Q.Copper ions-releasing IUDs are more efficient than non-medicated methods. Why?

Showing the 12 most recent of 72 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Study the following and choose the incorrect statements I. Schizocoelomates and enterocoelomates are together called acoelomates II. The regions characterized by high levels of species richness and high degree of endemism are called biodiversity hotspots III. The threatened species are listed in the Red Data Book IV. Cryopreservation is a type of in-situ conservation (A) I, IV (B) II, III (C) I, II (D) III, IV

   ›Reveal solutionSolution

   The question tests your understanding of key biology terms: coelom classification, biodiversity hotspots, Red Data Book, and conservation methods. The incorrect statements are I (schizocoelomates and enterocoelomates are coelomates, not acoelomates) and IV (cryopreservation is ex-situ, not in-situ conservation). So the correct option is (A) I, IV.

   Let’s break down each statement one by one.

   1. Statement I: "Schizocoelomates and enterocoelomates are together called acoelomates"

      This is fundamentally wrong. In animal classification based on body cavity:

      • Acoelomates have no coelom (e.g., flatworms).
      • Schizocoelomates develop a coelom by splitting of mesoderm (e.g., annelids, arthropods, molluscs).
      • Enterocoelomates develop a coelom from outpocketings of the archenteron (e.g., echinoderms, chordates). Both schizocoelomates and enterocoelomates are coelomates (true coelom animals), not acoelomates. So statement I is incorrect.

   2. Statement II: "The regions characterized by high levels of species richness and high degree of endemism are called biodiversity hotspots"

      This is correct. A biodiversity hotspot must meet two criteria: (a) at least 1,500 species of vascular plants as endemics, and (b) have lost at least 70% of its original habitat. High species richness and endemism are indeed the defining features.

   3. Statement III: "The threatened species are listed in the Red Data Book"

      This is correct. The Red Data Book (or Red List) published by IUCN categorizes species as extinct, extinct in the wild, critically endangered, endangered, vulnerable, etc. — all of which are threatened categories. …

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2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Vector for Plasmodium vivax is (A) Periplaneta (B) Musca (C) Anopheles (D) Culex

   ›Reveal solutionSolution

   The vector for Plasmodium vivax is the female Anopheles mosquito, which transmits the malaria parasite during a blood meal. The correct option is (C).

   The question asks about the biological vector — the organism that carries and transmits the pathogen from one host to another. For malaria, the pathogen is a protozoan of the genus Plasmodium, and P. vivax is one of the species that causes malaria in humans. The vector is not the pathogen itself, but the agent that spreads it.

   The key concept here is the life cycle of Plasmodium. The parasite requires two hosts: a human (where it reproduces asexually) and a female Anopheles mosquito (where it reproduces sexually). The mosquito picks up the parasite when it bites an infected human, and later injects it into a new human host during a subsequent bite. Only female mosquitoes feed on blood (needed for egg development), so only females act as vectors.

   Now, let's eliminate the other options:

   1. Periplaneta — This is the genus for cockroaches. Cockroaches are mechanical vectors for some bacteria (like Salmonella) but have no role in malaria transmission. They are not involved in the Plasmodium life cycle.

   2. Musca — This is the genus for houseflies. Houseflies can carry pathogens on their body parts (e.g., for typhoid, dysentery) but do not transmit malaria. Plasmodium cannot complete its life cycle inside a fly.

   3. Anopheles — This is the correct genus. Female Anopheles mosquitoes are the definitive host for Plasmodium (where sexual reproduction occurs) and the vector that transmits it to humans. Different Anopheles species are vectors for different Plasmodium species; Anopheles culicifacies and Anopheles stephensi are common vectors for P. vivax in India. …

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3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Consider the following statements Statement I: Use of cocaine produces a sense of euphoria and increased energy Statement II: Excessive use of cocaine cause hallucinations 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

   Cocaine is a stimulant that first produces euphoria and energy, but excessive use can lead to hallucinations — both statements are true.

   Cocaine belongs to the class of drugs called central nervous system stimulants. Unlike depressants (like alcohol) or hallucinogens (like LSD), stimulants primarily increase the activity of the brain's reward pathways and alertness centers. The key to this question is understanding the dose-dependent effects of stimulants: what happens at low/moderate use versus what happens with excessive or chronic use.

   Let's examine each statement carefully.

   1. Statement I: "Use of cocaine produces a sense of euphoria and increased energy"

      Cocaine works by blocking the reuptake of dopamine, norepinephrine, and serotonin in the brain. This leads to a rapid buildup of these neurotransmitters in the synaptic cleft. The surge of dopamine in the nucleus accumbens (the brain's pleasure center) produces intense euphoria, while norepinephrine increases heart rate, blood pressure, and alertness — giving a feeling of heightened energy. This is a well-documented, immediate effect of cocaine use. So Statement I is true.

   2. Statement II: "Excessive use of cocaine cause hallucinations" …

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4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Sexual phenotype of Drosophila with the karyotype AA+XXX is (A) Female (B) Metafemale (C) Inter sex (D) Triploid female

   ›Reveal solutionSolution

   Sex determination in Drosophila is governed by the ratio of X chromosomes to sets of autosomes (X:A ratio). For the karyotype AA+XXX, the X:A ratio is $1.5$, which results in a metafemale phenotype.

   In Drosophila melanogaster (fruit fly), sex is determined not by the mere presence or absence of a Y chromosome, but by the balance between the number of X chromosomes and the number of sets of autosomes. This is known as the X:A ratio mechanism. The Y chromosome in Drosophila is primarily involved in male fertility, not sex determination itself.

   Here's how the X:A ratio dictates the sexual phenotype:

   • X:A = 1.0: Normal female
   • X:A > 1.0: Metafemale (also called superfemale) — these individuals are often sterile and have developmental abnormalities.
   • X:A between 0.5 and 1.0: Intersex — individuals with a mixture of male and female characteristics.
   • X:A = 0.5: Normal male
   • X:A < 0.5: Metamale (also called supermale) — these individuals are often sterile and have developmental abnormalities.

   Let's apply this concept to the given karyotype.

   1. Identify the number of X chromosomes:

      The given karyotype is AA+XXX. This indicates there are three X chromosomes.

      So, number of X chromosomes ($N_X$) = $3$.

   2. Identify the number of sets of autosomes:

      The notation 'AA' signifies two complete sets of autosomes. Each 'A' represents one haploid set of autosomes.

      So, number of sets of autosomes ($N_A$) = $2$.

   3. Calculate the X:A ratio: …

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5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Match the following A Operator site I Binding site of mRNA polymerase B Promoter site II Binding site for repressor molecule C Regulator gene III Codes for protein / enzyme D Structural gene IV Codes for repressor molecule Options : (A) A – II, B – I, C – III, D – IV (B) A – II, B – I, C – IV, D – III (C) A – IV, B – III, C – I, D – II (D) A – II, B – III, C – I, D – IV

   ›Reveal solutionSolution

   This question tests your understanding of the components of an operon and their roles in gene regulation. The operator site binds the repressor, the promoter site binds RNA polymerase, the regulator gene codes for the repressor, and structural genes code for proteins/enzymes. The correct match is (B).

   The regulation of gene expression in prokaryotes, such as bacteria, often occurs through a mechanism called an operon. An operon is a unit of genetic material that functions in a coordinated manner to regulate the expression of a group of genes. It typically includes a promoter, an operator, and one or more structural genes, along with a separate regulatory gene. Understanding the specific role of each component is key to matching them correctly.

   Here's a breakdown of each component and its function:

   1. Operator site (A):

      The operator is a specific DNA sequence located adjacent to the promoter region, usually between the promoter and the structural genes. Its primary role is to act as a switch for transcription. It is the binding site for a repressor molecule. When a repressor protein binds to the operator, it physically obstructs the RNA polymerase from moving along the DNA and transcribing the structural genes, thereby switching off gene expression.

      Therefore, A (Operator site) matches II (Binding site for repressor molecule).

   2. Promoter site (B):

      The promoter is a DNA sequence located upstream of the structural genes. It serves as the recognition and binding site for RNA polymerase. RNA polymerase must bind to the promoter to initiate the transcription of the structural genes into mRNA. Without a functional promoter, transcription cannot begin.

      Therefore, B (Promoter site) matches I (Binding site of mRNA polymerase).

   3. Regulator gene (C): …

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6. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Commelina species produce two different types of bisexual flowers "A" and "B". Select the correct option regarding types of flowers and modes of pollination from the Table given below (A) I and II only (B) II and III only (C) I and III only (D) IV only

   ›Reveal solutionSolution

   The aerial flower of Commelina is chasmogamous (open $\Rightarrow$ self or cross pollination) and the underground flower is cleistogamous (never opens $\Rightarrow$ self-pollination only, and assured seed set). Only row IV states this — option (D).

   The concept: what "open" and "closed" really commit a flower to

   • Chasmogamous flowers ($chasmo$ = gaping) open normally: anthers dehisce and the stigma is exposed to the outside world. Pollen may land from the same flower (autogamy/geitonogamy) or be brought by an insect or the wind from another plant (xenogamy). So a chasmogamous flower permits both self- and cross-pollination — it does not guarantee either.
   • Cleistogamous flowers ($cleisto$ = closed) never open at all. The anthers and stigma lie in contact within the closed bud, so the pollen released must fall directly on the stigma of the same flower. Consequently:
     • pollination is compulsorily autogamous — self-pollination only;
     • no pollinator is required, and seed set is assured, which is the evolutionary pay-off (a fail-safe against years when pollinators are scarce);
     • the cost is complete loss of genetic recombination through outcrossing.

   Commelina is the standard textbook example of a plant carrying both kinds on the same individual: chasmogamous flowers in the air, cleistogamous ones borne underground.

   Step-by-step through the table

   1. Flower A is aerial. It must be chasmogamous — this eliminates row III, which calls the aerial flower cleistogamous. …

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7. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Study the following and choose the correct statements I. The anti-cancer drug vinblastin is obtained from the plant Digitalis II. The concept of biodiversity hot spots was proposed by Paul Ehrlich III. In protostomes cleavages are spiral and determinate IV. As the species share the same ecological niche, it is described on ecological unit. (A) III, IV (B) I, II (C) I, IV (D) II, III

   ›Reveal solutionSolution

   The question tests factual recall from biology — only statements III and IV are correct, so the answer is option (A).

   Let’s examine each statement carefully, one by one.

   1. Statement I: “The anti-cancer drug vinblastin is obtained from the plant Digitalis”

      This is false. Vinblastine is an anti-cancer drug derived from the Madagascar periwinkle (Catharanthus roseus), not from Digitalis (foxglove). Digitalis yields cardiac glycosides like digoxin, used for heart conditions. A classic mix-up.

   2. Statement II: “The concept of biodiversity hot spots was proposed by Paul Ehrlich”

      This is false. The biodiversity hotspot concept was developed by Norman Myers in 1988, not Paul Ehrlich (who is known for work on population and extinction). Myers identified regions with high endemism and severe habitat loss.

   3. Statement III: “In protostomes cleavages are spiral and determinate”

      This is true. Protostomes (e.g., annelids, molluscs, arthropods) typically exhibit spiral cleavage (cells divide at oblique angles) and determinate cleavage (the fate of each cell is fixed early). This contrasts with deuterostomes (e.g., chordates, echinoderms), which have radial and indeterminate cleavage.

   4. Statement IV: “As the species share the same ecological niche, it is described on ecological unit.” …

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8. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Ernst Haeckel proposed the (A) Theory of Natural selection (B) Germplasm theory (C) Theory of recapitulation (D) Mutation theory

   ›Reveal solutionSolution

   Haeckel is famous for the biogenetic law, which states that ontogeny recapitulates phylogeny — this is the Theory of recapitulation, option (C).

   The question asks you to match Ernst Haeckel, a 19th-century German biologist, with the correct theory he proposed. This is a classic fact from the history of evolutionary biology, often tested in Indian exams. The key is to know the specific contribution of each major figure — Haeckel’s name is tied to a single, distinctive idea that is not shared by Darwin, Weismann, or de Vries.

   Haeckel was a strong supporter of Darwin’s theory of evolution, but he went further and proposed his own law about how embryos develop. He observed that during development, an organism passes through stages that resemble the adult forms of its evolutionary ancestors. For example, a human embryo at one point has pharyngeal pouches (like a fish) and a tail. Haeckel called this the biogenetic law, or the theory of recapitulation: "ontogeny recapitulates phylogeny." Ontogeny is the development of an individual; phylogeny is the evolutionary history of the species. So, the developing embryo replays the evolutionary past.

   Let’s check each option to see why only one fits.

   1. Theory of Natural selection — This is Charles Darwin’s core idea, published in On the Origin of Species (1859). Haeckel popularized Darwin’s work in Germany but did not propose natural selection himself. So (A) is wrong.

   2. Germplasm theory — This was proposed by August Weismann in the late 1800s. It states that hereditary information is carried only in the germ cells (sperm and egg), and changes to the body (soma) are not inherited. Haeckel had a different view (he believed in inheritance of acquired characteristics, like Lamarck). So (B) is wrong. …

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9. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Choose the correct statement from the following (A) Cleistogamous flowers always exhibit autogamy (B) Chasmogamous flowers always exhibit geitonogamy (C) Cleistogamous flowers exhibits both autogamy and geitonogamy (D) Chasmogamous flowers never exhibit autogamy

   ›Reveal solutionSolution

   The key idea is that cleistogamous flowers never open, forcing self-pollination (autogamy), while chasmogamous flowers open and can use any mode. The correct statement is (A).

   The question tests your understanding of two flower types defined by their opening behaviour and how that dictates the mode of pollination. Let’s get the definitions straight first.

   Cleistogamous flowers are flowers that never open. Their anthers and stigma remain enclosed within the bud, so pollen from the same flower can only land on its own stigma. This makes autogamy (self-pollination within the same flower) the only possible mode — no external agent (wind, insect) can enter, and pollen cannot travel to another flower.

   Chasmogamous flowers are the typical flowers that open at maturity, exposing their reproductive parts. Once open, they can undergo autogamy (if pollen lands on the same flower’s stigma), geitonogamy (pollen from another flower on the same plant), or xenogamy (pollen from a different plant). There is no restriction — all three are possible.

   Now evaluate each option:

   1. Option (A): “Cleistogamous flowers always exhibit autogamy.”

      Since the flower never opens, the only pollen source is its own anthers. Self-pollination within the same flower is guaranteed. This is true.

   2. Option (B): “Chasmogamous flowers always exhibit geitonogamy.”

      “Always” is the trap. Chasmogamous flowers can also self-pollinate (autogamy) or cross-pollinate with another plant (xenogamy). Geitonogamy is just one possibility, not a certainty. This is false.

   3. Option (C): “Cleistogamous flowers exhibit both autogamy and geitonogamy.” …

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10. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
    Q.Family Fabaceae differs from Solanaceae and Liliaceae in relation to stamens. Identify the characteristics specific to family Fabaceae (A) Monadelphous and monothecous anthers (B) Epiphyllous and dithecous anthers (C) Diadelphous and dithecous anthers (D) Polyadelphous and epipetalous stamens

    ›Reveal solutionSolution

    The key difference is in stamen fusion: Fabaceae typically has diadelphous stamens (9 fused + 1 free) with dithecous anthers, which is not seen in Solanaceae or Liliaceae. The correct option is (C).

    The question asks you to pick the stamen characteristic that is specific to Fabaceae — meaning it should be present in Fabaceae but absent (or different) in Solanaceae and Liliaceae. So you need to recall the stamen morphology of each family and find the unique combination.

    Let’s break it down.

    1. Recall the stamen features of each family

       • Fabaceae (pea family): Stamens are typically diadelphous — 9 filaments fused into a tube and 1 free. Anthers are dithecous (two thecae, i.e., two pollen sacs per anther).
       • Solanaceae (potato family): Stamens are epipetalous (attached to petals) and usually polyadelphous or simply free; anthers are dithecous. No diadelphous condition.
       • Liliaceae (lily family): Stamens are epiphyllous (attached to the perianth) and free; anthers are dithecous. No diadelphous condition.

    2. Analyze each option

       • (A) Monadelphous and monothecous anthers: Monadelphous (all filaments fused into one bundle) occurs in Malvaceae, not Fabaceae. Monothecous anthers (single theca) are also not typical of Fabaceae. So this is wrong.
       • (B) Epiphyllous and dithecous anthers: Epiphyllous stamens are characteristic of Liliaceae, not Fabaceae. So this is not specific to Fabaceae.
       • (C) Diadelphous and dithecous anthers: Diadelphous is a hallmark of Fabaceae (especially subfamily Papilionoideae). Dithecous anthers are common to all three families, but the combination of diadelphous + dithecous is unique to Fabaceae among these three. This matches.
       • (D) Polyadelphous and epipetalous stamens: Polyadelphous (many bundles) occurs in some families like Rutaceae; epipetalous is typical of Solanaceae. Neither is specific to Fabaceae.

    3. Why the combination matters …

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11. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
    Q.Match the following List-1 List-2 A. Dehydrogenases I. $\mathrm{C-C}$ bearing X and Y on adjacent carbons $\rightarrow \mathrm{X-Y} + \mathrm{C=C}$ B. Transferases II. $\mathrm{CO_2 + H_2O \rightarrow H_2CO_3}$ C. Lyases III. $\mathrm{S-G + S'' \rightarrow S + S''-G}$ D. Carbonic anhydrase IV. S reduced + S$'$ oxidized $\rightarrow$ S oxidized + S$'$ reduced The correct answer is (A) A-III, B-II, C-I, D-IV (B) A-IV, B-III, C-I, D-II (C) A-IV, B-I, C-III, D-II (D) A-II, B-III, C-I, D-IV

    ›Reveal solutionSolution

    Each enzyme class is defined by the type of reaction it catalyses. Dehydrogenases catalyse redox reactions (IV), transferases move a group (III), lyases cleave C–C bonds without hydrolysis (I), and carbonic anhydrase hydrates CO₂ (II). The correct match is A‑IV, B‑III, C‑I, D‑II, which is option (B).

    The question tests your understanding of the IUBMB enzyme classification — the six major classes based on reaction type. You don’t need to memorise every subclass; you just need to recognise the kind of chemical change each description represents.

    1. Dehydrogenases are oxidoreductases. They catalyse the transfer of electrons (hydride ions or hydrogen atoms) from one substrate to another. The giveaway is the language “S reduced + S′ oxidized → S oxidized + S′ reduced” — that’s a textbook redox reaction. So A matches IV.

    2. Transferases move a functional group (like a glycosyl, methyl, or phosphate group) from a donor to an acceptor. The reaction shown, $\mathrm{S-G + S'' \rightarrow S + S''-G}$, is exactly that: the group G is transferred from S to S″. So B matches III.

    3. Lyases cleave C–C, C–O, C–N, or other bonds by means other than hydrolysis or oxidation, often leaving a double bond. The description “C–C bearing X and Y on adjacent carbons → X–Y + C=C” is a classic lyase elimination — two groups X and Y are removed from adjacent carbons, forming a double bond. So C matches I. …

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12. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.Mating between male parent and female offspring or female parent with male offspring is called (A) Line breeding (B) Cross breeding (C) Close breeding (D) Out crossing

    ›Reveal solutionSolution

    The question asks for the term used when mating involves a parent and its own offspring — this is the most intense form of inbreeding, known as close breeding. The correct option is (C).

    The concept here is inbreeding — the mating of individuals that are more closely related than the average of the population. In animal breeding, different levels of inbreeding have specific names depending on how close the relatives are. The most extreme case is when you mate a parent directly with its own offspring (father-daughter or mother-son). That is not just "line breeding" (which is milder, keeping a common ancestor a few generations back) nor "out crossing" (which is mating unrelated individuals). It is the tightest possible genetic pairing, and the standard term for it is close breeding.

    Let’s walk through the options:

    1. Line breeding — This is a form of inbreeding where you mate individuals that share a common ancestor, but the ancestor is usually a few generations back (e.g., mating a granddaughter back to her grandfather). It aims to concentrate the genes of that ancestor without going as far as parent-offspring mating. So this is not the answer.

    2. Cross breeding — This is the opposite of inbreeding: mating individuals from different breeds or distinct genetic lines. It increases genetic diversity. Clearly not what is described.

    3. Close breeding — This is the most intense form of inbreeding, where the mates are very close relatives — parent-offspring or full-sibling matings. The question specifically says "male parent and female offspring or female parent with male offspring", which is exactly parent-offspring mating. That is the textbook definition of close breeding. …

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