Chapter 6: Molecular Basis of Inheritance – Case-Based Questions with Answers
CBSE Class 12 Biology Case-Based Questions (NCERT): Molecular Basis of Inheritance
Course & Examination Details
- Course: CBSE Class 12 Biology
- Unit: Unit II – Genetics and Evolution
- Chapter: Chapter 6 – Molecular Basis of Inheritance
- Prescribed Textbook: NCERT Biology Class XII
- Examination: CBSE Class 12 Board Examination
- Question Type: Case-Based / Source-Based Questions
Section A: DNA as Genetic Material
Case 1
Griffith observed that non-virulent bacteria became virulent when mixed with heat-killed virulent bacteria.
Q. What conclusion was drawn from this experiment?
Answer: Griffith concluded that a “transforming principle” transferred genetic information from dead virulent bacteria to live non-virulent bacteria, altering their phenotype.
Case 2
A student studies Avery–MacLeod–McCarty experiments.
Q. How did these experiments establish DNA as genetic material?
Answer: They showed that destroying DNA prevented transformation, while destroying proteins or RNA did not, proving DNA is the genetic material.
Case 3
Hershey and Chase labelled bacteriophages with radioactive isotopes.
Q. Which molecule entered the bacterial cell?
Answer: Radioactively labelled DNA entered the bacterial cell, confirming DNA as the genetic material.
Case 4
RNA viruses show rapid mutation rates.
Q. Why is RNA suitable as genetic material in viruses?
Answer: RNA can replicate and mutate rapidly, helping viruses adapt quickly to host environments despite being less stable than DNA.
Case 5
DNA is described as chemically stable.
Q. Give one structural reason for this stability.
Answer: Absence of a hydroxyl group in deoxyribose sugar makes DNA less reactive and more stable.
Section B: Structure of DNA and RNA
Case 6
A researcher analyses base composition of DNA.
Q. Which rule explains base equality?
Answer: Chargaff’s rule explains that adenine equals thymine and guanine equals cytosine in DNA.
Case 7
Two DNA strands run in opposite directions.
Q. What is this arrangement called?
Answer: This arrangement is called antiparallel orientation of DNA strands.
Case 8
Hydrogen bonds break during replication.
Q. Which base pair requires more energy to separate?
Answer: Guanine–cytosine base pairs require more energy due to three hydrogen bonds.
Case 9
RNA contains uracil instead of thymine.
Q. How does this affect RNA structure?
Answer: Replacement of thymine with uracil makes RNA chemically less stable and suitable for short-term information transfer.
Case 10
A nucleotide is formed.
Q. Name its components.
Answer: A nucleotide consists of a nitrogenous base, pentose sugar, and phosphate group.
Section C: Replication of DNA
Case 11
DNA replication produces two identical DNA molecules.
Q. What type of replication is this?
Answer: It is semi-conservative replication where each daughter DNA contains one parental strand.
Case 12
DNA polymerase cannot start synthesis independently.
Q. Which enzyme helps initiate replication?
Answer: Primase synthesises RNA primers required to initiate DNA synthesis.
Case 13
Lagging strand synthesis occurs in fragments.
Q. Why does this happen?
Answer: DNA synthesis occurs only in the 5′→3′ direction, causing discontinuous synthesis on the lagging strand.
Case 14
Small DNA fragments are joined into a continuous strand.
Q. Name the enzyme involved.
Answer: DNA ligase joins Okazaki fragments into a continuous DNA strand.
Case 15
Replication fork progresses smoothly.
Q. Which enzyme relieves supercoiling stress?
Answer: Topoisomerase relieves supercoiling ahead of the replication fork.
Section D: Transcription
Case 16
Only one DNA strand is transcribed.
Q. Why is transcription asymmetric?
Answer: To prevent formation of double-stranded RNA that would interfere with translation.
Case 17
RNA polymerase binds to a specific DNA region.
Q. What is this region called?
Answer: The region is called the promoter.
Case 18
RNA synthesis stops at a particular site.
Q. What is this site known as?
Answer: This site is called the terminator region.
Case 19
RNA contains uracil instead of thymine.
Q. What advantage does this provide?
Answer: Uracil allows RNA to function as a temporary information carrier.
Case 20
mRNA is produced in prokaryotes.
Q. Why is processing not required?
Answer: Prokaryotic genes lack introns, so mRNA does not require splicing.
Section E: Genetic Code and Translation
Case 21
AUG initiates protein synthesis.
Q. What is its significance?
Answer: AUG acts as the start codon and codes for methionine, initiating translation.
Case 22
Multiple codons code for the same amino acid.
Q. What is this feature called?
Answer: This feature is called degeneracy of the genetic code.
Case 23
A stop codon is encountered during translation.
Q. What happens next?
Answer: Protein synthesis stops and the completed polypeptide chain is released.
Case 24
tRNA brings amino acids to ribosomes.
Q. What structural feature enables this?
Answer: The anticodon loop and acceptor arm of tRNA enable codon recognition and amino acid attachment.
Case 25
Ribosomes facilitate peptide bond formation.
Q. Which RNA plays a catalytic role?
Answer: rRNA plays a catalytic role in peptide bond formation.