Chapter 5: Molecular Basis of Inheritance Case Study Questions | biology Class 12 CBSE

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Chapter 5: Molecular Basis of Inheritance Case Study Questions | biology Class 12 CBSE | ByteNotes.in

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Ramesh, a Class 12 student, was studying the double helix structure of DNA. His teacher explained that James Watson and Francis Crick proposed the double helix model in 1953 based on X-ray diffraction data. The model showed that two polynucleotide chains are coiled in a right-handed fashion with a pitch of 3.4 nm and approximately 10 base pairs per turn. The distance between consecutive base pairs is 0.34 nm. The backbone is formed by sugar and phosphate groups, while nitrogenous bases project inward. Adenine pairs with Thymine via two hydrogen bonds and Guanine pairs with Cytosine via three hydrogen bonds. Erwin Chargaff's rule stated that in any double-stranded DNA, the ratio of A to T and G to C is always equal to one. This base pairing makes the two strands complementary to each other.

Question 1: State Chargaff's rule and explain its significance in DNA structure.

Chargaff's rule states that in double-stranded DNA, the amount of Adenine equals Thymine (A=T) and the amount of Guanine equals Cytosine (G=C).
This confirms complementary base pairing between the two antiparallel strands of the DNA double helix.

Question 2: Why does the distance between the two polynucleotide chains remain almost constant throughout the DNA double helix?

A purine always pairs with a pyrimidine — Adenine (purine) with Thymine (pyrimidine) and Guanine (purine) with Cytosine (pyrimidine).
Since purines are larger and pyrimidines are smaller, a purine-pyrimidine pairing maintains a uniform width of approximately 2 nm between the two strands.

Question 3: Describe the salient features of the double helix structure of DNA.

It is made of two polynucleotide chains with a sugar-phosphate backbone; bases project inward.
The two chains have antiparallel polarity — one runs 5'→3' and the other 3'→5'.
Bases in opposite strands are paired through hydrogen bonds: A-T (2 H-bonds) and G-C (3 H-bonds).
The chains are coiled in a right-handed fashion; pitch is 3.4 nm, ~10 bp per turn; distance between bp is 0.34 nm.
Base stacking in addition to H-bonds provides stability to the helical structure.

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Case Set 2

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A researcher was investigating the genetic material in bacteria. Frederick Griffith's 1928 experiment with Streptococcus pneumoniae showed that heat-killed S-strain bacteria could transform R-strain bacteria into virulent S-strain organisms. Later, Avery, MacLeod and McCarty (1933–44) purified biochemicals from heat-killed S cells and tested which one caused transformation. They found that only DNA caused transformation. Treatment with DNase inhibited transformation, while proteases and RNases had no effect. This conclusively pointed to DNA as the transforming principle. Although this strongly suggested DNA as the genetic material, complete proof came from the Hershey–Chase experiment in 1952 using radioactively labelled bacteriophages. Their experiment showed that only radioactive phosphorus (marking DNA) entered bacterial cells, while radioactive sulfur (marking protein) did not.

Question 1: What was the conclusion of Griffith's experiment and what term did he use for the phenomenon observed?

Griffith concluded that some 'transforming principle' from heat-killed S bacteria converted R bacteria into virulent S bacteria.
This phenomenon of change in character of one organism by the genetic material of another was called transformation.

Question 2: How did Avery, MacLeod and McCarty determine that DNA is the transforming principle?

They purified proteins, DNA, and RNA separately from heat-killed S-strain and tested each for transformation.
Only purified DNA caused transformation; DNase treatment inhibited it, while protease and RNase treatments did not — proving DNA is the transforming principle.

Question 3: Describe the Hershey–Chase experiment and explain how it conclusively proved DNA to be the genetic material.

Hershey and Chase grew bacteriophages on radioactive 32P (labels DNA) and 35S (labels protein) media separately.
Radioactive phages were allowed to infect E. coli; then viral coats were separated by blending and centrifugation.
Bacteria infected with 32P-labelled phages were radioactive, showing DNA entered the bacteria.
Bacteria infected with 35S-labelled phages were not radioactive, showing protein did not enter.
This proved unequivocally that DNA, not protein, is the genetic material passed from virus to bacteria.

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Case Set 3

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During a practical class, students observed that a human cell nucleus, roughly 10 micrometres in diameter, contains approximately 2.2 metres of DNA. Their teacher explained the mechanism of DNA packaging. In eukaryotes, positively charged histone proteins, rich in lysine and arginine, form an octamer. Negatively charged DNA wraps around the histone octamer to form a nucleosome, which contains about 200 bp of DNA. Nucleosomes appear as 'beads-on-string' under an electron microscope, forming chromatin. Chromatin is further compacted into chromatin fibers, and at metaphase, these condense into chromosomes. Non-histone chromosomal (NHC) proteins help in higher-level packaging. Loosely packed euchromatin is transcriptionally active, while densely packed heterochromatin is transcriptionally inactive.

Question 1: What is a nucleosome? Name the proteins around which DNA is wrapped.

A nucleosome is the basic unit of chromatin packaging, consisting of DNA wrapped around a histone octamer.
It is made of 8 histone molecules and contains approximately 200 bp of DNA.

Question 2: Differentiate between euchromatin and heterochromatin.

Euchromatin is loosely packed, stains lightly and is transcriptionally active.
Heterochromatin is densely packed, stains darkly and is transcriptionally inactive.

Question 3: Describe the levels of DNA packaging from nucleosome to metaphase chromosome.

DNA wraps around histone octamer (~200 bp per nucleosome) forming nucleosomes — appearing as 'beads-on-string'.
Nucleosomes are compacted further into chromatin fibers.
Chromatin fibers are coiled and condensed further, aided by NHC proteins.
At metaphase stage of cell division, chromatin is maximally condensed to form distinct chromosomes visible under a microscope.

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In 1958, Matthew Meselson and Franklin Stahl performed an elegant experiment to prove the mode of DNA replication. They grew E. coli cells in a medium containing 15N (heavy nitrogen) for several generations, so all DNA was heavy (15N-15N). Then cells were transferred to a 14N (normal nitrogen) medium. After one generation (20 min), DNA extracted and centrifuged in a CsCl density gradient showed a single band of intermediate density — a hybrid (15N-14N) DNA. After two generations (40 min), two equal bands appeared — one hybrid and one light (14N-14N). These results supported the semiconservative model of replication proposed by Watson and Crick, where each new DNA molecule retains one parental strand and one newly synthesised strand.

Question 1: What was the purpose of using 15N in the Meselson and Stahl experiment?

15N is a heavy isotope of nitrogen that gets incorporated into newly synthesised DNA.
This allowed heavy DNA to be distinguished from normal 14N-DNA by centrifugation in a CsCl density gradient based on differences in density.

Question 2: What result would be expected after three generations (80 minutes) in Meselson and Stahl's experiment?

After 80 minutes (3 generations), there would be 8 DNA molecules total: 2 hybrid (15N-14N) and 6 light (14N-14N).
The ratio of hybrid to light DNA would be 1:3, with both bands visible in CsCl gradient — hybrid band less intense than light band.

Question 3: Explain the semiconservative mode of DNA replication. How do the results of Meselson and Stahl's experiment support this model?

In semiconservative replication, the two strands of parental DNA separate and each acts as a template for a new complementary strand; each daughter DNA has one old and one new strand.
After one generation in 14N medium: one parental (15N) strand + one new (14N) strand → hybrid density DNA — this matched with one intermediate band seen in CsCl gradient.
After two generations: two hybrid and two light molecules — seen as equal hybrid and light bands — confirming each parental strand continued to be conserved in progeny.
This progressive shift in density bands provided strong experimental evidence for the semiconservative mechanism.

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Case Set 5

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In a biology lecture, students learnt about the process of transcription. The teacher described that in bacteria, a single DNA-dependent RNA polymerase binds to the promoter region with the help of sigma (σ) factor to initiate transcription, elongates the RNA chain and terminates with the help of rho (ρ) factor. However, in eukaryotes, there are three types of RNA polymerases with distinct roles: RNA polymerase I transcribes rRNA (28S, 18S, 5.8S); RNA polymerase II transcribes hnRNA (precursor of mRNA); and RNA polymerase III transcribes tRNA, 5S rRNA and snRNA. The primary transcript (hnRNA) in eukaryotes undergoes processing — capping (addition of 7-methyl guanosine triphosphate at 5' end), tailing (addition of 200–300 adenylate residues at 3' end) and splicing (removal of introns and joining of exons) — before becoming functional mRNA.

Question 1: What is the role of the sigma factor and rho factor during transcription in bacteria?

The sigma (σ) factor associates transiently with RNA polymerase to help it recognise the promoter and initiate transcription.
The rho (ρ) factor helps in termination of transcription by causing the nascent RNA and RNA polymerase to dissociate from the DNA.

Question 2: Name the three types of RNA polymerases in eukaryotes and state what each transcribes.

RNA polymerase I: transcribes rRNAs (28S, 18S, and 5.8S).
RNA polymerase II: transcribes hnRNA (precursor of mRNA); RNA polymerase III: transcribes tRNA, 5S rRNA, and snRNAs.

Question 3: Describe the post-transcriptional modifications that hnRNA undergoes to become a functional mRNA in eukaryotes.

Capping: An unusual nucleotide — 7-methyl guanosine triphosphate — is added to the 5'-end of hnRNA, protecting it from degradation.
Tailing: Adenylate residues (200–300) are added at the 3'-end in a template-independent manner (poly-A tail); aids in stability and export.
Splicing: Non-coding introns are removed and coding exons are joined in a specific order to produce the mature, functional mRNA.
The processed mRNA is then transported out of the nucleus to the cytoplasm for translation.

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Chapter 5: Molecular Basis of Inheritance | Case Study Questions

Passage

Question 1: State Chargaff's rule and explain its significance in DNA structure.

Question 2: Why does the distance between the two polynucleotide chains remain almost constant throughout the DNA double helix?

Question 3: Describe the salient features of the double helix structure of DNA.

Passage

Question 1: What was the conclusion of Griffith's experiment and what term did he use for the phenomenon observed?

Question 2: How did Avery, MacLeod and McCarty determine that DNA is the transforming principle?

Question 3: Describe the Hershey–Chase experiment and explain how it conclusively proved DNA to be the genetic material.

Passage

Question 1: What is a nucleosome? Name the proteins around which DNA is wrapped.

Question 2: Differentiate between euchromatin and heterochromatin.

Question 3: Describe the levels of DNA packaging from nucleosome to metaphase chromosome.

Passage

Question 1: What was the purpose of using 15N in the Meselson and Stahl experiment?

Question 2: What result would be expected after three generations (80 minutes) in Meselson and Stahl's experiment?

Question 3: Explain the semiconservative mode of DNA replication. How do the results of Meselson and Stahl's experiment support this model?

Passage

Question 1: What is the role of the sigma factor and rho factor during transcription in bacteria?

Question 2: Name the three types of RNA polymerases in eukaryotes and state what each transcribes.

Question 3: Describe the post-transcriptional modifications that hnRNA undergoes to become a functional mRNA in eukaryotes.

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