Q.What is heterogamety ? Explain the mechanism of sex determination in Drosophila.

Showing the 12 most recent of 34 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Consider the following statements Assertion (A): The RNAi can be introduced in an organism by insertion of gene encoding ssRNA only. Reason (R): RNAi takes place in all eukaryotic organisms as a method of cellular defence The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   RNA interference (RNAi) can be triggered by dsRNA, not just ssRNA, so Assertion (A) is false; RNAi does occur in many eukaryotes as a defence mechanism, so Reason (R) is true. The correct answer is (D).

   The question tests your understanding of RNA interference (RNAi), a fascinating gene-silencing mechanism. Let's break down the biology behind each statement.

   RNAi is a process where small RNA molecules (like siRNA or miRNA) bind to complementary mRNA, preventing its translation or marking it for destruction. This is a powerful tool for regulating gene expression and defending against viruses and transposons.

   The key trigger for RNAi is double-stranded RNA (dsRNA). When a cell encounters dsRNA, an enzyme called Dicer chops it into small fragments (siRNAs). These siRNAs then guide the RISC complex to complementary mRNA, silencing the gene. Single-stranded RNA (ssRNA) alone does not efficiently trigger this pathway — it needs to be double-stranded to be recognized by Dicer.

   Now, let's evaluate each statement.

   1. Assertion (A): "The RNAi can be introduced in an organism by insertion of gene encoding ssRNA only."

      This is false. To induce RNAi, you typically introduce a gene that produces dsRNA (often by designing an inverted repeat that forms a hairpin). A gene encoding only ssRNA would not produce the double-stranded trigger required for the RNAi machinery to act. While some ssRNA can form secondary structures that mimic dsRNA, the statement says "ssRNA only," which is misleading and incorrect in the standard sense.

   2. Reason (R): "RNAi takes place in all eukaryotic organisms as a method of cellular defence." …

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2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A DNA is having guanines and adenines in 2 : 1 ratio. Adenines are bonded with thymine by 80 hydrogen bonds. If this DNA transcribes information into m-RNA, how many t-RNA's are required to translate that. (A) 36 (B) 41 (C) 42 (D) 39

   ›Reveal solutionSolution

   The key is to use the hydrogen-bond count to find the number of adenine–thymine pairs, then the G:A ratio to find the total base pairs, and finally the number of codons in the mRNA to determine the number of tRNAs required. The answer is 39.

   The problem ties together DNA structure, transcription, and translation. You need to work backwards from the given hydrogen bonds to the number of base pairs, then to the number of codons in the mRNA, and finally to the number of tRNAs needed. Each tRNA brings one amino acid, and each amino acid corresponds to one codon (three bases) on the mRNA. The tricky part is that the stop codon does not code for any amino acid, so no tRNA binds to it.

   Let’s break it down step by step.

   1. Understand the hydrogen bonding in DNA. In a DNA double helix, adenine (A) pairs with thymine (T) via 2 hydrogen bonds, and guanine (G) pairs with cytosine (C) via 3 hydrogen bonds. The problem says adenines are bonded with thymine by 80 hydrogen bonds. Since each A–T pair contributes 2 bonds, the number of A–T pairs is:

   $\frac{80}{2} = 40$

   So there are 40 adenine bases and 40 thymine bases in the DNA.

   2. Use the given ratio of guanines to adenines. The ratio of guanines (G) to adenines (A) is 2:1. Since A = 40, we have:

   $\frac{G}{A} = \frac{2}{1} \quad \Rightarrow \quad G = 2 \times 40 = 80$

   So there are 80 guanine bases. Because G pairs with C, there are also 80 cytosine bases.

   3. Find the total number of base pairs in the DNA. Each base pair consists of one purine and one pyrimidine. The total number of base pairs is the sum of A–T pairs and G–C pairs:

   $40 + 80 = 120$

   So the DNA has 120 base pairs.

   4. Transcription produces an mRNA with the same number of bases as one strand of the DNA.

      During transcription, only one strand of DNA (the template strand) is used to synthesize mRNA. The mRNA will have a sequence complementary to that template strand, but the number of bases in the mRNA equals the number of bases in the DNA strand (which is the same as the number of base pairs). So the mRNA has 120 bases.

   5. Translate the mRNA into a protein: codons and tRNAs. …

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3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Damage to cerebellum of brain causes (A) Coma (B) Excessive hunger (C) Loss of speech (D) Loss of balance

   ›Reveal solutionSolution

   The cerebellum is vital for coordinating voluntary movements, maintaining posture, and ensuring balance. Damage to it primarily results in a loss of balance and coordination.

   The brain is an incredibly complex organ, and different regions are specialized for different functions. To understand the effect of damage to the cerebellum, we first need to understand its normal role.

   Concept and Intuition

   The cerebellum, Latin for "little brain," is located at the back of the brain, beneath the occipital and temporal lobes of the cerebrum, and posterior to the brainstem. Despite its relatively small size (about 10% of the brain's volume), it contains over half of the brain's neurons.

   Its primary functions are:

   • Coordination of voluntary movements: It fine-tunes motor activity, ensuring movements are smooth, precise, and well-timed. For example, reaching for an object or walking in a straight line.
   • Maintenance of posture and balance: The cerebellum receives sensory input from the vestibular system (inner ear, responsible for head position and movement) and proprioceptors (sensors in muscles and joints that tell the brain about body position). It integrates this information to adjust muscle activity, allowing us to stand upright and maintain equilibrium.
   • Motor learning: It plays a role in adapting and refining motor skills through practice, like learning to ride a bicycle or play a musical instrument.

   When the cerebellum is damaged, these functions are impaired. The specific symptoms depend on the extent and location of the damage, but they generally relate to problems with coordination and balance.

   Step-by-step analysis of the options:

   1. Analyze option (A) Coma:

      • A coma is a state of prolonged unconsciousness where a person is unresponsive to their environment.
      • This condition is typically caused by widespread damage to the cerebral hemispheres or, more commonly, to the brainstem, which contains the reticular activating system crucial for arousal and consciousness.
      • While severe brain trauma can affect multiple areas, damage primarily limited to the cerebellum is not the direct cause of a coma.

   2. Analyze option (B) Excessive hunger:

      • The regulation of hunger, satiety (feeling full), and other basic drives like thirst and body temperature is primarily controlled by the hypothalamus, a small but vital structure located deep within the brain, below the thalamus.
      • Damage to the cerebellum does not typically lead to disorders of appetite or hunger.

   3. Analyze option (C) Loss of speech: …

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4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Trisomy of $23^{\text{rd}}$ chromosome in male results in (A) Turner syndrome (B) Down syndrome (C) Edward syndrome (D) Klinefelter syndrome

   ›Reveal solutionSolution

   Trisomy of the 23rd chromosome in a male means an extra sex chromosome (XXY), which causes Klinefelter syndrome. The correct option is (D).

   The key here is understanding what "trisomy of the 23rd chromosome" means. Humans have 23 pairs of chromosomes — 22 pairs of autosomes and 1 pair of sex chromosomes (the 23rd pair). In a normal male, the sex chromosome pair is XY. Trisomy means having three copies of a particular chromosome instead of the usual two. So trisomy of the 23rd chromosome in a male means the sex chromosome constitution is XXY — two X chromosomes and one Y.

   Now, which syndrome does XXY produce? That is Klinefelter syndrome. Let’s walk through the options to see why the others are wrong.

   1. Turner syndrome results from monosomy of the X chromosome in females — a single X (45,X). That is a missing sex chromosome, not an extra one. So it cannot be trisomy of the 23rd.

   2. Down syndrome is trisomy of chromosome 21, not the 23rd. It is an autosomal trisomy, not a sex chromosome trisomy. So this is a common confusion — students sometimes mix up the chromosome numbers.

   3. Edward syndrome is trisomy of chromosome 18, another autosomal trisomy. Again, not the 23rd. …

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5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.If the blood groups of parents are homozygous for A and homozygous for B blood groups, these blood groups are not expected in their children (A) A, AB, O (B) A, B, O (C) B, AB, O (D) A, B, AB

   ›Reveal solutionSolution

   When one parent is homozygous A (genotype $I^A I^A$) and the other is homozygous B ($I^B I^B$), all children inherit one $I^A$ and one $I^B$ allele, giving only blood group AB. So groups A, B, and O are not expected.

   The key here is understanding how blood groups are inherited. The ABO system is controlled by a single gene with three alleles: $I^A$, $I^B$, and $i$. Both $I^A$ and $I^B$ are dominant over $i$, but they are codominant with each other — meaning if both are present, both are expressed, giving the AB blood group.

   When a parent is homozygous for a blood group, it means they carry two copies of the same allele. So a homozygous A parent has genotype $I^A I^A$, and a homozygous B parent has genotype $I^B I^B$.

   Now let’s work through what happens in their children.

   1. Gamete formation

      The $I^A I^A$ parent can only produce gametes carrying the $I^A$ allele.

      The $I^B I^B$ parent can only produce gametes carrying the $I^B$ allele.

   2. Fertilization

      Every child receives one allele from each parent. So every child gets exactly one $I^A$ and one $I^B$ — the only possible genotype is $I^A I^B$.

   3. Resulting blood group

      Since $I^A$ and $I^B$ are codominant, the $I^A I^B$ genotype expresses as blood group AB.

      No other combination is possible.

   4. Which groups are not expected?

      The children cannot have: …

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6. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.In several fungi and plants, the bisexual condition is denoted by (A) Homothallic and monoecious (B) Heterothallic and monoecious (C) Homothallic and dioecious (D) Heterothallic and dioecious

   ›Reveal solutionSolution

   The question asks for the terms that describe the bisexual condition in fungi and plants. In fungi, “homothallic” means both sexes on the same thallus (bisexual); in plants, “monoecious” means male and female flowers on the same plant (bisexual). So the correct pairing is homothallic and monoecious.

   The key here is to match the biological terminology to the concept of bisexuality — having both male and female reproductive structures in the same organism. Different kingdoms use different words for the same idea, and this question tests your ability to translate across them.

   1. In fungi, the term “thallus” refers to the body of the fungus. If a single thallus produces both male and female gametes (or gametangia), it is called homothallic (“same thallus”). If two different thalli are needed — one male, one female — it is heterothallic. So the bisexual condition in fungi is homothallic.

   2. In plants, the term “monoecious” comes from Greek monos (single) and oikos (house) — meaning “one house.” A monoecious plant has separate male and female flowers on the same individual (e.g., maize, cucumber). That is the bisexual condition at the organism level. “Dioecious” (“two houses”) means male and female flowers are on separate plants (e.g., papaya, date palm), which is unisexual at the organism level. …

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7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Consider the following statements Assertion (A): In human beings, Y-linked genes of non homologous part are called holandric genes. Reason (R): Y-chromosome is present in males only The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   Holandric genes are Y-linked genes found only in males since the Y chromosome is male-specific; the reason correctly explains why these genes show this pattern of inheritance. The correct option is (A).

   Understanding Sex Chromosomes and Gene Inheritance

   In human genetics, sex chromosomes (X and Y) determine biological sex and carry genes beyond just sex determination. The X and Y chromosomes have both homologous regions (where they can pair during meiosis) and non-homologous regions (unique to each chromosome).

   The key concept here involves understanding what happens with genes located in the non-homologous portion of the Y chromosome—the part that has no corresponding region on the X chromosome.

   Evaluating the Assertion

   Assertion (A): "In human beings, Y-linked genes of non homologous part are called holandric genes."

   Let's break this down:

   1. Holandric genes are indeed genes located on the Y chromosome, specifically in the non-homologous region (the region unique to Y that doesn't pair with X).

   2. These genes are passed exclusively from father to son, creating a strict patrilineal inheritance pattern.

   3. Examples include genes for testis development (SRY gene), some spermatogenesis factors, and certain genes controlling male-specific traits.

   This assertion is TRUE.

   Evaluating the Reason

   Reason (R): "Y-chromosome is present in males only"

   1. In humans, the typical sex chromosome composition is:

      • Males: XY (one X, one Y)
      • Females: XX (two X chromosomes, no Y)

   2. Therefore, the Y chromosome is indeed present only in males (barring rare chromosomal abnormalities).

   This reason is TRUE.

   Determining the Relationship

   Now the critical question: Does (R) correctly explain (A)? …

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8. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Study the following and choose the correct statements: I. Haemophilia-A and Haemophilia-B are X-linked disorders due to recessive genes II. Sickle cell anaemia is caused by a point mutation where valine at $6^{\text{th}}$ position is replaced by glutamic acid III. Protonopia is red colour blindness IV. Phenyl ketonuria is an allosomal metabolic genetic disorder (A) I, III (B) II, IV (C) I, II (D) III, IV

   ›Reveal solutionSolution

   The key idea is to recall the genetic basis and inheritance patterns of each disorder. Only statements I and III 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: Haemophilia-A and Haemophilia-B are X-linked disorders due to recessive genes

      Haemophilia-A (deficiency of factor VIII) and Haemophilia-B (deficiency of factor IX) are both caused by recessive mutations on the X chromosome. Males (XY) are affected if they inherit one defective X; females (XX) need two defective copies to show the disease. This is textbook X-linked recessive inheritance. So statement I is correct.

   2. Statement II: Sickle cell anaemia is caused by a point mutation where valine at $6^{\text{th}}$ position is replaced by glutamic acid

      This is a classic trap. In sickle cell anaemia, the mutation is in the $\beta$-globin gene: a single base change (GAG to GTG) leads to glutamic acid being replaced by valine at the $6^{\text{th}}$ position of the $\beta$-chain — exactly the opposite of what the statement says. The statement has the amino acids swapped. So statement II is incorrect.

      Watch out

      A common mistake is to reverse the amino acids. Remember: in sickle cell anaemia, glutamic acid (hydrophilic) is replaced by valine (hydrophobic), not the other way around.

   3. Statement III: Protonopia is red colour blindness …

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9. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Transduction was discovered by 'X' in 'Y'. Identify 'X' and 'Y' respectively? (A) X – Lederberg and Tatum, Y – E. coli (B) X – Lederberg and Zinder, Y – S. typhimurium (C) X – Lederberg and Ehrenberg, Y – S. Pneumoniae (D) X – Griffith and Zinder, Y – C. tetani

   ›Reveal solutionSolution

   Transduction was discovered by Lederberg and Zinder in Salmonella typhimurium. The correct option is (B).

   The question tests a specific fact from the history of molecular genetics: who discovered transduction and in which organism. This is a classic memory-based point, but understanding the context helps you lock it in without confusion.

   Transduction is the process by which bacterial DNA is transferred from one bacterium to another via a bacteriophage (a virus that infects bacteria). It was discovered in 1952 by Norton Zinder and Joshua Lederberg while they were studying genetic recombination in Salmonella typhimurium. They initially thought they were observing conjugation (like Lederberg and Tatum had found in E. coli), but experiments showed that the transfer was mediated by a filterable agent — later identified as a phage — and thus transduction was born.

   Let’s walk through the options:

   1. Option (A) – Lederberg and Tatum, E. coli: This pair discovered conjugation (not transduction) in E. coli in 1946. So this is wrong for transduction.

   2. Option (B) – Lederberg and Zinder, S. typhimurium: This is correct. Zinder was a graduate student in Lederberg’s lab, and together they demonstrated transduction in Salmonella typhimurium.

   3. Option (C) – Lederberg and Ehrenberg, S. Pneumoniae: Ehrenberg is not associated with this discovery. S. pneumoniae was used by Griffith (transformation) and later by Avery, MacLeod, and McCarty. So this is incorrect.

   4. Option (D) – Griffith and Zinder, C. tetani: Griffith discovered transformation (not transduction) in Streptococcus pneumoniae. Zinder worked on transduction, but not with C. tetani. So this is wrong. …

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10. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.Consider the following statements Assertion (A): In hymenopteran insects, sperms are produced by mitosis Reason (R): Male hymenopteran insects are haploid The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

    ›Reveal solutionSolution

    In hymenopteran insects (ants, bees, wasps), males are haploid and produce sperm by mitosis, not meiosis. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

    The key here is understanding haploidy and sex determination in hymenopterans. In these insects, females develop from fertilized (diploid) eggs, while males develop from unfertilized (haploid) eggs through a process called arrhenotokous parthenogenesis. This means males have only one set of chromosomes — they are haploid.

    Now, think about what happens when a haploid organism needs to produce gametes. Normal meiosis would halve the chromosome number, but a haploid cell cannot undergo reduction division — it would produce inviable gametes with zero chromosomes. So nature has a clever workaround: male hymenopterans produce sperm by mitosis instead. Each sperm is an exact copy of the male's haploid genome, which is exactly what's needed to fertilize an egg and restore diploidy in the female offspring.

    Let's break it down step by step.

    1. Why the Reason (R) is true

       Male hymenopterans (drones in honeybees, for example) develop from unfertilized eggs. Since no sperm contributed genetic material, the egg's nucleus alone forms the embryo — it is haploid ($n$). This is a well-established biological fact.

    2. Why the Assertion (A) is true

       Because the male is haploid, it cannot undergo meiosis to produce sperm — meiosis would reduce the chromosome number to $n/2$, which is not viable. Instead, the cells in the testes divide by mitosis, producing haploid sperm that are genetically identical to the male parent. This is a direct adaptation to haploidy.

    3. How (R) explains (A) …

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11. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.The following type of ribosome sub-units are present in an eukaryotic cell (A) 30S, 50S only (B) 40S, 60S only (C) 50S, 60S only (D) 30S, 40S, 50S and 60S

    ›Reveal solutionSolution

    A eukaryotic cell contains both 80S (40S+60S) and organellar 70S (30S+50S) ribosomes, so all of 30S, 40S, 50S, 60S occur — option (D).

    Concept: Ribosome types by sedimentation coefficient:

    • 80S ribosomes in the cytoplasm of eukaryotes = 60S + 40S subunits.
    • 70S ribosomes in prokaryotes and inside eukaryotic mitochondria and chloroplasts = 50S + 30S subunits. …

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12. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.Match the following

    List – I		List – II
    A. $\mathrm{p^{BR322}}$	I	To join DNA fragments
    B. T – DNA	II	Plasmid with ‘cos’ site
    C. COSMID	III	E. Coli
    D. DNA Ligase	IV	Agrobacterium tumifaciens

    The correct answer is (A) A-III, B-IV, C-I, D-II (B) A-III, B-IV, C-II, D-I (C) A-III, B-I, C-II, D-IV (D) A-III, B-II, C-I, D-IV

    ›Reveal solutionSolution

    pBR322 is an E. coli plasmid, T-DNA comes from Agrobacterium tumefaciens, a cosmid is a plasmid carrying a phage cos site, and DNA ligase is the enzyme that joins DNA fragments — giving A–III, B–IV, C–II, D–I, i.e. option (B).

    The concept first: the toolkit of genetic engineering

    Recombinant DNA work needs three kinds of tool, and this question samples all three.

    1. Vectors (the vehicles).

    • pBR322 — the first widely used artificial cloning vector, built from plasmid DNA of Escherichia coli. It carries an ori, two selectable antibiotic-resistance genes ($amp^R$ and $tet^R$) and unique restriction sites (PstI in $amp^R$, BamHI and SalI in $tet^R$) that allow insertional inactivation as a screen.
    • Cosmid — a hybrid of a plasmid and phage $\lambda$. It is simply a plasmid that has been given the $\lambda$ 'cos' (cohesive-end) site. The cos site lets the recombinant DNA be packaged inside a phage coat, so a cosmid can carry much larger inserts than an ordinary plasmid while still replicating as a plasmid inside the host.
    • Ti plasmid / T-DNA — Agrobacterium tumefaciens is a soil bacterium that naturally infects dicots and transfers a piece of its tumour-inducing (Ti) plasmid — the T-DNA — into the plant genome, causing crown gall. Genetic engineers disarm the tumour genes and use the T-DNA as the delivery vehicle for a gene of interest. So T-DNA's home is Agrobacterium.

    2. Enzymes (the tools).

    • Restriction endonucleases are the "molecular scissors" — they cut. …

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