Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 61: How many t-RNAs are present in the cytoplasmic translation system?

A. 20 (Correct Answer)
B. 26
C. 28
D. 31

Explanation: **Explanation:** The correct answer is **A (20)**. In the human cytoplasmic translation system, there are **20 distinct families of t-RNAs**, each corresponding to one of the 20 standard amino acids. While there are 61 sense codons in the genetic code, the "Wobble Hypothesis" allows a single t-RNA to recognize multiple codons. However, at a fundamental functional level, there is **one specific aminoacyl t-RNA synthetase enzyme for each amino acid**, which ensures that the correct amino acid is attached to its corresponding t-RNA(s). Therefore, in the context of basic cytoplasmic protein synthesis, we recognize 20 functional types of t-RNA. **Analysis of Incorrect Options:** * **Options B, C, and D (26, 28, 31):** These numbers do not represent the standard functional classification of cytoplasmic t-RNAs. While the human genome contains approximately 497 nuclear genes encoding t-RNAs with various anticodons (roughly 48-50 distinct anticodon types), the most high-yield answer for medical examinations regarding the *minimum* required types for the 20 amino acids in the cytoplasm is 20. Mitochondrial translation, by contrast, is more "economical" and uses only 22 t-RNAs. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Wobble Hypothesis:** Proposed by Francis Crick; it explains why we don't need 61 different t-RNAs. The 3rd base of the codon and the 1st base of the anticodon have non-standard pairing. * **Initiator t-RNA:** In eukaryotes (cytoplasm), it is **tRNAi^Met** (carrying methionine), whereas in prokaryotes and mitochondria, it is **tRNA^fMet** (formyl-methionine). * **Charging:** The attachment of an amino acid to t-RNA is catalyzed by **Aminoacyl t-RNA synthetase**, requiring **ATP**. This step provides the "proofreading" mechanism for translation. * **Structure:** t-RNA has a "cloverleaf" secondary structure and an "L-shaped" tertiary structure. The amino acid attaches to the **3' acceptor arm (CCA sequence)**.

Question 62: In the human leukocyte antigen (HLA) system, where is the cluster of genes located?

A. Chromosome 6 (Correct Answer)
B. Chromosome 21
C. X-Chromosome
D. Y-chromosome

Explanation: ### Explanation **Correct Answer: A. Chromosome 6** The **Human Leukocyte Antigen (HLA)** system is the human version of the **Major Histocompatibility Complex (MHC)**. It consists of a large cluster of genes located on the **short arm (p-arm) of Chromosome 6**. These genes encode cell surface proteins essential for the immune system to distinguish "self" from "non-self." * **MHC Class I (HLA-A, B, C):** Present on all nucleated cells; recognized by CD8+ T-cells. * **MHC Class II (HLA-DP, DQ, DR):** Present on antigen-presenting cells (APCs); recognized by CD4+ T-cells. * **MHC Class III:** Encodes components of the complement system (C2, C4) and cytokines (TNF-α). --- ### Why the other options are incorrect: * **B. Chromosome 21:** This is the smallest autosome. Trisomy 21 results in **Down Syndrome**. It does not house the HLA cluster. * **C. X-Chromosome:** Associated with sex-linked disorders like Hemophilia A/B and Duchenne Muscular Dystrophy. * **D. Y-Chromosome:** Contains the **SRY gene**, responsible for male sex determination. --- ### High-Yield Clinical Pearls for NEET-PG: 1. **Inheritance:** HLA genes are highly polymorphic and inherited as a **haplotype** (one set from each parent) in a **codominant** fashion. 2. **Disease Associations:** * **HLA-B27:** Strongly associated with **Ankylosing Spondylitis**, Reiter’s syndrome, and Acute Anterior Uveitis. * **HLA-DR3/DR4:** Associated with **Type 1 Diabetes Mellitus**. * **HLA-DQ2/DQ8:** Associated with **Celiac Disease**. * **HLA-B*5701:** Screening required before starting **Abacavir** (to prevent hypersensitivity). 3. **Linkage Disequilibrium:** HLA alleles are often inherited together more frequently than expected by chance, a classic example of linkage disequilibrium.

Question 63: At which of the following levels does the predominant regulation of gene expression take place in both prokaryotes and eukaryotes?

A. DNA replication
B. Transcription (Correct Answer)
C. Translation
D. Post-translational modification

Explanation: **Explanation:** **Why Transcription is the Correct Answer:** The regulation of gene expression is most efficient when it occurs at the earliest possible stage. **Transcription initiation** is the predominant level of control in both prokaryotes and eukaryotes because it prevents the cell from wasting energy and resources (nucleotides, ATP, and amino acids) on synthesizing RNA and proteins that are not currently required. By controlling the synthesis of mRNA, the cell dictates the entire downstream proteome. In prokaryotes, this is often managed via **operons** (e.g., Lac operon), while eukaryotes utilize complex **transcription factors**, enhancers, and promoters. **Why Other Options are Incorrect:** * **A. DNA Replication:** This is the process of copying the entire genome for cell division. It is a "template-copying" mechanism rather than a regulatory mechanism for specific gene expression. * **C. Translation:** While translational control exists (e.g., via heme-regulated inhibitors in RBCs or miRNA), it is a secondary level of regulation. Controlling expression here is less energy-efficient than stopping transcription. * **D. Post-translational modification:** This involves modifying proteins after they are synthesized (e.g., phosphorylation, glycosylation). This regulates **protein activity** and longevity rather than the primary expression of the gene itself. **High-Yield Clinical Pearls for NEET-PG:** * **The "Gold Standard" of Regulation:** Transcription initiation is the most common site of control. * **Prokaryotes:** Regulation is primarily at the level of **transcription initiation** (Sigma factors and Operons). * **Eukaryotes:** While transcription is predominant, eukaryotes also utilize significant **post-transcriptional** regulation (Alternative splicing, 5' capping, and 3' polyadenylation), which is absent in prokaryotes. * **Clinical Correlation:** Many antibiotics (e.g., Rifampicin) and toxins (e.g., Alpha-amanitin) target the transcriptional machinery (RNA Polymerase).

Question 64: ABO blood groups inheritance is an example of?

A. Mitochondrial inheritance
B. Allelic exclusion
C. Sex-linked inheritance
D. Co-dominance (Correct Answer)

Explanation: ### Explanation **Correct Option: D. Co-dominance** The ABO blood group system is a classic example of **Co-dominance** and **Multiple Allelism**. In co-dominance, both alleles in a heterozygote are fully expressed, and neither is dominant over the other. * The ABO system is governed by the *I* gene, which has three alleles: $I^A$, $I^B$, and $i$. * While $I^A$ and $I^B$ are both dominant over $i$ (complete dominance), they are **co-dominant** to each other. * In an individual with the genotype $I^AI^B$, both A and B antigens are expressed equally on the red blood cell surface, resulting in the AB blood group. **Why other options are incorrect:** * **A. Mitochondrial inheritance:** This refers to traits passed only from the mother to all offspring (e.g., LHON, MELAS). ABO genes are located on **Chromosome 9** (autosomal). * **B. Allelic exclusion:** This is a process where only one allele of a gene is expressed while the other is silenced (common in B-lymphocytes for immunoglobulin synthesis). In ABO, both alleles are expressed. * **C. Sex-linked inheritance:** These traits are carried on X or Y chromosomes (e.g., Hemophilia, Color blindness). ABO inheritance is **autosomal**. **High-Yield Clinical Pearls for NEET-PG:** * **Bombay Phenotype:** A rare condition where the individual lacks the **H-substance** (genotype *hh*). Even if they possess $I^A$ or $I^B$ genes, they phenotypically test as O-group because the precursor H-antigen is missing. * **Universal Donor/Recipient:** O negative is the universal donor (no antigens); AB positive is the universal recipient (no antibodies). * **Linkage:** The ABO gene locus is linked to the gene for **Nail-Patella Syndrome**.

Question 65: What method is used to identify the sequence in a long chain of protein?

A. Restriction fragment length polymorphism (RFLP)
B. Chromosome walking (Correct Answer)
C. Leucine Zipper
D. Southern Blot hybridization (SSOP)

Explanation: **Explanation:** **Chromosome walking** is a technique used to map and sequence long stretches of DNA by moving step-by-step along the chromosome. In the context of identifying a gene sequence that codes for a long protein chain, this method involves using a known DNA marker or a cloned fragment as a "probe" to find an overlapping clone from a genomic library. This process is repeated sequentially to "walk" along the chromosome, allowing researchers to characterize large genes or gene clusters that are too long to be sequenced in a single read. **Analysis of Incorrect Options:** * **A. Restriction Fragment Length Polymorphism (RFLP):** This technique detects variations in homologous DNA sequences (polymorphisms) based on different lengths of fragments after digestion with restriction enzymes. It is used for genetic mapping and forensic analysis, not for sequencing long chains. * **C. Leucine Zipper:** This is a common **structural motif** found in DNA-binding proteins (transcription factors). It facilitates protein-protein dimerization but is not a sequencing method. * **D. Southern Blot:** This is a laboratory method used to detect a **specific DNA sequence** in a blood or tissue sample using electrophoresis and probe hybridization. It identifies the presence or size of a gene but does not determine the sequence of a long chain. **High-Yield Pearls for NEET-PG:** * **Chromosome Jumping:** A variation used to bypass long repetitive sequences that are difficult to "walk" through. * **Sanger Sequencing:** The "gold standard" for sequencing small to medium DNA fragments. * **Edman Degradation:** The classic biochemical method for sequencing the **amino acids** of a protein directly (from the N-terminus). * **Zinc Finger & Helix-Turn-Helix:** Other high-yield DNA-binding motifs similar to the Leucine Zipper.

Question 66: Which of the following is true about the location of the immunoglobulin (Ig) heavy chain gene locus?

A. Chromosome 14 (Correct Answer)
B. Chromosome 2
C. Chromosome 6
D. All of the above

Explanation: ### Explanation The production of antibodies (immunoglobulins) involves specific genetic loci that undergo somatic recombination. In humans, the genes encoding the **heavy chain (IgH)** and the two types of **light chains (Kappa and Lambda)** are located on three distinct chromosomes. **1. Why Chromosome 14 is Correct:** The **Immunoglobulin Heavy Chain (IGH) locus** is located on the long arm of **Chromosome 14 (14q32)**. This locus contains the gene segments for the variable (V), diversity (D), and joining (J) regions, as well as the constant (C) regions (μ, δ, γ, ε, α) that determine the antibody isotype (IgM, IgD, IgG, IgE, IgA). **2. Analysis of Incorrect Options:** * **Chromosome 2:** This is the location of the **Kappa (κ) light chain** gene locus. * **Chromosome 22 (Not listed, but relevant):** This is the location of the **Lambda (λ) light chain** gene locus. * **Chromosome 6:** This chromosome houses the **Major Histocompatibility Complex (MHC)** genes, which encode HLA antigens, not immunoglobulin chains. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Burkitt Lymphoma:** A classic high-yield association is the **t(8;14)** translocation. This involves the *MYC* proto-oncogene on chromosome 8 moving to the Ig heavy chain locus on chromosome 14, leading to constitutive expression of MYC and oncogenesis. * **Follicular Lymphoma:** Associated with **t(14;18)**, where the *BCL-2* gene translocates to the Ig heavy chain locus. * **Mnemonic for Ig Loci:** * **H**eavy = **14** (H is the 8th letter, 1+4=5... better to remember "Heavy 14") * **K**appa = **2** * **L**ambda = **22** (L for 22)

Question 67: Which of the following are stop codons?

A. UAA
B. UAG
C. UGA
D. All of the above (Correct Answer)

Explanation: **Explanation:** In molecular biology, the genetic code consists of 64 codons. Out of these, 61 codons are "sense" codons that code for specific amino acids, while **3 codons** are "nonsense" or **Stop Codons**. These codons do not code for any amino acid; instead, they signal the termination of protein synthesis (translation) by binding to release factors. The three stop codons are: 1. **UAA (Ochre)** 2. **UAG (Amber)** 3. **UGA (Opal)** Since all three options (A, B, and C) represent the universal stop codons, the correct answer is **D (All of the above).** **Why other options are not "incorrect" but incomplete:** * **UAA, UAG, and UGA** are individual stop codons. Selecting only one would be partially correct but incomplete in the context of a "Multiple Choice Question" where all listed options serve the same biological function. **High-Yield Clinical Pearls for NEET-PG:** * **Nonsense Mutation:** A point mutation that changes a sense codon into a stop codon, leading to premature termination of the polypeptide chain and often resulting in a non-functional protein (e.g., in some forms of β-thalassemia). * **Exceptions to the Rule:** In human **mitochondria**, the genetic code differs slightly: **UGA** codes for Tryptophan (rather than stop), while **AGA and AGG** function as stop codons (rather than coding for Arginine). * **Initiation Codon:** Contrast these with **AUG**, which is the universal start codon (coding for Methionine in eukaryotes and Formyl-methionine in prokaryotes). * **Mnemonic:** To remember the stop codons: **U** **A**re **A**way (UAA), **U** **A**re **G**one (UAG), **U** **G**o **A**way (UGA).

Question 68: What is a required component for Polymerase Chain Reaction (PCR)?

A. Primer (Correct Answer)
B. DNA Polymerase
C. Deoxyribonucleotide triphosphate
D. Dideoxyribonucleotide triphosphate

Explanation: **Explanation:** Polymerase Chain Reaction (PCR) is an *in vitro* enzymatic method used to amplify specific DNA sequences. The correct answer is **Primer** because DNA polymerases cannot initiate DNA synthesis *de novo*; they require a pre-existing 3'-OH group to add nucleotides. In PCR, synthetic oligonucleotide primers provide this starting point by flanking the target sequence. **Analysis of Options:** * **A. Primer (Correct):** Essential for defining the target region and providing the 3'-OH terminus for elongation. * **B. DNA Polymerase:** While essential, the question likely focuses on the specific requirement for initiation. (Note: In many competitive exams, if multiple essential components are listed, the "primer" is often highlighted as the specific requirement for the *initiation* phase of the reaction). * **C. Deoxyribonucleotide triphosphates (dNTPs):** These are the building blocks (dATP, dCTP, dGTP, dTTP) required for synthesis. * **D. Dideoxyribonucleotide triphosphates (ddNTPs):** These are **not** used in standard PCR. They lack a 3'-OH group and act as chain terminators; they are a hallmark of **Sanger Sequencing**. **High-Yield NEET-PG Pearls:** 1. **Taq Polymerase:** Derived from *Thermus aquaticus*, it is heat-stable, allowing it to survive the denaturation step (94-96°C). 2. **Steps of PCR:** Denaturation (95°C) → Annealing (55-65°C) → Extension (72°C). 3. **Clinical Use:** PCR is the gold standard for diagnosing viral infections (e.g., COVID-19, HIV viral load), detecting genetic mutations, and forensic DNA profiling. 4. **RT-PCR:** Uses Reverse Transcriptase to convert RNA into cDNA before amplification (used for RNA viruses).

Question 69: Although 61 codons code for amino acids, only 20 amino acids are naturally occurring. This feature of the genetic code is known as:

A. Transcription
B. Degeneracy (Correct Answer)
C. Mutation
D. Unambiguity

Explanation: ### Explanation **Correct Option: B. Degeneracy** The genetic code is described as **degenerate** (or redundant) because a single amino acid can be coded for by multiple different codons. Since there are 64 possible triplet codons (4³) and only 20 standard amino acids, the system is mathematically "over-determined." * **Mechanism:** 61 codons specify amino acids, while 3 (UAA, UAG, UGA) are stop codons. * **Wobble Hypothesis:** This redundancy often occurs at the 3rd base of the codon (the "wobble position"), where non-traditional base pairing allows one tRNA to recognize multiple codons, protecting the cell against minor mutations. **Why other options are incorrect:** * **A. Transcription:** This is the biological process of synthesizing RNA from a DNA template; it is a mechanism of gene expression, not a property of the code itself. * **C. Mutation:** This refers to a permanent alteration in the DNA sequence. While degeneracy helps mitigate the effects of mutations (silent mutations), it is not the term for the coding ratio. * **D. Unambiguity:** This is the opposite concept. Unambiguity means that **one specific codon always codes for only one specific amino acid** (e.g., UUU always codes for Phenylalanine). The code is redundant but never ambiguous. **High-Yield Clinical Pearls for NEET-PG:** * **Universal Code:** The genetic code is the same in almost all organisms. **Exception:** Human mitochondrial DNA (e.g., UGA codes for Tryptophan instead of acting as a Stop codon). * **Non-overlapping & Commaless:** The code is read sequentially from a fixed starting point without skipping any bases. * **Initiation Codon:** **AUG** (Methionine) is the start codon in eukaryotes; in prokaryotes, it codes for N-formylmethionine (fMet). * **Clinical Relevance:** Degeneracy is the basis for **Silent Mutations**, where a base change (usually at the 3rd position) does not alter the resulting protein sequence.

Question 70: Primers are removed by which of the following enzymes, except?

A. Delta Polymerase
B. RNase H1
C. FEN1
D. None of the above (Correct Answer)

Explanation: In eukaryotic DNA replication, the removal of RNA primers is a coordinated process involving multiple enzymes. The question asks which enzyme does **not** participate in this process; since all listed enzymes are involved, the correct answer is **None of the above**. ### **Mechanism of Primer Removal (Eukaryotes)** Unlike prokaryotes (where DNA Polymerase I removes primers via 5’→3’ exonuclease activity), eukaryotes utilize a "strand displacement" mechanism: 1. **DNA Polymerase δ (Delta):** As it synthesizes the Okazaki fragment, it encounters the RNA primer of the preceding fragment. It displaces the 5’ end of the primer, creating a "flap." 2. **RNase H1:** This enzyme recognizes and degrades the RNA portion of the RNA-DNA hybrid primer, leaving a single ribonucleotide attached to the DNA. 3. **FEN1 (Flap Endonuclease 1):** This enzyme acts as an endonuclease to clip off the RNA "flaps" displaced by Polymerase δ. It is essential for removing the final ribonucleotide and ensuring the gap is ready for ligation. ### **Analysis of Options** * **Option A (Delta Polymerase):** Correctly involved; it provides the motor force to displace the primer. * **Option B (RNase H1):** Correctly involved; it degrades the bulk of the RNA primer. * **Option C (FEN1):** Correctly involved; it cleaves the displaced flap to complete primer removal. ### **High-Yield Clinical Pearls for NEET-PG** * **Prokaryotic Equivalent:** In *E. coli*, **DNA Polymerase I** is the sole enzyme responsible for primer removal due to its unique **5’ to 3’ exonuclease activity**. * **DNA Ligase:** Once primers are removed and the gap is filled by Pol δ, DNA Ligase I seals the phosphodiester bond. * **PCNA (Proliferating Cell Nuclear Antigen):** Acts as a sliding clamp that increases the processivity of Pol δ and recruits FEN1 to the replication fork.

Practice by Chapter

DNA Replication and Repair Mechanisms

Practice Questions

Transcription Factors and Gene Regulation

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Epigenetics and DNA Methylation

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RNA Processing and Splicing

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miRNA and RNA Interference

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Protein Synthesis and Post-Translational Modifications

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Genomics and Human Genome Project

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Single Nucleotide Polymorphisms

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Gene Therapy Approaches

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CRISPR-Cas9 and Genome Editing

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DNA Fingerprinting and Forensics

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Molecular Basis of Genetic Diseases

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