Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 341: What is true about DNA methylation?

A. It alters gene expression.
B. The genetic code remains intact.
C. It plays a role in carcinogenesis and acts as a protective mechanism against cleavage by restriction endonucleases.
D. All of the above. (Correct Answer)

Explanation: DNA methylation is a key **epigenetic modification** involving the addition of a methyl group to the 5th carbon of the cytosine ring, typically within **CpG islands**. **Explanation of Options:** * **A. Alters gene expression:** Methylation of promoter regions generally leads to **gene silencing**. It recruits methyl-CpG-binding domain proteins (MBDs) and histone deacetylases, which condense chromatin (heterochromatin), making it inaccessible to transcription factors. * **B. Genetic code remains intact:** This is the hallmark of epigenetics. Methylation changes the *phenotype* (expression) without altering the *genotype* (DNA sequence). The underlying nucleotide sequence remains unchanged. * **C. Carcinogenesis and Restriction Endonucleases:** * **In Cancer:** Global hypomethylation (causing genomic instability) and focal hypermethylation of **tumor suppressor genes** (e.g., *BRCA1, RB1*) are classic drivers of malignancy. * **In Bacteria:** The **Restriction-Modification (R-M) system** uses methylation to mark "self" DNA. This protects the bacterial genome from being degraded by its own restriction endonucleases, which are designed to cleave unmethylated "foreign" viral DNA. **High-Yield Clinical Pearls for NEET-PG:** * **Enzyme:** DNA methyltransferase (DNMT) uses **S-adenosylmethionine (SAM)** as the methyl donor. * **Genomic Imprinting:** Methylation is the mechanism behind Prader-Willi and Angelman syndromes (Chromosome 15). * **Fragile X Syndrome:** Characterized by hypermethylation of the *FMR1* gene due to CGG triplet repeat expansion. * **Lyonization:** X-chromosome inactivation in females occurs via extensive DNA methylation.

Question 342: What is true about next-generation sequencing?

A. The entire human genome can be sequenced within a single day.
B. It is more sensitive than Sanger sequencing.
C. It utilizes RNA sequencing to discover novel RNA variants.
D. All of the above. (Correct Answer)

Explanation: **Explanation:** Next-Generation Sequencing (NGS), also known as high-throughput sequencing, represents a paradigm shift from traditional Sanger sequencing by utilizing **massive parallel processing**. This allows millions of DNA fragments to be sequenced simultaneously rather than one at a time. * **Option A is correct:** Due to its high-throughput nature, NGS has drastically reduced the time required for genomic analysis. While the first Human Genome Project took 13 years, modern NGS platforms can sequence an entire human genome in **under 24 hours**. * **Option B is correct:** NGS is significantly more sensitive than Sanger sequencing. It can detect **low-frequency variants** (somatic mutations) and mosaicism that Sanger sequencing (which has a detection limit of ~15-20%) would typically miss. * **Option C is correct:** NGS is not limited to DNA. **RNA-Seq** (Transcriptome sequencing) allows for the discovery of novel RNA variants, alternative splicing patterns, and gene fusion events, which are critical in oncology and personalized medicine. Since all individual statements are accurate descriptions of NGS capabilities, **Option D (All of the above)** is the correct choice. **High-Yield Clinical Pearls for NEET-PG:** * **Sanger Sequencing:** Still the "Gold Standard" for clinical validation of single gene variants but is low-throughput. * **Library Preparation:** The first step in NGS involving fragmentation of DNA and ligation of adapters. * **Bridge Amplification:** A key feature of Illumina sequencing (the most common NGS technology). * **Clinical Application:** NGS is the preferred method for **Liquid Biopsies** (detecting circulating tumor DNA) and diagnosing rare genetic disorders via **Whole Exome Sequencing (WES)**.

Question 343: What part of mRNA is removed during protein synthesis?

A. Interon (Correct Answer)
B. Codon
C. Exon
D. Suistron

Explanation: ### Explanation The correct answer is **A. Intron** (spelled as "Interon" in some question banks). **1. Why Introns are the Correct Answer:** In eukaryotes, the initial product of transcription is **pre-mRNA** (hnRNA). This precursor contains both coding sequences (**Exons**) and non-coding intervening sequences (**Introns**). Before translation (protein synthesis) can occur, the pre-mRNA undergoes **Post-transcriptional modification**. A critical step is **Splicing**, where introns are excised and removed by the spliceosome complex, and exons are joined together to form the mature mRNA. Therefore, introns are the parts removed to ensure only the functional genetic code reaches the ribosome. **2. Analysis of Incorrect Options:** * **B. Codon:** These are sequences of three nucleotides that correspond to a specific amino acid. They are the fundamental units of the genetic code and are *read* during protein synthesis, not removed. * **C. Exon:** These are the "expressed" sequences. They carry the actual information for the protein sequence and are retained and joined together after splicing. * **D. Suistron (Cistron):** A cistron is a genetic unit synonymous with a gene (a segment of DNA that encodes a single polypeptide). It is a structural unit, not a part removed during processing. **3. NEET-PG High-Yield Clinical Pearls:** * **Spliceosome:** Composed of **snRNPs** (small nuclear ribonucleoproteins). Autoantibodies against these (Anti-Smith antibodies) are highly specific for **Systemic Lupus Erythematosus (SLE)**. * **Alternative Splicing:** A process where different combinations of exons are joined, allowing one single gene to code for multiple different proteins (e.g., membrane-bound vs. secreted antibodies). * **Splicing Mutations:** Mutations at splice sites are a common cause of genetic diseases like **β-Thalassemia**.

Question 344: Angelman syndrome is due to what genetic mechanism?

A. Mitochondrial inheritance
B. Chimerism
C. Mosaicism
D. Uniparental disomy (Correct Answer)

Explanation: **Explanation:** Angelman syndrome (AS) is a classic example of **Genomic Imprinting**, where the expression of a gene depends on whether it is inherited from the mother or the father. It is caused by the loss of function of the **UBE3A gene** on chromosome 15 (15q11-q13). While the most common cause is a maternal deletion, **Paternal Uniparental Disomy (UPD)**—where a child inherits two copies of chromosome 15 from the father and none from the mother—accounts for approximately 3–7% of cases. Since the paternal UBE3A gene is normally silenced (imprinted), UPD results in a total lack of active UBE3A expression in the brain. **Analysis of Options:** * **Mitochondrial inheritance (A):** This involves genes located in the mitochondrial DNA, passed exclusively from the mother (e.g., MELAS, LHON). AS involves nuclear DNA on chromosome 15. * **Chimerism (B):** This refers to an individual composed of cells from two different zygotes. It is not the mechanism for imprinting disorders. * **Mosaicism (C):** This is the presence of two or more populations of cells with different genotypes in one individual (e.g., some cells having Trisomy 21 and others being normal). While rare mosaic forms of AS exist, UPD is a primary recognized mechanism. **High-Yield Clinical Pearls for NEET-PG:** * **Prader-Willi Syndrome (PWS):** The "sister" condition caused by **Maternal UPD** (or paternal deletion) of the same 15q11-q13 region. * **Mnemonic:** **M**aternal deletion/Paternal UPD = **A**ngelman (**M**appy Child: **M**aternal **A**ngelman, **P**uppet-like gait, **P**aroxysms of laughter). * **Clinical Features of AS:** "Happy Puppet" syndrome—inappropriate laughter, seizures, ataxia, and severe intellectual disability.

Question 345: Which enzyme produces single-strand nicks in DNA?

A. DNA polymerase I
B. DNase I (Correct Answer)
C. Polynucleotide kinase
D. 5'-exonuclease

Explanation: **Explanation:** **DNase I (Deoxyribonuclease I)** is an endonuclease that cleaves phosphodiester bonds in DNA. Its primary characteristic is that it acts on double-stranded DNA (dsDNA) to produce **single-strand nicks** (random internal breaks) rather than cutting both strands simultaneously. In molecular biology, this property is exploited for "Nick Translation" and "DNase Footprinting" to study DNA-protein interactions. **Analysis of Incorrect Options:** * **DNA Polymerase I:** This enzyme is primarily involved in DNA replication and repair. While it has 5'→3' and 3'→5' exonuclease activity to remove RNA primers and proofread, its main function is **synthesis** of DNA, not the production of nicks. * **Polynucleotide Kinase (PNK):** This enzyme catalyzes the transfer of a phosphate group from ATP to the **5'-hydroxyl terminus** of DNA or RNA. It is used for end-labeling nucleic acids, not for cleaving the phosphodiester backbone. * **5'-Exonuclease:** Exonucleases remove nucleotides one by one from the **ends** of a DNA molecule. They cannot produce internal nicks in a circular or intact linear dsDNA molecule as they require a pre-existing free end to function. **High-Yield Clinical Pearls for NEET-PG:** * **DNase I Clinical Use:** Recombinant human DNase (Dornase alfa) is used in **Cystic Fibrosis** to reduce the viscosity of sputum by digesting the extracellular DNA released from degenerating neutrophils. * **Topoisomerase I:** Another enzyme that produces transient single-strand nicks to relieve torsional strain (supercoiling) during replication. * **Restriction Endonucleases:** Unlike DNase I, these produce specific double-strand breaks at palindromic sequences.

Question 346: Which of the following amino acid substitutions is considered a conservative mutation?

A. Glutamic acid-glutamine
B. Histidine-glycine
C. Alanine-leucine (Correct Answer)
D. Arginine-aspartic acid

Explanation: ### Explanation **Concept Overview:** A **conservative mutation** is a type of missense mutation where one amino acid is replaced by another with **similar physicochemical properties** (e.g., charge, hydrophobicity, or size). Because the new amino acid "mimics" the original, the overall tertiary structure and function of the protein often remain largely unaffected. **Why Option C is Correct:** * **Alanine and Leucine** are both **non-polar, hydrophobic, branched-chain aliphatic amino acids**. * Substituting Alanine with Leucine preserves the hydrophobic nature of the protein core. While the side chain size differs slightly, their chemical behavior is identical, making this a classic example of a conservative substitution. **Analysis of Incorrect Options:** * **A. Glutamic acid (Acidic/Negative) to Glutamine (Neutral/Polar):** This changes a charged residue to an uncharged one, potentially disrupting ionic bonds (salt bridges). * **B. Histidine (Basic/Positive) to Glycine (Smallest/Neutral):** This is a non-conservative change. Glycine provides extreme conformational flexibility and lacks the bulky, charged imidazole ring of Histidine. * **D. Arginine (Basic/Positive) to Aspartic acid (Acidic/Negative):** This is a **radical substitution**. Reversing the charge from positive to negative drastically alters protein folding and enzymatic activity. **NEET-PG High-Yield Pearls:** * **Sickle Cell Anemia:** A classic **non-conservative** mutation where Glutamic acid (polar) is replaced by Valine (non-polar) at the 6th position of the $\beta$-globin chain. * **Transition vs. Transversion:** Transitions (Purine to Purine) are more likely to result in conservative mutations than Transversions (Purine to Pyrimidine). * **Silent Mutation:** A point mutation that changes the codon but results in the *same* amino acid (due to degeneracy of the genetic code).

Question 347: What sequence is present at the 3' end of tRNA?

A. Poly 'A' tail
B. CCA sequence (Correct Answer)
C. Anticodon
D. D arm

Explanation: **Explanation:** The correct answer is **B. CCA sequence**. Transfer RNA (tRNA) is a cloverleaf-shaped molecule essential for protein synthesis. The **3' end** of all mature tRNA molecules terminates in the highly conserved sequence **5'-CCA-3'**. This is known as the **acceptor stem**. The amino acid is covalently attached to the hydroxyl (-OH) group of the adenosine (A) residue in this CCA sequence by the enzyme *aminoacyl-tRNA synthetase*. **Analysis of Incorrect Options:** * **A. Poly 'A' tail:** This is a long chain of adenine nucleotides added to the **3' end of mRNA** (not tRNA) to enhance stability and facilitate nuclear export. * **C. Anticodon:** This is a triplet of nucleotides located in the **Anticodon Loop** (at the opposite end of the acceptor stem). It base-pairs with the complementary codon on mRNA to ensure the correct amino acid is incorporated. * **D. D arm:** This is a lateral loop of the tRNA containing **Dihydrouridine**. Its primary function is the recognition of the specific aminoacyl-tRNA synthetase enzyme. **High-Yield NEET-PG Pearls:** 1. **Post-transcriptional modification:** In eukaryotes, the CCA sequence is not coded by the DNA but is added enzymatically by **tRNA nucleotidyltransferase**. 2. **TψC Arm:** Contains Pseudouridine and is responsible for binding the tRNA to the **ribosome** (specifically the 5S rRNA). 3. **Smallest RNA:** tRNA is the smallest (73–93 nucleotides) and is also known as **Soluble RNA (sRNA)**. 4. **Charging:** The process of adding an amino acid to the CCA end is called "charging" or aminoacylation, which requires ATP.

Question 348: Highly repetitive DNA is seen in which of the following locations?

A. Microsatellite DNA
B. Telomere
C. Centromere
D. All of these (Correct Answer)

Explanation: ### Explanation The human genome is categorized into **Single-copy DNA** (coding sequences) and **Repetitive DNA** (non-coding sequences). Repetitive DNA is further divided based on the length of the repeat unit and its location. **Highly repetitive DNA**, often referred to as **Satellite DNA**, consists of short sequences repeated thousands to millions of times. **Why "All of these" is correct:** Highly repetitive DNA is characteristically found in specific structural regions of the chromosome: * **Centromeres:** These contain **Alpha-satellite DNA** (171 bp repeats). This repetitive sequence is essential for kinetochore assembly and proper spindle attachment during cell division. * **Telomeres:** These consist of hexameric repeats (**TTAGGG** in humans). These repetitive "caps" protect the ends of chromosomes from degradation and fusion. * **Microsatellites (Short Tandem Repeats - STRs):** These are small repetitive sequences (2–6 bp) scattered throughout the genome. They are highly polymorphic between individuals. **Analysis of Options:** * **Option A (Microsatellites):** While these are scattered, they are a classic example of repetitive DNA used in DNA fingerprinting. * **Option B (Telomere):** These are specialized repetitive structures at chromosomal ends. * **Option C (Centromere):** This is the primary site for constitutive heterochromatin, which is composed almost entirely of highly repetitive DNA. **NEET-PG High-Yield Pearls:** 1. **Satellite DNA** is separated from bulk DNA during **density gradient centrifugation** as distinct "satellite" bands due to differing G-C content. 2. **Minisatellites (VNTRs):** Used in DNA fingerprinting (Southern Blotting). 3. **Microsatellites (STRs):** Used in forensic analysis and paternity testing via PCR. 4. **Telomerase:** A reverse transcriptase enzyme that maintains telomere length; its activity is high in cancer cells and germ cells but low in somatic cells.

Question 349: Radio-labeled DNA was allowed to replicate twice in a non-radioactive environment. Which of the following is true?

A. All the strands will have radioactivity
B. Half of the DNA will have no radioactivity (Correct Answer)
C. No strands will have radioactivity
D. Three-fourth of the DNA replicated will have radioactivity

Explanation: ### Explanation This question tests the fundamental concept of **Semiconservative DNA Replication**, as proven by the classic Meselson-Stahl experiment. **1. Why the Correct Answer is Right:** According to the semiconservative model, each strand of the parent DNA molecule acts as a template for the synthesis of a new complementary strand. * **Initial State:** You start with one double-stranded DNA (dsDNA) molecule where both strands are radio-labeled (**R-R**). * **First Replication (in non-radioactive environment):** The two radioactive strands separate. Each picks up a non-radioactive partner (**N**). This results in **two** hybrid dsDNA molecules (**R-N** and **R-N**). At this stage, 100% of the DNA molecules contain radioactivity. * **Second Replication:** The four strands from the first generation (2 **R** and 2 **N**) separate. * The 2 **R** strands pick up **N** partners $\rightarrow$ 2 **R-N** molecules. * The 2 **N** strands pick up **N** partners $\rightarrow$ 2 **N-N** molecules. * **Result:** Out of 4 DNA molecules, 2 are hybrid (**R-N**) and 2 are entirely non-radioactive (**N-N**). Thus, **half (50%) of the DNA molecules have no radioactivity.** **2. Why the Other Options are Wrong:** * **Option A:** Incorrect because the radioactive material is diluted; new strands are synthesized using non-radioactive nucleotides. * **Option C:** Incorrect because the original parent strands are stable and remain intact (though separated) across generations. * **Option D:** This would only occur if the replication mechanism were dispersive, which it is not. **3. High-Yield Clinical Pearls for NEET-PG:** * **Meselson-Stahl Experiment:** Used *E. coli* and heavy nitrogen isotopes ($^{15}N$ and $^{14}N$) to prove semiconservative replication. * **DNA Polymerase:** Synthesizes DNA in the **5' to 3' direction** only. * **Quinolones (e.g., Ciprofloxacin):** Act by inhibiting DNA Gyrase (Topoisomerase II), preventing the relaxation of supercoils during replication. * **5-Fluorouracil (5-FU):** Inhibits Thymidylate Synthase, depleting the dTMP pool required for DNA synthesis.

Question 350: Which of the following blood cells contains drumsticks?

A. Neutrophils (Correct Answer)
B. Lymphocytes
C. Monocytes
D. Platelets

Explanation: **Explanation:** The correct answer is **Neutrophils**. **Why Neutrophils are correct:** In females, the **drumstick** (also known as a Davidson body) is a small, pedunculated nuclear appendage found in approximately 1–6% of mature **neutrophils**. This structure represents the inactivated X chromosome, known as **sex chromatin** or a **Barr body**. While Barr bodies are typically seen as dense masses against the nuclear membrane in buccal mucosal cells, they appear as "drumsticks" in neutrophils due to the unique multilobed nature of the polymorphonuclear leukocyte nucleus. **Why the other options are incorrect:** * **Lymphocytes & Monocytes:** These are mononuclear cells. While they do contain the inactivated X chromosome in females, it does not manifest as a "drumstick" appendage. In these cells, the Barr body is usually indistinguishable from other chromatin clumps within the nucleus. * **Platelets:** These are anucleated cell fragments derived from megakaryocytes. Since they lack a nucleus, they cannot possess nuclear appendages like drumsticks or Barr bodies. **High-Yield Clinical Pearls for NEET-PG:** * **Lyon’s Hypothesis:** Explains that X-inactivation is random, fixed, and occurs early in embryonic development. * **Formula:** The number of Barr bodies/drumsticks = (Total number of X chromosomes – 1). * **Clinical Correlation:** * **Turner Syndrome (45, XO):** Zero drumsticks (Male-like pattern). * **Klinefelter Syndrome (47, XXY):** One drumstick present (Female-like pattern). * **Diagnostic Note:** While historically used for sex determination, chromosomal analysis (Karyotyping) or PCR for the SRY gene are now the gold standards.

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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