Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 131: What is true about restriction enzymes?

A. They recognize palindromic sequences. (Correct Answer)
B. They produce DNA sticky ends.
C. They restrict DNA replication.
D. Their restriction sites are not specific.

Explanation: **Explanation:** **Restriction Endonucleases (REs)**, often called "molecular scissors," are enzymes that cleave double-stranded DNA at specific recognition sites. **1. Why Option A is Correct:** Restriction enzymes recognize **palindromic sequences**. In genetics, a palindrome is a sequence where the 5' to 3' reading on one strand is identical to the 5' to 3' reading on the complementary strand (e.g., 5'-GAATTC-3' paired with 3'-CTTAAG-5'). These enzymes bind as homodimers to these symmetrical sites to execute precise cleavage. **2. Why Other Options are Incorrect:** * **Option B:** While many REs (like *EcoRI*) produce **sticky (cohesive) ends**, others (like *HpaI* or *SmaI*) produce **blunt ends**. Therefore, producing sticky ends is a common feature but not a universal truth for all restriction enzymes. * **Option C:** Their primary role is to "restrict" the entry of foreign (viral) DNA into bacteria, not to inhibit the process of DNA replication itself. * **Option D:** Restriction sites are **highly specific**. Each enzyme recognizes a unique sequence (usually 4–8 base pairs). A single base mutation in the recognition site can prevent cleavage, a principle used in RFLP (Restriction Fragment Length Polymorphism) analysis. **High-Yield Clinical Pearls for NEET-PG:** * **Source:** These enzymes are naturally found in bacteria as a defense mechanism against bacteriophages. * **Nomenclature:** The first letter is the Genus, the next two are the species, and the Roman numeral denotes the order of discovery (e.g., *EcoRI*: *Escherichia coli*, strain R, 1st enzyme). * **Type II REs:** These are the most commonly used in recombinant DNA technology because they cleave within or at a fixed distance from their specific recognition site and do not require ATP. * **Application:** Essential for gene cloning, DNA fingerprinting, and Southern Blotting.

Question 132: A decrease in the level of heme leads to a reduction in globin synthesis in reticulocytes. Which of the following best explains this phenomenon?

A. RNA polymerase activity is decreased in reticulocytes by low heme
B. The initiation factor eIF-2 becomes phosphorylated, reducing its level of activity (Correct Answer)
C. Heme normally activates peptidyl transferase in reticulocytes
D. A tRNA degrading enzyme is active in the absence of heme

Explanation: This question tests your understanding of the **translational control of protein synthesis**, specifically how reticulocytes coordinate the production of globin chains with the availability of heme. ### **Explanation of the Correct Answer (B)** In reticulocytes (which lack a nucleus), protein synthesis is regulated at the **initiation stage** of translation. The key regulator is an enzyme called **Heme-Regulated Inhibitor (HRI)**, also known as Heme-Controlled Inhibitor (HCI). * **When Heme is present:** It binds to HRI and keeps it inactive. Translation proceeds normally. * **When Heme is low:** HRI becomes active as a protein kinase. It **phosphorylates the alpha-subunit of eIF-2** (Eukaryotic Initiation Factor 2). * **The Result:** Phosphorylated eIF-2 forms an inactive complex with eIF-2B (the guanine nucleotide exchange factor). This prevents the recycling of GDP to GTP on eIF-2, effectively halting the initiation of translation. This ensures that globin is not synthesized in excess when heme is unavailable. ### **Why the Other Options are Incorrect** * **Option A:** Reticulocytes are enucleated cells; they do not perform significant transcription. Therefore, RNA polymerase activity is not the primary regulatory site for globin synthesis. * **Option C:** Peptidyl transferase is a ribozyme (28S rRNA) involved in the elongation phase (peptide bond formation). It is not regulated by heme levels. * **Option D:** While mRNA stability can be regulated, there is no specific "tRNA degrading enzyme" triggered by heme deficiency that controls this process. ### **High-Yield NEET-PG Pearls** * **eIF-2 Function:** It is responsible for bringing the initiator methionyl-tRNA (Met-tRNAi) to the 40S ribosomal subunit. * **Rate-Limiting Step:** The phosphorylation of eIF-2 is a classic example of the rate-limiting step in eukaryotic translation initiation. * **Clinical Correlation:** This mechanism prevents the accumulation of free globin chains, which are toxic to the cell and can lead to proteotoxicity (similar to the pathophysiology seen in Thalassemia).

Question 133: What does CRISPR stand for?

A. Clustered regularly interspaced short palindromic repeats (Correct Answer)
B. Clustered Regularly in short polymerase reaction
C. Cryptococcal immune short pole reaction
D. None of the above

Explanation: **Explanation:** **1. Why Option A is Correct:** CRISPR stands for **Clustered Regularly Interspaced Short Palindromic Repeats**. This term describes a specific segment of prokaryotic DNA (found in bacteria and archaea) that serves as an adaptive immune system. * **Clustered/Regularly Interspaced:** The DNA contains short, unique "spacer" sequences derived from previous viral infections. * **Palindromic Repeats:** These spacers are separated by identical, repeating sequences that read the same forward and backward. When paired with the **Cas9 enzyme** (CRISPR-associated protein 9), this system acts as "molecular scissors," allowing for precise genome editing by cutting specific DNA sequences. **2. Why Other Options are Incorrect:** * **Option B:** "Short polymerase reaction" is a distractor confusing CRISPR with PCR (Polymerase Chain Reaction). CRISPR is a gene-editing tool, not a thermal cycling amplification method. * **Option C:** "Cryptococcal immune short pole reaction" is a fabricated medical term. While CRISPR is an immune mechanism, it is unrelated to the fungus *Cryptococcus*. **3. NEET-PG High-Yield Pearls:** * **Nobel Prize Connection:** Emmanuelle Charpentier and Jennifer Doudna were awarded the Nobel Prize in Chemistry (2020) for developing the CRISPR-Cas9 method. * **Mechanism:** It uses a **guide RNA (gRNA)** to lead the Cas9 nuclease to a specific genomic locus. * **Clinical Application:** Currently being researched for "ex vivo" gene therapy in **Sickle Cell Anemia** and **Beta-Thalassemia** to induce fetal hemoglobin (HbF). * **PAM Sequence:** Cas9 requires a **Protospacer Adjacent Motif (PAM)**—usually 5'-NGG-3'—to bind and cut the target DNA.

Question 134: Biotechnology is used for which of the following?

A. Viral vaccine production (Correct Answer)
B. Curing genetic disorders
C. Developing genetically modified crops
D. Gene synthesis

Explanation: **Explanation:** Biotechnology is the broad application of biological systems, living organisms, or derivatives thereof to make or modify products for specific use. In the context of this question, **Viral vaccine production** is a primary and well-established application of biotechnology in medical science. **Why Option A is Correct:** Modern vaccinology relies heavily on recombinant DNA technology. For example, the **Hepatitis B vaccine** is produced by inserting the gene for the Hepatitis B surface antigen (HBsAg) into yeast cells (*Saccharomyces cerevisiae*). Similarly, viral vector vaccines (like those for COVID-19) and mRNA vaccines are direct products of biotechnological engineering. **Analysis of Incorrect Options:** * **Option B (Curing genetic disorders):** While biotechnology is used in **Gene Therapy** to *treat* disorders (e.g., SCID or Spinal Muscular Atrophy), "curing" implies a permanent, universal resolution which is currently experimental and not yet a standard clinical outcome for most genetic diseases. * **Option C (Developing GM crops):** This is an application of **Agricultural Biotechnology**, not typically the focus of medical biochemistry in a clinical NEET-PG context. * **Option D (Gene synthesis):** This is a **laboratory technique** or a tool used *within* biotechnology, rather than an end-goal application like vaccine production. **High-Yield Facts for NEET-PG:** * **First Recombinant Vaccine:** Hepatitis B vaccine (HBsAg). * **Humulin:** The first biosynthetic human insulin developed using *E. coli* (1982). * **PCR (Polymerase Chain Reaction):** The cornerstone of biotechnology, developed by Kary Mullis, used for amplifying DNA. * **Vectors:** Plasmids and bacteriophages are the most commonly used vectors in recombinant DNA technology. * **Restriction Endonucleases:** Known as "molecular scissors," they are essential for creating recombinant molecules.

Question 135: Synthesis of an immunoglobulin in membrane-bound or secretory form is determined by which process?

A. One turn to two turn joining rule
B. Class switching
C. Differential RNA processing (Correct Answer)
D. Allelic exclusion

Explanation: **Explanation:** The synthesis of an immunoglobulin (Ig) in either a **membrane-bound** or **secretory** form is determined by **Differential RNA processing** (specifically, alternative polyadenylation and splicing). A single B-cell gene contains two potential polyadenylation sites at the 3' end of the heavy chain gene. 1. If the first site is used, the mRNA codes for a shorter protein lacking the hydrophobic transmembrane domain, resulting in a **secretory antibody**. 2. If the second site is used, the mRNA includes exons encoding a hydrophobic anchor, resulting in a **membrane-bound BCR** (B-cell receptor). This process allows a B-cell to switch from expressing surface receptors to secreting antibodies without changing its antigen specificity. **Analysis of Incorrect Options:** * **A. One turn to two turn joining rule:** This refers to the **12/23 rule** in V(D)J recombination, ensuring that gene segments (V, D, and J) are joined in the correct order during initial DNA rearrangement. * **B. Class switching (Isotype switching):** This is a **DNA recombination** process that changes the constant region of the heavy chain (e.g., from IgM to IgG). It changes the *function* of the antibody but not its form (membrane vs. secretory). * **C. Allelic exclusion:** This process ensures that only one allele of a heavy chain and one allele of a light chain are expressed, ensuring **monospecificity** of the B-cell. **High-Yield Clinical Pearls for NEET-PG:** * **Alternative Splicing:** Also responsible for the simultaneous expression of **IgM and IgD** on the surface of mature, naive B-cells. * **Key Enzyme:** **RAG-1 and RAG-2** are essential for V(D)J recombination; their deficiency leads to Omenn Syndrome or SCID. * **AID (Activation-Induced Deaminase):** The key enzyme required for both **Class Switching** and **Somatic Hypermutation**.

Question 136: Which of the following statements is FALSE regarding cytosolic eukaryotic gene expression?

A. Capping helps in the attachment of mRNA to the 40S ribosome.
B. N-formyl methionine tRNA is the first tRNA to come into action. (Correct Answer)
C. EF2 shifts between GDP and GTP.
D. A releasing factor releases the polypeptide chain from the P site.

Explanation: ### Explanation The question focuses on the differences between **Prokaryotic** and **Eukaryotic** translation. **Why Option B is the Correct Answer (The False Statement):** In **Eukaryotes**, the initiator tRNA is **methionyl-tRNA (Met-tRNAi)**. The methionine is *not* formylated. In contrast, **Prokaryotes** (and mitochondria) use **N-formyl methionine (fMet-tRNA)** as the first amino acid. This is a high-yield distinction often tested in NEET-PG. **Analysis of Other Options (True Statements):** * **Option A:** The **7-methylguanosine cap** at the 5' end of eukaryotic mRNA is essential for recognition by the eIF4F complex, which facilitates the binding of the mRNA to the **40S ribosomal subunit**. * **Option C:** **Elongation Factor 2 (EF2)** is responsible for translocation. It is a G-protein that hydrolyzes **GTP to GDP** to provide the energy required to move the ribosome along the mRNA. * **Option D:** Translation terminates when a stop codon reaches the A-site. **Releasing Factors (eRF)** recognize the stop codon and promote the hydrolysis of the ester bond, releasing the completed polypeptide chain from the **P-site**. **Clinical Pearls for NEET-PG:** 1. **Diphtheria Toxin & Pseudomonas Exotoxin A:** Both inhibit protein synthesis by ADP-ribosylation of **EF2**, leading to cell death. 2. **Shiga Toxin:** Inhibits the **60S subunit** by removing adenine from rRNA. 3. **Kozak Sequence:** In eukaryotes, the initiation codon (AUG) is contained within the Kozak consensus sequence, which helps the ribosome identify the correct start site (analogous to the **Shine-Dalgarno** sequence in prokaryotes). 4. **Mitochondrial Translation:** Because mitochondria have their own DNA and ribosomes (70S), they utilize **fMet-tRNA**, similar to bacteria.

Question 137: The gene responsible for folic acid transport is situated on which chromosome?

A. Chromosome 10
B. Chromosome 5
C. X chromosome
D. Chromosome 21 (Correct Answer)

Explanation: **Explanation:** The correct answer is **Chromosome 21**. The gene responsible for the primary transport of folic acid into cells is the **SLC19A1 gene** (Solute Carrier Family 19 Member 1), which encodes the **Reduced Folate Carrier 1 (RFC1)**. This gene is located on the long arm of chromosome 21 (21q22.3). **Why Chromosome 21 is correct:** The RFC1 protein is the major transporter for 5-methyltetrahydrofolate (the predominant form of folate in plasma) into systemic cells. This is clinically significant in **Down Syndrome (Trisomy 21)**; individuals with this condition have three copies of the SLC19A1 gene, leading to altered folate metabolism and increased sensitivity to methotrexate (a folate antagonist). **Analysis of Incorrect Options:** * **Chromosome 10:** While many metabolic genes are located here, it does not harbor the primary RFC1 transporter. * **Chromosome 5:** This chromosome contains the gene for the *Folate Receptor Alpha (FOLR1)*, which is involved in folate uptake in specific tissues (like the choroid plexus), but the primary "folate transport gene" referred to in standard biochemistry texts is SLC19A1 on Chromosome 21. * **X Chromosome:** No major systemic folate transport genes are mapped to the X chromosome. **High-Yield Clinical Pearls for NEET-PG:** * **Hereditary Folate Malabsorption:** Caused by mutations in the **SLC46A1** gene (Proton-coupled folate transporter - PCFT) located on **Chromosome 17**. * **Methotrexate Toxicity:** Because the SLC19A1 gene is on Chromosome 21, children with Down Syndrome are at a higher risk of toxicity when treated for Leukemia due to increased intracellular drug accumulation. * **Folate vs. B12:** Remember that folate deficiency leads to increased homocysteine but **normal** methylmalonic acid (MMA) levels, distinguishing it from B12 deficiency.

Question 138: Which of the following best describes the centromere position of the human X chromosome?

A. Metacentric
B. Submetacentric (Correct Answer)
C. Acrocentric
D. Telocentric

Explanation: **Explanation:** The classification of chromosomes is based on the position of the **centromere**, which divides the chromosome into a short arm (**p arm**) and a long arm (**q arm**). **1. Why Submetacentric is correct:** The human **X chromosome** is a large **submetacentric** chromosome (specifically belonging to Group C in the Denver classification). In submetacentric chromosomes, the centromere is located slightly away from the center, resulting in an unequal length of arms where the p arm is distinctly shorter than the q arm. **2. Analysis of Incorrect Options:** * **Metacentric:** The centromere is in the exact middle, creating arms of equal length. Examples include chromosomes **1, 3, 16, 19, and 20**. * **Acrocentric:** The centromere is located very near one end. These chromosomes possess "satellites" and stalks containing rDNA. Examples include chromosomes **13, 14, 15, 21, 22, and the Y chromosome**. * **Telocentric:** The centromere is at the very tip (telomere). **Telocentric chromosomes do not occur in humans**; they are found in other species like mice. **Clinical Pearls & High-Yield Facts:** * **Denver Classification:** Chromosomes are grouped A-G based on size and centromere position. The X chromosome is in **Group C**, while the Y chromosome is in **Group G** (Acrocentric). * **Robertsonian Translocation:** This occurs only in **acrocentric** chromosomes (13, 14, 15, 21, 22) because the loss of their short arms (satellites) does not result in the loss of essential genetic material. * **Lyonization:** In females, one X chromosome is randomly inactivated to form a **Barr body** (facultative heterochromatin) to ensure dosage compensation.

Question 139: What is the role of the H2 histone?

A. Stabilize the 30-nm chromatin fiber
B. Central role in the formation of the nucleosome
C. Stabilizes the primary particle
D. Firmly binds two additional half-turns of DNA (Correct Answer)

Explanation: **Explanation:** The nucleosome is the fundamental repeating unit of chromatin. It consists of a **nucleosome core particle** and **linker DNA**. **1. Why the Correct Answer is Right:** The nucleosome core particle consists of an octamer of histones (two each of H2A, H2B, H3, and H4) around which **1.46 turns** (approximately 147 base pairs) of DNA are wrapped. **Histone H1** (often referred to as the linker histone) sits outside the core particle. It functions by "locking" the DNA in place as it enters and exits the core, effectively binding **two additional half-turns** of DNA. This brings the total DNA associated with the histone complex to approximately 166 base pairs (two full turns). **2. Analysis of Incorrect Options:** * **Option A:** While H1 is essential for the higher-order folding of chromatin into the **30-nm fiber** (solenoid structure), its primary biochemical role at the nucleosomal level is the stabilization of the extra DNA turns. * **Option B:** The **histone octamer** (H2A, H2B, H3, H4) plays the central role in forming the nucleosome core. H1 is not part of the core octamer. * **Option C:** The "primary particle" or core particle is stabilized by the interactions between the DNA phosphate backbone and the basic amino acids of the octamer histones, not H1. **3. High-Yield Facts for NEET-PG:** * **Linker Histone:** H1 is the most tissue-specific and species-specific histone. * **Basic Nature:** Histones are rich in **Lysine and Arginine**, giving them a positive charge to bind the negatively charged DNA. * **Acetylation:** Occurs on Lysine residues; it decreases the positive charge, leading to relaxed chromatin (**Euchromatin**) and increased transcription. * **Smallest Histone:** H4 is the most conserved and among the smallest histones.

Question 140: Which of the following processes allows apolipoprotein B to be synthesized in the liver as a 100-kDa protein and in the intestine as a 48-kDa protein?

A. RNA splicing
B. DNA rearrangement
C. Proteolytic cleavage
D. RNA editing (Correct Answer)

Explanation: **Explanation:** The correct answer is **RNA editing**. This process involves post-transcriptional modifications to the mRNA sequence that change the coding information before translation. **Why RNA Editing is Correct:** The *APOB* gene is transcribed into a single mRNA transcript in both the liver and the intestine. In the **intestine**, a specific enzyme called **cytidine deaminase** acts on the mRNA, converting a cytosine (C) to a uracil (U) at codon 2153. This change converts the glutamine codon (**CAA**) into a premature stop codon (**UAA**). Consequently, translation terminates early, producing the shorter **Apo B-48** (48% of the protein). In the **liver**, this enzyme is absent; the mRNA remains unedited, allowing full translation into the 100-kDa **Apo B-100**. **Why Other Options are Incorrect:** * **A. RNA Splicing:** This involves removing introns and joining exons. While alternative splicing creates protein diversity, it is not the mechanism for Apo B truncation. * **B. DNA Rearrangement:** This involves physical movement of DNA segments (e.g., in immunoglobulin genes). The DNA sequence for *APOB* remains identical in both tissues. * **C. Proteolytic Cleavage:** This is a post-translational modification where a protein is cut by enzymes (e.g., proinsulin to insulin). Apo B-48 is not a cleaved product of Apo B-100; it is synthesized as a shorter chain from the start. **Clinical Pearls for NEET-PG:** * **Apo B-100:** Found in VLDL, IDL, and LDL. It serves as the ligand for the **LDL receptor**. * **Apo B-48:** Found exclusively in **Chylomicrons**. It lacks the LDL receptor-binding domain found in the C-terminal half of Apo B-100. * **Mnemonic:** **L**iver = **L**ong (B-100); **I**ntestine = **I**ncomplete (B-48).

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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