Q.(a) 'Insertional inactivation' is a method to detect recombinant DNA. Explain the method.

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A mixture contains DNA fragments a, b, c and d with molecular weights of $a+b=c$, $a>b$ and $d>c$ was subjected to agarose gel electrophoresis. The positions of these fragments from cathode to anode sides of the gel plate (A) b, a, c, d (B) a, b, c, d (C) c, b, a, d (D) b, a, d, c

   ›Reveal solutionSolution

   Ranking the fragments by size ($b<a<c<d$) and remembering that smaller DNA migrates faster in agarose gel gives the order b, a, c, d — option (A).

   Order the sizes. From the given relations:

   • $a+b=c$, so $c$ is larger than either $a$ or $b$.
   • $a>b$.
   • $d>c$.

   Combining these, the molecular-weight order from lightest to heaviest is

   $b < a < c < d.$ …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.How many meiotic divisions are required to produce 1600 seeds in an angiospermic plant? (A) 1600 (B) 1800 (C) 3200 (D) 2000

   ›Reveal solutionSolution

   To produce 1600 seeds, we need 1600 functional egg cells and 1600 functional pollen grains (each containing male gametes). This requires 1600 meiotic divisions for egg cells and 400 meiotic divisions for pollen grains, totaling $\boxed{2000}$ meiotic divisions.

   In angiosperms, a seed develops from a fertilized ovule. For each seed to form, one egg cell (female gamete) must be fertilized by one male gamete. Both male and female gametes are produced through meiosis. Understanding how many gametes result from each meiotic division for both sexes is key to solving this problem.

   Here's the breakdown:

   Production of Female Gametes (Egg Cells)

   1. Megaspore Mother Cell (MMC) to Egg Cell: In the ovule, a single diploid megaspore mother cell (MMC) undergoes meiosis.
   2. Meiosis in MMC: This meiotic division produces four haploid megaspores.
   3. Functional Megaspore: In most angiosperms, only one of these four megaspores is functional, while the other three degenerate.
   4. Embryo Sac Formation: The functional megaspore then undergoes three mitotic divisions to form an eight-nucleate, seven-celled embryo sac, which contains one egg cell.
   5. Conclusion for Female Gametes: Therefore, one meiotic division in a megaspore mother cell ultimately leads to the formation of one functional egg cell.

   Production of Male Gametes (Pollen Grains)

   1. Microspore Mother Cell (PMC) to Pollen Grains: In the anther, diploid microspore mother cells (PMCs) undergo meiosis.
   2. Meiosis in PMC: Each PMC undergoes one meiotic division to produce four haploid microspores.
   3. Pollen Grain Development: Each microspore develops into a pollen grain. Each mature pollen grain contains two male gametes (formed by mitosis of the generative cell).
   4. Fertilization Potential: Since each pollen grain contains two male gametes, it can participate in double fertilization, but for the formation of one seed, only one male gamete is required to fertilize the egg cell. Thus, one pollen grain contributes to one seed.
   5. Conclusion for Male Gametes: Therefore, one meiotic division in a microspore mother cell produces four functional pollen grains, each capable of fertilizing one ovule.

   Calculating Total Meiotic Divisions for 1600 Seeds

   We need to produce 1600 seeds. This means we need 1600 successful fertilizations, which in turn requires 1600 functional egg cells and 1600 functional pollen grains.

   1. Meiotic Divisions for Female Gametes:
      • Since 1 meiotic division in an MMC produces 1 egg cell, to produce 1600 egg cells, we need 1600 meiotic divisions. …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.Identify the correct statements among the following A. RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence B. Bt-toxin gene, CryIIAb controls cotton bollworms C. Mycorrhizae helps the plants for more absorption of potassium D. Elusion is a technique of extracting separated bands of DNA from agarose gel (A) A, B, C (B) B, C, D (C) A, C, D (D) A, B, D

   ›Reveal solutionSolution

   The question tests factual recall of four biotech concepts. RNAi is not universal in all eukaryotes; CryIIAb targets corn borers, not cotton bollworms; mycorrhizae primarily aid phosphorus, not potassium; elution (not "elusion") is the correct term for DNA band extraction. Only statements A, B, D are correct.

   Let’s examine each statement carefully, because this is a classic trap — mixing one plausible-sounding wrong fact with three correct ones.

   1. Statement A: RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.

      RNAi is a gene-silencing mechanism triggered by double-stranded RNA. It is indeed a natural defence against viruses and transposons in many eukaryotes — but not all. For example, some unicellular eukaryotes like Saccharomyces cerevisiae (budding yeast) lack the key RNAi machinery (Dicer, Argonaute). So the word “all” makes this statement false.

      Watch out

      The trap is that RNAi is widespread, but “all” is an absolute that rarely holds in biology. Always check for such qualifiers.

   2. Statement B: Bt-toxin gene, CryIIAb controls cotton bollworms.

      Bt cotton uses CryIAc and CryIIAb genes. CryIIAb is indeed effective against lepidopteran pests like the cotton bollworm (Helicoverpa armigera). This statement is true.

      Tip

      Remember: CryIAc and CryIIAb are the two main genes used in Indian Bt cotton. CryIIAb also controls corn borers, but its use in cotton is well-established.

   3. Statement C: Mycorrhizae helps the plants for more absorption of potassium.

      Mycorrhizal fungi form symbiotic associations with plant roots. Their primary benefit is enhanced uptake of phosphorus (which is immobile in soil), not potassium. While they can indirectly help with other nutrients, potassium absorption is not their hallmark. This statement is false.

   4. Statement D: Elusion is a technique of extracting separated bands of DNA from agarose gel.

      The correct term is elution (not “elusion”). Elution is the process of cutting out a DNA band from the gel and extracting the DNA — a standard step in gel electrophoresis. The statement uses the wrong spelling, but the technique described is correct. In exam contexts, such minor spelling errors are usually ignored if the concept is right. So this is true. …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.The separated DNA fragments can be visualised after staining with ‘X’ and exposure to ‘Y’. The ‘X’ and ‘Y’ are (A) X Ethidium iodide, Y X-Rays (B) X Ethidium bromide, Y UV-Rays (C) X Ethidium chloride, Y $\gamma$-Rays (D) X Ethidium fluoride, Y $\beta$-Rays

   ›Reveal solutionSolution

   DNA fragments separated by gel electrophoresis are visualised using ethidium bromide as the stain and UV light for detection through fluorescence. The answer is (B).

   Why this technique works

   When DNA fragments are separated by gel electrophoresis, they remain invisible to the naked eye because DNA is colourless and transparent. We need a method to locate these fragments on the gel. The solution lies in using a fluorescent dye that intercalates between DNA base pairs and becomes visible under specific wavelengths of light.

   Ethidium bromide is the classic choice because it inserts itself between the stacked bases of the DNA double helix (a process called intercalation). Once bound to DNA, its fluorescent properties change dramatically: when exposed to ultraviolet light (typically 260–360 nm), it fluoresces bright orange, making the DNA bands glow against a dark background.

   Step-by-step reasoning

   1. The staining agent (X)

      After electrophoresis, the gel is soaked in a solution containing ethidium bromide. The dye molecules diffuse through the gel and bind to DNA wherever it is present. The amount of fluorescence is roughly proportional to the amount of DNA in each band.

   2. Why ethidium bromide specifically?

      Ethidium bromide has a planar aromatic ring system that slips neatly between the base pairs of double-stranded DNA. This intercalation increases the fluorescence quantum yield of the dye by about 20–30 fold compared to free dye in solution. This dramatic enhancement makes even small amounts of DNA visible.

   3. The visualization method (Y) …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.Separated bands of DNA are cut out from the agrose gel and extracted from gel piece. This is known as (A) Elution (B) Spooling (C) Centrifugation (D) Detergent lysis

   ›Reveal solutionSolution

   The process of cutting out and extracting DNA from an agarose gel piece is called elution. The correct option is (A).

   The question tests your understanding of a specific laboratory technique used in molecular biology and genetic engineering. When you run DNA fragments on an agarose gel, they separate by size. To use a particular fragment for further experiments (like cloning or sequencing), you need to recover it from the gel. The term for this recovery is elution.

   Let’s break down why each option fits or doesn’t.

   1. Elution – This is the correct term. In gel electrophoresis, after the DNA bands are separated, you physically cut out the band of interest with a scalpel or blade. The DNA is then extracted from the gel slice using methods like freeze-squeeze, electroelution, or commercial gel extraction kits (which often use a chaotropic salt and a silica membrane). The general word for “washing out” or “extracting” a substance from a solid matrix is elution. So, cutting out the band and extracting the DNA is precisely elution.

   2. Spooling – This refers to a method of isolating DNA from a solution by winding it onto a glass rod or pipette tip after precipitation with ethanol or isopropanol. It is used for bulk DNA isolation (e.g., from bacterial cultures or plant tissue), not for recovering a specific band from a gel. Spooling gives you long, visible threads of DNA, but it cannot target a single fragment from a gel. …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.Sonalika is high yielding and disease resistant variety of this crop (A) Rice (B) Wheat (C) Ground nut (D) Maize

   ›Reveal solutionSolution

   Sonalika is a high-yielding, disease-resistant variety of wheat, developed during the Green Revolution in India.

   The question tests your knowledge of crop varieties developed in India, especially those linked to the Green Revolution. Sonalika is a famous wheat variety — it was introduced in the 1960s and became widely popular because it gave high yields and resisted rust diseases (a major problem for wheat farmers). Many students confuse it with rice varieties like IR-8 or Jaya, but Sonalika is specifically wheat.

   1. Recall the context of the Green Revolution

      In the 1960s, India faced food shortages. High-yielding varieties (HYVs) of wheat and rice were introduced. For wheat, the key varieties were Kalyan Sona and Sonalika. Both were developed from Mexican dwarf wheat strains and were resistant to rust diseases.

   2. Identify the crop for Sonalika

      Sonalika is a semi-dwarf wheat variety. It matures early, produces more grain per plant, and is resistant to yellow and brown rust. It was released for commercial cultivation in India in the late 1960s. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
   Q.The first artificial rDNA was constructed by using a native plasmid of which of the following organisms? (A) Salmonella typhimurium (B) Escherichia coli (C) Agrobacterium tumifaciens (D) Propionibacterium sharmani

   ›Reveal solutionSolution

   The first recombinant DNA molecule was created in 1972 by Stanley Cohen and Herbert Boyer using the plasmid pSC101 from Salmonella typhimurium. The answer is (A).

   The construction of the first artificial recombinant DNA marks the birth of modern genetic engineering. Understanding which organism contributed the plasmid requires knowing the historical context of this breakthrough experiment.

   In 1972, Stanley Cohen and Herbert Boyer pioneered the technique of cutting DNA with restriction enzymes and joining fragments from different sources using DNA ligase. The key innovation was using a plasmid vector — a small, circular DNA molecule that replicates independently inside bacterial cells — to carry foreign DNA.

   For their landmark experiment, they needed a plasmid with two critical features: the ability to replicate autonomously and a selectable marker (typically antibiotic resistance) to identify cells that had taken up the recombinant DNA.

   The plasmid they chose was pSC101, isolated from Salmonella typhimurium. This plasmid carried a tetracycline-resistance gene, making it ideal for selection. They cut this plasmid with the restriction enzyme EcoRI, inserted a foreign DNA fragment (from another plasmid, pSC102), and sealed the construct with DNA ligase. When introduced into E. coli, the recombinant plasmid replicated successfully, proving that DNA from different sources could be combined and propagated. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
   Q.Which of the following is required to precipitate purified DNA from solution. (A) Chilled calcium chloride (B) Chilled ethanol (C) Chitinase (D) Ethidium bromide

   ›Reveal solutionSolution

   To precipitate purified DNA from a solution, chilled ethanol is used. The alcohol dehydrates the DNA, and in the presence of salt, the DNA aggregates and becomes insoluble, allowing it to be collected. The correct option is (B).

   Concept and Intuition

   DNA precipitation is a crucial step in molecular biology to concentrate and purify DNA from a solution. The underlying principle relies on altering the solubility of DNA.

   DNA molecules are long polymers with a negatively charged phosphate backbone. In an aqueous solution, these negative charges interact with water molecules, forming a hydration shell around the DNA, which keeps it soluble.

   To precipitate DNA, we need to disrupt this hydration shell and neutralize the negative charges on the DNA backbone. This allows individual DNA molecules to come closer, aggregate, and fall out of solution as a visible pellet.

   The process typically involves two main components:

   1. Alcohol (e.g., ethanol or isopropanol): Alcohol reduces the dielectric constant of the solution, which weakens the electrostatic interactions between DNA and water. More importantly, alcohol molecules compete with water molecules for binding to the DNA, effectively dehydrating the DNA. As the DNA loses its hydration shell, its solubility decreases significantly.
   2. Salt (e.g., sodium acetate, ammonium acetate, or lithium chloride): The positive ions from the salt (e.g., $Na^+$ from sodium acetate) neutralize the negative charges on the DNA phosphate backbone. This neutralization is critical because it eliminates the electrostatic repulsion between individual DNA molecules, allowing them to aggregate and form a precipitate.

   Chilling the solution (typically to $-20^\circ C$ or $-80^\circ C$) further reduces the solubility of DNA in the alcohol-salt mixture, enhancing precipitation efficiency and helping to stabilize the DNA.

   Step-by-Step Analysis

   Let's evaluate each option based on this understanding:

   1. Chilled calcium chloride: Calcium chloride is a salt, and $Ca^{2+}$ ions can neutralize the negative charges on DNA. While a salt is necessary for DNA precipitation, calcium chloride alone is not the primary precipitating agent in the way alcohol is. The main function of the salt is to provide cations to neutralize the DNA's charge, allowing it to aggregate once dehydrated by alcohol. Without alcohol, DNA would remain soluble in a salt solution. …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
   Q.Which one of the following is a correct part of methodology used in plant DNA isolation? (A) Digestion of tissues with chitinase (B) Dissolution histones with detergents (C) Addition of ethyl alcohol to precipitate the genetic material having uracil (D) Removing polymer of amino acids with proteases

   ›Reveal solutionSolution

   Plant DNA isolation requires breaking down protein-DNA complexes and removing contaminating proteins. Proteases digest the histone and non-histone proteins that package DNA, leaving purified nucleic acid. The answer is (D).

   DNA in living cells doesn't float freely—it exists tightly wound around histone proteins and associated with many other proteins. To isolate pure DNA, you need to systematically dismantle this protein scaffold while keeping the DNA intact. The methodology hinges on understanding what you're removing at each step.

   Let's examine each option against the actual biochemistry of plant DNA isolation:

   1. Option A claims chitinase digests plant tissues.

      Chitin is the structural polysaccharide of fungal cell walls and arthropod exoskeletons, not plants. Plant cell walls are made of cellulose, hemicellulose, and pectin. To break down plant tissue, you would use cellulase or pectinase, or simply mechanical grinding in liquid nitrogen. Chitinase has no role here.

   2. Option B suggests detergents dissolve histones.

      Detergents (like SDS or CTAB) do play a critical role in DNA isolation—they lyse membranes by disrupting lipid bilayers and denature some proteins. However, histones are not "dissolved" by detergents in the sense of being removed from DNA. Histones are proteins, and while detergents help denature them, the actual removal of histones from DNA requires proteolytic digestion. The phrasing here conflates membrane lysis with protein removal.

   3. Option C mentions precipitating genetic material containing uracil with ethanol.

      Ethanol (or isopropanol) does precipitate DNA by neutralizing its phosphate backbone and making it insoluble. However, DNA contains thymine, not uracil. Uracil is found in RNA. While ethanol precipitates both DNA and RNA, calling DNA "genetic material having uracil" is biochemically incorrect. This option contains a fundamental error about nucleic acid composition.

   4. Option D states proteases remove the polymer of amino acids. …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
    Q.Assertion (A): Chromogenic substrates give blue coloured colonies if the plasmid does not have an insert. Reason (R): Recombinant colonies produce colour in presence of the chromogenic substrates due to the presence of $\beta$ galactosidase. The correct option among the following is: (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

    ›Reveal solutionSolution

    The assertion is true (blue colonies = no insert) but the reason is false (recombinant colonies do NOT produce colour; it’s the non‑recombinants that do). So (A) is true, (R) is false → option (C).

    Concept & Intuition

    This question tests the classic blue‑white screening used in molecular cloning. The key idea: a plasmid carries the lacZ gene (coding for β‑galactosidase). When you insert foreign DNA into the multiple cloning site (MCS) inside lacZ, you disrupt the gene — this is called insertional inactivation.

    • If the plasmid lacks an insert, lacZ is intact, β‑galactosidase is produced, and it cleaves the chromogenic substrate X‑gal to give a blue colony.
    • If the plasmid has an insert, lacZ is broken, no functional β‑galactosidase is made, and the colony remains white (or colourless).

    So the Assertion (A) is correct: chromogenic substrates give blue colonies when the plasmid has no insert.

    But the Reason (R) says “recombinant colonies produce colour … due to β‑galactosidase” — that’s backwards. Recombinants (with insert) do not produce colour; it’s the non‑recombinants that turn blue. Hence (R) is false.

    Step‑by‑step reasoning

    1. Understand the system

       The plasmid contains the lacZ gene, which encodes β‑galactosidase. The chromogenic substrate X‑gal (5‑bromo‑4‑chloro‑3‑indolyl‑β‑D‑galactopyranoside) is colourless until cleaved by β‑galactosidase, releasing an indigo dye → blue colour.

    2. What happens without an insert?

       The lacZ gene is intact → functional β‑galactosidase is produced → X‑gal is cleaved → blue colonies. This matches Assertion (A).

    3. What happens with an insert (recombinant plasmid)? …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
    Q.Assertion (A): Bacteriophages have the ability to replicate within bacterial cells Reason (R): Bacterial plasmids have very high copy numbers of their genome in the bacterial cells The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

    ›Reveal solutionSolution

    Bacteriophages really do replicate inside bacterial cells (A true), and some plasmids really do reach high copy numbers (R true) — but they are wholly independent entities, so the plasmid's copy number cannot explain the phage's replication. Option (B).

    The concept first: two very different pieces of DNA in a bacterium

    When we study cloning vectors, we meet two things that multiply inside a bacterial cell, and it is easy to blur them. They are not related.

    1. Bacteriophage — a virus.

    • It is an external agent, not part of the bacterium.
    • Its life cycle: adsorb to the cell wall $\rightarrow$ inject its nucleic acid $\rightarrow$ commandeer the host's polymerases, ribosomes and nucleotide pool $\rightarrow$ make many copies of its genome and coat proteins $\rightarrow$ assemble progeny $\rightarrow$ lyse the cell (lytic cycle), or integrate as a prophage (lysogenic cycle).
    • Because it takes over the whole cell, a phage can reach several copies per cell — in fact hundreds — which is exactly why phage vectors give high yields of cloned DNA.

    2. Plasmid — the bacterium's own accessory DNA.

    • Extra-chromosomal, small, circular, double-stranded DNA.
    • It replicates autonomously using its own origin of replication (ori) — independently of the bacterial chromosome.
    • Copy number varies with the plasmid: some have only one or two copies per cell, others 15–100 copies.
    • It confers accessory traits (antibiotic resistance, toxin genes) but is not essential to the cell.

    Step-by-step

    1. Assertion: "Bacteriophages have the ability to replicate within bacterial cells." That is the definition of a phage. A is TRUE.
    2. Reason: "Bacterial plasmids have very high copy numbers of their genome in the bacterial cells." Many plasmids do — up to 15–100 copies per cell. Taken as a general statement about high-copy plasmids, R is TRUE. …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
    Q.Identify the correct statements. (A) In gel electrophoresis, DNA fragments resolve according to their size. (B) It is not necessary to digest the bacterial cells with lysozyme to isolate DNA (C) In PCR, DNA polymerase extends the primers using the nucleotides. (D) During the isolation of DNA, ribonuclease is added along with protease (E) Agarose gel electrophoresis is employed to check the progression of restriction enzyme digestion. (A) B, C, D, E (B) A, B, C, D (C) A, C, D, E (D) A, B, D, E

    ›Reveal solutionSolution

    The key idea is to evaluate each statement about molecular biology techniques (gel electrophoresis, DNA isolation, PCR) for factual correctness. The correct set is A, C, D, E, which corresponds to option (C).

    Let’s go through each statement one by one, with the reasoning rooted in standard molecular biology protocols.

    1. Statement (A): "In gel electrophoresis, DNA fragments resolve according to their size."

       This is correct. Agarose gel electrophoresis separates DNA fragments based on their length (size). The gel acts as a molecular sieve: smaller fragments move faster through the pores, while larger ones lag behind. The result is a ladder-like pattern where position correlates with size.

    2. Statement (B): "It is not necessary to digest the bacterial cells with lysozyme to isolate DNA."

       This is false. For bacterial cells, lysozyme is almost always required to break down the peptidoglycan cell wall. Without it, the cell wall remains intact and prevents efficient lysis and DNA release. In standard DNA isolation protocols (e.g., from E. coli), lysozyme treatment is a key early step.

    3. Statement (C): "In PCR, DNA polymerase extends the primers using the nucleotides."

       This is correct. During the extension step of PCR, a thermostable DNA polymerase (like Taq polymerase) adds deoxynucleotides (dNTPs) to the 3' end of each primer, synthesizing the complementary strand. The primers provide the starting point, and the polymerase does the elongation.

    4. Statement (D): "During the isolation of DNA, ribonuclease is added along with protease."

       This is correct. In DNA purification, RNase (ribonuclease) is added to digest RNA, and protease (e.g., proteinase K) is added to digest proteins. Both are used together to remove contaminants and obtain clean DNA. The order can vary, but they are often added in the same step. …

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