Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 561: What determines the function of a gene?

A. Southern blot
B. Western blot
C. Inoculating in transgenic mice
D. Inserting or knocking out a gene (Correct Answer)

Explanation: **Explanation:** The function of a gene is best determined by observing the phenotypic consequences of its presence or absence within a living system. This is the principle of **Reverse Genetics**. * **Why Option D is Correct:** * **Gene Knockout:** By "knocking out" (inactivating) a specific gene, researchers can observe what biological processes fail or what pathologies develop. If removing Gene X leads to insulin deficiency, the function of Gene X is determined to be related to insulin production. * **Gene Insertion (Transgenesis/Knock-in):** Adding a gene to an organism to see if it gains a specific function further confirms the gene's role. These methods provide a direct causal link between a DNA sequence and its biological function. * **Why Other Options are Incorrect:** * **Southern Blot (A):** Used for the detection of a specific **DNA sequence** in a sample. It tells you if a gene is present, but not what it does. * **Western Blot (B):** Used for the detection and quantification of specific **proteins**. While it confirms gene expression (translation), it does not define the functional role of that protein in the organism. * **Inoculating in Transgenic Mice (C):** This is a distractor. Transgenic mice are the *tools* used, but the *process* that determines function is the actual insertion or deletion of the gene within them. **High-Yield NEET-PG Pearls:** * **Knockout Mice:** Created using homologous recombination in embryonic stem cells. * **CRISPR-Cas9:** The most modern and efficient tool for gene editing (knocking out or inserting genes). * **North-Western Blot:** Used to detect DNA-binding proteins. * **South-Western Blot:** Used to detect protein-DNA interactions. * **Reporter Genes (e.g., GFP):** Used to study the *expression pattern* and localization of a gene rather than its physiological function.

Question 562: Transfer RNA (tRNA) primarily interacts with which of the following molecules during protein synthesis?

A. Deoxyribonucleic acid (DNA)
B. Messenger ribonucleic acid (mRNA) (Correct Answer)
C. Micro ribonucleic acid (miRNA)
D. Small interfering ribonucleic acid (siRNA)

Explanation: ### Explanation **Why mRNA is the Correct Answer:** Protein synthesis (translation) occurs on the ribosome, where **tRNA** acts as an "adapter molecule." Each tRNA molecule carries a specific amino acid and possesses an **anticodon** loop. This anticodon recognizes and binds to a complementary **codon** on the **messenger RNA (mRNA)** through base pairing. This interaction ensures that the genetic code carried by the mRNA is accurately translated into a specific sequence of amino acids to form a polypeptide chain. **Analysis of Incorrect Options:** * **A. DNA:** tRNA does not interact with DNA during translation. DNA serves as the template for transcription (forming mRNA, tRNA, and rRNA) in the nucleus, but the functional role of tRNA is restricted to the cytoplasm/ribosomes. * **C & D. miRNA and siRNA:** These are small non-coding RNAs involved in **gene silencing** and RNA interference (RNAi). They typically interact with mRNA to degrade it or inhibit its translation, but they do not directly interact with tRNA during the assembly of proteins. **High-Yield NEET-PG Pearls:** * **Wobble Hypothesis:** Proposed by Francis Crick, it explains why the 3rd base of the mRNA codon can have non-traditional pairing with the 1st base of the tRNA anticodon, allowing one tRNA to recognize multiple codons. * **Aminoacyl-tRNA Synthetase:** The enzyme responsible for "charging" tRNA by attaching the correct amino acid (requires ATP). This is the true "translator" of the genetic code. * **CCA Tail:** All tRNA molecules have a **CCA sequence at the 3' end**, which is the attachment site for the amino acid. * **DHU Arm:** Responsible for recognition by the aminoacyl-tRNA synthetase enzyme. * **TψC Arm:** Involved in binding the tRNA to the ribosomal surface.

Question 563: Which of the following is NOT required for Polymerase Chain Reaction (PCR)?

A. Deoxyribonucleotides
B. Taq polymerase
C. Dideoxyribonucleotides (Correct Answer)
D. Template DNA

Explanation: ### Explanation **1. Why Dideoxyribonucleotides (ddNTPs) is the correct answer:** Polymerase Chain Reaction (PCR) is an *in vitro* method used to amplify specific DNA sequences. It requires **Deoxyribonucleotides (dNTPs)**—dATP, dCTP, dGTP, and dTTP—to build the new DNA strands. **Dideoxyribonucleotides (ddNTPs)**, however, lack a 3'-OH group. If incorporated, they cause **chain termination** because no further phosphodiester bonds can be formed. Therefore, ddNTPs are the hallmark of **Sanger Sequencing**, not standard PCR. **2. Why the other options are incorrect:** * **Deoxyribonucleotides (dNTPs):** These are the essential "building blocks" required by the DNA polymerase to synthesize the complementary strand. * **Taq polymerase:** This is a heat-stable DNA polymerase (isolated from *Thermus aquaticus*). It is vital because it remains functional during the high-temperature denaturation step (94-96°C) of PCR. * **Template DNA:** This is the target sample containing the specific region of interest that needs to be amplified. Without a template, there is no sequence for the primers to bind to. **3. High-Yield Clinical Pearls for NEET-PG:** * **Steps of PCR:** Denaturation (95°C) → Annealing (55-65°C) → Extension (72°C). * **RT-PCR:** Uses Reverse Transcriptase to convert RNA into cDNA before amplification; it is the gold standard for diagnosing **COVID-19 (SARS-CoV-2)**. * **Real-Time PCR (qPCR):** Allows for the quantification of DNA in real-time using fluorescent dyes (e.g., SYBR Green) or probes (TaqMan). * **Components of PCR Mix:** Template, Primers (forward and reverse), dNTPs, Taq Polymerase, and $Mg^{2+}$ (cofactor).

Question 564: What is the result of a transition mutation in the DNA sequence GATCCT?

A. GGTCCT (Correct Answer)
B. GTTCCT
C. GABCCT
D. GrUGGT

Explanation: ### Explanation **1. Understanding the Correct Answer (Option A: GGTCCT)** A **transition mutation** is a point mutation where a **purine is replaced by another purine** (A ↔ G) or a **pyrimidine is replaced by another pyrimidine** (C ↔ T). In the original sequence **GATCCT**, the second base is Adenine (A), which is a purine. In Option A (**GGTCCT**), the Adenine has been replaced by Guanine (G), which is also a purine. Since a purine is replaced by a purine, this is a classic transition mutation. **2. Analysis of Incorrect Options** * **Option B (GTTCCT):** Here, Adenine (Purine) is replaced by Thymine (Pyrimidine). This is a **transversion mutation** (Purine ↔ Pyrimidine). * **Option C (GABCCT):** "B" is not a standard nitrogenous base in DNA. This option is biologically invalid. * **Option D (GrUGGT):** This sequence contains "U" (Uracil), which is found in RNA, not DNA. Furthermore, it involves multiple base changes, whereas the question implies a single point mutation. **3. High-Yield Clinical Pearls for NEET-PG** * **Transition vs. Transversion:** Transitions are more common in the genome than transversions (ratio approx. 2:1), despite there being more possible transversion pathways. * **Silent Mutations:** Often occur at the 3rd position of a codon (wobble hypothesis) and do not change the amino acid. * **Missense Mutation:** Results in a different amino acid (e.g., Sickle Cell Anemia: Glutamate → Valine). * **Nonsense Mutation:** Results in a premature stop codon (UAA, UAG, UGA), leading to a truncated protein. * **Frameshift Mutation:** Caused by insertions or deletions (indels) not in multiples of three; these are usually more deleterious than point mutations.

Question 565: What does a Poly(A) tail translate into?

A. Polyproline
B. Polysine (Correct Answer)
C. Polyalanine
D. Polyglycine

Explanation: ### Explanation The correct answer is **Polylysine**. **The Underlying Concept:** In eukaryotic mRNA, the **Poly(A) tail** consists of a long chain of adenine nucleotides added to the 3' end during post-transcriptional modification. In the genetic code, the triplet codon for the amino acid **Lysine is AAA**. Therefore, if a ribosome were to translate a continuous sequence of adenines (Poly-A), it would result in a polypeptide chain consisting entirely of lysine residues (**Polylysine**). **Analysis of Options:** * **B. Polylysine (Correct):** As established, the codon **AAA** codes for Lysine. This is a high-yield fact often tested in the context of translation and mRNA processing. * **A. Polyproline:** Proline is encoded by codons starting with CC (e.g., **CCC**). A Poly(C) tail would translate into Polyproline. * **C. Polyalanine:** Alanine is encoded by codons starting with GC (e.g., **GCC**). * **D. Polyglycine:** Glycine is encoded by codons starting with GG (e.g., **GGG**). A Poly(G) tail would translate into Polyglycine. **High-Yield Clinical Pearls for NEET-PG:** * **Polyadenylation:** Occurs in the nucleus; it is catalyzed by the enzyme **Poly(A) Polymerase**, which does *not* require a DNA template. * **Function:** The Poly(A) tail increases mRNA stability, facilitates nuclear export, and enhances translation efficiency. * **Non-stop Decay (NSD):** In vivo, if a ribosome translates into the Poly(A) tail (due to a missing stop codon), the resulting polylysine stretch acts as a signal for the **Exosome complex** to degrade the faulty mRNA and the stalled ribosome. * **Nirenberg’s Experiment:** Marshall Nirenberg used synthetic homopolymers (like Poly-U, Poly-A, and Poly-C) to decipher the genetic code. He discovered Poly-U translates to Polyphenenylalanine, Poly-C to Polyproline, and **Poly-A to Polylysine**.

Question 566: Okazaki segments are required for:

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

Explanation: **Explanation:** **Why DNA Synthesis is Correct:** DNA replication is **semidiscontinuous**. The DNA polymerase enzyme can only synthesize new strands in the **5' to 3' direction**. During replication, the "Leading Strand" is synthesized continuously toward the replication fork. However, the "Lagging Strand" runs in the opposite direction (3' to 5' relative to the fork). To overcome this, the cell synthesizes short fragments of DNA in a 5' to 3' direction away from the fork. These short sequences (100–200 nucleotides in eukaryotes) are called **Okazaki fragments**. They are eventually joined together by the enzyme **DNA Ligase** to form a continuous strand. **Why Other Options are Incorrect:** * **RNA Synthesis (Transcription):** This process involves RNA polymerase synthesizing a single-stranded RNA molecule from a DNA template. It does not involve discontinuous fragments or Okazaki segments. * **Protein Synthesis (Translation):** This occurs in the ribosomes where mRNA is decoded into amino acids. It involves tRNA and rRNA, not DNA fragments. **High-Yield NEET-PG Pearls:** * **Directionality:** DNA synthesis always occurs in the **5' → 3'** direction. * **Enzymology:** Okazaki fragments are initiated by **RNA Primers** (synthesized by DNA Primase). * **The "Glue":** **DNA Ligase** is the enzyme responsible for joining Okazaki fragments by forming phosphodiester bonds. * **Clinical Correlation:** Deficiencies in DNA ligase or proteins involved in lagging strand synthesis (like the FEN1 endonuclease) can lead to genomic instability and are linked to conditions like **Bloom Syndrome** or increased cancer predisposition.

Question 567: What is the estimated total number of genes in the human genome?

A. 12,000
B. 19,000-20,000 (Correct Answer)
C. 24,000-25,000
D. 30,000

Explanation: **Explanation:** The human genome consists of approximately 3.2 billion base pairs. While initial estimates from the Human Genome Project (HGP) predicted over 100,000 genes based on protein diversity, current genomic sequencing and proteomic data have refined this number significantly. **1. Why Option B is Correct:** The current consensus, supported by databases like GENCODE and RefSeq, identifies approximately **19,000 to 20,000 protein-coding genes**. This accounts for only about **1.5% to 2%** of the total genome. The vast complexity of humans compared to simpler organisms (like *C. elegans*, which has a similar gene count) is attributed to **alternative splicing** and post-translational modifications, which allow one gene to produce multiple distinct proteins. **2. Analysis of Incorrect Options:** * **Option A (12,000):** This is an underestimate; even simpler eukaryotes like *Drosophila* have approximately 14,000 genes. * **Option C (24,000-25,000):** This was the widely accepted estimate in the mid-2000s (post-HGP completion) before more stringent criteria for "protein-coding" status were applied. * **Option D (30,000):** This was an early 2001 estimate. As sequencing technology improved, many sequences previously thought to be genes were reclassified as non-coding RNAs or pseudogenes. **High-Yield NEET-PG Pearls:** * **Exome:** The total collection of exons (coding regions); it is the target for **Whole Exome Sequencing (WES)**. * **Non-coding DNA:** Makes up ~98% of the genome, including introns, regulatory sequences, and "junk DNA" (transposons). * **Gene Density:** Chromosome 19 has the highest gene density, while Chromosome 13 and the Y chromosome have the lowest. * **Largest Gene:** Dystrophin (*DMD* gene), spanning 2.4 million base pairs.

Question 568: Which of the following statements is true regarding the genetics of cancer?

A. Topoisomerase II is useful for breaking double-stranded DNA. (Correct Answer)
B. Increased telomerase activity enhances the anti-tumour effect.
C. Maximum synthesis of DNA occurs during the G2 phase.
D. Transition from G2 to M phase is important in the blockage of cancer spread.

Explanation: ### Explanation **1. Why Option A is Correct:** **Topoisomerase II** is a critical enzyme that manages DNA tangles and supercoiling. It functions by creating a transient **double-stranded break** in the DNA helix, allowing another segment of the DNA duplex to pass through before resealing the break. This ATP-dependent process is essential for DNA replication and chromosome segregation. In oncology, this enzyme is the target of several chemotherapeutic agents (e.g., Etoposide, Teniposide, and Anthracyclines like Doxorubicin), which stabilize the DNA-enzyme complex, preventing ligation and leading to apoptosis of cancer cells. **2. Why the Other Options are Incorrect:** * **Option B:** Telomerase is an enzyme that maintains telomere length, granting cells "immortality." **Increased** telomerase activity is a hallmark of cancer cells, allowing them to bypass senescence. Therefore, it **promotes** tumor growth rather than enhancing an anti-tumor effect. * **Option C:** Maximum DNA synthesis occurs during the **S phase** (Synthesis phase) of the cell cycle. The G2 phase is characterized by protein synthesis and preparation for mitosis. * **Option D:** The transition from **G1 to S phase** (regulated by Cyclin D/CDK4 and Rb protein) is the most critical checkpoint in the cell cycle. Loss of control at this "restriction point" is a primary driver of oncogenesis. **3. High-Yield Clinical Pearls for NEET-PG:** * **Topoisomerase I** creates single-stranded breaks (Targeted by Irinotecan/Topotecan). * **Quinolones** (Ciprofloxacin) inhibit bacterial DNA Gyrase (a type of Topoisomerase II). * **Li-Fraumeni Syndrome:** Caused by a mutation in the **p53 gene**, which normally acts at the G1/S checkpoint to allow for DNA repair or trigger apoptosis. * **Telomerase** is a specialized **reverse transcriptase** (RNA-dependent DNA polymerase).

Question 569: What is a chromosome inversion?

A. A portion of the chromosome is missing or deleted.
B. A portion of the chromosome is duplicated, resulting in extragenetic material.
C. A single chromosome undergoes breakage and rearranges within itself. (Correct Answer)
D. A portion of one chromosome has been deleted from its normal place and inserted into another chromosome.

Explanation: **Explanation:** **Chromosome Inversion** is a structural chromosomal aberration where a single chromosome undergoes two breaks, the segment between the breaks flips **180 degrees**, and then reattaches. Because the genetic material is rearranged within the same chromosome without loss or gain of DNA, it is considered a **balanced rearrangement**. 1. **Why Option C is Correct:** Inversion is defined by internal breakage and rearrangement. It is classified into two types based on the involvement of the centromere: * **Paracentric:** Does not include the centromere (breaks are in one arm). * **Pericentric:** Includes the centromere (breaks in both p and q arms), which can change the arm ratio. 2. **Analysis of Incorrect Options:** * **Option A (Deletion):** Refers to the loss of a chromosomal segment (e.g., 5p deletion in Cri-du-chat syndrome). * **Option B (Duplication):** Results in extra genetic material, often leading to developmental abnormalities (e.g., Charcot-Marie-Tooth disease type 1A). * **Option D (Translocation/Insertion):** Describes the movement of a segment between two different chromosomes (interchromosomal), whereas inversion is strictly **intrachromosomal**. **High-Yield Clinical Pearls for NEET-PG:** * **Phenotype:** Most inversion carriers are phenotypically normal because the rearrangement is balanced. * **Reproductive Risk:** Problems arise during meiosis. Inversion carriers are at high risk for **infertility, recurrent spontaneous abortions, or offspring with unbalanced karyotypes** due to the formation of "inversion loops" during crossing over. * **Chromosome 9 Inversion:** The most common inversion seen in the general human population is a pericentric inversion of chromosome 9, usually considered a normal variant.

Question 570: What type of mutation involves the substitution of a purine for a pyrimidine or vice versa?

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

Explanation: ### Explanation **1. Why Transversion is Correct:** Point mutations (single base substitutions) are classified into two categories: transitions and transversions. A **transversion** occurs when a **purine (A, G)** is replaced by a **pyrimidine (C, T)** or vice versa. Structurally, this involves swapping a two-ringed structure for a single-ringed structure (or vice versa), which causes a more significant distortion of the DNA helix than a transition. **2. Why the Other Options are Incorrect:** * **B. Transition:** This is a substitution where a purine is replaced by another purine (A ↔ G) or a pyrimidine is replaced by another pyrimidine (C ↔ T). Transitions are more common in the genome than transversions. * **C. Insertion:** This involves the addition of one or more extra nucleotides into the DNA sequence. If not in multiples of three, it leads to a **frameshift mutation**. * **D. Deletion:** This involves the removal of one or more nucleotides. Like insertions, these are categorized as "indels" and often result in truncated or non-functional proteins. **3. High-Yield Clinical Pearls for NEET-PG:** * **Frequency:** Although there are twice as many possible transversion pathways as transitions, **transitions** occur more frequently in nature due to the spontaneous deamination of 5-methylcytosine to thymine. * **Sickle Cell Anemia:** A classic example of a transversion mutation (GAG → GTG), where Adenine (purine) is replaced by Thymine (pyrimidine) in the β-globin gene, leading to Glutamate being replaced by Valine. * **Silent vs. Missense:** Substitutions can be **silent** (no amino acid change), **missense** (different amino acid), or **nonsense** (creates a premature stop codon: UAA, UAG, UGA).

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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