Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 411: The human genome contains approximately how many base pairs?

A. 3 x 10^9 (Correct Answer)
B. 3 x 10^8
C. 3 x 10^7
D. 3 x 10^6

Explanation: **Explanation:** The human genome consists of the total genetic material stored within 23 pairs of chromosomes in the nucleus, plus the mitochondrial DNA. The correct answer is **3 x 10⁹ base pairs (3 billion bp)** per haploid set of chromosomes. **1. Why Option A is Correct:** The haploid human genome (n) contains approximately **3.2 billion base pairs**. In a diploid cell (2n), such as a somatic cell, this number doubles to approximately 6.4 x 10⁹ bp. In the context of standard medical examinations like NEET-PG, the value is typically rounded to **3 x 10⁹ bp**. This vast amount of data is packaged into the nucleus via high-order folding involving histones to form nucleosomes. **2. Why Other Options are Incorrect:** * **Option B (3 x 10⁸):** This is ten times smaller than the actual human genome. For comparison, some smaller vertebrates or large plant genomes might fall into this range, but it does not represent human complexity. * **Options C & D (3 x 10⁷ and 3 x 10⁶):** These values are significantly lower. For perspective, the genome of *Escherichia coli* is approximately **4.6 x 10⁶ bp**, which is roughly 1,000 times smaller than the human genome. **High-Yield Clinical Pearls for NEET-PG:** * **Coding vs. Non-coding:** Only about **1–2%** of the human genome actually codes for proteins (exons). * **Mitochondrial DNA (mtDNA):** Unlike the nuclear genome, mtDNA is circular, double-stranded, and contains only **16,569 base pairs** encoding 37 genes. * **Repeat Sequences:** Nearly 50% of the human genome consists of repetitive sequences (e.g., SINEs, LINEs, and satellite DNA), which are crucial for chromosomal structural integrity and genetic regulation. * **Gene Count:** The human genome contains approximately **20,000–25,000 protein-coding genes**.

Question 412: Which of the following is a reverse transcriptase?

A. Topoisomerase 2
B. Telomerase (Correct Answer)
C. RNA polymerase 2
D. DNA polymerase alpha

Explanation: **Explanation:** **Why Telomerase is the Correct Answer:** Telomerase is a specialized **ribonucleoprotein enzyme** that functions as a **DNA-dependent DNA polymerase** (specifically, a **reverse transcriptase**). It contains an intrinsic RNA template (hTR) which it uses to synthesize repetitive DNA sequences (TTAGGG in humans) at the 3' ends of linear chromosomes. This process prevents the "end-replication problem," where chromosomes shorten with each cell division, thereby maintaining genomic stability and cellular immortality in germ cells and cancer cells. **Analysis of Incorrect Options:** * **A. Topoisomerase 2:** This enzyme manages DNA tangling by creating double-stranded breaks to relieve torsional strain (supercoiling) during replication and transcription. It does not synthesize DNA from an RNA template. * **C. RNA polymerase 2:** This enzyme is responsible for the synthesis of **mRNA** (and some snRNA/miRNA) from a DNA template (DNA-dependent RNA polymerase). * **D. DNA polymerase alpha:** This is a eukaryotic DNA polymerase that initiates DNA replication by synthesizing an RNA primer followed by a short string of DNA nucleotides. It is a DNA-dependent DNA polymerase. **High-Yield Clinical Pearls for NEET-PG:** * **Telomerase & Cancer:** Telomerase activity is upregulated in ~90% of cancer cells, making them "immortal." * **Reverse Transcriptase Examples:** Other key examples include **HIV Reverse Transcriptase** (RNA-dependent DNA polymerase) and **Hepatitis B Virus** (which uses reverse transcription in its replication cycle). * **Inhibitors:** Topoisomerase 2 is the target of anticancer drugs like **Etoposide** and **Teniposide**, and the antibacterial **Fluoroquinolones** (targeting DNA Gyrase).

Question 413: Excessive ultraviolet (UV) radiation is harmful to life. What is the primary mechanism by which ultraviolet radiation causes damage to biological systems?

A. Inhibition of DNA synthesis
B. Formation of thymidine dimers (Correct Answer)
C. Ionization
D. DNA fragmentation

Explanation: ### Explanation **Correct Option: B. Formation of thymidine dimers** Ultraviolet (UV) radiation, specifically UV-B (280–320 nm), is non-ionizing radiation that is absorbed by the nitrogenous bases of DNA. The primary photochemical reaction involves the formation of **cyclobutane pyrimidine dimers**, most commonly between two adjacent **thymine** residues on the same DNA strand. This covalent cross-linking creates a "bulge" in the DNA helix, which distorts the structure and interferes with both transcription and DNA replication. **Analysis of Incorrect Options:** * **A. Inhibition of DNA synthesis:** While UV damage eventually leads to the arrest of the replication fork, this is a *consequence* of the damage, not the primary mechanism of the radiation itself. * **C. Ionization:** UV radiation is **non-ionizing**. Ionizing radiation (like X-rays or Gamma rays) works by ejecting electrons from atoms, creating free radicals and causing double-strand breaks. * **D. DNA fragmentation:** This is typically a result of ionizing radiation or late-stage apoptosis. UV radiation primarily causes localized point lesions (dimers). **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Repair Mechanism:** Pyrimidine dimers are repaired by **Nucleotide Excision Repair (NER)**. This involves the "cut and patch" mechanism using endonucleases (UV-specific), DNA polymerase, and ligase. * **Clinical Correlation:** A defect in the NER pathway (specifically the excinuclease enzyme) leads to **Xeroderma Pigmentosum**. Patients present with extreme photosensitivity and a high risk of skin cancers (Basal Cell Carcinoma, Squamous Cell Carcinoma, and Melanoma). * **Prokaryotic Repair:** Bacteria can also use **Photoreactivation** (via the enzyme DNA photolyase), a process not found in humans.

Question 414: Given the DNA template strand 5'-TTACGTAC-3', what will be the resulting mRNA sequence after transcription?

A. 5'-TTACGTAC-3'
B. 3'-TTACGTAC-5'
C. 5'-CATGCATT-3'
D. 5'-GUACGUAA-3' (Correct Answer)

Explanation: ### Explanation **1. Why Option D is Correct:** Transcription is the process where RNA polymerase synthesizes mRNA using a **DNA template strand**. This process follows two fundamental rules: * **Antiparallel Orientation:** The mRNA is synthesized in the 5' to 3' direction, meaning it must be complementary and antiparallel to the template strand. * **Base Pairing:** Adenine (A) pairs with Uracil (U) in RNA, and Cytosine (C) pairs with Guanine (G). Given Template: **5'- T T A C G T A C -3'** To find the mRNA, we read the template from 3' to 5' and pair accordingly: * Template 3' → 5' is: **C-A-T-G-C-A-T-T** * Complementary mRNA 5' → 3' is: **G-U-A-C-G-U-A-A** **2. Why Other Options are Incorrect:** * **Option A:** This is identical to the template DNA. mRNA cannot have Thymine (T) and must be complementary, not identical. * **Option B:** This is simply the template strand written in reverse. It ignores base-pairing rules and the presence of Uracil. * **Option C:** This is the "Coding Strand" sequence (substituting T for U). While the mRNA sequence matches the coding (non-template) strand, this option incorrectly retains Thymine (T) instead of Uracil (U). **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Coding vs. Template:** The **Coding Strand** (5'→3') has the same sequence as mRNA (except T is replaced by U). The **Template Strand** (3'→5') is what the RNA polymerase actually "reads." * **Directionality:** RNA synthesis always occurs in the **5' to 3' direction**. * **Inhibitors:** **Rifampicin** inhibits bacterial DNA-dependent RNA polymerase (used in TB), while **Actinomycin D** inhibits transcription in both prokaryotes and eukaryotes (used in chemotherapy). * **Alpha-amanitin:** Found in *Amanita phalloides* mushrooms; it specifically inhibits **RNA Polymerase II**, leading to severe liver failure.

Question 415: Apolipoprotein B48 is synthesized via which of the following mechanisms?

A. RNA alternate splicing
B. RNA editing (Correct Answer)
C. DNA editing
D. RNA inference

Explanation: **Explanation:** The synthesis of **Apolipoprotein B48 (Apo B48)** is a classic example of **post-transcriptional RNA editing**, specifically involving **site-specific deamination**. In humans, a single gene codes for both Apo B100 and Apo B48. In the liver, the full gene is transcribed and translated into **Apo B100** (the full-length protein). However, in the **intestine**, the enzyme **Cytidine Deaminase** acts on the mRNA. It converts a specific Cytosine (C) to Uracil (U) at codon 2153. This change converts the original glutamine codon (**CAA**) into a premature stop codon (**UAA**). As a result, translation terminates early, producing a protein that represents only the N-terminal **48%** of the full sequence—hence the name **Apo B48**. **Analysis of Incorrect Options:** * **A. RNA alternate splicing:** This involves joining different exons to create protein isoforms (e.g., Calcitonin vs. CGRP). While it creates diversity, it is not the mechanism for Apo B48. * **C. DNA editing:** This is not a standard physiological process for protein diversity; genetic information is typically altered at the RNA level to preserve the integrity of the genome. * **D. RNA interference (RNAi):** This is a regulatory mechanism where small RNA molecules (siRNA/miRNA) inhibit gene expression by neutralizing targeted mRNA molecules. **High-Yield Clinical Pearls for NEET-PG:** * **Apo B100:** Found in VLDL, IDL, and LDL; acts as a ligand for the LDL receptor. * **Apo B48:** Found exclusively in **Chylomicrons** and chylomicron remnants; it lacks the C-terminal LDL receptor-binding domain. * **Enzyme:** Remember **Cytidine Deaminase** (specifically APOBEC-1) as the mediator for this C-to-U editing. * **Location:** Liver = B100; Intestine = B48.

Question 416: A prokaryotic regulatory gene synthesizes a protein known as:

A. The promoter
B. The operator
C. The repressor (Correct Answer)
D. The enhancer

Explanation: **Explanation:** In prokaryotic gene regulation, specifically within the **Operon model** (e.g., the *lac* operon), genes are organized into functional units. A **regulatory gene** (such as the *i* gene) is located upstream of the operon and is constitutively expressed. It synthesizes a specific protein called the **Repressor**. 1. **Why the Repressor is correct:** The repressor protein is the functional product of the regulatory gene. Its primary role is to bind to the **operator** site. When bound, it physically blocks RNA polymerase from moving forward, thereby inhibiting the transcription of structural genes. This is a classic example of **negative regulation**. 2. **Why other options are incorrect:** * **The Promoter (A):** This is a DNA sequence, not a protein. It is the binding site for RNA polymerase to initiate transcription. * **The Operator (B):** This is also a DNA sequence located between the promoter and structural genes. It acts as a "switch" where the repressor protein binds. * **The Enhancer (D):** Enhancers are regulatory DNA sequences that increase the rate of transcription. They are predominantly found in **eukaryotes**, not prokaryotes, and are not proteins. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Inducer:** In the *lac* operon, **allolactose** acts as an inducer by binding to the repressor, changing its conformation so it can no longer bind to the operator. * **Polycistronic mRNA:** Prokaryotic operons produce a single mRNA that codes for multiple proteins, unlike the monocistronic mRNA typical of eukaryotes. * **Constitutive Genes:** Also known as "housekeeping genes," these are expressed at a constant rate regardless of environmental conditions. The regulatory gene for the repressor is an example.

Question 417: On which of the following does tRNA act specifically?

A. ATP
B. Golgi body
C. Specific amino acid (Correct Answer)
D. Ribosome

Explanation: **Explanation:** The core function of **tRNA (transfer RNA)** is to act as an "adapter molecule" during protein synthesis (translation). Its primary specificity lies in its ability to link a specific genetic code (codon) to a **specific amino acid**. **Why the correct answer is right:** Each tRNA molecule is "charged" with a specific amino acid at its 3' end (the CCA tail). This process is catalyzed by the enzyme **aminoacyl-tRNA synthetase**, which ensures high fidelity by matching the tRNA's anticodon with the correct amino acid. This specificity is crucial because it ensures that the ribosome incorporates the exact amino acid dictated by the mRNA template. **Analysis of incorrect options:** * **ATP (A):** While ATP is required for the activation of amino acids (forming aminoacyl-adenylate), tRNA does not act "specifically" on ATP; ATP is a universal energy donor used by many enzymes. * **Golgi body (B):** The Golgi apparatus is involved in post-translational modification and sorting of proteins, not the translation process where tRNA functions. * **Ribosome (D):** tRNA interacts with the ribosome (at the A, P, and E sites), but the ribosome serves as the physical workbench. The tRNA's *specificity* is defined by the amino acid it carries, not the ribosome itself, which remains the same for all tRNAs. **NEET-PG High-Yield Pearls:** * **Wobble Hypothesis:** Explains why there are fewer tRNAs (approx. 30-40) than codons (61), allowing the 3rd base of the anticodon to have non-traditional pairing. * **Charging:** The attachment of an amino acid to tRNA is called "charging," and the enzyme aminoacyl-tRNA synthetase is often referred to as the "true genetic decoder." * **Structure:** tRNA has a secondary **cloverleaf structure** and a tertiary **L-shaped structure**.

Question 418: Which of the following statements is not true regarding the 5' cap of mRNA?

A. The 5' terminal of mRNA is capped by 7-methylguanosine triphosphate.
B. The 5' cap contains 2'-O-methyl ribonucleoside adjacent to it.
C. The cap is attached to the 5' end of mRNA through a 5' to 5' phosphodiester bond. (Correct Answer)
D. 7-methylguanosine is attached to mRNA through three intervening phosphate groups.

Explanation: ### Explanation The 5' cap is a critical post-transcriptional modification of eukaryotic mRNA. Understanding its structure is high-yield for NEET-PG. **1. Why Option C is the Correct (False) Statement:** The 5' cap is attached to the mRNA via a **5' to 5' triphosphate bridge**, not a standard phosphodiester bond. While a phosphodiester bond typically connects the 3' carbon of one sugar to the 5' carbon of the next, the cap involves a unique linkage between the 5' carbon of the 7-methylguanosine and the 5' carbon of the first transcribed nucleotide. This unconventional linkage protects the mRNA from degradation by 5' exonucleases. **2. Analysis of Other Options:** * **Option A (True):** The cap consists of a guanine residue methylated at the N7 position (**7-methylguanosine**). * **Option B (True):** In many eukaryotes, the ribose sugars of the first one or two nucleosides adjacent to the cap are methylated at the **2'-OH position** (Cap 1 and Cap 2 structures), which helps the cell distinguish "self" mRNA from viral RNA. * **Option D (True):** The linkage involves **three phosphate groups** (a triphosphate bridge), which are added by the enzyme guanylyltransferase. **3. Clinical Pearls & High-Yield Facts:** * **Functions of the Cap:** 1) Protection from exonucleases, 2) Facilitation of nuclear export, and 3) Recognition by the **eIF4F complex** for initiation of translation. * **Enzymatic Steps:** Capping occurs in the nucleus and involves three steps: RNA triphosphatase (removes γ-phosphate), Guanylyltransferase (adds GMP), and Guanine-7-methyltransferase (adds methyl group from **S-adenosylmethionine**). * **SAM (S-adenosylmethionine):** It is the universal methyl donor for the capping process. Deficiency in folate or Vitamin B12 can indirectly impair methylation reactions.

Question 419: Ultraviolet light can damage a DNA strand causing what type of alteration?

A. Two adjacent purine residues to form a covalently bonded dimer.
B. Two adjacent pyrimidine residues to form a covalently bonded dimer. (Correct Answer)
C. Disruption of phosphodiesterase linkage.
D. Disruption of non-covalent linkage.

Explanation: ### Explanation **Correct Answer: B. Two adjacent pyrimidine residues to form a covalently bonded dimer.** **Mechanism:** Ultraviolet (UV) radiation, specifically UV-B (280–320 nm), is a potent physical mutagen. When DNA is exposed to UV light, it induces the formation of **covalent bonds** between two adjacent pyrimidine bases (usually **Thymine-Thymine**, but sometimes Cytosine-Cytosine) on the same DNA strand. The most common lesion is the **Cyclobutane Pyrimidine Dimer (CPD)**. This dimer creates a "kink" or "bulge" in the DNA helix, which inhibits proper base pairing and stalls DNA polymerase during replication, potentially leading to mutations. **Analysis of Incorrect Options:** * **Option A:** UV light specifically targets pyrimidines (T, C), not purines (A, G). Purines are significantly more resistant to UV-induced dimerization. * **Option B:** While UV light provides energy to break bonds, its primary mutagenic effect is the *formation* of new covalent bonds (dimers) rather than the non-specific disruption of the sugar-phosphate backbone (phosphodiester linkage). Ionizing radiation (X-rays) is more likely to cause backbone breaks. * **Option D:** UV damage involves the formation of a stable **covalent** bond (a strong chemical bond), not just the disruption of weak non-covalent forces like hydrogen bonds. **Clinical Pearls for NEET-PG:** 1. **Repair Mechanism:** Pyrimidine dimers are repaired by **Nucleotide Excision Repair (NER)**. 2. **Clinical Correlation:** A deficiency in the NER pathway (specifically the UV-specific endonuclease) leads to **Xeroderma Pigmentosum**, characterized by extreme photosensitivity and a high risk of skin cancers (Basal Cell Carcinoma, Squamous Cell Carcinoma, and Melanoma). 3. **Direct Reversal:** In some organisms (not humans), the enzyme **Photolyase** can directly break these dimers using visible light (Photoreactivation).

Question 420: Which of the following is an example of post-translational modification?

A. Gamma carboxylation of glutamate residues
B. N-glycosylation of proteins (Correct Answer)
C. 7-methylguanosine capping
D. Poly(A) tail addition

Explanation: ### Explanation **Post-translational modification (PTM)** refers to the covalent and generally enzymatic modification of proteins following protein biosynthesis. These modifications occur after the messenger RNA (mRNA) has been translated into a polypeptide chain. **Correct Option: B. N-glycosylation of proteins** Glycosylation is the process of adding carbohydrate chains to proteins. **N-glycosylation** occurs in the **Lumen of the Rough Endoplasmic Reticulum (RER)**, where a carbohydrate is attached to the nitrogen atom of an **Asparagine** residue. This is a classic example of a PTM essential for protein folding, stability, and cell signaling. **Analysis of Incorrect Options:** * **A. Gamma carboxylation of glutamate residues:** While this is indeed a post-translational modification (crucial for clotting factors II, VII, IX, and X), the question likely seeks the most representative or "textbook" example of PTM in a general cellular context. *Note: In many standardized exams, if multiple PTMs are listed, focus on the site of occurrence (ER/Golgi).* * **C. 7-methylguanosine capping:** This is a **post-transcriptional** modification. It occurs in the nucleus on the 5' end of the pre-mRNA to protect it from nuclease degradation. * **D. Poly(A) tail addition:** This is also a **post-transcriptional** modification where a tail of adenine nucleotides is added to the 3' end of the mRNA. **High-Yield Clinical Pearls for NEET-PG:** * **N-glycosylation** occurs in the **RER**, while **O-glycosylation** (attachment to Serine/Threonine) occurs in the **Golgi apparatus**. * **I-Cell Disease:** Caused by a deficiency in the enzyme phosphotransferase, leading to a failure of mannose-6-phosphate tagging (a PTM) on lysosomal enzymes. * **Vitamin K** is a co-factor for the **gamma-carboxylation** of glutamate residues; inhibition of this PTM is the mechanism of action for Warfarin.

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

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