Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 41: Which of the following has the function of a peptidyl transferase?

A. Enzymes
B. Single stranded DNA
C. Elongation Factor
D. Ribozyme (Correct Answer)

Explanation: **Explanation:** **1. Why Ribozyme is Correct:** Peptidyl transferase is the primary enzyme responsible for peptide bond formation during the elongation phase of translation. In humans and other organisms, this catalytic activity is not performed by a protein, but by the **28S rRNA** (in eukaryotes) or **23S rRNA** (in prokaryotes) of the large ribosomal subunit. RNA molecules that possess catalytic/enzymatic activity are termed **Ribozymes**. This discovery shifted the biological paradigm that all enzymes are proteins. **2. Why Other Options are Incorrect:** * **Enzymes:** While most catalysts are enzymes (proteins), the specific function of peptidyl transferase is unique to catalytic RNA. Selecting "Enzymes" is less specific than "Ribozyme" in this context. * **Single-stranded DNA:** DNA serves as the genetic blueprint (template) but does not possess the catalytic ability to form peptide bonds during protein synthesis. * **Elongation Factors (EF):** EFs (like EF-Tu or EF-G) are proteins that facilitate the movement of tRNA and mRNA through the ribosome and assist in GTP hydrolysis, but they do not catalyze the actual formation of the peptide bond. **3. NEET-PG High-Yield Pearls:** * **The Ribozyme Concept:** The ribosome is essentially a "ribozyme" because its active site is composed entirely of RNA. * **Specific rRNA:** Remember **23S rRNA** for Prokaryotes and **28S rRNA** for Eukaryotes as the specific peptidyl transferase. * **Clinical Correlation:** Certain antibiotics, like **Chloramphenicol**, act by inhibiting the peptidyl transferase activity of the bacterial 50S subunit, thereby stopping bacterial protein synthesis. * **Other Ribozymes:** Examples include **SnRNAs** (involved in splicing) and **Ribonuclease P** (cleaves tRNA precursors).

Question 42: The strand of DNA from which mRNA is formed by transcription is called:

A. Template (Correct Answer)
B. Anti template
C. Coding
D. Transcript

Explanation: **Explanation:** In molecular biology, transcription is the process where a specific segment of DNA is copied into RNA by the enzyme **RNA polymerase**. **1. Why 'Template' is correct:** The DNA molecule is double-stranded, but only one strand serves as the guide for RNA synthesis. This strand is called the **Template strand** (also known as the **Antisense** or **Non-coding** strand). RNA polymerase reads this strand in the **3' to 5' direction** to synthesize a complementary mRNA molecule in the **5' to 3' direction**. Because of complementary base pairing, the mRNA sequence is a mirror image of the template strand. **2. Why other options are incorrect:** * **Coding Strand (Option C):** This is the DNA strand complementary to the template. Its sequence is identical to the synthesized mRNA (except T is replaced by U). It is called "coding" because its sequence represents the actual genetic code, but it is **not** used as a pattern by RNA polymerase. * **Anti-template (Option B):** This is a synonymous term for the Coding strand. Since it does not serve as the pattern for mRNA synthesis, it is incorrect. * **Transcript (Option D):** This refers to the product of transcription (the mRNA molecule itself), not the DNA strand from which it originated. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Directionality:** RNA synthesis always occurs in the **5' → 3'** direction. * **Promoters:** These are specific DNA sequences (like the TATA box) located on the coding strand that signal RNA polymerase where to begin transcription. * **Alpha-Amanitin:** A toxin from the *Amanita phalloides* mushroom that inhibits **RNA Polymerase II**, preventing mRNA synthesis—a classic high-yield toxicology/biochemistry correlation. * **Primary Transcript:** In eukaryotes, the initial product is **hnRNA** (heterogeneous nuclear RNA), which must undergo splicing, capping, and tailing to become mature mRNA.

Question 43: The Shine-Dalgarno sequence in bacterial mRNA is located near which codon?

A. AUG codon (Correct Answer)
B. UAA codon
C. UAG codon
D. UGA codon

Explanation: ### Explanation **1. Why Option A is Correct:** The **Shine-Dalgarno (SD) sequence** is a ribosomal binding site in bacterial (prokaryotic) mRNA. It is a purine-rich sequence (typically **AGGAGG**) located approximately **8 base pairs upstream (5' end)** of the **AUG start codon**. The underlying mechanism involves the **16S rRNA** (part of the 30S small ribosomal subunit), which contains a complementary pyrimidine-rich sequence at its 3' end. The base-pairing between the SD sequence and the 16S rRNA ensures that the ribosome is correctly positioned and aligned on the mRNA so that the AUG start codon is placed exactly at the **P-site**, allowing translation initiation to begin. **2. Why Other Options are Incorrect:** * **Options B, C, and D (UAA, UAG, UGA):** These are **Stop Codons** (nonsense codons). They signal the termination of translation, not initiation. The ribosome dissociates at these sequences with the help of Release Factors (RFs). There is no requirement for a Shine-Dalgarno-like sequence to position the ribosome at the 3' end of the mRNA. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Kozak Sequence:** This is the eukaryotic equivalent of the Shine-Dalgarno sequence. It helps the 40S ribosomal subunit identify the correct AUG start codon via a "scanning" mechanism. * **Mechanism of Action of Aminoglycosides:** Drugs like Streptomycin bind to the **16S rRNA** of the 30S subunit, interfering with the Shine-Dalgarno interaction and causing mRNA misreading. * **Polycistronic mRNA:** In bacteria, a single mRNA can have multiple SD sequences, allowing for the translation of several different proteins from one transcript. * **Formyl-methionine (fMet):** In prokaryotes, the AUG codon recruits N-formylmethionine as the first amino acid, whereas eukaryotes use unmodified methionine.

Question 44: Which of the following is NOT a circular DNA?

A. Nuclear DNA (Correct Answer)
B. Plasmid DNA
C. Mitochondrial DNA
D. Yeast DNA

Explanation: **Explanation:** The shape of DNA is primarily determined by the organism's complexity and the cellular compartment in which it resides. In eukaryotes, the genetic material is organized differently across various organelles. **1. Why Nuclear DNA is the Correct Answer:** In humans and other eukaryotes, **Nuclear DNA** is **linear**, not circular. It is organized into multiple chromosomes, each consisting of a single long linear molecule of double-stranded DNA associated with histone proteins. These linear strands have distinct ends called **telomeres**, which are essential for maintaining genomic stability. **2. Analysis of Incorrect Options:** * **Plasmid DNA:** These are small, extrachromosomal, **circular** DNA molecules found predominantly in bacteria and some eukaryotes (like yeast). They are widely used in recombinant DNA technology. * **Mitochondrial DNA (mtDNA):** According to the endosymbiotic theory, mitochondria originated from prokaryotes. Consequently, mtDNA is **double-stranded and circular**, resembling bacterial DNA. It lacks histones. * **Yeast DNA:** While yeast (Saccharomyces cerevisiae) has linear nuclear chromosomes, it is unique among eukaryotes for naturally harboring the **2-micron circle**, a stable **circular plasmid**. Since the question asks which is *not* circular, and yeast contains significant circular DNA elements, Nuclear DNA remains the most definitive answer. **Clinical Pearls for NEET-PG:** * **Prokaryotes:** Generally possess a single, circular chromosome. * **Telomerase:** This enzyme is unique to linear DNA (Nuclear DNA) to prevent the "end-replication problem." Circular DNA does not require telomerase. * **Maternal Inheritance:** Mitochondrial DNA is inherited exclusively from the mother; mutations here lead to "Mitochondrial Myopathies" (e.g., MELAS, LHON). * **Histones:** Present in Nuclear DNA but absent in Mitochondrial and Plasmid DNA.

Question 45: Which of the following is true about a silent mutation in a gene?

A. No change in mRNA.
B. No change in amino acid sequence in protein. (Correct Answer)
C. No expression of protein.
D. No change in the expression of protein.

Explanation: ### Explanation **1. Why Option B is Correct:** A **silent mutation** (also known as a synonymous mutation) is a type of point mutation where a single nucleotide base is substituted, but the resulting codon still codes for the **same amino acid**. This phenomenon is possible due to the **degeneracy of the genetic code**, where multiple codons can encode a single amino acid (e.g., GAA and GAG both code for Glutamic acid). Because the primary structure of the protein remains unchanged, the protein's function typically remains unaffected. **2. Why Other Options are Wrong:** * **Option A (No change in mRNA):** Incorrect. A mutation involves a change in the DNA sequence, which is directly transcribed into mRNA. Therefore, the mRNA sequence *will* be different at that specific codon. * **Option C (No expression of protein):** Incorrect. This describes a **nonsense mutation** (where a premature stop codon is created) or a promoter mutation that prevents transcription. In silent mutations, translation proceeds normally. * **Option D (No change in expression):** While often true, this is not the *definition* of a silent mutation. Some silent mutations can actually affect mRNA stability or splicing efficiency, potentially altering the *amount* of protein produced (expression levels), even if the amino acid sequence remains identical. **3. NEET-PG High-Yield Pearls:** * **Degeneracy/Redundancy:** Most silent mutations occur at the **3rd position** of the codon (the "Wobble" position). * **Transition vs. Transversion:** Transitions (Purine to Purine) are more common than transversions. * **Missense Mutation:** Results in a *different* amino acid (e.g., Sickle Cell Anemia: Glutamate → Valine). * **Nonsense Mutation:** Results in a *Stop* codon (UAA, UAG, UGA), leading to truncated proteins. * **Frameshift Mutation:** Insertion or deletion of bases (not a multiple of 3), altering the entire downstream reading frame.

Question 46: Which enzyme is used in Polymerase Chain Reaction (PCR)?

A. Reverse transcriptase
B. Taq polymerase (Correct Answer)
C. RNA polymerase
D. None of the above

Explanation: **Explanation:** **Taq polymerase** is the correct answer because PCR involves repeated cycles of high-temperature heating (denaturation at ~95°C) to separate DNA strands. Standard DNA polymerases would denature and lose function at these temperatures. Taq polymerase, isolated from the thermophilic bacterium *Thermus aquaticus*, is **thermostable**, meaning it remains active despite repeated exposure to high heat, allowing it to catalyze the synthesis of new DNA strands during the extension phase. **Analysis of Incorrect Options:** * **Reverse transcriptase:** This enzyme synthesizes DNA from an RNA template. While used in **RT-PCR** (Reverse Transcription PCR) to amplify RNA sequences (like the SARS-CoV-2 virus), it is not the core enzyme used for the standard PCR amplification cycles. * **RNA polymerase:** This enzyme synthesizes RNA from a DNA template during transcription. It is not involved in DNA replication or the PCR process. **High-Yield Clinical Pearls for NEET-PG:** * **Components of PCR:** Template DNA, Primers (forward and reverse), dNTPs (deoxynucleotide triphosphates), and Taq polymerase. * **Steps of PCR:** 1. Denaturation (94-96°C), 2. Annealing (50-65°C), 3. Extension (72°C). * **Applications:** Diagnosis of infectious diseases (HIV, TB), prenatal diagnosis of genetic mutations (Sickle cell anemia), and forensic medicine (DNA fingerprinting). * **Pfx/Pfu Polymerase:** These are newer thermostable polymerases with "proofreading" (3'-5' exonuclease) activity, offering higher fidelity than Taq polymerase.

Question 47: What is a rapid method for chromosome identification in intersex individuals?

A. Fluorescence in situ hybridization (FISH) (Correct Answer)
B. Polymerase chain reaction (PCR)
C. Single-strand conformation polymorphism (SSCP)
D. Karyotyping

Explanation: ### Explanation **Correct Answer: A. Fluorescence in situ hybridization (FISH)** **Why FISH is the correct answer:** In cases of intersex disorders (Ambiguous Genitalia), rapid identification of the sex chromosomes (X and Y) is critical for clinical management and parental counseling. **FISH** uses fluorescently labeled DNA probes that bind to specific sequences on chromosomes. It is considered a **rapid method** because it can be performed on **interphase nuclei** (non-dividing cells), providing results within 24 hours. It quickly detects the presence or absence of the SRY gene or specific sex chromosomes without waiting for cell culture. **Analysis of Incorrect Options:** * **B. Polymerase Chain Reaction (PCR):** While PCR can amplify specific genes (like SRY), it does not provide a visual structural overview of the chromosomes or detect numerical abnormalities (aneuploidies) as reliably as FISH in a clinical cytogenetic context. * **C. Single-strand conformation polymorphism (SSCP):** This is a screening method used to detect **point mutations** or small polymorphisms based on the conformational changes of single-stranded DNA. It is not used for chromosome identification. * **D. Karyotyping:** This is the "Gold Standard" for chromosomal analysis. However, it requires culturing cells and arresting them in **metaphase**, which typically takes **48 to 72 hours (or up to 1-2 weeks)**. Therefore, it is not the "rapid" method of choice compared to FISH. **High-Yield Clinical Pearls for NEET-PG:** * **FISH** is the bridge between cytogenetics and molecular biology (Molecular Cytogenetics). * **Interphase FISH** does not require cell culture, making it the fastest tool for diagnosing trisomies (13, 18, 21) and sex chromosome status in newborns. * **Barr Body:** A condensed, inactive X chromosome seen in females. The number of Barr bodies = (Total X chromosomes - 1). * **Gold Standard for Intersex:** While FISH is the *fastest*, a formal Karyotype is always eventually performed to confirm the diagnosis and rule out structural rearrangements.

Question 48: Which of the following is true about xeroderma pigmentosa?

A. Autosomal dominant
B. Good long-term prognosis
C. Purine dimers
D. DNA repair defect (Correct Answer)

Explanation: **Explanation:** **Xeroderma Pigmentosum (XP)** is a rare genetic disorder characterized by an extreme sensitivity to ultraviolet (UV) radiation. 1. **Why the correct answer is right (DNA repair defect):** The fundamental defect in XP is a deficiency in **Nucleotide Excision Repair (NER)**. Normally, UV light causes the formation of **pyrimidine dimers** (specifically thymine dimers) in DNA. The NER pathway utilizes specific endonucleases to excise these damaged segments. In XP patients, these enzymes are defective, leading to the accumulation of mutations, which results in skin cancers at a very young age. 2. **Why the incorrect options are wrong:** * **A. Autosomal dominant:** XP is an **Autosomal Recessive** disorder. Both parents must be carriers for the offspring to be affected. * **B. Good long-term prognosis:** The prognosis is generally **poor**. Patients have a 1000-fold increased risk of developing cutaneous malignancies (Basal Cell Carcinoma, Squamous Cell Carcinoma, and Melanoma) and often die young due to metastasis. * **C. Purine dimers:** UV radiation specifically causes **Pyrimidine dimers** (Thymine-Thymine or Cytosine-Thymine), not purine dimers. **High-Yield Clinical Pearls for NEET-PG:** * **Enzyme involved:** UV-specific endonuclease (also known as XP proteins). * **Clinical Presentation:** Severe sunburn after minimal sun exposure, "parchment-like" skin, excessive freckling, and photophobia. * **Neurological involvement:** Some variants (e.g., De Sanctis-Cacchione syndrome) present with progressive neurological degeneration and dwarfism. * **Management:** Strict UV protection (sunscreen, protective clothing) is the only management.

Question 49: Which of the following is a strategy used to increase the yield of protein produced in recombinant protein synthesis?

A. Promoter induction
B. Genes for protease inhibitors (Correct Answer)
C. Translation initiation
D. Translation and transcription termination

Explanation: In recombinant DNA technology, the primary goal is to maximize the production and stability of a functional protein. **Explanation of the Correct Answer:** **B. Genes for protease inhibitors:** Once a recombinant protein is synthesized in a host cell (like *E. coli* or yeast), it is highly susceptible to degradation by the host's endogenous **proteases**. These enzymes recognize the foreign protein as a target for proteolysis, significantly reducing the final yield. By co-expressing **genes for protease inhibitors** or using "protease-deficient" host strains, the degradation process is blocked, ensuring the protein accumulates to high levels. This is a critical strategy for the industrial production of therapeutic proteins like insulin or growth hormone. **Explanation of Incorrect Options:** * **A. Promoter induction:** While promoters are necessary to *start* transcription, "induction" is a standard step in the process, not a specific strategy to maximize yield beyond the cell's natural capacity. * **C. Translation initiation:** This is a fundamental biological process. While optimizing the Shine-Dalgarno sequence can help, "translation initiation" itself is a general term and not a specific yield-enhancement strategy compared to preventing protein loss. * **D. Translation and transcription termination:** Termination is required to release the mRNA or protein; however, enhancing termination does not increase the total quantity of the protein produced. **High-Yield Facts for NEET-PG:** * **Host Cells:** *E. coli* is the most common host, but eukaryotic hosts (like CHO cells) are used for proteins requiring **post-translational modifications** (e.g., glycosylation). * **Codon Optimization:** Another strategy to increase yield is replacing rare codons in the target gene with those frequently used by the host cell to prevent "ribosomal stalling." * **Inclusion Bodies:** Sometimes, high-yield production leads to insoluble protein aggregates called inclusion bodies, which require denaturation and refolding.

Question 50: A hypothetical gene contains a coding strand with the sequence 5'-TGGAATTGTATG-3'. What would be the sequence of the resultant mRNA following transcription?

A. 3'-ACCTTAACATAC-5'
B. 5'-ACCUUTTAACAUAC-3'
C. 5'-UGGAAUUGUAUG-3' (Correct Answer)
D. 3'-UCCTTUUCAUAC-5'

Explanation: ### Explanation **1. Why the Correct Answer is Right:** In molecular biology, transcription involves synthesizing RNA from a DNA template. To understand the relationship between DNA and mRNA, one must distinguish between the two DNA strands: * **Coding Strand (Sense Strand):** This strand has the same sequence and polarity (5' to 3') as the resultant mRNA, with the sole exception that **Thymine (T)** in DNA is replaced by **Uracil (U)** in RNA. * **Template Strand (Antisense Strand):** This is the strand actually read by RNA polymerase (3' to 5') to create a complementary RNA molecule. Since the question provides the **Coding Strand (5'-TGGAATTGTATG-3')**, the mRNA sequence will be identical in sequence and direction, simply substituting 'T' for 'U'. Thus, the mRNA is **5'-UGGAAUUGUAUG-3'**. **2. Why Incorrect Options are Wrong:** * **Option A:** This represents the complementary sequence to the coding strand (the template strand) in DNA form, but in the wrong orientation. * **Option B:** This contains an incorrect sequence length and incorrect base substitutions. * **Option D:** This is a random sequence that does not follow the rules of complementarity or transcription. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Directionality:** RNA polymerase always synthesizes mRNA in the **5' → 3' direction**, reading the template strand in the **3' → 5' direction**. * **The "T to U" Rule:** In any exam question, if the "Coding Strand" is given, do not perform complementary base pairing; just swap T for U. If the "Template Strand" is given, you must find the complement. * **Promoter Region:** Transcription starts upstream of the coding sequence at the promoter site (e.g., TATA box in eukaryotes, Pribnow box in prokaryotes). * **Post-transcriptional modification:** In eukaryotes, the initial transcript (hnRNA) undergoes 5' capping, 3' polyadenylation, and splicing before becoming mature mRNA.

Practice by Chapter

DNA Replication and Repair Mechanisms

Practice Questions

Transcription Factors and Gene Regulation

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Epigenetics and DNA Methylation

Practice Questions

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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