Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 281: Which enzyme is used for cDNA synthesis?

A. DNA dependent RNA polymerase
B. DNA dependent DNA polymerase
C. RNA dependent RNA polymerase
D. RNA dependent DNA polymerase (Correct Answer)

Explanation: **Explanation:** **1. Why the Correct Answer is Right:** The synthesis of **cDNA (complementary DNA)** involves creating a DNA strand from a mature mRNA template. This process is known as **Reverse Transcription**. The enzyme responsible for this is **RNA-dependent DNA polymerase**, commonly referred to as **Reverse Transcriptase**. It "reads" an RNA sequence and "writes" a complementary DNA sequence. This is a crucial step in molecular biology techniques like RT-PCR and in the life cycle of retroviruses. **2. Analysis of Incorrect Options:** * **A. DNA-dependent RNA polymerase:** This enzyme is responsible for **Transcription** (synthesizing RNA from a DNA template), such as the production of mRNA, tRNA, and rRNA in the nucleus. * **B. DNA-dependent DNA polymerase:** This is the primary enzyme for **DNA Replication** (e.g., DNA Polymerase III or δ/ε). It synthesizes a new DNA strand using an existing DNA template. * **C. RNA-dependent RNA polymerase:** Also known as **RNA Replicase**, this enzyme is used by certain RNA viruses (like Poliovirus or SARS-CoV-2) to replicate their RNA genome directly without a DNA intermediate. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Source:** Reverse transcriptase was originally isolated from retroviruses like **HIV**. * **cDNA Characteristics:** Unlike genomic DNA, cDNA **lacks introns** because it is synthesized from processed mRNA. This makes it essential for cloning eukaryotic genes into prokaryotic vectors. * **Diagnostic Use:** **RT-PCR** (Reverse Transcription Polymerase Chain Reaction) is the gold standard for detecting RNA viruses (e.g., HIV viral load, COVID-19). * **Telomerase:** A specialized human version of RNA-dependent DNA polymerase that maintains chromosomal ends (telomeres) using its own internal RNA template.

Question 282: During DNA replication, Okazaki fragments are observed in relation to which strand?

A. Leading strand
B. Lagging strand (Correct Answer)
C. Both strands
D. Helicase

Explanation: **Explanation:** DNA replication is **semi-discontinuous** because the DNA polymerase enzyme can only synthesize DNA in the **5' to 3' direction**. Since the two strands of the DNA double helix are antiparallel, they must be replicated differently as the replication fork opens. 1. **Why the Lagging Strand is Correct:** The lagging strand has a 3' to 5' orientation relative to the fork's movement. To maintain the mandatory 5' to 3' synthesis, DNA polymerase must work in short bursts moving *away* from the replication fork. These short, discontinuous segments of DNA are called **Okazaki fragments**. They are later joined together by **DNA ligase**. 2. **Why Other Options are Incorrect:** * **Leading Strand:** This strand is oriented 5' to 3' toward the replication fork. Synthesis occurs **continuously** in the same direction as the fork movement, requiring only a single RNA primer. * **Both Strands:** Only the lagging strand is synthesized discontinuously; the leading strand is continuous. * **Helicase:** This is an enzyme responsible for unwinding the DNA double helix by breaking hydrogen bonds; it is not a DNA strand. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **DNA Ligase:** The "molecular glue" that joins Okazaki fragments. It requires **ATP** in eukaryotes and **NAD+** in some bacteria. * **RNA Primers:** Each Okazaki fragment requires its own RNA primer (synthesized by **Primase/Pol α**), whereas the leading strand requires only one. * **Directionality:** Remember: Synthesis is always **5' → 3'**, while the template is read **3' → 5'**. * **Clinical Correlation:** Deficiencies in DNA ligase I can lead to **immunodeficiency and sun sensitivity** (Ligase I deficiency syndrome).

Question 283: Which of the following enzymatic activities is possessed by the ribosome?

A. Peptidyl transferase (Correct Answer)
B. Peptidase
C. Aminoacyl tRNA synthetase
D. GTPase

Explanation: **Explanation:** The ribosome is a complex molecular machine composed of ribosomal RNA (rRNA) and proteins. Its primary enzymatic function is **Peptidyl transferase** activity, which catalyzes the formation of peptide bonds between adjacent amino acids during translation. 1. **Why Peptidyl Transferase is correct:** In eukaryotes, this activity is located in the **28S rRNA** of the large (60S) ribosomal subunit; in prokaryotes, it is in the **23S rRNA** of the 50S subunit. Because the catalytic activity resides in the RNA rather than a protein, the ribosome is classified as a **Ribozyme**. It facilitates the transfer of the growing polypeptide chain from the P-site tRNA to the amino acid on the A-site tRNA. 2. **Why other options are incorrect:** * **Peptidase:** These enzymes break peptide bonds (proteolysis). The ribosome builds them. * **Aminoacyl tRNA synthetase:** These are separate cytoplasmic enzymes that "charge" tRNA by attaching the correct amino acid to its 3' end. This occurs *before* the tRNA reaches the ribosome. * **GTPase:** While translation factors (like EF-Tu and EF-G) possess GTPase activity to provide energy, this is not an intrinsic enzymatic property of the ribosome itself. **Clinical Pearls for NEET-PG:** * **Antibiotic Target:** Several antibiotics inhibit peptidyl transferase. **Chloramphenicol** specifically binds to the 50S subunit and inhibits this enzyme in bacteria. * **Ribozyme Concept:** Remember that the ribosome is the most prominent example of a ribozyme. Another high-yield ribozyme is **RNase P**, involved in tRNA processing. * **Shine-Dalgarno Sequence:** In prokaryotes, the 16S rRNA (small subunit) recognizes the mRNA to initiate translation, while the 23S (large subunit) handles the peptidyl transferase activity.

Question 284: To synthesize insulin on a large scale, the most suitable staining material obtained from the beta cells of the pancreas is:

A. Genomic DNA
B. Total cellular RNA
C. cDNA of insulin
D. mRNA of insulin (Correct Answer)

Explanation: ### **Explanation** The synthesis of human insulin on a large scale (Recombinant DNA Technology) requires the insertion of the human insulin gene into a bacterial expression system (like *E. coli*). **Why mRNA is the correct starting material:** In eukaryotic cells, **Genomic DNA** contains both **exons** (coding regions) and **introns** (non-coding regions). Bacteria lack the post-transcriptional machinery (spliceosomes) required to remove introns. Therefore, if genomic DNA is used, the bacteria will translate the introns, resulting in a non-functional protein. To bypass this, scientists isolate **mature mRNA** from pancreatic beta cells. This mRNA has already undergone splicing and contains only the continuous coding sequence for insulin. This mRNA is then used as a template to create **cDNA** via reverse transcription for cloning. **Analysis of Incorrect Options:** * **A. Genomic DNA:** Contains introns which bacteria cannot process; the resulting protein would be junk. * **B. Total cellular RNA:** This includes rRNA and tRNA, which do not code for the insulin protein. It is too non-specific. * **C. cDNA of insulin:** While cDNA is the final product inserted into the vector, it is not the "material obtained" directly from the beta cells. cDNA is synthesized *in vitro* from the isolated mRNA. **High-Yield Clinical Pearls for NEET-PG:** * **Reverse Transcriptase:** The enzyme used to convert mRNA into cDNA (originally discovered in retroviruses). * **Humulin:** The first recombinant DNA drug approved by the FDA (1982). * **Proinsulin vs. Insulin:** Human insulin consists of two chains (A and B) linked by disulfide bonds. In recombinant production, the C-peptide is removed to form active insulin. * **Expression Vectors:** Must contain a promoter, antibiotic resistance gene, and an origin of replication (ori).

Question 285: Which of the following genetic elements does not undergo recombination during gametogenesis?

A. X chromosome
B. Y chromosome
C. Autosome 21
D. Mitochondrial chromosome (Correct Answer)

Explanation: **Explanation:** The correct answer is **Mitochondrial chromosome (Option D)**. **Why it is correct:** Recombination (crossing over) occurs during **Prophase I of Meiosis**, where homologous chromosomes pair up and exchange genetic material. Mitochondrial DNA (mtDNA) is unique because it is **inherited exclusively from the mother** (maternal inheritance). Since mitochondria are not part of the nuclear meiotic process and do not have a homologous partner to pair with during gametogenesis, they do not undergo meiotic recombination. They replicate via binary fission and are passed directly through the ooplasm to the zygote. **Why the other options are incorrect:** * **Autosome 21 (Option C):** All autosomes (1–22) exist as homologous pairs. During meiosis, they undergo synapsis and crossing over to ensure genetic diversity. * **X and Y chromosomes (Options A & B):** Although they are heterologous, the X and Y chromosomes contain **Pseudoautosomal Regions (PAR)** at their tips. These regions are homologous and undergo obligatory recombination during male meiosis to ensure proper segregation. **High-Yield Clinical Pearls for NEET-PG:** * **Maternal Inheritance:** All offspring of an affected mother will inherit a mitochondrial disease, but an affected father will never pass it on. * **Heteroplasmy:** The presence of a mixture of wild-type and mutant mtDNA within a single cell, which explains the variable clinical severity of mitochondrial diseases (e.g., MELAS, LHON). * **Recombination Hotspots:** These are specific regions on nuclear chromosomes where recombination occurs more frequently, often mediated by the protein **PRDM9**.

Question 286: Restriction endonuclease cleaves:

A. dsDNA (Correct Answer)
B. RNA
C. Histone
D. Protein

Explanation: **Explanation:** **Restriction Endonucleases (REs)**, often referred to as "molecular scissors," are enzymes primarily derived from bacteria. Their physiological role is to protect bacteria from viral (bacteriophage) infections by identifying and cleaving foreign DNA. **Why Option A is Correct:** Restriction endonucleases specifically recognize short, symmetrical sequences (usually 4–8 base pairs long) called **palindromic sequences** on **double-stranded DNA (dsDNA)**. They catalyze the hydrolysis of the phosphodiester bond within both strands of the DNA helix, resulting in either "sticky ends" (overhangs) or "blunt ends." This specificity makes them indispensable tools in Recombinant DNA technology. **Why Other Options are Incorrect:** * **B. RNA:** Enzymes that cleave RNA are called **Ribonucleases (RNases)**. REs are highly specific for the deoxyribose sugar backbone of DNA. * **C & D. Histones/Proteins:** Enzymes that degrade proteins are called **Proteases** or **Peptidases**. Histones are basic proteins around which DNA is wrapped; they are not the substrate for REs. **High-Yield Clinical Pearls for NEET-PG:** * **Nomenclature:** The first letter comes from the Genus, the next two from the Species, and the Roman numeral denotes the order of discovery (e.g., *EcoRI* from *Escherichia coli*). * **Methylation:** Bacteria protect their own DNA from being cleaved by these enzymes through **DNA methylation** (via methyltransferases). * **Applications:** REs are used in **Restriction Fragment Length Polymorphism (RFLP)** for DNA fingerprinting, prenatal diagnosis of sickle cell anemia, and gene cloning. * **Type II REs** are the most commonly used in labs because they cleave DNA within or at specific sites and do not require ATP.

Question 287: Which RNA polymerase transcribes micro RNA?

A. RNA polymerase I
B. RNA polymerase II (Correct Answer)
C. RNA polymerase III
D. DNA polymerase

Explanation: **Explanation:** The transcription of microRNA (miRNA) is primarily mediated by **RNA polymerase II**. Most miRNAs are transcribed from DNA sequences into long primary transcripts called **pri-miRNAs**. These transcripts possess a 5' cap and a 3' poly-A tail, which are characteristic features of RNA polymerase II products (similar to mRNA). While a small subset of miRNAs associated with repetitive elements can be transcribed by RNA polymerase III, the standard consensus for medical examinations is RNA polymerase II. **Analysis of Options:** * **Option A (RNA Pol I):** Located in the nucleolus, it exclusively transcribes the **45S pre-rRNA**, which is processed into 5.8S, 18S, and 28S ribosomal RNA. * **Option B (RNA Pol II):** Correct. It transcribes all protein-coding genes (**mRNA**), most **snRNA**, and **miRNA**. * **Option C (RNA Pol III):** Transcribes small "housekeeping" RNAs, including **tRNA**, **5S rRNA**, and U6 snRNA. * **Option D (DNA Polymerase):** This enzyme is involved in DNA replication and repair, not transcription (RNA synthesis). **High-Yield Clinical Pearls for NEET-PG:** 1. **Amanita phalloides (Death Cap Mushroom):** Contains **α-amanitin**, which potently inhibits RNA Polymerase II, leading to severe hepatotoxicity. 2. **miRNA Function:** They regulate gene expression post-transcriptionally by binding to the 3' UTR of target mRNA, leading to mRNA degradation or translational repression. 3. **Processing Enzymes:** Remember the "MicroRNA Pathway": **Drosha** (nuclear processing) → **Dicer** (cytoplasmic processing) → **RISC complex** (silencing). 4. **Mnemonic for RNA Pol I, II, III:** **R-M-T** (1-rRNA, 2-mRNA/miRNA, 3-tRNA).

Question 288: What is the approximate total number of base pairs in the human haploid set of chromosomes?

A. 3 million
B. 3 billion (Correct Answer)
C. 33 billion
D. 5 million

Explanation: **Explanation:** The human genome is organized into chromosomes within the cell nucleus. The term **haploid genome** refers to a single set of 23 chromosomes (22 autosomes and 1 sex chromosome), typically found in germ cells (sperm and egg). **1. Why Option B is Correct:** According to the Human Genome Project, the human haploid genome consists of approximately **3.2 billion base pairs (3.2 × 10⁹ bp)**. In a standard diploid somatic cell (containing 46 chromosomes), this number doubles to approximately 6.4 billion base pairs. For examination purposes, "3 billion" is the standard high-yield figure used to represent the haploid set. **2. Why the Other Options are Incorrect:** * **Option A (3 million):** This is far too small. For context, the bacterium *E. coli* has a genome size of about 4.6 million base pairs. * **Option C (33 billion):** This is an overestimation by a factor of ten. While some plants and amphibians have genomes of this size (C-value paradox), it does not apply to humans. * **Option D (5 million):** This is roughly the size of a typical bacterial genome, not a complex eukaryotic genome. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Coding vs. Non-coding:** Only about **1.5% to 2%** of the human genome actually codes for proteins (exons). * **Mitochondrial DNA (mtDNA):** Unlike the nuclear genome, mtDNA is circular, double-stranded, and contains only **16,569 base pairs** coding for 37 genes. * **Chargaff’s Rule:** In any double-stranded DNA, the number of Adenine (A) equals Thymine (T), and Guanine (G) equals Cytosine (C). * **Packaging:** This 3 billion bp DNA string is approximately 1 meter long per haploid set, necessitating highly organized packaging into **nucleosomes** (DNA wrapped around histone octamers).

Question 289: Xeroderma pigmentosum is produced as a result of a defect in:

A. DNA polymerase III
B. DNA polymerase I and DNA ligase (Correct Answer)
C. DNA exonuclease
D. DNA ligase

Explanation: **Explanation:** Xeroderma Pigmentosum (XP) is an autosomal recessive genetic disorder characterized by an extreme sensitivity to ultraviolet (UV) radiation. The underlying molecular defect lies in the **Nucleotide Excision Repair (NER)** pathway. **Why Option B is Correct:** When UV light causes the formation of **pyrimidine dimers** (usually thymine dimers), the NER pathway is activated. The process involves: 1. Recognition of damage by specific proteins (XP proteins). 2. Cleavage of the damaged strand by **UV-specific endonucleases** (excision). 3. Filling the resulting gap with new nucleotides by **DNA Polymerase I** (in prokaryotes) or Pol $\delta$/$\epsilon$ (in eukaryotes). 4. Sealing the "nick" to restore continuity by **DNA Ligase**. A deficiency in these repair enzymes, particularly the inability to complete the resynthesis and ligation steps, leads to the accumulation of mutations, resulting in skin cancers. **Analysis of Incorrect Options:** * **Option A (DNA Polymerase III):** This is the primary enzyme for prokaryotic DNA replication, not primarily involved in the repair of UV-induced dimers. * **Option C (DNA Exonuclease):** While exonucleases remove nucleotides from the ends of DNA, the specific defect in XP is initiated by an *endonuclease* (uvrABC complex) and completed by the polymerase-ligase duo. * **Option D (DNA Ligase):** While ligase is involved, the pathology of XP is more broadly associated with the entire multi-enzyme NER complex; Option B is more comprehensive in the context of the repair synthesis phase. **Clinical Pearls for NEET-PG:** * **Key Presentation:** Photosensitivity, "parchment-like" skin, hyperpigmentation, and a 1000-fold increased risk of **Basal Cell Carcinoma** and **Squamous Cell Carcinoma**. * **Enzyme Deficit:** Most commonly associated with **UV-specific endonuclease** deficiency. * **Associated Condition:** Cockayne Syndrome also involves NER defects but presents with "bird-like" facies and dwarfism without increased cancer risk.

Question 290: A number of structural chromosome abnormalities are seen clinically. No loss of genetic material occurs in which of the following?

A. Deletion
B. Insertion
C. Substitution
D. Inversion (Correct Answer)

Explanation: ### Explanation **Correct Answer: D. Inversion** **1. Why Inversion is Correct:** An inversion occurs when a single chromosome undergoes two breaks, the internal segment flips 180 degrees, and then reinserts into the same location. Because the genetic material is simply rearranged rather than lost or gained, it is considered a **balanced structural rearrangement**. In most cases, individuals with inversions are phenotypically normal because the gene dosage remains constant (unless a breakpoint disrupts a functional gene or creates a position effect). **2. Why the Other Options are Incorrect:** * **A. Deletion:** This involves the loss of a segment of DNA. It results in **unbalanced** genetic material (monosomy for that segment), leading to clinical syndromes like Cri-du-chat (5p deletion) or DiGeorge syndrome (22q11.2 deletion). * **B. Insertion:** This occurs when a segment of DNA from one chromosome is integrated into a non-homologous chromosome. While the total genetic material in the individual might remain the same (balanced), the specific chromosome receiving the segment undergoes a gain of material, and the donor chromosome undergoes a loss. * **C. Substitution:** In the context of chromosomal mutations, this usually refers to the replacement of one nucleotide base for another (point mutation). While it doesn't "lose" a segment like a deletion, it is a change in the genetic code. However, in the context of *structural chromosomal abnormalities* (large scale), the term is less standard than "Inversion." **3. High-Yield Clinical Pearls for NEET-PG:** * **Paracentric Inversion:** Does NOT include the centromere. * **Pericentric Inversion:** Includes the centromere (can change the arm ratio/shape of the chromosome). * **Balanced vs. Unbalanced:** Inversions and Translocations (Reciprocal) are generally "balanced" (no loss/gain). Deletions, Duplications, and Isochromosomes are "unbalanced." * **Clinical Risk:** Carriers of balanced inversions are often asymptomatic but have a high risk of producing **unbalanced gametes**, leading to recurrent spontaneous abortions or offspring with congenital anomalies.

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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