Q.In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself?

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.The sum of the roots of the equation $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^{t} + 4 = 0$ is (A) $\log_e 10$ (B) $2 \log_e 2$ (C) $\log_e 2$ (D) $2 \log_e 10$

   ›Reveal solutionSolution

   The equation is a quartic in $e^t$; substituting $x = e^t$ gives a polynomial whose roots sum to 10, so the sum of the original $t$-roots is $\log_e 10$. The correct option is (A).

   We are asked for the sum of the roots of

   $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^{t} + 4 = 0.$

   The variable $t$ appears only inside exponentials, so a natural substitution is $x = e^t$. Then $e^{2t} = x^2$, $e^{3t} = x^3$, $e^{4t} = x^4$, and the equation becomes

   $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0.$

   Now each positive root $x$ of this polynomial corresponds to a root $t = \ln x$ of the original equation. The sum of the $t$-roots is therefore $\sum \ln x_i = \ln\!\left(\prod x_i\right)$. But the product of the roots of a quartic $x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0$ is $(-1)^4 a_0 = a_0$ (here $a_0 = 4$), so $\prod x_i = 4$. That would give $\ln 4 = 2\ln 2$, which is option (B). That seems too quick — but wait: we must check that every $x$-root is positive. If any root is negative or complex, its logarithm is not a real number, and the problem likely expects real roots only. Let’s examine the polynomial more carefully.

   1. Factor the polynomial in $x$. The quartic $x^4 - 10x^3 + 29x^2 - 20x + 4$ looks like it might factor into quadratics. Try $(x^2 + ax + b)(x^2 + cx + d)$. Expanding:

   $x^4 + (a+c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.$

   Matching coefficients:

   $a + c = -10,\quad ac + b + d = 29,\quad ad + bc = -20,\quad bd = 4.$

   Since $bd = 4$, possible integer pairs: (1,4), (2,2), (4,1), and negatives. Trying $b = 2, d = 2$: then $a + c = -10$ and $ac + 4 = 29 \Rightarrow ac = 25$. So $a$ and $c$ are roots of $u^2 + 10u + 25 = 0$, i.e., $u = -5$ (double). Then $ad + bc = a\cdot2 + 2c = 2(a+c) = -20$, which matches. So the factorization is

   $(x^2 - 5x + 2)(x^2 - 5x + 2) = (x^2 - 5x + 2)^2.$

   Indeed, the quartic is a perfect square:

   $(x^2 - 5x + 2)^2 = 0.$

   2. Solve for $x$. So $x^2 - 5x + 2 = 0$ gives

   $x = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}.$

   Both roots are positive because $\sqrt{17} \approx 4.123$, so $5 - \sqrt{17} > 0$. Each root is repeated (double root), but for the sum of the $t$-roots we count each distinct $t$ once? The problem says “the sum of the roots” — in an equation, repeated roots are still roots, so we count multiplicities. The polynomial in $x$ has two distinct roots, each of multiplicity 2. So the four $x$-roots (with multiplicity) are:

   $\frac{5 + \sqrt{17}}{2},\; \frac{5 + \sqrt{17}}{2},\; \frac{5 - \sqrt{17}}{2},\; \frac{5 - \sqrt{17}}{2}.$

   Their product is $\left(\frac{5+\sqrt{17}}{2}\right)^2 \left(\frac{5-\sqrt{17}}{2}\right)^2 = \left(\frac{25 - 17}{4}\right)^2 = \left(\frac{8}{4}\right)^2 = 2^2 = 4$, consistent.

   3. Sum of the $t$-roots. Each $x$-root gives $t = \ln x$. So the four $t$-roots (with multiplicity) are: $$$$

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
   Q.The change in the frequency of a gene that occurs merely by chance and not by selection in small population is called (A) Founder's effect (B) Genetic load (C) Gene flow (D) Sewall Wright effect

   ›Reveal solutionSolution

   Random fluctuations in allele frequencies in small populations, independent of natural selection, are called genetic drift or the Sewall Wright effect. The answer is (D).

   When populations are large, natural selection is the dominant force shaping allele frequencies. But in small populations, chance events can cause gene frequencies to drift randomly from generation to generation, even when those alleles confer no advantage or disadvantage. This phenomenon is genetic drift.

   Imagine flipping a coin ten times versus a thousand times. With ten flips, you might easily get seven heads and three tails purely by chance. With a thousand flips, you'll almost certainly land close to 50-50. Small populations behave like the ten-flip scenario: random sampling of gametes during reproduction can produce large swings in allele frequency that have nothing to do with fitness.

   Sewall Wright, one of the founders of population genetics, extensively studied this random sampling effect in small populations. The phenomenon is named after him: the Sewall Wright effect is synonymous with genetic drift.

   Now let's see why the other options don't fit:

   1. Founder's effect (A) is a special case of genetic drift. It occurs when a small group breaks off from a larger population to establish a new colony. The founders carry only a subset of the genetic variation from the original population, so the new population starts with different allele frequencies. While it involves drift, it specifically describes the initial sampling event, not the ongoing random fluctuation in any small population. …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Identify the correct statement with reference to the diagram given below [FIGURE: Three adjoining cells drawn as boxes. Cell A (upper left) has solute concentration 10%; cell B (upper right) has 5%; cell C (below, spanning between them) has 15%. An arrow labelled "Solute" points into cell A.] (A) water moves from A to B (B) water moves from C to A (C) B has highest water potential (D) C has highest water potential

   ›Reveal solutionSolution

   Adding solute lowers water potential, so among cells at 10 %, 5 % and 15 % solute, the most dilute one — cell B at 5 % — has the highest water potential. Water therefore flows from B, not to it. The only statement that survives is "B has highest water potential" — option (C).

   The concept first: what water potential is, and which way water moves

   The single idea that decides this whole question is:

   $\boxed{\ \Psi_w = \Psi_s + \Psi_p\ }$

   where

   • $\Psi_w$ = water potential (the free energy of water per unit volume, measured in pascals),
   • $\Psi_s$ = solute potential (also called osmotic potential),
   • $\Psi_p$ = pressure potential (turgor pressure).

   Three facts follow, and they are the whole of osmosis:

   Fact 1 — Pure water has the highest possible water potential, defined as $\Psi_w = 0$ at standard conditions.

   Fact 2 — Dissolving a solute always LOWERS water potential. The solute potential $\Psi_s$ is always negative, and it grows more negative as you add more solute. Physically: solute molecules bind and immobilise water molecules, reducing the fraction of water that is free to move. Hence a solution always has $\Psi_w < 0$.

   $\text{more solute} \ \Longrightarrow \ \Psi_s \ \text{more negative} \ \Longrightarrow \ \Psi_w \ \text{lower}$

   Fact 3 — Water moves DOWN the water-potential gradient, i.e. from a region of higher $\Psi_w$ to a region of lower $\Psi_w$ (from high free energy to low free energy). Nothing else decides the direction of osmosis.

   A very common student error is to think "water moves towards more water" or to confuse concentration of solute with concentration of water. Say it to yourself the safe way: water moves towards the more concentrated (saltier) solution, because that is the solution with the lower water potential.

   Step-by-step solution

   Step 1 — Read the solute concentrations off the diagram.

   Cell	Solute concentration
   A	10 %
   B	5 %
   C	15 %

   Step 2 — Convert concentration into a water-potential ranking. Assuming (as such diagrams intend) that the pressure potential $\Psi_p$ contributions are comparable, the ordering of $\Psi_w$ is set entirely by $\Psi_s$:

   • B (5 %) — least solute → $\Psi_s$ least negative → $\Psi_w$ HIGHEST
   • A (10 %) — intermediate → $\Psi_w$ intermediate
   • C (15 %) — most solute → $\Psi_s$ most negative → $\Psi_w$ LOWEST

   $\Psi_w(\text{B}) \;>\; \Psi_w(\text{A}) \;>\; \Psi_w(\text{C})$

   Step 3 — Deduce the direction of water flow. Water runs downhill in $\Psi_w$:

   $\text{B} \ \longrightarrow \ \text{A} \ \longrightarrow \ \text{C}$

   Cell C, being the most concentrated, is the sink: water flows into it from both A and B. Cell B, being the most dilute, is the source: water flows out of it.

   Step 4 — Test each printed statement against this picture.

   • (A) "water moves from A to B" — FALSE. A (10 %) has lower $\Psi_w$ than B (5 %). Water moves the other way, B → A. …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
   Q.Reservoir host for Trypanosoma gambiense. (A) Monkey (B) Rabbit (C) African antelope (D) Female Culex mosquito

   ›Reveal solutionSolution

   A reservoir host is an animal that harbors a pathogen without suffering ill effects, serving as a source of infection for other susceptible hosts. For Trypanosoma gambiense, the causative agent of West African sleeping sickness, humans are the primary reservoir, but various wild and domestic animals, including African antelopes, can also act as reservoir hosts. The correct option is (C).

   Concept and Intuition

   To understand the role of a reservoir host, imagine a hidden pool where a disease-causing organism (pathogen) can live and multiply without causing severe illness to its host. This host, called a reservoir host, then acts as a continuous source of infection for other susceptible individuals or species. This is crucial for the pathogen's survival and spread, especially when its primary host population is low or when the pathogen needs to persist in the environment.

   For parasitic diseases like trypanosomiasis, understanding the reservoir host is vital for disease control. If a pathogen can survive in animals, eradicating the disease in humans becomes much harder, as the animals can re-introduce the infection.

   Important

   A reservoir host is a long-term host of a pathogen that does not get sick from the pathogen but can transmit it to other species.

   Step-by-step Explanation

   1. Identify the Pathogen and Disease: The question refers to Trypanosoma gambiense. This is a protozoan parasite that causes West African sleeping sickness, also known as Gambian trypanosomiasis. This disease is characterized by a chronic course, often leading to neurological symptoms if untreated.

   2. Understand the Transmission: Trypanosoma gambiense is transmitted to humans and animals through the bite of an infected tsetse fly (genus Glossina). The tsetse fly acts as a biological vector, meaning the parasite undergoes part of its life cycle within the fly.

   3. Primary Host and Reservoir Hosts:

      • Humans are the primary hosts and the main reservoir for Trypanosoma gambiense. This means that human-to-human transmission (via the tsetse fly) is the most significant epidemiological cycle.
      • However, various animals can also harbor the parasite without showing severe symptoms, thus acting as reservoir hosts. These animal reservoirs can maintain the parasite in nature, making disease control more challenging.

   4. Evaluate the Options for Reservoir Hosts:

      • (A) Monkey: Monkeys can be infected with Trypanosoma gambiense and can serve as reservoir hosts, particularly in sylvatic (forest) cycles.
      • (B) Rabbit: Rabbits are not typically recognized as significant natural reservoir hosts for Trypanosoma gambiense. …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
   Q.Match the following lists. LIST-I A) Atropa B) Computer usage in taxonomy C) Gloriosa D) Odd sepal anterior LIST-II

   i) Ornament

   ii) Medicine

   iii) Fabaceae

   iv) Numerical Taxonomy LIST-III I) Number and codes II) Pterocarpus III) solanaceae IV) Rhizome The correct match is: (A) ii, III (B) iv, I (C) i, IV (D) iii, II (B) ii, IV (B) iv, II (C) iii, III (D) i, I (C) iii, IV (B) i, III (C) ii, I (D) iv, II (D) i, I (B) ii, II (C) iii, IV (D) iv, III

   ›Reveal solutionSolution

   This is a matching problem linking plants (LIST-I) to their uses (LIST-II) and families/features (LIST-III). The correct pairings are: Atropa → Medicine → Solanaceae; Computer usage in taxonomy → Numerical Taxonomy → Number and codes; Gloriosa → Ornament → Rhizome; Odd sepal anterior → Fabaceae → Pterocarpus. The only option that matches all four is (C).

   The key is to recognize that each item in LIST-I must be matched with one item from LIST-II and one from LIST-III, forming a consistent triple. The question gives several rows of possible matches; we need to find the row where all four triples are correct.

   Why this approach works: Instead of memorizing every plant, we use botanical knowledge: Atropa belladonna (deadly nightshade) is a medicinal plant in the Solanaceae family; Gloriosa superba (glory lily) is an ornamental plant with a rhizome; "Odd sepal anterior" is a floral characteristic of Fabaceae (pea family), which includes Pterocarpus; and computer usage in taxonomy refers to numerical taxonomy, which uses numbers and codes.

   Step-by-step reasoning:

   1. Atropa (A): This genus includes Atropa belladonna, a source of the drug atropine. It belongs to the family Solanaceae (LIST-III, III). Its use is Medicine (LIST-II, ii). So A → ii, III.

   2. Computer usage in taxonomy (B): This is the definition of Numerical Taxonomy (LIST-II, iv), which relies on Number and codes (LIST-III, I). So B → iv, I.

   3. Gloriosa (C): Gloriosa superba is a climbing lily grown as an Ornament (LIST-II, i). It propagates via a Rhizome (LIST-III, IV). So C → i, IV.

   4. Odd sepal anterior (D): This floral term describes a condition where the odd sepal is in the anterior (lower) position, a characteristic of the Fabaceae family (LIST-III, II). A common example is Pterocarpus (LIST-II, iii). So D → iii, II.

   Now we check the given options. The correct triplets are:

   • A: ii, III
   • B: iv, I
   • C: i, IV
   • D: iii, II

   Looking at the rows, only option (C) lists these exact pairings in order:

   (C) iii, IV (B) i, III (C) ii, I (D) iv, II — Wait, that’s not matching. Let’s read the options carefully. The question presents four rows of four matches each. The row that matches our triplets is the third row: …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.The view of inheritance of acquired characters of Lamarck was opposed by (A) Spencer (B) Kammarer (C) Weisman (D) Hugo de Vries

   ›Reveal solutionSolution

   Lamarck’s idea that traits acquired during an organism’s lifetime can be inherited was directly challenged by Weismann’s germ-plasm theory, which showed that only changes in reproductive cells (germ cells) are passed on. The correct opponent is (C) Weisman.

   The core of this question lies in the history of evolutionary thought. Lamarck proposed that if a giraffe stretches its neck to reach leaves, that stretched neck is passed to its offspring. This is the “inheritance of acquired characters.” The key scientific counterargument came from August Weismann, who conducted a famous experiment: he cut off the tails of mice for many generations and observed that every new generation was born with full-length tails. This proved that a bodily change (acquired character) does not affect the germ cells (sperm and eggs), so it cannot be inherited.

   Let’s see why each option fits or fails:

   1. Herbert Spencer was a philosopher who actually supported Lamarckian ideas and coined “survival of the fittest.” He did not oppose Lamarck — he extended the concept to social evolution. So (A) is wrong.

   2. Paul Kammerer was a biologist who claimed to have found evidence for Lamarckism (e.g., in midwife toads). His work was later discredited. He was a supporter, not an opponent. So (B) is wrong.

   3. August Weismann is the correct answer. He proposed the germ-plasm theory, which states that hereditary information is carried only in germ cells, and somatic (body) changes are not transmitted. His tail-cutting experiment directly refuted Lamarck. So (C) is correct. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.To destroy cancer cells that have moved to other parts of the body the following therapy is useful (A) Radio therapy (B) Bio therapy (C) Chemo therapy (D) Biotechno therapy

   ›Reveal solutionSolution

   When cancer cells spread to multiple parts of the body (metastasize), a systemic treatment is needed to reach them all. Chemotherapy, which uses drugs that travel throughout the bloodstream, is the most effective therapy for destroying these widespread cancer cells. The correct option is (C).

   When cancer cells detach from the primary tumor and travel through the bloodstream or lymphatic system to establish new tumors in distant organs, this process is called metastasis. Once cancer has metastasized, it is no longer confined to a single location. To effectively treat such widespread cancer, a therapy that can reach and destroy cancer cells throughout the entire body is required. Localized treatments, which target only a specific area, would be insufficient.

   Here's a breakdown of the given options:

   1. Understanding the need for systemic treatment: The question describes cancer cells that "have moved to other parts of the body." This means the cancer is no longer localized to one area; it has spread. Therefore, any effective therapy must be able to reach these disseminated cells throughout the body.

   2. Evaluating Radiotherapy (A): Radiotherapy uses high-energy radiation to kill cancer cells and shrink tumors. It is a localized treatment, meaning it is directed at a specific part of the body. While highly effective for treating primary tumors or isolated metastatic sites, it cannot be used to treat cancer cells spread throughout the entire body without causing severe damage to healthy tissues. Thus, it is not suitable for widespread destruction of metastatic cells.

   3. Evaluating Biotherapy (B): Biotherapy, often referred to as immunotherapy or biological therapy, uses substances made from living organisms to treat cancer. These substances can work by boosting the body's immune system to fight cancer, or by directly targeting specific cancer cells or pathways. While some forms of biotherapy can have systemic effects and are used for metastatic cancer, the most general and widely applicable therapy for destroying widespread cancer cells by directly killing them is chemotherapy. Biotherapy often focuses on modulating the immune response or targeting specific molecular pathways rather than broad cellular destruction. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.If a gene has four alleles, possible number of genotypes are (A) 8 (B) 10 (C) 20 (D) 6

   ›Reveal solutionSolution

   When a gene has multiple alleles, the total number of possible genotypes is the sum of homozygous and heterozygous combinations. For 4 alleles, there are 10 possible genotypes.

   Concept and Intuition

   A gene is a segment of DNA that codes for a specific trait. Alleles are different forms of the same gene. For example, a gene for flower color might have an allele for red flowers and an allele for white flowers.

   An individual inherits two alleles for each gene, one from each parent. These two alleles together constitute the individual's genotype.

   There are two main types of genotypes:

   1. Homozygous genotypes: Both inherited alleles are identical (e.g., $AA$ or $aa$).
   2. Heterozygous genotypes: The two inherited alleles are different (e.g., $Aa$).

   To find the total number of possible genotypes, we need to consider all unique combinations of two alleles that can be formed from the given set of alleles.

   Step-by-step Derivation

   Let's denote the four alleles as $A_1, A_2, A_3,$ and $A_4$.

   1. Identify Homozygous Genotypes:

      A homozygous genotype occurs when an individual inherits two identical alleles. With four distinct alleles, we can form four homozygous genotypes:

      • $A_1A_1$
      • $A_2A_2$
      • $A_3A_3$
      • $A_4A_4$ Thus, there are 4 possible homozygous genotypes.

   2. Identify Heterozygous Genotypes:

      A heterozygous genotype occurs when an individual inherits two different alleles. The order in which the alleles are listed does not matter (e.g., $A_1A_2$ is the same genotype as $A_2A_1$). This is a combination problem: we need to choose 2 different alleles from the 4 available alleles.

      The number of ways to choose 2 distinct alleles from 4 is given by the combination formula $^nC_k = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items and $k$ is the number of items to choose.

      In this case, $n=4$ (number of alleles) and $k=2$ (number of alleles in a genotype).

   $^4C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$

   The 6 possible heterozygous genotypes are:
   *   $A_1A_2$
   *   $A_1A_3$
   *   $A_1A_4$
   *   $A_2A_3$
   *   $A_2A_4$
   *   $A_3A_4$ …
   

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
   Q.The following phenotypic results are obtained when a pea plant with round and yellow seeds is crossed to another pea plant with wrinkled and yellow seeds.

   [!FORMULA] $\begin{array}{|c|c|c|c|} \hline \text{Round and Yellow} & \text{Round and green} & \text{Wrinkled and green} & \text{Wrinkled and Yellow} \\ \hline 3 & 1 & 1 & 3 \\ \hline \end{array}$

   Identify the genotypes of the plants involved in the cross. (A) $Rr\ Yy\ \times\ rrYY$ (B) $Rr\ Yy\ \times\ rrYy$ (C) $Rr\ YY\ \times\ rrYy$ (D) $RRYy\ \times\ rrYy$

   ›Reveal solutionSolution

   The observed 3:1:1:3 ratio (round:wrinkled ≈ 1:1, yellow:green ≈ 3:1) points to one parent heterozygous for both traits and the other heterozygous for seed color but homozygous recessive for shape — so the cross is Rr Yy × rr Yy, option (B).

   We are given a cross between two pea plants: one with round, yellow seeds and the other with wrinkled, yellow seeds. The offspring show four phenotypes in the ratio 3 round yellow : 1 round green : 1 wrinkled green : 3 wrinkled yellow. Our job is to deduce the genotypes of the parents.

   Key concept: In a dihybrid cross, the ratio of phenotypes tells us the mode of inheritance for each trait separately. Here, seed shape (round vs. wrinkled) and seed color (yellow vs. green) are independent (Mendelian). Round (R) is dominant over wrinkled (r); yellow (Y) is dominant over green (y). The observed ratio is not the classic 9:3:3:1, so at least one parent is not fully heterozygous for both traits. We analyze each trait independently.

   1. Analyze seed shape (round vs. wrinkled):

      Count the offspring: Round = 3 (RY) + 1 (R green) = 4; Wrinkled = 1 (wrinkled green) + 3 (wrinkled yellow) = 4.

      Ratio round : wrinkled = 4 : 4 = 1 : 1.

      A 1:1 ratio for a single trait is the classic testcross ratio: one parent is heterozygous (Rr) and the other is homozygous recessive (rr).

      Since the first parent has round seeds, it must be Rr (cannot be RR, because that would give all round). The second parent has wrinkled seeds, so it must be rr.

   2. Analyze seed color (yellow vs. green):

      Count: Yellow = 3 (RY) + 3 (wrinkled yellow) = 6; Green = 1 (round green) + 1 (wrinkled green) = 2.

      Ratio yellow : green = 6 : 2 = 3 : 1.

      A 3:1 ratio for a single trait indicates both parents are heterozygous (Yy × Yy).

      Both parents have yellow seeds, so each must carry one recessive green allele: both are Yy.

   3. Combine the two traits:

      First parent: round and yellow → genotype Rr Yy.

      Second parent: wrinkled and yellow → genotype rr Yy.

   4. Check the cross: …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.Identify the enzyme of the following reaction Phosphoglyceric acid + ATP $\xrightarrow{\text{Enzyme}}$ Bisphosphoglyceric acid + ADP (A) Phosphoglycero carboxylase (B) Phosphoglycero mutase (C) Phosphoglycero oxidase (D) Phosphoglycerokinase

    ›Reveal solutionSolution

    The reaction involves the transfer of a phosphate group from ATP to phosphoglyceric acid, forming bisphosphoglyceric acid and ADP. Enzymes that catalyze such phosphate group transfers are called kinases. Therefore, the enzyme is Phosphoglycerokinase.

    Enzymes are biological catalysts, and their names often reflect the substrate they act upon and the type of reaction they catalyze. Understanding this naming convention is key to identifying the enzyme in a given biochemical reaction.

    In general, enzymes ending in "-ase" indicate an enzyme. The prefix often refers to the substrate or the type of reaction. For reactions involving the transfer of a phosphate group, a specific class of enzymes called kinases is involved. Kinases are responsible for phosphorylating molecules, typically using ATP as the phosphate donor.

    1. Analyze the reactants and products:

       • Reactants: Phosphoglyceric acid and ATP (Adenosine Triphosphate).
       • Products: Bisphosphoglyceric acid and ADP (Adenosine Diphosphate).

    2. Identify the change occurring:

       • ATP loses one phosphate group to become ADP.
       • Phosphoglyceric acid gains one phosphate group to become Bisphosphoglyceric acid (meaning it now has two phosphate groups).
       • This clearly indicates the transfer of a phosphate group from ATP to phosphoglyceric acid.

    3. Relate the change to enzyme classification:

       • Enzymes that catalyze the transfer of a functional group from one molecule to another are classified as transferases.
       • Specifically, enzymes that transfer phosphate groups are known as kinases. They typically use ATP (or sometimes GTP) as the phosphate donor.

       Substrate + ATP $\xrightarrow{\text{Kinase}}$ Substrate-P + ADP

       Where 'P' represents a phosphate group. …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
    Q.Multiple fission in Amoeba is called (A) Schizogony (B) Gametogony (C) Sporogony (D) Sporulation

    ›Reveal solutionSolution

    Multiple fission in Amoeba is a form of asexual reproduction where the parent cell divides into many daughter cells simultaneously, and the correct term for this process is sporulation.

    The question asks for the specific name given to multiple fission in Amoeba. To answer this, we need to understand what multiple fission means and how it differs from other types of cell division. In Amoeba, multiple fission occurs under unfavorable conditions (like lack of food or drying up). The Amoeba forms a protective cyst around itself, and inside this cyst, its nucleus divides repeatedly without the cytoplasm dividing. Later, the cytoplasm divides all at once, producing many tiny daughter cells (called spores or amoebulae). This entire process is what we need to name.

    Let’s look at each option and see which one fits.

    1. Schizogony – This is a type of multiple fission seen in some parasites like Plasmodium (the malaria parasite). In schizogony, the nucleus divides many times, and then the cytoplasm divides to form many daughter cells (merozoites). While it is a form of multiple fission, it is not the term used for Amoeba. Schizogony is specific to certain protozoans, especially those in the phylum Apicomplexa.

    2. Gametogony – This refers to the formation of gametes (sex cells). In some parasites, gametogony produces male and female gametes that fuse to form a zygote. This is a sexual process, not an asexual one like multiple fission. So this is not correct for Amoeba.

    3. Sporogony – This is another type of multiple fission that produces spores, but it is typically associated with the sexual cycle of parasites like Plasmodium (after fertilization, the zygote undergoes sporogony to produce sporozoites). Again, this is not the term used for Amoeba. …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
    Q.Excessive consumption of this drug causes hallucinations (A) Cocaine (B) Morphine (C) Heroin (D) Ganja

    ›Reveal solutionSolution

    Excessive consumption of Cocaine, a powerful stimulant, is well-known to induce stimulant psychosis, which often includes paranoia, delusions, and characteristic tactile hallucinations. The correct option is (A).

    The Concept: Understanding Drug Effects and Classifications

    To answer this question, we need to understand the primary pharmacological effects of different drug classes, especially when consumed in excessive amounts. Drugs interact with our brain chemistry in various ways, leading to distinct effects. Hallucinations are sensory experiences that appear real but are created by the mind, often a sign of significant neurological disruption or altered consciousness. We'll look at each option and classify it to determine which one most characteristically causes hallucinations with excessive use.

    Step-by-Step Analysis

    1. Classifying the Drugs:

       • (A) Cocaine: This is a potent stimulant. It primarily affects the central nervous system by increasing levels of dopamine, norepinephrine, and serotonin.
       • (B) Morphine: This is an opioid. It acts as a powerful analgesic (painkiller) and central nervous system depressant, primarily by binding to opioid receptors.
       • (C) Heroin: Also an opioid, derived from morphine. It shares similar depressant and analgesic properties, acting on the same opioid receptors.
       • (D) Ganja (Cannabis/Marijuana): This is a cannabinoid. Its effects are complex and can include stimulant, depressant, and mild hallucinogenic properties, primarily due to compounds like THC interacting with cannabinoid receptors.

    2. Analyzing the Effects of Excessive Consumption:

       • Cocaine (A):

         • Excessive consumption of cocaine is notorious for leading to a condition known as stimulant psychosis. This severe mental state is characterized by extreme paranoia, delusions (false beliefs), and often vivid hallucinations.
         • A classic type of hallucination associated with cocaine abuse is formication, a tactile hallucination where the user feels insects crawling under or on their skin (often referred to as "coke bugs"). Visual and auditory hallucinations can also occur.
         • The intense overstimulation of dopamine pathways is a key factor in inducing these psychotic symptoms.

       • Morphine (B) and Heroin (C):

         • As opioids, their primary effects are pain relief, euphoria, and significant central nervous system depression.
         • Excessive consumption or overdose primarily leads to respiratory depression (slowed or stopped breathing), pinpoint pupils, and loss of consciousness, which can be fatal.
         • While severe overdose or withdrawal can sometimes lead to delirium or confusion, hallucinations are not a characteristic or primary effect of excessive opioid consumption in the same way they are for stimulants or classic hallucinogens. Their main danger lies in suppressing vital bodily functions.

       • Ganja (D):

         • Cannabis can indeed alter perception, and at high doses, or in individuals predisposed to mental health issues, it can induce paranoia, anxiety, and even transient psychotic episodes or hallucinations. …

    🔒 Start your 14-day free trial to see the full solution →