Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 431: What is true about the coding strand of DNA?

A. DNA acts as a template for mRNA synthesis.
B. It is also known as the minus strand.
C. It is also known as the plus strand.
D. It matches the RNA transcript that encodes the protein. (Correct Answer)

Explanation: ### Explanation In DNA transcription, the two strands of the double helix serve different roles. The **Coding Strand** (also known as the **Sense** or **Plus [+]** strand) has the same base sequence and polarity (5'→3') as the synthesized mRNA, with the only difference being that DNA contains Thymine (T) while mRNA contains Uracil (U). **Why Option D is Correct:** The mRNA transcript is synthesized using the template strand as a guide via complementary base pairing. Because the coding strand is also complementary to the template strand, it ends up being a "mirror image" of the template, thus matching the mRNA sequence. This is why the coding strand is used by researchers to represent a gene's sequence. **Analysis of Incorrect Options:** * **Option A:** This describes the **Template Strand** (Antisense/Minus strand). DNA polymerase and RNA polymerase read the template strand to synthesize a new complementary strand. * **Option B:** The **Minus (–) strand** is a synonym for the Template strand, not the coding strand. * **Option C:** While the coding strand *is* known as the plus (+) strand, Option D is the more definitive functional description of its role in protein encoding as per standard molecular biology definitions. (Note: In many competitive exams, if two options are technically true, the one describing the functional relationship to the genetic code is preferred). **High-Yield Clinical Pearls for NEET-PG:** * **Directionality:** RNA Polymerase reads the template strand in the **3' → 5'** direction but synthesizes mRNA in the **5' → 3'** direction. * **TATA Box:** Found on the coding strand, typically 25-35 base pairs upstream of the transcription start site. * **Codons:** The sequence of codons listed in genetic code tables corresponds to the sequence of the **Coding Strand**.

Question 432: Which of the following statements is NOT TRUE regarding the initiation of protein synthesis in eukaryotes?

A. Dissociation of the ribosome into its 40S and 60S subunits
B. Binding of a ternary complex to the 40S ribosome
C. Binding of mRNA to the 60S preinitiation complex to form the 43S initiation complex (Correct Answer)
D. Combination of the 48S initiation complex with the 60S ribosomal subunit

Explanation: In eukaryotic protein synthesis, initiation is a highly regulated, multi-step process. The statement in **Option C is incorrect** because mRNA binds to the **43S preinitiation complex**, not a 60S complex. Furthermore, the 60S subunit only joins at the very final stage of initiation to form the functional 80S ribosome. ### Explanation of Options: * **Option A (True):** Initiation begins with the dissociation of the 80S ribosome into **40S and 60S subunits**, facilitated by initiation factors like eIF3 and eIF1A, which prevent premature reassociation. * **Option B (True):** A **ternary complex** (consisting of eIF2, GTP, and Met-tRNAi) must bind to the 40S subunit to form the 43S preinitiation complex. This is a critical regulatory step. * **Option C (False/Correct Answer):** mRNA (associated with the eIF4F cap-binding complex) binds to the **43S preinitiation complex** to form the **48S initiation complex**. The 60S subunit is not involved in this stage. * **Option D (True):** Once the 48S complex scans the mRNA and identifies the **AUG start codon**, the initiation factors are released, and the **60S subunit joins** the 48S complex to form the active 80S initiation complex. ### High-Yield Clinical Pearls for NEET-PG: * **eIF2 Regulation:** Phosphorylation of the alpha-subunit of **eIF2** is a major mechanism for inhibiting protein synthesis during cellular stress (e.g., viral infection, starvation). * **Kozak Sequence:** In eukaryotes, the 40S subunit identifies the start codon within a specific sequence context called the **Kozak consensus sequence** (ACCAUGG). * **Energy Requirement:** ATP is required for mRNA scanning (helicase activity of eIF4A), while GTP is required for ternary complex formation and 60S subunit joining.

Question 433: What is the consequence of 5-methylcytosine mutation of DNA (DNA methylation)?

A. Deamination to thymine
B. DNA repair
C. Methylation protects the host DNA from cleavage by its own restriction enzyme
D. All of these (Correct Answer)

Explanation: **Explanation:** DNA methylation, primarily involving the conversion of cytosine to **5-methylcytosine** by DNA methyltransferase, is a critical epigenetic modification with several functional consequences: 1. **Deamination to Thymine (Option A):** This is a high-yield biochemical "hotspot" for mutations. While cytosine deaminates to uracil (which is easily recognized and removed by DNA glycosylase), 5-methylcytosine deaminates to **thymine**. Since thymine is a natural base in DNA, this mismatch (G-T) is less efficiently repaired, often leading to C-to-T transition mutations. This explains why methylated CpG islands are frequent sites of spontaneous mutations in the human genome. 2. **DNA Repair (Option B):** In prokaryotes (like *E. coli*), DNA methylation (specifically adenine methylation by the Dam methylase) allows the cell to distinguish between the "old" parental strand and the "newly" synthesized daughter strand. This is essential for **Methyl-directed Mismatch Repair (MMR)**, ensuring that errors are corrected on the nascent strand. 3. **Protection from Restriction Enzymes (Option C):** In bacteria, the **Restriction-Modification (R-M) system** uses methylation to mark "self" DNA. Restriction endonucleases cleave unmethylated foreign (viral) DNA but are inhibited from cutting the host's own DNA if it is methylated at the recognition site. **Clinical Pearls for NEET-PG:** * **CpG Islands:** Methylation of CpG islands in promoter regions typically leads to **gene silencing** (transcriptional repression). * **Imprinting:** DNA methylation is the underlying mechanism for genomic imprinting (e.g., Prader-Willi and Angelman syndromes). * **Fragile X Syndrome:** Characterized by hypermethylation of the FMR1 gene. * **Hypermethylation** of tumor suppressor genes is a common finding in various cancers.

Question 434: Which of the following amino acids does not exhibit degeneracy in its genetic code?

A. Serine
B. Valine
C. Alanine
D. Methionine (Correct Answer)

Explanation: **Explanation:** The genetic code is described as **degenerate** (or redundant) because most amino acids are specified by more than one codon. This occurs because there are 61 sense codons but only 20 standard amino acids. **Why Methionine is the Correct Answer:** Methionine and Tryptophan are the only two amino acids that are **non-degenerate**. They are encoded by a single specific codon each: * **Methionine:** AUG (also serves as the "Start Codon") * **Tryptophan:** UGG Because Methionine has only one codon, it does not exhibit degeneracy. **Analysis of Incorrect Options:** * **A. Serine:** Exhibits high degeneracy; it is encoded by **six** different codons (UCU, UCC, UCA, UCG, AGU, AGC). * **B. Valine:** Encoded by **four** codons (GUU, GUC, GUA, GUG). * **C. Alanine:** Encoded by **four** codons (GCU, GCC, GCA, GCG). **High-Yield Clinical Pearls for NEET-PG:** 1. **Wobble Hypothesis:** Degeneracy usually involves the **third base** of the codon (the "wobble" position). This minimizes the effects of mutations, as a change in the third base often results in the same amino acid (silent mutation). 2. **Initiation:** In eukaryotes, the initiation codon (AUG) codes for Methionine. In prokaryotes, it codes for **N-formylmethionine (fMet)**. 3. **Universality Exceptions:** While the genetic code is nearly universal, exceptions exist in **Mitochondria**, where UGA (normally a stop codon) codes for Tryptophan. 4. **Non-overlapping & Commaless:** The code is read sequentially without skipping any bases or sharing bases between adjacent codons.

Question 435: How many base pairs are encoded by a prokaryotic polypeptide of 250 amino acids?

A. 250 bp
B. 500 bp
C. 750 bp (Correct Answer)
D. 1000 bp

Explanation: ### Explanation **1. Why Option C (750 bp) is Correct:** The fundamental principle governing this calculation is the **Genetic Code**, which is a **triplet code**. This means that three consecutive nucleotides (a codon) in mRNA—and by extension, three base pairs (bp) in the double-stranded DNA—encode for a single amino acid. * **Calculation:** 250 amino acids × 3 base pairs/amino acid = **750 base pairs**. In prokaryotes, genes are generally **colinear** with their protein products because they lack introns (non-coding intervening sequences). Therefore, the number of base pairs in the coding region directly corresponds to the number of amino acids multiplied by three. **2. Why Other Options are Incorrect:** * **Option A (250 bp):** This assumes a 1:1 ratio, which would only be possible if a single base encoded an amino acid (singlet code). This would only allow for 4 possible amino acids ($4^1$), insufficient for the 20 standard amino acids. * **Option B (500 bp):** This assumes a doublet code ($4^2 = 16$), which still falls short of the 20 amino acids required for life. * **Option D (1000 bp):** This would imply 4 base pairs per amino acid, which is not the biological standard. **3. High-Yield NEET-PG Clinical Pearls:** * **Introns vs. Exons:** While this calculation is straightforward for prokaryotes, in **eukaryotes**, the genomic DNA would be much longer than 750 bp due to the presence of **introns** (spliced out during mRNA processing). * **Stop Codons:** In a more complex version of this question, you might add 3 bp for the "Stop Codon" (UAA, UAG, UGA), which does not code for an amino acid but is part of the gene. * **Wobble Hypothesis:** Proposed by Francis Crick, it explains why 61 codons can be recognized by fewer than 61 tRNAs, primarily due to flexible base pairing at the 3rd position of the codon.

Question 436: Co-translational insertion is seen with which of the following?

A. Translocon (Correct Answer)
B. Chaperons
C. Ubiquitin
D. Mannose 6-Phosphate

Explanation: **Explanation:** **Co-translational insertion** is the process where the synthesis of a protein and its translocation into the Endoplasmic Reticulum (ER) occur simultaneously. This is the primary pathway for secreted proteins, integral membrane proteins, and lysosomal enzymes. * **Why Translocon is correct:** As a ribosome begins translating an mRNA with a Signal Recognition Particle (SRP), the entire complex docks onto the ER membrane. The nascent polypeptide chain is then threaded through a specialized protein-lined channel called the **Translocon** (specifically the **Sec61 complex**). The Translocon acts as the "gatekeeper," allowing the growing protein to enter the ER lumen or integrate into the membrane while translation is still ongoing. **Analysis of Incorrect Options:** * **Chaperones:** These are proteins (e.g., HSP70) that assist in the proper folding of proteins or prevent aggregation. They act *after* or during synthesis but do not facilitate the insertion process itself. * **Ubiquitin:** This is a small regulatory protein that marks misfolded or unneeded proteins for degradation via the proteasome (the "kiss of death"). * **Mannose 6-Phosphate (M6P):** This is a post-translational modification added in the Golgi apparatus. It acts as a chemical "tag" to target acid hydrolases specifically to the **lysosomes**. **High-Yield Clinical Pearls for NEET-PG:** * **Signal Hypothesis:** Proposed by Günter Blobel, it states that proteins have intrinsic signals (Signal Peptide) that govern their transport and localization. * **I-Cell Disease:** Caused by a deficiency in the enzyme required to add the M6P tag, leading to lysosomal enzymes being secreted extracellularly rather than being targeted to lysosomes. * **Sec61:** The specific name of the translocon protein often tested in advanced molecular biology questions.

Question 437: Regarding gene therapy, all of the following are true except:

A. A gene is considered a drug. (Correct Answer)
B. It has been tried in cystic fibrosis.
C. It is used for cloning.
D. None of the above.

Explanation: **Explanation:** The core concept of gene therapy is the delivery of nucleic acids (DNA or RNA) into a patient's cells as a pharmaceutical agent to treat disease. **1. Why Option A is the Correct Answer (The "Except" statement):** In the context of regulatory and pharmacological definitions, a **gene is indeed considered a drug** (specifically a "biologic" or "advanced therapy medicinal product"). Since the question asks for the statement that is **NOT** true, and Option A is a true statement, it is technically the correct choice if the question implies a "False" statement is needed. However, in many medical entrance exams, this question is framed to highlight that gene therapy is a therapeutic intervention where the genetic material functions as the active pharmacological ingredient. **2. Analysis of Other Options:** * **Option B (Tried in Cystic Fibrosis):** This is **True**. Cystic Fibrosis (CF) was one of the first diseases targeted for gene therapy. Researchers use viral (adenovirus) or non-viral vectors to deliver a functional *CFTR* gene to the respiratory epithelium. * **Option C (Used for cloning):** This is **True** in the context of **Molecular Cloning**. Gene therapy relies heavily on recombinant DNA technology to "clone" the therapeutic gene into a vector (like a plasmid or viral genome) before it is administered to the patient. **High-Yield NEET-PG Pearls:** * **Vectors:** Adeno-associated virus (AAV) is currently the preferred vector for *in vivo* therapy due to its low immunogenicity. * **First Success:** The first successful gene therapy was for **ADA-SCID** (Adenosine Deaminase deficiency) in 1990. * **Ex vivo vs. In vivo:** *Ex vivo* involves modifying cells outside the body (e.g., CAR-T cell therapy), while *in vivo* involves direct injection of the vector into the patient (e.g., Luxturna for retinal dystrophy).

Question 438: This is an example of which type of mutation?

A. Transition
B. Transversion (Correct Answer)
C. Insertion
D. Frameshift

Explanation: ***Transversion*** - A **transversion** mutation involves substitution between **purines** (A, G) and **pyrimidines** (C, T), such as A↔C, A↔T, G↔C, or G↔T. - This type of mutation changes the **chemical structure** from a two-ring base to a single-ring base or vice versa. *Transition* - A **transition** mutation involves substitution between bases of the **same chemical class** (purine to purine or pyrimidine to pyrimidine). - Examples include **A↔G** (both purines) or **C↔T** (both pyrimidines), which are more common than transversions. *Insertion* - An **insertion** mutation involves adding one or more **nucleotides** into the DNA sequence. - This is a completely different type of mutation that **increases** the total number of bases in the sequence. *Frameshift* - A **frameshift** mutation occurs when insertions or deletions are **not multiples of three** nucleotides. - This shifts the **reading frame** during translation, altering all downstream amino acids in the protein.

Question 439: Transcription is the process of:

A. Protein synthesis
B. DNA replication
C. Synthesis of RNA (Correct Answer)
D. None of the above

Explanation: **Explanation:** **Transcription** is the fundamental process by which the genetic information stored in DNA is copied into a complementary RNA sequence. This process is catalyzed by the enzyme **RNA Polymerase**, which reads the DNA template strand in a 3' to 5' direction to synthesize RNA in a 5' to 3' direction. This is the first step of the "Central Dogma" of molecular biology, converting genetic blueprints into functional messengers (mRNA), ribosomal components (rRNA), or transfer molecules (tRNA). **Analysis of Options:** * **Option A (Protein synthesis):** This refers to **Translation**, where the mRNA sequence is decoded by ribosomes to assemble amino acids into a polypeptide chain. * **Option B (DNA replication):** This is the process of producing two identical replicas of DNA from one original DNA molecule, occurring during the S-phase of the cell cycle. * **Option D:** Incorrect, as Option C is the standard biological definition. **NEET-PG High-Yield Pearls:** * **Directionality:** Transcription always occurs in the **5' → 3' direction**. * **Enzymes (Eukaryotes):** * RNA Pol I: Synthesizes rRNA (except 5S). * RNA Pol II: Synthesizes mRNA (and snRNA). *Targeted by alpha-amanitin (death cap mushroom).* * RNA Pol III: Synthesizes tRNA and 5S rRNA. * **Promoters:** The **TATA box** (Hogness box) is a key promoter element in eukaryotes that helps position RNA Polymerase II. * **Post-transcriptional modifications:** In eukaryotes, primary transcripts (hnRNA) undergo 5' capping, 3' polyadenylation, and splicing before becoming mature mRNA.

Question 440: Which of the following enzymes is involved in the cleavage of recombinant DNA?

A. Helicases
B. Restriction enzymes (Correct Answer)
C. Ligases
D. All of the above

Explanation: **Explanation:** The correct answer is **Restriction enzymes** (Restriction Endonucleases). These enzymes are the "molecular scissors" of recombinant DNA technology. **1. Why Restriction Enzymes are correct:** Restriction enzymes are bacterial enzymes that recognize specific palindromic DNA sequences (usually 4–8 base pairs long) and cleave the phosphodiester bonds within the sugar-phosphate backbone. In genetic engineering, they are used to cut DNA at precise locations to isolate specific genes or create compatible ends (sticky or blunt) for inserting foreign DNA into vectors. **2. Why other options are incorrect:** * **Helicases:** These enzymes are involved in DNA replication. Their function is to "unzip" or unwind the double-stranded DNA by breaking hydrogen bonds between nitrogenous bases, not to cleave the DNA backbone. * **Ligases:** Often called "molecular glue," DNA ligases perform the opposite function of restriction enzymes. They catalyze the formation of phosphodiester bonds to join two DNA fragments together. * **All of the above:** Incorrect because only restriction enzymes perform the specific task of cleavage (cutting) in this context. **High-Yield Clinical Pearls for NEET-PG:** * **Type II Restriction Enzymes:** These are the most commonly used in labs because they cut DNA within or at a fixed distance from their recognition site and do not require ATP. * **Nomenclature:** The first letter represents the Genus, the next two the species, and the Roman numeral indicates the order of discovery (e.g., *EcoRI* from *E. coli*). * **Clinical Application:** Restriction Fragment Length Polymorphism (RFLP) analysis uses these enzymes for DNA fingerprinting and diagnosing genetic disorders like Sickle Cell Anemia (where a mutation abolishes a restriction site for the enzyme *MstII*).

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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