Molecular Biology and Genomics Practice Questions

Q: The BRCA1 gene is located on which chromosome?

Chromosome 17. **Explanation:** The **BRCA1 (Breast Cancer 1)** gene is a tumor suppressor gene located on the **long (q) arm of Chromosome 17 (specifically 17q21)**. It plays a critical role in maintaining genomic stability by encoding a protein involved in the repair of double-stranded DNA breaks via **homologous recombination**. Mutations in this gene significantly increase the lifetime risk of breast, ovarian, and fallopian tube cancers. **Analysis of Options:** * **Option C (Correct):** BRCA1 is located on **Chromosome 17**. A helpful mnemonic is "BRCA**1** is on **17**" (both have a '7' or '1' association). * **Option A:** **Chromosome 13** is the location of the **BRCA2** gene (specifically 13q12.3). While both genes are involved in DNA repair, they are located on different chromosomes. * **Option B:** **Chromosome 11** houses several important genes like the Beta-globin gene cluster and the WT1 (Wilms tumor) gene, but not BRCA1. * **Option D:** **Chromosome 22** is associated with the NF2 gene and the "Philadelphia Chromosome" (translocation with Chromosome 9), but not BRCA1. **High-Yield Clinical Pearls for NEET-PG:** * **Inheritance:** Autosomal Dominant with variable expressivity. * **DNA Repair Mechanism:** BRCA1/2 are essential for **Homologous Recombination** (error-free repair). Deficiency leads to "BRCAness," making tumors sensitive to **PARP inhibitors** (e.g., Olaparib) via synthetic lethality. * **Cancer Risks:** BRCA1 carries a higher risk of **Ovarian Cancer** (up to 40%) compared to BRCA2 (up to 20%). BRCA2 is more strongly associated with **Male Breast Cancer**. * **Li-Fraumeni Syndrome:** Also involves Chromosome 17, but is due to a mutation in the **TP53** gene (17p13).

Q: What is true regarding cytoplasmic messenger RNA (mRNA)?

It is chiefly translated from nuclear DNA.. ### Explanation **Correct Option: A (It is chiefly translated from nuclear DNA)** In eukaryotic cells, the genetic blueprint is stored in the nucleus as DNA. mRNA is synthesized through the process of **transcription**, where an RNA polymerase enzyme uses the nuclear DNA as a template. Once processed, this mRNA is exported to the cytoplasm to serve as a template for protein synthesis (translation). Thus, cytoplasmic mRNA is the direct transcript of nuclear genetic information. **Analysis of Incorrect Options:** * **B & C:** These options describe the characteristics of **DNA**, not RNA. mRNA contains **ribose** sugar (not deoxyribose) and the nitrogenous base **uracil** (not thymine). * **D:** mRNA is actually **smaller** than heterogeneous nuclear RNA (hnRNA). hnRNA is the primary transcript (pre-mRNA) that contains both exons (coding regions) and introns (non-coding regions). During post-transcriptional modification (splicing), introns are removed and a 5' cap and 3' poly-A tail are added, resulting in a shorter, mature mRNA molecule. **Clinical Pearls & High-Yield Facts for NEET-PG:** * **Post-transcriptional modifications:** Occur in the nucleus and include 5’ capping (7-methylguanosine), 3’ polyadenylation (Poly-A tail), and splicing. * **Splicing:** Carried out by **snRNPs** (small nuclear ribonucleoproteins). Autoantibodies against snRNPs (Anti-Smith antibodies) are highly specific for **Systemic Lupus Erythematosus (SLE)**. * **mRNA Stability:** The poly-A tail at the 3' end determines the stability and lifespan of the mRNA in the cytoplasm. * **Codons:** mRNA is read in triplets called codons; the start codon is almost always **AUG** (coding for Methionine).

Q: What is an Okazaki fragment?

A DNA fragment with an RNA primer. ### Explanation **Correct Option: C (A DNA fragment with an RNA primer)** **Concept:** DNA replication is **semidiscontinuous**. While the leading strand is synthesized continuously, the **lagging strand** is synthesized in short, discontinuous segments known as **Okazaki fragments**. DNA polymerase requires a free 3'-OH group to initiate synthesis, which is provided by an **RNA primer** (synthesized by the enzyme Primase). Therefore, each Okazaki fragment begins with a short RNA primer (approx. 10 nucleotides) followed by a stretch of newly synthesized DNA (approx. 100–200 nucleotides in eukaryotes). Eventually, the RNA primers are removed by DNA Polymerase I (in prokaryotes) or RNase H/FEN1 (in eukaryotes) and replaced with DNA. **Why other options are incorrect:** * **Option A & B:** These are incomplete. An Okazaki fragment is a hybrid molecule during the replication process; it is not purely DNA or purely RNA until the maturation phase. * **Option D:** This is biologically reversed. Primers are always RNA (in natural replication), and the elongation is always DNA. --- ### High-Yield Clinical Pearls for NEET-PG: * **Directionality:** Okazaki fragments are synthesized in the **5' → 3' direction**, even though the lagging strand overall grows in the 3' → 5' direction relative to the replication fork. * **DNA Ligase:** This enzyme is responsible for joining Okazaki fragments by forming phosphodiester bonds. It is the "glue" of the lagging strand. * **Clinical Correlation:** Deficiencies in enzymes involved in Okazaki fragment processing (like **DNA Ligase I**) can lead to clinical conditions like **Bloom Syndrome** (characterized by genomic instability and sister chromatid exchanges). * **Length:** Okazaki fragments are significantly shorter in eukaryotes (100–200 bp) compared to prokaryotes (1000–2000 bp).

Q: Lac operon transcription is induced by what substance?

An inducer without glucose. The **Lac Operon** is a classic model of prokaryotic gene regulation, functioning as an inducible system that ensures bacteria only expend energy to metabolize lactose when it is present and glucose is absent. ### **Why "An inducer without glucose" is correct:** Transcription of the Lac operon requires two simultaneous conditions: 1. **Presence of an Inducer (Allolactose):** The inducer binds to the **repressor protein**, causing it to detach from the operator site. This "unlocks" the gene for transcription. 2. **Absence of Glucose (Positive Control):** When glucose levels are low, **Adenylate Cyclase** is active, increasing **cAMP** levels. cAMP binds to the **Catabolite Activator Protein (CAP)**. The cAMP-CAP complex then binds to the promoter, acting as a "gas pedal" to recruit RNA polymerase for high-level transcription. ### **Why other options are incorrect:** * **A & B (Glucose present):** Glucose is the preferred energy source. If glucose is present, cAMP levels remain low. Without the cAMP-CAP complex, RNA polymerase cannot bind efficiently to the promoter (Catabolite Repression), even if an inducer is present. * **D:** Transcription is highly specific; it is inhibited by the presence of glucose and the absence of an inducer. ### **High-Yield Clinical Pearls for NEET-PG:** * **Inducer:** The natural inducer is **allolactose** (a lactose metabolite). In laboratory settings, **IPTG** (Isopropyl β-D-1-thiogalactopyranoside) is used as a "gratuitous inducer" because it isn't metabolized. * **Diauxic Growth:** If both glucose and lactose are provided, the bacteria show a biphasic growth curve—using glucose first, followed by a lag phase, then using lactose. * **Constitutive Mutations:** Mutations in the **i gene** (repressor) or the **operator** can lead to "constitutive expression," where the operon is always "on" regardless of inducer presence.

Q: In DNA, the coding region reads 5'-CGT-3'. This would code in the RNA as:

5'-CGU-3'. ### Explanation **1. Why Option A is Correct:** In molecular biology, DNA consists of two strands: the **coding (sense) strand** and the **template (antisense) strand**. * The **coding strand** (5' to 3') has the same sequence and polarity as the resulting mRNA, with the sole exception that **Thymine (T)** in DNA is replaced by **Uracil (U)** in RNA. * The **template strand** (3' to 5') is the one actually read by RNA polymerase to synthesize mRNA via complementary base pairing. Since the question provides the **coding region** as 5'-CGT-3', the mRNA will be identical in sequence and direction, replacing T with U. Thus, 5'-CGT-3' (DNA) becomes **5'-CGU-3' (RNA)**. **2. Why Other Options are Incorrect:** * **Option B (5'-GCA-3'):** This is the sequence of the **template strand** (complementary to the coding strand). While RNA polymerase uses this as a template, the resulting RNA is complementary to it, not identical to it. * **Option C (5'-ACG-3'):** This represents the sequence if read backward or incorrectly transcribed without maintaining polarity. * **Option D (5'-UGC-3'):** This is the **anticodon** sequence (found on tRNA) that would pair with the mRNA codon 5'-CGU-3'. **3. NEET-PG High-Yield Pearls:** * **The "Golden Rule":** mRNA sequence = Coding strand sequence (T $\rightarrow$ U). * **Directionality:** RNA synthesis always occurs in the **5' $\rightarrow$ 3' direction**. * **Template vs. Coding:** The template strand is also called the **non-coding** or **antisense** strand. The coding strand is also called the **sense** strand. * **Clinical Correlation:** Many antibiotics (like Rifampicin) and toxins (like $\alpha$-amanitin) act by inhibiting RNA polymerase, preventing this transcription process.

Q: A dideoxynucleotide does not contain which of the following?

2' and 3' OH groups. **Explanation:** The core concept here is the structure of nucleotides used in DNA synthesis versus DNA sequencing. A standard **Deoxynucleotide (dNTP)** lacks an -OH group at the 2' position but possesses a hydroxyl (-OH) group at the 3' position, which is essential for forming phosphodiester bonds. **1. Why Option A is Correct:** A **Dideoxynucleotide (ddNTP)** is a synthetic nucleotide that lacks hydroxyl groups at **both the 2' and 3' positions** (hence "di-deoxy"). In DNA synthesis, the 3' -OH group acts as a nucleophile to attack the incoming nucleotide. Without the 3' -OH group, no further nucleotides can be added to the chain. This leads to **obligatory chain termination**, a principle utilized in Sanger sequencing. **2. Analysis of Incorrect Options:** * **Options B, C, and D:** These are incorrect because all nucleotides (including ddNTPs) must retain their **5' group** (usually attached to phosphates) to initiate a bond with the previous nucleotide, and the **4' position** in the pentose sugar ring does not typically carry a free hydroxyl group in this context. The specific modification that defines a "dideoxy" nucleotide is strictly at the 2' and 3' carbons. **3. NEET-PG High-Yield Pearls:** * **Sanger Sequencing:** Also known as the "Dideoxy chain termination method." It relies on ddNTPs to stop synthesis at specific bases. * **Mechanism of Action:** Drugs like **Zidovudine (AZT)** and **Didanosine (ddI)** used in HIV treatment are nucleoside reverse transcriptase inhibitors (NRTIs). They function similarly to ddNTPs by lacking a 3' -OH group, thereby terminating the viral DNA chain. * **Structure:** Remember: Ribose (RNA) has 2' and 3' -OH; Deoxyribose (DNA) has only 3' -OH; Dideoxyribose (Sequencing/Drugs) has neither.

Question 1

The BRCA1 gene is located on which chromosome?

Accepted Answer

Chromosome 17

Answer

Chromosome 13

Answer

Chromosome 11

Answer

Chromosome 22

Question 2

What is true regarding cytoplasmic messenger RNA (mRNA)?

Accepted Answer

It is chiefly translated from nuclear DNA.

Answer

The sugar component is deoxyribose.

Answer

Thymine is present in place of uracil.

Answer

Its molecular weight is greater than heterogeneous nuclear RNA (hnRNA).

Question 3

What is an Okazaki fragment?

Accepted Answer

A DNA fragment with an RNA primer

Answer

A DNA fragment

Answer

An RNA fragment

Answer

An RNA fragment with a DNA primer

Question 4

Lac operon transcription is induced by what substance?

Accepted Answer

An inducer without glucose

Answer

Glucose

Answer

Glucose with an inducer

Answer

All of the above

Question 5

Which of the following conditions is associated with hnRNA?

Accepted Answer

Spinal muscular atrophy

Answer

Sickle cell disease

Answer

Huntington's chorea

Answer

Alpha-thalassemia

Question 6

In DNA, the coding region reads 5'-CGT-3'. This would code in the RNA as:

Accepted Answer

5'-CGU-3'

Answer

5'-GCA-3'

Answer

5'-ACG-3'

Answer

5'-UGC-3'

Question 7

Thalassemia occurs due to which type of mutations?

Accepted Answer

Splicing mutations

Answer

Missense mutations

Answer

Transition mutations

Answer

All of the above

Question 8

In a DNA molecule, one strand contains 20 thymine (T), 25 cytosine (C), 30 guanine (G), and 22 adenine (A) residues. How many of each base pair are found in the complete double-stranded molecule?

Accepted Answer

T=42, C=55, G=55, A=42

Answer

T=44, C=60, G=50, A=40

Answer

T=22, C=30, G=25, A=20

Answer

T=40, C=50, G=60, A=44

Question 9

Which statement regarding nucleotide repair mechanisms is FALSE?

Accepted Answer

Mismatch repair primarily occurs during the G1 phase of the cell cycle.

Answer

UV radiation exposure leads to the formation of thymine-thymine dimers.

Answer

MUTYH-associated polyposis is associated with base excision repair.

Answer

Base excision repair mainly occurs in the G1 phase.

Question 10

A dideoxynucleotide does not contain which of the following?

Accepted Answer

2' and 3' OH groups

Answer

3' and 4' OH groups

Answer

4' and 5' OH groups

Answer

2' and 5' OH groups

Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics — MCQs

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