Q.Identify the correct statement from the following:

Showing the 12 most recent of 36 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Match the following: List-1 List-2 A. Ligaments I. Dense irregular connective tissue B. Valves of heart II. Transitional epithelium C. Epiglottis III. Dense regular connective tissue D. Wall of urinary bladder IV. Vascular tissue V. Gristle The correct answer is (A) A – III, B – IV, C – I, D – V (B) A – I, B – II, C – III, D – IV (C) A – IV, B – I, C – V, D – III (D) A – III, B – I, C – V, D – II

   ›Reveal solutionSolution

   Match each structure to its tissue type by understanding the functional demands: ligaments need parallel collagen for tensile strength (dense regular), heart valves need multidirectional strength (dense irregular), epiglottis needs flexible support (cartilage/gristle), and the bladder wall needs stretch accommodation (transitional epithelium). The correct answer is (D).

   The key to matching anatomical structures with tissue types lies in understanding what each structure does and what tissue properties support that function.

   Concept: Structure follows function in tissue organization

   Dense regular connective tissue has collagen fibers arranged in parallel bundles, providing maximum tensile strength in one direction—perfect for structures that experience unidirectional pulling forces. Dense irregular connective tissue has collagen fibers arranged in multiple directions, resisting tension from various angles. Transitional epithelium is specialized stratified tissue that can stretch and recoil. Cartilage (gristle) provides firm yet flexible support.

   Let's analyze each structure:

   1. Ligaments (A) connect bone to bone across joints and must resist strong pulling forces along their length. The collagen fibers run parallel to the direction of stress, making this dense regular connective tissue (III). The organized, unidirectional arrangement maximizes tensile strength where it's needed.

   2. Valves of the heart (B) are thin flaps that must withstand blood pressure from multiple directions as they open and close. They need strength in all directions within their plane, which requires dense irregular connective tissue (I). The randomly oriented collagen network prevents tearing regardless of pressure direction.

   3. Epiglottis (C) is the flap of tissue that covers the trachea during swallowing. It must be firm enough to hold its shape yet flexible enough to move quickly. This is elastic cartilage, commonly called gristle (V). The cartilage matrix provides the perfect balance of rigidity and flexibility.

   4. Wall of the urinary bladder (D) must accommodate dramatic volume changes as urine accumulates and is expelled. The inner lining is transitional epithelium (II), a specialized tissue whose cells can slide over one another and change shape, allowing the bladder to expand from empty to full without tearing.

   Watch out

   Don't confuse dense regular with dense irregular. "Regular" means the fibers are regularly arranged (parallel), not that it's the "regular" or common type. The arrangement directly reflects the mechanical demands on the tissue.

   Tip

   Remember: Ligaments and tendons both use dense regular tissue because both transmit force in one direction. Valves, dermis, and organ capsules use dense irregular because they face multidirectional stress.

   Matching these up: A–III, B–I, C–V, D–II.

   ✓Final answer

   The correct option is (D) A – III, B – I, C – V, D – II.

2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Consider the following statements Statement I: Dryopithecus is a stage in human evolution and was more ape like Statement II: In human evolution Homo erectus was followed by Homo habilis The correct answer is (A) Both statement I and statement II are true (B) Both statement I and statement 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

   Dryopithecus was an ape-like ancestor in the lineage leading to humans and great apes, making Statement I true. However, Homo habilis appeared before Homo erectus in human evolution, making Statement II false. The correct option is (C).

   The question asks us to evaluate two statements regarding human evolution. To do this, we need to recall the general timeline and characteristics of various hominid species. Understanding the sequence and key features of these stages is crucial for correctly assessing each statement.

   1. Analyze Statement I: Dryopithecus is a stage in human evolution and was more ape like

      • Dryopithecus is an extinct genus of hominoids that lived during the Miocene epoch, approximately 9 to 12 million years ago.
      • Fossils of Dryopithecus have been found in Europe and Asia. They are considered to be among the earliest ancestors of both modern humans and great apes (chimpanzees, gorillas, orangutans).
      • Their skeletal features, particularly their limb structure and dental characteristics, indicate that they were primarily arboreal (tree-dwelling) and exhibited many ape-like traits. For example, they had relatively long arms and short legs, and their molars had thin enamel, similar to modern apes.
      • Therefore, Dryopithecus is indeed considered a significant stage in the broader evolutionary lineage that eventually led to humans, and it was distinctly more ape-like in its morphology and lifestyle.
      • Conclusion for Statement I: Statement I is true.

   2. Analyze Statement II: In human evolution Homo erectus was followed by Homo habilis

      • Let's recall the general sequence of the genus Homo in human evolution:
        • Homo habilis ("handy man") lived approximately 2.4 to 1.4 million years ago. They are known for being the first to make and use stone tools.
        • Homo erectus ("upright man") lived approximately 1.9 million to 110,000 years ago. They are known for migrating out of Africa, using more sophisticated tools, and possibly controlling fire.
        • Homo sapiens (modern humans) evolved from earlier Homo species.
      • From this timeline, it is clear that Homo habilis appeared before Homo erectus. In fact, Homo habilis is generally considered an ancestor of Homo erectus.
      • The statement claims that Homo erectus was followed by Homo habilis, which means Homo habilis came after Homo erectus. This is incorrect.
      • Conclusion for Statement II: Statement II is false.

   Important

   The general evolutionary sequence for the genus Homo is: Homo habilis $\rightarrow$ Homo erectus $\rightarrow$ Homo sapiens.

   3. Combine the conclusions
      • Statement I is true.
      • Statement II is false.
      • This matches option (C).

   ✓Final answer

   Statement I is true, but Statement II is false, so the correct option is (C).

3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Consider the following statements Statement I: Barbiturates cause sleeplessness Statement II: Benzodiazepines are sleeping pills The correct answer is Options : (A) Both statement I and statement II are true (B) Both statement I and statement 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

   Barbiturates cause sleep, not sleeplessness (I false); benzodiazepines are sleeping pills (II true).

   Barbiturates depress the central nervous system and are classic sedative-hypnotics — they promote sleep, so 'cause sleeplessness' is incorrect. Benzodiazepines (e.g. diazepam, nitrazepam) are widely prescribed as anti-anxiety and sleep-inducing drugs, so Statement II is correct.

   ✓Final answer

   Statement I is false, Statement II is true. The correct option is (D).

   ANSWER: D

4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Accumulation of iron particles in tissues lead to (A) Asbestosis (B) Silicosis (C) Siderosis (D) Black lung disease

   ›Reveal solutionSolution

   The accumulation of iron particles in tissues is medically termed siderosis. The correct option is (C).

   The human body has mechanisms to clear foreign particles, especially in the respiratory system. However, prolonged or intense exposure to certain types of dust or particles can overwhelm these mechanisms, leading to the deposition and accumulation of these particles in tissues. This accumulation can trigger inflammatory responses, fibrosis, and impaired organ function, leading to specific diseases depending on the nature of the inhaled particle.

   Let's examine the given options:

   1. Siderosis: This term specifically refers to the accumulation of iron in tissues. When iron particles, such as iron dust or fumes, are inhaled over time, they can deposit in the lungs, leading to a condition called pulmonary siderosis. This is commonly observed in occupations involving iron exposure, like welding or iron mining. The iron particles themselves are relatively inert, but their presence can cause changes in lung tissue.

   2. Asbestosis: This is a chronic lung disease caused by inhaling asbestos fibers. Asbestos is a group of naturally occurring fibrous minerals. When these fibers are inhaled, they can become lodged in the lungs, leading to inflammation and scarring (fibrosis) of the lung tissue. This results in symptoms like shortness of breath and a persistent cough, and it increases the risk of lung cancer and mesothelioma.

   3. Silicosis: This is a lung disease caused by inhaling crystalline silica dust. Silica is a common mineral found in sand, rock, and quartz. Occupations such as mining, quarrying, construction, and sandblasting carry a high risk of silica exposure. The inhaled silica particles cause inflammation and nodular fibrosis in the lungs, leading to progressive shortness of breath and other respiratory problems.

   4. Black lung disease (Coal Workers' Pneumoconiosis - CWP): This is a lung disease caused by inhaling coal dust. It is prevalent among coal miners. The coal dust accumulates in the lungs, leading to inflammation and the formation of scar tissue. There are two forms: simple CWP, which usually causes few symptoms, and complicated CWP (progressive massive fibrosis), which can lead to severe breathing difficulties and lung damage.

   Based on these definitions, the accumulation of iron particles in tissues directly corresponds to siderosis.

   ✓Final answer

   The accumulation of iron particles in tissues leads to (C) Siderosis.

5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Study the following and choose the correct statement I. Sickle cell anemia is due to replacement of glutamic acid in $6^{\text{th}}$ position by valine II. Thalassemia is sex linked disorder due to recessive genes III. Down syndrome is an allosomal disorder IV. Phenylketonuria is due to mutation of the gene-PAH in chromosome 12 (A) I, IV (B) II, III (C) I, III (D) III, IV

   ›Reveal solutionSolution

   The question tests your knowledge of specific genetic disorders and their causes. Only statements I and IV are correct, so the answer is option (A).

   Let’s go through each statement one by one, understanding the biology behind it.

   1. Statement I: Sickle cell anemia is due to replacement of glutamic acid in the 6th position by valine.

      This is a classic point from molecular genetics. Sickle cell anemia results from a point mutation in the beta-globin gene on chromosome 11. The mutation changes the codon GAG to GUG, which substitutes valine (a hydrophobic amino acid) for glutamic acid (a hydrophilic one) at the 6th position of the beta chain. This alters hemoglobin’s shape, causing sickling of red blood cells. This statement is correct.

   2. Statement II: Thalassemia is a sex-linked disorder due to recessive genes.

      Thalassemia is actually an autosomal recessive disorder, not sex-linked. It involves reduced or absent synthesis of globin chains (alpha or beta) due to mutations in genes on chromosomes 16 (alpha) or 11 (beta). It is not linked to the X or Y chromosome. So this statement is incorrect.

   3. Statement III: Down syndrome is an allosomal disorder.

      “Allosomal” refers to sex chromosomes (X and Y). Down syndrome is caused by trisomy of chromosome 21, which is an autosome (non-sex chromosome). It is an autosomal aneuploidy, not an allosomal one. So this statement is incorrect.

   4. Statement IV: Phenylketonuria is due to mutation of the gene PAH on chromosome 12.

      Phenylketonuria (PKU) is an autosomal recessive disorder caused by mutations in the PAH gene, which codes for the enzyme phenylalanine hydroxylase. This gene is located on chromosome 12. The deficiency leads to accumulation of phenylalanine, causing intellectual disability if untreated. This statement is correct.

   Watch out

   A common mistake is confusing “allosomal” (sex chromosome) with “autosomal” (non-sex chromosome). Down syndrome is autosomal, not allosomal.

   Thus, the correct statements are I and IV.

   ✓Final answer

   The correct option is (A) I, IV.

6. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Identify the correct statements related to the class Phaeophyceae A. Asexual reproduction occurs usually by biflagellate zoospores B. Sexual reproduction is by oogamous method only C. Food is stored as simple carbohydrates in the form of mannitol or laminarin D. Major pigments are chlorophyll a, c, xanthophylls and carotenoids E. Cellulose cell wall has outer gelatinous coating of algin (A) A, B, C and D only (B) B, C, D and E only (C) A, D and E only (D) A, B, C and E only

   ›Reveal solutionSolution

   Phaeophyceae (brown algae) have biflagellate zoospores, store food as mannitol/laminarin, contain chlorophyll a, c, xanthophylls, and carotenoids, and have a cellulose wall with algin coating — but sexual reproduction is not only oogamous; it can also be isogamous or anisogamous. The correct set is A, C, D, E, which matches option (C).

   The question tests your knowledge of the brown algae class Phaeophyceae — a group you’ve likely studied under plant kingdom diversity. The trick is to recall that while many brown algae show oogamy (e.g., Fucus), the class as a whole includes all three types of sexual reproduction: isogamous, anisogamous, and oogamous. Statement B says “only oogamous” — that’s the trap.

   Let’s go statement by statement.

   1. Statement A: Asexual reproduction occurs usually by biflagellate zoospores

      This is correct. In Phaeophyceae, asexual reproduction commonly happens through zoospores that are pear-shaped and have two flagella — one pointing forward (tinsel type) and one trailing (whiplash type). These are produced in sporangia.

   2. Statement B: Sexual reproduction is by oogamous method only

      This is false. While oogamy is present (e.g., in Fucus and Sargassum), many brown algae also reproduce by isogamy (e.g., Ectocarpus) or anisogamy (e.g., Dictyota). So “only” makes this statement incorrect.

   3. Statement C: Food is stored as simple carbohydrates in the form of mannitol or laminarin

      Correct. Brown algae do not store starch like green plants. Their reserve food is mannitol (a sugar alcohol) and laminarin (a β-glucan polymer). These are indeed simple carbohydrates.

   4. Statement D: Major pigments are chlorophyll a, c, xanthophylls and carotenoids

      Correct. The brown colour comes from the dominance of fucoxanthin (a xanthophyll), which masks chlorophyll a and c. Carotenoids like β-carotene are also present. Note: chlorophyll b is absent — that’s a key distinction from green algae.

   5. Statement E: Cellulose cell wall has outer gelatinous coating of algin

      Correct. The cell wall is made of cellulose and algin (alginic acid), a gelatinous polysaccharide that gives the thallus its slippery, flexible texture. This is a characteristic feature of brown algae.

   Watch out

   A common mistake is to think all brown algae are oogamous because Fucus is a well-known example. But the class includes simpler forms too — always check for “only” or “always” in such statements.

   Now, tallying the correct ones: A, C, D, and E are true. B is false. That set corresponds to option (C).

   ✓Final answer

   The correct option is (C) A, D and E only.

7. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Consider the following statements Statement I: Cannabinoids are being abused by sports persons (doping) Statement II: Cocaine is involved in the transport of neurotransmitter dopamine The correct answer is (A) Both statement I and statement II are true (B) Both statement I and statement 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

   We need to verify two independent statements about drugs: cannabinoids in sports doping and cocaine's interaction with dopamine transport. Both statements are factually correct, making (A) the right answer.

   Let me evaluate each statement based on established pharmacological and sports medicine facts.

   Statement I: Cannabinoids are being abused by sports persons (doping)

   Evaluating the claim:

   Cannabinoids, particularly THC (tetrahydrocannabinol) from cannabis, are indeed on the World Anti-Doping Agency's (WADA) prohibited substances list. While the performance-enhancing effects are debated, cannabinoids are monitored and restricted in competitive sports for several reasons:

   • They may reduce anxiety and pain perception
   • They can affect reaction time and coordination
   • WADA prohibits them in-competition
   • Numerous athletes have faced sanctions for cannabinoid use

   Conclusion for Statement I: This statement is TRUE. Cannabinoids are classified as substances of abuse in sports, and their use constitutes doping violations.

   Statement II: Cocaine is involved in the transport of neurotransmitter dopamine

   Evaluating the claim:

   This requires understanding cocaine's mechanism of action at the neurochemical level:

   1. Normal dopamine transport: After dopamine is released into the synaptic cleft, it's normally reabsorbed by the presynaptic neuron through dopamine transporter (DAT) proteins—a process called reuptake.

   2. Cocaine's mechanism: Cocaine binds to and blocks dopamine transporters, preventing the reuptake of dopamine from the synapse.

   3. Result: This blockade causes dopamine to accumulate in the synaptic cleft, prolonging and intensifying its effects—this is the primary mechanism behind cocaine's euphoric and addictive properties.

   Cocaine's Action: Cocaine acts as a dopamine reuptake inhibitor by binding to DAT (dopamine transporter) proteins, thereby interfering with normal dopamine transport mechanisms.

   Conclusion for Statement II: This statement is TRUE. Cocaine is directly involved in (specifically, it interferes with) the transport of dopamine by blocking its reuptake.

   Final Evaluation

   • Statement I: TRUE ✓
   • Statement II: TRUE ✓

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Study the lists I, II, III & IV given below and identify the mismatch.

   	List-1 (Type of inflorescence)	List-2 (Family)	List-3 (Example)	List-4 (Character)
   I	Verticellaster	Lamiaceae	Leucas	Begin as a monochasial cyme
   II	Cyathium	Euphorbiaceae	Euphorbia	Bisexual flowers arranged in cymose fashion
   III	Hypanthodium	Moraceae	Ficus	Male, female and gall flowers are present
   (A) I & III
   (B) II & III
   (C) I & II
   (D) III only

   

   ›Reveal solutionSolution

   Row I is wrong because a verticillaster begins as a dichasial cyme (and only later becomes monochasial/scorpioid), and Row II is wrong because a cyathium contains unisexual flowers — many male flowers around one female — not bisexual ones. Row III (hypanthodium in Ficus, with male, female and gall flowers) is perfectly correct. The mismatches are I & II, i.e. option (C).

   The concept first

   "Special" inflorescences are the ones that don't fit the plain racemose/cymose scheme. Three of them are examination favourites, and each is defined by a precise developmental description. Learn the description, not just the family name, because that is exactly what the question tests.

   Inflorescence	Family	Example	The defining description
   Verticillaster	Lamiaceae	Leucas, Ocimum	A condensed cluster in the axil of opposite leaves that starts as a dichasial (biparous) cyme and is continued by monochasial (scorpioid, uniparous) cymes
   Cyathium	Euphorbiaceae	Euphorbia, Poinsettia	A cup-like involucre enclosing one central, stalked female flower (a naked pistil) surrounded by many male flowers, each reduced to a single stamen — all flowers unisexual
   Hypanthodium	Moraceae	Ficus	A hollow, pear/flask-shaped fleshy receptacle with a small apical pore (ostiole), lined internally by male, female and gall flowers

   Step-by-step check of each printed row

   Step 1 — Row I: Verticillaster / Lamiaceae / Leucas / "Begin as a monochasial cyme".

   The family and example are right. But the character is not: a verticillaster begins as a dichasial cyme at the centre, and afterwards the lateral branches continue in the monochasial (scorpioid) manner. The printed statement reverses the order of development, so Row I is a mismatch.

   Step 2 — Row II: Cyathium / Euphorbiaceae / Euphorbia / "Bisexual flowers arranged in cymose fashion".

   Again the family and example are right, but a cyathium is famous precisely for the extreme reduction of unisexual flowers: many male flowers (each just one stamen) surround a single female flower (just one pistil), all inside a cup-shaped involucre of bracts. The whole point of the structure is that no flower is bisexual — the inflorescence as a whole merely looks like a single bisexual flower. So Row II is a mismatch.

   Step 3 — Row III: Hypanthodium / Moraceae / Ficus / "Male, female and gall flowers are present".

   This is textbook-exact. Inside the hollow receptacle of a fig you find male flowers near the ostiole, female flowers lower down, and gall flowers in which the fig wasp lays its eggs. Row III is correctly matched.

   Step 4 — Collect the mismatches.

   Mismatched rows $=\{\text{I},\ \text{II}\}$. Scanning the printed options: (A) I & III — no; (B) II & III — no; (C) I & II — yes; (D) III only — no.

   ✓Final answer

   Rows I (verticillaster begins as a dichasial, not monochasial, cyme) and II (cyathium has unisexual, not bisexual, flowers) are the mismatches, so the correct option is (C).

   ANSWER: C

9. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Karyotype of Turner's syndrome is (A) 45, X [AA + XO] (B) 47, XXY [AA + XXY] (C) 47, XX + 13 (D) 47, XX + 18

   ›Reveal solutionSolution

   Turner's syndrome results from the loss of one X chromosome in females, giving a total of 45 chromosomes with the sex chromosome constitution XO. The correct karyotype notation is 45, X.

   Turner's syndrome is a classic example of a sex chromosome aneuploidy — a condition where the number of sex chromosomes deviates from the normal pair. In humans, the normal female karyotype is 46, XX (22 pairs of autosomes + two X chromosomes). Turner's syndrome arises when one of the X chromosomes is completely or partially missing, leaving a single X. This is written as 45, X (or sometimes 45, XO), where "O" denotes the absence of a second sex chromosome.

   The key idea is that the total chromosome count drops to 45 because one sex chromosome is absent, while all 44 autosomes (22 pairs) remain intact. The notation "45, X" is the standard way to express this: the number before the comma is the total chromosome count, and the symbols after the comma describe the sex chromosome constitution.

   Let's examine each option:

   1. Option (A): 45, X [AA + XO]

      This correctly states the total as 45, with the sex chromosome constitution as X (or XO). The "AA" stands for the 44 autosomes (22 pairs). This matches Turner's syndrome exactly.

   2. Option (B): 47, XXY [AA + XXY]

      This describes Klinefelter's syndrome — a male with an extra X chromosome. The total is 47, and the sex chromosomes are XXY. Not Turner's.

   3. Option (C): 47, XX + 13

      This indicates trisomy 13 (Patau syndrome), where there is an extra copy of chromosome 13. The sex chromosomes are normal (XX), but the total is 47 due to the autosomal trisomy. Not Turner's.

   4. Option (D): 47, XX + 18

      This is trisomy 18 (Edwards syndrome), with an extra chromosome 18. Again, not Turner's.

   Watch out

   A common mistake is to think Turner's syndrome involves a missing Y chromosome in males. Remember: Turner's syndrome occurs only in females (phenotypically female) and involves a missing X, not a missing Y. The karyotype is 45, X, not 45, Y (which is non-viable).

   Tip

   To quickly recall:

   • Turner's: 45, X (female, missing one X)
   • Klinefelter's: 47, XXY (male, extra X)
   • Down's: 47, +21 (trisomy 21)
   • Patau's: 47, +13
   • Edwards': 47, +18

   ✓Final answer

   The correct option is (A) 45, X [AA + XO].

10. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.Match the following List-1 List-2 A. Physical barrier I. Lysozyme B. Physiological barriers II. Interferons C. Cellular barriers III. Monocytes D. Cytokine barriers IV. Colostrum V. Mucus membrane The correct answer is (A) A-V, B-I, C-IV, D-II (B) A-V, B-I, C-III, D-II (C) A-III, B-II, C-I, D-V (D) A-I, B-V, C-II, D-III

    ›Reveal solutionSolution

    This question tests your understanding of the different types of innate immune barriers in the human body. The correct match is A-V, B-I, C-III, D-II, which corresponds to option (B).

    The immune system has multiple layers of defense. The first line of defense consists of physical and physiological barriers that prevent pathogen entry. The second line includes cellular and cytokine barriers that act if a pathogen does get in. This question asks you to correctly classify specific examples into these four categories.

    Let’s go through each item in List-1 and match it with the correct example from List-2.

    1. Physical barrier (A) — These are structures that physically block pathogens from entering the body. The mucus membrane (V) lines the respiratory, digestive, and urogenital tracts. Mucus traps microbes, and cilia sweep them out. This is a classic physical barrier. So A matches V.

    2. Physiological barriers (B) — These are conditions or secretions that create an inhospitable environment for pathogens. Lysozyme (I) is an enzyme found in tears, saliva, and mucus that breaks down bacterial cell walls. It is a physiological barrier because it acts chemically, not physically. So B matches I.

    3. Cellular barriers (C) — These are immune cells that directly attack pathogens. Monocytes (III) are a type of white blood cell that can differentiate into macrophages and engulf microbes. They are a cellular component of innate immunity. So C matches III.

    4. Cytokine barriers (D) — These are signaling proteins released by infected cells to alert and activate other immune cells. Interferons (II) are cytokines produced by virus-infected cells; they warn neighboring cells to prepare antiviral defenses. So D matches II.

    Watch out

    A common mistake is to confuse lysozyme (a physiological barrier) with a physical barrier, or to think colostrum (IV) belongs here. Colostrum provides passive immunity via antibodies, not innate barriers — it is not listed in the correct matches for this question.

    Tip

    Remember the mnemonic: Physical = Prevent entry (mucus), Physiological = Produce hostile chemistry (lysozyme), Cellular = Cells that eat (monocytes), Cytokine = Chemical messengers (interferons).

    ✓Final answer

    The correct option is (B) A-V, B-I, C-III, D-II.

11. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.Statement I: Jean Baptiste de Lamarck stated that acquired characters are inherited to the next generations Statement II: August Weisman strongly supported the view of inheritance of acquired characters (A) Both statements I and II are correct (B) Both statements I and II are false (C) Statement I is true, but II is false (D) Statement I is false. But II is true

    ›Reveal solutionSolution

    Lamarck proposed the inheritance of acquired characters; Weismann disproved it with his germ-plasm theory. Statement I is correct, Statement II is false.

    The core of this question lies in the history of evolutionary thought — specifically, the clash between two major ideas about how traits are passed on. Lamarck’s theory, often summarized as “use and disuse,” claimed that changes an organism acquires during its lifetime (like a giraffe stretching its neck) could be inherited by its offspring. Weismann, on the other hand, is famous for refuting this idea through his germ-plasm theory, which showed that only changes in the reproductive cells (germ cells) are heritable, not changes in the body cells (soma).

    Let’s break down each statement.

    1. Statement I: Lamarck and acquired characters. Jean-Baptiste de Lamarck did indeed propose that acquired characteristics — features developed through an organism’s own efforts or environmental influences during its life — could be passed to the next generation. This was a cornerstone of his theory of evolution, published in Philosophie Zoologique (1809). For example, he argued that the long necks of giraffes resulted from generations of stretching to reach high leaves. So, Statement I is historically accurate.

    2. Statement II: Weismann’s support. August Weismann did the opposite of supporting Lamarck. In the late 1800s, he conducted a famous experiment: he cut off the tails of mice for 22 generations and observed that each new generation was born with full-length tails. This directly contradicted Lamarck’s idea. Weismann then proposed the germ-plasm theory, arguing that hereditary information flows only from germ cells (eggs and sperm) to the next generation, and that changes to the body (soma) are not inherited. He was a strong opponent of the inheritance of acquired characters, not a supporter.

    Watch out

    A common mistake is to confuse Weismann’s role. He is often remembered for disproving Lamarckism, not for supporting it. Remember the tail-cutting experiment — it’s the classic evidence against acquired characters being inherited.

    Since Statement I is true and Statement II is false, the correct pairing is option (C).

    ✓Final answer

    The correct option is (C).

12. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.Identify the correct matching I Nerium = Verticillaster = Sunken stomata II Vallisnaria = Free floating hydrophyte = Epiphylophy III Tribulus = Ephemeral = Xerophyte IV Rhizophora = Halophyte = Vivipary (A) I and III (B) II and IV (C) I and II (D) III and IV

    ›Reveal solutionSolution

    Vallisneria is rooted-submerged (not free-floating) and Nerium does not have a verticillaster, so I and II fail. Tribulus (an ephemeral xerophyte) and Rhizophora (a viviparous halophyte) are both correct — option (D).

    The concept first — the ecological classification of plants

    Plants are grouped by the water regime they live in:

    • Xerophytes — dry habitats. Sub-types: ephemerals (drought escapers, e.g. Tribulus, Argemone), succulents (drought evaders, e.g. Opuntia) and true xerophytes (drought endurers, e.g. Nerium, Casuarina), which show sunken stomata, thick cuticle, multiple epidermis.
    • Hydrophytes — water. Sub-types: free-floating (Pistia, Eichhornia, Wolffia), rooted-floating (Nelumbo, Nymphaea), rooted-submerged (Vallisneria, Hydrilla), suspended submerged (Ceratophyllum).
    • Halophytes — saline/marshy soil (mangroves such as Rhizophora, Avicennia), showing pneumatophores and vivipary.

    Now test each statement.

    Step-by-step

    1. I. Nerium = Verticillaster = Sunken stomata. Nerium (oleander) is a classic true xerophyte with a thick cuticle, multilayered epidermis and stomata sunk in pits lined with hairs — that part is right. But its leaves are whorled (verticillate phyllotaxy), and a verticillaster is a cymose inflorescence characteristic of the Lamiaceae (Ocimum, Salvia). Nerium belongs to Apocynaceae and has a cymose inflorescence, not a verticillaster. → I is not a fully correct match.
    2. II. Vallisneria = Free floating hydrophyte = Epiphyllophy. Vallisneria is rooted and submerged, with long ribbon-like leaves; its flowers are the famous water-pollinated (hydrophilous) ones that surface on a long stalk. It is definitely not free-floating. → II is wrong.
    3. III. Tribulus = Ephemeral = Xerophyte. Tribulus germinates, flowers and sets seed within the short wet season, passing the dry season as seed — the textbook definition of an ephemeral (drought-escaping) xerophyte. → III is correct.
    4. IV. Rhizophora = Halophyte = Vivipary. Rhizophora is a mangrove of saline coastal mud (halophyte) that shows vivipary: the seed germinates inside the fruit while attached to the parent, forming a hypocotyl that drops and anchors in the mud — an adaptation to the anaerobic, saline substratum. → IV is correct.
    5. Correct statements = III and IV → option (D).

    ✓Final answer

    Only III (Tribulus — ephemeral xerophyte) and IV (Rhizophora — viviparous halophyte) are correctly matched.

    ANSWER: D