Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 71: Which of the following types of mutations generally leads to a truncated protein?

A. Deletion
B. Frameshift mutation (Correct Answer)
C. Insertion
D. Missense mutation

Explanation: ### Explanation **Why Frameshift Mutation is Correct:** A **frameshift mutation** occurs when a number of nucleotides (not divisible by three) are inserted into or deleted from a DNA sequence. Since the genetic code is read in non-overlapping triplets (codons), this shifts the "reading frame" of the mRNA. This shift almost inevitably leads to the premature encounter of a **stop codon** (UAA, UAG, or UGA) further downstream. The premature termination of translation results in a **truncated (shortened) protein**, which is usually non-functional and often degraded via nonsense-mediated decay. **Analysis of Incorrect Options:** * **Deletion/Insertion (Options A & C):** While these *can* cause a frameshift, they only do so if the number of bases involved is not a multiple of three. If 3, 6, or 9 bases are deleted/inserted (in-frame mutations), the reading frame remains intact, resulting in the loss or gain of specific amino acids without truncating the entire protein. * **Missense Mutation (Option D):** This is a point mutation where a single nucleotide change results in a codon that codes for a *different* amino acid. The protein length remains the same, though its function may be altered (e.g., Hemoglobin S in Sickle Cell Anemia). **High-Yield Clinical Pearls for NEET-PG:** * **Duchenne Muscular Dystrophy (DMD):** Caused by **frameshift mutations** leading to a complete absence of functional dystrophin (severe). * **Becker Muscular Dystrophy (BMD):** Caused by **in-frame mutations**, resulting in a truncated but partially functional dystrophin (milder). * **Nonsense Mutation:** Another key cause of protein truncation where a point mutation directly changes an amino acid codon into a stop codon. * **Tay-Sachs Disease:** Often caused by a 4-base pair insertion (frameshift) in the *HEXA* gene.

Question 72: Which of the following DNA types are found in mitochondria?

A. Circular DNA and Single stranded DNA (Correct Answer)
B. Double stranded DNA
C. Plasmid DNA and Circular DNA
D. Plasmid DNA and Double stranded DNA

Explanation: **Explanation:** The correct answer is **Circular DNA and Single stranded DNA**. **Underlying Concept:** Mitochondrial DNA (mtDNA) is unique because it follows the **Endosymbiotic Theory**, suggesting mitochondria originated from ancient prokaryotes. Consequently, mtDNA is primarily **double-stranded, closed-circular DNA**. However, during the process of replication, mitochondria utilize a "D-loop" (Displacement loop) mechanism. In this process, the synthesis of the new strand displaces one of the parental strands, leaving a portion of the DNA in a **single-stranded** state. Thus, both circular and single-stranded forms are characteristic features of the mitochondrial genome. **Analysis of Options:** * **Option A (Correct):** Reflects the circular nature of the genome and the single-stranded intermediates formed during D-loop replication. * **Option B (Incorrect):** While mtDNA is double-stranded, this option is incomplete as it ignores the circularity and the specific single-stranded phases essential for its replication. * **Option C & D (Incorrect):** **Plasmids** are extrachromosomal DNA molecules found in bacteria and some fungi/plants, but they are **not** a structural component of human mitochondrial genetics. **High-Yield Clinical Pearls for NEET-PG:** * **Maternal Inheritance:** mtDNA is inherited exclusively from the mother (sperm mitochondria are degraded in the zygote). * **Heteroplasmy:** The presence of a mixture of more than one type of organellar genome (mutant vs. wild-type) within a cell. * **Lack of Introns:** Unlike nuclear DNA, mtDNA is highly compact and lacks introns. * **High Mutation Rate:** mtDNA lacks robust repair mechanisms (like histones), making it 10 times more prone to mutations than nuclear DNA. * **Key Diseases:** MELAS, MERRF, and Leber’s Hereditary Optic Neuropathy (LHON).

Question 73: In which of the following conditions does a trinucleotide repeat mutation occur in the coding regions of DNA?

A. Myotonic dystrophy
B. Huntington's disease (Correct Answer)
C. Fragile X syndrome
D. Friedreich's ataxia

Explanation: ### Explanation Trinucleotide repeat expansion disorders are characterized by an increase in the number of specific three-nucleotide sequences. The clinical manifestation depends on whether the expansion occurs in a **coding region** (exons) or a **non-coding region** (introns or UTRs). #### Why Huntington’s Disease is Correct: **Huntington’s disease** is caused by a **CAG** repeat expansion within the **coding region** (exon 1) of the *HTT* gene on chromosome 4. Since CAG codes for the amino acid **Glutamine**, these are known as **Polyglutamine (PolyQ) diseases**. The expansion leads to a "gain-of-function" mutation, resulting in an abnormally long polyglutamine tract that causes the huntingtin protein to misfold and become neurotoxic. #### Why Other Options are Incorrect: * **Myotonic Dystrophy (Type 1):** The **CTG** repeat occurs in the **3' Untranslated Region (UTR)** of the *DMPK* gene. It is not translated into protein but causes RNA toxicity. * **Fragile X Syndrome:** The **CGG** repeat occurs in the **5' Untranslated Region (UTR)** of the *FMR1* gene. This leads to hypermethylation of the promoter, gene silencing, and a loss of protein function. * **Friedreich’s Ataxia:** The **GAA** repeat occurs within the **first intron** (non-coding) of the *FXN* gene, leading to impaired transcription and decreased levels of the protein frataxin. #### High-Yield Clinical Pearls for NEET-PG: * **Anticipation:** The phenomenon where the disease becomes more severe or has an earlier onset in successive generations (most prominent in Huntington’s when inherited from the father). * **Mnemonic for Huntington's:** "**C**AG: **C**audate atrophy, **A**cetylcholine decrease, **G**ABA decrease." * **Location Summary:** * **Fragile X:** 5' UTR (CGG) * **Friedreich’s Ataxia:** Intron (GAA) * **Huntington’s:** Exon (CAG) * **Myotonic Dystrophy:** 3' UTR (CTG)

Question 74: Which enzyme catalyzes the hydrolytic step leading to the release of the polypeptide chain from ribosomes?

A. Stop codons
B. Peptidyl transferase (Correct Answer)
C. AUG codon
D. All of the above

Explanation: **Explanation:** The process of translation termination occurs when a ribosome encounters a **stop codon** (UAA, UAG, or UGA) in the A-site. Since there are no aminoacyl-tRNAs that recognize these codons, **Release Factors (RFs)** bind to the ribosome instead. **1. Why Peptidyl Transferase is correct:** The enzyme **peptidyl transferase** (a ribozyme activity of the 28S rRNA in eukaryotes and 23S rRNA in prokaryotes) normally catalyzes peptide bond formation. However, during termination, the binding of Release Factors alters its specificity. Instead of transferring the growing polypeptide chain to an amino acid, it catalyzes the **hydrolysis** of the ester bond between the tRNA and the polypeptide chain using a water molecule. This hydrolytic step releases the completed protein into the cytosol. **2. Why other options are incorrect:** * **Stop codons (A):** These are the *signals* for termination, not the enzymes that catalyze the chemical reaction of release. * **AUG codon (C):** This is the initiation codon that codes for Methionine; it plays no role in the termination or release of the polypeptide. **High-Yield Clinical Pearls for NEET-PG:** * **Ribozyme:** Peptidyl transferase is a classic example of a ribozyme (an RNA molecule with catalytic activity). * **Energy Requirement:** Termination and ribosome dissociation require energy in the form of **GTP hydrolysis**. * **Antibiotic Link:** Macrolides (e.g., Erythromycin) and Chloramphenicol act by inhibiting the peptidyl transferase center, thereby halting bacterial protein synthesis. * **Release Factors:** In eukaryotes, a single factor (**eRF1**) recognizes all three stop codons.

Question 75: All of the following are true about mitochondrial DNA except:

A. It contains 37 genes.
B. It is transmitted from mother to offspring.
C. It is transmitted in a classical Mendelian fashion. (Correct Answer)
D. It can cause Leber hereditary optic neuropathy.

Explanation: **Explanation:** The correct answer is **C**, as mitochondrial DNA (mtDNA) follows **Non-Mendelian (Maternal) inheritance**, not classical Mendelian patterns. **1. Why Option C is correct (The Concept):** Mendelian inheritance involves the equal contribution of alleles from both parents. However, mtDNA is inherited exclusively from the mother because the mitochondria in a zygote are derived almost entirely from the oocyte; the few mitochondria in the sperm's neck are destroyed during fertilization. This is known as **Matrilineal inheritance**. **2. Analysis of Incorrect Options:** * **Option A:** mtDNA is a circular, double-stranded molecule containing **37 genes**. These encode 13 polypeptides (subunits of the respiratory chain), 22 tRNAs, and 2 rRNAs. * **Option B:** As established, mtDNA is transmitted from the **mother to all her offspring** (both sons and daughters), but only daughters can pass it to the next generation. * **Option D:** **Leber Hereditary Optic Neuropathy (LHON)** is the classic example of a mitochondrial disorder caused by mutations in mtDNA, leading to bilateral loss of central vision. **Clinical Pearls for NEET-PG:** * **Heteroplasmy:** The presence of a mixture of both normal and mutated mtDNA within a single cell. This explains the variable clinical severity of mitochondrial diseases. * **High Mutation Rate:** mtDNA lacks histones and has limited repair mechanisms, making its mutation rate 10 times higher than nuclear DNA. * **Other Examples:** MERRF (Myoclonic Epilepsy with Ragged Red Fibers) and MELAS (Mitochondrial Encephalopathy, Lactic Acidosis, and Stroke-like episodes). * **Replication:** mtDNA replicates throughout the cell cycle and is independent of the S-phase of nuclear DNA.

Question 76: During DNA replication, which enzyme produces a short strand of RNA complementary to the template DNA with a free 3'-OH end?

A. DNA ligase
B. DNA Polymerase I
C. DNA Polymerase III
D. Primase (Correct Answer)

Explanation: **Explanation:** **1. Why Primase is Correct:** DNA polymerases are incapable of initiating DNA synthesis *de novo*; they require a pre-existing free **3'-OH group** to add nucleotides. To overcome this, **Primase** (a specialized RNA polymerase) synthesizes a short RNA sequence (approx. 10 nucleotides) called a **primer**. This primer provides the essential 3'-OH terminus that DNA Polymerase III uses to begin elongation. **2. Why the Other Options are Incorrect:** * **DNA Ligase:** This enzyme acts as "molecular glue." It catalyzes the formation of phosphodiester bonds to seal nicks between DNA fragments (like Okazaki fragments) but does not synthesize new strands. * **DNA Polymerase I:** In prokaryotes, this enzyme primarily functions in **primer removal** (via 5'→3' exonuclease activity) and filling the resulting gaps with DNA. It does not create the initial RNA strand. * **DNA Polymerase III:** This is the primary enzyme for **elongation**. While it synthesizes the bulk of the new DNA strand, it cannot function without the primer provided by Primase. **3. High-Yield Clinical Pearls for NEET-PG:** * **DnaG:** In *E. coli*, the primase enzyme is specifically known as DnaG. * **Primosome:** This is a complex consisting of Primase and Helicase (DnaB) that works together to unwind DNA and lay down primers. * **Eukaryotic Equivalent:** In humans, **DNA Polymerase α** (alpha) possesses primase activity. * **Directionality:** Synthesis always occurs in the **5' to 3' direction**. On the lagging strand, multiple primers are required to initiate each Okazaki fragment.

Question 77: Which of the following statements regarding complementary DNA (cDNA) is NOT true?

A. It is DNA complementary to messenger RNA (mRNA).
B. Reverse transcriptase is required for its synthesis.
C. Synthesis of human insulin involves cDNA.
D. It contains both noncoding and coding sequences. (Correct Answer)

Explanation: ### Explanation **Why Option D is the Correct Answer (The "NOT True" Statement):** The fundamental characteristic of **cDNA (complementary DNA)** is that it is synthesized from a mature mRNA template. In eukaryotes, mature mRNA has already undergone **post-transcriptional modification (splicing)**, where introns (non-coding sequences) are removed and exons (coding sequences) are joined together. Therefore, cDNA contains **only coding sequences (exons)** and lacks the introns found in genomic DNA. This makes it essential for expressing eukaryotic genes in prokaryotic systems (like *E. coli*), which cannot perform splicing. **Analysis of Incorrect Options:** * **Option A:** This is a definition. cDNA is synthesized by using mRNA as a template, making it complementary to the RNA sequence. * **Option B:** The enzyme **Reverse Transcriptase** (an RNA-dependent DNA polymerase) is strictly required to transcribe the single-stranded RNA into a DNA strand. * **Option C:** To produce human insulin commercially, the human insulin cDNA is inserted into bacterial plasmids. Since bacteria cannot splice human pre-mRNA, using cDNA (which lacks introns) is the only way to ensure the correct protein is synthesized. **NEET-PG High-Yield Pearls:** * **cDNA Library vs. Genomic Library:** A cDNA library represents only the genes being actively expressed (the transcriptome) in a specific tissue, whereas a genomic library contains the entire genome (introns, exons, promoters). * **Reverse Transcriptase:** Originally derived from retroviruses (like HIV). It requires an **oligo-dT primer** that binds to the poly-A tail of mRNA to initiate synthesis. * **Clinical Use:** cDNA is used in **RT-PCR** (Reverse Transcription Polymerase Chain Reaction) to detect RNA viruses like SARS-CoV-2 or to measure gene expression levels in oncology.

Question 78: Which of the following protein structures contains a DNA binding domain?

A. Zinc finger
B. Helix-turn-helix
C. Leucine zipper
D. All of these (Correct Answer)

Explanation: ### Explanation The correct answer is **D. All of these**. This question tests the knowledge of **DNA-binding motifs**, which are specific structural configurations within transcription factors that allow them to bind to the major or minor grooves of DNA to regulate gene expression. 1. **Zinc Finger:** This is the most common DNA-binding motif in humans. It consists of an $\alpha$-helix and a $\beta$-sheet held together by a zinc ion coordinated by Cysteine and Histidine residues (e.g., Steroid hormone receptors). 2. **Helix-Turn-Helix (HTH):** This motif consists of two $\alpha$-helices connected by a short turn of amino acids. One helix (the recognition helix) fits into the major groove of DNA. It is characteristic of homeodomain proteins involved in development. 3. **Leucine Zipper:** This motif features a periodic repetition of Leucine residues every seventh position (heptad repeat), forming an amphipathic $\alpha$-helix that facilitates dimerization. The basic region of the protein then binds to the DNA (e.g., CREB, c-fos, c-jun). **Why "All of these" is correct:** All three structures are specialized protein domains that facilitate the interaction between regulatory proteins and the double helix. Since each option functions as a DNA-binding domain, the collective answer is correct. ### High-Yield Clinical Pearls for NEET-PG: * **Zinc Finger:** Mutations in the zinc finger domain of the **Vitamin D Receptor (VDR)** lead to Hereditary Vitamin D-resistant Rickets. * **Leucine Zipper:** The **c-myc** oncogene (associated with Burkitt Lymphoma) utilizes a Leucine Zipper for dimerization. * **Homeobox (HOX) Genes:** These contain the **Helix-Turn-Helix** motif and are crucial for craniofacial and limb development; mutations lead to synpolydactyly or hand-foot-genital syndrome. * **TATA-binding protein (TBP):** Uses a unique "Beta-sheet" motif to bind DNA.

Question 79: During unwinding of DNA, relief of supercoil is made possible by?

A. Topoisomerase (Correct Answer)
B. Gyrase
C. Helicase
D. Polymerase

Explanation: ### Explanation **Correct Answer: A. Topoisomerase** **Mechanism:** As DNA helicase unwinds the double helix, the DNA ahead of the replication fork becomes overwound, leading to **positive supercoiling** (torsional strain). If not relieved, this strain would halt replication. **Topoisomerases** are enzymes that relieve this tension by creating transient breaks in the DNA backbone. * **Topoisomerase I** cuts a single strand (no ATP required). * **Topoisomerase II** cuts both strands (requires ATP). Once the supercoil is relaxed, the enzyme ligates the phosphodiester bonds back together. **Why other options are incorrect:** * **B. Gyrase:** While DNA Gyrase is a specific type of Topoisomerase II found in **prokaryotes**, "Topoisomerase" is the more general and accurate term for the process of relieving supercoils across all organisms. In many exam contexts, if both are present, Topoisomerase is the broader physiological answer. * **C. Helicase:** This enzyme is responsible for **unwinding** the DNA double helix by breaking hydrogen bonds between nitrogenous bases. It *causes* the supercoiling ahead of the fork rather than relieving it. * **D. Polymerase:** DNA Polymerases are responsible for **synthesizing** new DNA strands by adding nucleotides; they do not possess the nuclease activity required to relax supercoils. **High-Yield Clinical Pearls for NEET-PG:** 1. **Fluoroquinolones** (e.g., Ciprofloxacin) inhibit bacterial **DNA Gyrase** and Topoisomerase IV. 2. **Anticancer Drugs:** * **Etoposide/Teniposide** inhibit Topoisomerase II. * **Irinotecan/Topotecan** inhibit Topoisomerase I. 3. **DNA Gyrase** is unique because it can introduce *negative* supercoils, a feature specific to bacteria.

Question 80: Which of the following statements regarding gene regulation is false?

A. The repressor protein binds to the operator gene.
B. Regulator genes produce repressor subunits.
C. Isopropyl thiogalactoside (IPTG) acts as an inducer but not a substrate.
D. The regulator gene is typically inducible. (Correct Answer)

Explanation: ### Explanation **1. Why Option D is the Correct (False) Statement:** In the *lac* operon model, the **regulator gene (*lacI*) is constitutively expressed**, meaning it is transcribed continuously at a low rate to ensure the repressor protein is always present in the cell. It is **not inducible**; its expression does not change in response to the presence or absence of lactose. Inducibility refers to the structural genes (*lacZ, lacY, lacA*), which are turned "on" only when an inducer is present. **2. Analysis of Incorrect Options:** * **Option A:** The **repressor protein** (product of the regulator gene) has a high affinity for the **operator site**. When it binds, it physically blocks RNA polymerase from transcribing the structural genes. * **Option B:** The regulator gene (*lacI*) codes for mRNA that is translated into **repressor subunits**. Four of these subunits assemble to form the functional homotetrameric repressor. * **Option C:** **IPTG** is a classic example of a **gratuitous inducer**. It mimics allolactose and binds to the repressor to initiate transcription, but because it is not a substrate for $\beta$-galactosidase, it is not metabolized, maintaining a constant level of induction in experimental settings. ### High-Yield Clinical Pearls for NEET-PG * **The Lac Operon is a Negative Inducible System:** It is "off" by default (due to the repressor) but can be turned "on" (by an inducer). * **Catabolite Repression:** Even if lactose is present, the operon is not fully active if **glucose** is available. Glucose lowers **cAMP** levels, preventing the CAP-cAMP complex from binding to the promoter, a process known as "glucose effect." * **Key Enzyme:** *lacZ* encodes **$\beta$-galactosidase**, which cleaves lactose into glucose and galactose. * **Clinical Correlation:** Understanding gene regulation is fundamental to molecular medicine, including the mechanism of certain antibiotics and the pathogenesis of metabolic disorders.

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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