CBSE Class 12 Biology Molecular Basis of Inheritance Question Bank

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BIOLOGY

MOLECULAR BASIS OF INHERITANCE

1. If adenine constitutes 30% of an isolated DNA fragment, then what is the expected % of the base cytosine in it? [1 Mark]

Answer:
1. According to Chargaff's rules, the amount of adenine is equal to thymine (A = T) and the amount of guanine is equal to cytosine (G = C).
2. If adenine is 30%, thymine is also 30%, making a total of 60% for A and T.
3. The remaining 40% consists of guanine and cytosine combined, so cytosine constitutes 20%.

Teacher's Note:
a) Apply Chargaff's equivalence rule where purines equal pyrimidines.
b) Remember that total percentages of all four bases always add up to 100.

2. How does the flow of genetic information in HIV deviate from the central dogma proposed by Francis Crick? [1 Mark]

Answer:
1. Francis Crick's central dogma states that genetic information flows unidirectionally from DNA to mRNA to protein.
2. In HIV, which is a retrovirus, genetic information flows in reverse from RNA to DNA by the process of reverse transcription.

Teacher's Note:
a) Retroviruses use the enzyme reverse transcriptase to synthesize DNA from an RNA template.
b) This discovery proved that genetic information transfer is not strictly unidirectional.

3. How does HIV differ from a bacteriophage? [1 Mark]

Answer:
1. HIV is an animal retrovirus that infects human host cells (specifically helper T-lymphocytes) and contains single-stranded RNA.
2. A bacteriophage is a virus that infects bacterial cells and typically contains double-stranded DNA.

Teacher's Note:
a) Contrast host specificity and genetic material types clearly.
b) Bacteriophages have a protein coat with a tail structure, unlike retroviruses.

4. What is a nucleoid? [1 Mark]

Answer:
1. The nucleoid is the irregularly shaped region within a prokaryotic cell that contains all or most of the genetic material.
2. It is not surrounded by a nuclear membrane.

Teacher's Note:
a) Prokaryotes lack a true membrane-bound nucleus, hence the DNA is localized in the cytoplasm as a nucleoid.
b) The bacterial DNA is typically a single circular double-stranded molecule.

5. Name two amino acids that provide +ve charge to histone proteins. [1 Mark]

Answer:
1. Lysine and arginine.

Teacher's Note:
a) Histones are rich in basic amino acids containing positive side chains.
b) These positive charges help them bind tightly to the negatively charged DNA phosphate backbone.

6. Why do RNA viruses undergo mutation and evolution faster than most of the other viruses? [1 Mark]

Answer:
1. RNA is unstable compared to DNA and lacks a proofreading enzyme mechanism during replication.
2. Consequently, RNA viruses mutate at a much faster rate, leading to rapid evolution.

Teacher's Note:
a) Mention the lack of 2'-OH group stability and absence of proofreading activity.
b) This explains the rapid emergence of new viral strains like influenza or HIV.

7. Name the enzyme involved in the continuous replication of DNA strand. Mention the polarity of the template strand. [1 Mark]

Answer:
1. Enzyme: DNA polymerase.
2. Polarity of the template strand: 3' to 5'.

Teacher's Note:
a) DNA replication proceeds only in the 5' to 3' direction.
b) Therefore, the continuous strand is synthesized on the template with 3' to 5' polarity.

8. What is a cistron? [1 Mark]

Answer:
1. A cistron is a segment of DNA encoding a polypeptide.
2. It is functionally equivalent to a gene.

Teacher's Note:
a) Prokaryotic genes are often polycistronic, while eukaryotic genes are monocistronic.
b) Structural genes in a transcription unit are referred to as cistrons.

9. Which one out of Rho factor and sigma factor, acts as initiation factor during transcription in prokaryote? [1 Mark]

Answer:
1. Sigma factor.

Teacher's Note:
a) The sigma factor associates with RNA polymerase to recognize the promoter site for initiation.
b) The Rho factor is responsible for the termination of transcription.

10. Write the function of RNA polymerase II. [1 Mark]

Answer:
1. RNA polymerase II transcribes precursor of mRNA, which is heterogeneous nuclear RNA (hnRNA).

Teacher's Note:
a) Eukaryotes have three distinct nuclear RNA polymerases with specialized roles.
b) RNA polymerase I transcribes rRNA, and RNA polymerase III transcribes tRNA and 5S rRNA.

11. What is meant by hnRNA? [1 Mark]

Answer:
1. Heterogeneous nuclear RNA (hnRNA) is the primary transcript synthesized by RNA polymerase II in eukaryotes.
2. It undergoes processing such as splicing, capping, and tailing to become functional mRNA.

Teacher's Note:
a) hnRNA contains both non-coding introns and coding exons.
b) Splicing removes introns and joins exons together.

12. (a) .Differentiate between euchromatin and heterochromatin.
(b) What is the function of non-histone chromosomal proteins? [2 Marks]

Answer:
1. Euchromatin is loosely packed, stains lightly, and is transcriptionally active, whereas heterochromatin is densely packed, stains dark, and is transcriptionally inactive.
2. Non-histone chromosomal (NHC) proteins are required for packaging chromatin at a higher level and assist in gene regulation.

Teacher's Note:
a) Contrast packing density and transcriptional activity clearly.
b) NHC proteins include structural and regulatory proteins.

13. Why is DNA a better genetic material when compared to RNA? [2 Marks]

Answer:
1. DNA is chemically less reactive and structurally more stable due to the presence of thymine instead of uracil and lack of a 2'-OH group on its sugar.
2. DNA has a double-stranded nature and repair mechanisms that protect genetic information from mutation, making it superior for long-term storage.

Teacher's Note:
a) Highlight chemical stability and structural robustness as key factors.
b) RNA is catalytic and labile, making it prone to degradation.

14. Answer the following questions based on Meselson and Stahl's experiment:
a) Write the name of the chemical substance used as a source of nitrogen in the experiment by them.
b) Why did the scientists synthesize the light and the heavy DNA molecules in the organism in the experiment?
c) How did the scientists synthesize make it possible to distinguish the heavy DNA molecule from the light DNA molecule? Explain.
d) Write the conclusion the scientists arrived at, after completing the experiment. [4 Marks]

Answer:
1. (a) Ammonium chloride (\(NH_4Cl\)) containing \({}^{15}N\) and \({}^{14}N\).
2. (b) To track the mode of DNA replication by distinguishing parental strands from newly synthesized progeny strands.
3. (c) They used \({}^{15}N\), a heavy isotope of nitrogen, which gets incorporated into newly synthesized bacterial DNA, making it denser than normal \({}^{14}N\) DNA, allowing separation by CsCl density gradient centrifugation.
4. (d) They concluded that DNA replication is semiconservative, where each newly formed DNA molecule has one parental and one newly synthesized strand.

Teacher's Note:
a) This classic experiment used Escherichia coli and density gradient centrifugation.
b) Emphasize the semi-conservative nature of DNA replication as the core takeaway.

15. Monocistronic structural genes in eukaryotes have interrupted coding sequences, Explain. How are they different in prokaryotes? [2 Marks]

Answer:
1. Eukaryotic genes are interrupted because they contain coding regions (exons) interspersed with non-coding regions (introns).
2. In contrast, prokaryotic genes are continuous and lack introns, meaning their coding sequences are uninterrupted.

Teacher's Note:
a) Interrupted genes require post-transcriptional splicing to remove introns.
b) Prokaryotes lack splicing machinery because their genes are uninterrupted.

16 Describe the initiation process of transcription in bacteria? [2 Marks]

Answer:
1. The RNA polymerase enzyme binds to the promoter site on the DNA template with the help of the sigma factor.
2. This binding initiates the opening of the DNA double helix to form a transcription bubble for RNA synthesis.

Teacher's Note:
a) Focus on the role of the promoter and the sigma factor.
b) Transcription begins only after successful enzyme-promoter complex formation.

17 Explain the elongation process of transcription in bacteria? [2 Marks]

Answer:
1. RNA polymerase uses nucleoside triphosphates as substrates and polymerizes them in a template-dependent fashion following the rule of complementarity.
2. The enzyme moves along the DNA template, causing the elongation of the RNA chain in the 5' to 3' direction.

Teacher's Note:
a) Sigma factor separates shortly after initiation begins, leaving the core enzyme to handle elongation.
b) The process is highly processive until termination signals are reached.

18 Describe the termination process of transcription in bacterium? [2 Marks]

Answer:
1. When RNA polymerase reaches the terminator region on the DNA template, the process halts.
2. The newly synthesized RNA molecule and the core enzyme are released, assisted by the Rho factor in specific instances.

Teacher's Note:
a) Termination can be rho-dependent or rho-independent.
b) The release of RNA concludes the transcription cycle in prokaryotes.

19 Explain the role of \({}^{35}S\) and \({}^{32}P\) in the experiments conducted by Hershey and Chase? [3 Marks]

Answer:
1. Radioactive \({}^{35}S\) was used to label viral proteins because amino acids containing sulfur are found in proteins but not in DNA.
2. Radioactive \({}^{32}P\) was used to label viral DNA because phosphorus is present in nucleic acids but not in proteins.
3. The experiment proved that DNA enters the host cell and serves as the genetic material, while protein coats remain outside.

Teacher's Note:
a) This famous Hershey-Chase blender experiment definitively proved DNA is the genetic material.
b) Differentiate clearly between protein labelling with sulfur and DNA labelling with phosphorus.

20 It is established that RNA is the first genetic material. Explain giving three reasons? [3 Marks]

Answer:
1. Essential life processes such as metabolism, translation, and splicing evolved around RNA.
2. RNA has the ability to act as both a genetic material and a biological catalyst (ribozyme).
3. Genetic information can be directly stored and replicated by RNA molecules, as seen in many viruses.

Teacher's Note:
a) Discuss catalytic RNA and metabolic dependency.
b) Mention that RNA was unstable and eventually gave rise to more stable DNA.

21 (a) What are the transcriptional products of RNA polymerase III?
(b) Differentiate between 'capping' and 'tailing'? [2 Marks]

Answer:
1. (a) RNA polymerase III transcribes tRNA, 5S rRNA, and snRNAs.
2. (b) Capping involves adding a methylguanosine triphosphate cap to the 5' end of hnRNA, whereas tailing involves adding an adenylate residue tail to the 3' end.

Teacher's Note:
a) Both capping and tailing are essential post-transcriptional modifications for eukaryotic mRNA stability.
b) Make sure to distinguish the 5' cap from the 3' poly-A tail.

22. DNA is the genetic material in most of the organisms, while RNA is the genetic material in a few viruses/ What are the four general/common functions performed by RNA in other organisms? [2 Marks]

Answer:
1. RNA acts as a messenger carrying genetic instructions from DNA to ribosomes (mRNA).
2. RNA serves as an adaptor molecule bringing amino acids during translation (tRNA) and provides structural and catalytic support in ribosomes (rRNA).

Teacher's Note:
a) Enumerate the major types and roles of cellular RNA.
b) Mention catalytic functions like splicing and translation.

23 Write short notes on RNA polymerases of eukaryotic cells? [3 Marks]

Answer:
1. RNA polymerase I transcribes rRNAs (5.8S, 18S, and 28S).
2. RNA polymerase II transcribes the precursor of mRNA, which is heterogeneous nuclear RNA (hnRNA).
3. RNA polymerase III is responsible for transcribing tRNA, 5S rRNA, and small nuclear RNAs (snRNAs).

Teacher's Note:
a) Clearly separate the three eukaryotic polymerases by their specific targets.
b) Note that eukaryotic transcription is much more complex than prokaryotic transcription.

24 How do histones acquire positive charge? [1 Mark]

Answer:
1. Histones acquire a positive charge due to the high abundance of basic amino acid residues, specifically lysine and arginine, which carry positive charges in their side chains at physiological pH.

Teacher's Note:
a) Emphasize the contribution of basic amino acid side chains.
b) This positive charge is crucial for electrostatic interaction with negatively charged DNA.

25 Explain the role of RNA polymerase in transcription in bacteria. [2 Marks]

Answer:
1. RNA polymerase catalyses the polymerization of ribonucleotides to construct an RNA strand complementary to the DNA template.
2. It interacts with the sigma factor for promoter recognition at initiation and single-handedly manages elongation.

Teacher's Note:
a) Bacteria possess a single type of RNA polymerase for all types of transcription.
b) Mention its key roles across initiation and elongation stages.

26. Identify giving reasons, the salient features of genetic code by studying the following nucleotide sequences of mRNA strand and the polypeptide translated from it? [3 Marks]

Answer:
1. The genetic code is a triplet code, as three consecutive nucleotides code for one specific amino acid.
2. It is degenerate, meaning some amino acids are coded by more than one codon.
3. It is universal and unambiguous, as a specific codon always codes for the same amino acid across organisms without overlapping.

Teacher's Note:
a) Explain triplet nature, degeneracy, and non-overlapping features.
b) Reference codon properties comprehensively.

27. Ratan was a known sportsman in his school. While returning home he found some unknown miscreants beating a young fellow. He tried to drive them off but by that time the fellow died of injury. The police arrested Ratan and he was put on trial. The judge being convinced by Ratan's plea, ordered for DNA finger printing reports.
a) Ratan's fingerprints on the dead body were sufficient to convict him but the judge asked for authentic proof? What values can be observed?
b) What is the basis of DNA finger printing?
c) Explain the steps in DNA finger printing. [4 Marks]

Answer:
1. (a) Values observed include justice, logical reasoning, fairness, and scientific validation over circumstantial evidence.
2. (b) The basis of DNA fingerprinting is identifying differences in repetitive DNA sequences known as Variable Number of Tandem Repeats (VNTRs).
3. (c) Steps include isolation of DNA, digestion of DNA by restriction endonucleases, separation of DNA fragments by agarose gel electrophoresis, transferring separated fragments to nitrocellulose membranes (Southern blotting), and hybridisation using labelled VNTR probes followed by autoradiography.
4. (d) The resulting autoradiogram provides a unique genetic profile to establish identity with absolute certainty.

Teacher's Note:
a) Connect value-based questions with scientific reasoning and ethics.
b) Detail the standard steps of VNTR-based DNA fingerprinting methodology clearly.

1. 20% [1 Mark]

Answer:
1. According to Chargaff's rules, the amount of adenine equals thymine (%A = %T) and cytosine equals guanine (%C = %G).
2. If guanine is 30%, cytosine is also 30%, making the total for G and C equal to 60%.
3. The remaining 40% is shared equally between adenine and thymine, so adenine is 20%.

Teacher's Note:
a) Apply Chargaff's equivalence rule for double-stranded DNA.
b) Remember that total purines equal total pyrimidines.

2. HIV shows reverse transcription, i.e., formation of DNA on RNA template. [1 Mark]

Answer:
1. Retroviruses like HIV possess an RNA genome that replicates through a DNA intermediate.
2. The enzyme reverse transcriptase synthesizes a complementary DNA strand using the viral RNA as a template.

Teacher's Note:
a) Understand the central dogma exception shown by retroviruses.
b) Note the function of the enzyme reverse transcriptase.

3. HIV has RNA as its genome, while bacteriophage has double-stranded DNA as its genome. [1 Mark]

Answer:
1. HIV is a retrovirus containing single-stranded RNA genetic material.
2. Common bacteriophages such as lambda phage or T2 phage contain double-stranded DNA.

Teacher's Note:
a) Compare genetic materials across different virus types.
b) Recognize viral diversity in nucleic acid types.

4. Nucleoid is the region in prokaryotic cells where DNA is held with prositively-charged proteins [1 Mark]

Answer:
1. Prokaryotes lack a true membrane-bound nucleus.
2. Their negatively charged circular DNA is organized and held in the cytoplasm by positively charged non-histone proteins in a region called the nucleoid.

Teacher's Note:
a) Differentiate between eukaryotic nucleus and prokaryotic nucleoid.
b) Note the interaction between negatively charged DNA and positively charged proteins.

5. Lysine and Arginine [1 Mark]

Answer:
1. Histones are rich in basic amino acid residues.
2. Specifically, lysine and arginine carry positive side chains at physiological pH, allowing tight binding to negatively charged DNA.

Teacher's Note:
a) Memorize the specific basic amino acids present in histones.
b) Understand the biochemical basis of DNA-histone binding.

6. The 2' OH-group in the nucleotides of RNA is a reactive group,that makes RNA labile and easily degradable [1 Mark]

Answer:
1. RNA contains a 2'-hydroxyl group on every nucleotide ribose sugar.
2. This reactive group makes RNA chemically labile and susceptible to catalytic cleavage, unlike DNA which lacks the 2'-OH group.

Teacher's Note:
a) Explain why DNA is a better genetic material than RNA.
b) Focus on the chemical difference at the 2' position of the pentose sugar.

7. DNA polymerase. Template strand has 3'->5' polarity. [1 Mark]

Answer:
1. DNA-dependent DNA polymerase is the key enzyme for replication.
2. It catalyzes polymerization strictly in the 5' to 3' direction, requiring the template strand to run in the 3' to 5' direction.

Teacher's Note:
a) State the polarity requirement for DNA polymerase activity.
b) Connect enzyme directionality with template orientation.

8. A ciston is a segment of DNA, coding for a polypeptide. [1 Mark]

Answer:
1. A cistron is functionally equivalent to a gene.
2. It represents a specific nucleotide sequence that codes for a functional polypeptide chain or protein.

Teacher's Note:
a) Define cistron in the context of molecular genetics.
b) Distinguish between monocistronic and polycistronic transcription units.

9. Sigma factor. [1 Mark]

Answer:
1. Transcription initiation in bacteria requires RNA polymerase holoenzyme.
2. The sigma factor specifically recognizes and binds to the promoter region on the DNA template.

Teacher's Note:
a) Identify the role of transcription factors in prokaryotes.
b) Note that sigma factor is responsible for specific initiation.

10. It transcribes hnRNA, the precursor of mRNA, in eukaryotes. [1 Mark]

Answer:
1. Eukaryotes possess three distinct RNA polymerases.
2. RNA polymerase II is specifically responsible for synthesizing heterogeneous nuclear RNA (hnRNA), which undergoes processing to become mRNA.

Teacher's Note:
a) List the functions of eukaryotic RNA polymerases I, II, and III.
b) Focus on RNA polymerase II and its primary product.

11. The precursor of mRNA or the primary transcript transcribed by RNA polymerase II is called Heterogenous nuclear RNA. [1 Mark]

Answer:
1. Newly transcribed RNA in eukaryotes contains both exons and introns.
2. This initial transcript is termed heterogeneous nuclear RNA (hnRNA) before undergoing splicing, capping, and tailing.

Teacher's Note:
a) Define hnRNA and its relation to functional mRNA.
b) Recall post-transcriptional modifications in eukaryotic systems.

12. (a) and (b) [2 Marks]

Answer:

(b) Non-histone chromosomal proteins are involved in the packaging of chromatin fibers at the higher levels.

Teacher's Note:
a) Contrast structural packaging and transcriptional activity between euchromatin and heterochromatin.
b) Mention the role of non-histone chromosomal proteins in higher-order chromatin organization.

13. DNA is a better genetic material for the following reasons: [2 Marks]
(I) The genetic material should be stable and should not change with age or change in physiology; this stability is given to DNA by its
(b) Double stranded nature,
(c) Presence of thymine.
(ii) Because the 2'- OH group of RNA nucleotid3es is a reactive group that makes RNA labile and easily degradable
(iii) RNA (23S RNA) is also catalytic, i.e., it is reactive.

Answer:
1. DNA provides chemical and structural stability due to its double-stranded nature and thymine substitution instead of uracil.
2. RNA is more reactive and labile due to the presence of a 2'-hydroxyl group on its ribose sugar.
3. RNA exhibits catalytic properties (ribozymes), making it inherently less stable than DNA as a long-term information repository.

Teacher's Note:
a) Outline criteria that define an ideal genetic material.
b) Compare chemical properties of DNA and RNA to justify why DNA evolved as the primary genetic material.

14. [2 Marks]
a) NH4Cl (ammonium chloride).
b) It was done to show that after on generation of Escherichia coli with 15N-DNA, in medium of 14 N, the DNA was of intermediate density between the lightly and heavy DNAs; it shows that of the two strands, only one strand is synthesized newly, using the 14N-nitrogen source in the medium.
c)The heavy and light DNA molecules can be distinguished by centrifugation in a cesium chloride (Cycle) density gradient; the 15N-DNA was heavier than 14N-DNA and the hybrid 15N-14N-DNA was intermediate between the two.
d) They concluded that DNA replication is semiconservative, i.e., of the two strands of DNA, one is the parental strand while the other is synthesized new.

Answer:
1. Meselson and Stahl used heavy isotope ammonium chloride (\(^{15}\text{NH}_4\text{Cl}\)) to label bacterial DNA.
2. Cells transferred to a normal medium (\(^{14}\text{N}\)) produced hybrid DNA after one generation, proving semiconservative replication.
3. Cesium chloride density gradient centrifugation separated DNA molecules based on density differences.
4. Each resulting daughter DNA molecule contains one parental strand and one newly synthesized strand.

Teacher's Note:
a) Describe the experimental design and isotope labeling used by Meselson and Stahl.
b) Interpret cesium chloride density gradient results to validate semiconservative replication.

15. In eukaryotes, the hnRNA (primary transcript of mRNA) has coding sequences, called exons as well as non-coding sequences, called introns, i.e., the information is split
- If undergoes a process, called splicing, in which the introns are remove and the exons are joined together in a particular manner, to form the functional mRNA.
- In prokaryotes, the mRNA is polycistronic, i.e., codes for more than one polypeptide.
- The information is continuous and no splicing is required. [2 Marks]

Answer:
1. Eukaryotic primary transcripts (hnRNA) contain both coding exons and non-coding introns.
2. Splicing removes introns and joins exons to produce functional mRNA.
3. Prokaryotic mRNA is polycistronic and lacks introns, meaning no splicing is required.

Teacher's Note:
a) Splicing is a post-transcriptional modification unique to eukaryotes.
b) Polycistronic mRNA codes for multiple proteins under the control of a single promoter.
c) Mentioning both eukaryotic split genes and prokaryotic continuous coding is essential for full credit.

16. Initiation process of transcription in bacteria
- The RNA polymerase binds transiently to an initiation factor (sigma \(\sigma\) factor) and binds to a specific sequence on the DNA, called promoter, to initiate transcription.
-The DNA strand with 3'\(\rightarrow\)5' polarity acts as the template. [2 Marks]

Answer:
1. RNA polymerase associates with the sigma factor to recognize and bind to the promoter region.
2. This binding initiates the transcription process at the appropriate start site.
3. The DNA strand oriented in the 3' to 5' direction serves as the template strand.

Teacher's Note:
a) The sigma factor confers specificity to RNA polymerase for promoter binding.
b) Transcription always proceeds in the 5' to 3' direction along the newly synthesized RNA strand.

17. Elongation process of Transcription in bacteria
- After binding to the promoter, the RNA polymerase facilitates the opening of the DNA.
- It uses nucleoside triphosphates as substrate and polymerises the nucleotides in a templatdependent fashion following complementarity of bases, in 5'\(\rightarrow\)3 direction
The process continues till the RNA polymerase reaches the terminator region on the DNA stran [2 Marks]

Answer:
1. RNA polymerase unwinds the DNA duplex to expose the template strand.
2. Ribonucleoside triphosphates are polymerized in the 5' to 3' direction based on base complementarity.
3. Elongation continues until the enzyme encounters the terminator sequence.

Teacher's Note:
a) Energy for polymerization is derived from nucleoside triphosphates.
b) No external primer is required for RNA polymerase to initiate RNA synthesis.

18. Once the RNA polymerase reaches the terminator sequences of the DNA, it binds transiently to the transiently to the termination factor (rho \(\rho\))
-The RNA strand synthesized falls off, followed by the RNA polymerase, i.e termination has occurred [2 Marks]

Answer:
1. RNA polymerase reaches the terminator region and interacts with the rho factor.
2. The newly synthesized RNA transcript is released from the DNA template.
3. The RNA polymerase enzyme dissociates from the DNA, completing termination.

Teacher's Note:
a) Termination can be rho-dependent or rho-independent in bacteria.
b) The release of both the RNA molecule and the enzyme marks the end of transcription.

19. -The viruses/ bacteriophages grown on radioactive Sulphur (35S) contained radio active proteins but not radioactive DNA, because DNA does not contain Sulphur.
- When these viruses were allowed to infect bacteria, the bacteria did not contain radioactivity, because proteins did not enter the bacteria; hence protein is n0ot the genetic material.
- The viruses grown on radioactive phosphorous (32P) contained radioactive DNA, because DNA contains phosphorus and not proteins.
- When these viruses were allowed to infect the bacteria, the bacteria were radioactive, indicating that DNA is the genetic material that has passed from the virus into bacteria. [3 Marks]

Answer:
1. Hershey and Chase used radioactive sulfur to label protein coats and radioactive phosphorus to label DNA.
2. Infection with sulfur-labeled phage showed no radioactivity in bacterial cells, proving protein does not enter.
3. Infection with phosphorus-labeled phage showed radioactivity inside bacterial cells, proving DNA is the genetic material.

Teacher's Note:
a) Sulfur is found in proteins (amino acids like methionine and cysteine) but absent in DNA.
b) Phosphorus is present in the DNA backbone but absent in standard proteins.
c) This experiment conclusively established DNA as the genetic material.

20. RNA is the first genetic material because:
(i) RNA can directly code for the synthesis of proteins and hence can easily express the character; it is the genetic material in many viruses.
(ii) RNA can also act as a catalyst; there are some important biochemical reactions in living systems that are catalyzed by RNAs and not proteins.
Many essential life processes like splicing, translation, etc. have evolved around RNA. [3 Marks]

Answer:
1. RNA can code for proteins directly and act as genetic material in several viruses.
2. RNA possesses catalytic properties, as seen in ribozymes functioning in biological systems.
3. Essential metabolic processes like splicing and translation evolved around RNA machinery.

Teacher's Note:
a) The presence of reactive 2'-OH groups makes RNA unstable compared to DNA.
b) Catalytic RNA molecules demonstrate that life could have originated without proteins initially.
c) Essential life processes rely heavily on RNA-based mechanisms.

21. (a) RNA polymerase III transcribes tRNA, 5S rRNA and SnRNA, i.e., small nuclear RNA [3 Marks]

(b) Heterogenous nuclear ribonucleic acid (hnRNA).

Answer:
1. RNA polymerase III is responsible for transcribing tRNA, 5S rRNA, and snRNAs.
2. Capping adds methyl guanosine triphosphate at the 5' end, while tailing adds adenylate residues at the 3' end.
3. The abbreviation hnRNA stands for heterogeneous nuclear ribonucleic acid.

Teacher's Note:
a) Eukaryotes have three distinct RNA polymerases with specialized transcriptional roles.
b) Both capping and tailing protect mRNA from enzymatic degradation in the cytoplasm.
c) hnRNA serves as the immediate precursor to functional messenger RNA.

22. Functions of RNA [3 Marks]
- It functions as a messenger.
- It is an adapter.
- It is a structural component of ribosome
- It also acts as a catalysts.

Answer:
1. Messenger RNA (mRNA) provides the template for protein synthesis.
2. Transfer RNA (tRNA) acts as an adapter molecule that reads the code and brings amino acids.
3. Ribosomal RNA (rRNA) plays a structural and catalytic role (ribozyme) during translation.

Teacher's Note:
a) RNA performs diverse roles in genetic information flow.
b) Mention all major types of RNA and their specific functions clearly to secure full marks.
c) Ribosomal RNA acts as a catalyst in bacteria for peptide bond formation.

23. In eukaryotes, there are three RNA-polymerases [3 Marks]
- RNA-polymerase I catalyses transcription of r RNAs (28 S, 18 A and 5.8 S).
- RNA-polymerase II catalyses transcription of precursor of mRNA; it is called hnRNA.
- RNA-polymerase III catalyses tRNA, 5srRNA and hnRNA.

Answer:
1. RNA polymerase I is located in the nucleolus and transcribes rRNAs.
2. RNA polymerase II is responsible for transcription of heterogeneous nuclear RNA (hnRNA), which is the precursor of mRNA.
3. RNA polymerase III transcribes tRNA, 5S rRNA, and snRNAs.

Teacher's Note:
a) Eukaryotic transcription is more complex than prokaryotic transcription due to multiple polymerases.
b) Do not confuse the types of RNA transcribed by each specific polymerase.
c) RNA polymerase III products are crucial for translation machinery assembly.

24. A proteins acquires a charge depending on the abundance of amino acid residues with charged side chains. [3 Marks]
- Histones are rich is basic amino acids, lysine's and arginine's, which carry positive charges in their side chains; hence histones are positively charged.

Answer:
1. The overall charge on a protein molecule is determined by the side chains of its constituent amino acids.
2. Histones are structural proteins associated with DNA in eukaryotic chromosomes.
3. They are rich in basic amino acids such as lysine and arginine, which carry positive charges, enabling them to bind tightly to negatively charged DNA.

Teacher's Note:
a) Basic amino acids contain extra amino groups in their side chains.
b) Electrostatic interactions between positively charged histones and negatively charged DNA stabilize chromatin.
c) High concentration of basic residues is a signature feature of all core histone proteins.

25. A single DNA dependent RNA polymerase catalyses the formation of mRNA, tRNA and rRNA in bacteria. [5 Marks]
- The enzyme is capable of catalyzing only the elongation step of transcription
- It combines transiently to the initiation or sigma factor and binds to the promoter and initiates transcription
- It somehow facilitates the opening of the DNA helix and catalyses the polymerisation of ribonucleoside triphosphates in a template-dependent fashion, i.e., elongation.
- When it reaches the terminator sequence, the enzyme associates transiently with the termination or rho (p) factor and terminates transcription, the RNA and the enzyme fall off the template.

Answer:
1. Bacterial transcription is carried out by a single DNA-dependent RNA polymerase core enzyme.
2. The core enzyme requires a initiation factor (sigma factor) to recognize the promoter site and initiate transcription.
3. During elongation, the enzyme uses ribonucleoside triphosphates and moves along the template strand.
4. Termination occurs when the core enzyme associates with a termination factor (rho factor) to release the RNA transcript.

Teacher's Note:
a) Prokaryotes have a simpler transcription mechanism involving transient factor binding.
b) Sigma factor and rho factor are not permanently bound to the RNA polymerase.
c) Transcription and translation can be coupled in bacteria since both occur in the cytoplasm.

26. Prosperities of genetic code: [5 Marks]
- AUG is the initiation codon and codes for methionine; methionine is the first amino acid in the given polypeptide.
- Genetic code is unambiguous and specific, i.e., each codon codes only for a particular amino acid.
- Genetic code is degenerate, i.e., more than one codon code on one amino acid.
- Each codon is a triplet, i.e., made of three nucleotides e.g., AUG, UUU, etc.
- UAG does not code for any amino acid, it is a termination codon.

Answer:
1. The genetic code is a triplet codon, where 64 codons specify 20 amino acids.
2. It is degenerate, meaning some amino acids are coded by more than one codon.
3. It is unambiguous and specific, as one codon specifies only one particular amino acid.
4. It is universal across organisms and has punctuation codons like start (AUG) and stop (UAG, UAA, UGA).

Teacher's Note:
a) Understanding codon properties is essential for deciphering mutations and protein synthesis.
b) Degeneracy prevents deleterious effects of point mutations in many cases.
c) AUG has dual functions: it codes for methionine and acts as an initiator.

27.
Ans-a) The judge showed wisdom and intention to search for the truth.
Ans-b)
1. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA.
2. Depending on base composition (A: T rich or G:C rich), length of segment, and number of repetitive units, the satellite DNA is classified into many categories, (Micro-satellites, mini-satellites etc).
3. These sequences normally do not code for any proteins, but they form a large portion of human genome. These sequence show high degree of polymorphism and form the basis of DNA fingerprinting.
Ans-c) (i) isolation of DNA, (ii) digestion of DNA by restriction endonucleases,
(iii) separation of DNA fragments by electrophoresis, (iv) transferring (blotting)
of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
(v) hybridisation using labelled VNTR probe, and (vi) detection of hybridised DNA fragments by autoradiography. [1+1+2 Marks]

Answer:
1. Part (a): The judge applied scientific reasoning and forensic evidence to deliver justice.
2. Part (b): Repetitive DNA sequences show high polymorphism, forming the foundation of DNA fingerprinting technology.
3. Part (c): The steps include DNA isolation, restriction digestion, agarose gel electrophoresis, Southern blotting, hybridization with VNTR probe, and autoradiography.

Teacher's Note:
a) Case-based questions integrate core biological principles with real-world applications like forensics.
b) Mentioning all sequential steps of DNA fingerprinting is vital for full credit in part (c).
c) VNTR (Variable Number of Tandem Repeats) shows a high degree of polymorphism.