Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 161: Which type of mutation usually involves mutation of a gene coding for the following molecule?

A. Silent mutation
B. Nonsense mutation
C. Missense mutation
D. Nonsense suppressor mutation (Correct Answer)

Explanation: ***Nonsense suppressor mutation*** - Involves mutations in **tRNA genes** that alter the **anticodon sequence** to read through **stop codons** (UAG, UAA, UGA). - The **tRNA structure** shown is directly involved in this mechanism, where mutant tRNA can insert an amino acid at a **nonsense codon** instead of terminating translation. *Silent mutation* - Changes a **codon** without altering the **amino acid sequence** due to the **degeneracy of the genetic code**. - Does not typically involve **tRNA gene mutations** but rather changes in protein-coding genes that don't affect protein function. *Nonsense mutation* - Creates a **premature stop codon** (UAG, UAA, UGA) in a **protein-coding gene**, leading to **truncated proteins**. - This type of mutation occurs in **mRNA-coding sequences**, not in **tRNA genes** like the molecule shown. *Missense mutation* - Results in a **single amino acid substitution** in the protein sequence due to a **point mutation** in the coding sequence. - Involves changes in **protein-coding genes** rather than **tRNA genes**, and doesn't affect the tRNA structure or anticodon recognition.

Question 162: What is the most likely effect of a 2bp insertion in the middle of an intron in a typical eukaryotic gene?

A. Normal transcription, altered translation
B. Defective termination of transcription, normal translation
C. Normal transcription, defective mRNA splicing
D. Normal transcription, normal translation (Correct Answer)

Explanation: ### Explanation **1. Why Option D is Correct:** In eukaryotic gene expression, **introns** are non-coding sequences that are transcribed into pre-mRNA but are subsequently removed during **splicing**. For splicing to occur correctly, the cellular machinery (spliceosome) primarily recognizes specific conserved sequences: the **5' donor site (GT)**, the **3' acceptor site (AG)**, and the **internal branch point (Adenine)**. A 2bp insertion in the **middle** of an intron—away from these critical splice sites—does not interfere with the recognition of exon-intron boundaries. Consequently, the intron is spliced out normally, the mature mRNA remains unchanged, and the resulting protein (translation) is identical to the wild-type. **2. Why Other Options are Incorrect:** * **Option A:** Translation would only be altered if the mutation occurred in an **exon** (causing a frameshift) or if it created a new, cryptic splice site that led to the inclusion of intronic sequences in the mature mRNA. * **Option B:** Transcription termination is governed by specific signals (like the polyadenylation signal AAUAAA) located in the **3' Untranslated Region (UTR)**, not by sequences in the middle of an intron. * **Option C:** Splicing is only defective if the mutation affects the **highly conserved** splice donor/acceptor sites or the branch point. Most "junk" DNA within the middle of an intron can tolerate small insertions or deletions without functional consequences. **3. High-Yield Clinical Pearls for NEET-PG:** * **Splice Site Mutations:** Mutations at the GT/AG boundaries are a common cause of genetic diseases, such as **β-thalassemia**, where they lead to exon skipping or intron retention. * **The Rule of Three:** Frameshift mutations (like a 2bp insertion) are devastating in **exons** because they shift the reading frame, but they are functionally silent in the middle of **introns**. * **Introns vs. Exons:** Remember that introns are "Intervening" (removed), while exons are "Expressed" (retained).

Question 163: Which of the following is the cofactor for prokaryotic DNA ligase?

A. Tetrahydrobiopterin
B. ATP
C. NAD (Correct Answer)
D. FAD

Explanation: **Explanation:** DNA ligase is the essential enzyme responsible for sealing "nicks" in the phosphodiester backbone by catalyzing the formation of a bond between a 3'-hydroxyl group and a 5'-phosphate group. This process requires an energy source to activate the 5' phosphate. **1. Why NAD+ is correct:** In **prokaryotes** (such as *E. coli*), DNA ligase specifically utilizes **Nicotinamide Adenine Dinucleotide (NAD+)** as the cofactor. The enzyme cleaves NAD+ into Nicotinamide Mononucleotide (NMN) and AMP; the AMP is then transferred to the enzyme to initiate the ligation reaction. **2. Why the other options are incorrect:** * **ATP:** This is the cofactor for **Eukaryotic DNA ligase** and T4 bacteriophage ligase. While prokaryotes use NAD+, humans and other eukaryotes use ATP. * **Tetrahydrobiopterin (BH4):** This is a cofactor for hydroxylation reactions involving amino acids (e.g., Phenylalanine hydroxylase, Tyrosine hydroxylase). It is not involved in DNA replication. * **FAD:** This is a redox cofactor involved in the Electron Transport Chain and various metabolic dehydrogenation reactions (e.g., Succinate dehydrogenase), but not in DNA ligation. **High-Yield Clinical Pearls for NEET-PG:** * **DNA Ligase Function:** Essential for joining **Okazaki fragments** on the lagging strand during replication and for DNA repair mechanisms (like Nucleotide Excision Repair). * **The "Glue":** DNA ligase is often referred to as "molecular glue" in recombinant DNA technology. * **Deficiency:** A deficiency in DNA Ligase I in humans leads to **46BR syndrome**, characterized by growth retardation, immunodeficiency, and sensitivity to DNA-damaging agents. * **Distinction:** Remember the mnemonic: **P**rokaryotes = **P**rimitive = **N**AD+; **E**ukaryotes = **E**volved = **A**TP.

Question 164: Which of the following is NOT a characteristic of the genetic code?

A. Overlapping (Correct Answer)
B. Nonambiguous
C. Universal
D. Degeneracy

Explanation: The genetic code is a set of rules by which information encoded in genetic material is translated into proteins. Understanding its properties is fundamental for molecular biology and genetics. ### **Why "Overlapping" is the Correct Answer** The genetic code is **non-overlapping**. This means that in a sequence like ABCDEF, the first codon is ABC, the second is DEF, and so on. Each nucleotide is part of only one codon. If the code were overlapping, a single mutation could affect multiple amino acids in a protein chain, which is not what occurs in nature. ### **Explanation of Incorrect Options** * **Non-ambiguous:** This means that one specific codon always codes for the same amino acid (e.g., UGG always codes for Tryptophan). It is "specific." * **Universal:** The genetic code is nearly identical in all known living organisms, from bacteria to humans. (Note: Minor exceptions exist in mitochondrial DNA). * **Degeneracy (Redundancy):** Most amino acids are coded by more than one codon (e.g., Leucine is coded by six different codons). This acts as a protective mechanism against minor mutations. ### **High-Yield Clinical Pearls for NEET-PG** * **Commaless:** There are no "punctuations" or gaps between codons; the mRNA is read continuously. * **Wobble Hypothesis:** Proposed by Francis Crick, it explains why the third base of a codon can sometimes vary without changing the amino acid, accounting for degeneracy. * **Initiation Codon:** **AUG** (Methionine in eukaryotes, Formyl-methionine in prokaryotes). * **Stop Codons (Nonsense Codons):** **UAA** (Ochre), **UAG** (Amber), and **UGA** (Opal). These do not code for any amino acid. * **Exceptions to Universality:** In human mitochondria, **UGA** codes for Tryptophan (instead of Stop) and **AUA** codes for Methionine (instead of Isoleucine).

Question 165: Gene therapy is most effectively given in which of the following conditions?

A. Cystic fibrosis
B. Thalassemia
C. Sickle cell anemia
D. Severe combined immunodeficiency (Correct Answer)

Explanation: **Explanation:** **Severe Combined Immunodeficiency (SCID)** is considered the "poster child" for successful gene therapy because it fulfills the ideal criteria for genetic intervention: it is caused by a single gene defect, affects hematopoietic stem cells (which are easily accessible), and provides a **selective survival advantage** to corrected cells. 1. **Why SCID is the Correct Answer:** The most common form treated via gene therapy is **ADA-SCID** (Adenosine Deaminase deficiency). When the functional ADA gene is inserted into the patient's autologous hematopoietic stem cells and re-infused, the corrected T-cells have a natural proliferative advantage over the defective ones. This allows for the restoration of the immune system without the high risks associated with allogeneic bone marrow transplants. 2. **Why Other Options are Less Effective:** * **Cystic Fibrosis:** Delivery is the main hurdle. The thick mucus in the lungs prevents viral vectors from reaching the target epithelial cells, and the turnover of these cells means the effect is often transient. * **Thalassemia & Sickle Cell Anemia:** These involve the hemoglobin molecule. Achieving the precise, high-level expression of the globin gene required to balance the alpha/beta chain ratio is biochemically complex compared to the simple enzyme replacement needed in SCID. **High-Yield Clinical Pearls for NEET-PG:** * **First Gene Therapy:** Performed in 1990 by William French Anderson on a 4-year-old girl (Ashanthi DeSilva) with **ADA-SCID**. * **Vectors:** Retroviruses and Lentiviruses are commonly used for SCID to ensure integration into the host genome. * **Recent Advance:** CRISPR-Cas9 is currently being explored for Sickle Cell Anemia, but SCID remains the classic historical and clinical success story in medical textbooks.

Question 166: Single gene defect causing multiple unrelated problems is known as:

A. Pleiotropism (Correct Answer)
B. Pseudodominance
C. Penetrance
D. Anticipation

Explanation: ### Explanation **Correct Answer: A. Pleiotropism** **Pleiotropism** occurs when a **single gene mutation** results in multiple, seemingly unrelated phenotypic effects across different organ systems. This happens because the gene product (usually a protein or enzyme) is utilized in various tissues or biochemical pathways. * **Classic Example:** **Phenylketonuria (PKU)**. A defect in the *PAH* gene leads to mental retardation, reduced hair/skin pigmentation, and a mousy body odor. * **Other Examples:** Marfan Syndrome (fibrillin-1 defect affecting eyes, heart, and skeleton) and Sickle Cell Anemia. --- ### Why the other options are incorrect: * **B. Pseudodominance:** This occurs when a recessive trait appears to be inherited in a dominant fashion. This typically happens when a homozygous recessive individual mates with a heterozygous carrier, resulting in a 50% chance of affected offspring in every generation. * **C. Penetrance:** This refers to the **percentage** of individuals with a specific genotype who actually express the associated phenotype. If 100 people have the gene but only 80 show symptoms, the gene has 80% penetrance. * **D. Anticipation:** This is the phenomenon where a genetic disorder becomes more severe or appears at an earlier age in successive generations. It is classically associated with **Trinucleotide Repeat Disorders** (e.g., Huntington’s Disease, Fragile X Syndrome). --- ### High-Yield Clinical Pearls for NEET-PG: * **Variable Expressivity:** Unlike pleiotropy (multiple traits), this refers to the *degree* or severity of the phenotype among individuals with the same genotype (e.g., two people with Neurofibromatosis Type 1 having different numbers of café-au-lait spots). * **Locus Heterogeneity:** Mutations at different loci (different genes) produce the same phenotype (e.g., Albinism). * **Allelic Heterogeneity:** Different mutations within the same gene produce the same phenotype (e.g., Beta-thalassemia).

Question 167: What percentage of mitochondrial DNA constitutes total cellular DNA?

A. 1% (Correct Answer)
B. 1.30%
C. 3%
D. 5%

Explanation: **Explanation:** **1. Why Option A is Correct:** In human cells, the vast majority of genetic material is housed within the nucleus (Nuclear DNA). Mitochondrial DNA (mtDNA) is a small, circular, double-stranded molecule located within the mitochondria. Despite there being hundreds to thousands of mitochondria per cell (and multiple copies of mtDNA per mitochondrion), the actual size of the mitochondrial genome is very small—approximately 16.5 kb compared to the 3.2 billion base pairs of the nuclear genome. Consequently, mtDNA accounts for approximately **1% of the total cellular DNA**. **2. Why Other Options are Incorrect:** * **Option B (1.30%):** This is a distractor often confused with the percentage of the human genome that codes for proteins (exons), which is approximately 1.5%. * **Options C & D (3% and 5%):** These values significantly overestimate the mass of mitochondrial DNA. While mitochondria are numerous, their individual genomes are too compact to reach these percentages of total cellular DNA mass. **3. High-Yield Clinical Pearls for NEET-PG:** * **Maternal Inheritance:** mtDNA is inherited exclusively from the mother because the sperm's mitochondria are degraded during fertilization. * **Lack of Introns:** Unlike nuclear DNA, mtDNA is highly economical; it contains no introns and very little non-coding sequences. * **Mutation Rate:** The mutation rate in mtDNA is **10–20 times higher** than in nuclear DNA due to the lack of protective histones and proximity to reactive oxygen species (ROS) generated during oxidative phosphorylation. * **Heteroplasmy:** This refers to the presence of a mixture of more than one type of organellar genome (mutant vs. wild-type) within a cell, which explains the variable clinical severity of mitochondrial diseases (e.g., MELAS, LHON).

Question 168: Okazaki fragments formed during DNA replication are structurally:

A. DNA (Correct Answer)
B. mRNA
C. rRNA
D. tRNA

Explanation: ### Explanation **1. Why the Correct Answer is Right (DNA):** DNA replication is **semi-discontinuous**. While the leading strand is synthesized continuously, the lagging strand is synthesized in short, discrete segments called **Okazaki fragments**. These fragments are composed of **deoxyribonucleotides (DNA)**. The process begins with a short RNA primer (synthesized by Primase), which provides a 3'-OH group. DNA Polymerase III then adds DNA nucleotides to this primer. Although the fragment *starts* with a tiny RNA stretch, the bulk of the fragment—and its functional identity—is **DNA**. Eventually, DNA Polymerase I removes the RNA primer and replaces it with DNA, and DNA Ligase joins the fragments to form a continuous DNA strand. **2. Why the Incorrect Options are Wrong:** * **mRNA (Messenger RNA):** This is the product of **transcription**, used as a template for protein synthesis (translation). It is not a structural component of DNA replication fragments. * **rRNA (Ribosomal RNA):** This forms the structural and catalytic core of ribosomes. It is involved in translation, not DNA synthesis. * **tRNA (Transfer RNA):** This acts as an adapter molecule that carries amino acids to the ribosome during translation. It has no role in the structure of Okazaki fragments. **3. High-Yield Clinical Pearls for NEET-PG:** * **Directionality:** Okazaki fragments are always synthesized in the **5' to 3' direction**, even though the lagging strand overall grows in the 3' to 5' direction relative to the replication fork. * **Enzyme Key:** **DNA Ligase** is the "glue" that joins Okazaki fragments by forming phosphodiester bonds. Deficiencies in proteins involved in this process (like the FEN1 protein) can lead to genomic instability. * **Length:** In eukaryotes, Okazaki fragments are typically shorter (100–200 nucleotides) than in prokaryotes (1000–2000 nucleotides). * **Clinical Correlation:** Inhibitors of DNA replication (like Cytarabine or Methotrexate) disrupt the formation or elongation of these fragments, making them potent anti-cancer agents.

Question 169: The DNA nucleotide sequence to which RNA polymerase binds to initiate transcription is called the:

A. Exon
B. Enhancer region
C. Promoter region (Correct Answer)
D. Repressor region

Explanation: ### Explanation **Correct Answer: C. Promoter region** **1. Why the Promoter region is correct:** Transcription initiation is a highly regulated process. The **Promoter** is a specific DNA sequence located upstream (5') of the gene that serves as the recognition and binding site for **RNA polymerase**. In prokaryotes, this involves the sigma factor identifying sequences like the Pribnow box (-10). In eukaryotes, RNA polymerase II binds to the promoter with the help of general transcription factors, often recognizing the **TATA box** (Hogness box) located approximately 25 base pairs upstream of the start site. **2. Why the other options are incorrect:** * **Exon (A):** These are the coding regions of a gene that remain in the mature mRNA after splicing. They do not facilitate the binding of RNA polymerase. * **Enhancer region (B):** These are regulatory DNA sequences that increase the rate of transcription. While they can be located far from the gene, they function by looping DNA to interact with the promoter complex, rather than being the primary binding site for RNA polymerase itself. * **Repressor region (D):** This term is often confused with the **Operator** or the **Repressor protein**. A repressor is a protein that binds to an operator sequence to *block* RNA polymerase, thereby inhibiting transcription. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **TATA Box:** The most conserved eukaryotic promoter element; mutations here can significantly decrease gene expression (e.g., certain types of **β-thalassemia**). * **CAAT Box:** Located further upstream (-70 to -80); it determines the frequency of transcription. * **Rifampicin:** A key antitubercular drug that acts by inhibiting the **β-subunit of bacterial DNA-dependent RNA polymerase**, preventing transcription initiation. * **α-Amanitin:** Found in *Amanita phalloides* (death cap mushroom); it specifically inhibits **RNA polymerase II**, leading to severe liver failure.

Question 170: In humans, what is the sequence of telomeres?

A. 5'-GGCTTG-3'
B. 5'-TTAGGG-3' (Correct Answer)
C. 5'-TAACGT-3'
D. 5'-GTAGGC-3'

Explanation: **Explanation:** **1. Why the Correct Answer is Right:** Telomeres are specialized nucleoprotein structures located at the ends of linear eukaryotic chromosomes. They consist of tandem repeats of a specific hexanucleotide sequence. In humans (and all vertebrates), this conserved sequence is **5'-TTAGGG-3'**. The primary function of telomeres is to protect the chromosome ends from degradation, fusion, and being recognized as double-stranded DNA breaks. Because DNA polymerase cannot fully replicate the 3' end of a linear DNA molecule (the "end-replication problem"), telomeres shorten with each cell division. The enzyme **Telomerase**, a ribonucleoprotein, uses its internal RNA template to add these TTAGGG repeats back to the ends, thereby maintaining chromosomal stability. **2. Analysis of Incorrect Options:** * **Option A, C, and D:** These are arbitrary sequences that do not correspond to the highly conserved human telomeric repeat. While different species have different telomeric sequences (e.g., *Tetrahymena* uses TTGGGG), the sequence **TTAGGG** is specific and universal for human genetics. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Hayflick Limit:** The finite number of times a normal somatic cell population will divide before cell division stops (senescence), dictated by telomere shortening. * **Cancer Connection:** Approximately 85–90% of cancer cells upregulate **Telomerase**, allowing them to achieve "replicative immortality." * **Shelterin Complex:** A group of six proteins (including TRF1 and TRF2) that binds to TTAGGG repeats to form a protective "T-loop," preventing DNA repair machinery from attacking the chromosome ends. * **Disease Link:** Mutations in telomerase components (TERT or TERC) lead to **Dyskeratosis Congenita**, characterized by premature aging, bone marrow failure, and mucosal leukoplakia.

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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