Every living organism is built of cells. This module examines the microscopic structure of prokaryotic and eukaryotic cells, the endomembrane system, the energy-converting organelles, and the nucleus. It is a high-yield NEET unit (6–8 questions/year) — NTA frequently tests the fluid mosaic model, endomembrane organelles, and 70S vs 80S ribosomes.
Estimated Reading Time: 30 minutes | Difficulty: Moderate | Prerequisites: Basic cell terminology.
Prokaryotes (bacteria, cyanobacteria) lack a nuclear envelope and membrane-bound organelles; their genetic material sits in a nucleoid and they carry 70S ribosomes. Eukaryotes compartmentalise labour into membrane-bound organelles and carry 80S ribosomes (though their mitochondria and plastids retain 70S ribosomes — a clue to their endosymbiotic origin).
The plasma membrane is described by the Fluid Mosaic Model (Singer & Nicolson): a phospholipid bilayer (polar heads out, non-polar tails in) with proteins that float like a "mosaic"; integral proteins are embedded/span the bilayer, peripheral proteins sit on the surface. The quasi-fluid nature allows lateral movement and is key to endocytosis, secretion and cell division.
| Transport | Energy? | How |
|---|---|---|
| Simple diffusion | Passive | Down the gradient, directly through the bilayer |
| Facilitated diffusion | Passive | Down the gradient via carrier/channel proteins |
| Osmosis | Passive | Water moves across a semipermeable membrane toward higher solute |
| Active transport | ATP | Against the gradient via pump proteins (e.g. Na⁺/K⁺ pump) |
| Feature | Prokaryote | Eukaryote |
|---|---|---|
| Nuclear envelope | Absent (nucleoid) | Present |
| Membrane-bound organelles | Absent | Present |
| Ribosome | 70S (50S + 30S) | 80S (60S + 40S) |
| Cell wall | Peptidoglycan (bacteria) | Cellulose (plants) / absent (animals) |
| Single membrane | Double membrane | No membrane |
|---|---|---|
| ER, Golgi, lysosome, vacuole | Mitochondria, chloroplast, nucleus | Ribosome, nucleolus, centriole |
Trace a secreted protein: Rough ER (synthesis) \(\rightarrow\) transport vesicle \(\rightarrow\) Golgi cis-to-trans (glycosylation) \(\rightarrow\) secretory vesicle \(\rightarrow\) plasma membrane (exocytosis). This is the endomembrane system working as one production line.
1. Which organelle is bound by a single membrane?
Correct: C. Lysosome. Lysosomes, ER and Golgi are single-membrane bound; mitochondria, chloroplasts and the nucleus are double-membrane bound.
2. The subunits of a prokaryotic ribosome are:
Correct: A. A 70S prokaryotic ribosome = 50S (large) + 30S (small). Subunits do not add arithmetically because S is a sedimentation coefficient.
3. The primary site for synthesis of glycoproteins and glycolipids is:
Correct: C. The Golgi carries out post-translational modification (glycosylation) to form glycoproteins and glycolipids.
Monomers are monosaccharides (glucose, ribose). A glycosidic bond forms by dehydration between two hydroxyl groups:
$$ \text{R–OH} + \text{HO–R'} \rightarrow \text{R–O–R'} + \text{H}_2\text{O} $$
Polysaccharides: starch (helical, plant energy store), glycogen (branched, animal energy store), cellulose (unbranched \(\beta\)-1,4 glucose, structural). Chitin (a polymer of N-acetylglucosamine) forms fungal cell walls.
Monomers are amino acids, each with an amino group \(\text{–NH}_2\), a carboxyl group \(\text{–COOH}\), an H, and a variable R group on the \(\alpha\)-carbon. At physiological pH they exist as dipolar zwitterions \(\text{H}_3\text{N}^+\text{–CH(R)–COO}^-\). A peptide bond links the carboxyl of one to the amino of the next:
$$ \text{R–COOH} + \text{H}_2\text{N–R'} \rightarrow \text{R–CO–NH–R'} + \text{H}_2\text{O} $$
Structural levels: primary (sequence) → secondary (\(\alpha\)-helix / \(\beta\)-sheet via H-bonds) → tertiary (3-D fold, disulfide \(\text{–S–S–}\) bridges) → quaternary (multiple subunits, e.g. haemoglobin = \(2\alpha + 2\beta\)).
| Biomolecule | Monomer | Linkage |
|---|---|---|
| Polysaccharide | Monosaccharide | Glycosidic bond |
| Protein | Amino acid | Peptide bond |
| Nucleic acid | Nucleotide | Phosphodiester bond |
| Triglyceride | Glycerol + fatty acid | Ester bond |
A double-stranded DNA has 2000 base pairs, with 30% of bases adenine. Find its length and total hydrogen bonds.
Answer: length \(= 680\ \text{nm}\); hydrogen bonds \(= 4800\).
At neutral pH an amino acid is a dipolar zwitterion \(\text{H}_3\text{N}^+\text{–CH(R)–COO}^-\) (net charge 0). Lowering pH floods the solution with \(\text{H}^+\); the carboxylate accepts a proton to give \(\text{H}_3\text{N}^+\text{–CH(R)–COOH}\), a cation (net \(+1\)).
1. The bond between the phosphate group and the sugar hydroxyl in a nucleic acid is a:
Correct: C. Nucleotides are joined by 3′–5′ phosphodiester bonds linking one sugar to the next phosphate.
2. An amino acid zwitterion has:
Correct: C. It carries \(\text{–NH}_3^+\) and \(\text{–COO}^-\) simultaneously, so the net charge is zero.
3. The nitrogenous base found only in RNA and not in DNA is:
Correct: B. Uracil (a pyrimidine) replaces thymine in RNA.
Enzymes are biological catalysts (almost all proteins, except ribozymes) that lower activation energy. Reaction rate depends on substrate concentration \([S]\):
$$ V = \frac{V_{\text{max}}\,[S]}{K_m + [S]} $$
| Type | \(K_m\) | \(V_{\text{max}}\) | Overcome by more \([S]\)? |
|---|---|---|---|
| Competitive | Increases | Unchanged | Yes |
| Non-competitive | Unchanged | Decreases | No |
| Uncompetitive | Decreases | Decreases | No |
$$ \frac{1}{V} = \left(\frac{K_m}{V_{\text{max}}}\right)\frac{1}{[S]} + \frac{1}{V_{\text{max}}} $$
A straight line of the form \(y = mx + c\): \(y\)-intercept \(= 1/V_{\text{max}}\), \(x\)-intercept \(= -1/K_m\), slope \(= K_m/V_{\text{max}}\).
1. A lower \(K_m\) value for an enzyme indicates:
Correct: B. A low \(K_m\) means little substrate is needed to reach half-maximal velocity — high affinity.
2. Non-competitive inhibition of an enzyme leads to:
Correct: B. A non-competitive inhibitor binds an allosteric site, lowering \(V_{\text{max}}\) without changing substrate affinity (\(K_m\)).
3. An enzyme without its non-protein cofactor is called a:
Correct: B. The complete active enzyme is the holoenzyme; its protein-only part is the apoenzyme, which needs a cofactor to function.
The cycle is regulated by cyclins and cyclin-dependent kinases (CDKs) at three checkpoints:
DNA replication happens in the S phase (not G2). Mitosis is equational (\(2n \rightarrow 2n\)); meiosis I is reductional (\(2n \rightarrow n\)).
| Stage | Key cytological event |
|---|---|
| Leptotene | Chromosome condensation |
| Zygotene | Synapsis (synaptonemal complex) |
| Pachytene | Crossing over / recombination |
| Diplotene | Chiasmata become visible |
| Diakinesis | Terminalisation of chiasmata |
| Phase | Chromosomes | DNA |
|---|---|---|
| G1 | 12 | 2C |
| S / G2 | 12 | 4C |
| Anaphase (mitosis) | 24 | 4C |
Cancer is a disease of uncontrolled cell-cycle progression, driven by mutations in checkpoint proteins (p53, Rb) that normally gate the G1→S transition.
A cell with \(2n = 16\) undergoes meiosis. Number of bivalents at Metaphase I \(= n = 8\). Each bivalent has 2 homologues × 2 chromatids \(= 4\) chromatids, so total \(= 8 \times 4 = 32\) chromatids.
1. Synaptonemal complex formation occurs during which stage of Prophase I?
Correct: B. Synapsis (pairing) and the synaptonemal complex form during zygotene; crossing over follows in pachytene.
2. During the S phase of the cell cycle:
Correct: B. S phase doubles the DNA content per cell (each chromosome becomes two sister chromatids); the chromosome number is unchanged.
3. In which stage of mitosis do sister chromatids separate toward opposite poles?
Correct: C. At mitotic anaphase, centromeres split and sister chromatids move to opposite poles.