Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 21: Which of the following is not a stop codon?

A. UAG
B. UAA
C. UGA
D. UGG (Correct Answer)

Explanation: ### Explanation In the genetic code, **stop codons** (also known as nonsense codons) are specific nucleotide triplets within messenger RNA (mRNA) that signal the termination of translation. When a ribosome encounters a stop codon, no corresponding transfer RNA (tRNA) binds; instead, **release factors** bind to the ribosome, causing the newly synthesized polypeptide chain to be released. **Why UGG is the correct answer:** **UGG** is not a stop codon; it is the specific codon that codes for the amino acid **Tryptophan**. Tryptophan and Methionine (AUG) are unique because they are the only two amino acids encoded by a single codon each. **Analysis of incorrect options:** There are three standard stop codons in the universal genetic code: * **UAG (Amber):** A stop codon that signals the end of translation. * **UAA (Ochre):** The most common stop codon used in bacteria. * **UGA (Opal):** A stop codon in the universal code (though it notably codes for Selenocysteine in specific contexts). **High-Yield Clinical Pearls for NEET-PG:** * **Mnemonic:** To remember the stop codons, use: **U** **A**re **G**one (UAG), **U** **A**re **A**way (UAA), and **U** **G**o **A**way (UGA). * **Nonsense Mutation:** A point mutation that changes a functional codon into a stop codon, leading to a truncated, usually non-functional protein (e.g., in some forms of β-thalassemia). * **Mitochondrial Exception:** In human mitochondria, the genetic code differs slightly; **UGA** codes for Tryptophan instead of acting as a stop codon, and **AGA/AGG** act as stop codons instead of coding for Arginine. * **21st Amino Acid:** **Selenocysteine** is often referred to as the 21st amino acid and is encoded by the UGA codon when a specific insertion sequence (SECIS) is present.

Question 22: A cell is placed in a medium containing radioactively labelled thymidine. After the cells undergo replication 3 times, what percentage of the cells will have both strands of DNA labelled?

A. 25%
B. 50%
C. 75% (Correct Answer)
D. 100%

Explanation: ### Explanation **Concept: Semiconservative Mode of DNA Replication** According to the semiconservative model (demonstrated by Meselson and Stahl), each strand of the parent DNA molecule acts as a template for the synthesis of a new complementary strand. When a cell replicates in a medium containing **radioactively labelled thymidine**, every *newly synthesized* strand will be radioactive. **Step-by-Step Calculation:** 1. **Initial State:** 1 cell with 2 non-labeled (cold) strands. 2. **1st Generation (2 cells):** Each cell gets one old strand and one new labeled strand. (0% fully labeled). 3. **2nd Generation (4 cells):** The 2 labeled strands from the 1st generation act as templates for 2 new labeled strands (creating 2 fully labeled cells). The 2 old "cold" strands act as templates for 2 new labeled strands (creating 2 hybrid cells). (50% fully labeled). 4. **3rd Generation (8 cells):** * The 4 "cold" strands (from the original parent) will always remain in hybrid cells (1 labeled + 1 unlabeled). * The remaining 12 strands in the pool are all labeled. * Total cells = $2^3 = 8$. * Cells with one "cold" strand (Hybrid) = 2. * Cells with both strands labeled = $8 - 2 = 6$. * **Percentage:** $(6/8) \times 100 = \mathbf{75\%}$. **Analysis of Incorrect Options:** * **A (25%):** This would be the result if we were looking for the percentage of "hybrid" cells after 4 generations. * **B (50%):** This is the percentage of fully labeled cells after only 2 generations. * **D (100%):** This is impossible in a semiconservative model as long as the original unlabeled parent strands persist in the population. **NEET-PG High-Yield Pearls:** * **Meselson-Stahl Experiment:** Used $N^{15}$ (heavy isotope) and $N^{14}$ to prove semiconservative replication. * **Taylor’s Experiment:** Used tritiated thymidine in *Vicia faba* to prove semiconservative replication at the chromosome level. * **Rule of Thumb:** After '$n$' generations, the number of hybrid cells (one old strand) always remains **2**, while the total cells are $2^n$. All other cells ($2^n - 2$) will be fully labeled.

Question 23: Which of the following formulas represents Chargaff's rule?

A. A + G = T + C
B. A/T = G/C (Correct Answer)
C. A = U = T = G = C
D. A + T = G + C

Explanation: **Explanation:** **Chargaff’s Rules** are fundamental principles of DNA structure established by Erwin Chargaff. The rule states that in double-stranded DNA (dsDNA), the amount of purines is always equal to the amount of pyrimidines. Specifically, Adenine (A) pairs with Thymine (T), and Guanine (G) pairs with Cytosine (C). 1. **Why Option B is Correct:** Since A pairs with T and G pairs with C, their molar ratios are equal (A=T and G=C). Therefore, **A/T = 1** and **G/C = 1**. Mathematically, this means **A/T = G/C**. 2. **Why Option A is Correct (Alternative Form):** While Option B is mathematically correct, Option A (**A + G = T + C**) is the most common representation of the rule (Purines = Pyrimidines). In many competitive contexts, both are valid; however, Option B specifically highlights the 1:1 stoichiometric ratio of base pairing. 3. **Why Options C and D are Incorrect:** * **Option C:** Bases are not present in equal proportions (A does not equal G). * **Option D:** The sum of A+T does not necessarily equal G+C. This ratio (A+T/G+C) varies between species and determines the DNA's melting temperature (Tm). **High-Yield Facts for NEET-PG:** * **Applicability:** Chargaff’s rules apply **only to double-stranded DNA**. They do not apply to single-stranded DNA (ssDNA) or RNA (where Uracil replaces Thymine). * **Base Pairing:** A-T pairs have **2 hydrogen bonds**, while G-C pairs have **3 hydrogen bonds**. * **Clinical Correlation:** DNA with high G-C content is more stable and has a higher **Melting Temperature (Tm)** due to the extra hydrogen bond and stronger base-stacking interactions. * **Calculation Tip:** If a question states DNA contains 20% Adenine, you can conclude it has 20% Thymine, leaving 60% for G-C (meaning 30% Guanine and 30% Cytosine).

Question 24: Which of the following is false about telomerase?

A. Has reverse transcriptase activity.
B. Has an intrinsic primer.
C. It is a ribonucleoprotein.
D. Telomerase leads to premature aging. (Correct Answer)

Explanation: **Explanation:** Telomerase is a specialized enzyme responsible for maintaining the length of telomeres (repetitive DNA sequences at the ends of linear chromosomes). **Why Option D is the correct (false) statement:** Telomerase activity is associated with **immortality and cellular longevity**, not premature aging. It prevents the "end-replication problem" by adding TTAGGG repeats to the 3' end of DNA. **Premature aging syndromes** (like Progeria or Werner syndrome) are actually caused by **shortened telomeres** or a **deficiency/absence** of telomerase activity. When telomeres become critically short, the cell enters senescence (the Hayflick limit). **Analysis of other options:** * **Option A (True):** Telomerase is a specialized **reverse transcriptase** (specifically TERT: Telomerase Reverse Transcriptase). It uses an RNA template to synthesize DNA. * **Option B (True):** It contains an **intrinsic RNA template** (TERC: Telomerase RNA Component) that acts as a guide/primer for the synthesis of telomeric repeats. * **Option C (True):** It is a **ribonucleoprotein** because it is a complex consisting of both a protein catalytic subunit (TERT) and an RNA molecule (TERC). **Clinical Pearls for NEET-PG:** * **Cancer Connection:** Telomerase is highly active in **85-90% of cancer cells**, allowing them to divide indefinitely (immortality). * **Stem Cells:** It is normally active in germ cells, stem cells, and lymphocytes, but absent in most somatic cells. * **Shelterin Complex:** A group of proteins that protects telomeres from being recognized as DNA damage (double-strand breaks).

Question 25: A 10-year-old female patient with a known history of beta-thalassemia complains of tiredness and shortness of breath. Beta-thalassemia is caused by faulty splicing of which of the following?

A. hnRNA (Correct Answer)
B. snRNA
C. scRNA
D. snoRNA

Explanation: **Explanation:** **1. Why hnRNA is Correct:** Beta-thalassemia is a genetic blood disorder characterized by reduced or absent synthesis of the beta-globin chains of hemoglobin. In many cases, the underlying molecular defect is a mutation in the **introns** or at the **splice site junctions** of the beta-globin gene. The primary transcript produced immediately after transcription is **hnRNA (heterogeneous nuclear RNA)**, also known as pre-mRNA. This hnRNA contains both exons (coding) and introns (non-coding). For functional mRNA to be formed, introns must be removed via **splicing**. Mutations in beta-thalassemia often create "cryptic" splice sites, leading to incorrect splicing of the hnRNA. This results in defective mRNA, which is either degraded or translated into non-functional proteins, leading to a deficiency of beta-globin. **2. Why Other Options are Incorrect:** * **snRNA (small nuclear RNA):** These molecules complex with proteins to form **snRNPs** ("snurps"), which are the *machinery* (spliceosome) that performs the splicing. While they facilitate the process, they are not the molecule being spliced. * **scRNA (small cytoplasmic RNA):** These are involved in protein targeting (e.g., Signal Recognition Particle) and are not involved in the splicing of globin genes. * **snoRNA (small nucleolar RNA):** These play a role in the post-transcriptional modification of **rRNA** (ribosomal RNA) within the nucleolus, not mRNA splicing. **Clinical Pearls for NEET-PG:** * **Post-transcriptional modifications** of hnRNA include: 5' Capping, 3' Polyadenylation, and Splicing. * **Splice site mutations** are a classic cause of $\beta^+$-thalassemia (reduced synthesis) or $\beta^0$-thalassemia (total absence). * **High-yield association:** Systemic Lupus Erythematosus (SLE) involves antibodies against snRNPs (**Anti-Smith antibodies**).

Question 26: Which of the following is a true statement about Ribozyme?

A. RNA molecule that acts catalytically to change itself or another RNA molecule. (Correct Answer)
B. Transfer RNA (t-RNA).
C. Messenger RNA (m-RNA).
D. Ribosome.

Explanation: **Explanation:** **1. Why the Correct Answer is Right:** Ribozymes are specialized RNA molecules that possess **catalytic activity**, similar to protein enzymes. Unlike most enzymes which are proteins, ribozymes use their complex three-dimensional folding to catalyze biochemical reactions, most commonly the cleavage or ligation of phosphodiester bonds. They can act on themselves (cis-acting/autocatalytic) or on other RNA molecules (trans-acting). This discovery challenged the "central dogma" that all biological catalysts are proteins. **2. Why the Other Options are Incorrect:** * **B. Transfer RNA (t-RNA):** tRNA acts as an "adapter" molecule that carries specific amino acids to the ribosome during translation. While essential for protein synthesis, it does not possess intrinsic catalytic activity. * **C. Messenger RNA (m-RNA):** mRNA serves as a temporary template that carries genetic information from DNA to the ribosome. It is a substrate for translation, not a catalyst. * **D. Ribosome:** While the ribosome *contains* a ribozyme (the 23S rRNA in prokaryotes or 28S rRNA in eukaryotes), the ribosome as a whole is a large ribonucleoprotein complex (RNA + Protein). The specific catalytic component is the ribozyme, but the term "Ribosome" refers to the entire organelle. **3. High-Yield Clinical Pearls for NEET-PG:** * **Peptidyl Transferase:** The most clinically relevant ribozyme is the **23S rRNA** (prokaryotes) or **28S rRNA** (eukaryotes), which catalyzes peptide bond formation during translation. * **RNase P:** An enzyme responsible for processing tRNA precursors; it is a classic example of a ribozyme. * **Spliceosome:** Small nuclear RNAs (snRNAs) within the spliceosome act as ribozymes to remove introns from pre-mRNA. * **Medical Application:** Ribozymes are being researched as "molecular scissors" for gene therapy to cleave viral RNA (e.g., HIV) or oncogenic mRNA.

Question 27: Which of the following is true about X chromosome inactivation?

A. X gene (Correct Answer)
B. RNA interference
C. Seen in males
D. All of the above

Explanation: **Explanation:** **X-chromosome inactivation (Lyonization)** is a dosage compensation mechanism in females where one of the two X chromosomes is epigenetically silenced to ensure that females, like males, have only one functional copy of X-linked genes. **Why Option A is correct:** The process is mediated by the **X-inactivation center (XIC)**, which contains the **Xist gene** (*X-inactive specific transcript*). This gene does not encode a protein; instead, it produces a large **long non-coding RNA (lncRNA)**. This RNA "coats" the X chromosome from which it is transcribed, triggering heterochromatin formation and silencing. Therefore, the "X gene" (specifically Xist) is the fundamental driver of this process. **Why other options are incorrect:** * **Option B (RNA interference):** While X-inactivation involves non-coding RNA, it is not mediated by the RNA interference (RNAi) pathway (which typically involves siRNA/miRNA and the RISC complex). It is an epigenetic process involving chromatin remodeling (methylation and acetylation). * **Option C (Seen in males):** Males (46,XY) have only one X chromosome, which remains active. X-inactivation only occurs when more than one X chromosome is present (e.g., normal females or Klinefelter syndrome 47,XXY). **Clinical Pearls for NEET-PG:** 1. **Barr Body:** The inactivated X chromosome is visible as a dense mass of heterochromatin against the nuclear membrane, known as a Barr Body. 2. **Formula:** Number of Barr bodies = **(Total X chromosomes - 1)**. 3. **Mosaicism:** Because inactivation is random in each cell during early embryonic life, females are "genetic mosaics" (e.g., explains the presentation of G6PD deficiency or Hemophilia carriers). 4. **Lyon Hypothesis:** States that inactivation is random, fixed, and occurs early in development.

Question 28: Polypeptide chain termination is enhanced by which of the following?

A. Stop codon
B. Peptidyl transferase
C. UAA
D. All of these (Correct Answer)

Explanation: Translation (protein synthesis) termination is a highly regulated process that occurs when a ribosome encounters a termination signal on the mRNA. **Explanation of the Correct Answer:** Termination is "enhanced" or facilitated by the collective action of specific sequences and enzymes: * **Stop Codons (UAA, UAG, UGA):** These are the primary signals for termination. They do not code for any amino acid and are not recognized by tRNA but by **Release Factors (RFs)**. * **UAA (Ochre):** This is one of the three specific stop codons. In many organisms, UAA is the most frequently used termination signal, making it a specific example of a stop codon. * **Peptidyl Transferase:** Traditionally known for forming peptide bonds, during termination, this ribozyme (part of the 28S rRNA in eukaryotes) changes its activity. In the presence of Release Factors, it catalyzes the **hydrolysis** of the bond between the completed polypeptide chain and the tRNA in the P-site, effectively releasing the protein. Since all three components are essential for the efficient release of the polypeptide chain, **Option D** is the correct answer. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Release Factors:** In prokaryotes, RF1 recognizes UAA/UAG, and RF2 recognizes UAA/UGA. In eukaryotes, a single factor, **eRF1**, recognizes all three. * **Nonsense Mutations:** A mutation that creates a premature stop codon (UAA, UAG, or UGA) leads to truncated, often non-functional proteins (e.g., in β-thalassemia). * **Energy Requirement:** Termination is an energy-dependent process requiring **GTP hydrolysis**. * **Mnemonic for Stop Codons:** * **UAA:** **U** Are **A**way * **UAG:** **U** Are **G**one * **UGA:** **U** Go **A**way

Question 29: Which of the following procedures is a routine technique for karyotyping using light microscopy?

A. C-banding
B. G-banding (Correct Answer)
C. Q-banding
D. V-staining

Explanation: **Explanation:** **G-banding (Giemsa banding)** is the gold standard and most common routine technique used in cytogenetics for karyotyping. The process involves treating chromosomes with a protease (typically **trypsin**) followed by staining with **Giemsa stain**. This produces a distinct pattern of dark and light bands: * **Dark bands (G-positive):** Represent AT-rich, gene-poor, heterochromatic regions that replicate late. * **Light bands (G-negative):** Represent GC-rich, gene-dense, euchromatic regions that replicate early. This banding pattern allows for the identification of individual chromosomes and the detection of structural abnormalities like deletions, translocations, or inversions. **Analysis of Incorrect Options:** * **C-banding (Constitutive heterochromatin):** Specifically stains the centromeres and regions containing constitutive heterochromatin (e.g., chromosomes 1, 9, 16, and Y). It is not used for routine whole-genome karyotyping. * **Q-banding (Quinacrine):** The first banding method developed; it uses fluorescent microscopy. Because it requires a fluorescence microscope and the stains fade quickly (photobleaching), it is not the routine choice for light microscopy. * **V-staining:** This is a distractor and is not a recognized standard technique in clinical cytogenetics. **High-Yield Clinical Pearls for NEET-PG:** * **Sample Source:** For routine postnatal karyotyping, **peripheral blood lymphocytes** (stimulated by Phytohemagglutinin) are used. * **Cell Cycle Stage:** Karyotyping is performed during **Metaphase** (when chromosomes are most condensed). * **Resolution:** Standard G-banding identifies changes at a resolution of 5–10 Mb. For smaller microdeletions, FISH or Chromosomal Microarray is preferred. * **R-banding:** The "Reverse" of G-banding; useful for looking at GC-rich regions near telomeres.

Question 30: Actinomycin D interferes with enzyme induction by combining with which of the following?

A. Transfer RNA (tRNA)
B. Deoxyribonucleic acid (DNA) (Correct Answer)
C. Ribosomal RNA (rRNA)
D. Repressor protein

Explanation: ### Explanation **Correct Answer: B. Deoxyribonucleic acid (DNA)** **Mechanism of Action:** Actinomycin D (also known as Dactinomycin) is a potent polypeptide antibiotic that functions as a **transcription inhibitor**. It exerts its effect by binding specifically to **double-stranded DNA**. It intercalates between adjacent **Guanine-Cytosine (G-C) base pairs**, forming a stable complex. This physical obstruction prevents the movement of **RNA Polymerase** along the DNA template, thereby blocking the synthesis of messenger RNA (mRNA). Since enzyme induction requires the transcription of new mRNA to produce specific proteins, Actinomycin D effectively interferes with this process. **Analysis of Incorrect Options:** * **A. Transfer RNA (tRNA):** Actinomycin D does not bind to tRNA. tRNA is involved in the translation phase (carrying amino acids), whereas this drug acts at the transcriptional level. * **C. Ribosomal RNA (rRNA):** While Actinomycin D can inhibit the synthesis of rRNA (especially at low doses in the nucleolus), its primary site of binding is the DNA template itself, not the RNA product. * **D. Repressor protein:** Repressor proteins bind to the operator region of DNA to inhibit transcription naturally (e.g., in the Lac operon). Actinomycin D is an exogenous chemical agent that binds to DNA directly, not a protein-protein regulator. **High-Yield Clinical Pearls for NEET-PG:** * **Clinical Use:** Actinomycin D is a key chemotherapeutic agent used in pediatric oncology, specifically for **Wilms tumor**, **Ewing sarcoma**, and **Rhabdomyosarcoma**. It is also used in **Gestational Trophoblastic Neoplasia (Choriocarcinoma)**. * **Key Distinction:** Unlike **Rifampicin** (which binds directly to bacterial RNA Polymerase), Actinomycin D binds to the **DNA template**. * **Toxicity:** It is a potent vesicant and can cause significant bone marrow suppression and "radiation recall" phenomenon.

Practice by Chapter

DNA Replication and Repair Mechanisms

Practice Questions

Transcription Factors and Gene Regulation

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Epigenetics and DNA Methylation

Practice Questions

RNA Processing and Splicing

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miRNA and RNA Interference

Practice Questions

Protein Synthesis and Post-Translational Modifications

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Genomics and Human Genome Project

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Single Nucleotide Polymorphisms

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Gene Therapy Approaches

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CRISPR-Cas9 and Genome Editing

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DNA Fingerprinting and Forensics

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Molecular Basis of Genetic Diseases

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