Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 171: Which enzyme is known to prevent cellular aging?

A. DNA ligase
B. DNA polymerase alpha
C. Telomerase (Correct Answer)
D. RNA polymerase II

Explanation: **Explanation:** **Telomerase** is the correct answer because it is a specialized ribonucleoprotein reverse transcriptase that maintains chromosomal stability. In somatic cells, DNA polymerase cannot replicate the extreme 3' ends of linear chromosomes (the **"End Replication Problem"**), leading to progressive shortening of telomeres with each cell division. When telomeres reach a critical minimum length, the cell enters **senescence** (cellular aging) or apoptosis. Telomerase prevents this by adding repetitive TTAGGG sequences to the ends of chromosomes, effectively acting as a "cellular fountain of youth." It is highly active in germ cells, stem cells, and cancer cells, but inactive in most mature somatic cells. **Incorrect Options:** * **DNA ligase:** Responsible for joining Okazaki fragments by creating phosphodiester bonds; it is essential for DNA repair and replication but does not prevent telomere shortening. * **DNA polymerase alpha:** Involved in the initiation of DNA replication by synthesizing an RNA primer; it cannot solve the end-replication problem. * **RNA polymerase II:** Responsible for the transcription of DNA into mRNA; it has no role in maintaining chromosomal length or preventing aging. **High-Yield Clinical Pearls for NEET-PG:** * **Composition:** Telomerase contains an RNA template (**TERC**) and a catalytic protein subunit (**TERT**). * **Cancer Link:** Approximately 85–90% of cancer cells upregulate telomerase to achieve **replicative immortality**. * **Shelterin Complex:** A protein complex that protects telomeres from being recognized as DNA double-strand breaks. * **Dyskeratosis Congenita:** A genetic disorder caused by telomerase deficiency, leading to premature aging signs, bone marrow failure, and mucosal leukoplakia.

Question 172: A mutation in which of the following sequences in eukaryotic mRNA will affect the process by which the poly-A tail is added to the 3' end of mRNA?

A. CCA
B. CAAT
C. AAUAAA (Correct Answer)
D. GU...A...AG

Explanation: **Explanation:** The correct answer is **AAUAAA**. This sequence is known as the **Polyadenylation Signal**, located approximately 10–30 nucleotides upstream of the cleavage site at the 3' end of eukaryotic pre-mRNA. **1. Why AAUAAA is correct:** In eukaryotes, the maturation of mRNA involves the addition of a poly-A tail (200–250 adenine residues). This process is triggered when the **Cleavage and Polyadenylation Specificity Factor (CPSF)** recognizes the highly conserved **AAUAAA** sequence. Once recognized, the pre-mRNA is cleaved, and the enzyme **Poly-A Polymerase (PAP)** adds the adenine residues. A mutation in this sequence prevents proper cleavage and polyadenylation, leading to unstable mRNA that is rapidly degraded. **2. Why other options are incorrect:** * **CCA:** This sequence is added to the **3' end of tRNA** (not mRNA) post-transcriptionally. it is the site where amino acids attach. * **CAAT:** This is the **CAAT box**, a conserved promoter sequence located upstream of the transcription start site that regulates the frequency of transcription initiation. * **GU...A...AG:** These are conserved sequences at the **splice sites**. GU is the 5' donor site, AG is the 3' acceptor site, and 'A' represents the branch point. They are essential for splicing, not polyadenylation. **Clinical Pearls for NEET-PG:** * **Poly-A Tail Function:** Increases mRNA stability, facilitates nucleocytoplasmic transport, and enhances translation efficiency. * **Poly-A Polymerase:** Unlike RNA polymerase, it does **not** require a DNA template. * **Clinical Correlation:** Mutations in the AAUAAA signal are associated with certain types of **α-thalassemia**, where failure to polyadenylate α-globin mRNA leads to its deficiency.

Question 173: What is the number of Barr bodies in XXY males?

A. 1 (Correct Answer)
B. 2
C. 3
D. 0

Explanation: ### Explanation **1. The Correct Answer (A):** The number of Barr bodies is determined by the formula: **n – 1**, where ‘n’ is the total number of X chromosomes present in the cell. This phenomenon is known as the **Lyon Hypothesis**. In a male with Klinefelter syndrome (47, XXY), there are two X chromosomes. Applying the formula (2 – 1 = 1), there is **one Barr body**. The Barr body represents a highly condensed, transcriptionally inactive form of the X chromosome called **facultative heterochromatin**. This inactivation occurs early in embryonic development to ensure **dosage compensation**, preventing females (or individuals with extra X chromosomes) from having double the amount of X-linked gene products compared to XY males. **2. Why Other Options are Incorrect:** * **Option B (2):** This would be seen in individuals with three X chromosomes, such as **47, XXX (Triple X syndrome)** or **48, XXXY**. * **Option C (3):** This would be seen in individuals with four X chromosomes, such as **48, XXXX** or **49, XXXXY**. * **Option D (0):** This is the normal finding in a **46, XY male** or a female with **45, XO (Turner syndrome)**, as they possess only one X chromosome. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Lyonization:** The process of X-inactivation is random, fixed, and incomplete (some genes on the short arm of the X chromosome escape inactivation). * **Xist Gene:** The *X-inactive specific transcript* (Xist) gene, located on the X inactivation center (Xic), produces a non-coding RNA that coats and silences the chromosome. * **Staining:** Barr bodies are typically visualized as dark-staining masses at the periphery of the nucleus in buccal smear preparations or as "drumsticks" in polymorphonuclear leukocytes (neutrophils). * **Rule of Thumb:** Regardless of the number of Y chromosomes, the cell always maintains only **one active X chromosome**; all others become Barr bodies.

Question 174: What is the approximate number of genes contained in the human genome?

A. 40,000
B. 30,000 (Correct Answer)
C. 80,000
D. 1,00,000

Explanation: ### Explanation The human genome consists of approximately **3.2 billion base pairs**, but only a small fraction (about 1.5%) codes for proteins. According to the findings of the **Human Genome Project (HGP)**, the estimated number of protein-coding genes is significantly lower than historical predictions, currently placed between **20,000 and 30,000**. **1. Why Option B is Correct:** While the exact number is refined as technology improves (recent estimates suggest ~21,000), **30,000** remains the standard benchmark for medical examinations like NEET-PG. This figure highlights the "G-value paradox," where the complexity of an organism does not necessarily correlate with the total number of genes, but rather with how those genes are regulated and spliced. **2. Why Other Options are Incorrect:** * **Option A (40,000):** This was an early post-genomic estimate that has since been revised downward as researchers identified many sequences as non-coding pseudogenes. * **Options C & D (80,000 and 1,00,000):** Prior to the completion of the HGP in 2003, scientists predicted much higher numbers based on the vast diversity of human proteins. We now know that **alternative splicing** allows a single gene to produce multiple protein isoforms, explaining how 30,000 genes can generate over 100,000 distinct proteins. **High-Yield Clinical Pearls for NEET-PG:** * **Coding vs. Non-coding:** Only **1.5%** of the genome is exonic (protein-coding). * **Largest Gene:** **Dystrophin** (2.4 million bases). * **Smallest Gene:** **TATA-binding protein** (or SRY on the Y chromosome). * **Chromosome with most genes:** Chromosome 1 (approx. 2,968 genes). * **Chromosome with fewest genes:** Y chromosome (approx. 231 genes). * **Single Nucleotide Polymorphisms (SNPs):** Occur at about 10 million locations; these account for most human genetic variation.

Question 175: Histone undergoes post-translational modification by which of the following mechanisms, except?

A. Acetylation
B. Methylation
C. Phosphorylation
D. Glycosylation (Correct Answer)

Explanation: **Explanation:** Histones are highly basic proteins that package DNA into structural units called nucleosomes. Their N-terminal "tails" extend out from the nucleosome core and are subject to various **Post-Translational Modifications (PTMs)** that regulate chromatin structure and gene expression (the "Histone Code"). **Why Glycosylation is the Correct Answer:** While histones can undergo a specific type of sugar modification called *O-GlcNAcylation*, traditional **Glycosylation** (the complex addition of oligosaccharides typically seen in membrane-bound or secreted proteins in the ER/Golgi) is **not** a standard regulatory mechanism for histones. Histones are nuclear proteins, and their modifications are primarily small chemical groups that alter charge or recruit specific binding proteins. **Analysis of Incorrect Options:** * **Acetylation (Option A):** Occurs on Lysine residues via Histone Acetyltransferases (HATs). It neutralizes the positive charge of histones, weakening their bond with negatively charged DNA, leading to **euchromatin** (transcriptionally active). * **Methylation (Option B):** Occurs on Lysine and Arginine residues. Unlike acetylation, it does not change the charge. Depending on the site, it can lead to either activation or repression (e.g., H3K4 methylation activates, while H3K9 methylation silences/forms **heterochromatin**). * **Phosphorylation (Option C):** Occurs on Serine, Threonine, and Tyrosine residues. It adds a negative charge and is heavily involved in chromosome condensation during mitosis and the DNA damage response. **High-Yield Clinical Pearls for NEET-PG:** * **Histone Deacetylase (HDAC) Inhibitors:** Drugs like **Vorinostat** and **Valproic acid** are used in oncology and psychiatry to alter gene expression by keeping chromatin in an open (acetylated) state. * **Linker Histone:** **H1** is the only histone not part of the nucleosome octamer; it facilitates the folding of the "beads-on-a-string" structure into the 30nm fiber. * **Amino Acid Richness:** Histones are rich in **Lysine and Arginine**, giving them the positive charge necessary to bind DNA.

Question 176: Which of the following trinucleotide repeat mutations is present in Huntington's disease?

A. CGG
B. GAA
C. CTG
D. CAG (Correct Answer)

Explanation: **Explanation:** **Huntington’s Disease (HD)** is an autosomal dominant neurodegenerative disorder characterized by chorea, psychiatric symptoms, and dementia. The underlying molecular defect is the expansion of the **CAG** trinucleotide repeat within the *Huntingtin (HTT)* gene on chromosome 4. 1. **Why CAG is correct:** The CAG sequence codes for the amino acid **Glutamine**. In HD, the expansion leads to a "Polyglutamine (polyQ) tract" in the huntingtin protein. This results in a **toxic gain-of-function**, leading to neuronal death, particularly in the caudate nucleus and putamen (striatum). 2. **Analysis of Incorrect Options:** * **CGG (Option A):** Associated with **Fragile X Syndrome**. It occurs in the 5' untranslated region (UTR) of the *FMR1* gene, leading to gene silencing via hypermethylation. * **GAA (Option B):** Associated with **Friedreich Ataxia**. This is an intronic expansion in the *FXN* gene (encoding Frataxin), leading to impaired mitochondrial function. * **CTG (Option C):** Associated with **Myotonic Dystrophy (Type 1)**. It occurs in the 3' UTR of the *DMPK* gene. **High-Yield Clinical Pearls for NEET-PG:** * **Anticipation:** HD exhibits anticipation (earlier onset/increased severity in successive generations), typically when inherited from the **father** (paternal transmission), due to instability during spermatogenesis. * **Threshold:** Normal repeats are <26; >40 repeats are fully penetrant for the disease. * **Neuroimaging:** Classic finding is **atrophy of the caudate nucleus**, leading to "boxcar ventricles" (enlargement of the frontal horns of lateral ventricles). * **Mnemonic:** "Huntington's **CAG**es **4** **C**audate" (CAG repeat, Chromosome 4, Caudate atrophy).

Question 177: What is the effect of UV radiation on cells?

A. Stimulates formation of purine dimers
B. Stimulates formation of pyrimidine dimers (Correct Answer)
C. Prevents formation of purine dimers
D. All of the above

Explanation: ### Explanation **Correct Option: B. Stimulates formation of pyrimidine dimers** Ultraviolet (UV) radiation, specifically UV-B (280–320 nm), is a potent mutagen. When DNA is exposed to UV light, it causes the formation of **covalent bonds** between adjacent pyrimidine bases (Cytosine or Thymine) on the same strand. The most common lesion is the **Thymine-Thymine (T-T) dimer**, also known as a cyclobutane pyrimidine dimer. These dimers create a "kink" in the DNA backbone, which obstructs DNA polymerase during replication and RNA polymerase during transcription, leading to mutations or cell death if left unrepaired. **Analysis of Incorrect Options:** * **Option A & C:** UV radiation specifically targets pyrimidines because their chemical structure (single-ring) is more susceptible to photochemical excitation than the double-ring structure of purines (Adenine and Guanine). Purine dimers are not a standard pathological feature of UV damage. * **Option D:** Since only pyrimidine dimer formation is the characteristic mechanism, "All of the above" is incorrect. **Clinical Pearls for NEET-PG:** 1. **Repair Mechanism:** Pyrimidine dimers are normally repaired by the **Nucleotide Excision Repair (NER)** pathway. 2. **Clinical Correlation:** A defect in the NER pathway (specifically the UV-specific endonuclease) leads to **Xeroderma Pigmentosum**. Patients present with extreme photosensitivity and a high risk of skin cancers (Basal Cell Carcinoma, Squamous Cell Carcinoma, and Melanoma). 3. **Enzyme involved:** In bacteria, the enzyme **DNA Photolyase** can directly reverse this damage (Photoreactivation), but this enzyme is absent in humans.

Question 178: Which of the following is NOT a method used in gene therapy?

A. Fluorescence in situ hybridization (FISH) (Correct Answer)
B. Transfection
C. Electroporation
D. Bacteriophage transduction

Explanation: **Explanation:** The core objective of **gene therapy** is the delivery of therapeutic genetic material into a patient's cells to treat or prevent disease. This requires specific **vector systems** or physical methods to transport DNA across the cell membrane. **Why Option A is the correct answer:** **Fluorescence in situ hybridization (FISH)** is a **cytogenetic diagnostic technique**, not a therapeutic one. It uses fluorescent probes that bind to specific parts of a chromosome to detect and locate specific DNA sequences. It is used for diagnosing chromosomal abnormalities (e.g., Trisomy 21), microdeletions (e.g., DiGeorge syndrome), or gene amplifications (e.g., HER2/neu in breast cancer). It does not involve the transfer of genetic material for treatment. **Why the other options are incorrect:** * **B. Transfection:** This is a non-viral method of introducing nucleic acids into eukaryotic cells using chemical means (like calcium phosphate or liposomes). It is a fundamental process in *ex vivo* gene therapy. * **C. Electroporation:** A physical method where high-voltage electrical pulses create temporary pores in the cell membrane, allowing large molecules like DNA to enter. It is widely used for gene delivery. * **D. Bacteriophage transduction:** This involves using viral vectors (modified viruses) to inject genetic material into a host cell. Viral vectors (Retrovirus, Adenovirus, AAV) are the most common vehicles used in clinical gene therapy. **High-Yield Clinical Pearls for NEET-PG:** * **Vectors:** Adeno-associated virus (AAV) is currently preferred for *in vivo* therapy because it is non-integrating and has low immunogenicity. * **First Gene Therapy:** Successfully performed in 1990 for **ADA-SCID** (Adenosine Deaminase deficiency). * **Liposomes:** These are "artificial vesicles" used in chemical transfection to bypass the lipid bilayer. * **FISH vs. Karyotyping:** FISH has higher resolution than standard karyotyping and can be performed on non-dividing (interphase) cells.

Question 179: What is the melting temperature of DNA proportional to?

A. Proportional to AT pairs
B. Proportional to GC pairs (Correct Answer)
C. No relation with base pair composition
D. Directly proportional to length of DNA

Explanation: **Explanation:** The melting temperature ($T_m$) of DNA is defined as the temperature at which 50% of the DNA double helix is denatured into single strands. This process involves breaking the hydrogen bonds between complementary base pairs. **1. Why Option B is Correct:** The stability of the DNA duplex is primarily determined by the base composition. **Guanine-Cytosine (GC) pairs** are held together by **three hydrogen bonds**, whereas Adenine-Thymine (AT) pairs are held by only two. Because more energy (heat) is required to break three bonds than two, DNA with a higher GC content has a higher $T_m$. Therefore, $T_m$ is directly proportional to the GC content. **2. Why Other Options are Incorrect:** * **Option A:** $T_m$ is inversely proportional to AT content. Higher AT pairs mean fewer hydrogen bonds, leading to a lower melting temperature. * **Option C:** Base pair composition is the primary intrinsic factor determining $T_m$; claiming no relation is biochemically incorrect. * **Option D:** While length does influence stability in very short oligonucleotides, for genomic DNA, the **base ratio (GC%)** and **ionic strength** of the solution are the dominant factors. $T_m$ is not a simple linear function of length in long DNA molecules. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Hyperchromicity:** Denaturation of DNA leads to an increase in UV light absorption at **260 nm**. This is known as the hyperchromic effect. * **Formamide & Urea:** These are chemical denaturants that lower the $T_m$ by disrupting hydrogen bonds. * **Ionic Strength:** High salt concentration ($Na^+$ ions) increases $T_m$ by neutralizing the negatively charged phosphate backbone, reducing repulsion between strands. * **TATA Box:** Promoters are often rich in AT pairs (like the TATA box) because they need to be easily "melted" or opened by RNA polymerase to initiate transcription.

Question 180: The two strands of DNA are held together by:

A. Van-der-Waal bond
B. Hydrogen bond (Correct Answer)
C. Covalent bond
D. Ionic interaction

Explanation: **Explanation:** The stability and structure of the DNA double helix are primarily maintained by **Hydrogen bonds** between complementary nitrogenous bases. According to Watson-Crick base pairing, Adenine (A) pairs with Thymine (T) via **two** hydrogen bonds, while Guanine (G) pairs with Cytosine (C) via **three** hydrogen bonds. These bonds are weak enough to allow "unzipping" during replication and transcription but strong enough to hold the strands together under physiological conditions. **Analysis of Incorrect Options:** * **Van-der-Waal bonds:** While these forces contribute to the "base-stacking" stability between adjacent bases on the *same* strand, they are not the primary force holding the two strands together. * **Covalent bonds:** These are strong bonds found in the **phosphodiester backbone** (linking the 3' carbon of one sugar to the 5' carbon of the next). If strands were held by covalent bonds, they could not be easily separated for biological processes. * **Ionic interaction:** DNA is negatively charged due to phosphate groups, but these charges actually cause repulsion between strands, which is neutralized by magnesium ions ($Mg^{2+}$) rather than holding the strands together. **High-Yield Clinical Pearls for NEET-PG:** * **GC Content & $T_m$:** DNA with higher G-C content has a higher **Melting Temperature ($T_m$)** because G-C pairs have three hydrogen bonds compared to the two in A-T pairs. * **Chargaff’s Rule:** In double-stranded DNA, the amount of A = T and G = C; therefore, Purines = Pyrimidines. * **Denaturation:** Agents like heat, formamide, and urea disrupt hydrogen bonds, leading to DNA denaturation.

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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