Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 651: Which of the following enzymes is part of DNA dependent RNA polymerase?

A. DNA ligase
B. Primase (Correct Answer)
C. DNA polymerase III
D. RNA transcriptase

Explanation: **Explanation:** The correct answer is **Primase**. **Why Primase is the correct answer:** DNA replication cannot be initiated *de novo* because DNA polymerases require a free 3'-OH group to add nucleotides. To overcome this, the enzyme **Primase** (specifically DnaG in prokaryotes) synthesizes a short stretch of RNA (approximately 10 nucleotides) called a **primer**. Since Primase uses a DNA template to synthesize an RNA strand, it is biochemically classified as a **DNA-dependent RNA polymerase**. Once the primer is in place, DNA polymerase III can begin elongation. **Analysis of Incorrect Options:** * **DNA ligase:** This enzyme joins DNA fragments (like Okazaki fragments) by catalyzing the formation of phosphodiester bonds. It does not synthesize RNA. * **DNA polymerase III:** This is a DNA-dependent **DNA** polymerase. It synthesizes DNA by adding deoxyribonucleotides, but it cannot initiate synthesis without a primer. * **RNA transcriptase:** This term is often confused with RNA polymerase used in transcription. While RNA polymerase is also a DNA-dependent RNA polymerase, "Primase" is the specific enzyme integrated into the replisome during DNA replication. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Prokaryotes vs. Eukaryotes:** In eukaryotes, the primase activity is associated with **DNA Polymerase α** (alpha). * **Directionality:** Primase synthesizes the primer in the **5' to 3'** direction. * **Requirement:** The leading strand requires only one primer, whereas the lagging strand requires multiple primers (one for each Okazaki fragment). * **Removal:** In prokaryotes, RNA primers are removed by **DNA Polymerase I** (via its 5' to 3' exonuclease activity), not by Primase.

Question 652: Which of the following is a protein that is rich in basic amino acids and that functions in the packaging of DNA in chromosomes?

A. Histone (Correct Answer)
B. Collagen
C. Fibrinogen
D. Hyaluronic acid binding protein

Explanation: **Explanation:** **Histones** are the correct answer because they are specialized proteins responsible for the first level of DNA organization. DNA is negatively charged due to its phosphate backbone. To package this long molecule into the compact nucleus, it wraps around histones, which are highly **basic proteins** rich in **Arginine and Lysine**. These basic amino acids carry positive charges at physiological pH, allowing for high-affinity ionic bonding with the negatively charged DNA. This DNA-protein complex forms the **nucleosome**, the fundamental repeating unit of chromatin. **Analysis of Incorrect Options:** * **Collagen:** This is the most abundant structural protein in the extracellular matrix (ECM). While it is rich in Glycine, Proline, and Hydroxyproline, it does not bind DNA or function in genomic packaging. * **Fibrinogen:** A high-molecular-weight plasma glycoprotein synthesized in the liver. It is a clotting factor (Factor I) converted to fibrin during the coagulation cascade; it has no role in nuclear organization. * **Hyaluronic acid binding protein:** These are proteins (like CD44) that interact with hyaluronan in the ECM to regulate cell adhesion and signaling, rather than DNA packaging. **High-Yield NEET-PG Pearls:** * **Nucleosome Core:** Consists of an octamer of two molecules each of **H2A, H2B, H3, and H4**. * **Linker Histone:** **H1** is the "linker histone" that binds the entry/exit sites of DNA on the nucleosome to stabilize higher-order chromatin structures. * **Epigenetics:** Histone tails undergo post-translational modifications (Acetylation, Methylation, Phosphorylation). **Acetylation** (by HATs) neutralizes the positive charge, relaxing chromatin (Euchromatin) and increasing transcription.

Question 653: The APO-B100 and APO-B48 lipoproteins differ due to which molecular mechanism?

A. DNA duplication
B. RNA splicing
C. RNA editing (Correct Answer)
D. Misreading of mRNA

Explanation: **Explanation:** The correct answer is **RNA editing**. This is a post-transcriptional process where the nucleotide sequence of an mRNA molecule is altered after it has been transcribed from the DNA. **Why RNA editing is correct:** Both APO-B100 and APO-B48 are encoded by the same gene (*APOB*). In the **liver**, the gene is transcribed normally to produce **APO-B100** (full-length protein). However, in the **small intestine**, an enzyme called **Cytidine Deaminase** acts on the mRNA. It converts a specific Cytosine (C) to Uracil (U) at codon 2153. This changes the codon **CAA** (which codes for Glutamine) into **UAA** (a **stop codon**). This results in premature translation termination, producing a truncated protein that is 48% of the original length—hence, **APO-B48**. **Why other options are incorrect:** * **DNA duplication:** This involves copying the genetic material; both proteins arise from a single copy of the gene. * **RNA splicing:** This involves removing introns and joining exons. While it creates protein diversity (isoforms), it is not the mechanism for the B100/B48 distinction. * **Misreading of mRNA:** This refers to translation errors (e.g., due to antibiotics or mutations), whereas RNA editing is a deliberate, site-specific physiological process. **High-Yield Clinical Pearls for NEET-PG:** * **APO-B100:** Found in VLDL, IDL, and LDL. It serves as the ligand for the LDL receptor. * **APO-B48:** Found exclusively in Chylomicrons and Chylomicron remnants. It lacks the LDL-receptor binding domain. * **Mnemonic:** **L**iver = **L**ong (B100); **S**mall Intestine = **S**hort (B48). * **Enzyme:** Remember **Cytidine Deaminase** for the C $\rightarrow$ U conversion.

Question 654: If the molar percentage of guanine in a human DNA is 30%, what is the molar percentage of adenine in that molecule?

A. 10%
B. 20% (Correct Answer)
C. 30%
D. 40%

Explanation: ### Explanation The correct answer is **20%**. This question is based on **Chargaff’s Rules**, which are fundamental to understanding the structure of double-stranded DNA (dsDNA). #### 1. Why the Correct Answer is Right According to Chargaff’s Rule of Base Pairing: * The amount of Purines equals the amount of Pyrimidines ($A+G = T+C$). * Specifically, **Adenine (A) always pairs with Thymine (T)**, and **Guanine (G) always pairs with Cytosine (C)**. Therefore, $\%A = \%T$ and $\%G = \%C$. **Step-by-step Calculation:** 1. If **G = 30%**, then its partner **C must also be 30%**. 2. Total G + C = $30\% + 30\% = 60\%$. 3. The remaining percentage for A + T is $100\% - 60\% = 40\%$. 4. Since A and T must be equal, **A = 40% / 2 = 20%** (and T = 20%). #### 2. Why Incorrect Options are Wrong * **Option A (10%):** This would imply G+C equals 80%, which contradicts the given data. * **Option C (30%):** This assumes A = G, which is not a rule. A only equals G if the GC content is exactly 50%. * **Option D (40%):** This represents the total sum of A+T, not the individual percentage of Adenine. #### 3. Clinical Pearls & High-Yield Facts for NEET-PG * **Applicability:** Chargaff’s rules apply **only to double-stranded DNA**. They do not apply to single-stranded DNA (ssDNA) or RNA (e.g., Parvovirus B19 or HIV genome). * **Bonding:** A-T pairs are held by **2 hydrogen bonds**, while G-C pairs are held by **3 hydrogen bonds**. * **Melting Temperature ($T_m$):** DNA with higher G-C content has a higher $T_m$ because triple hydrogen bonds require more energy to denature. * **Human Genome:** The human genome is generally "AT-rich" (~59-60% A+T and ~40-41% G+C).

Question 655: Which of the following is not part of the genetic code?

A. Adenine
B. Guanine
C. Thymine (Correct Answer)
D. Cytosine

Explanation: **Explanation:** The **genetic code** is the set of rules by which information encoded in genetic material (DNA or RNA sequences) is translated into proteins by living cells. Specifically, the genetic code refers to the **codons** found on **Messenger RNA (mRNA)**. **Why Thymine is the correct answer:** The genetic code is "read" from mRNA during the process of translation. While DNA contains the nitrogenous bases Adenine (A), Guanine (G), Cytosine (C), and **Thymine (T)**, the process of transcription replaces Thymine with **Uracil (U)** in RNA. Therefore, the triplets (codons) that constitute the genetic code consist of A, G, C, and U. Thymine is never a component of a codon in the genetic code. **Analysis of incorrect options:** * **Adenine (A), Guanine (G), and Cytosine (C):** These three nitrogenous bases are common to both DNA and RNA. They are essential components of the 64 possible mRNA codons (e.g., AUG for Methionine, GGG for Glycine). Since they are part of the mRNA sequence used during translation, they are considered part of the genetic code. **High-Yield Clinical Pearls for NEET-PG:** * **The Start Codon:** **AUG** (codes for Methionine in eukaryotes and N-formylmethionine in prokaryotes). * **Stop Codons (Nonsense Codons):** **UAA** (Ochre), **UAG** (Amber), and **UGA** (Opal). These do not code for any amino acid. * **Degeneracy/Redundancy:** Most amino acids are coded by more than one codon (except Methionine and Tryptophan). * **Universality:** The genetic code is the same across almost all organisms, with minor exceptions in mitochondria (e.g., UGA codes for Tryptophan in mitochondria instead of a Stop signal).

Question 656: What is the complementary sequence to 5' GCACC 3'?

A. 5' CCACG 3'
B. 5' CGTGG 3'
C. 3' GGTGC 5'
D. 5' GGTGC 3' (Correct Answer)

Explanation: ### Explanation **1. Understanding the Concept** To find the complementary sequence of DNA, two fundamental rules of molecular biology must be applied: * **Base Pairing Rule:** Adenine (A) pairs with Thymine (T), and Guanine (G) pairs with Cytosine (C). * **Antiparallel Orientation:** DNA strands run in opposite directions. If the template is **5' to 3'**, the complement must be **3' to 5'**. **Step-by-step derivation:** 1. **Template:** 5' G-C-A-C-C 3' 2. **Complement (Antiparallel):** 3' C-G-T-G-G 5' 3. **Standard Notation:** By convention, DNA sequences are always written in the **5' to 3' direction** unless specified otherwise. Flipping 3' CGTGG 5' gives us **5' GGTGC 3'**. **2. Analysis of Options** * **Option D (Correct):** Correct base pairing and correctly reversed to the 5' → 3' orientation. * **Option A:** Incorrect; it simply reverses the original sequence without applying base-pairing rules. * **Option B:** Incorrect; while it applies base pairing, it maintains the same polarity (5' to 3'), violating the antiparallel rule. * **Option C:** Incorrect; it uses the correct bases but labels the ends incorrectly (it is actually the 5' to 3' sequence). **3. NEET-PG High-Yield Clinical Pearls** * **Chargaff’s Rule:** In any double-stranded DNA, %A = %T and %G = %C. Therefore, Purines (A+G) = Pyrimidines (T+C). * **Bond Strength:** G-C pairs have **three hydrogen bonds**, while A-T pairs have **two**. Sequences with high G-C content have a higher melting temperature (Tm). * **Clinical Relevance:** Understanding complementarity is essential for **PCR (Polymerase Chain Reaction)**, where primers must be complementary to the 3' end of the target DNA strand to initiate synthesis.

Question 657: Arrange the cyclins and CDKs in the cell cycle from G1 to S phase progression.

A. CDK6/Cyclin E (Correct Answer)
B. CDK4/Cyclin D
C. CDK1/Cyclin B
D. CDK2/Cyclin A

Explanation: The cell cycle is a highly regulated process governed by the sequential activation of **Cyclin-Dependent Kinases (CDKs)** and their regulatory subunits, **Cyclins**. ### **Explanation of the Correct Answer** The progression from **G1 to S phase** (the G1/S checkpoint) is primarily mediated by the **Cyclin E/CDK2** complex and the **Cyclin D/CDK4/6** complex. * **Early G1:** Cyclin D binds to CDK4 or CDK6. This complex phosphorylates the Retinoblastoma (Rb) protein, releasing E2F transcription factors. * **Late G1 to S Transition:** E2F induces the expression of **Cyclin E**, which binds to **CDK2** (and sometimes CDK6 in specific contexts). This complex completes the hyperphosphorylation of Rb, committing the cell to DNA replication (S phase). *Note: While CDK2/Cyclin E is the classic textbook driver for the G1/S transition, in many competitive exams, CDK4/6 and CDK2 are grouped as the G1-phase regulators.* ### **Analysis of Incorrect Options** * **B. CDK4/Cyclin D:** These act in **Early G1 phase**. They are the first to respond to growth factors but precede the actual transition into the S phase. * **C. CDK1/Cyclin B:** This complex (also known as Mitosis-Promoting Factor or MPF) regulates the **G2 to M phase** transition. * **D. CDK2/Cyclin A:** This complex is primarily responsible for the **progression through the S phase** and the transition into G2. ### **High-Yield Clinical Pearls for NEET-PG** 1. **The "Restriction Point":** The point in late G1 where the cell becomes independent of external growth factors; it is controlled by the Rb protein. 2. **CKIs (CDK Inhibitors):** **p21** (induced by p53) inhibits all CDKs, while the **INK4 family (p16)** specifically inhibits CDK4/6. 3. **Pharmacology Link:** **Palbociclib** is a CDK4/6 inhibitor used in the treatment of HR-positive breast cancer to arrest the cell cycle in G1. 4. **Mnemonic:** **D**ogs **E**at **A**pples **B**efore (Cyclins D $\rightarrow$ E $\rightarrow$ A $\rightarrow$ B) / **4-2-2-1** (CDKs 4 $\rightarrow$ 2 $\rightarrow$ 2 $\rightarrow$ 1).

Question 658: Prenatal diagnosis of Hemophilia is best done by?

A. PCR (Correct Answer)
B. Linkage analysis
C. Cytometry
D. Microarray

Explanation: **Explanation:** **Hemophilia A and B** are X-linked recessive disorders caused by mutations in the Factor VIII and Factor IX genes, respectively. **Why PCR is the Correct Answer:** Polymerase Chain Reaction (PCR) is the gold standard for prenatal diagnosis because it allows for the **rapid amplification of fetal DNA** obtained via Chorionic Villus Sampling (CVS) or amniocentesis. Once amplified, specific techniques like **Reverse Transcriptase-PCR (RT-PCR)** or **Direct Mutation Analysis** can identify the exact genetic defect (e.g., the common Intron 22 inversion in Hemophilia A). It is preferred due to its high sensitivity, specificity, and the small amount of fetal sample required. **Analysis of Other Options:** * **Linkage Analysis:** This was historically used when the specific mutation was unknown. It tracks the disease gene using polymorphic markers (RFLPs). However, it requires DNA from multiple family members and is prone to errors due to genetic recombination. * **Cytometry (Flow Cytometry):** This is used for analyzing physical and chemical characteristics of cells (e.g., immunophenotyping in leukemia). It cannot detect single-gene mutations required for hemophilia diagnosis. * **Microarray:** While useful for detecting chromosomal imbalances (Copy Number Variations), it is not the primary tool for the point mutations or inversions typically seen in Hemophilia. **Clinical Pearls for NEET-PG:** * **Most common mutation in severe Hemophilia A:** Intron 22 inversion. * **Gold standard for carrier detection:** DNA analysis (PCR-based). * **Prenatal sampling timing:** CVS is usually done at 10–12 weeks; Amniocentesis at 15–18 weeks. * **Factor levels:** Prenatal diagnosis via fetal blood sampling (cordocentesis) to check factor levels is rarely done now due to the high risk of fetal loss compared to PCR.

Question 659: Where does miRNA typically bind to facilitate gene knockdown?

A. 3' untranslated region (Correct Answer)
B. 5' untranslated region
C. Introns
D. Both 3' and 5' untranslated regions

Explanation: ### Explanation **1. Why Option A is Correct:** MicroRNAs (miRNAs) are small, non-coding RNA molecules (typically 21–25 nucleotides) that play a crucial role in post-transcriptional gene regulation. To facilitate gene knockdown, the miRNA incorporates into the **RNA-induced silencing complex (RISC)**. The "seed sequence" (nucleotides 2–7) of the miRNA then undergoes complementary base pairing primarily with the **3' Untranslated Region (3' UTR)** of the target messenger RNA (mRNA). This binding leads to either **translational repression** (if binding is partially complementary) or **mRNA degradation** (if binding is perfectly complementary), effectively silencing the gene. **2. Why Other Options are Incorrect:** * **Option B (5' UTR):** While some rare interactions occur here, the 5' UTR is primarily involved in the initiation of translation (ribosome loading). miRNA binding here is not the standard mechanism for gene knockdown. * **Option C (Introns):** Introns are non-coding sequences removed during splicing in the nucleus. miRNA-mediated silencing occurs in the **cytoplasm** on mature, processed mRNA. * **Option D:** Although experimental evidence shows occasional binding at the 5' UTR, the **canonical and most efficient site** for miRNA-mediated regulation in humans is the 3' UTR. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Origin:** miRNAs are transcribed by **RNA Polymerase II** as primary-miRNA (pri-miRNA). * **Processing Enzymes:** **Drosha** (in the nucleus) and **Dicer** (in the cytoplasm) are the key endonucleases that process miRNA. * **OncomiRs:** miRNAs that regulate oncogenes or tumor suppressor genes. For example, a decrease in **miRNA-15 and miRNA-16** is associated with Chronic Lymphocytic Leukemia (CLL). * **Therapeutic Potential:** miRNA mimics and antagomirs (inhibitors) are being researched as targeted therapies for cancer and viral infections (e.g., Hepatitis C).

Question 660: What cellular component carries the genetic information for hereditary diseases?

A. DNA (Correct Answer)
B. Ribosome
C. RNA
D. Membrane

Explanation: **Explanation:** The correct answer is **DNA (Deoxyribonucleic Acid)**. DNA serves as the primary repository of genetic information in almost all living organisms. It is organized into genes, which are the functional units of heredity. Mutations or alterations in the nucleotide sequence of DNA lead to the synthesis of defective proteins or regulatory errors, which manifest as **hereditary diseases** (e.g., Sickle Cell Anemia, Cystic Fibrosis). **Analysis of Options:** * **DNA (Correct):** It contains the "blueprint" of life. In eukaryotes, it is primarily located in the nucleus (genomic DNA) and mitochondria (mtDNA). * **Ribosome:** These are the cellular "protein factories." While they are essential for translating genetic code into proteins, they do not store or transmit hereditary information. * **RNA:** While RNA carries genetic information in some viruses (e.g., Retroviruses like HIV), in human cells, it primarily acts as an intermediary (mRNA, tRNA, rRNA) to convert DNA instructions into proteins. It is not the primary carrier of human hereditary traits. * **Membrane:** Cell membranes provide structural integrity and regulate transport but play no role in storing genetic information. **NEET-PG High-Yield Pearls:** * **Mitochondrial DNA (mtDNA):** Inherited exclusively from the **mother** (Maternal Inheritance). Mutations here cause diseases like Leber’s Hereditary Optic Neuropathy (LHON). * **Central Dogma:** The flow of genetic information is DNA → RNA → Protein. * **Nucleosome:** The basic unit of DNA packaging, consisting of DNA wrapped around **Histone octamers** (H2A, H2B, H3, H4). H1 is the linker histone.

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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