Showing the 12 most recent of 70 on this concept.

1. KCET 2025Set A-11 markMCQ
   Q.The number of diagonals that can be drawn in an octagon is (A) 15 (B) 20 (C) 28 (D) 30

   ›Reveal solutionSolution

   Count all segments joining pairs of vertices, then subtract the $n$ sides — that leaves exactly the diagonals.

   Step 1 — Why a combination.

   A diagonal is determined by which two vertices it joins, and order does not matter. So the total number of segments joining two of the $n$ vertices is the number of $2$-element subsets:

   $\binom{n}{2} = \frac{n(n-1)}{2}.$

   Step 2 — Remove the sides.

   Of those segments, exactly $n$ are the sides of the polygon (adjacent-vertex pairs). Everything else is a diagonal:

   $D = \binom{n}{2} - n = \frac{n(n-1)}{2} - n = \frac{n(n-3)}{2}.$

   Step 3 — Substitute $n = 8$ (octagon).

   $\binom{8}{2} = \frac{8 \times 7}{2} = 28, \qquad D = 28 - 8 = 20.$ …

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2. KCET 2025Set C-41 markMCQ
   Q.RNA polymerase II is responsible for the transcription of ___ (A) rRNA (B) hnRNA (C) snRNA (D) tRNA

   ›Reveal solutionSolution

   Recall the division of labour among the three eukaryotic nuclear RNA polymerases; polymerase II is the mRNA-precursor (hnRNA) enzyme.

   Step 1 — Why there are three polymerases.

   Unlike prokaryotes (a single RNA polymerase transcribing all RNA), the eukaryotic nucleus uses three distinct RNA polymerases, each dedicated to a class of RNA:

   Enzyme	Transcribes
   RNA polymerase I	rRNAs — 28S, 18S and 5.8S
   RNA polymerase II	hnRNA (heterogeneous nuclear RNA), the precursor of mRNA
   RNA polymerase III	tRNA, 5S rRNA and snRNAs (small nuclear RNAs)

   Step 2 — Match the options.

   • (A) rRNA — mainly RNA polymerase I (and the 5S by III). ✗
   • (B) hnRNA — RNA polymerase II. ✓
   • (C) snRNA — RNA polymerase III. ✗
   • (D) tRNA — RNA polymerase III. ✗

   Step 3 — The biology behind it. …

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3. KCET 2025Set C-41 markMCQ
   Q.When a change in the gene frequency of a population occurs by chance, it is called ___ (A) Gene migration (B) Genetic recombination (C) Genetic drift (D) Founder effect

   ›Reveal solutionSolution

   The word doing the work in the stem is "by chance" — that is the definition of genetic drift.

   Step 1 — What the stem is asking.

   Hardy–Weinberg equilibrium says allele frequencies stay constant unless disturbed. The five disturbing agents are: gene migration (gene flow), genetic drift, mutation, genetic recombination and natural selection. The stem specifies that the change happens by chance — a random, non-adaptive, non-directional event.

   Step 2 — Test each option.

   • (A) Gene migration — allele frequencies change because individuals (and their genes) physically move between populations. This is a real cause, but it is not a chance sampling event; it is directional gene flow. ✗
   • (B) Genetic recombination — reshuffles existing alleles into new combinations during meiosis; it changes genotype combinations, not the population's allele frequencies. ✗
   • (C) Genetic drift — a change in gene frequency that occurs purely by chance, i.e. by random sampling of which individuals happen to reproduce. Its effect is strongest in small populations, where it can even fix or eliminate an allele. ✓ …

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4. KCET 2025Set C-41 markMCQ
   Q.Read the following statements and select the correct option Statement I: Biocontrol refers to the use of biological methods for controlling plant diseases and pests. Statement II: Trichoderma species are effective biocontrol agents for several plant pathogens (A) Both statement I and statement II are incorrect (B) Statement I is incorrect but statement II is correct (C) Both statement I and statement II are correct (D) Statement I is correct and statement II is incorrect

   ›Reveal solutionSolution

   Evaluate the two statements independently; both are textbook-correct, so the option asserting both is the answer.

   Step 1 — Statement I: "Biocontrol refers to the use of biological methods for controlling plant diseases and pests."

   This is the standard definition. Biocontrol replaces chemical insecticides and pesticides with living organisms / biological agents, treating pests and pathogens as parts of a balanced ecosystem rather than something to be eradicated with chemicals. TRUE. ✓

   Step 2 — Statement II: "Trichoderma species are effective biocontrol agents for several plant pathogens."

   Trichoderma are free-living fungi, very common in the root ecosystems (rhizosphere), and they are indeed effective biocontrol agents of several plant pathogens — a standard example alongside Baculovirus (genus Nucleopolyhedrovirus) for insect pests and Bacillus thuringiensis for caterpillars, and the ladybird/dragonfly examples for aphids and mosquitoes. TRUE. ✓ …

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5. KCET 2025Set C-41 markMCQ
   Q.In mature insulin, which of the peptide is not present? (A) B-peptide (B) C-peptide (C) A and B peptides (D) A-peptide

   ›Reveal solutionSolution

   Mature insulin = A chain + B chain held by disulphide bridges; the connecting C-peptide of pro-insulin is cut out during maturation, so it is the one peptide absent.

   Step 1 — The pro-hormone concept.

   Many hormones are synthesised as inactive precursors (pro-hormones) and are activated by removing a piece. Insulin is the classic example. In the human pancreas it is first made as pro-insulin, a single polypeptide with three stretches:

   $\text{Pro-insulin} = \underbrace{\text{B chain}}_{30\ \text{aa}} - \underbrace{\text{C peptide}}_{\text{connecting}} - \underbrace{\text{A chain}}_{21\ \text{aa}}$

   Step 2 — What maturation does.

   The C-peptide's job is purely structural: it holds A and B in the right geometry so that the inter-chain disulphide (—S—S—) bridges can form correctly. Once those bridges are in place the C-peptide is redundant, and it is enzymatically cleaved out. What is secreted — mature insulin — is therefore only the A chain and B chain, still linked by disulphide bonds.

   Step 3 — Answer the question.

   The peptide not present in mature insulin is the C-peptide. …

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6. KCET 2025Set C-41 markMCQ
   Q.A student observed the slide of mitosis under the microscope and observed that the chromosomes were placed at the opposite poles. Which stage was the student observing? (A) Anaphase (B) Metaphase (C) Telophase (D) Prophase

   ›Reveal solutionSolution

   Chromosomes actively moving to / positioned at opposite poles (mid-separation) is the hallmark of anaphase, not telophase (which is the settled, post-separation stage with reforming nuclei).

   Why not telophase. In telophase, the chromosomes have already reached the poles and begin decondensing back into chromatin as new nuclear envelopes form — a student describing chromosomes simply "placed at the opposite poles" (without mentioning decondensation or new nuclear membranes) is describing the earlier, defining moment of separation. …

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7. KCET 2025Set C-41 markMCQ
   Q.In nephron, transport of substances: like sodium chloride and urea is facilitated by the special arrangement called counter current mechanism that comprises of (A) Henle's loop and glomerulus (B) Vasa Recta and collecting duct (C) Ascending limb and collecting duct (D) Henle’s loop and Vasa Recta

   ›Reveal solutionSolution

   The counter current system in the kidney is a pair of hairpin loops with flow in opposite directions — the Henle's loop and the vasa recta — which together build and preserve the medullary osmotic gradient.

   Step 1 — What “counter current” means

   “Counter current” simply means two adjacent tubes with fluid flowing in opposite directions, so that exchange between them is maximised. In the kidney medulla there are exactly two such hairpin structures, lying close and parallel:

   • Henle's loop — the tubular hairpin: filtrate runs down the descending limb and up the ascending limb.
   • Vasa recta — the vascular hairpin: blood runs down and back up alongside the loop.

   Because the flow in the descending limb is countercurrent to that in the ascending limb (and likewise in the two limbs of the vasa recta), the system is called a counter current mechanism.

   Step 2 — How it moves NaCl and urea

   • The ascending limb of Henle's loop is impermeable to water but actively transports out $\mathrm{Na^{+}}$ and $\mathrm{Cl^{-}}$, so the interstitium becomes progressively concentrated toward the inner medulla.
   • The descending limb is permeable to water, so water leaves and the filtrate becomes concentrated — a self-reinforcing (multiplier) effect.
   • The collecting duct allows a small amount of urea to diffuse into the medullary interstitium, and this urea is returned to the interstitium by the vasa recta — the classic “urea recycling”.
   • Vitally, the vasa recta carries away the water removed from the filtrate without washing out the solute gradient, because its own countercurrent flow lets solute picked up on the way down be given back on the way up. …

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8. KCET 2025Set C-41 markMCQ
   Q.Identify the correct sequence of action potential as it arrives at the axon terminal from the choices given below: (A) Axon terminal $\rightarrow$ Synaptic cleft $\rightarrow$ Synaptic vesicles $\rightarrow$ Post-synaptic neuron $\rightarrow$ Post-synaptic membrane (B) Axon terminal $\rightarrow$ Post-synaptic membrane $\rightarrow$ Synaptic cleft $\rightarrow$ Synaptic vesicles $\rightarrow$ Post-synaptic neuron (C) Axon terminal $\rightarrow$ Synaptic vesicles $\rightarrow$ Post-synaptic membrane $\rightarrow$ Synaptic cleft $\rightarrow$ Post-synaptic neuron (D) Axon terminal $\rightarrow$ Synaptic vesicles $\rightarrow$ Synaptic cleft $\rightarrow$ Post-synaptic membrane $\rightarrow$ Post-synaptic neuron

   ›Reveal solutionSolution

   Follow the anatomy of chemical synaptic transmission in order: terminal → vesicles → cleft → post-synaptic membrane → post-synaptic neuron.

   Step 1 — Arrival at the axon terminal. The action potential travelling down the axon reaches the axon terminal (pre-synaptic knob) and depolarises it, opening voltage-gated $\mathrm{Ca^{2+}}$ channels.

   Step 2 — Synaptic vesicles. The $\mathrm{Ca^{2+}}$ influx makes the synaptic vesicles (which are filled with neurotransmitter) move towards and fuse with the pre-synaptic plasma membrane.

   Step 3 — Synaptic cleft. Fusion releases the neurotransmitter by exocytosis into the synaptic cleft — the fluid-filled space between the pre- and post-synaptic neurons.

   Step 4 — Post-synaptic membrane. The transmitter diffuses across and binds specific receptors on the post-synaptic membrane, opening ion channels. …

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9. KCET 2024Set B-41 markMCQ
   Q.Match the parts of the brain given in List I with their functions given in List II. List I (Parts of the brain)

   1. Medulla oblongata
   2. Hypothalamus
   3. Cerebral cortex
   4. Limbic system

   List II (Functions) p. Body temperature q. Olfaction r. Respiration s. Motor function Choose the correct option from the following : (A) 1-p, 2-r, 3-s, 4-q (B) 1-q, 2-s, 3-r, 4-p (C) 1-s, 2-p, 3-q, 4-r (D) 1-r, 2-p, 3-s, 4-q

   ›Reveal solutionSolution

   Match each brain part to the centre it houses: medulla → respiration, hypothalamus → temperature, cerebral cortex → motor function, limbic system → olfaction.

   1. Medulla oblongata → r (Respiration). The hindbrain's medulla contains the vital cardiovascular and respiratory rhythm centres; it drives involuntary breathing (also vomiting, coughing). So 1–r.

   2. Hypothalamus → p (Body temperature). The hypothalamus, in the forebrain floor of the diencephalon, holds centres for thermoregulation, hunger, thirst and the endocrine link to the pituitary. So 2–p.

   3. Cerebral cortex → s (Motor function). The cortex is divided into motor, sensory and association areas; the precentral gyrus (motor area) initiates voluntary movement. So 3–s. …

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10. KCET 2024Set B-41 markMCQ
    Q.Match the content of List I with List II : List I

    1. Polyembryony
    2. Perisperm
    3. False fruit
    4. Parthenocarpy

    List II p. Black pepper q. Banana r. Lemon s. Apple Choose the correct option from the following : (A) 1-r, 2-p, 3-s, 4-q (B) 1-p, 2-r, 3-s, 4-q (C) 1-q, 2-p, 3-s, 4-r (D) 1-r, 2-s, 3-p, 4-q

    ›Reveal solutionSolution

    Recall the NCERT type-examples: polyembryony → citrus/lemon, perisperm → black pepper (and beet), false fruit → apple, parthenocarpy → banana.

    1. Polyembryony → r (Lemon). The occurrence of more than one embryo in a single seed — typically from nucellar cells (apomixis) — is seen in many Citrus species and in mango. Lemon is the citrus example given. So 1–r.

    2. Perisperm → p (Black pepper). The perisperm is the persistent, unused remnant of the nucellus in the mature seed. It is retained in black pepper and beet. So 2–p.

    3. False fruit → s (Apple). A true fruit develops only from the ovary; in a false (pseudocarp) fruit the fleshy edible part develops from the thalamus. Apple, strawberry and cashew are the standard examples. So 3–s. …

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11. KCET 2024Set B-41 markMCQ
    Q.DNA polymerase of $Thermus\ aquaticus$ is (A) Thermolabile (B) Thermophobic (C) Exonuclease (D) Thermostable

    ›Reveal solutionSolution

    Thermus aquaticus lives in hot springs, so its polymerase is heat-stable — the property that makes automated PCR possible.

    Step 1 — Read the organism's name.

    Thermus aquaticus — thermus = heat, aquaticus = of water. It is a thermophilic bacterium isolated from hot springs. An organism living at near-boiling temperatures must have proteins that do not denature there; natural selection guarantees its enzymes are heat-resistant.

    Step 2 — Why PCR needs exactly this property.

    Each PCR cycle has three steps at three temperatures:

    1. Denaturation — $\approx 94^\circ\text{C}$, to separate the two template strands.
    2. Annealing — $\approx 50\text{–}60^\circ\text{C}$, for the two primers to bind.
    3. Extension — $\approx 72^\circ\text{C}$, the optimum for Taq polymerase, which extends the primers using dNTPs.

    The cycle is repeated ~30 times. A normal E. coli DNA polymerase is thermolabile: it would be irreversibly denatured during the very first $94^\circ\text{C}$ step, and fresh enzyme would have to be added after every single cycle. Taq polymerase remains active despite the high temperature-induced denaturation of double-stranded DNA — so it is added once, and the whole reaction can be automated in a thermal cycler.

    $\text{Taq stable at }94^\circ\text{C} \;\Longrightarrow\; \text{add enzyme once} \;\Longrightarrow\; \text{automated }\sim 10^9\text{-fold amplification}$

    Step 3 — Eliminate the other options. …

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12. KCET 2024Set B-41 markMCQ
    Q.Match the pigments given in List I with their colour in chromatogram given in List II. List I (Pigments)

    1. Chlorophyll 'b'
    2. Carotenoids ,
    3. Chlorophyll 'a'
    4. Xanthophylls.

    List II (Colour in chromatogram) p. Yellow orange q. Orange red r. Yellow s. Blue green t. Yellow green Choose the correct option from the following : (A) 1-s, 2-t, 3-r, 4-q (B) 1-t, 2-p, 3-s, 4- r (C) l-p, 2-q, 3-r, 4-t (D) 1-t, 2-p, 3-r, 4-s

    ›Reveal solutionSolution

    Use the four chromatogram bands of leaf pigments: chl b = yellow green, carotenoids = yellow-orange, chl a = blue green, xanthophylls = yellow.

    Concept — Separation of photosynthetic pigments by paper chromatography (Photosynthesis in Higher Plants). When a leaf extract is run on a chromatogram, four pigments separate out and each shows a characteristic colour:

    • Chlorophyll a — bright or blue green
    • Chlorophyll b — yellow green
    • Xanthophylls — yellow
    • Carotenoids — yellow to yellow-orange

    Now read the list:

    1. Chlorophyll 'b' → yellow green → t
    2. Carotenoids → yellow orange → p
    3. Chlorophyll 'a' → blue green → s
    4. Xanthophylls → yellow → r …

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