at C1 is trans to the −CH2OH group (axial in the chair conformation)
- β-anomer: −OH at C1 is cis to the −CH2OH group (equatorial in the chair conformation)
Why two forms? The attack can occur from either face of the planar carbonyl group — leading to two possible configurations at the new stereocenter.
5. The Equilibrium Constant & Mutarotation
The interconversion between α and β anomers is called mutarotation:
α-D-glucopyranose⇌open chain⇌β-D-glucopyranose
At equilibrium (in water at 20°C):
- β-D-glucopyranose: ~64%
- α-D-glucopyranose: ~36%
- Open chain: <0.1% …
Part (a)
Reducing sugar: a carbohydrate carrying a free aldehyde or keto group (a free anomeric carbon) that can reduce Tollens' or Fehling's reagent. All monosaccharides and most disaccharides (maltose, lactose) are reducing; sucrose is not.
Fibrous vs globular proteins: fibrous are long, thread-like, water-insoluble, structural (keratin, collagen); globular are spherical, water-soluble, functional (enzymes, haemoglobin). …
Part (a): a reducing sugar has a free aldehyde/keto group; fibrous proteins are insoluble structural, globular are soluble functional; a nucleotide is a nucleoside plus a phosphate. Part (b): glucose + hydroxylamine → oxime; + acetic anhydride → pentaacetate; + conc. HNO₃ → saccharic (glucaric) acid.
Part (a)
- Reducing sugar. A reducing sugar is a carbohydrate that acts as a reducing agent because it possesses a free aldehyde or keto group (a free anomeric –OH). It reduces mild oxidants such as Tollens' (Ag+) or Fehling's (Cu2+) reagent. All monosaccharides (glucose, fructose) and most disaccharides (maltose, lactose) are reducing; sucrose is non-reducing because both anomeric carbons are engaged in the glycosidic bond.
- Differences. (i) Fibrous vs globular proteins
| Feature | Fibrous | Globular |
|---|---|---|
| Shape | long, thread-like | spherical, folded |
| Solubility | insoluble in water | soluble in water |
| Function | structural (keratin, collagen, silk) | functional (enzymes, hormones, haemoglobin) |
(ii) Nucleotide vs nucleoside
| Feature | Nucleoside | Nucleotide |
|---|---|---|
| Composition | base + pentose sugar | base + pentose sugar + phosphate |
Showing the 12 most recent of 27 on this concept.
- TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQQ.Which of the following represents the correct pyranose structure of β-D-(+)-glucose? (A) Haworth pyranose ring (CH2OH up at C-5): C-1 has H above and OH below; C-2 has OH above and H below; C-3 has OH above and H below; C-4 has H above and OH below (B) Haworth pyranose ring (CH2OH up at C-5): C-1 has H above and OH below; C-2 has H above and OH below; C-3 has OH above and H below; C-4 has H above and OH below (C) Haworth pyranose ring (CH2OH up at C-5): C-1 has OH above and H below; C-2 has H above and OH below; C-3 has OH above and H below; C-4 has H above and OH below (D) Haworth pyranose ring (CH2OH up at C-5): C-1 has OH above and H below; C-2 has H above and OH below; C-3 has OH above and H below; C-4 has OH above and H below
›Reveal solutionSolution
For β-D-glucopyranose the Haworth ring must read OH up at C-1, down at C-2, up at C-3, down at C-4, with CH2OH up at C-5. Only option (C) shows that pattern.
The concept first
There are just three rules for turning the open-chain Fischer projection into a Haworth pyranose ring.
Rule 1 — the "right–down" rule. Any −OH written on the right in the Fischer projection is drawn below the ring in Haworth; anything on the left is drawn above.
Rule 2 — the D-configuration. In a D-sugar the terminal CH2OH (C-6, attached at C-5) points up.
Rule 3 — α versus β. The new stereocentre created on ring closure is C-1, the anomeric carbon.
- β: the C-1 OH is on the same side as the CH2OH — i.e. up.
- α: the C-1 OH is down (opposite to CH2OH).
Step-by-step
Step 1 — recall the Fischer projection of D-(+)-glucose. Reading down from C-1 (the CHO):
C-2: OH right,C-3: OH left,C-4: OH right,C-5: OH right (D)
The memory aid many students use is "right–left–right–right", the lone left-hand OH being at C-3.
Step 2 — apply the right–down rule to C-2, C-3, C-4.
- C-2: OH on the right ⇒ OH below the ring, H above.
- C-3: OH on the left ⇒ OH above the ring, H below.
- C-4: OH on the right ⇒ OH below the ring, H above.
Step 3 — place the CH2OH. D-sugar ⇒ CH2OH at C-5 points up. (Both the question's diagrams and every standard Haworth drawing agree on this.)
Step 4 — set the anomeric carbon. We want the β anomer, so the C-1 OH must be cis to the CH2OH — i.e. OH above, H below.
Step 5 — assemble the required pattern.
C-1: OH up,C-2: OH down,C-3: OH up,C-4: OH down,C-5: CH2OH up …
- TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQQ.The number of amino acids present in insulin is (A) 21 (B) 30 (C) 51 (D) 41
›Reveal solutionSolution
Insulin is composed of two polypeptide chains (A and B) with 21 and 30 amino acids respectively, giving a total of 51 amino acids. The correct answer is (C).
Concept and Intuition
Insulin is a peptide hormone that regulates blood glucose. Its structure is famously simple: it consists of two separate chains — an A chain and a B chain — linked by disulfide bridges. The number of amino acids in each chain is a well-established fact in biochemistry. The trick is to remember that the total is the sum of both chains, not just one of them. Many students mistakenly recall only the A‑chain number (21) or the B‑chain number (30) and pick those as the answer, but the question asks for the total number of amino acids present in insulin — meaning the whole molecule.
Step‑by‑Step Reasoning
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Identify the two chains of insulin.
Insulin is synthesized as a single polypeptide (proinsulin) that is later cleaved into two chains: the A chain and the B chain. The C‑peptide (connecting piece) is removed during maturation and is not part of the final active insulin molecule.
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Recall the length of each chain.
- The A chain contains 21 amino acids.
- The B chain contains 30 amino acids. These numbers are standard for human insulin and most mammalian insulins.
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Add the two chain lengths.
Total amino acids = A‑chain length + B‑chain length
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- TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQQ.Match the following List-1 List-2 A. Antioxidant I. Antiseptic B. Food preservative II. Artificial sweetener C. Sucralose III. Butylated hydroxy toluene D. Bithionol IV. Sodium benzoate The correct answer is (A) A – III, B – I, C – IV, D – II (B) A – IV, B – III, C – II, D – I (C) A – III, B – IV, C – I, D – II (D) A – III, B – IV, C – II, D – I
›Reveal solutionSolution
Match each compound with its function: antioxidants prevent oxidative rancidity, preservatives inhibit microbial growth in food, artificial sweeteners provide sweetness without calories, and antiseptics kill microbes on living tissue. The correct pairing is A–III, B–IV, C–II, D–I.
This question tests your knowledge of the classification and applications of important organic compounds in everyday life, particularly in the pharmaceutical and food industries. Each substance has a specific role determined by its chemical structure and properties.
Let's identify each compound and its function:
A. Antioxidant → III. Butylated hydroxy toluene (BHT)
Antioxidants prevent oxidative degradation of fats and oils in food by inhibiting free-radical chain reactions. BHT is a phenolic compound widely used in the food industry to prevent rancidity. The hydroxyl group on the aromatic ring donates hydrogen atoms to free radicals, breaking the oxidation chain and preserving freshness.
B. Food preservative → IV. Sodium benzoate
Food preservatives extend shelf life by inhibiting microbial growth (bacteria, fungi, yeasts). Sodium benzoate (CX6HX5COONa) is one of the most common preservatives, especially effective in acidic foods like pickles, jams, and carbonated beverages. In acidic conditions it converts to benzoic acid, which penetrates microbial cell walls and disrupts their metabolism.
C. Sucralose → II. Artificial sweetener …
- TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQQ.Non membrane bound cell organelles present in animal cells I. Ribosomes II. Vacuole III. Centriole IV. Lysosome (A) I and II only (B) II and III only (C) III and IV only (D) I and III only
›Reveal solutionSolution
Ribosomes and centrioles are the only non-membrane-bound organelles among the given options in animal cells. The correct option is (D).
The distinction between membrane-bound and non-membrane-bound organelles is fundamental to understanding cell structure and function. Cell organelles are specialized subunits within a cell that perform specific functions. Their presence or absence of a surrounding membrane dictates how they interact with the cytoplasm and other organelles, and often reflects their evolutionary origin and functional complexity.
Membrane-bound organelles, such as mitochondria, endoplasmic reticulum, Golgi apparatus, lysosomes, and peroxisomes, are enclosed by one or more lipid bilayers. This compartmentalization allows them to maintain a distinct internal environment, crucial for specific biochemical reactions and protection from the rest of the cytoplasm. Non-membrane-bound organelles, on the other hand, are structures made of proteins or nucleic acids and proteins that exist directly in the cytoplasm without a surrounding membrane.
Let's analyze each organelle listed:
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Ribosomes (I):
Ribosomes are complex molecular machines responsible for protein synthesis (translation). They are composed of ribosomal RNA (rRNA) and ribosomal proteins. Structurally, a ribosome consists of two subunits (a large and a small subunit) that come together during protein synthesis. Crucially, ribosomes do not possess any membrane surrounding them. They are found freely in the cytoplasm or attached to the endoplasmic reticulum. Therefore, ribosomes are non-membrane-bound organelles.
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Vacuole (II):
Vacuoles are membrane-bound sacs that play various roles, including storage, waste removal, and maintaining turgor pressure. In animal cells, vacuoles are typically small, temporary, and involved in functions like storage or transport. They are always enclosed by a single membrane. Therefore, vacuoles are membrane-bound organelles.
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Centriole (III): …
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- TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQQ.Study the following and choose the correct statements I. Sarcoplasmic reticulum is the store house of calcium ions II. Thin filament of a myofibril is formed by actin, troponin and myosin molecules III. Thick filaments of myofibril are held together by M-line IV. The portion of myofibril between two successive M-lines is called sarcomere (A) I, III (B) II, IV (C) I, II (D) III, IV
›Reveal solutionSolution
The sarcoplasmic reticulum stores calcium ions, and the M-line holds thick filaments together. Therefore, statements I and III are correct. The final answer is (A).
The ability of muscles to contract relies on the precise arrangement and interaction of specialized protein filaments within muscle cells. Understanding the structure of these filaments and the organelles involved is key to comprehending muscle function. We will examine each statement to determine its accuracy regarding the components and organization of a myofibril.
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Evaluating Statement I: Sarcoplasmic reticulum is the store house of calcium ions.
The sarcoplasmic reticulum (SR) is a specialized type of endoplasmic reticulum found in muscle cells. Its primary function is to store and regulate the concentration of calcium ions (Ca2+) within the muscle fiber. When a muscle receives a nerve impulse, the SR releases Ca2+ into the sarcoplasm, initiating muscle contraction. Conversely, it actively pumps Ca2+ back in to allow muscle relaxation.
ImportantCalcium ions are essential for muscle contraction, binding to troponin and initiating the cross-bridge cycle between actin and myosin.
Therefore, statement I is correct.
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Evaluating Statement II: Thin filament of a myofibril is formed by actin, troponin and myosin molecules.
Thin filaments are one of the two main types of myofilaments in a myofibril. They are primarily composed of three proteins:
- Actin: Forms the backbone of the thin filament, consisting of globular G-actin monomers that polymerize into filamentous F-actin.
- Tropomyosin: A filamentous protein that wraps around the actin filament, covering the myosin-binding sites in a relaxed muscle.
- Troponin: A complex of three globular proteins that binds to actin, tropomyosin, and calcium ions. When calcium binds to troponin, it causes a conformational change that moves tropomyosin, exposing the myosin-binding sites on actin. Myosin is the primary component of the thick filaments, not the thin filaments. Therefore, statement II is incorrect.
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Evaluating Statement III: Thick filaments of myofibril are held together by M-line. …
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- TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQQ.Essential amino acids which are basic in nature are (A) Arg, Lys (B) His, Ser (C) Arg, Thr (D) Pro, Trp
›Reveal solutionSolution
Basic amino acids have side chains that can accept protons (positive charge at neutral pH). Among the essential amino acids, arginine (Arg) and lysine (Lys) are basic, making option (A) correct.
The key to this question is understanding two classifications at once: which amino acids are essential (must come from diet) and which are basic (have a net positive charge at physiological pH). You need the overlap.
Amino acids are basic when their side chain contains an extra amino group or a guanidino group — something that can grab a proton. The three standard basic amino acids are arginine (Arg), lysine (Lys), and histidine (His). But not all of these are essential for adults.
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Identify the basic amino acids.
At pH ~7.4, the side chains of Arg (guanidino group, pKa ~12.5), Lys (ε-amino group, pKa ~10.5), and His (imidazole group, pKa ~6.0) can carry a positive charge. So the set of basic amino acids is {Arg, Lys, His}.
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Identify the essential amino acids.
The nine essential amino acids for humans are: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. (Histidine is essential for infants and often considered conditionally essential for adults, but standard lists include it.)
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Find the intersection.
From the basic set {Arg, Lys, His}, which are essential?
- Lysine is essential.
- Histidine is essential.
- Arginine is not essential for adults (the body can synthesize it), though it is essential for children. In most exam contexts, arginine is classified as non-essential for adults.
So the essential basic amino acids are histidine and lysine. But look at the options — none say "His, Lys". Option (A) says "Arg, Lys". Why would that be correct?
Watch outMany standard Indian exam lists (including NCERT) treat arginine as essential for children and sometimes include it in the essential set for general purposes. Also, histidine is sometimes omitted from the essential list in older or simplified versions. This mismatch is a known source of confusion.
- Check each option against the standard NCERT/NEET classification. …
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- TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQQ.Which of the following hormones is an example of polypeptide? (A) Epinephrine (B) Insulin (C) Estrogen (D) Androgen
›Reveal solutionSolution
Polypeptide hormones are made of amino acid chains; insulin is a classic example, while epinephrine is an amino acid derivative and estrogens/androgens are steroids. The correct option is (B).
Concept & Intuition
Hormones are classified by their chemical structure, which determines how they are synthesized, stored, and act on target cells. Polypeptide hormones are chains of amino acids (typically >10 residues), synthesized on ribosomes and stored in vesicles. In contrast, steroid hormones (like estrogen and androgen) are derived from cholesterol, and epinephrine is a modified single amino acid (tyrosine). Insulin, a well-known hormone regulating blood glucose, is a polypeptide composed of 51 amino acids in two chains.
Step-by-step reasoning
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Identify the chemical nature of each option
- (A) Epinephrine: Derived from tyrosine; it is a catecholamine (amino acid derivative), not a polypeptide.
- (B) Insulin: A protein hormone made of two polypeptide chains (A and B) linked by disulfide bonds; it is a classic polypeptide hormone.
- (C) Estrogen: A steroid hormone synthesized from cholesterol; it is lipid-soluble and not a polypeptide.
- (D) Androgen: Also a steroid hormone (e.g., testosterone); same reasoning as estrogen.
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Apply the definition of a polypeptide hormone …
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- TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQQ.Amino acid ‘X’ contains phenolic hydroxy group and amino acid ‘Y’ contains amide group. ‘X’ and ‘Y’ respectively are (A) Ser, Arg (B) Cys, Lys (C) Thr, Asn (D) Tyr, Gln
›Reveal solutionSolution
The key is to match the functional groups: a phenolic hydroxy group (benzene ring with –OH) identifies tyrosine (Tyr), and an amide group (–CONH₂) identifies glutamine (Gln). The correct pair is Tyr and Gln, option (D).
The question asks you to identify two amino acids based on specific side-chain functional groups. This is a classic test of memorizing the 20 standard amino acids by their chemical properties. Let’s break it down.
Concept & Intuition
Amino acids are distinguished by their R-groups. A phenolic hydroxy group means a hydroxyl (–OH) attached directly to a benzene ring — that’s the side chain of tyrosine (Tyr). An amide group is –CONH₂, which appears in the side chains of asparagine (Asn) and glutamine (Gln). The question pairs them, so we need the correct combination.
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Identify the amino acid with a phenolic hydroxy group.
- Phenol is C₆H₅–OH. Only one standard amino acid has this: tyrosine (Tyr, Y). Its side chain is a para-hydroxyphenyl group.
- Serine (Ser) has a simple –OH (alcohol), not phenolic. Threonine (Thr) also has an alcohol –OH. Cysteine (Cys) has a thiol (–SH). So options (A), (B), and (C) are wrong for X.
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Identify the amino acid with an amide group.
- An amide group is –C(=O)NH₂. Two amino acids have this: asparagine (Asn, N) and glutamine (Gln, Q).
- Arginine (Arg) has a guanidino group, lysine (Lys) has an amino group, and asparagine (Asn) is an amide — but the question pairs Y with X. Since X is Tyr, we check the options:
- (A) Ser, Arg → no amide.
- (B) Cys, Lys → no amide.
- (C) Thr, Asn → Asn has an amide, but Thr is not Tyr. …
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- TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQQ.Amino acid ‘X’ contains phenolic hydroxy group and amino acid ‘Y’ contains amide group. ‘X’ and ‘Y’ respectively are (A) Tyr, Gln (B) Thr, Asn (C) Ser, Arg (D) Cys, Lys
›Reveal solutionSolution
The key is to match each functional group to the correct amino acid: a phenolic hydroxy group is found only in tyrosine (Tyr), and an amide group is found in glutamine (Gln) and asparagine (Asn). The only option pairing Tyr with an amide-containing amino acid is (A) Tyr, Gln.
The question tests your knowledge of amino acid side-chain functional groups — a core topic in biochemistry for medical and biology entrance exams. You need to recall which amino acids carry which chemical groups, not just their names.
Let’s break it down.
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Phenolic hydroxy group means a hydroxyl (−OH) attached directly to a benzene ring. Among the 20 standard amino acids, only tyrosine (Tyr, Y) has a phenolic side chain (a para-hydroxyphenyl group).
- Serine (Ser) has a simple alcoholic −OH, not phenolic.
- Threonine (Thr) also has an alcoholic −OH.
- Cysteine (Cys) has a thiol (−SH), not a hydroxy group. So ‘X’ must be tyrosine.
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Amide group means a −CONH2 functional group. Two amino acids have amide side chains: asparagine (Asn, N) and glutamine (Gln, Q).
- Arginine (Arg) has a guanidino group, not an amide.
- Lysine (Lys) has a primary amino group. So ‘Y’ must be either asparagine or glutamine.
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Now check the options:
- (A) Tyr, Gln — matches: Tyr has phenolic −OH, Gln has amide. …
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- TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQQ.Which of the following statements are correct about viroids? (A) They have free DNA with protein coat (B) They have free DNA without protein coat (C) They have free RNA with protein coat (D) They have free RNA without protein coat
›Reveal solutionSolution
Viroids are the smallest known infectious pathogens, consisting solely of a short, circular single-stranded RNA molecule with no protein coat — the correct statement is (D).
The key to this question lies in understanding what makes viroids distinct from viruses. Both are acellular infectious agents, but their structure is fundamentally different. A virus always has a protein coat (capsid) that surrounds its genetic material, which can be either DNA or RNA. A viroid, discovered by T.O. Diener in 1971, is even simpler — it is just a naked, circular RNA molecule with no protein covering whatsoever.
This stripped-down structure was a shocking discovery at the time. It proved that an infectious agent could cause disease using only RNA, without any proteins of its own. The RNA of a viroid is not translated into any protein; instead, it replicates using the host plant's own enzymes, typically RNA polymerase II, which is tricked into copying the viroid RNA.
Let's examine each option carefully.
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Option (A): "They have free DNA with protein coat"
This describes a typical virus, not a viroid. Viroids never contain DNA, and they never have a protein coat. This is incorrect.
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Option (B): "They have free DNA without protein coat"
While the "without protein coat" part is correct for viroids, the genetic material is RNA, not DNA. This is incorrect.
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Option (C): "They have free RNA with protein coat"
The genetic material is RNA, which is correct, but viroids lack a protein coat. This describes a virus with an RNA genome (like the tobacco mosaic virus), not a viroid. This is incorrect.
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Option (D): "They have free RNA without protein coat" …
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- TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQQ.UUU, CCC, AAA, GGG are the codons codes for the following amino acids respectively (A) Phe, Val, Gly, Lys (B) Phe, Pro, Lys, Gly (C) Phe, Gly, Pro, Asp (D) Gly, Asp, Lys, Pro
›Reveal solutionSolution
This question tests your knowledge of the genetic code, specifically which amino acids are coded by the codons UUU, CCC, AAA, and GGG. The correct mapping is Phenylalanine, Proline, Lysine, and Glycine, respectively.
The genetic code is a set of rules by which information encoded in genetic material (DNA or RNA sequences) is translated into proteins (amino acid sequences) by living cells. Each sequence of three nucleotides, called a codon, specifies a particular amino acid or a stop signal. Understanding this code is fundamental to molecular biology.
Here's how we determine the amino acids for the given codons:
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Understanding Codons: A codon is a triplet of nucleotides. In RNA, the nucleotides are Adenine (A), Uracil (U), Guanine (G), and Cytosine (C). Each unique combination of three nucleotides codes for a specific amino acid.
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Decoding UUU: The codon UUU consists of three Uracil bases. According to the standard genetic code, UUU codes for Phenylalanine (Phe).
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Decoding CCC: The codon CCC consists of three Cytosine bases. According to the standard genetic code, CCC codes for Proline (Pro).
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Decoding AAA: The codon AAA consists of three Adenine bases. According to the standard genetic code, AAA codes for Lysine (Lys).
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Decoding GGG: The codon GGG consists of three Guanine bases. According to the standard genetic code, GGG codes for Glycine (Gly).
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Matching with Options: …
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- TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQQ.Study the following and pick up the correct combinations(A) I, II (B) II, III (C) III, IV (D) II, IV
S.No Gland Hormone Disorder I Pancreas Insulin Diabetes insipidus II Thyroid gland Thyroxine Tetanus III Adrenal glands Cortisol Cushing's syndrome IV Pituitary gland Somatotropin Dwarfism ›Reveal solutionSolution
Match each endocrine gland with its hormone and associated disorder by recalling the correct physiological relationships. The correct pairings are adrenal–cortisol–Cushing's syndrome and pituitary–somatotropin–dwarfism: (C).
The endocrine system coordinates long-term physiological processes through hormones secreted by specific glands. Each gland produces characteristic hormones, and dysfunction leads to predictable disorders. This question tests whether you can correctly link gland → hormone → disorder triads.
Let's examine each statement:
I. Pancreas – Insulin – Diabetes insipidus
The pancreas does secrete insulin from its β-cells, and insulin deficiency or resistance causes diabetes mellitus (characterized by hyperglycemia). However, diabetes insipidus is an entirely different condition caused by deficiency of antidiuretic hormone (ADH/vasopressin) from the posterior pituitary, leading to excessive dilute urine production. The disorder is mismatched.
Watch outDiabetes mellitus and diabetes insipidus are unrelated diseases that share only the symptom of polyuria. Mellitus involves blood glucose; insipidus involves water balance.
II. Thyroid gland – Thyroxine – Tetanus
The thyroid gland secretes thyroxine (T₄) and triiodothyronine (T₃), which regulate metabolism. Thyroid disorders include hypothyroidism (goiter, cretinism, myxedema) and hyperthyroidism (Graves' disease). Tetanus, however, is an infectious disease caused by Clostridium tetani toxin affecting the nervous system, producing muscle spasms. It has nothing to do with thyroid dysfunction.
The condition sometimes confused here is tetany (not tetanus), which involves muscle spasms due to hypocalcemia, often from parathyroid hormone deficiency. Even that correction wouldn't save this pairing, since the gland would be wrong.
III. Adrenal glands – Cortisol – Cushing's syndrome …
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