CBSE Class 12 Biology Molecular Basis of Inheritance Assignment Set 01

Anticodon : A sequence of three nitrogenous bases on tRNA which is complementary to the codon on mRNA.
Transformation : The phenomenon by which the DNA isolated from one type of a cell, when introduced into another type, is able to express some of the properties of the former into the latter.
Transcription : The process of copying genetic information from one strand of DNA into RNA.
Translation : The process of polymerisation of amino-acids to form a polypeptide as dictated by mRNA.
Nucleosome : The structure formed when negatively charged DNA is wrapped around positively charged histone octamer.
DNA Polymorphism : The variations at genetic level, where an inheritable mutation is observed.
Satellite DNA : The repetitive DNA sequences which form a large portion of genome and have high degree of polymorphism but do not code for any proteins.
Operon : A group of genes which control a metabolic pathway.
Exons : The regions of a gene which become part of mRNA and code for different regions of proteins.
Introns : The regions of a gene which are removed during the processing of mRNA.
Euchromatin : The region of chromatin which is loosely packed and transcriptionally active.
Heterochromatin : The chromatin that is more densely packed, stains dark and is transcriptionally inactive.
Capping : Adding of methyl guanosine triphosphate to the 5´ end of hnRNA.
Splicing : The process in eukaryotic genes in which introns are removed and the exons are joined together to form mRNA.

Question. Spliceosomes are not found in cells of 
(a) plants
(b) fungi
(c) animals
(d) bacteria
Answer : D

Question. Name the nucleotide added to 5¢ end of hnRNA in capping.
(a) Ethyl cytosine triphosphate
(b) Ethyl guanosine triphosphate
(c) Methyl guanosine triphosphate
(d) Methyl cytosine triphosphate
Answer : C

Question. Which is present at 5¢ end of eukaryotic mRNA?
(a) Poly-A tail
(b) Modified C at 5¢
(c) 7 mG
(d) Poly-C
Answer : C

Question. What happens in the tailing process of transcription?
(a) Adenylate residues added at 5¢ end of RNA
(b) Adenylate residues added at 3¢ end of RNA
(c) Guanylate residues added at 5¢ end of RNA
(d) Guanylate residues added at 3¢ end of RNA
Answer : B

Question. Fully processed hnRNA to undergo translation is called
(a) mRNA
(b) rRNA
(c) tRNA
(d) sRNA
Answer : A

Question. The following diagram refers to the process of transcription in eukaryotes. Identify A, B, C and D. 
(a) A–RNA polymerase-II, B–Exon, C– Intron, D–Poly-A tail
(b) A–RNA polymerase-III, B–Intron, C– Exon, D–Poly-A tail
(c) A–RNA polymerase-II, B–Intron, C– Exon, D–Poly-A tail
(d) A–RNA polymerase-III, B–Intron, C– Exon, D–Poly-G tail
Answer : C

Question. Choose the correct option.
(a) Splicing represent the dominance of RNA world
(b) The presence of introns is reminiscent of antiquity
(c) Split gene arrangements represent an ancient feature of the genome
(d) All of the above
Answer : D

Question. Identify the correct pair of mRNA type and its function.
(a) Messenger RNA — Provides the template
(b) Transfer RNA — Brings amino acids and reads genetic code
(c) Ribosomal RNA—Plays catalytic role during translation
(d) All of the above
Answer : D

Question. Which of the following RNAs should be most abundant in animals cell? 
(a) rRNA
(b) tRNA
(c) mRNA
(d) miRNA
Answer : A

Question. DNA-dependent RNA polymerase catalyses transcription on one strand of the DNA which is called the
(a) template strand
(b) coding strand
(c) alpha strand
(d) anti-strand
Answer : A

Question. Genetic code
(a) is a relationship between sequence of DNA or mRNA to polypeptide
(b) triplet base on mRNA
(c) determines the sequence of amino acid in polypeptide
(d) All of the above
Answer : D

Question. In order to code for all the 20 amino acids, the code should be made up of three nucleotides.
This statement was suggested by
(a) Har Gobind Khorana
(b) George Gamow
(c) Marshall Nirenberg
(d) Servo Ochoa
Answer : B

Question. The codon AUG codes for (in eukaryotes)
(a) methionine
(b) histidine
(c) tryptophan
(d) alanine
Answer : A

Question. Which of the following is not a stop condon?
(a) UAA
(b) UAC
(c) UAG
(d) UGA
Answer : B

Question. The one aspect, which is not a salient feature of genetic code is, its being
(a) degenerate
(b) ambiguous
(c) universal
(d) specific
Answer : B

Question. Codons are non-ambiguous, which means that one codon codes for
(a) more than one amino acid
(b) two amino acids
(c) Only one amino acid
(d) non-sense amino acid
Answer : C

Question. From the following, identify the correct combination of salient features of genetic code.
(a) Universal, non-ambiguous, overlapping
(b) Degenerate, overlapping, commaless
(c) Universal, ambiguous, degenerate
(d) Degenerate, non-overlapping, non-ambiguous
Answer : D

Question. Degeneracy refers to
(a) one amino acid has more than one code triplet
(b) one amino acid has only one code triplet
(c) codons which specify the same amino acids differ only in the third base of the triplet
(d) Both (a) and (c)
Answer : D

Question. Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology?
(a) Genetic code is redundant
(b) Genetic code is nearly universal
(c) Genetic code is specific
(d) Genetic code is not ambiguous
Answer : B

Question. Select the correct match.
(a) Matthew Meselson and F Stahl : Pisum sativum
(b) Alfred Hershey and Martha Chase : TMV
(c) Alec Jeffreys : Streptococcus pneumoniae
(d) Francois Jacob and Jacques Monod : Lac operon
Answer : D

Question. What is the length and constituent base of tail in functional m-RNA
(a) Poly U - 200-300 bp
(b) Poly A - 200-300 bp
(c) Poly C - 200-300 nucleotides
(d) Poly A - 200-300 nucleotides
Answer : D

Question. DNA dependent RNA polymerases mediated synthesis of RNA over DNA called transcription.
About it which of the following statement is wrong
(a) In bacteria m-RNA doesnot required any processing to become active
(b) In eukaryotes there is clearcut division of labour in RNA polymerases
(c) Absence of introns in RNA of eukaryotes is reminiscent of antiquity
(d) RNA polymerase - III is responsible for synthesis of sn-RNA
Answer : C

Question. Which of the folloiwng cell cycle event is responsible for polyploidy phenomenon
(a) Failure of karyokinesis
(b) Failure of cytokinesis
(c) Failure of segregation
(d) Failure of non-disjunction
Answer : B

Question. Semiconservative replication of DNA in chromosomes was proved by
(a) Meselson & Stahl by using 15NH4Cl
(b) Taylor by using 15NH4Cl
(c) Meselson & Stahl by using tritiated thymidine
(d) Taylor by using tritiated thymidine
Answer : D

Question. Which of the following was not involved in deciphering of genetic code
(a) Physicist george Gamow's permutation combination of 43 bases
(b) H.G. Khorana's based synthesis of RNA molecules with defined combination of bases
(c) Severo ochoa enzyme for polymerising DNA with defined sequences
(d) Marshall Nirenberg's cell free system for protein synthesis
Answer : C

Question. Which of the following mutation forms the genetic basis of proof that codon is a triplet and it is read in a continuous manner
(a) Chromosomal structural mutations
(b) Chromosomal numerical mutations
(c) Substitutional mutation
(d) Frame shift insertion or deletion mutation
Answer : D

Question. An adapter molecule that would on one hand read the code and on the other hand would bind to specific amino acids is
(a) m-RNA
(b) r-RNA
(c) t-RNA
(d) hm-RNA
Answer : C

Question. Which of the following r-RNA show structural as well as functional role in bacteria :-
(a) 16s rRNA
(b) 23s rRNA
(c) 5s rRNA
(d) 28s rRNA
Answer : B

Question. During transcription only one of the strand of DNA get transcribed. Which of the following reason explain it
(a) Otherwise one segment of DNA would be coding for two different proteins
(b) Otherwise dsRNA comes in existance
(c) Otherwise antisense RNA arise which donot participate in Translation
(d) All the above
Answer : D

Ques. Anticodon occurs in
(a) tRNA
(b) mRNA
(c) rRNA
(d) DNA. 
Answer: A

Ques. In three dimensional view the molecule of tRNA is
(a) L-shaped
(b) S-shaped
(c) Y-shaped
(d) E-shaped. 
Answer: A

Ques. Genes that are involved in turning on or off the transcription of a set of structural genes are called
(a) redundant genes
(b) regulatory genes
(c) polymorphic genes
(d) operator genes. 
Answer: D

Ques. DNA elements, which can switch their position, are called
(a) cistrons
(b) transposons
(c) exons
(d) introns.
Answer: B

Ques. The maximum formation of mRNA occurs in
(a) ribosome
(b) nucleoplasm
(c) cytoplasm
(d) nucleolus. 
Answer: D

Ques. If the sequence of bases in DNA is ATTCGATG, then the sequence of bases in its transcript will be
(a) GUAGCUUA
(b) AUUCGAUG
(c) CAUCGAAU
(d) UAAGCUAC. 
Answer: D

Ques. In split genes, the coding sequences are called
(a) exons
(b) cistrons
(c) introns
(d) operons. 
Answer: A

Ques. The process of transfer of genetic information from DNA to RNA/formation of RNA from DNA is
(a) transversion
(b) transcription
(c) translation
(d) translocation. 
Answer: B

Ques. If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6 × 109 bp, then the length of the DNA is approximately
(a) 2.0 meters
(b) 2.5 meters
(c) 2.2 meters
(d) 2.7 meters.
Answer: C

Ques. Under which of the following conditions there will be no change in the reading frame of following mRNA?
5’ AACAGCGGUGCUAUU 3’
(a) Deletion of GGU from 7th, 8th and 9th positions
(b) Insertion of G at 5th position
(c) Deletion of G from 5th position
(d) Insertion of A and G at 4th and 5th position respectively 
Answer: A

Ques. Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology?
(a) Genetic code is specific.
(b) Genetic code is not ambiguous.
(c) Genetic code is redundant.
(d) Genetic code is nearly universal. 
Answer: D

Ques. If there are 999 bases in an RNA that code for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered?
(a) 11
(b) 33
(c) 333
(d) 1 
Answer: B

Ques. Which one of the following is the starter codon?
(a) UAA
(b) UAG
(c) AUG
(d) UGA 
Answer: C

Ques. Which of the following is not a property of the genetic code?
(a) Non-overlapping
(b) Ambiguous
(c) Degeneracy
(d) Universal
Answer: B

Ques. The one aspect which is not a salient feature of genetic code, is its being
(a) degenerate
(b) ambiguous
(c) universal
(d) specific. 
Answer: B

Ques. Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a “triplet”?
(a) Hershey and Chase
(b) Morgan and Sturtevant
(c) Beadle and Tatum
(d) Nirenberg and Mathaei 
Answer: D

Ques. What is not true for genetic code?
(a) It is nearly universal.
(b) It is degenerate.
(c) It is unambiguous.
(d) A codon in mRNA is read in a non-contiguous fashion. 
Answer: D

Ques. Which one of the following pairs of codons is correctly matched with their function or the signal for the particular amino acid?
(a) AUG, ACG - Start/methionine
(b) UUA, UCA - Leucine
(c) GUU, GCU - Alanine
(d) UAG, UGA - Stop 
Answer: D

Ques. After a mutation at a genetic locus the character of an organism changes due to change in
(a) protein structure (b) DNA replication
(c) protein synthesis pattern
(d) RNA transcription pattern. 
Answer: A

Ques. In mutational event, when adenine is replaced by guanine, it is a case of
(a) frame shift mutation
(b) transcription
(c) transition
(d) transversion.
Answer: C

Important Questions for NCERT Class 12 Biology Molecular Basis of Inheritance

Question. There is only one possible sequence of amino acids when deduced from a given nucleotides. But multiple nucleotides sequence can be deduced from a single amino acid sequence. Explain this phenomena.
Answer : Some amino acids are coded by more then one codon (known as degeneracy of codons), hence, on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequence will be obtained, e.g., Ile (Isoleucine) has three codons AUU, AUC, AUA. Hence, a dipeptide Met-Ile can have the following nucleotide sequence. (i) AUG-AUU (ii) AUG-AUC (iii) AUG-AUA And if, we deduce amion acid sequence from the above nucleotide sequences, all the three will code for Met-Ile.

Question. A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defined your answer.
Answer : The statement is correct. Because of degeneracy of codons, mutations at third base of codon, usually does not result into any change is phenotype. This is called silent mutations. On other hand, if codon is changed in away that now it specifies another amino acid, it may other the protein function as it happens in cse of b-globulin of haemoglobin protein. Where a substitution of valine instead of glutamic acid causes change in its structure and function, and resulting into sickle-cell trait.

Question. A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomena.
Answer : In the complete absence of expression of lac operon, permease will not be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducers. Hence, cannot relieve the lac operon from its repressed state.

Question. How has the sequencing of human genome opened new windows for treatment of various genetic disorders. Discuss amongst your classmates.
Answer : The sequencing of human genome helped in enhancing the basic understanding of genetics and immunity to various disorders. Various genes that cause genetic disorders were identified with the help of this project. It was found that more than 1200 genes are responsible for common human cardiovascular diseases, endocrine diseases (like diabetes), neurological, disorders (like Alzeimer’s disease, cancers and many more. These diseases can be treated easily by knowing the particular gene responsible for the particular disease.

Question. The total number of genes in humans is far less (< 25000) than the previous estimate (up to 140000 gene). Comment.
Answer : The total number of genes is estimated at 25000 much lower than previous estimates of 140000 that had been based on extrapolations from gene-rich areas as opposed to a composite of gene-rich and gene-poor areas. Almost all (99.9%) nucleotide bases are exactly the same in all people. Functions for over 50% discovered genes are not known yet. Scientist have identified about 1.4 million locations where single-base DNA difference (SNPs or Single Nucleotide Polymorphisms) occur in humans This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequence and tracing human history.

Question. Now, sequencing of total genomes is getting less expensive day by day. Soon it may be affordable for a common man to get his genome sequenced. What in your opinion could be the advantage and disadvantage of this development?
Answer : Human genome helps to find out the complete genome sequence of the human. It has many advantages and disadvantages. Some important advantages It provides the knowledge of the effects of variations of DNA among individuals can revolutionise the ways to diagnose, treat and prevent many diseases that affect humans It also provides clues to the understanding of human biology. It helps to find out the human evolution. Identification through DNA forensics is also possible. Some important disadvantages People might discover and untreatable genetic disease. People may abuse the knowledge obtained from the HGP. Problem can occur for the ownership of the genetic test result and the patenting of human genes and DNA. People believe that they are special and unique in their own ways and may wish to remain like that.

Question. Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage?
Answer : Bacteriophage does not have repetitive sequences such as VNTRs in its genome, as its genome is very small and have all the coding sequence. DNA finger printing is not done for phages.

Question. During in vitro synthesis of DNA, a researcher used 2’, 3’-dideoxy cytidine triphosphate as raw nucleotide in place of 2’-deoxy cytidine. What would be the consequence?
Answer : Further polymerisation would not occur, as the 3¢ OH on sugar is not there to add a new nucleotide for forming ester bond.

Question. That background information did Watson and Crick have made available for developing a model of DNA? What was their contribution?
Answer : Watson and Crick had the following informations which helped them to develop a model of DNA.
(i) Chargaff’s Law suggesting A = T and C= G
(ii) Wilkins and Rosalind Franklin’s work on DNA crystal’s X-ray diffraction studies about DNAs physical structure. Watson and Crick proposed
(a) Pattern of complementary bases pair
(b) Semi-conservative replication
(c) Mutation through tautomerism

QUESTIONS

VSA (1 MARK)

Question 1. Name the factors for RNA polymerase enzyme which recognises the start and termination signals on DNA for transcription process in Bacteria.
Answer: Sigma (\(\sigma\)) factor recognises the start signal (promoter) and Rho (\(\rho\)) factor recognises the termination signal.
In simple words: Sigma helps start transcription and Rho stops it.

Exam Tip: Remember sigma for initiation and rho for termination in prokaryotes.

Question 2. Mention the function of non-histone protein.
Answer: Non-histone chromosomal (NHC) proteins are responsible for the higher-level packaging of chromatin inside the nucleus.
In simple words: They help pack and organize DNA tightly into chromosomes.

Exam Tip: Distinguish between histone proteins (basic unit packaging) and non-histone proteins (higher-level packaging).

Question 3. During translation what role is performed by tRNA
Answer: (i) It plays a structural role by acting as an adapter molecule.
(ii) It reads the genetic code on mRNA using its anticodon loop and transfers the specific amino acid to the site of protein synthesis.
In simple words: tRNA brings the correct amino acids to the ribosome by reading the mRNA code.

Exam Tip: Always mention both adapter function and amino acid carrier function for full marks.

Question 4. RNA viruses mutate and evolve faster than other viruses. Why?
Answer: RNA is unstable because of the presence of a reactive \(2^{\prime}\text{-OH}\) group at every nucleotide in its sugar-phosphate backbone. Being unstable, RNA is prone to rapid mutations, allowing RNA viruses to evolve much faster.
In simple words: The extra reactive chemical group in RNA makes it unstable and prone to quick changes.

Exam Tip: Highlight the term \(2^{\prime}\text{-OH}\) group and instability as key reasons for rapid mutation.

Question 5. Name the parts X and Y of the transcription unit given below.
Answer: X - Template strand (or Promoter depending on diagram context), Y - Terminator.
In simple words: X points to the template strand and Y points to the terminator region.

Exam Tip: Label transcription units clearly with promoter, structural gene, and terminator.

Question 6. Mention the dual functions of AUG.
Answer: (i) It acts as the initiation codon for protein synthesis (start codon).
(ii) It codes for the amino acid methionine.
In simple words: AUG starts the translation process and brings methionine.

Exam Tip: Mention both starting translation and coding for methionine to secure full marks.

Question 7. Write the segment of RNA transcribed from the given DNA - 3´ - A T G C A G T A C G T C G T A - 5´ - Template Strand, 5´ - T A C G T C A T G C A G C A T - 3´ - Coding Strand.
Answer: 5´ - U A C G U C A U G C A G C A U - 3´
In simple words: The mRNA sequence is complementary to the template strand, with uracil replacing thymine.

Exam Tip: Pay close attention to 5' and 3' polarities and substitute T with U when writing RNA.

SA-II (2 MARKS)

Question 8. The process of termination during transcription in a prokaryotic cell is being represented here. Name the label a, b, c and d.
Answer: (a) DNA molecule
(b) mRNA transcript
(c) RNA polymerase
(d) Rho factor
In simple words: The labels identify the core components involved in stopping transcription.

Exam Tip: Practice labeling transcription and replication forks accurately from NCERT diagrams.

Question 9. Complete the blanks a, b, c and d on the basis of Frederick Griffith Experiment. S Strain - inject into mice - (a), R strain - inject into mice - (b), S strain (heat killed) - inject into mice - (c), S strain (heat killed) + R strain (live) - inject into mice - (d)
Answer: (a) Mice die
(b) Mice live
(c) Mice live
(d) Mice die
In simple words: Live virulent strains and mixtures containing transformed cells kill the mice.

Exam Tip: Memorize the outcome of each injection in Griffith's transformation experiment.

Question 10. Give two reasons why both the strands of DNA are not copied during transcription.
Answer: (i) If both strands code for RNA, two different complementary RNA molecules would form, which would pair with each other to form double-stranded RNA, preventing translation.
(ii) One segment of DNA would code for two different polypeptides, complicating the genetic information machinery.
In simple words: Copying both strands would create messy, double-stranded RNA and produce confusing protein codes.

Exam Tip: Write any two clear biological consequences of dual-strand transcription.

Question 11. Mention any two applications of DNA fingerprinting.
Answer: (i) To identify criminals in forensic laboratories.
(ii) To determine biological parents in paternity disputes.
In simple words: It is used in crime investigation and resolving family disputes.

Exam Tip: Forensic science and paternity testing are the most common expected applications.

Question 12. State the 4 criteria which a molecule must fulfill to act as a genetic material.
Answer: (i) It should be able to generate its replica (replication).
(ii) It should be chemically and structurally stable.
(iii) It should provide scope for slow changes (mutations) that are required for evolution.
(iv) It should be able to express itself in the form of Mendelian characters.
In simple words: A genetic material must copy itself, stay stable, mutate slowly, and express traits.

Exam Tip: Learn all four criteria strictly as they are frequently asked in board exams.

SA-I (3 MARKS)

Question 13. Give six points of difference between DNA and RNA in their structure/chemistry and function.
Answer:

In simple words: DNA is double-stranded and stable for storage, while RNA is single-stranded and active in protein synthesis.

Exam Tip: Present differences in a tabular format with clear headings for full credit.

Question 14. Explain how does the hnRNA becomes the mRNA. OR Explain the process of splicing, capping and tailing which occur during transcription in Eukaryotes.
Answer: Heterogeneous nuclear RNA (hnRNA) undergoes three major modifications to become functional mRNA:
(i) Splicing: Introns (non-coding regions) are removed and exons (coding regions) are joined together.
(ii) Capping: An unusual nucleotide (methylguanosine triphosphate) is added to the \(5^{\prime}\text{-end}\).
(iii) Tailing: Adenylate residues (\(200 - 300\) adenylate residues) are added at the \(3^{\prime}\text{-end}\).
In simple words: Raw RNA gets cleaned by removing useless parts, adding a cap at the front, and a tail at the back.

Exam Tip: Clearly define splicing, capping, and tailing with their specific chemical additions.

Question 15. Name the three major types of RNAs, specifying the function of each in the synthesis of polypeptide.
Answer: (i) mRNA (Messenger RNA): Determines the amino acid sequence of the polypeptide chain.
(ii) tRNA (Transfer RNA): Acts as an adapter molecule that recognizes codons on mRNA and brings specific amino acids to the ribosome.
(iii) rRNA (Ribosomal RNA): Plays structural and catalytic roles during translation.
In simple words: mRNA carries the recipe, tRNA brings the ingredients, and rRNA builds the protein structure.

Exam Tip: Specify both structural and functional roles for each RNA type.

Question 16. Enlist the goals of Human genome project.
Answer: (i) Identify all the approximately \(20,000 - 25,000\) genes in human DNA.
(ii) Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
(iii) Store this information in databases.
(iv) Improve tools for data analysis and transfer related technologies to industries.
(v) Address ethical, legal, and social issues (ELSI) arising from the project.
In simple words: HGP aimed to map all human genes, sequence base pairs, store data, and handle ethical concerns.

Question 17. A tRNA is charged with the amino acid methionine. (i) Give the anti-codon of this tRNA. (ii) Write the Codon for methionine. (iii) Name the enzyme responsible for binding of amino acid to tRNA.
Answer: (i) Anti-codon: \(3^{\prime}\text{-UAC-5}^{\prime}\)
(ii) Codon: \(5^{\prime}\text{-AUG-3}^{\prime}\)
(iii) Enzyme: Aminoacyl-tRNA synthetase
In simple words: The anticodon is UAC, the start codon is AUG, and the linking enzyme is aminoacyl-tRNA synthetase.

Question 18. Illustrate schematically the process of initiation, elongation and termination during transcription of a gene in a bacterium.
Answer: Transcription in bacteria involves three phases:
1. Initiation: RNA polymerase binds to the promoter site with the help of the sigma (\(\sigma\)) factor.
2. Elongation: RNA polymerase separates from the sigma factor and uses nucleoside triphosphates to polymerize RNA in the \(5^{\prime} \to 3^{\prime}\) direction.
3. Termination: RNA polymerase reaches the terminator region, binds with the rho (\(\rho\)) factor, and releases the nascent RNA transcript.
In simple words: Transcription starts with sigma, grows along the template, and stops with rho.

Exam Tip: Mention the specific roles of sigma and rho factors during initiation and termination.

LA (5 MARKS)

Question 19. What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally?
Answer: Semi-conservative replication means that in each newly synthesized DNA molecule, one strand is parental (original) and the other strand is newly synthesized.
Experimental Proof by Meselson and Stahl (1958):
- They grew \(\text{E. coli}\) in a medium containing \(^{15}\text{NH}_4\text{Cl}\) (heavy isotope of nitrogen) for many generations so that DNA became heavy.
- These bacteria were transferred to a medium with normal \(^{14}\text{N}\) (light isotope).
- DNA extracted after one generation showed an intermediate density in a \(\text{CsCl}\) density gradient.
- DNA extracted after two generations consisted of equal amounts of light (\(^{14}\text{N}-^{14}\text{N}\)) and hybrid (\(^{15}\text{N}-^{14}\text{N}\)) DNA.
- This conclusively proved that DNA replication is semi-conservative.
In simple words: Meselson and Stahl used heavy and light nitrogen isotopes in bacteria to prove that every new DNA copy keeps one original strand.

Exam Tip: Clearly state the generations, centrifuge results (intermediate, light, hybrid), and conclusion.

Question 20. What does the lac operon consist of? How is the operator switch turned on and off in the expression of genes in this operon? Explain.
Answer: Lac operon consists of:
- One regulatory gene (\(i\)-gene) coding for the repressor.
- Three structural genes (\(z, y, a\)) coding for \(\beta\)-galactosidase, permease, and transacetylase respectively.
- Promoter site and Operator site.
Switching Mechanism:
- Absence of Inducer (Off condition): The repressor protein synthesized by the \(i\)-gene binds to the operator region, blocking RNA polymerase from transcribing the structural genes.
- Presence of Inducer (On condition): Lactose acts as an inducer and binds to the repressor protein, inactivating it. The operator turns ON, allowing RNA polymerase to transcribe the structural genes (\(z, y, a\)).
In simple words: Without lactose, a repressor blocks the operon. With lactose, the repressor is turned off, allowing enzyme production.

Question 21. State salient features of genetic code.
Answer: (i) The codon is a triplet (3 bases code for one amino acid). 61 codons code for amino acids and 3 are stop codons (UAA, UAG, UGA).
(ii) The code is unambiguous and specific (one codon codes for only one specific amino acid).
(iii) The code is degenerate (some amino acids are coded by more than one codon).
(iv) The codon is read in mRNA in a contiguous fashion without punctuations.
(v) The code is nearly universal (from bacteria to humans, a codon codes for the same amino acid).
(vi) AUG has dual functions: it acts as a start codon and codes for methionine.
In simple words: Genetic codes are triplets, specific, degenerate, universal, and read continuously.

Exam Tip: List at least 5 to 6 key features with proper terminology like degenerate and universal.

Question 22. Describe the process of transcription of mRNA is an eukaryotic cell.
Answer: In eukaryotes, transcription involves complex steps in the nucleus:
1. Three RNA polymerases are involved: RNA Polymerase I transcribes rRNA, RNA Polymerase II transcribes hnRNA (precursor of mRNA), and RNA Polymerase III transcribes tRNA and 5S rRNA.
2. Transcription of structural genes produces primary transcript (hnRNA) containing both exons and introns.
3. Post-transcriptional modification (RNA processing):
- Splicing: Introns are removed and exons are joined.
- Capping: Methylguanosine triphosphate is added at the \(5^{\prime}\text{-end}\).
- Tailing: Adenylate residues are added at the \(3^{\prime}\text{-end}\).
4. Fully processed hnRNA becomes mRNA and is transported to the cytoplasm for translation.
In simple words: Eukaryotic transcription produces a raw RNA transcript that undergoes splicing, capping, and tailing before becoming mature mRNA.

Exam Tip: Mention all three types of RNA polymerases and the three post-transcriptional modifications.

Question 23. Describe the various steps involved in the technique of DNA fingerprinting.
Answer: DNA fingerprinting involves the following sequential steps:
1. Extraction of DNA: Isolation of DNA from cells using high-speed refrigerated centrifuge.
2. Amplification: Making multiple copies of target DNA using Polymerase Chain Reaction (PCR).
3. Restriction Digestion: Cutting DNA into smaller fragments using restriction endonuclease enzymes.
4. Separation of DNA fragments: Separation based on size using agarose gel electrophoresis.
5. Southern Blotting: Transferring separated DNA fragments from gel onto synthetic membranes like nylon or nitrocellulose.
6. Hybridisation: Incubating the membrane with radioactive VNTR probes.
7. Autoradiography: Exposing the hybridized membrane to X-ray film to detect dark bands representing VNTR patterns.
In simple words: DNA is extracted, copied, cut, separated on gel, blotted, probed, and photographed on X-ray film.

Exam Tip: Memorize the steps in exact chronological order: Extraction, PCR, Digestion, Electrophoresis, Blotting, Hybridization, Autoradiography.