Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 641: Which type of nucleic acid is detected by the Northern blot technique?

A. DNA
B. RNA (Correct Answer)
C. Protein
D. DNA-protein interaction

Explanation: **Explanation:** The **Northern blot** is a classical molecular biology technique used specifically for the detection and quantification of **RNA** (Option B). It allows researchers to study gene expression by measuring the abundance of specific mRNA sequences in a sample. **Why RNA is the correct answer:** In Northern blotting, RNA fragments are separated by size via gel electrophoresis (usually under denaturing conditions to prevent secondary structures). These fragments are then transferred (blotted) onto a nylon or nitrocellulose membrane and hybridized with a labeled nucleic acid probe (DNA or RNA) complementary to the target sequence. **Analysis of Incorrect Options:** * **Option A (DNA):** DNA is detected using the **Southern blot**. (Mnemonic: **S**outhern = **D**NA; **N**orthern = **R**NA). * **Option C (Protein):** Proteins are detected using the **Western blot**, which utilizes antibodies for identification. * **Option D (DNA-protein interaction):** These interactions are typically studied using **Electrophoretic Mobility Shift Assay (EMSA)** or **ChIP (Chromatin Immunoprecipitation)**. **High-Yield Clinical Pearls for NEET-PG:** * **SNOW DROP Mnemonic:** * **S**outhern — **D**NA * **N**orthern — **R**NA * **O** (ignore) — **O** (ignore) * **W**estern — **P**rotein * **Southwestern Blot:** Used to detect **DNA-binding proteins** (e.g., transcription factors like c-Jun or c-Fos). * **Clinical Application:** While Northern blotting is largely replaced by Real-Time PCR (RT-qPCR) and Microarrays in modern clinics, it remains the "gold standard" for measuring mRNA size and alternative splicing patterns.

Question 642: Which of the following enzymes can polymerize deoxyribonucleotides into DNA?

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

Explanation: **Explanation:** The core concept here is the ability of an enzyme to act as a **DNA polymerase**. A DNA polymerase is defined by its ability to catalyze the formation of phosphodiester bonds between deoxyribonucleotides (dNTPs) to synthesize a DNA strand. **1. Why Reverse Transcriptase is correct:** Reverse transcriptase (found in retroviruses like HIV and in human telomerase) is an **RNA-dependent DNA polymerase**. It uses an RNA template to synthesize a complementary DNA (cDNA) strand. Since it builds a polymer of deoxyribonucleotides, it is functionally a DNA polymerase. **2. Why the other options are incorrect:** * **DNA Ligase:** This enzyme does not polymerize new nucleotides. Instead, it "seals" nicks in the sugar-phosphate backbone by creating a phosphodiester bond between existing DNA fragments (e.g., joining Okazaki fragments). * **DNA Gyrase (Topoisomerase II):** This enzyme relieves torsional strain (supercoiling) during DNA replication by cutting and resealing DNA strands. It has no polymerizing activity. * **RNA Polymerase III:** This enzyme polymerizes **ribonucleotides** (NTPs) to synthesize RNA (specifically tRNA and 5S rRNA), not deoxyribonucleotides. **High-Yield Clinical Pearls for NEET-PG:** * **Telomerase** is a specialized reverse transcriptase (hTRT) that carries its own RNA template to maintain chromosomal ends. * **DNA Polymerase γ (Gamma)** is the only DNA polymerase found in **mitochondria**. * **DNA Polymerase III** (in prokaryotes) has the highest processivity and is the primary enzyme for elongation. * **Zidovudine (AZT)**, used in HIV treatment, works by inhibiting Reverse Transcriptase, thereby halting the polymerization of viral DNA.

Question 643: Which of the following statements about transgenic mice is true?

A. Developed from DNA insertion into a fertilized egg.
B. Have the same genome as parents except for one or more genes.
C. Homozygous strains are selected.
D. All of the above. (Correct Answer)

Explanation: Transgenic mice are essential tools in molecular biology and medical research, used to study gene function and model human diseases. **Explanation of the Correct Answer:** The correct answer is **D (All of the above)** because each statement describes a fundamental step in the production of a transgenic line: * **Option A:** The most common method for creating transgenic mice is **microinjection**. Foreign DNA (the transgene) is injected directly into the male pronucleus of a **fertilized egg (zygote)**. This DNA integrates randomly into the genome, ensuring the transgene is present in every cell of the developing organism. * **Option B:** Transgenic mice are genetically engineered to be **congenic**. They are designed to be genetically identical to the parental strain (isogenic) except for the specific addition of the transgene(s) of interest. * **Option C:** Initial offspring (founders) are usually hemizygous. To ensure stability and consistent expression for experimental research, these mice are crossbred to produce **homozygous strains**, where both alleles carry the transgene. **High-Yield Clinical Pearls for NEET-PG:** * **Transgenic vs. Knockout:** Transgenic mice involve the *addition* of genetic material (gain of function), whereas Knockout mice involve the *deletion* or inactivation of a specific gene (loss of function) using embryonic stem cells. * **Methods of Gene Transfer:** Apart from microinjection, other methods include viral vectors (retroviruses) and the use of **Embryonic Stem (ES) cells**. * **Applications:** They are used to produce "biopharmaceuticals" (e.g., producing human insulin in milk) and to create models for diseases like Alzheimer’s or Oncomice for cancer research. * **Key Enzyme:** Cre-Lox recombination is a high-yield site-specific recombinase technology used to create *conditional* transgenic models.

Question 644: Which molecular biology technique uses an oligomer with a single base pair substitution as a primer?

A. Polymerase Chain Reaction (PCR)
B. Restriction Fragment Length Polymorphism (RFLP)
C. Error coded mutation analysis with PCR
D. Site-directed mutagenesis (Correct Answer)

Explanation: **Explanation:** **Site-directed mutagenesis** is a molecular biology technique used to create specific, targeted changes in a double-stranded DNA sequence. The process involves using a synthetic **oligonucleotide primer** that is complementary to the target DNA but contains a **single base pair substitution** (mismatch) at the desired site. During DNA synthesis (often via PCR-based methods), the DNA polymerase incorporates this mismatch into the new strand. This technique is fundamental in protein engineering to study the effect of specific amino acid changes on protein function. **Analysis of Incorrect Options:** * **A. Polymerase Chain Reaction (PCR):** Standard PCR uses primers that are perfectly complementary to the flanking regions of the target DNA to amplify a specific sequence. It does not inherently require a base pair substitution. * **B. Restriction Fragment Length Polymorphism (RFLP):** This technique relies on variations in DNA sequences that change the recognition sites for restriction enzymes. It involves digestion of DNA with enzymes, not the use of mismatched primers. * **C. Error coded mutation analysis:** While this uses PCR, it focuses on detecting existing mutations rather than intentionally inducing a specific point mutation using a mismatched primer. **Clinical Pearls for NEET-PG:** * **Site-directed mutagenesis** is the gold standard for studying **Structure-Function relationships** of enzymes and receptors. * It is used to create "knock-in" models and to study the pathogenicity of **Single Nucleotide Polymorphisms (SNPs)**. * **High-yield fact:** To perform site-directed mutagenesis, the DNA is often cloned into a plasmid, and a heat-stable DNA polymerase (like *Pfu* polymerase) is used for its high fidelity.

Question 645: Identify the bonds labeled X, Y, and Z in the structure of DNA.

A. X-Hydrogen bond, Y-beta N-glycosidic bond, Z-3' 5' phosphodiester bond (Correct Answer)
B. X-Hydrogen bond, Y-beta O-glycosidic bond, Z-2' 5' phosphodiester bond
C. X-Hydrogen bond, Y-beta N-glycosidic bond, Z-2' 3' phosphodiester bond
D. X-Hydrogen bond, Y-phosphodiester bond, Z-3' 5' beta N-glycosidic bond

Explanation: ### Explanation The structure of DNA is maintained by specific covalent and non-covalent interactions between its three components: a nitrogenous base, a pentose sugar (deoxyribose), and a phosphate group. 1. **X (Hydrogen Bonds):** These are non-covalent interactions between complementary nitrogenous bases (Adenine-Thymine: 2 bonds; Guanine-Cytosine: 3 bonds). They stabilize the double helix while allowing "unzipping" during replication and transcription. 2. **Y (β-N-glycosidic Bond):** This covalent bond connects the **C1'** of the deoxyribose sugar to the **N9** of purines or **N1** of pyrimidines. It is specifically in the **beta (β)** configuration in DNA. 3. **Z (3'–5' Phosphodiester Bond):** This forms the "sugar-phosphate backbone." It connects the **3' hydroxyl (-OH)** group of one nucleotide to the **5' phosphate** group of the adjacent nucleotide. #### Analysis of Incorrect Options: * **Option B:** Incorrect because the bond between sugar and base is **N-glycosidic**, not O-glycosidic (found in carbohydrates). Also, DNA uses 3'–5' linkages, not 2'–5'. * **Option C:** Incorrect because 2'–3' phosphodiester bonds do not exist in the linear backbone of DNA; the 2' position in DNA lacks an oxygen atom (deoxyribose). * **Option D:** Incorrectly swaps the nomenclature of the glycosidic and phosphodiester bonds. #### High-Yield Clinical Pearls for NEET-PG: * **Chargaff’s Rule:** In double-stranded DNA, A+G (Purines) = T+C (Pyrimidines). * **Denaturation:** The "Melting Temperature" (Tm) is higher in DNA with high **G-C content** due to the triple hydrogen bonds. * **Drug Target:** Zidovudine (AZT), an anti-retroviral drug, acts by inhibiting the formation of the **3'–5' phosphodiester bond** because it lacks a 3'-OH group (chain termination). * **Orientation:** DNA strands are **antiparallel**; one runs 5'→3' and the other 3'→5'.

Question 646: Okazaki fragments are joined to form continuous strands of DNA by which of the following enzymes?

A. Helicase
B. Topoisomerase
C. DNA Ligase (Correct Answer)
D. DNA primase

Explanation: **Explanation:** DNA replication is semi-discontinuous. While the leading strand is synthesized continuously, the **lagging strand** is synthesized in short segments known as **Okazaki fragments**. **1. Why DNA Ligase is correct:** DNA Ligase is the "molecular glue." Once DNA Polymerase I (in prokaryotes) or Pol $\delta$ (in eukaryotes) replaces RNA primers with DNA, a "nick" remains in the phosphodiester backbone. **DNA Ligase** catalyzes the formation of a phosphodiester bond between the 3'-OH end of one fragment and the 5'-phosphate end of the next, thereby joining Okazaki fragments into a continuous strand. This process requires energy (ATP in eukaryotes/humans; NAD+ in bacteria). **2. Why other options are incorrect:** * **Helicase:** Responsible for unwinding the DNA double helix at the replication fork by breaking hydrogen bonds. * **Topoisomerase:** Relieves torsional strain (supercoiling) ahead of the replication fork by cutting and resealing DNA strands. * **DNA Primase:** An RNA polymerase that synthesizes short RNA primers, providing the essential 3'-OH group required for DNA polymerase to initiate synthesis. **High-Yield Clinical Pearls for NEET-PG:** * **Directionality:** DNA synthesis always occurs in the **5' to 3'** direction. * **Bloom Syndrome:** Caused by a mutation in the *BLM* gene (encoding **DNA Helicase**), leading to genomic instability and sister chromatid exchanges. * **Fluoroquinolones:** These antibiotics (e.g., Ciprofloxacin) act by inhibiting bacterial **DNA Gyrase** (Topoisomerase II) and Topoisomerase IV. * **Etoposide/Teniposide:** Anticancer drugs that inhibit human **Topoisomerase II**, leading to DNA breaks and apoptosis.

Question 647: Okazaki fragments are formed during the synthesis of which type of nucleic acid?

A. Double-stranded DNA (ds DNA) (Correct Answer)
B. Single-stranded DNA (ss DNA)
C. Messenger RNA (mRNA)
D. Transfer RNA (tRNA)

Explanation: ### Explanation **Correct Answer: A. Double-stranded DNA (ds DNA)** **Why it is correct:** Okazaki fragments are a fundamental feature of **DNA replication**, the process by which a cell duplicates its double-stranded DNA (dsDNA). DNA polymerase can only synthesize new strands in the **5' to 3' direction**. Because the two strands of the DNA helix are **antiparallel**, only one strand (the leading strand) can be synthesized continuously. The other strand (the **lagging strand**) must be synthesized discontinuously in short segments known as **Okazaki fragments**. These fragments are later joined by the enzyme **DNA ligase** to form a continuous double-stranded molecule. **Why the other options are incorrect:** * **B. Single-stranded DNA (ss DNA):** While the template is briefly single-stranded during replication, the *product* of the process involving Okazaki fragments is a double-stranded DNA molecule. * **C & D. mRNA and tRNA:** These are products of **transcription**. Transcription involves RNA polymerase, which synthesizes a single-stranded RNA molecule from a DNA template. This process is continuous and does not involve the formation of Okazaki fragments. **High-Yield Clinical Pearls for NEET-PG:** * **Enzymology:** Okazaki fragments are initiated by **RNA Primers** (synthesized by DNA Primase). In eukaryotes, **DNA Polymerase δ** (delta) is primarily responsible for lagging strand synthesis. * **DNA Ligase:** This enzyme creates the final phosphodiester bond. A deficiency or inhibition of ligase prevents the joining of Okazaki fragments. * **Size Difference:** Okazaki fragments are significantly shorter in eukaryotes (100–200 nucleotides) compared to prokaryotes (1000–2000 nucleotides). * **Clinical Correlation:** Defective DNA ligase I is associated with **Bloom Syndrome**, characterized by genomic instability and increased sister chromatid exchange.

Question 648: Oncogenes can be best studied by which of the following methods?

A. Transfection (Correct Answer)
B. Transduction
C. Transformation
D. Conjugation

Explanation: ### Explanation **Why Transfection is the Correct Answer:** Transfection is the process of deliberately introducing naked or purified nucleic acids (DNA or RNA) into **eukaryotic cells**. In cancer research, this is the gold-standard method for studying oncogenes. By "transfecting" specific DNA sequences into normal mammalian cell lines (like NIH 3T3 cells), researchers can observe if the cells undergo malignant transformation (e.g., loss of contact inhibition, increased proliferation). This technique was instrumental in the discovery of the first human oncogene, *RAS*. **Analysis of Incorrect Options:** * **Transduction:** This involves the transfer of genetic material from one bacterium to another via a **bacteriophage** (virus). While viral vectors are used in gene therapy, "transduction" as a classical genetic term refers primarily to prokaryotic horizontal gene transfer. * **Transformation:** In a classical microbiology context, this is the uptake of naked DNA by **prokaryotic (bacterial) cells**. While the term is also used to describe a normal cell becoming cancerous, as a *method* of genetic study, it refers to bacteria. * **Conjugation:** This is "bacterial mating," where genetic material is transferred between two bacteria through direct cell-to-cell contact (via a sex pilus). It has no application in studying human oncogenes. **High-Yield Clinical Pearls for NEET-PG:** * **NIH 3T3 Assay:** The classic experiment where DNA from human bladder cancer was transfected into mouse fibroblasts, leading to the identification of the *H-RAS* oncogene. * **Transfection Methods:** Can be chemical (calcium phosphate, liposomes) or physical (electroporation). * **Key Distinction:** Remember: **Transformation** (Bacteria), **Transfection** (Eukaryotes), **Transduction** (Viruses).

Question 649: Which nitrogenous base is not found in DNA?

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

Explanation: **Explanation:** The fundamental difference between DNA (Deoxyribonucleic acid) and RNA (Ribonucleic acid) lies in their pentose sugars and their nitrogenous base composition. Nitrogenous bases are categorized into **Purines** (Adenine and Guanine) and **Pyrimidines** (Cytosine, Thymine, and Uracil). **Why Uracil is the Correct Answer:** Uracil is a pyrimidine base found exclusively in **RNA**. In DNA, the corresponding pyrimidine is **Thymine** (5-methyluracil). The presence of Thymine instead of Uracil in DNA is a crucial evolutionary mechanism for genetic stability; it allows DNA repair enzymes to recognize the spontaneous deamination of Cytosine (which turns into Uracil) as an error that needs fixing. **Analysis of Incorrect Options:** * **A. Adenine:** A purine base found in both DNA and RNA. It pairs with Thymine in DNA and Uracil in RNA via two hydrogen bonds. * **B. Guanine:** A purine base found in both DNA and RNA. It pairs with Cytosine via three hydrogen bonds. * **C. Cytosine:** A pyrimidine base found in both DNA and RNA. **High-Yield Clinical Pearls for NEET-PG:** 1. **Chargaff’s Rule:** In double-stranded DNA, the amount of Purines equals Pyrimidines (A+G = T+C). This rule does *not* apply to RNA as it is usually single-stranded. 2. **5-Fluorouracil (5-FU):** A common chemotherapy agent that acts as a pyrimidine analog, inhibiting thymidylate synthase and disrupting DNA synthesis. 3. **Deamination:** Spontaneous deamination of Cytosine yields Uracil, while deamination of 5-methylcytosine yields Thymine (a common cause of "hotspot" mutations). 4. **Bond Strength:** G-C pairs have three hydrogen bonds, making DNA sequences with high G-C content more stable and harder to denature (higher Melting Temperature, Tm).

Question 650: Which of the following can be used as a vector for gene therapy?

A. Viruses
B. Liposomes
C. Plasmids
D. All of these (Correct Answer)

Explanation: **Explanation:** Gene therapy involves the delivery of a functional gene into a patient’s cells to treat or prevent disease. To achieve this, a **vector** (delivery vehicle) is required to transport the genetic material across the cell membrane and into the nucleus. 1. **Viruses (Viral Vectors):** These are the most common and efficient vectors. They exploit the natural ability of viruses to infect cells. Examples include **Retroviruses** (integrate into the host genome), **Adenoviruses** (remain episomal), and **Adeno-associated viruses (AAV)** (preferred for long-term expression with low immunogenicity). 2. **Liposomes (Non-viral):** These are synthetic lipid bilayers that encapsulate DNA. They are safer and less immunogenic than viruses but generally have lower transfection efficiency. 3. **Plasmids (Physicochemical):** Naked DNA or plasmid DNA can be introduced directly into tissues (e.g., via electroporation or "gene guns"). While simple to produce, they are susceptible to degradation by nucleases. **Why "All of these" is correct:** Gene therapy utilizes a diverse toolkit. Since viruses, liposomes, and plasmids are all established methods for transporting therapeutic DNA into target cells, option D is the correct choice. **High-Yield Clinical Pearls for NEET-PG:** * **ADA Deficiency:** The first successful human gene therapy (1990) was for Severe Combined Immunodeficiency (SCID) due to Adenosine Deaminase deficiency. * **Vector of Choice:** **AAV (Adeno-associated virus)** is currently the "gold standard" for many in-vivo therapies because it is non-pathogenic and provides stable expression. * **CRISPR-Cas9:** A revolutionary gene-editing tool (Nobel Prize 2020) that allows for precise "cut and paste" editing of the genome, often delivered via these same vectors.

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

Want unlimited practice?

Get full access to all questions, explanations, and performance tracking.

Start For Free