Q.Does moonlight support photosynthesis?

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Inulin is a polymer of (A) Glucose (B) Fructose (C) Galactose (D) Raffinose

   ›Reveal solutionSolution

   Inulin is a polysaccharide composed primarily of repeating units of fructose. The correct option is (B).

   Inulin is a type of natural polysaccharide, which means it is a large molecule made up of many smaller, identical or similar, repeating units called monomers. Understanding what these monomer units are is key to identifying the correct option.

   1. Identify Inulin's Classification: Inulin belongs to a class of carbohydrates known as fructans. Fructans are polymers where the primary monomer unit is fructose.
   2. Determine the Monomer: Specifically, inulin is a linear polymer consisting mainly of $\beta-(2 \to 1)$ linked D-fructose units. It often has a terminal glucose unit, but the bulk of the polymer chain is made of fructose. Therefore, it is considered a polymer of fructose.
   3. Consider Other Options:
      • (A) Glucose: Glucose is the monomer for many other important polysaccharides like starch, glycogen, and cellulose, but not inulin. …

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2. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Consider the following statements Statement-I: Salmon and Hilsa are catadromous fishes. Statement-II: In limnetic zone rate of photosynthesis and rate of respiration are almost equal. The correct answer is (A) Both statements I and II are true (B) Both statements I and II are false (C) Statement I is true, but statement II is false (D) Statement I is false, but statement II is true

   ›Reveal solutionSolution

   Salmon and Hilsa are anadromous, not catadromous, so Statement-I is false; the limnetic zone does reach down to the compensation level where photosynthesis $\approx$ respiration, so Statement-II is true. Option (D).

   Statement-I — the concept: two directions of breeding migration

   Many fishes cross the salt/fresh boundary to breed, and the two words for it are easy to mix up. Fix them by their Greek roots:

   • Ana- = upward. An anadromous fish spends its adult life in the sea and swims up a river into fresh water to spawn. The textbook examples are the Salmon and the Hilsa (Tenualosa ilisha), which ascends the Ganga and Godavari to breed.
   • Cata- = downward. A catadromous fish lives in fresh water and migrates down to the sea to spawn. The textbook example is the freshwater eel, Anguilla, which travels to the Sargasso Sea.

   (Fishes that never leave one medium are potamodromous — freshwater only — or oceanodromous — marine only.)

   Statement-I labels Salmon and Hilsa catadromous. That is exactly the wrong way round — both are anadromous. $\Rightarrow$ Statement-I is FALSE.

   Statement-II — the concept: the zones of a lake

   A lake is divided vertically and horizontally by light penetration, because light sets the rate of photosynthesis while respiration continues everywhere, all the time:

   1. Littoral zone — the shallow, well-lit water at the margin, where light reaches the bottom; rooted plants grow here and photosynthesis greatly exceeds respiration. …

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3. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.In a food chain, primary carnivores occupy (A) First trophic level (B) Second trophic level (C) Third trophic level (D) Fourth trophic level

   ›Reveal solutionSolution

   Primary carnivores eat herbivores, so they sit at the third trophic level — the first level of consumers that feed on plant-eaters.

   The key here is understanding how trophic levels are defined in a food chain. A trophic level is a step in the flow of energy from producers upward. The chain always starts with producers (plants) at the first level, then moves to consumers in order of what they eat.

   Let’s walk through it:

   1. First trophic level — Producers (autotrophs like green plants). They make their own food using sunlight. No one eats anyone else here.

   2. Second trophic level — Primary consumers (herbivores). These are animals that eat plants directly. Think of a rabbit eating grass, or a deer eating leaves.

   3. Third trophic level — Secondary consumers (primary carnivores). These are animals that eat herbivores. A cat eating a mouse, a snake eating a frog, or a lion eating a zebra — all are primary carnivores. They are the first meat-eaters in the chain. …

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4. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Based on the figure given below which of the following statements are not correct [FIGURE: A rectangular vessel split by a vertical semi-permeable membrane into chamber A (left) and chamber B (right). Chamber A holds water with only a few solute molecules; chamber B holds many more solute molecules. Labels on the figure: "Chamber", "A", "B", "Solute molecule", "Water", "Semi permeable membrane".] (A) Movement of solvent molecules occur from chamber A to B (B) Movement of solute molecules occur from chamber A to B (C) Semipermeable membrane is prerequisite for the process (D) Rate of osmosis depends on pressure gradient and concentration gradient of solutes

   ›Reveal solutionSolution

   Osmosis involves the movement of solvent (water) from a region of higher water concentration to a region of lower water concentration across a semi-permeable membrane, while solute molecules are restricted. The incorrect statement is that solute molecules move from chamber A to B.

   Osmosis is a specific type of diffusion involving the movement of solvent molecules, typically water, across a selectively permeable membrane. This process is driven by differences in solute concentration, which in turn create a difference in water potential. Understanding the roles of the solvent, solute, and the membrane is key to evaluating the given statements.

   Here's a breakdown of the concept and an analysis of each statement:

   Concept of Osmosis

   1. Semi-permeable Membrane: This is a crucial component. A semi-permeable membrane allows solvent molecules (like water) to pass through freely but restricts the passage of larger solute molecules.
   2. Concentration Gradient: Osmosis occurs when there is a difference in the concentration of solutes across the semi-permeable membrane.
      • A region with a lower solute concentration has a higher water concentration (or higher water potential).
      • A region with a higher solute concentration has a lower water concentration (or lower water potential).
   3. Movement of Solvent: Solvent molecules move from the region of higher water concentration (lower solute concentration) to the region of lower water concentration (higher solute concentration) through the semi-permeable membrane. This movement continues until equilibrium is reached or until an opposing pressure prevents further net movement.
   4. Movement of Solute: Due to the nature of the semi-permeable membrane, solute molecules are generally unable to pass through it. Therefore, there is no net movement of solute molecules across the membrane during osmosis.

   Analysis of the Figure

   The figure shows:

   • Chamber A: Contains water with a few solute molecules. This represents a region of higher water concentration (lower solute concentration).
   • Chamber B: Contains water with many more solute molecules. This represents a region of lower water concentration (higher solute concentration).
   • Semi-permeable membrane: Separates Chamber A and Chamber B.

   Evaluating the Statements

   1. Statement (A): Movement of solvent molecules occur from chamber A to B

      • Chamber A has a higher concentration of water (fewer solute molecules) compared to Chamber B (more solute molecules).
      • According to the principle of osmosis, solvent (water) moves from a region of higher water concentration to a region of lower water concentration across a semi-permeable membrane.
      • Therefore, water molecules will move from Chamber A to Chamber B.
      • This statement is correct.

   2. Statement (B): Movement of solute molecules occur from chamber A to B

      • A semi-permeable membrane is defined by its property of allowing solvent molecules to pass but restricting solute molecules.
      • Therefore, solute molecules cannot pass through the semi-permeable membrane. …

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5. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Parazoans exhibit (A) Tissue level organization (B) Organ level organization (C) Cellular level organization (D) Organ system level organization

   ›Reveal solutionSolution

   Parazoans are the simplest multicellular animals that lack true tissues and organs — they function at the cellular level of organization, making option (C) the correct answer.

   The key to this question lies in understanding what "parazoans" actually are. The name itself gives it away: para means "beside" or "alongside," and zoa means "animals." These are animals that sit just beside the true tissue-bearing animals (Eumetazoa). The only living examples are the sponges (phylum Porifera).

   In sponges, cells are relatively independent. They can change their functions, move around, and even re-aggregate if the sponge is broken apart. There is no coordination through nerve cells or muscle cells, and no real tissues form. Instead, the body is a loose aggregation of cell types — choanocytes, pinacocytes, amoebocytes — each doing its job but without forming the organized layers (like ectoderm, endoderm) that define true tissues. This is why their organization is called cellular level: the cell is the highest functional unit, not the tissue.

   Now let's walk through the options one by one.

   1. Tissue level organization — This is the hallmark of eumetazoans like cnidarians (jellyfish, hydra). In these animals, cells are grouped into true tissues (e.g., nerve net, muscle fibers) that work together. Parazoans lack this entirely. So (A) is wrong.

   2. Organ level organization — Organs are made of multiple tissues working together for a specific function. This appears in flatworms (platyhelminthes) and above. Sponges have no organs — no heart, no gut, no excretory system. So (B) is wrong.

   3. Cellular level of organization — This is exactly what parazoans show. Each cell type performs its function independently. For example, choanocytes create water currents and trap food, but they don't form a coordinated digestive tissue. Amoebocytes carry nutrients, but they move freely. There is no germ layer formation, no true tissue differentiation. This is the correct description. …

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6. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.World Malaria day celebrated on (A) 25th March (B) 25th April (C) 20th August (D) 20th September

   ›Reveal solutionSolution

   World Malaria Day is observed on 25th April each year, as declared by the World Health Organization (WHO) to raise awareness about malaria prevention and control.

   The question is a straightforward factual recall from general knowledge, often appearing in competitive exams and current affairs sections. The key is to remember that several global health awareness days are clustered in April — World Health Day (7th April), World Malaria Day (25th April), and World Immunization Week (last week of April). The date was chosen by the WHO in 2007 to mark the global effort against malaria, replacing the earlier Africa Malaria Day.

   1. Eliminate unrelated months.

      March 25th is not associated with any major WHO health day. August and September have their own awareness days (e.g., World Mosquito Day on 20th August, but that is not Malaria Day). So options (A), (C), and (D) are incorrect.

   2. Confirm the correct date. …

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7. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
   Q.Four cells A, B, C and D are placed in different water potentials solutions given below. Choose the pair of cells which exhibit same type of osmic flow Cell Solution A: $\Psi = -0.05$ $\Psi = -0.2$ B: $\Psi = -0.2$ $\Psi = -0.001$ C: $\Psi = -0.001$ $\Psi = -0.3$ D: $\Psi = -0.6$ $\Psi = -0.05$ (A) A and C : Exosmosis (B) B and C : Endosmosis (C) B and D : Exosmosis (D) A and D : Endosmosis

   ›Reveal solutionSolution

   Water flows down the water-potential gradient, from less negative $\Psi$ to more negative $\Psi$. Cells A and C are each placed in a solution more negative than themselves, so both lose water: both show exosmosis. The correct option is (A).

   The concept first: why the sign trips everyone up

   Water potential $\Psi$ measures the free energy of water per unit volume, relative to pure water, which is defined as $\Psi = 0$. Adding solute lowers it, so all real solutions have negative $\Psi$. That single convention causes most of the errors in this topic, so fix the rule once:

   $\text{Water moves from HIGHER } \Psi \text{ (closer to zero) } \longrightarrow \text{ LOWER } \Psi \text{ (more negative)}$

   In other words, water moves towards the more negative value — towards the more concentrated solution — because that is where its free energy is lower. Then:

   • Water goes into the cell $\Rightarrow$ endosmosis (the cell swells, may become turgid).
   • Water goes out of the cell $\Rightarrow$ exosmosis (the cell shrinks, may plasmolyse).

   Practical shortcut: if the solution's $\Psi$ is more negative than the cell's $\Psi$, the cell loses water.

   Step-by-step, cell by cell

   1. Cell A: cell $-0.05$ vs solution $-0.2$. $\Psi_{soln} (-0.2) < \Psi_{cell} (-0.05)$, so water flows cell $\rightarrow$ solution. Exosmosis.
   2. Cell B: cell $-0.2$ vs solution $-0.001$. $\Psi_{soln} (-0.001) > \Psi_{cell} (-0.2)$, so water flows solution $\rightarrow$ cell. Endosmosis.
   3. Cell C: cell $-0.001$ vs solution $-0.3$. $\Psi_{soln} (-0.3) < \Psi_{cell} (-0.001)$ — a big drop — so water rushes out. Exosmosis. …

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8. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
   Q.Study the following and pick up the correct statements: I. Oriented locomotor movement of an organism towards or away from light is called photokinesis. II. Influence of light on non directional movement of organism is known as phototaxis. III. The orientation to light of a non-motile organism (such as plant or plant part) is known as phototropism. IV. The response of organisms for the photoperiod is known as photoperiodism. (A) I, II (B) III, IV (C) I, III (D) II, IV

   ›Reveal solutionSolution

   This question tests understanding of biological responses to light. Statements III and IV correctly define phototropism and photoperiodism, respectively, while statements I and II incorrectly define photokinesis and phototaxis. The correct option is (B).

   The natural world is full of organisms responding to their environment, and light is one of the most fundamental stimuli. Organisms have evolved diverse ways to react to light, from moving towards it to timing their life cycles based on day length. Understanding these specific terms — taxis, kinesis, tropism, and photoperiodism — is key to distinguishing between different types of responses.

   Let's break down each statement to determine its correctness:

   1. Analyze Statement I: "Oriented locomotor movement of an organism towards or away from light is called photokinesis."

      • Locomotor movement refers to movement from one place to another.
      • Oriented movement means the movement has a specific direction relative to the stimulus (towards or away).
      • This description, "oriented locomotor movement... towards or away from light," accurately defines phototaxis. Phototaxis is a directional movement of a motile organism in response to light.
      • Photokinesis, on the other hand, refers to a non-directional change in the rate or frequency of movement of an organism in response to light intensity. For example, an organism might move faster in brighter light but without a specific direction.
      • Since the statement describes phototaxis but labels it photokinesis, Statement I is incorrect.

   2. Analyze Statement II: "Influence of light on non directional movement of organism is known as phototaxis."

      • As established above, phototaxis is a directional movement.
      • Non-directional movement influenced by light is characteristic of photokinesis.
      • Since the statement incorrectly associates non-directional movement with phototaxis, Statement II is incorrect.

   3. Analyze Statement III: "The orientation to light of a non-motile organism (such as plant or plant part) is known as phototropism."

      • Non-motile organisms are those that cannot move from place to place, like plants.
      • Orientation to light for a plant typically involves growth or bending. …

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9. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
   Q.Hensen’s disc in a myofibril is less darker than the edges of the A-band due to (A) Presence of thick filaments only (B) Presence of thin filaments only (C) Presence of both thick and thin filaments (D) Presence of Krause’s membrane

   ›Reveal solutionSolution

   Hensen's disc (H-zone) appears lighter than the edges of the A-band because it contains only thick (myosin) filaments, while the A-band edges have overlapping thick and thin filaments that increase optical density. The answer is (A).

   The key to this question lies in understanding the molecular architecture of a sarcomere and how different regions appear under a microscope based on protein density.

   A sarcomere is the functional unit of a myofibril, bounded by two Z-lines. Within it, thick filaments (myosin) and thin filaments (actin) are arranged in a precise, overlapping pattern. When you examine a sarcomere under polarized light or electron microscopy, different regions show different optical densities—darker regions have more protein overlap, lighter regions have less.

   The A-band represents the entire length of the thick filaments. It appears dark because thick filaments are present throughout. However, the A-band is not uniformly dark:

   • At the edges of the A-band, thin filaments from adjacent sarcomeres slide in and overlap with the thick filaments. This double layer of proteins (both thick and thin) creates maximum optical density—the darkest appearance.

   • In the center of the A-band lies the H-zone (Hensen's disc). During muscle relaxation, this is the region where thick filaments exist alone, without any overlapping thin filaments. With only one type of filament present, the protein density is lower than at the edges.

   This difference in overlap explains the optical appearance: fewer proteins per unit volume means less light absorption and a lighter appearance.

   1. The A-band contains thick filaments throughout its length. This gives it an overall dark appearance compared to the I-band (which has only thin filaments).

   2. At the periphery of the A-band, thin filaments interdigitate with thick filaments, creating a zone of overlap where both filament types coexist. This double occupancy maximizes protein density. …

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10. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
    Q.The following is extracted from Laminaria (A) Algin (B) Agar (C) Iodine (D) Carrageen

    ›Reveal solutionSolution

    The question asks which substance is extracted from the brown alga Laminaria. The correct answer is Iodine, as Laminaria is a major commercial source of iodine, though it also yields algin. The correct option is (C).

    Concept & Intuition

    Laminaria is a genus of brown algae (Phaeophyceae). These seaweeds are known for producing two economically important products: algin (a polysaccharide used as a thickener) and iodine (a trace element concentrated from seawater). While agar comes from red algae (e.g., Gelidium) and carrageen comes from red algae (e.g., Chondrus), iodine is historically extracted from brown kelps like Laminaria. The key is to match the organism to its characteristic extract.

    Step-by-Step Reasoning

    1. Identify the organism – Laminaria is a brown alga (kelp), common in cold coastal waters. It is not a red or green alga.

    2. Recall the major products of brown algae – Brown algae produce algin (alginic acid) as a cell-wall component and are also known to accumulate iodine from seawater, sometimes up to 1% of dry weight. Red algae produce agar and carrageen.

    3. Evaluate each option:

       • (A) Algin – Yes, Laminaria is a source of algin, but this is not the only or most distinctive extract; many brown algae yield algin.
       • (B) Agar – Agar is derived from red algae (e.g., Gracilaria, Gelidium), not from Laminaria.
       • (C) Iodine – Historically, Laminaria was a primary source for iodine extraction (especially in 19th-century Europe). This is a classic fact in phycology. …

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11. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
    Q.The mineral elements required for carbohydrate translocation, photolysis of water and for chlorophyll synthesis are respectively (A) Boron, Manganese, Zinc (B) Boron, Chlorine, Iron (C) Molybdenum, Copper, Magnesium (D) Molybdenum, Copper, Iron

    ›Reveal solutionSolution

    Boron → translocation of carbohydrates; Chlorine (with Mn) → photolysis of water; Iron → chlorophyll formation. In that order the answer is Boron, Chlorine, Iron — option (B).

    The concept first

    Mineral nutrition questions are pure recall, but the recall becomes easy if you attach one signature role to each micronutrient:

    Element	Signature role
    Boron (B)	Uptake and utilisation of Ca$^{2+}$, translocation of carbohydrates, pollen germination, cell elongation
    Manganese (Mn)	Activates photosynthetic enzymes; splitting of water to liberate O$_2$
    Chlorine (Cl)	With Mn and Ca, essential in the water-splitting reaction of PS II; also maintains solute balance
    Iron (Fe)	Required in largest amount of all micronutrients; component of ferredoxin/cytochromes; activates catalase; essential for chlorophyll formation
    Zinc (Zn)	Activates carboxylases; needed for auxin (IAA) synthesis
    Copper (Cu)	Component of redox enzymes (plastocyanin, cytochrome oxidase)
    Molybdenum (Mo)	Component of nitrogenase and nitrate reductase — nitrogen metabolism

    Note the distinction that trips people up: magnesium is a constituent of the chlorophyll molecule (it sits in the centre of the porphyrin ring), whereas iron is required for chlorophyll's synthesis (it is needed to build the ring). The question asks about synthesis.

    Step-by-step

    Step 1 — Role 1: carbohydrate translocation. Sucrose loading and movement in the phloem require Boron. None of Molybdenum, Copper or Zinc has this role. This alone eliminates options (C) and (D), which both begin with Molybdenum. …

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12. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
    Q.In a pond, there were 20 frogs last year and through reproduction 8 new frogs were added, taking current population to 28. The birth rate is (offspring per frog per year) (A) 0.1 (B) 0.4 (C) 2.5 (D) 10

    ›Reveal solutionSolution

    Birth rate is the number of offspring produced per frog per year. With 8 new frogs from 20 original frogs, the rate is $8/20 = 0.4$ offspring per frog per year. The correct option is (B).

    The concept here is per capita birth rate — a fundamental idea in population ecology. It tells us how many new individuals are produced, on average, by each existing individual in a given time period. This is not the same as the total number of births; it normalises the births by the population size so we can compare across populations of different sizes.

    The trap many students fall into is dividing the new frogs by the current population (28) instead of the original population (20). But the birth rate is defined relative to the population that did the reproducing — the frogs that were there at the start of the year. The new frogs are the result of that reproduction, not the ones doing it.

    Let’s work through it step by step.

    1. Identify the relevant quantities.

       Last year’s population (the reproducing individuals) = 20 frogs.

       Number of new frogs added through reproduction = 8 frogs.

       Time period = 1 year.

    2. Recall the definition of per capita birth rate.

       It is the number of offspring produced per individual per unit time. Mathematically:

    $\text{Birth rate} = \frac{\text{Number of births}}{\text{Initial population size} \times \text{Time}}$

    Here, time is 1 year, so it simplifies to:

    $\text{Birth rate} = \frac{8}{20}$

    3. Perform the calculation.

    $\frac{8}{20} = 0.4$

    4. Interpret the result. …

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