Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 321: What is the function of the sigma (s) subunit of prokaryotic RNA polymerase?

A. Binds the antibiotic rifampicin
B. Is inhibited by a-amanitin
C. Specifically recognizes the promoter site (Correct Answer)
D. Is part of the core enzyme

Explanation: ### Explanation **1. Why Option C is Correct:** In prokaryotes, the RNA polymerase (RNAP) exists in two forms: the **core enzyme** ($\alpha_2\beta\beta'\omega$) and the **holoenzyme** ($\alpha_2\beta\beta'\omega\sigma$). The core enzyme has 5'→3' polymerase activity but lacks the ability to identify where a gene begins. The **sigma ($\sigma$) subunit** is essential for **promoter recognition**. It specifically binds to the -10 (Pribnow box) and -35 sequences of the DNA, ensuring that transcription initiates at the correct site. Once the first few phosphodiester bonds are formed (initiation), the sigma factor dissociates, allowing the core enzyme to proceed with elongation. **2. Why Other Options are Incorrect:** * **Option A:** Rifampicin binds to the **beta ($\beta$) subunit** of bacterial RNA polymerase, inhibiting the initiation of transcription. It does not bind to the sigma subunit. * **Option B:** $\alpha$-amanitin (from the *Amanita phalloides* mushroom) is a potent inhibitor of **Eukaryotic RNA Polymerase II**. Prokaryotic RNA polymerase is resistant to $\alpha$-amanitin. * **Option C:** The sigma subunit is **not** part of the core enzyme. The core enzyme consists of $\alpha_2\beta\beta'\omega$. The addition of the sigma factor to the core enzyme creates the "Holoenzyme." **3. High-Yield Clinical Pearls for NEET-PG:** * **Rifampicin:** A first-line anti-tubercular drug. Mechanism: Inhibits the $\beta$-subunit of prokaryotic RNAP. * **Promoter Sequences:** In prokaryotes, these are the **Pribnow Box (-10)** and the **-35 sequence**. In eukaryotes, the equivalent is the **TATA box (Hogness box)** at -25. * **Sigma 70:** The most common sigma factor in *E. coli* used for general transcription. * **Actinomycin D:** Inhibits transcription in both prokaryotes and eukaryotes by intercalating into DNA.

Question 322: What is the action of telomerase?

A. DNA polymerase 1
B. DNA polymerase 2
C. DNA polymerase 3
D. Reverse transcriptase (Correct Answer)

Explanation: **Explanation:** **Telomerase** is a specialized ribonucleoprotein complex responsible for maintaining the length of telomeres (the repetitive TTAGGG sequences at the ends of eukaryotic chromosomes). 1. **Why Reverse Transcriptase is Correct:** Telomerase contains an internal RNA template (hTR) and a catalytic protein subunit called **hTERT (human Telomerase Reverse Transcriptase)**. It functions as an **RNA-dependent DNA polymerase**. It uses its own integral RNA molecule as a template to synthesize DNA, extending the 3' end of the chromosome to prevent the "end-replication problem." This process of converting an RNA code into DNA is the definition of reverse transcription. 2. **Why Other Options are Incorrect:** * **DNA Polymerase 1:** Primarily involved in prokaryotic DNA replication for removing RNA primers (5' to 3' exonuclease activity) and filling gaps. * **DNA Polymerase 2:** Involved in DNA repair mechanisms in prokaryotes. * **DNA Polymerase 3:** The primary enzyme responsible for the elongation of the leading and lagging strands during prokaryotic DNA replication. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **The "End-Replication Problem":** Conventional DNA polymerases cannot replicate the extreme 5' end of the lagging strand, leading to progressive chromosomal shortening. Telomerase solves this. * **Cellular Aging (Senescence):** Telomerase activity is low or absent in somatic cells, leading to aging. * **Cancer Association:** Approximately **85-90% of cancer cells** upregulate telomerase, granting them "replicative immortality." * **Stem Cells:** High telomerase activity is normally found in germ cells, stem cells, and hematopoietic cells. * **Disease Link:** Mutations in telomerase components lead to **Dyskeratosis Congenita** (characterized by bone marrow failure and mucosal leukoplakia).

Question 323: Which factor performs the termination process of protein synthesis?

A. Releasing factor
B. Stop codon
C. UAA codon
D. All of the above (Correct Answer)

Explanation: ### Explanation The termination of protein synthesis (translation) is a coordinated process that requires both genetic signals and specialized protein factors. **1. Why "All of the above" is correct:** Termination occurs when the ribosome encounters a **Stop Codon** (UAA, UAG, or UGA) in the A-site. These codons do not code for any amino acid and are not recognized by tRNA. Instead, they are recognized by **Releasing Factors (RFs)**. Therefore, both the specific codons (UAA) and the protein factors (RF) are essential components that perform the termination process. * **Stop Codons (B & C):** Also known as "nonsense codons," these act as the signal to halt elongation. **UAA (Ochre)** is one of the three primary stop codons (alongside UAG-Amber and UGA-Opal). * **Releasing Factors (A):** In eukaryotes (eRF) and prokaryotes (RF1, RF2, RF3), these proteins mimic the shape of tRNA. They bind to the stop codon and trigger the peptidyl transferase to hydrolyze the bond between the completed polypeptide chain and the tRNA, releasing the protein. **2. Analysis of Options:** * **Releasing Factor:** The actual protein machinery that executes the release of the polypeptide. * **Stop Codon:** The genetic "red light" that initiates the termination sequence. * **UAA Codon:** A specific example of a stop codon; its presence is a prerequisite for termination. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Energy Requirement:** Termination is an energy-dependent process requiring **GTP hydrolysis**. * **Prokaryotic RFs:** RF1 recognizes UAA/UAG; RF2 recognizes UAA/UGA; RF3 is a GTPase that helps RF1/RF2 dissociate. * **Eukaryotic RF:** A single factor, **eRF1**, recognizes all three stop codons. * **PTC (Premature Termination Codon):** Mutations creating a stop codon mid-sequence lead to truncated proteins, often seen in diseases like **Beta-thalassemia** or **Duchenne Muscular Dystrophy**. * **Aminoglycosides:** At high doses, these can interfere with termination by causing "read-through" of stop codons.

Question 324: Linker DNA is bound to which of the following histone?

A. H1 (Correct Answer)
B. H2A
C. H2B
D. H3

Explanation: ### Explanation The fundamental unit of chromatin is the **nucleosome**, which consists of a protein core wrapped by DNA. **1. Why H1 is the Correct Answer:** Histone **H1** is known as the **linker histone**. Unlike the other histones, it is not part of the nucleosome core particle. Instead, it binds to the "linker DNA" (the segment of DNA between adjacent nucleosomes) and the site where DNA enters and exits the core. Its primary role is to stabilize the nucleosome structure and facilitate the folding of the "beads-on-a-string" chromatin into more complex, higher-order structures like the **30-nm fiber**. **2. Why the Other Options are Incorrect:** * **H2A, H2B, H3, and H4** are known as **Core Histones**. * Two molecules of each (H2A, H2B, H3, and H4) combine to form an **octamer**. * Approximately 146 base pairs of DNA wrap around this octamer core to form the nucleosome core particle. These histones are involved in the internal structural integrity of the core, not the binding of linker DNA. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Charge:** Histones are highly **basic** proteins, rich in **Lysine and Arginine**, which allows them to bind strongly to the negatively charged phosphate backbone of DNA. * **Epigenetics:** Histone modification (Acetylation/Methylation) occurs primarily at the N-terminal tails. **Acetylation** (by HATs) neutralizes the positive charge, relaxing chromatin and increasing transcription (**Euchromatin**). * **Drug Connection:** **Hydralazine** and **Procainamide** can induce Systemic Lupus Erythematosus (Drug-induced SLE), where **Anti-histone antibodies** are a characteristic diagnostic marker. * **Evolutionary Conservation:** H3 and H4 are among the most highly conserved proteins across species.

Question 325: Which of the following diseases is NOT characterized by a defect in DNA repair mechanisms?

A. Xeroderma Pigmentosa
B. Fanconi syndrome
C. Huntington's disease (Correct Answer)
D. Hereditary nonpolyposis colon cancer

Explanation: **Explanation:** The correct answer is **Huntington’s disease**. Unlike the other options, Huntington’s disease is a **Trinucleotide Repeat Expansion disorder** (specifically CAG repeats on chromosome 4). It is characterized by protein misfolding and aggregation (Huntingtin protein) rather than a primary defect in DNA repair. **Analysis of Options:** * **Xeroderma Pigmentosa (XP):** A classic DNA repair disorder caused by a defect in **Nucleotide Excision Repair (NER)**. Patients cannot repair pyrimidine dimers formed by UV light, leading to extreme photosensitivity and early skin cancers. * **Fanconi Anemia (Note: Option says Fanconi syndrome, but in the context of DNA repair, Fanconi Anemia is implied):** This is caused by defects in the **interstrand cross-link repair** mechanism. It presents with bone marrow failure, physical abnormalities, and high cancer risk. (Note: *Fanconi Syndrome* is a renal proximal tubule defect; *Fanconi Anemia* is the DNA repair defect). * **Hereditary Nonpolyposis Colon Cancer (HNPCC/Lynch Syndrome):** Caused by mutations in **Mismatch Repair (MMR)** genes (e.g., MSH2, MLH1). This leads to microsatellite instability and increased risk of colorectal and endometrial cancers. **High-Yield Clinical Pearls for NEET-PG:** * **Ataxia Telangiectasia:** Defect in **Non-homologous end joining (NHEJ)** or ATM gene (double-strand break repair). * **Bloom Syndrome / Cockayne Syndrome:** Also categorized as DNA repair/stability defects. * **Huntington’s Disease Hallmark:** Shows **"Anticipation"** (earlier onset in successive generations) and is inherited in an Autosomal Dominant pattern.

Question 326: What are the components of the 50S ribosomal subunit?

A. 5S RNA, 5.8S & 28S RNA
B. 16S RNA & 23S RNA
C. 23S RNA & 5.8S RNA
D. 23S RNA & 5S RNA (Correct Answer)

Explanation: ### Explanation The ribosome is the cellular machinery responsible for protein synthesis (translation). In prokaryotes (like bacteria), the ribosome is **70S**, consisting of a small **30S** subunit and a large **50S** subunit. **1. Why Option D is Correct:** The **50S (large) subunit** of the prokaryotic ribosome is composed of two types of ribosomal RNA (rRNA) and approximately 34 proteins (L1–L34). The specific rRNA components are: * **23S rRNA:** Acts as a ribozyme (peptidyl transferase) to catalyze peptide bond formation. * **5S rRNA:** Provides structural stability. **2. Analysis of Incorrect Options:** * **Option A (5S, 5.8S, & 28S):** These are the components of the **60S (large) subunit of Eukaryotic ribosomes**. Eukaryotes have an 80S ribosome (40S + 60S). * **Option B (16S & 23S):** 16S rRNA is the signature component of the **30S (small) subunit** in prokaryotes, not the 50S subunit. * **Option C (23S & 5.8S):** This is a mismatch. 5.8S is exclusively eukaryotic, while 23S is exclusively prokaryotic. **3. High-Yield Clinical Pearls for NEET-PG:** * **Antibiotic Targets:** Many antibiotics selectively target the 50S subunit (e.g., **Macrolides, Chloramphenicol, Clindamycin, and Linezolid**) or the 30S subunit (e.g., **Aminoglycosides, Tetracyclines**). This selectivity is based on the structural differences between prokaryotic (70S) and eukaryotic (80S) ribosomes. * **Shine-Dalgarno Sequence:** The **16S rRNA** (of the 30S subunit) contains a sequence complementary to the mRNA's Shine-Dalgarno sequence, which is essential for initiation of translation in bacteria. * **Svedberg Unit (S):** It measures the sedimentation rate, which depends on both mass and shape (which is why 50S + 30S = 70S, not 80S).

Question 327: Mitochondrial DNA is:

A. Closed circular (Correct Answer)
B. Nicked circular
C. Linear
D. Open circular

Explanation: **Explanation:** **1. Why "Closed Circular" is Correct:** Mitochondrial DNA (mtDNA) is a double-stranded, **closed circular** molecule. Unlike nuclear DNA, which is organized into linear chromosomes, mtDNA resembles the genomic structure of prokaryotes (supporting the endosymbiotic theory). It is "closed" because the ends are covalently linked, forming a continuous loop without any free 5' or 3' ends. Each mitochondrion contains multiple copies of this 16.5 kb genome, which encodes 13 polypeptides of the respiratory chain, 22 tRNAs, and 2 rRNAs. **2. Why Other Options are Incorrect:** * **B. Nicked circular:** A "nicked" circle has a break in one of the phosphodiester backbones. While this can occur during DNA damage or replication, the native, functional state of mtDNA is intact and closed. * **C. Linear:** Nuclear DNA in eukaryotes is linear. Linear DNA requires telomeres to prevent degradation; mtDNA lacks telomeres and exists as a circle to maintain stability within the mitochondrial matrix. * **D. Open circular:** This term usually refers to a plasmid or circular DNA that has been relaxed by a single-strand cut (nick). Native mtDNA is supercoiled and closed. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Maternal Inheritance:** mtDNA is inherited exclusively from the mother because sperm mitochondria are degraded in the zygote. * **Lack of Introns:** Unlike nuclear DNA, mtDNA is highly compact and contains almost no non-coding sequences (introns). * **High Mutation Rate:** mtDNA lacks robust repair mechanisms and histones, making it 10 times more prone to mutations than nuclear DNA. * **Heteroplasmy:** This refers to the presence of a mixture of more than one type of organellar genome (normal and mutant) within a single cell, which explains the clinical variability in mitochondrial diseases like **MELAS** or **Leber’s Hereditary Optic Neuropathy (LHON)**.

Question 328: In sickle cell anaemia, what is the genetic defect?

A. A base insertion in DNA
B. A base deletion in DNA
C. A base substitution in DNA (Correct Answer)
D. None of the above

Explanation: Sickle cell anemia is a classic example of a **point mutation**, specifically a **missense mutation** involving a **base substitution**. ### Why the correct answer is right: The genetic defect occurs in the **$\beta$-globin gene** on chromosome 11. A single base substitution occurs where **Adenine (A) is replaced by Thymine (T)** in the DNA (GAG $\rightarrow$ GTG). This changes the mRNA codon from GAG to GUG. Consequently, the amino acid **Glutamic acid** (polar/acidic) is replaced by **Valine** (non-polar/hydrophobic) at the **6th position** of the $\beta$-chain. This substitution creates a "sticky patch" on the hemoglobin molecule, leading to polymerization under deoxygenated conditions and the characteristic "sickle" shape of RBCs. ### Why the incorrect options are wrong: * **A & B (Insertion/Deletion):** These mutations typically cause a **frameshift**, altering the entire reading frame downstream of the mutation. This usually results in a non-functional protein or a premature stop codon (nonsense mutation). Sickle cell anemia involves a single amino acid change, not a shift in the reading frame. * **D (None of the above):** Incorrect, as the mechanism is a well-defined molecular substitution. ### NEET-PG High-Yield Pearls: * **Type of Substitution:** It is a **Transversion** (Purine 'A' replaced by Pyrimidine 'T'). * **Electrophoresis:** On alkaline electrophoresis, HbS moves **slower** than HbA towards the anode because the loss of Glutamic acid makes the molecule less negatively charged. * **Protective Effect:** Heterozygotes (Sickle cell trait) show resistance to *Plasmodium falciparum* malaria. * **Diagnosis:** The gold standard is **Hemoglobin Electrophoresis** or **HPLC**. Screening is done via the Solubility test (Sodium dithionite).

Question 329: Which gene product is transcribed but not translated?

A. Glycosyl transferase
B. t-RNA (Correct Answer)
C. Keratin
D. Histone

Explanation: ### Explanation The central dogma of molecular biology involves two primary steps: **Transcription** (DNA to RNA) and **Translation** (RNA to Protein). While most genes code for proteins, a significant portion of the genome codes for **Non-coding RNAs (ncRNAs)**. These molecules are transcribed but function directly as RNA without ever being translated into proteins. **Why t-RNA is the Correct Answer:** Transfer RNA (t-RNA) is a classic example of a non-coding RNA. It is transcribed by **RNA Polymerase III** in eukaryotes. Its primary function is to act as an "adapter" molecule during protein synthesis, carrying specific amino acids to the ribosome. Because the t-RNA molecule itself is the final functional product, it does not undergo translation. Other examples include r-RNA, snRNA, and miRNA. **Analysis of Incorrect Options:** * **A. Glycosyl transferase:** This is an enzyme (protein) responsible for attaching carbohydrate chains to proteins or lipids. Being a protein, it must be translated from mRNA. * **C. Keratin:** This is a structural intermediate filament protein found in hair, skin, and nails. It is the product of mRNA translation. * **D. Histone:** Histones are highly alkaline proteins that package DNA into nucleosomes. They are translated from specific histone mRNAs (which are unique because they lack a poly-A tail). **NEET-PG High-Yield Pearls:** * **RNA Polymerase Roles:** Remember **1, 2, 3 → R, M, T** (Pol I for r-RNA, Pol II for m-RNA, Pol III for t-RNA). * **t-RNA Structure:** Contains unusual bases like **Pseudouridine** and **Dihydrouridine**. The 3' end always carries the sequence **CCA**, which is the amino acid attachment site. * **Small Nuclear RNA (snRNA):** Another transcribed-only molecule; it is essential for **splicing** (removal of introns).

Question 330: Which of the following statements is true about DNA structure?

A. Purines are adenine and guanine, and pyrimidines are uracil and cytosine.
B. Watson and Crick discovered the DNA structure in 1973.
C. It has a deoxyribose-phosphate backbone with bases stacked inside. (Correct Answer)
D. It primarily consists of a left-handed helix.

Explanation: **Explanation:** The correct answer is **C**. DNA (Deoxyribonucleic acid) is a double-stranded helical molecule. Its structural integrity is maintained by a **hydrophilic sugar-phosphate backbone** (composed of deoxyribose sugar and phosphodiester bonds) located on the exterior, while the **hydrophobic nitrogenous bases** are stacked on the interior. These bases are perpendicular to the helical axis, stabilized by hydrogen bonding between complementary strands and van der Waals forces between stacked bases. **Analysis of Incorrect Options:** * **Option A:** While Adenine (A) and Guanine (G) are purines, the pyrimidines in DNA are **Cytosine (C) and Thymine (T)**. Uracil (U) is found exclusively in RNA, replacing Thymine. * **Option B:** James Watson and Francis Crick proposed the double-helix model in **1953** (not 1973), based on Rosalind Franklin’s X-ray diffraction data. They were awarded the Nobel Prize in 1962. * **Option D:** The physiological form of DNA (B-DNA) is a **right-handed helix**. Left-handed helices (Z-DNA) are rare and typically occur in specific genomic regions with alternating purine-pyrimidine sequences. **High-Yield Clinical Pearls for NEET-PG:** * **Chargaff’s Rule:** In double-stranded DNA, the amount of A = T and G = C; therefore, Purines = Pyrimidines. * **Bonding:** A-T pairs have 2 hydrogen bonds, while G-C pairs have 3. Higher G-C content increases the **Melting Temperature (Tm)** of DNA. * **DNA Forms:** * **B-DNA:** Most common, right-handed, 10 base pairs (bp) per turn. * **A-DNA:** Right-handed, dehydrated form, 11 bp per turn. * **Z-DNA:** Left-handed, zigzag backbone, 12 bp per turn.

Practice by Chapter

DNA Replication and Repair Mechanisms

Practice Questions

Transcription Factors and Gene Regulation

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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

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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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