Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 121: Paternal disomy is found in which condition?

A. Prader-Willi syndrome
B. Angelman syndrome (Correct Answer)
C. Fragile X syndrome
D. Hydatiform mole

Explanation: ### Explanation The correct answer is **Angelman syndrome**. This question tests the concept of **Genomic Imprinting** and **Uniparental Disomy (UPD)**. #### 1. Why Angelman Syndrome is Correct Angelman syndrome is caused by the loss of the maternal expression of the **UBE3A gene** on chromosome 15 (15q11-q13). In a normal individual, the paternal copy of this gene is silenced (imprinted). Disease occurs when the maternal contribution is lost via: * Maternal deletion (70%) * **Paternal Uniparental Disomy (UPD):** The child inherits two copies of chromosome 15 from the father and none from the mother. Since both paternal copies are silenced, no functional UBE3A protein is produced. #### 2. Analysis of Incorrect Options * **A. Prader-Willi Syndrome:** This is the "opposite" of Angelman. It results from the loss of the *paternal* 15q11-q13 region. It is associated with **Maternal Disomy** (two copies from the mother, both of which are silenced). * **C. Fragile X Syndrome:** This is a **Trinucleotide Repeat Disorder (CGG)** involving the FMR1 gene on the X chromosome. It does not involve uniparental disomy. * **D. Hydatidiform Mole:** A **complete mole** involves paternal disomy of the *entire genome* (46,XX or 46,XY derived entirely from sperm), but in the context of specific clinical syndromes involving single chromosome pairs, Angelman is the classic association for paternal UPD. #### 3. High-Yield Clinical Pearls for NEET-PG * **Angelman Syndrome (Happy Puppet):** Characterized by inappropriate laughter, seizures, ataxia, and severe intellectual disability. * **Prader-Willi Syndrome:** Characterized by hyperphagia (obesity), hypogonadism, and hypotonia. * **Mnemonic:** **P**rader-Willi = **P**aternal deletion / **M**aternal disomy. **A**ngelman = **M**aternal deletion / **P**aternal disomy. (Remember: **P**ader-Willi has **P**aternal loss).

Question 122: What is the most important enzyme in DNA replication for chain elongation?

A. Helicase
B. DNA polymerase I
C. DNA polymerase III (Correct Answer)
D. Topoisomerase III

Explanation: **Explanation:** In prokaryotic DNA replication, **DNA Polymerase III** is the primary enzyme responsible for the synthesis of both the leading and lagging strands. It is highly processive, meaning it can add thousands of nucleotides without dissociating from the DNA template. Its high catalytic rate makes it the "workhorse" of **chain elongation**. It possesses 5' to 3' polymerase activity and 3' to 5' exonuclease activity (proofreading). **Analysis of Incorrect Options:** * **Helicase:** This enzyme is responsible for unwinding the DNA double helix at the replication fork by breaking hydrogen bonds. It initiates the process but does not elongate the chain. * **DNA Polymerase I:** While it has polymerase activity, its primary roles are removing RNA primers (via 5' to 3' exonuclease activity) and filling the resulting gaps. It is more involved in "clean-up" and repair than bulk elongation. * **Topoisomerase III:** Topoisomerases (like DNA Gyrase/Topoisomerase II) function to relieve torsional strain and supercoiling ahead of the replication fork. They do not synthesize DNA chains. **High-Yield Clinical Pearls for NEET-PG:** * **Eukaryotic Counterpart:** In humans, **DNA Polymerase δ (delta)** and **ε (epsilon)** perform the elongation roles equivalent to prokaryotic Pol III. * **Processivity:** The **Beta-clamp** (sliding clamp) is the subunit of Pol III that ensures it stays attached to the DNA, enabling high processivity. * **Drug Target:** Fluoroquinolones (e.g., Ciprofloxacin) target bacterial **DNA Gyrase** (Topoisomerase II) and Topoisomerase IV, preventing replication. * **Proofreading:** The 3' to 5' exonuclease activity is essential for maintaining high fidelity; defects in similar eukaryotic repair mechanisms lead to conditions like **HNPCC (Lynch Syndrome)**.

Question 123: Bromodeoxyuridine is structurally related to which of the following DNA components?

A. Uracil
B. Adenosine
C. Cytosine
D. Thymidine (Correct Answer)

Explanation: **Explanation:** **1. Why Thymidine is Correct:** Bromodeoxyuridine (BrdU) is a synthetic nucleoside that acts as a **structural analogue of Thymidine**. In its chemical structure, the methyl group at the 5th position of the uracil ring in thymidine is replaced by a **bromine atom**. Because of this structural similarity, DNA polymerase cannot distinguish between the two, and BrdU is incorporated into the DNA during the S-phase of the cell cycle in place of thymidine. This property makes it a vital tool in research to detect proliferating cells. **2. Why Other Options are Incorrect:** * **Uracil:** While BrdU is a derivative of uracil (5-bromouracil), it is specifically a **deoxyribonucleoside** (containing deoxyribose sugar). Uracil is a nitrogenous base found in RNA, not a DNA component. * **Adenosine:** This is a purine nucleoside. BrdU is a pyrimidine analogue; their ring structures (double vs. single ring) are entirely different. * **Cytosine:** Although cytosine is a pyrimidine, it lacks the specific substitution site at the 5th position that mimics the thymidine-adenine base pairing. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **BrdU Staining:** Used to measure **cell proliferation rates**. Incorporated BrdU is detected using fluorescently labeled monoclonal antibodies. * **Mechanism of Mutation:** 5-Bromouracil (the base in BrdU) can undergo a **tautomeric shift** from a keto form to an enol form. The enol form base-pairs with Guanine instead of Adenine, leading to **transition mutations** (T-A to C-G). * **Zidovudine (AZT):** Another high-yield thymidine analogue used in HIV treatment; it acts as a chain terminator during reverse transcription.

Question 124: Sickle cell anemia is the clinical manifestation of homozygous genes for an abnormal haemoglobin molecule. What event is responsible for the mutation in the b chain?

A. Insertion
B. Deletion
C. Non-disjunction
D. Point mutation (Correct Answer)

Explanation: **Explanation:** Sickle cell anemia is a classic example of a **Point Mutation**, specifically a **missense mutation** resulting from a single nucleotide substitution. 1. **Why Point Mutation is correct:** The molecular basis of Sickle Cell Anemia involves a single base change in the DNA sequence of the **$\beta$-globin gene** located on chromosome 11. Specifically, there is a transversion where **Adenine is replaced by Thymine (GAG $\rightarrow$ GTG)** at the 6th codon. This results in the substitution of the amino acid **Glutamic acid** (polar/hydrophilic) with **Valine** (non-polar/hydrophobic) at the 6th position of the $\beta$-polypeptide chain. This single change causes the hemoglobin (HbS) to polymerize under deoxygenated conditions, leading to the characteristic "sickling" of RBCs. 2. **Why other options are incorrect:** * **Insertion/Deletion:** These involve the addition or loss of nucleotides, which usually cause a **frameshift mutation**, altering the entire downstream reading frame. This is seen in certain types of Thalassemia, not Sickle Cell Anemia. * **Non-disjunction:** This is a failure of homologous chromosomes or sister chromatids to separate during cell division, leading to **aneuploidy** (e.g., Trisomy 21/Down Syndrome), not single-gene point mutations. **High-Yield Clinical Pearls for NEET-PG:** * **Inheritance:** Autosomal Recessive. * **Electrophoresis:** HbS moves **slower** than HbA toward the anode (+) because the loss of Glutamic acid reduces the negative charge of the molecule. * **Protective Effect:** Heterozygotes (Sickle cell trait) show resistance to *Plasmodium falciparum* malaria. * **Precipitating factors for sickling:** Hypoxia, acidosis, dehydration, and increased 2,3-BPG.

Question 125: In SCID, what type of DNA repair is defective?

A. Double strand break repair
B. Nucleotide excision repair
C. End Joining Repair (Correct Answer)
D. Mismatch repair

Explanation: **Explanation:** The correct answer is **End Joining Repair**, specifically **Non-Homologous End Joining (NHEJ)**. Severe Combined Immunodeficiency (SCID) can be caused by mutations in genes involved in **V(D)J recombination**, such as *RAG1/RAG2* or components of the NHEJ pathway (e.g., *Artemis*). During the development of B and T cells, the body intentionally creates double-strand breaks in DNA to rearrange antigen receptor genes. These breaks must be repaired by the NHEJ machinery. If this repair mechanism is defective, lymphocytes cannot mature, leading to a total lack of adaptive immunity (SCID). **Analysis of Options:** * **Option A (Double strand break repair):** While NHEJ is a *form* of double-strand break repair, "End Joining Repair" is the more specific and accurate term for the defect in SCID. General double-strand break repair via homologous recombination is defective in conditions like **Ataxia-telangiectasia** (ATM gene) or **BRCA1/2** mutations. * **Option B (Nucleotide excision repair):** This pathway repairs bulky lesions and pyrimidine dimers caused by UV light. Defects lead to **Xeroderma Pigmentosum**. * **Option D (Mismatch repair):** This pathway corrects errors during DNA replication. Defects lead to **Lynch Syndrome (HNPCC)**. **High-Yield Clinical Pearls for NEET-PG:** * **NHEJ** does not require a homologous template and is "error-prone." * **Adenosine Deaminase (ADA) deficiency** is the second most common cause of SCID (autosomal recessive), leading to the accumulation of dATP, which is toxic to lymphocytes. * **X-linked SCID** is the most common form, caused by a defect in the **IL-2 receptor gamma chain**. * **Radiosensitivity:** Patients with NHEJ-defective SCID are hypersensitive to ionizing radiation.

Question 126: Which of the following statements is NOT true about eukaryotic genes?

A. Polycistronic mRNA (Correct Answer)
B. Noncoding intron
C. Contain nuclear gene and pseudogene
D. Modification of mRNA before transportation from the nucleus

Explanation: **Explanation:** In eukaryotic genetics, the organization and expression of genes differ significantly from prokaryotes. The correct answer is **A (Polycistronic mRNA)** because this is a characteristic feature of **prokaryotes**, not eukaryotes. **1. Why Polycistronic mRNA is NOT true for Eukaryotes:** Eukaryotic genes are typically **monocistronic**, meaning one mRNA molecule carries the genetic information to encode only a **single polypeptide**. In contrast, prokaryotic mRNA is often polycistronic, where one mRNA strand contains multiple open reading frames (ORFs) that code for several different proteins (e.g., the Lac Operon). **2. Analysis of Incorrect Options:** * **B. Noncoding introns:** Eukaryotic genes are "split genes." They contain **exons** (coding regions) and **introns** (non-coding regions). Introns are removed during splicing. * **C. Nuclear genes and pseudogenes:** Eukaryotes contain functional nuclear genes and **pseudogenes** (sequences that resemble functional genes but have lost their protein-coding ability due to mutations). * **D. Modification of mRNA:** Eukaryotic pre-mRNA undergoes extensive **post-transcriptional modification** (5' capping, 3' polyadenylation, and splicing) within the nucleus before being exported to the cytoplasm for translation. **High-Yield Clinical Pearls for NEET-PG:** * **Splicing Errors:** Mutations at splice sites are responsible for diseases like **β-Thalassemia**. * **Cap & Tail:** The 7-methylguanosine cap (5') and Poly-A tail (3') protect mRNA from exonucleases and are essential for translation initiation. * **Exception:** While most eukaryotic mRNA is monocistronic, some viruses (like Poliovirus) produce a single polyprotein that is later cleaved, mimicking a polycistronic-like effect.

Question 127: Sigma (σ) subunit is present in which of the following?

A. 80S ribosomes
B. RNA polymerase (Correct Answer)
C. Initiation of DNA replication
D. DNA polymerase

Explanation: **Explanation:** The **Sigma (σ) subunit** is a critical component of the **Prokaryotic RNA polymerase holoenzyme**. In bacteria, the core enzyme (consisting of subunits α₂ββ'ω) can synthesize RNA but lacks the ability to recognize specific promoter sequences. The addition of the sigma factor converts the core enzyme into the **holoenzyme**, which specifically recognizes and binds to the promoter region (Pribnow box/-10 and -35 sequences), thereby initiating transcription. Once the RNA chain reaches about 10 nucleotides in length, the sigma factor dissociates, and the core enzyme continues elongation. **Analysis of Options:** * **80S Ribosomes (Option A):** These are eukaryotic ribosomal complexes involved in translation. They consist of 40S and 60S subunits, not sigma factors. * **Initiation of DNA replication (Option C):** This process involves proteins like DnaA (recognition), Helicase (unwinding), and Primase, but not the sigma subunit. * **DNA Polymerase (Option D):** This enzyme is responsible for DNA synthesis. While it requires a primer and various accessory proteins (like the sliding clamp), it does not utilize a sigma subunit for initiation. **High-Yield Clinical Pearls for NEET-PG:** * **Rifampicin:** This key antitubercular drug acts by inhibiting the **β-subunit** of bacterial DNA-dependent RNA polymerase, preventing transcription initiation. * **Promoter Recognition:** In eukaryotes, the function of the sigma factor is analogous to **General Transcription Factors (GTFs)** like TFIID. * **Alpha-amanitin:** Found in *Amanita phalloides* mushrooms, it specifically inhibits **RNA Polymerase II**, leading to severe hepatotoxicity.

Question 128: What constitutes the driving force for the transport of proteins into and out of the nucleus?

A. ATP hydrolysis within the cytosol
B. ATP hydrolysis within the nucleus
C. GTP hydrolysis within the cytosol (Correct Answer)
D. GTP hydrolysis within the nucleus

Explanation: ### Explanation The transport of proteins across the nuclear envelope (through Nuclear Pore Complexes) is an active process regulated by the **Ran-GTPase cycle**. This cycle provides the directionality and energy required for both nuclear import and export. **1. Why Option C is Correct:** The driving force for nuclear transport is the **hydrolysis of GTP to GDP**, which occurs specifically in the **cytosol**. * **The Mechanism:** The enzyme **Ran-GAP** (GTPase Activating Protein) is localized in the cytosol. It triggers Ran-GTP to hydrolyze its bound GTP into GDP. * **The Result:** This hydrolysis maintains a high concentration of Ran-GDP in the cytosol and Ran-GTP in the nucleus. This steep **Ran-GTP gradient** is what powers the release of cargo into the nucleus (import) or the movement of cargo out of the nucleus (export). **2. Why Other Options are Incorrect:** * **Options A & B (ATP):** While many cellular transport mechanisms use ATP, nuclear transport specifically utilizes the energy from **GTP**. * **Option D (GTP hydrolysis in the nucleus):** The nucleus contains **Ran-GEF** (Guanine Nucleotide Exchange Factor), which replaces GDP with GTP. Therefore, the nucleus is a site of **GTP loading**, not hydrolysis. Hydrolysis in the nucleus would destroy the gradient necessary for transport. **3. High-Yield NEET-PG Pearls:** * **Ran-GTP Gradient:** High in the Nucleus; Low in the Cytosol. * **Importins:** Bind cargo in the cytosol (where Ran-GTP is low) and release it in the nucleus (where Ran-GTP is high). * **Exportins:** Bind cargo in the nucleus only when Ran-GTP is present, forming a ternary complex. * **NLS & NES:** Proteins destined for the nucleus have a **Nuclear Localization Signal** (rich in basic amino acids like Lysine and Arginine), while those leaving have a **Nuclear Export Signal**.

Question 129: What is the primary function of microRNA?

A. Gene Regulation (Correct Answer)
B. Splicing of pre mRNA
C. Initiation of Translation
D. Carrying messages for protein synthesis

Explanation: **Explanation:** **MicroRNAs (miRNAs)** are small, non-coding RNA molecules (typically 21–25 nucleotides long) that play a pivotal role in **post-transcriptional gene regulation**. They function by binding to the 3' untranslated region (3' UTR) of specific target messenger RNAs (mRNAs). This binding usually leads to either **translational repression** or **mRNA degradation**, effectively "silencing" the gene expression. **Analysis of Options:** * **A (Correct):** miRNAs regulate approximately 30–60% of human genes, acting as rheostats to fine-tune protein levels within the cell. * **B (Incorrect):** Splicing of pre-mRNA is primarily the function of **snRNAs** (small nuclear RNAs) which form the spliceosome complex. * **C (Incorrect):** Initiation of translation is mediated by eukaryotic initiation factors (eIFs) and the 40S ribosomal subunit. miRNAs generally *inhibit* rather than initiate this process. * **D (Incorrect):** Carrying messages for protein synthesis is the primary function of **mRNA** (messenger RNA). **High-Yield Clinical Pearls for NEET-PG:** * **Biogenesis:** miRNAs are transcribed by **RNA Polymerase II** as primary-miRNA (pri-miRNA), processed in the nucleus by the **Drosha** enzyme, and further cleaved in the cytoplasm by the **Dicer** enzyme. * **RISC Complex:** To function, miRNA must be loaded into the **RNA-induced silencing complex (RISC)**, which contains the **Argonaute** protein. * **OncomiRs:** miRNAs that are dysregulated in cancer. For example, *miR-21* often acts as an oncogene, while *let-7* acts as a tumor suppressor. * **Therapeutics:** miRNA mimics and antagomirs (antisense oligonucleotides) are being researched as targeted molecular therapies.

Question 130: Which of the following is NOT true about the genetic code?

A. Degenerate
B. Overlapping (Correct Answer)
C. Ambiguous
D. Universal

Explanation: **Explanation:** The genetic code is a set of rules used by living cells to translate information encoded within genetic material into proteins. Understanding its properties is fundamental to molecular biology. **Why "Overlapping" is the correct answer:** The genetic code is **non-overlapping**. This means that in a sequence of nucleotides (e.g., ABCDEF), the first codon is ABC, the second is DEF, and so on. A single nucleotide is part of only one codon. If the code were overlapping, a single mutation could affect multiple amino acids in a protein sequence, which is not what occurs in nature. **Analysis of incorrect options:** * **A. Degenerate (Redundant):** This is a true property. Most amino acids are coded by more than one codon (e.g., Leucine has six different codons). This provides a "buffer" against mutations, particularly at the third nucleotide position (**Wobble hypothesis**). * **C. Ambiguous:** This is **NOT** a property of the genetic code. The code is **Unambiguous**, meaning one specific codon always codes for the same specific amino acid (e.g., UGG always codes for Tryptophan). *Note: The question asks what is NOT true; since the code is non-ambiguous, "Ambiguous" is technically also a false statement, but in standard medical exams, "Overlapping" is the classic distractor used to test the fundamental structure of the triplet code.* * **D. Universal:** The code is nearly the same in all organisms, from bacteria to humans. (Exceptions exist in mitochondrial DNA). **High-Yield Clinical Pearls for NEET-PG:** * **Initiation Codon:** AUG (codes for Methionine in eukaryotes and N-formylmethionine in prokaryotes). * **Stop Codons (Nonsense Codons):** UAA (Ochre), UAG (Amber), UGA (Opal). * **Frameshift Mutations:** Occur because the code is **commaless** (no punctuation). Deleting or inserting a base shifts the entire reading frame downstream. * **Mitochondrial Exception:** In mitochondria, UGA codes for Tryptophan instead of acting as a stop codon.

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