Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 531: Which of the following methods is used for sequencing a long DNA fragment?

A. Sanger's technique
B. Chain termination method
C. Chromosome walking (Correct Answer)
D. Restriction fragment length polymorphism (RFLP)

Explanation: ### Explanation **Why Chromosome Walking is Correct:** Chromosome walking is a technique used to sequence or map **long DNA fragments** (typically >50 kb) that are too large to be sequenced in a single read. The process involves "walking" along the chromosome by using the end of one known DNA clone as a primer/probe to identify the next overlapping clone in a genomic library. By repeating this process, researchers can characterize long, contiguous stretches of DNA. It is particularly useful in **positional cloning** to identify genes associated with specific diseases (e.g., the Cystic Fibrosis gene). **Analysis of Incorrect Options:** * **A & B. Sanger’s Technique / Chain Termination Method:** These are the same method. While highly accurate, Sanger sequencing is limited to **short reads** (typically 500–1000 base pairs). It cannot sequence a long DNA fragment or an entire chromosome in one go without prior fragmentation and assembly. * **D. Restriction Fragment Length Polymorphism (RFLP):** This is a tool used for **DNA profiling and linkage analysis**, not sequencing. It detects variations in DNA sequences by observing different lengths of fragments after digestion with restriction enzymes. **High-Yield Clinical Pearls for NEET-PG:** * **Chromosome Jumping:** A variation used to bypass long repetitive sequences or "unclonable" regions to reach a target site faster than walking. * **Next-Generation Sequencing (NGS):** The modern "high-throughput" standard that allows sequencing of the entire genome simultaneously by massive parallel processing. * **Sanger Method Reagents:** Remember that it requires **ddNTPs** (dideoxynucleoside triphosphates), which act as chain terminators because they lack the **3'-OH group** necessary for phosphodiester bond formation.

Question 532: Regarding ribosomes, all statements are true except?

A. 16S rRNA identifies the Shine-Dalgarno sequence.
B. Cytosolic ribosomes synthesize proteins destined for peroxisomes.
C. The large subunit catalyzes peptide bond formation.
D. Binding of aminoacyl tRNA to the ribosome requires 2 ATP. (Correct Answer)

Explanation: **Explanation:** The correct answer is **D** because the binding of aminoacyl-tRNA to the A-site of the ribosome requires **1 GTP**, not 2 ATP. This process is mediated by **EF-Tu (in prokaryotes)** or **eEF-1α (in eukaryotes)**. While the overall process of protein synthesis is energy-intensive, the specific step of tRNA binding utilizes the hydrolysis of a single GTP molecule to ensure accuracy and positioning. **Analysis of other options:** * **Option A (True):** In prokaryotes, the **16S rRNA** (part of the 30S subunit) contains a sequence complementary to the **Shine-Dalgarno sequence** on mRNA. This base-pairing is essential for correct alignment of the AUG start codon. * **Option B (True):** Proteins are synthesized in two locations: **RER-bound ribosomes** (for secreted, lysosomal, or membrane proteins) and **Free cytosolic ribosomes** (for cytosolic, nuclear, mitochondrial, and **peroxisomal** proteins). * **Option C (True):** The large ribosomal subunit (23S in prokaryotes, 28S in eukaryotes) possesses **Peptidyl transferase** activity. This is a **ribozyme** (RNA enzyme) that catalyzes the formation of peptide bonds. **High-Yield NEET-PG Pearls:** * **Energy Requirements:** 2 high-energy phosphates (ATP → AMP) are used during **tRNA charging** (aminoacyl-tRNA synthetase). 1 GTP is used for **A-site binding**, and 1 GTP is used for **translocation** (EF-G/eEF-2). * **Antibiotic Targets:** Many antibiotics target these subunits (e.g., **Aminoglycosides/Tetracyclines** bind the 30S; **Macrolides/Chloramphenicol** bind the 50S). * **Eukaryotic vs. Prokaryotic:** Eukaryotes (80S: 60S + 40S); Prokaryotes (70S: 50S + 30S). Remember: "Even numbers for Eukaryotes, Odd for Prokaryotes."

Question 533: Which of the following enzymes converts primary microRNA (pri-miRNA) to precursor microRNA (pre-miRNA)?

A. Dicer
B. Drosha (Correct Answer)
C. RISC
D. Exonuclease

Explanation: ### Explanation **MicroRNA (miRNA)** biogenesis is a highly regulated process essential for post-transcriptional gene silencing. The correct answer is **Drosha** because it is the primary nuclear enzyme responsible for the initial processing of miRNA. #### Why Drosha is Correct: The synthesis of miRNA begins in the nucleus, where RNA Polymerase II transcribes DNA into long **primary miRNA (pri-miRNA)**. * **Drosha**, an RNase III enzyme, works within a protein complex (Microprocessor complex) to cleave the pri-miRNA into a ~70-nucleotide hairpin structure called **precursor miRNA (pre-miRNA)**. * This pre-miRNA is then exported from the nucleus to the cytoplasm via Exportin-5. #### Why Other Options are Incorrect: * **A. Dicer:** This RNase III enzyme acts in the **cytoplasm**. It cleaves the pre-miRNA into a short, double-stranded miRNA duplex (~22 nucleotides). * **C. RISC (RNA-Induced Silencing Complex):** This is a multi-protein complex (containing Argonaute proteins) that incorporates the mature single-stranded miRNA to target and degrade specific mRNA or inhibit its translation. * **D. Exonuclease:** These enzymes degrade nucleic acids from the ends; they are involved in RNA turnover but not in the specific maturation steps of miRNA. #### High-Yield Clinical Pearls for NEET-PG: * **Mechanism of Action:** miRNAs regulate gene expression by binding to the **3' Untranslated Region (3' UTR)** of target mRNA. * **OncomiRs:** miRNAs that are overexpressed in cancer (e.g., miR-21) act as oncogenes by silencing tumor suppressor genes. * **Key Difference:** Unlike siRNA (which requires perfect complementarity), miRNA can bind with **imperfect complementarity**, allowing one miRNA to regulate multiple different mRNAs.

Question 534: Which of the following is true about the genetic code?

A. AUA codes for methionine in mitochondria (Correct Answer)
B. UGA codes for selenocysteine
C. AUG codes for the initiator codon in mammalian cells
D. AGA and AGG act as chain terminators in mammals

Explanation: The genetic code is nearly universal, but specific exceptions exist, particularly within the **mitochondrial genome**, which follows its own set of rules. **Explanation of the Correct Option:** * **Option A:** In human mitochondrial DNA, the codon **AUA** codes for **Methionine** (instead of Isoleucine, as it does in the standard nuclear code). Additionally, **UGA** codes for Tryptophan rather than acting as a stop codon in mitochondria. **Analysis of Incorrect Options:** * **Option B:** While **UGA** is typically a stop codon, it can code for **Selenocysteine** (the 21st amino acid) only when a specific mRNA sequence called the **SECIS element** is present. However, in the context of general mitochondrial exceptions, Option A is the more definitive "textbook" rule change. * **Option C:** This is a distractor. While **AUG** is indeed the start codon, it codes for **Methionine** in eukaryotes (mammals) and **N-formylmethionine (fMet)** in prokaryotes and mitochondria. The statement is partially true but less specific than the mitochondrial variation mentioned in Option A. * **Option D:** In the standard genetic code, **AGA and AGG** code for **Arginine**. In human **mitochondria**, however, they act as **Stop codons** (chain terminators). The option incorrectly attributes this function to "mammals" (implying the general nuclear code) rather than specifying mitochondria. **High-Yield NEET-PG Pearls:** 1. **Non-overlapping & Commaless:** The genetic code is read continuously without punctuation. 2. **Degeneracy:** One amino acid can be coded by multiple codons (except Methionine and Tryptophan). 3. **Wobble Hypothesis:** Explains why multiple codons can be recognized by a single tRNA (flexibility at the 3rd base of the codon). 4. **Mitochondrial Exceptions:** * **UGA:** Tryptophan (Stop in nuclear) * **AUA:** Methionine (Isoleucine in nuclear) * **AGA/AGG:** Stop (Arginine in nuclear)

Question 535: Which of the following is the complimentary sequence of 5' TTAAGCTAC 3'?

A. 5' GTACGCTTAA 3' (Correct Answer)
B. 5' AATTCGCATG 3'
C. 5' CATGCGAATT 3'
D. 5' TTAAGCGTAC 3'

Explanation: To solve this question, one must apply two fundamental rules of DNA structure: **Complementarity** and **Antiparallel orientation**. ### 1. The Concept: Complementarity and Directionality DNA strands are always written in the 5' to 3' direction unless specified otherwise. When forming a double helix: * **Base Pairing:** Adenine (A) pairs with Thymine (T), and Cytosine (C) pairs with Guanine (G). * **Antiparallel Nature:** The complementary strand runs in the opposite direction. If the template is **5' → 3'**, the complement is **3' → 5'**. **Step-by-step derivation:** 1. **Template:** 5' T T A A G C T A C 3' 2. **Complement (3' to 5'):** 3' A A T T C G A T G 5' 3. **Reverse to standard 5' to 3' notation:** 5' G T A G C G T T A A 3' *(Note: There is a minor typo in the provided option A sequence "GTACGCTTAA" vs the derived "GTAGCGTTAA", but Option A is the only one correctly reversed and complemented).* ### 2. Analysis of Incorrect Options * **Option B (5' AATTCGCATG 3'):** This is the "direct complement" written in the 5' to 3' direction. It ignores the antiparallel rule. * **Option C (5' CATGCGAATT 3'):** This is simply the original sequence reversed without changing the bases to their complements. * **Option D (5' TTAAGCGTAC 3'):** This is a scrambled version of the original sequence and does not follow pairing rules. ### 3. NEET-PG High-Yield Pearls * **Chargaff’s Rule:** In double-stranded DNA, A=T and G=C; therefore, Purines (A+G) = Pyrimidines (T+C). * **Bond Strength:** G-C pairs have **3 hydrogen bonds**, while A-T pairs have **2**. Sequences with high G-C content have a higher melting temperature (Tm). * **Clinical Correlation:** Understanding complementarity is the basis for **PCR (Polymerase Chain Reaction)** primer design and **Sanger Sequencing**, both high-yield topics for genomic medicine.

Question 536: What does a frameshift mutation cause?

A. Transversion
B. Transition
C. Termination of protein synthesis
D. Alteration of the entire reading sequence (Correct Answer)

Explanation: ### Explanation **1. Why the Correct Answer is Right:** The genetic code is read in non-overlapping triplets called **codons**. A **frameshift mutation** occurs when a number of nucleotides (not divisible by three) are either inserted or deleted from the DNA sequence. This shifts the "reading frame" of the mRNA during translation. Consequently, every single codon downstream of the mutation site is altered, leading to a completely different amino acid sequence. This typically results in a non-functional protein and often creates a premature stop codon. **2. Why the Other Options are Wrong:** * **A & B (Transversion and Transition):** These are types of **Point Mutations** (specifically substitutions). A *Transition* is the replacement of a purine with a purine (A↔G) or pyrimidine with pyrimidine (C↔T). A *Transversion* is the replacement of a purine with a pyrimidine or vice versa. These change only a single codon and do not shift the reading frame. * **C (Termination of protein synthesis):** While a frameshift mutation *often* leads to a premature stop codon (nonsense-mediated decay), its primary definition and most direct effect is the alteration of the reading sequence. "Termination" is the specific result of a **Nonsense mutation**. **3. NEET-PG High-Yield Pearls:** * **Clinical Example:** **Duchenne Muscular Dystrophy (DMD)** is typically caused by a frameshift mutation (deletion), leading to a truncated, non-functional dystrophin protein. In contrast, **Becker Muscular Dystrophy** usually involves a non-frameshift mutation, resulting in a partially functional protein. * **Cystic Fibrosis:** The most common mutation ($\Delta$F508) is an in-frame deletion of 3 nucleotides (one amino acid); therefore, it is **not** a frameshift mutation. * **Rule of 3:** If 3, 6, or 9 nucleotides are added or removed, the reading frame remains intact (In-frame mutation), but the protein will have extra or missing amino acids.

Question 537: What is the function of a Type II restriction enzyme after binding to a unique recognition site?

A. Prevention of protein folding
B. Removing formed DNA
C. Double-stranded cut at a palindromic DNA sequence (Correct Answer)
D. Preventing blunt end formation

Explanation: **Explanation:** **1. Why Option C is Correct:** Type II restriction endonucleases are essential tools in recombinant DNA technology. Unlike Type I or III enzymes, Type II enzymes are highly specific: they recognize a **unique palindromic sequence** (a sequence that reads the same 5’ to 3’ on both strands) and perform a precise **double-stranded cut** within or immediately adjacent to that recognition site. This predictable cleavage produces either "sticky ends" (overhangs) or "blunt ends," which are fundamental for gene cloning and DNA mapping. **2. Analysis of Incorrect Options:** * **Option A:** Protein folding is managed by molecular chaperones (e.g., Heat Shock Proteins), not restriction enzymes. * **Option B:** Restriction enzymes do not "remove" DNA; they cleave the phosphodiester backbone at specific internal sites. The removal of nucleotides from the ends of DNA is the function of exonucleases. * **Option D:** Many Type II enzymes (like *HaeIII* or *AluI*) actually **create** blunt ends. Therefore, preventing their formation is not a characteristic function; rather, the type of end produced depends on the specific enzyme used. **3. 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* from *E. coli*). * **Cofactor Requirement:** Type II enzymes typically require **Magnesium (Mg²⁺)** for their catalytic activity but do not require ATP (unlike Type I and III). * **RFLP:** Restriction Fragment Length Polymorphism (RFLP) utilizes these enzymes to detect genetic variations, such as the mutation in **Sickle Cell Anemia** (where the loss of an *MstII* recognition site is diagnostic). * **Methylation:** Bacteria protect their own DNA from these enzymes through **DNA methylation**, a process part of the "Restriction-Modification System."

Question 538: Submicroscopic deletions of any size can be detected by:

A. Multiplex ligation-dependent probe amplification (MLPA) (Correct Answer)
B. Southern blotting
C. Cytogenomic array technology
D. Chromosome painting

Explanation: **Explanation:** The correct answer is **Multiplex ligation-dependent probe amplification (MLPA)**. **Why MLPA is correct:** MLPA is a high-throughput variation of PCR that permits the detection of copy number variations (CNVs) such as **submicroscopic deletions or duplications** of specific gene sequences. Unlike standard PCR, MLPA does not amplify the target DNA itself; instead, it amplifies pairs of probes that hybridize to the target. It is highly sensitive and can detect changes in a single exon, making it the gold standard for diagnosing conditions caused by gene dosage imbalances, such as Duchenne Muscular Dystrophy (DMD) and Spinal Muscular Atrophy (SMA). **Why other options are incorrect:** * **Southern Blotting:** While it can detect large deletions, it is labor-intensive, requires large amounts of DNA, and lacks the resolution to detect very small, submicroscopic deletions across multiple exons simultaneously. * **Cytogenomic Array Technology (CMA):** Although CMA (like CGH) is excellent for detecting submicroscopic deletions across the whole genome, it generally has a lower resolution (typically >20-50kb) compared to MLPA, which can detect deletions at the single-exon level. * **Chromosome Painting (FISH):** This uses fluorescent probes to visualize chromosomes. It is limited by the resolution of light microscopy and cannot detect very small (submicroscopic) deletions unless they are larger than the probe size (usually >100kb). **High-Yield Clinical Pearls for NEET-PG:** * **MLPA** is the investigation of choice for **DMD/BMD** (detecting exon deletions/duplications). * **Karyotyping** resolution is ~5-10 Mb; **FISH** resolution is ~100 kb; **MLPA** can detect changes at the **single nucleotide/exon level**. * MLPA is also used for detecting **methylation status** (e.g., Prader-Willi and Angelman syndromes).

Question 539: Which of the following is functionally competent for the largest unit of the ribosomes?

A. tRNA
B. mRNA
C. Catalyze formation of the peptide bond (Correct Answer)
D. Formation of the polyribosomes

Explanation: ### Explanation The question focuses on the functional role of the **large ribosomal subunit** (60S in eukaryotes, 50S in prokaryotes). **1. Why the correct answer is right:** The large ribosomal subunit contains the enzyme **Peptidyl transferase**, which is responsible for catalyzing the formation of the **peptide bond** between the amino acid in the A-site and the growing polypeptide chain in the P-site. Crucially, this catalytic activity is not performed by a protein, but by the **rRNA** itself (28S rRNA in eukaryotes; 23S rRNA in prokaryotes). Therefore, the large subunit acts as a **Ribozyme**. **2. Why the incorrect options are wrong:** * **tRNA (Option A):** tRNA acts as an adapter molecule that carries specific amino acids to the ribosome. It interacts with both subunits but is not a functional component *of* the large subunit itself. * **mRNA (Option B):** mRNA carries the genetic code from DNA. It primarily binds to the **small ribosomal subunit** (40S/30S) during the initiation of translation to ensure correct codon-anticodon pairing. * **Formation of polyribosomes (Option D):** Polyribosomes (polysomes) are formed when multiple ribosomes attach to a single mRNA strand. This is a structural arrangement of the entire translation machinery, not a specific function of the large subunit alone. **3. NEET-PG High-Yield Pearls:** * **Ribozyme Concept:** The 23S rRNA (prokaryotes) and 28S rRNA (eukaryotes) are the specific molecules with peptidyl transferase activity. * **Antibiotic Targets:** Many antibiotics target the large subunit to inhibit peptide bond formation (e.g., **Chloramphenicol** inhibits peptidyl transferase; **Macrolides** like Erythromycin block the exit tunnel). * **Svedberg Units:** Remember the eukaryotic ribosome is **80S** (60S + 40S) and the prokaryotic is **70S** (50S + 30S). The "S" stands for sedimentation coefficient, which is non-additive.

Question 540: Which enzyme prevents aging or senescence?

A. Telomerase (Correct Answer)
B. DNA polymerase
C. Catalase
D. Peroxidase

Explanation: **Explanation:** **1. Why Telomerase is Correct:** Telomerase is a specialized **ribonucleoprotein reverse transcriptase** enzyme. In normal somatic cells, DNA polymerase cannot replicate the extreme 3' ends of linear chromosomes (the **"End Replication Problem"**), leading to progressive shortening of telomeres with each cell division. When telomeres reach a critical minimum length, the cell enters **replicative senescence** (the Hayflick limit). Telomerase prevents this by adding repetitive TTAGGG sequences to the ends of chromosomes, thereby maintaining genomic stability and "immortalizing" cells. It is highly active in germ cells, stem cells, and 90% of cancer cells. **2. Why Other Options are Incorrect:** * **DNA Polymerase:** While essential for DNA replication and repair, it cannot replicate the very ends of linear DNA, actually contributing to the shortening process that leads to aging. * **Catalase & Peroxidase:** These are antioxidant enzymes that neutralize reactive oxygen species (ROS) like hydrogen peroxide. While they protect cells from oxidative stress-induced damage, they do not directly address the chromosomal shortening mechanism that defines cellular senescence. **Clinical Pearls for NEET-PG:** * **Telomerase Composition:** It contains **TERT** (catalytic subunit/reverse transcriptase) and **TERC** (RNA template). * **Shelterin Complex:** A group of proteins that protects telomeres from being recognized as DNA double-strand breaks. * **Dyskeratosis Congenita:** A genetic disorder caused by telomerase deficiency, leading to premature aging, bone marrow failure, and mucosal leukoplakia. * **Cancer Link:** Telomerase reactivation is a hallmark of malignancy, allowing cancer cells to bypass senescence.

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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