Competency-Based Question Bank: Molecular Basis of Inheritance

Section A: Competency-Based Multiple Choice Questions (Application & Analysis)

1. Analyze the Experiment: In the Hershey-Chase experiment, bacteriophages were labeled with radioactive sulfur (³⁵S) and allowed to infect bacteria. After blending and centrifugation, where would you expect to find the radioactivity? a) Inside the bacterial pellet. b) In the supernatant fluid. c) Inside the bacterial DNA. d) Attached to the bacterial cell wall. Answer: b) In the supernatant fluid. Explanation: Sulfur is part of proteins (viral coat). The viral coats do not enter the bacteria; they remain outside. Centrifugation separates the heavy bacteria (pellet) from the light liquid containing the viral coats (supernatant).

2. Calculate: A double-stranded DNA molecule has 20% Adenine. According to Chargaff's rule, what is the percentage of Cytosine? a) 20% b) 30% c) 40% d) 80% Answer: b) 30% Explanation: A=T, so T=20%. A+T=40%. The remaining 60% must be G+C. Since G=C, Cytosine is 60/2 = 30%.

3. Predict the Consequence: A mutation changes a codon in mRNA from UAC (Tyrosine) to UAA. What will be the immediate effect on translation? a) The protein will be longer than normal. b) The protein synthesis will stop prematurely (Nonsense mutation). c) A different amino acid will be incorporated (Missense mutation). d) There will be no change in the protein (Silent mutation). Answer: b) The protein synthesis will stop prematurely (Nonsense mutation). Explanation: UAA is a stop codon. If a sense codon is changed to a stop codon, translation ends there, resulting in a truncated, usually non-functional protein.

4. Assertion (A): The leading strand in DNA replication is synthesized continuously, while the lagging strand is synthesized discontinuously. Reason (R): DNA Polymerase can only synthesize DNA in the 5' → 3' direction. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Because DNA polymerase adds nucleotides only to the 3' end, one strand can follow the replication fork (Leading), while the other must be made in short pieces (Lagging/Okazaki) as the fork opens.

5. Molecular Mechanism: Why is RNA less stable than DNA? a) RNA is single-stranded and lacks thymine. b) RNA has a 2'-OH group in ribose which makes it reactive and catalytic. c) RNA contains Uracil which pairs with Adenine less strongly. d) RNA polymerase is error-prone. Answer: b) RNA has a 2'-OH group in ribose which makes it reactive and catalytic. Explanation: The extra hydroxyl group in ribose makes the sugar-phosphate backbone vulnerable to hydrolysis, making RNA a transient molecule compared to the stable DNA.

6. Gene Regulation: In the Lac Operon, what happens if both Glucose and Lactose are present in the medium? a) The operon is fully active. b) The operon is inactive or expressed at a very low level (Catabolite repression). c) The repressor binds tightly to the operator. d) The lacZ gene is mutated. Answer: b) The operon is inactive or expressed at a very low level. Explanation: This is "Catabolite Repression". Bacteria prefer glucose. When glucose is present, cAMP levels are low, so the activator protein (CAP) cannot bind to the promoter, keeping transcription low.

7. Identify the Process: A researcher isolates a mature mRNA from a eukaryote and compares it with the genomic DNA gene sequence. She notices that the mRNA is much shorter than the gene. This is due to: a) Capping b) Tailing c) Splicing (removal of introns) d) Mutation Answer: c) Splicing (removal of introns) Explanation: Eukaryotic genes are "split". They contain non-coding segments called Introns which are removed during processing, leaving only the coding Exons in the mRNA.

8. Sequence Analysis: If the sequence of the coding strand of DNA is 5'-ATGCCTAG-3', what will be the sequence of the transcribed mRNA? a) 3'-TACGGATC-5' b) 5'-AUGCCUAG-3' c) 5'-TACGGATC-3' d) 3'-AUGCCUAG-5' Answer: b) 5'-AUGCCUAG-3' Explanation: The mRNA sequence is identical to the Coding Strand (Sense strand), with the only difference being that Thymine (T) is replaced by Uracil (U).

9. Forensic Application: Two individuals are suspected in a crime. DNA fingerprinting using VNTRs is performed. Why are VNTRs suitable for identification? a) They code for essential proteins. b) They are highly conserved across the population. c) They show high degree of polymorphism (variation in repeat number) between individuals. d) They are mitochondrial DNA sequences. Answer: c) They show high degree of polymorphism... Explanation: VNTRs are repetitive sequences where the number of repeats varies significantly between people, creating a unique "barcode" or fingerprint.

10. Central Dogma: Which of the following violates the classic Central Dogma of molecular biology? a) Replication of DNA in E. coli. b) Transcription of tRNA in humans. c) Reverse transcription in HIV. d) Translation of insulin protein. Answer: c) Reverse transcription in HIV. Explanation: The Central Dogma usually flows DNA -> RNA. HIV uses Reverse Transcriptase to flow RNA -> DNA.

11. Structural Biology: The 'beads-on-string' structure seen under an electron microscope refers to: a) Ribosomes on mRNA (Polysome). b) Nucleosomes in chromatin. c) Spliceosomes on hnRNA. d) DNA polymerase on the replication fork. Answer: b) Nucleosomes in chromatin. Explanation: DNA wraps around histone octamers to form nucleosomes ("beads"), connected by linker DNA ("string").

12. Genetic Code: The genetic code is "degenerate". This means: a) One codon codes for multiple amino acids. b) One amino acid is coded by multiple codons. c) The code is ambiguous. d) The code varies between species. Answer: b) One amino acid is coded by multiple codons. Explanation: There are 61 sense codons but only 20 amino acids. For example, Leucine has 6 different codons. This helps protect against mutations.

13. Experimental Logic: Meselson and Stahl used ¹⁵N and ¹⁴N to prove semi-conservative replication. If they had allowed the bacteria to divide for two generations in ¹⁴N medium (after growing in ¹⁵N), what ratio of DNA bands would they observe in the CsCl gradient? a) 100% Hybrid (¹⁵N-¹⁴N). b) 50% Hybrid, 50% Light (¹⁴N-¹⁴N). c) 25% Heavy, 75% Light. d) 50% Heavy, 50% Hybrid. Answer: b) 50% Hybrid, 50% Light. Explanation:

• Gen 0: 100% Heavy (15/15).
• Gen 1: 100% Hybrid (15/14).
• Gen 2: Two hybrid molecules split into four. Two get the '15' strand (become Hybrid 15/14) and two get the '14' strand (become Light 14/14).

14. Enzyme Function: Which enzyme is responsible for removing the RNA primer and replacing it with DNA nucleotides during replication in prokaryotes? a) DNA Polymerase I b) DNA Polymerase III c) Primase d) Ligase Answer: a) DNA Polymerase I Explanation: Pol I has 5' to 3' exonuclease activity which allows it to "chew" the RNA primer and fill the gap with DNA.

15. HGP: One of the surprising findings of the Human Genome Project was: a) Humans have over 100,000 genes. b) More than 98% of the genome does not code for proteins. c) Every individual has a completely unique genome sequence (0% similarity). d) Chromosome Y has the most genes. Answer: b) More than 98% of the genome does not code for proteins. Explanation: Only about 1.5% - 2% of our DNA actually codes for proteins (Exons); the rest is repetitive DNA, introns, and regulatory regions.

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Section B: Case-Study & Source-Based Questions

Case Study 1: The Inducible System

E. coli bacteria are growing in a culture medium containing glucose. A researcher suddenly adds lactose to the medium and removes glucose.

16. Explain: Before the addition of lactose, where was the Repressor protein located? a) Bound to the Promoter. b) Bound to the Operator. c) Bound to the structural genes. d) Free in the cytoplasm. Answer: b) Bound to the Operator. Explanation: When lactose is absent, the repressor protein binds to the operator, physically blocking RNA polymerase from transcribing the genes.

17. Mechanism: How did the addition of lactose trigger the production of the enzyme? a) Lactose bound to the DNA polymerase. b) Lactose (Allolactose) bound to the Repressor, causing it to change shape and release the Operator. c) Lactose mutated the lacI gene. d) Lactose acted as a co-factor for the enzyme. Answer: b) Lactose (Allolactose) bound to the Repressor... Explanation: Lactose acts as an Inducer. It binds to the repressor, changing its conformation so it can no longer bind to the operator. This "unblocks" the genes.

18. Application: If a mutation occurred in the lacI gene (regulatory gene) such that the Repressor protein could no longer bind to the Operator, what would be the status of the lac operon? a) Constitutively ON (always active). b) Permanently OFF. c) Inducible only by high concentrations of lactose. d) It would transcribe only lacZ but not lacY. Answer: a) Constitutively ON (always active). Explanation: If the "lock" (operator) cannot be "blocked" by the "gatekeeper" (repressor), the "factory" (structural genes) runs non-stop.

Case Study 2: The Mystery of the Shorter Protein

A scientist is studying a gene "X" in eukaryotic cells. The gene DNA is 5000 base pairs long. However, the mature mRNA found in the cytoplasm is only 1500 bases long.

19. Identify: What biological process accounts for the size difference between the gene and the mRNA? a) Replication error. b) RNA Splicing (Removal of Introns). c) Protein degradation. d) Translation termination. Answer: b) RNA Splicing. Explanation: Initial transcripts (hnRNA) contain both introns and exons. Splicing removes the large introns, leaving only the smaller exons.

20. Reasoning: Are the 3500 missing bases "junk" DNA? a) Yes, they have no function. b) No, they were Introns, which may have regulatory roles or allow for alternative splicing (evolutionary flexibility). c) They were signal peptides. d) They were destroyed by DNase. Answer: b) No, they were Introns... Explanation: Introns are not "junk"; they allow one gene to code for multiple proteins (Alternative splicing) and are involved in regulating expression.

21. Contrast: If this gene "X" were taken from the eukaryote and inserted directly into a bacterium (E. coli), would the bacterium produce the correct functional protein? a) Yes, the genetic code is universal. b) No, bacteria lack the machinery to splice out introns, so the protein would be incorrect/non-functional. c) Yes, but at a slower rate. d) No, bacteria cannot transcribe eukaryotic DNA. Answer: b) No, bacteria lack the machinery to splice out introns... Explanation: Bacteria are prokaryotes and don't have spliceosomes. They would translate the introns as well, creating a completely different and likely non-functional protein.

Case Study 3: DNA Fingerprinting in Forensics

A crime scene sample (blood) is collected. DNA is isolated and subjected to DNA Fingerprinting. Suspect B's banding pattern matches the crime scene sample perfectly.

22. Analyze: Which specific regions of the DNA are responsible for these unique banding patterns? a) The coding regions (Exons). b) The promoter regions. c) Variable Number Tandem Repeats (VNTRs) / Satellite DNA. d) The mitochondrial DNA. Answer: c) VNTRs / Satellite DNA. Explanation: These are repetitive non-coding DNA segments where the repeat count is highly individual-specific.

23. Technique: Which technique was likely used to multiply the small amount of DNA from the blood sample before testing? a) Gel Electrophoresis. b) PCR (Polymerase Chain Reaction). c) Southern Blotting. d) Centrifugation. Answer: b) PCR (Polymerase Chain Reaction). Explanation: PCR is the standard method for amplifying DNA from trace samples.

24. Evaluate: Can this technique distinguish between identical twins? a) Yes, their VNTRs are different. b) No, identical twins have identical DNA/VNTR patterns. c) Yes, but only using mitochondrial DNA. d) No, unless one has a mutation. Answer: b) No, identical twins have identical DNA/VNTR patterns. Explanation: Since identical twins come from the same zygote, their DNA sequences are identical. DNA fingerprinting cannot distinguish them (though epigenetics or environmental factors might differ).

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Section C: Creating, Designing & Critical Thinking

25. Designing an Experiment: You want to prove that DNA replication is semi-conservative in a human cell line (not bacteria).

• Task: Design an experiment using an analogue of Thymidine (e.g., BrdU). Answer:
• Protocol:
  1. Labeling: Grow human cells for one generation in a medium containing BrdU (which takes the place of Thymidine).
  2. Chase: Transfer cells to a normal medium (without BrdU).
  3. Visualization: Use a specific fluorescent stain that differentiates BrdU-containing strands from normal ones.
• Results: If semi-conservative, after one round, each chromosome will consist of one "old" (normal) and one "new" (BrdU) strand. Under the microscope, this appears as differential staining of sister chromatids (Harlequin chromosomes).

26. Visualizing the Process: Create a flowchart showing the "Central Dogma" in a eukaryotic cell vs. a retrovirus. Answer:

• Eukaryote: DNA --(Transcription)--> mRNA --(Translation)--> Protein.
• Retrovirus: viral RNA --(Reverse Transcription)--> viral DNA --(Integration into Host DNA) --(Transcription)--> mRNA --(Translation)--> Protein.
• Enzyme: The extra step uses Reverse Transcriptase.

27. Formulating a Hypothesis: Observation: The genetic code is "degenerate" (e.g., Leucine is coded by 6 different codons).

• Hypothesis: Formulate a hypothesis regarding the "protective" nature of degeneracy against mutations. Answer:
• Hypothesis: Degeneracy acts as a biological "safety buffer". If a mutation occurs in the 3rd base of a codon (the Wobble position), the degenerate code ensures that there is a high probability the same amino acid will still be incorporated (Silent mutation). This prevents changes in the protein's primary structure, maintaining its function despite DNA damage.

28. Decoding a Sequence: Template DNA strand: 3' – T A C C G A T C C A T T – 5'

• Task 1: Write the coding strand sequence.
• Task 2: Write the mRNA transcript.
• Task 3: Write the amino acid sequence. (AUG=Met, GCU=Ala, AGG=Arg, UAA=Stop). Answer:
• Task 1 (Coding): 5' - A T G G C T A G G T A A - 3'
• Task 2 (mRNA): 5' - A U G G C U A G G U A A - 3'
• Task 3 (Protein): Methionine - Alanine - Arginine - STOP.

29. Critical Analysis: "The Human Genome Project revealed that 99.9% of nucleotide bases are exactly the same in all people."

• Analysis: If we are so similar, why are we so different? Answer:
• immense diversity comes from:
  1. SNPs: Single Nucleotide Polymorphisms. The 0.1% difference represents ~3 million locations where individuals vary.
  2. Repetitive DNA: Variations in the number of repeats (VNTRs).
  3. Epigenetics: Genes might be the same, but they are "switched on/off" differently in different people due to environmental factors.

30. Scenario Analysis (Mutation): A frameshift mutation occurs near the end of a gene versus the beginning. Answer:

• Beginning: Much more deleterious. It shifts the reading frame for almost the entire protein, changing every subsequent amino acid and usually creating a premature stop codon. The protein will be completely different and non-functional.
• End: Less deleterious. Only the last few amino acids are changed. The majority of the protein remains intact and may still retain partial function.

31. Developing a Model: Design a physical analogy for the Lac Operon. Answer:

• The System: A locked library door (Operator) and a librarian (Repressor).
• OFF State: The librarian stands in front of the door, blocking students (RNA Pol) from entering to read books (Structural genes).
• ON State: A bowl of free pizza (Inducer/Lactose) is placed in the hall. The librarian leaves the door to go eat the pizza. Now the door is unblocked, and students can enter to read.