Molecular Biology and Genomics Practice Questions

Q: What is the approximate number of base pairs associated with a single nucleosome?

146. ### Explanation **1. Why Option A is Correct:** The nucleosome is the fundamental structural unit of chromatin. It consists of a **nucleosome core particle** and linker DNA. The core particle is composed of an octamer of histone proteins (two each of H2A, H2B, H3, and H4) around which DNA is wrapped. Specifically, **146 base pairs (bp)** of DNA wrap approximately **1.75 times** around this histone octamer in a left-handed superhelix. This "beads-on-a-string" structure is essential for compacting the large eukaryotic genome into the cell nucleus. **2. Why Other Options are Incorrect:** * **Option B (292):** This is exactly double the core DNA length. While a "chromatosome" (core + H1 histone + linker DNA) involves about 160–200 bp, 292 bp does not represent a standard structural unit. * **Option C (73):** This is half of 146. While 146 bp completes nearly two turns, 73 bp would represent roughly a single turn around the octamer, which is not the complete functional unit. * **Option D (1460):** This is a ten-fold overestimation and does not correspond to any specific repeating unit in chromatin architecture. **3. NEET-PG High-Yield Clinical Pearls:** * **Histone Composition:** Histones are rich in **Lysine and Arginine**, giving them a positive charge to bind the negatively charged phosphate backbone of DNA. * **Linker Histone:** **Histone H1** is not part of the core octamer; it sits outside the core to stabilize the entry and exit of DNA (forming the chromatosome). * **Epigenetics:** Acetylation of histones (by HATs) neutralizes the positive charge, leading to relaxed chromatin (**Euchromatin**) and increased transcription. Deacetylation (by HDACs) leads to condensed chromatin (**Heterochromatin**). * **Clinical Link:** In **Systemic Lupus Erythematosus (SLE)**, patients often develop antibodies against histones (Anti-histone antibodies), especially in drug-induced SLE (e.g., Hydralazine, Procainamide).

Q: A 10-year-old female patient with a known history of beta-thalassemia presents with complaints of tiredness and shortness of breath. Which of the following genetic mechanisms can cause beta-thalassemia?

Faulty splicing. ### Explanation **Correct Answer: B. Faulty splicing** **Mechanism:** Beta-thalassemia is a heterogeneous group of genetic disorders characterized by reduced ($\beta^+$) or absent ($\beta^0$) synthesis of the beta-globin chains of hemoglobin. While various mutations can occur, **faulty splicing** is one of the most common molecular mechanisms. Most often, point mutations occur within **introns** or at the **exon-intron junctions** (splice sites). These mutations can: 1. Destroy existing splice sites. 2. Create "cryptic" splice sites. This leads to the production of abnormal mRNA transcripts that are either degraded or translated into non-functional proteins, resulting in a deficiency of beta-globin chains. **Analysis of Incorrect Options:** * **A & C (Terminal base addition/deletion):** These are not standard mechanisms for beta-thalassemia. While frame-shift mutations (insertions or deletions within the coding sequence) can cause the disease, they are typically internal, not "terminal." Terminal modifications usually refer to post-transcriptional polyadenylation or capping, which are rarely the primary cause of this pathology. * **D (None of the above):** Incorrect, as splicing defects are a hallmark of the molecular pathology of beta-thalassemia. **NEET-PG High-Yield Pearls:** * **Most common mutation in India:** The most common mutation causing beta-thalassemia in the Indian population is **IVS 1-5 (G→C)**, which is a splice-site mutation. * **Alpha vs. Beta Thalassemia:** Remember that **Alpha-thalassemia** is most commonly caused by **large gene deletions**, whereas **Beta-thalassemia** is most commonly caused by **point mutations** (including splicing, nonsense, and promoter mutations). * **Clinical Presentation:** Look for microcytic hypochromic anemia, increased HbA2 (>3.5%), and "crew-cut" appearance on skull X-ray due to extramedullary hematopoiesis.

Q: All of the following are true regarding DNA repair defect syndromes EXCEPT?

Fragile X Syndrome. **Explanation:** The correct answer is **Fragile X Syndrome** because it is a **Trinucleotide Repeat Expansion disorder**, not a primary defect in DNA repair mechanisms. It is caused by the expansion of CGG repeats in the *FMR1* gene, leading to hypermethylation and gene silencing. **Analysis of Options:** * **Hereditary Non-polyposis Colorectal Cancer (HNPCC/Lynch Syndrome):** This is caused by a defect in the **Mismatch Repair (MMR)** genes (primarily *MSH2* and *MLH1*). This leads to microsatellite instability (MSI). * **Fanconi Anemia:** This is a defect in **Interstrand Cross-link (ICL) repair**. Patients exhibit bone marrow failure, physical anomalies, and a high risk of AML. * **Ataxia-Telangiectasia (AT):** This is caused by a mutation in the *ATM* gene, which is responsible for detecting **Double-Strand Breaks (DSB)** and initiating the repair process via Non-Homologous End Joining (NHEJ) or Homologous Recombination. **High-Yield Clinical Pearls for NEET-PG:** 1. **Xeroderma Pigmentosum:** Defect in **Nucleotide Excision Repair (NER)**; inability to repair pyrimidine dimers caused by UV light. 2. **Cockayne Syndrome:** Defect in transcription-coupled DNA repair (a subtype of NER); characterized by "bird-like" facies and dwarfism but *no* increased risk of skin cancer. 3. **Bloom Syndrome:** Defect in **DNA Helicase** (*BLM* gene); presents with sister chromatid exchanges and "butterfly" rash. 4. **BRCA 1 & 2:** Defects in **Homologous Recombination** (Double-strand break repair).

Q: Paternal disomy is found in which condition?

Angelman syndrome. ### 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).

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

DNA polymerase III. **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)**.

Q: 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?

Point mutation. **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.

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

Polycistronic mRNA. **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.

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

RNA polymerase. **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.

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

GTP hydrolysis within the cytosol. ### 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 1

What is the approximate number of base pairs associated with a single nucleosome?

Accepted Answer

146

Answer

292

Answer

73

Answer

1460

Question 2

A 10-year-old female patient with a known history of beta-thalassemia presents with complaints of tiredness and shortness of breath. Which of the following genetic mechanisms can cause beta-thalassemia?

Accepted Answer

Faulty splicing

Answer

Terminal base addition

Answer

Terminal base deletion

Answer

None of the above

Question 3

All of the following are true regarding DNA repair defect syndromes EXCEPT?

Accepted Answer

Fragile X Syndrome

Answer

Hereditary Non-polyposis Colorectal Cancer (HNPCC)

Answer

Fanconi Anemia

Answer

Ataxia-Telangiectasia (AT)

Question 4

Paternal disomy is found in which condition?

Accepted Answer

Angelman syndrome

Answer

Prader-Willi syndrome

Answer

Fragile X syndrome

Answer

Hydatiform mole

Question 5

A 10-year-old child presents with recurrent skin lesions on the face and upper extremities that rapidly progress to cancer. Xeroderma pigmentosa is considered a possible differential. Which of the following enzymes is defective in this condition?

Accepted Answer

Endonuclease

Answer

Topoisomerase

Answer

3' 	→ 5' Exonuclease

Answer

Helicase

Question 6

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

Accepted Answer

DNA polymerase III

Answer

Helicase

Answer

DNA polymerase I

Answer

Topoisomerase III

Question 7

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?

Accepted Answer

Point mutation

Answer

Insertion

Answer

Deletion

Answer

Non-disjunction

Question 8

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

Accepted Answer

Polycistronic mRNA

Answer

Noncoding intron

Answer

Contain nuclear gene and pseudogene

Answer

Modification of mRNA before transportation from the nucleus

Question 9

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

Accepted Answer

RNA polymerase

Answer

80S ribosomes

Answer

Initiation of DNA replication

Answer

DNA polymerase

Question 10

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

Accepted Answer

GTP hydrolysis within the cytosol

Answer

ATP hydrolysis within the cytosol

Answer

ATP hydrolysis within the nucleus

Answer

GTP hydrolysis within the nucleus

Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics — MCQs

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