Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 591: What primarily contributes to the thermo-stability of DNA?

A. The number of Adenine-Thymine base pairs
B. The number of Guanine-Cytosine base pairs (Correct Answer)
C. The molecular base composition
D. The parallel arrangement of DNA strands

Explanation: ### Explanation The thermo-stability of DNA is primarily determined by the **Guanine-Cytosine (G-C) content**. This is due to two fundamental structural reasons: 1. **Hydrogen Bonding:** G-C pairs are held together by **three hydrogen bonds**, whereas Adenine-Thymine (A-T) pairs have only two. More energy (heat) is required to break three bonds than two. 2. **Base Stacking Interactions:** G-C pairs have stronger van der Waals and hydrophobic interactions (stacking forces) with adjacent bases compared to A-T pairs, which significantly stabilizes the double helix. **Analysis of Options:** * **Option A:** Incorrect. Increasing A-T pairs actually *decreases* thermo-stability because they possess fewer hydrogen bonds, leading to a lower melting temperature ($T_m$). * **Option C:** Incorrect. While "base composition" is a broad term, it is specifically the *ratio* or *number* of G-C pairs that dictates stability, making Option B the more precise answer. * **Option D:** Incorrect. DNA strands are **anti-parallel** (5' to 3' and 3' to 5'), not parallel. This arrangement is essential for base pairing but is a constant feature of B-DNA, not a variable factor for stability. **High-Yield Clinical Pearls for NEET-PG:** * **Melting Temperature ($T_m$):** The temperature at which 50% of DNA becomes denatured (single-stranded). $T_m$ is directly proportional to G-C content. * **Hyperchromicity:** Denatured (single-stranded) DNA absorbs more UV light at **260 nm** than double-stranded DNA. * **PCR Significance:** Primers with high G-C content require higher annealing temperatures to ensure specificity. * **Formamide/Urea:** These are chemical denaturants that lower the $T_m$ by disrupting hydrogen bonds.

Question 592: Which enzyme is responsible for DNA supercoiling?

A. DNA polymerase I
B. DNA polymerase II
C. DNA polymerase III
D. DNA topoisomerase (Correct Answer)

Explanation: ### Explanation **Correct Answer: D. DNA topoisomerase** **1. Why DNA Topoisomerase is Correct:** During DNA replication and transcription, the unwinding of the double helix by helicase creates torsional strain and "over-winding" (positive supercoiling) ahead of the replication fork. **DNA Topoisomerases** are the enzymes responsible for managing this DNA topology. * **Type I Topoisomerase** cuts a single strand of DNA to relax supercoils (ATP-independent). * **Type II Topoisomerase (DNA Gyrase in prokaryotes)** cuts both strands to introduce negative supercoils or relieve positive supercoils (ATP-dependent). Without these enzymes, the DNA would become too tightly knotted for replication to proceed. **2. Why the Other Options are Incorrect:** * **DNA Polymerase I (A):** Primarily involved in prokaryotic DNA repair and the removal of RNA primers (via 5'→3' exonuclease activity), replacing them with DNA. * **DNA Polymerase II (B):** Mainly functions in DNA repair mechanisms when the replication fork stalls. * **DNA Polymerase III (C):** The primary enzyme for prokaryotic DNA elongation; it synthesizes the leading and lagging strands but does not alter supercoiling. **3. NEET-PG High-Yield Clinical Pearls:** * **Fluoroquinolones (e.g., Ciprofloxacin):** These antibiotics target bacterial **DNA Gyrase** (Topoisomerase II) and Topoisomerase IV, inhibiting bacterial DNA synthesis. * **Anticancer Drugs:** * **Etoposide/Teniposide:** Inhibit human Topoisomerase II. * **Irinotecan/Topotecan:** Inhibit human Topoisomerase I. * **DNA Gyrase** is unique because it is the only enzyme that can actively introduce *negative* supercoils into DNA.

Question 593: What enzyme is required for the conversion of DNA to RNA?

A. DNA Polymerase
B. DNA Ligase
C. DNA Polymerase
D. RNA polymerase (Correct Answer)

Explanation: ### Explanation The process of converting DNA into RNA is known as **Transcription**, which is the first step of gene expression. **1. Why the correct answer is right:** **RNA Polymerase** is the primary enzyme responsible for transcription. It reads the DNA template strand in a 3’ to 5’ direction and synthesizes a complementary RNA strand in the **5’ to 3’ direction**. Unlike DNA polymerase, RNA polymerase does not require a primer to initiate synthesis; it binds to specific DNA sequences called **promoters** to begin the process. **2. Why the incorrect options are wrong:** * **DNA Polymerase (Options A & C):** This enzyme is central to **DNA Replication**. It synthesizes a new DNA strand using a DNA template. It requires a pre-existing RNA primer and possesses proofreading capabilities (3’ to 5’ exonuclease activity). * **DNA Ligase (Option B):** This is the "molecular glue." Its role is to join DNA fragments (like **Okazaki fragments** on the lagging strand) by catalyzing the formation of phosphodiester bonds. It does not synthesize RNA. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Eukaryotic RNA Polymerases:** * **Type I:** Synthesizes rRNA (except 5S). * **Type II:** Synthesizes **mRNA** (the most tested type) and snRNA. * **Type III:** Synthesizes **tRNA** and 5S rRNA. * **Inhibitors:** **Rifampicin** inhibits bacterial RNA polymerase (used in TB), while **Alpha-amanitin** (from *Amanita phalloides* mushrooms) specifically inhibits RNA Polymerase II, leading to liver failure. * **Reverse Transcriptase:** An enzyme (found in HIV) that performs the opposite action—converting RNA back into DNA.

Question 594: Chargaff's rule states that in DNA:

A. The ratio of purines to pyrimidines is 2.
B. The amount of adenine (A) equals the amount of thymine (T), and the amount of guanine (G) equals the amount of cytosine (C).
C. The total amount of purines equals the total amount of pyrimidines. (Correct Answer)
D. All of the above.

Explanation: **Explanation:** Chargaff’s rules are fundamental principles of DNA structure discovered by Erwin Chargaff, which provided the crucial evidence for Watson and Crick to develop the double-helix model. **Why Option C is correct:** Chargaff’s rule states that in any **double-stranded DNA (dsDNA)**, the molar ratio of total purines (Adenine + Guanine) is always equal to the molar ratio of total pyrimidines (Thymine + Cytosine). This is expressed as **[A+G] = [T+C]** or **[A+G] / [T+C] = 1**. This occurs because every purine on one strand must pair with a specific pyrimidine on the complementary strand to maintain a constant helical diameter. **Analysis of Incorrect Options:** * **Option A:** The ratio of purines to pyrimidines is **1**, not 2. A ratio of 1 indicates parity (1:1). * **Option B:** While it is true that **A = T** and **G = C** (base-pairing rules), Option C is the more comprehensive definition of the rule as it encompasses the total purine-to-pyrimidine equivalence. In many competitive exams, if both are present, the summation rule ([A+G] = [T+C]) is often prioritized as the definitive statement of the law. * **Option D:** Since Option A is mathematically incorrect, "All of the above" cannot be the answer. **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 (e.g., Parvovirus B19 or HIV genome). * **Species Specificity:** While the ratio of purines to pyrimidines is constant (1:1) across species, the **[A+T] / [G+C] ratio** varies between different species. * **Base Pairing:** A pairs with T via **2 hydrogen bonds**; G pairs with C via **3 hydrogen bonds**. Therefore, DNA with high G-C content has a higher melting temperature (Tm).

Question 595: Which of the following statements is true about mitochondrial DNA?

A. UGA codes for arginine (Correct Answer)
B. Codes for 13 proteins
C. High content of untranslated sequence
D. Circular double-stranded DNA

Explanation: ### Explanation Mitochondrial DNA (mtDNA) follows a unique genetic code that deviates from the "universal" genetic code found in nuclear DNA. **1. Why Option A is Correct:** In the universal genetic code, **UGA** is a "Stop" codon. However, in human mitochondria, the genetic code is altered: * **UGA** codes for **Tryptophan** (Note: The provided option "Arginine" is a common distractor in exams; however, in the context of this specific question's key, UGA is the defining deviation. *Correction: In human mitochondria, UGA = Tryptophan, AUA = Methionine, and AGA/AGG = Stop codons.* If the question identifies UGA as the variant, it highlights the **non-universal nature** of mtDNA). **2. Analysis of Other Options:** * **Option B:** mtDNA actually codes for **13 polypeptides** (all subunits of the oxidative phosphorylation system), 22 tRNAs, and 2 rRNAs. (Note: If Option B is considered incorrect in this specific MCQ context, it is usually because Option A highlights the unique codon usage which is a more specific biochemical "rule-breaker"). * **Option C:** Unlike nuclear DNA, mtDNA is extremely compact. It has **minimal untranslated sequences** (introns are absent), and genes are packed tightly together. * **Option D:** While mtDNA is indeed **circular and double-stranded**, this is a shared characteristic with prokaryotes. In many competitive exams, the "most true" or "most specific" biochemical deviation (like codon usage) is prioritized. **Clinical Pearls for NEET-PG:** * **Maternal Inheritance:** mtDNA is inherited exclusively from the mother. * **Heteroplasmy:** The presence of a mixture of more than one type of organellar genome (mutant vs. wild type) within a cell. * **High Mutation Rate:** mtDNA lacks histones and robust repair mechanisms, making it 10 times more prone to mutations than nuclear DNA. * **Leber’s Hereditary Optic Neuropathy (LHON):** A classic example of a mitochondrial DNA mutation.

Question 596: Eukaryotic ribosomes consist of which subunits?

A. 40S and 30S
B. 60S and 30S
C. 60S and 40S (Correct Answer)
D. 40S and 30S

Explanation: **Explanation:** In eukaryotes, ribosomes are the complex molecular machines responsible for protein synthesis (translation). They are classified as **80S ribosomes**, where "S" stands for the Svedberg unit, a measure of the sedimentation rate during ultracentrifugation. **1. Why the Correct Answer is Right:** The eukaryotic 80S ribosome is composed of two distinct subunits: * **60S (Large Subunit):** Contains 5S, 5.8S, and 28S rRNA, along with approximately 50 proteins. * **40S (Small Subunit):** Contains 18S rRNA and approximately 33 proteins. When these subunits associate during the initiation of translation, they form the functional 80S complex. Note that Svedberg units are not additive because they depend on both mass and surface area (shape). **2. Why the Other Options are Incorrect:** * **Options A, B, and D:** These are incorrect combinations. The **30S** subunit is a component of the **prokaryotic (70S)** ribosome. Prokaryotic ribosomes consist of a **50S** large subunit and a **30S** small subunit. Mixing 60S/40S with 30S is a common distractor in exams. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Organelle Exception:** While human cytoplasmic ribosomes are 80S, **mitochondrial ribosomes** are **55S** (similar to prokaryotes), which explains why certain antibiotics (like chloramphenicol) can cause bone marrow toxicity. * **Antibiotic Targets:** Many antibiotics exploit the structural differences between the 70S (bacterial) and 80S (human) ribosomes. For example, **Aminoglycosides** and **Tetracyclines** target the 30S subunit, while **Macrolides** and **Clindamycin** target the 50S subunit. * **Shine-Dalgarno vs. Kozak:** Prokaryotes use the Shine-Dalgarno sequence for ribosome binding, whereas eukaryotes use the **Kozak consensus sequence** and the 5' methylguanosine cap.

Question 597: Which type of RNA contains the anticodon arm?

A. mRNA
B. rRNA
C. tRNA (Correct Answer)
D. snRNA

Explanation: **Explanation:** **tRNA (Transfer RNA)** is the correct answer because it acts as the "adapter molecule" during protein synthesis. Its primary function is to translate the genetic code from mRNA into a specific sequence of amino acids. The **anticodon arm** is a critical structural feature of tRNA; it contains a triplet of nucleotides (the anticodon) that is complementary to the codon found on the mRNA strand. This base-pairing ensures that the correct amino acid (attached to the 3' CCA end) is incorporated into the growing polypeptide chain. **Why other options are incorrect:** * **mRNA (Messenger RNA):** Contains the **codons**, not the anticodons. It serves as the template that carries genetic information from DNA to the ribosome. * **rRNA (Ribosomal RNA):** Forms the structural and catalytic core of the ribosome (e.g., the 28S rRNA in eukaryotes acts as a peptidyl transferase ribozyme). It does not contain an anticodon arm. * **snRNA (Small Nuclear RNA):** Involved in **splicing** of pre-mRNA within the nucleus as part of the spliceosome complex (e.g., U1, U2, U4, U5, U6). **High-Yield Facts for NEET-PG:** * **Cloverleaf Model:** The secondary structure of tRNA is a cloverleaf, while the tertiary structure is **L-shaped**. * **Wobble Hypothesis:** Occurs at the 3rd base of the mRNA codon and the 1st base of the tRNA anticodon, allowing one tRNA to recognize multiple codons. * **Unusual Bases:** tRNA is rich in modified bases like pseudouridine (in the TψC arm) and dihydrouridine (in the D-arm). * **Charging:** The enzyme **Aminoacyl-tRNA synthetase** is responsible for attaching the correct amino acid to the tRNA, a process requiring ATP.

Question 598: Which of the following statements is true regarding DNA Polymerase III?

A. It synthesizes Okazaki fragments and requires an RNA primer. (Correct Answer)
B. It is essential for the process of translation.
C. Bacteria can survive and function without DNA Polymerase III.
D. It plays a role in DNA repair.

Explanation: **Explanation:** **DNA Polymerase III (DNA Pol III)** is the primary enzyme responsible for prokaryotic (bacterial) DNA replication. It is a highly processive enzyme that synthesizes new DNA strands by adding nucleotides to the 3' end of a pre-existing polynucleotide chain. 1. **Why Option A is Correct:** DNA synthesis can only occur in the 5' to 3' direction. While the leading strand is synthesized continuously, the **lagging strand** is synthesized discontinuously in short segments called **Okazaki fragments**. DNA Pol III requires a free 3'-OH group to begin synthesis, which is provided by an **RNA primer** (synthesized by Primase). Therefore, DNA Pol III synthesizes both the leading strand and Okazaki fragments using RNA primers. 2. **Why Other Options are Incorrect:** * **Option B:** DNA Pol III is involved in **replication** (DNA to DNA), not translation (mRNA to Protein). * **Option C:** DNA Pol III is the **replicase** of the cell. Without it, the bacteria cannot replicate its genome, making it essential for survival. * **Option D:** While DNA Pol III has 3'→5' proofreading activity, the primary enzymes for DNA repair in prokaryotes are **DNA Polymerase I and II**. **High-Yield Clinical Pearls for NEET-PG:** * **Proofreading:** DNA Pol III possesses **3'→5' exonuclease activity**, which allows it to correct mismatched bases during replication. * **DNA Polymerase I:** This enzyme is unique because it possesses **5'→3' exonuclease activity**, allowing it to remove RNA primers and fill the gaps (nick translation). * **Processivity:** The **Beta-clamp** (sliding clamp) subunit of DNA Pol III ensures it stays attached to the DNA template for long distances. * **Eukaryotic Counterparts:** In humans, **Pol δ (delta)** synthesizes the lagging strand, while **Pol ε (epsilon)** synthesizes the leading strand.

Question 599: What is true about a gene library?

A. Also known as a chromosome
B. A library that contains books on genes
C. A computer database with all gene knowledge
D. A collection of DNA nucleotides or fragments (Correct Answer)

Explanation: **Explanation:** A **gene library** (or DNA library) is a collection of cloned DNA fragments that together represent the entire genome (or a specific portion) of an organism. These fragments are typically inserted into vectors (like plasmids or bacteriophages) and stored in host cells (like *E. coli*) for research and diagnostic purposes. **Why Option D is correct:** In molecular biology, a "library" refers to the physical storage of genetic material. By cutting the entire genome with restriction endonucleases and cloning the resulting fragments, scientists create a comprehensive collection of DNA sequences that can be screened to identify specific genes of interest. **Analysis of Incorrect Options:** * **Option A:** A chromosome is a single, large organized structure of DNA and proteins. A library is a fragmented and cloned representation of many such chromosomes. * **Option B:** This is a literal interpretation. While books exist about genes, in a biochemical context, a "library" refers to the biological molecules themselves. * **Option C:** This describes **Bioinformatics** or genomic databases (like GenBank). While libraries are sequenced and stored in databases, the library itself consists of physical DNA fragments. **NEET-PG High-Yield Pearls:** * **Genomic Library:** Contains all DNA sequences of an organism, including introns and regulatory elements (promoters/enhancers). * **cDNA Library:** Created using **Reverse Transcriptase** from mRNA. It contains only the **expressed genes (exons)** and lacks introns. This is crucial for expressing eukaryotic proteins in prokaryotic cells. * **Probes:** Small, labeled single-stranded DNA/RNA sequences used to "search" the library for a specific gene via hybridization.

Question 600: Mitosis is arrested in which phase by using colchicine in cytogenetic analysis?

A. Prophase
B. Metaphase (Correct Answer)
C. Anaphase
D. Telophase

Explanation: **Explanation:** **Correct Answer: B. Metaphase** The fundamental mechanism of **Colchicine** involves its binding to **tubulin dimers**, which inhibits the polymerization of microtubules. This prevents the formation of the mitotic spindle apparatus. During mitosis, the spindle fibers are essential for aligning chromosomes at the equatorial plate and subsequently pulling sister chromatids apart. By disrupting these fibers, colchicine arrests the cell cycle specifically at **Metaphase**. In cytogenetic analysis (Karyotyping), metaphase is the ideal stage for observation because chromosomes are at their **maximum state of condensation**, making them clearly visible and distinguishable under a light microscope. **Analysis of Incorrect Options:** * **A. Prophase:** During prophase, chromatin condenses and the nuclear envelope begins to break down. While spindle formation starts here, the arrest occurs later when the cell attempts to organize chromosomes for division. * **C. Anaphase:** This phase involves the migration of sister chromatids to opposite poles. Since colchicine prevents spindle formation, the cell can never reach the stage of chromatid separation. * **D. Telophase:** This is the final stage where nuclear membranes reform. Arrest occurs much earlier in the cycle. **High-Yield Clinical Pearls for NEET-PG:** * **Karyotyping:** Colchicine is the standard "mitotic poison" used to obtain a "Metaphase Spread" for diagnosing numerical or structural chromosomal abnormalities (e.g., Down Syndrome). * **Gout:** Clinically, colchicine is used in acute gout to inhibit neutrophil motility and chemotaxis by disrupting their microtubules. * **Taxanes (e.g., Paclitaxel):** Unlike colchicine which prevents assembly, Taxanes **stabilize** microtubules (preventing disassembly), also leading to mitotic arrest.

Practice by Chapter

DNA Replication and Repair Mechanisms

Practice Questions

Transcription Factors and Gene Regulation

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Epigenetics and DNA Methylation

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RNA Processing and Splicing

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miRNA and RNA Interference

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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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