Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 671: In gene cloning, which vector can incorporate the largest DNA fragment?

A. Plasmid
B. Bacteriophage
C. Cosmid (Correct Answer)
D. Retrovirus

Explanation: **Explanation:** In molecular biology, the choice of a cloning vector depends primarily on the size of the DNA insert it can accommodate. **Why Cosmid is correct:** A **Cosmid** is a hybrid vector containing a plasmid’s origin of replication and the *cos* sites (cohesive ends) from a lambda bacteriophage. This unique combination allows it to package DNA into phage heads while maintaining the ability to replicate like a plasmid in *E. coli*. Cosmids can carry DNA fragments ranging from **30 to 45 kb**, which is significantly larger than the capacity of standard plasmids or bacteriophages. **Analysis of Incorrect Options:** * **Plasmid (A):** These are small, circular, extra-chromosomal DNA molecules. They are easy to manipulate but have a limited carrying capacity, typically accommodating inserts of **<10 kb**. * **Bacteriophage (B):** Specifically the Lambda ($\lambda$) phage, these vectors can carry fragments of approximately **10 to 25 kb**. Their capacity is limited by the physical space available within the viral capsid. * **Retrovirus (D):** Used primarily in gene therapy for stable integration into host genomes, their payload capacity is generally limited to **<8 kb** to ensure successful packaging into viral particles. **High-Yield Facts for NEET-PG:** * **Vector Capacity Hierarchy (Smallest to Largest):** Plasmid < Bacteriophage < Cosmid < BAC (Bacterial Artificial Chromosome) < YAC (Yeast Artificial Chromosome). * **YACs** have the largest capacity of all, carrying up to **1000–2000 kb**, and were crucial for the Human Genome Project. * **BACs** carry **100–300 kb** and are preferred over YACs for genomic libraries due to better stability. * **Expression Vectors:** Unlike cloning vectors, these are specifically designed for the synthesis of proteins from the inserted gene (e.g., producing recombinant insulin).

Question 672: UV light damage to DNA leads to which of the following?

A. Formation of pyrimidine dimers (Correct Answer)
B. No damage to DNA
C. DNA hydrolysis
D. Double-stranded breaks

Explanation: **Explanation:** **1. Why Option A is correct:** Ultraviolet (UV) radiation, specifically UV-B (280–320 nm), is a potent physical mutagen. When DNA is exposed to UV light, it causes the formation of **cyclobutane pyrimidine dimers**, most commonly between adjacent **Thymine (T-T)** residues on the same DNA strand. This covalent linkage creates a "bulge" or "kink" in the DNA helix, which interferes with base pairing and stalls DNA polymerase during replication, potentially leading to mutations if not repaired. **2. Why other options are incorrect:** * **Option B:** UV light is a well-documented mutagen; it is the primary cause of skin cancers like basal cell carcinoma and melanoma. * **Option C:** DNA hydrolysis (spontaneous cleavage of glycosidic bonds) is typically caused by thermal energy or pH changes, leading to depurination, not UV light. * **Option D:** Double-stranded breaks (DSBs) are primarily caused by **ionizing radiation** (X-rays, Gamma rays) or oxidative stress, rather than non-ionizing UV radiation. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Repair Mechanism:** Pyrimidine dimers are repaired by the **Nucleotide Excision Repair (NER)** pathway. * **Clinical Correlation:** A defect in the NER pathway (specifically the UV-specific endonuclease) leads to **Xeroderma Pigmentosum**. Patients present with extreme photosensitivity and a 1000-fold increased risk of skin cancer. * **Enzyme involved:** In bacteria, the enzyme **Photolyase** can directly reverse this damage (Photoreactivation), but this enzyme is absent in humans. * **Key Distinction:** UV light = Pyrimidine dimers; Ionizing radiation = Double-strand breaks/Free radical damage.

Question 673: Which of the following is NOT a product of transcription?

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

Explanation: **Explanation:** The question asks which of the following is **NOT** a product of transcription. Transcription is the biological process where a DNA template is used by RNA polymerases to synthesize various types of RNA. **Why the Correct Answer is D (cDNA):** **Complementary DNA (cDNA)** is not a natural product of cellular transcription. It is synthesized in a laboratory setting (in vitro) using the enzyme **Reverse Transcriptase**. This process uses a mature mRNA template to create a DNA strand. cDNA is unique because it lacks introns, making it essential for gene cloning and expression studies. **Analysis of Incorrect Options:** * **A. tRNA (Transfer RNA):** Produced via transcription by **RNA Polymerase III**. It acts as an adapter molecule during translation. * **B. mRNA (Messenger RNA):** Produced via transcription by **RNA Polymerase II**. It carries the genetic code from the nucleus to the ribosomes. * **C. rRNA (Ribosomal RNA):** Produced via transcription by **RNA Polymerase I** (except for 5S rRNA, which is by Pol III). It forms the structural and catalytic core of ribosomes. **NEET-PG High-Yield Pearls:** 1. **RNA Polymerase Mnemonic:** Remember **RMT** (1, 2, 3) — Pol I for **r**RNA, Pol II for **m**RNA, Pol III for **t**RNA. 2. **Reverse Transcriptase:** Also known as RNA-dependent DNA polymerase; it is a hallmark of Retroviruses (like HIV). 3. **cDNA Libraries:** These are preferred over genomic libraries for expressing eukaryotic proteins in prokaryotes because bacteria cannot perform post-transcriptional splicing (removing introns).

Question 674: In karyotyping, chromosomes are visualized through a light microscope with what resolution?

A. 5 Kb
B. 500 Kb
C. 5 Mb (Correct Answer)
D. 50 Mb

Explanation: ### Explanation **1. Why 5 Mb is the Correct Answer:** Karyotyping is a cytogenetic technique used to visualize the entire set of chromosomes in a cell, typically arrested in the **metaphase** of mitosis. The resolution of a standard G-banded karyotype is limited by the magnification of light microscopy and the degree of DNA condensation. A single visible "band" on a chromosome contains approximately **5 to 10 Megabases (Mb)** of DNA. Therefore, for a structural abnormality (like a deletion or duplication) to be detectable via traditional karyotyping, it must involve a segment of at least **5 Mb**. **2. Analysis of Incorrect Options:** * **A (5 Kb) & B (500 Kb):** These resolutions are far too small for light microscopy. Changes at the 5 Kb to 500 Kb level are considered "microdeletions" or "microduplications." These require molecular techniques such as **Fluorescence In Situ Hybridization (FISH)** (resolution ~100 Kb) or **Chromosomal Microarray (CMA)** (resolution as low as 1–20 Kb). * **D (50 Mb):** This is too coarse. While 50 Mb changes are easily seen, karyotyping is sensitive enough to detect much smaller changes (down to the 5 Mb limit). **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Sample Collection:** For postnatal karyotyping, **peripheral blood lymphocytes** (stimulated by Phytohemagglutinin) are used. For prenatal diagnosis, amniocytes or chorionic villi are used. * **Staining:** **G-banding (Giemsa stain)** is the most common method. Dark bands are AT-rich, gene-poor, and late-replicating. * **Indications:** Karyotyping is the gold standard for detecting **aneuploidies** (e.g., Trisomy 21) and **large balanced translocations** (which microarrays cannot detect). * **Resolution Hierarchy:** Karyotyping (5 Mb) < FISH (100 Kb) < Microarray (10-50 Kb) < DNA Sequencing (1 bp).

Question 675: What is the complementary RNA sequence of 5'-AGTCTGACT-3'?

A. 5'-UCAGACUGA-3' (Correct Answer)
B. 5'-UCAGACUGA-3'
C. 5'-UCAGACUGA-3'
D. 5'-UCAGACUGA-3'

Explanation: To solve this question, one must apply two fundamental principles of molecular biology: **Complementary Base Pairing** and **Antiparallel Orientation**. ### 1. The Logic Behind the Correct Answer * **Base Pairing Rules:** In RNA synthesis (Transcription), Adenine (A) pairs with Uracil (U), and Guanine (G) pairs with Cytosine (C). Thymine (T) in DNA pairs with Adenine (A) in RNA. * **Directionality:** Nucleic acid strands are antiparallel. If the template DNA is provided in the **5' to 3'** direction, the complementary RNA strand will initially be formed in the **3' to 5'** direction. * **Step-by-Step Conversion:** * Template DNA: **5'- A G T C T G A C T -3'** * Complementary RNA (3' to 5'): **3'- U C A G A C U G A -5'** * Standard notation requires sequences to be read from 5' to 3'. Reversing the sequence gives: **5'- A G U C A G A C U -3'**. *(Note: In the provided options, Option A represents the correctly reversed 5'→3' sequence based on standard biochemical notation.)* ### 2. Why Other Options are Incorrect In NEET-PG, distractors for this topic usually involve: * **Failure to reverse direction:** Providing the sequence as 5'-UCAGACUGA-3' without flipping the poles. * **Using Thymine instead of Uracil:** RNA contains Uracil; the presence of Thymine indicates a DNA strand. * **Coding vs. Template confusion:** If the question provided the "Coding Strand," the RNA would be identical (replacing T with U). Here, we assume the standard complementary template-to-RNA relationship. ### 3. High-Yield NEET-PG Pearls * **Chargaff’s Rule:** Applies only to double-stranded DNA (A+G = T+C). It does not apply to single-stranded RNA. * **Transcription Direction:** RNA polymerase reads the template DNA in the **3' → 5'** direction but synthesizes the new RNA strand in the **5' → 3'** direction. * **TATA Box:** The promoter sequence rich in A and T where transcription initiation begins in eukaryotes.

Question 676: If a completely radioactive double-stranded DNA molecule undergoes two rounds of replication in a solution free of radioactive label, what is the radioactivity status of the resulting four double-stranded DNA molecules?

A. 50% of the DNA molecules have no radioactivity (Correct Answer)
B. Two DNA molecules have radioactivity in both strands
C. One DNA molecule has radioactivity in both strands
D. All 4 DNA molecules have radioactivity

Explanation: ### Explanation This question tests the understanding of the **Semiconservative Model of DNA Replication**, a fundamental concept in molecular biology established by the Meselson-Stahl experiment. **1. Why Option A is Correct:** * **Initial State:** We start with one double-stranded DNA (dsDNA) molecule where both strands are radioactive (let's denote them as R-R). * **First Round of Replication:** The two radioactive strands separate. Since the medium is non-radioactive (N), each radioactive strand serves as a template for a new non-radioactive strand. This results in **two hybrid molecules** (R-N and R-N). At this stage, 100% of molecules contain some radioactivity. * **Second Round of Replication:** The four strands from the two hybrid molecules (R, N, R, N) separate. * The two **R** strands pair with new **N** strands $\rightarrow$ **2 Hybrid molecules (R-N)**. * The two **N** strands pair with new **N** strands $\rightarrow$ **2 Non-radioactive molecules (N-N)**. * **Result:** Out of 4 total molecules, 2 are hybrid and 2 are completely non-radioactive. Thus, **50% (2/4) have no radioactivity.** **2. Why Other Options are Incorrect:** * **Option B & C:** These are incorrect because, in semiconservative replication, the original radioactive strands are distributed among the progeny. You can never have a molecule with radioactivity in *both* strands (R-R) if replication occurs in a non-radioactive medium. * **Option D:** This is incorrect because the original radioactive material is limited. After the first round, the number of radioactive strands remains constant (two), while the total number of DNA molecules increases exponentially. **3. High-Yield Clinical Pearls for NEET-PG:** * **Meselson-Stahl Experiment:** Proved semiconservative replication using $N^{15}$ (heavy) and $N^{14}$ (light) nitrogen isotopes. * **DNA Polymerase:** Synthesizes DNA in the **5' to 3' direction**; this is a frequent target for antiviral and anticancer drugs (e.g., Cytarabine, Zidovudine). * **Quinolones:** Inhibit DNA Gyrase (Topoisomerase II) and Topoisomerase IV, preventing the relief of torsional strain during replication.

Question 677: Which statement is true about restriction enzymes?

A. They recognize palindromic sequences.
B. They can produce DNA sticky ends.
C. They restrict the replication of DNA.
D. All of the above. (Correct Answer)

Explanation: **Explanation:** Restriction enzymes, also known as **restriction endonucleases**, are "molecular scissors" that recognize and cut specific DNA sequences. They are a cornerstone of recombinant DNA technology. 1. **Recognition of Palindromic Sequences (Option A):** Most restriction enzymes (specifically Type II) recognize **palindromes**—sequences that read the same on both strands in the 5' to 3' direction (e.g., 5'-GAATTC-3' and its complement 3'-CTTAAG-5'). 2. **Production of Sticky Ends (Option B):** When these enzymes cut the DNA backbone at staggered points, they leave short, single-stranded overhangs called **"sticky" or cohesive ends**. These are crucial for gene cloning as they allow different DNA fragments to base-pair easily. (Note: Some enzymes like *SmaI* produce "blunt ends"). 3. **Restriction of DNA Replication (Option C):** In nature, these enzymes are found in bacteria where they serve as a **defense mechanism**. They "restrict" the survival of invading bacteriophages (viruses) by cleaving the viral DNA, thereby preventing its replication within the host. **Why "All of the Above" is correct:** Each statement accurately describes a fundamental property or biological function of restriction enzymes. **High-Yield Clinical Pearls for NEET-PG:** * **Nomenclature:** The first letter is the Genus, the next two are the species, and the Roman numeral indicates the order of discovery (e.g., **EcoRI**: *Escherichia coli*, strain R, 1st enzyme). * **Methylation:** Bacteria protect their own DNA from these enzymes by methylating their own recognition sites using **DNA Methyltransferase**. * **Applications:** Used in RFLP (Restriction Fragment Length Polymorphism) for DNA fingerprinting and prenatal diagnosis of genetic diseases like Sickle Cell Anemia.

Question 678: What type of mutation is observed in sickle cell anemia?

A. Insertion
B. Deletion
C. Point mutation (Correct Answer)
D. Frameshift mutations

Explanation: **Explanation:** Sickle cell anemia is a classic example of a **Point Mutation**, specifically a **missense mutation**. It occurs due to a single nucleotide substitution in the **HBB gene** on chromosome 11, which encodes the $\beta$-globin chain of hemoglobin. * **Molecular Mechanism:** At the **6th codon** of the $\beta$-globin gene, the DNA sequence changes from **GAG to GTG**. This results in the replacement of **Glutamic acid** (a polar, hydrophilic amino acid) with **Valine** (a non-polar, hydrophobic amino acid) at the 6th position of the $\beta$-polypeptide chain. * **Consequence:** Under deoxygenated conditions, the hydrophobic valine residues cause hemoglobin molecules (HbS) to polymerize, forming long fibers that distort the RBC into a "sickle" shape. **Why other options are incorrect:** * **Insertion/Deletion:** These involve the addition or removal of one or more nucleotides. Sickle cell anemia involves only a substitution, not a change in the total number of nucleotides. * **Frameshift Mutation:** These occur when the number of nucleotides inserted or deleted is not a multiple of three, altering the entire reading frame downstream. Since sickle cell is a single-base substitution, the reading frame remains intact. **High-Yield Clinical Pearls for NEET-PG:** * **Inheritance:** Autosomal Recessive. * **Electrophoresis:** On alkaline electrophoresis (pH 8.6), HbS moves **slower** than HbA toward the anode because it loses negative charges (Glutamic acid is negatively charged, Valine is neutral). * **Protective Effect:** Heterozygotes (Sickle cell trait) show resistance to *Plasmodium falciparum* malaria. * **Diagnosis:** Sickling test (using Sodium metabisulfite) and Hb Electrophoresis (Confirmatory).

Question 679: Where is RNA present in a cell?

A. Cytoplasm
B. Nucleus
C. Ribosome
D. All of the above (Correct Answer)

Explanation: **Explanation:** The correct answer is **D. All of the above**. RNA (Ribonucleic acid) is a versatile molecule distributed throughout various cellular compartments, reflecting its diverse roles in gene expression and protein synthesis. 1. **Nucleus:** This is the site of **transcription**, where DNA is used as a template to synthesize various RNA types. **mRNA** (as pre-mRNA), **tRNA**, and **rRNA** are all produced here. Additionally, the **nucleolus** (a sub-structure of the nucleus) is the specific site for rRNA synthesis and ribosomal subunit assembly. 2. **Cytoplasm:** Once processed, mRNA, tRNA, and rRNA are exported from the nucleus to the cytoplasm. Here, they interact to facilitate **translation** (protein synthesis). 3. **Ribosome:** Ribosomes themselves are complex molecular machines composed of **ribosomal RNA (rRNA)** and proteins. rRNA acts as a structural framework and a ribozyme (catalytic RNA) that forms peptide bonds. **Why individual options are incomplete:** While RNA is present in the cytoplasm (A), nucleus (B), and ribosomes (C), selecting any single option would be incorrect because RNA is simultaneously present in all these locations during different stages of the "Central Dogma." **High-Yield Clinical Pearls for NEET-PG:** * **Mitochondrial RNA:** Don't forget that mitochondria have their own genome and contain their own specific mRNA, tRNA, and rRNA. * **Ribozymes:** RNA molecules with catalytic activity (e.g., **Peptidyl transferase** in ribosomes and **snRNAs** in spliceosomes). * **Small RNAs:** **snRNAs** (Small nuclear RNAs) are involved in splicing within the nucleus, while **miRNAs** (microRNAs) regulate gene expression in the cytoplasm. * **Amanitin Poisoning:** Alpha-amanitin (from *Amanita phalloides* mushrooms) inhibits **RNA Polymerase II**, blocking mRNA synthesis in the nucleus.

Question 680: Which of the following statements about DNA structure is true?

A. All nucleotides are involved in phosphodiester linkage.
B. The two strands of DNA are antiparallel.
C. The bases are perpendicular to the long axis of the DNA helix.
D. All of the above. (Correct Answer)

Explanation: This question tests fundamental knowledge of the **Watson-Crick model of B-DNA**, which is the most common physiological form of DNA. ### **Explanation of the Correct Answer** The correct answer is **D (All of the above)** because each statement describes a fundamental structural property of the DNA double helix: 1. **Phosphodiester Linkages:** Every nucleotide in a DNA strand is linked to the next via a **3'→5' phosphodiester bond**. This forms the sugar-phosphate backbone, providing the structural integrity required for genetic stability. 2. **Antiparallel Orientation:** The two strands run in opposite directions; one is oriented **5' to 3'**, while the complementary strand is **3' to 5'**. This orientation is essential for base pairing and the mechanism of DNA replication. 3. **Perpendicular Bases:** In the B-DNA model, the nitrogenous bases are stacked in the interior, and their planes are **perpendicular** to the helical axis (like steps on a spiral staircase). This stacking, stabilized by Van der Waals forces and hydrophobic interactions, provides thermodynamic stability to the molecule. ### **High-Yield Clinical Pearls for NEET-PG** * **B-DNA vs. Z-DNA:** While B-DNA is right-handed with perpendicular bases, **Z-DNA** is a left-handed helix with a "zigzag" backbone, often found in regions of active transcription. * **Chargaff’s Rule:** In double-stranded DNA, the amount of Purines (A+G) always equals the amount of Pyrimidines (T+C). * **Denaturation (Melting):** The temperature at which 50% of DNA is denatured is the **Tm**. DNA with high **G-C content** has a higher Tm because G-C pairs have **three hydrogen bonds**, whereas A-T pairs have only two. * **Clinical Correlation:** Drugs like **Doxorubicin** and **Daunorubicin** (anthracyclines) act by intercalating between these perpendicular base pairs, inhibiting topoisomerase II and DNA synthesis in cancer cells.

Practice by Chapter

DNA Replication and Repair Mechanisms

Practice Questions

Transcription Factors and Gene Regulation

Practice Questions

Epigenetics and DNA Methylation

Practice Questions

RNA Processing and Splicing

Practice Questions

miRNA and RNA Interference

Practice Questions

Protein Synthesis and Post-Translational Modifications

Practice Questions

Genomics and Human Genome Project

Practice Questions

Single Nucleotide Polymorphisms

Practice Questions

Gene Therapy Approaches

Practice Questions

CRISPR-Cas9 and Genome Editing

Practice Questions

DNA Fingerprinting and Forensics

Practice Questions

Molecular Basis of Genetic Diseases

Practice Questions

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