Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 611: Termination of RNA molecule synthesis is signaled by a specific sequence in the DNA template strand, which is recognized by which termination protein?

A. s factor
B. d factor
C. e factor
D. Rho (r) factor (Correct Answer)

Explanation: **Explanation:** In prokaryotic transcription, the termination of RNA synthesis occurs via two primary mechanisms: **Rho-independent** (intrinsic) and **Rho-dependent** termination. 1. **Why the Correct Answer is Right:** The **Rho (ρ) factor** is an ATP-dependent hexameric helicase protein. It binds to a specific C-rich sequence on the nascent RNA chain known as the **rut (Rho utilization) site**. Once bound, it moves along the RNA toward the RNA polymerase. When the polymerase pauses at a termination sequence on the DNA template, the Rho factor catches up, uses its ATPase activity to unwind the RNA-DNA hybrid, and facilitates the release of the RNA transcript. 2. **Analysis of Incorrect Options:** * **s factor (Sigma factor):** This is involved in the **initiation** of transcription. It helps RNA polymerase recognize and bind to the promoter sequence. It dissociates shortly after transcription begins. * **d factor (Delta) and e factor (Epsilon):** These are not standard nomenclature for transcription termination factors in prokaryotes. In eukaryotes, various subunits exist for DNA polymerases (like Pol δ and ε), but they are involved in DNA replication, not RNA termination. 3. **High-Yield Clinical Pearls for NEET-PG:** * **Rifampicin:** Inhibits the **initiation** of transcription by binding to the β-subunit of bacterial RNA polymerase (used in TB treatment). * **Actinomycin D:** Inhibits transcription by intercalating into the DNA template, preventing elongation (used in oncology). * **Alpha-amanitin:** Found in *Amanita phalloides* (death cap mushroom); it specifically inhibits **RNA Polymerase II** in eukaryotes, leading to severe liver failure.

Question 612: What is the bond between the strands in the given diagram?

A. Hydrogen bond (Correct Answer)
B. Phosphodiester bond
C. Covalent bond
D. Glycosidic bond

Explanation: ***Hydrogen bond*** - The **two strands** of the DNA double helix are held together by **hydrogen bonds** between complementary base pairs (**A-T** forms 2 H-bonds, **G-C** forms 3 H-bonds). - These **non-covalent interactions** allow for **strand separation** during processes like replication and transcription while maintaining structural stability. *Phosphodiester bond* - This bond connects the **sugar-phosphate backbone** within a single DNA strand, not between strands. - It forms the **covalent linkage** between the 3' carbon of one sugar and the 5' phosphate of the next nucleotide. *Covalent bond* - While **phosphodiester** and **glycosidic bonds** are types of covalent bonds, they don't directly connect the two DNA strands. - The inter-strand connections are specifically **weak hydrogen bonds**, not strong covalent bonds. *Glycosidic bond* - This bond connects the **nitrogenous base** to the **sugar (deoxyribose)** within each nucleotide. - It's an **intra-nucleotide bond**, not an inter-strand connection in the DNA double helix.

Question 613: Presence of which of the following in the expression vector ensures an increase in the yield of recombinant protein produced?

A. Inducible promoter (Correct Answer)
B. Gene coding for protease inhibitor
C. Translation initiation signals
D. Transcription and translation termination signals

Explanation: ### Explanation **1. Why Inducible Promoter is Correct:** In recombinant DNA technology, the goal is to maximize the yield of a specific protein. However, high levels of a foreign protein can often be **toxic** to the host cell (e.g., *E. coli*), inhibiting its growth and leading to low biomass. An **inducible promoter** (like the *lac* promoter) allows researchers to decouple the growth phase from the production phase. The host cells are first grown to a high density; then, an inducer (like IPTG) is added to "switch on" the gene. This ensures that the protein is produced rapidly and in large quantities only after a healthy population of host cells has been established, thereby maximizing the final yield. **2. Analysis of Incorrect Options:** * **B. Gene coding for protease inhibitor:** While preventing protein degradation is helpful, it is not a standard component of an expression vector. Usually, protease-deficient host strains are used instead. * **C. Translation initiation signals (e.g., Shine-Dalgarno sequence):** These are essential for the protein to be synthesized at all, but they do not "increase" the yield in the same regulatory manner as a promoter that controls the timing of expression. * **D. Transcription and translation termination signals:** These ensure the production of a discrete, functional protein and prevent "read-through," but they do not drive the overproduction or yield of the protein. **3. High-Yield Clinical Pearls for NEET-PG:** * **Expression Vector vs. Cloning Vector:** An expression vector must contain a promoter, a ribosome-binding site, and a termination signal, whereas a cloning vector only requires an Origin of Replication (ori), a selectable marker, and a Multiple Cloning Site (MCS). * **Common Inducer:** **IPTG** (Isopropyl β-D-1-thiogalactopyranoside) is a non-metabolizable analogue of lactose used to induce the *lac* operon in expression vectors. * **Medical Application:** Recombinant human insulin (Humulin) and growth hormone are produced using these expression vectors in *E. coli* or yeast.

Question 614: True about telomerase?

A. Has RNA polymerase activity
B. Causes carcinogenesis (Correct Answer)
C. Present in somatic cells
D. Absent in germ cells

Explanation: **Explanation:** **Telomerase** is a specialized ribonucleoprotein complex (a reverse transcriptase) that maintains chromosomal stability by adding repetitive DNA sequences (TTAGGG) to the 3' ends of chromosomes, preventing the "end-replication problem." **Why Option B is Correct:** In normal somatic cells, telomerase is inactive, leading to progressive telomere shortening and eventual cellular senescence. However, **90% of cancer cells** reactivate telomerase. This allows them to maintain telomere length indefinitely, bypassing the Hayflick limit and achieving **replicative immortality**, a hallmark of carcinogenesis. **Analysis of Incorrect Options:** * **Option A:** Telomerase has **RNA-dependent DNA polymerase** activity (Reverse Transcriptase), not RNA polymerase activity. It uses its own internal RNA template to synthesize DNA. * **Option C:** Telomerase is generally **absent or at very low levels in mature somatic cells**, which is why these cells have a finite lifespan. * **Option D:** Telomerase is **highly active in germ cells**, stem cells, and embryonic cells to ensure that the full length of the genome is passed on to the next generation. **High-Yield Clinical Pearls for NEET-PG:** * **Components:** It consists of **TERT** (Telomerase Reverse Transcriptase - the catalytic protein) and **TERC** (Telomerase RNA component - the template). * **Shelterin Complex:** A protein complex that protects telomeres from being recognized as DNA double-strand breaks. * **Progeria (Hutchinson-Gilford Syndrome):** Characterized by accelerated telomere shortening. * **Diagnostic Marker:** Telomerase activity is a potential biomarker for malignancy and a target for anti-cancer therapies (e.g., Imetelstat).

Question 615: Which of the following statements is incorrect?

A. Mutations within an exon can be detrimental, and mutations in introns will not change the protein.
B. Mutation of the TATA box markedly reduces transcription.
C. A change of a purine by a pyrimidine is known as a transversion. (Correct Answer)
D. A silent mutation does not lead to a change in amino acid.

Explanation: ### Explanation **1. Why Option C is the Correct Answer (The "Incorrect" Statement)** The question asks for the **incorrect** statement. Option C is actually a **correct** definition: a transversion is the substitution of a purine (A, G) for a pyrimidine (C, T) or vice versa. However, in the context of this specific question's key, **Option A** is the scientifically incorrect statement that should be identified. *Note: If the provided key marks C as the answer, it is likely a typographical error in the source material, as C is a factually correct definition. In NEET-PG, **Option A** is the classic "incorrect" statement because it falsely claims mutations in introns have no effect.* **2. Analysis of Options** * **Option A (Incorrect Statement):** While exons code for proteins, mutations in **introns** can be highly detrimental. They can disrupt **splice donor or acceptor sites**, leading to exon skipping or intron retention, which results in truncated or dysfunctional proteins (e.g., β-thalassemia). * **Option B (Correct Statement):** The TATA box is a core promoter element. Mutations here impair the binding of RNA Polymerase II and transcription factors, significantly reducing the rate of transcription. * **Option C (Correct Statement):** Transition = Purine to Purine (A↔G) or Pyrimidine to Pyrimidine (C↔T). Transversion = Purine to Pyrimidine (or vice versa). * **Option D (Correct Statement):** Silent mutations occur due to the **degeneracy** of the genetic code; a base change results in a codon that codes for the same amino acid. **3. Clinical Pearls & High-Yield Facts** * **Splice Site Mutations:** A common cause of genetic diseases. If a mutation occurs at the conserved **GT (5' donor)** or **AG (3' acceptor)** sequences of an intron, splicing fails. * **Transition vs. Transversion:** Transitions are more common in the genome than transversions. * **Frameshift Mutation:** Caused by insertion/deletion of bases not divisible by three; usually results in a premature stop codon.

Question 616: What is the function of reverse transcriptase?

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

Explanation: ### Explanation **1. Why Option B is Correct:** Reverse transcriptase (RT) is an enzyme that synthesizes a complementary DNA (cDNA) strand using an RNA template. In biochemical nomenclature, the first part of the name refers to the **template** used, and the second part refers to the **product** synthesized. Since RT uses **RNA** as a template to produce **DNA**, it is classified as an **RNA-dependent DNA polymerase**. This process reverses the "Central Dogma" of molecular biology (DNA → RNA). **2. Analysis of Incorrect Options:** * **Option A (DNA-dependent RNA polymerase):** This enzyme uses DNA as a template to synthesize RNA. This is the standard enzyme for **Transcription** (e.g., RNA Polymerase I, II, and III). * **Option C (DNA-dependent DNA polymerase):** This enzyme uses DNA as a template to synthesize a new DNA strand. This is the primary enzyme for **DNA Replication** (e.g., DNA Polymerase α, δ, ε). * **Option D (RNA-dependent RNA polymerase):** This enzyme uses RNA as a template to synthesize RNA. It is found in certain RNA viruses (like Poliovirus or SARS-CoV-2) to replicate their genetic material. **3. NEET-PG High-Yield Clinical Pearls:** * **Retroviruses:** Reverse transcriptase is a hallmark of Retroviridae (e.g., **HIV**). It allows the viral RNA genome to be integrated into the host's double-stranded DNA. * **Telomerase:** A specialized reverse transcriptase (TERT) that maintains chromosomal ends (telomeres) using an internal RNA template. * **Pharmacology Link:** Nucleoside Reverse Transcriptase Inhibitors (**NRTIs**) like Zidovudine (AZT) and Non-Nucleoside Reverse Transcriptase Inhibitors (**NNRTIs**) like Efavirenz are key components of HAART for HIV treatment. * **Laboratory Use:** RT is essential in **RT-PCR**, where it converts viral RNA into cDNA before amplification.

Question 617: During the translation process, which enzyme is responsible for the proofreading of mRNA?

A. RNA polymerase
B. Aminoacyl-tRNA synthetase (Correct Answer)
C. Leucine zipper
D. DNA

Explanation: **Explanation:** The fidelity of protein synthesis depends on the high specificity of **Aminoacyl-tRNA synthetases (aaRS)**. This enzyme is responsible for "charging" tRNA by attaching the correct amino acid to its corresponding tRNA. 1. **Why Option B is Correct:** Aminoacyl-tRNA synthetase performs a critical **proofreading (editing) function**. It possesses two active sites: an *activation site* (where the amino acid is linked to AMP) and an *editing site*. If an incorrect amino acid (usually one with a similar size or charge) is attached, the enzyme recognizes the mismatch and hydrolyzes the bond, removing the wrong amino acid before the tRNA is released. This ensures that the genetic code is translated accurately. 2. **Why Other Options are Incorrect:** * **A. RNA Polymerase:** This enzyme is involved in **transcription** (DNA to RNA), not translation. While it has some inherent error-correction capability, it does not proofread mRNA during the translation process. * **C. Leucine Zipper:** This is a common **structural motif** found in DNA-binding proteins (transcription factors), characterized by a periodic repetition of leucine residues. It is involved in gene regulation, not enzymatic proofreading. * **D. DNA:** DNA is the genetic template. While DNA polymerases have 3'→5' exonuclease activity for proofreading during replication, DNA itself has no enzymatic role in translation. **Clinical Pearls & High-Yield Facts:** * **Double Sieve Mechanism:** aaRS uses a "double sieve" method—the first sieve excludes amino acids that are too large, and the second (editing) site hydrolyzes amino acids that are too small or chemically incorrect. * **Energy Requirement:** The charging of tRNA requires **ATP**, which is converted to AMP and inorganic pyrophosphate (PPi). * **Mnemonic:** "Translation fidelity is the 'Synthetase's' responsibility."

Question 618: In which direction does DNA replication and transcription occur?

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

Explanation: **Explanation:** The synthesis of new nucleic acid strands (both DNA and RNA) occurs exclusively in the **5' to 3' direction**. **Why 5' to 3' is correct:** This directionality is determined by the biochemical requirement of the enzymes involved (DNA Polymerase and RNA Polymerase). These enzymes can only add a new nucleotide to the **free 3'-hydroxyl (-OH) group** of the preceding sugar molecule. During this process, the 5'-phosphate group of the incoming deoxynucleoside triphosphate (dNTP) or nucleoside triphosphate (NTP) reacts with the 3'-OH group of the growing chain, forming a phosphodiester bond. While the template strand is read in the 3' to 5' direction, the new strand is always synthesized 5' to 3'. **Why other options are incorrect:** * **3' to 5' (Option C):** This is the direction in which the **template** strand is read, but not the direction of synthesis. No known biological polymerase can add nucleotides to a 5' end. * **5' to 5' and 3' to 3' (Options B & D):** These are biochemically impossible as they would violate the antiparallel nature of nucleic acid pairing and the specific enzymatic mechanism of phosphodiester bond formation. **High-Yield Clinical Pearls for NEET-PG:** * **Okazaki Fragments:** Because DNA synthesis is restricted to the 5'-3' direction, the "lagging strand" must be synthesized discontinuously in short segments. * **Proofreading:** DNA Polymerase has **3'-5' exonuclease activity**, which allows it to move backward to remove mismatched bases—this is the "editing" function. * **Reverse Transcriptase:** Even in retroviruses (like HIV), the synthesis of DNA from an RNA template follows the universal **5' to 3'** rule. * **Z-DNA:** While most DNA is a right-handed helix (B-DNA), Z-DNA is a left-handed helix, but the chemical synthesis direction remains 5'-3'.

Question 619: Apolipoprotein B-48 and apolipoprotein B-100 are expressed as two different apoproteins due to a difference in which of the following?

A. RNA editing (Correct Answer)
B. RNA splicing
C. Chromosomal loci
D. The Apo-B gene

Explanation: **Explanation:** The correct answer is **RNA editing**. This is a post-transcriptional modification where the nucleotide sequence of the mRNA is altered after transcription but before translation. **Why RNA Editing is Correct:** Both Apo B-48 and Apo B-100 are encoded by the **same gene** located on chromosome 2. The difference in their expression is tissue-specific: * **Liver:** The full-length mRNA is translated into **Apo B-100**, which is essential for VLDL and LDL synthesis. * **Intestine:** An enzyme called **cytidine deaminase** acts on the mRNA, converting a specific Cytosine (C) to Uracil (U). This changes the codon **CAA** (coding for Glutamine) into **UAA**, which is a **stop codon**. This results in premature termination of translation, producing a protein that is only 48% of the original length—hence, **Apo B-48**, which is essential for chylomicron formation. **Why Other Options are Incorrect:** * **B. RNA Splicing:** This involves the removal of introns and joining of exons. While alternative splicing creates protein diversity, it is not the mechanism for Apo B-48/100. * **C & D. Chromosomal Loci/The Apo-B Gene:** These are incorrect because both proteins originate from the **same single gene** on the same locus. There is no genomic difference; the variation occurs at the mRNA level. **High-Yield Clinical Pearls for NEET-PG:** * **Apo B-100:** Found in VLDL, IDL, and LDL. It acts as a ligand for the **LDL receptor**. * **Apo B-48:** Found in Chylomicrons and Chylomicron remnants. It lacks the LDL receptor-binding domain. * **Mnemonic:** **L**iver = **L**ong (B-100); **I**ntestine = **I**ncomplete/Short (B-48). * **Abetalipoproteinemia:** A condition caused by a deficiency in Microsomal Triglyceride Transfer Protein (MTP), leading to an absence of both Apo B-48 and B-100.

Question 620: The Shine-Dalgarno sequence in prokaryotes is primarily associated with which molecular process?

A. Transcription
B. Translation (Correct Answer)
C. Replication
D. Translocation

Explanation: **Explanation:** The **Shine-Dalgarno (SD) sequence** is a critical regulatory element in prokaryotic **Translation** (Option B). It is a purine-rich sequence (typically AGGAGG) located approximately 8 base pairs upstream of the AUG start codon on the mRNA. Its primary function is to serve as the **ribosomal binding site**. The SD sequence base-pairs with a complementary pyrimidine-rich sequence at the **3' end of the 16S rRNA** (part of the 30S small ribosomal subunit). This interaction ensures the correct alignment of the ribosome on the mRNA, allowing translation to begin at the proper initiation codon. **Analysis of Incorrect Options:** * **A. Transcription:** This is the synthesis of RNA from DNA. The key regulatory sequences here are **Promoters** (e.g., Pribnow box/TATA box), not the SD sequence. * **C. Replication:** This involves DNA synthesis. Key elements include the **Origin of Replication (OriC)** and primers, unrelated to mRNA-ribosome binding. * **D. Translocation:** This is a specific step *within* translation where the ribosome moves along the mRNA. While related to the overall process, the SD sequence is specifically involved in **Initiation**, not the translocation phase. **High-Yield Clinical Pearls for NEET-PG:** * **Kozak Sequence:** The eukaryotic functional equivalent of the Shine-Dalgarno sequence. * **16S rRNA:** The specific component of the prokaryotic 30S subunit that recognizes the SD sequence. * **Aminoglycosides:** This class of antibiotics (e.g., Streptomycin) acts by binding to the 30S subunit, interfering with the initiation and fidelity of translation. * **Polycistronic mRNA:** In bacteria, multiple SD sequences can exist on a single mRNA strand, allowing for the translation of multiple proteins from one transcript.

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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