CBSE Class 12 Biology Molecular Basis of Inheritance Assignment Set 02

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.

Important Questions for NCERT Class 12 Biology Molecular Basis of Inheritance

Question. A transcription unit in DNA is defined primarily by three regions in DNA. These regions are
(a) Promoter, regulator and structural gene
(b) Promoter, structural gene and terminator
(c) Promoter, regulator and terminator
(d) Promoter, regulator and operator gene

Answer : B

Question. At which of the following levels, regulation of gene expression in eukaryotes donot occur
(a) Transcription level
(b) Processing level
(c) Transport of ribosomal subunits from nucleus to cytoplasm level
(d) Translation level

Answer : C

Question. Which of the following is not a goal of HGP
(a) Identify all the approximately 20,000-25,000 genes
(b) Store this information in database
(c) Restrict the related technologies so that other sector donot benefited with it
(d) Address the ethical, legal and social issues

Answer : C

Question. The human genome project was coordinated by
(a) U.S. department of energy
(b) National institute of health
(c) Sanger centre
(d) Both 1 and 2

Answer : D

Question. Which of the following organism was not used as model organism is human genome project
(a) Arabidopsis
(b) Caenorhabdities elegans
(c) Rice
(d) Hyacinthus orientalis

Answer : D

Question. Regarding to salient features of human genome select out the incorrect one
(a) Human genome contains 3164.7 million nucleotide bases
(b) Human genome contain 30,000 genes
(c) y-chromosome has largest number of genes
(d) 1.4 million locations are associated with SNPs

Answer : C

Question. If an inheritable mutation is observed in a population at high frequency, it is referred as
(a) DNA polyploidy
(b) DNA polymorphism
(c) DNA redundancy
(d) Sequence annotation

Answer : B

Question. Due to high degree of polymorphism, size of VNTR varies in size from
(a) 0.1 - 2 kb
(b) 0.1 - 2000 kb
(c) 0.1 - 20 kb
(d) 0.1 - 200 kb

Answer : C

Question. Which of the following statements is correct?
(a) Adenine pairs with thymine through two H-bonds.
(b) Adenine pairs with thymine through one H-bond.
(c) Adenine pairs with thymine through three H-bonds.
(d) Adenine does not pair with thymine.

Answer: A

Question. Purines found both in DNA and RNA are
(a) cytosine and thymine
(b) adenine and thymine
(c) adenine and guanine
(d) guanine and cytosine. 

Answer: C

Question. The association of histone H1 with a nucleosome indicates that
(a) DNA replication is occurring
(b) the DNA is condensed into a chromatin fibre
(c) the DNA double helix is exposed
(d) transcription is occurring.

Answer: B

Question. In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in this DNA are
(a) G 17%, A 33%, T 33%
(b) G 8.5%, A 50%, T 24.5%
(c) G 34%, A 24.5%, T 24.5%
(d) G 17%, A 16.5%, T 32.5%.

Answer: A

Question. What are the structures called that give an appearance as ‘beads-on-string’ in the chromosomes when viewed under electron microscope?
(a) Genes     
(b) Nucleotides
(c) Nucleosomes   
(d) Base pairs 

Answer: C

Question. Which one of the following does not follow the central dogma of molecular biology?
(a) Pea
(b) Mucor
(c) Chlamydomonas
(d) HIV 

Answer: D

Question. The 3′ - 5′ phosphodiester linkages inside a polynucleotide chain serve to join
(a) one DNA strand with the other DNA strand
(b) one nucleoside with another nucleoside
(c) one nucleotide with another nucleotide
(d) one nitrogenous base with pentose sugar.

Answer: C

Question. Which one of the following statements about the particular entity is true ?
(a) Centromere is found in animal cells, which produces aster during cell division.
(b) The gene for producing insulin is present in every body cell.
(c) Nucleosome is formed of nucleotides.
(d) DNA consists of core of eight histones.

Answer: B

Question. In the DNA molecule,
(a) the proportion of adenine in relation to thymine varies with the organism
(b) there are two strands which run antiparallel- one in 5′ → 3′ direction and other in 3′ → 5′
(c) the total amount of purine nucleotides and pyrimidine nucleotides is not always equal
(d) there are two strands which run parallel in the 5′ → 3′ direction. 

Answer: B

Question. Which one of the following pairs of nitrogenous bases of nucleic acids, is wrongly matched with the category mentioned against it?
(a) Guanine, Adenine - Purines
(b) Adenine, Thymine - Purines
(c) Thymine, Uracil - Pyrimidines
(d) Uracil, Cytosine - Pyrimidines 

Answer: B

Question. One turn of the helix in a B-form DNA is approximately
(a) 2 nm         
(b) 20 nm
(c) 0.34 nm   
(d) 3.4 nm. 

Answer: D

Question. Antiparallel strands of a DNA molecule means that
(a) one strand turns clockwise
(b) one strand turns anti-clockwise
(c) the phosphate groups of two DNA strands, at their ends, share the same position
(d) the phosphate groups at the start of two DNA strands are in opposite position (pole).

Answer: D

Question. Which one of the following makes use of RNA template to synthesise DNA?
(a) DNA polymerase
(b) RNA polymerase
(c) Reverse transcriptase
(d) DNA dependant RNA polymerase 

Answer: C

Question. Which one of the following hydrolyses internal phosphodiester bonds in a polynucleotide chain?
(a) Lipase     
(b) Protease
(c) Endonuclease   
(d) Exonuclease 

Answer: C

Question. The following ratio is generally constant for a given species:
(a) A + G / C + T
(b) T + C / G + A
(c) G + C / A + T
(d) A + C / T + G. 

Answer: C

Question. In a DNA percentage of thymine is 20% then what will be percentage of guanine?
(a) 20%
(b) 40%
(c) 30%
(d) 60% 

Answer: C

Question. Length of one loop of B-DNA
(a) 3.4 nm
(b) 0.34 nm
(c) 20 nm
(d) 10 nm.

Answer: A

Question. DNA is mainly found in
(a) nucleolus
(b) nucleus only
(c) cytoplasm only
(d) none of these. 

Answer: B

Question. In prokaryotes, the genetic material is
(a) linear DNA without histones
(b) circular DNA without histones
(c) linear DNA with histones
(d) circular DNA with histones.

Answer: B

Very Short Answer Questions 

Question. Calculate the length of the DNA of bacteriophage lambda that has 48502 base pairs.
Answer. Distance between two consecutive base pairs = 0.34 × 10^–9 m
The length of DNA in bacteriophage lambda = 48502 × 0.34 × 10^–9 m
= 16.49 × 10^–6 m

Question. Write the function of RNA polymerase II.
Answer. RNA polymerase II transcribes precursor of mRNA or hnRNA.

Question. Mention the carbon positions to which the nitrogen-s base and the phosphate molecule are respectively linked in the nucleotide given below:

Answer. Nitrogenus base is linked to first carbon.
Phosphate is linked to fifth carbon.

Question. How is repetitive/satellite DNA separated from bulk genomic DNA for vari-s genetic experiments? 
Answer. By density gradient centrifugation.

Question. State which human chromosome has
(i) the maximum number of genes and
(ii) the one which has the least number of genes?
Answer. (i) Chromosome-1
(ii) Y-Chromosome

Question. Given below is a schematic representation of a lac operon in the absence of an inducer. Identify ‘a’ and ‘b’ in it.

Answer. a–Repressor
b–Repressor b-nd to the operator and prevents transcription of structural genes.

Question. Mention the contribution of genetic maps in human genome project. 
Answer. Genetic maps have played an important role in se-encing of genes, DNA fingerprinting, tracing human history, chromosomal location for disease associated se-ences (Any one).

Question. What are ‘a’ and ‘b’ in the nucleotide with purine represented below?

Answer. ‘a’ is phosphate gr-p and ‘b’ is purine (adenine/guanine).

Question. Name the negatively charged and positively charged components of a nucleosome.
Answer. In a nucleosome, the negatively charged component is DNA and positively charged component is histone octamer.

Question. Why is lactose considered an inducer in lac operon? 
Answer. Lactose binds to repressor and prevents it from binding with the operator, as a result RNA polymerase binds to promoter–operator region to transcribe the structural genes.

Question. Mention the two additional processing which hnRNA needs to undergo after splicing so as to become functional.
Answer. Capping and tailing.

Question. At which ends do ‘capping’ and ‘tailing’ of hnRNA occur, respectively?
Answer. Capping occurs at 5′-end and tailing occurs at 3′-end.

Question. In a nucleus, the number of RNA nucleoside triphosphates is 10 times more than the number of DNA nucleoside triphosphates, still only DNA nucleotides are added during the DNA replication, and not the RNA nucleotides. Why?  
Answer. DNA polymerase is highly specific to recognise only deoxyribonucleoside triphosphates.
Therefore it cannot hold RNA nucleotides.

Question. What is cistron? 
Answer. A cistron is a segment of DNA coding for a polypeptide.

Short Answer Questions 

Question. A DNA segment has a total of 2,000 nucleotides, -t of which 520 are adenine containing nucleotides. How many purine bases this DNA segment possesses? 
Answer. [A] = [T]
A + T = 520 + 520
= 1040
Total number of nucleotides = 2000
∴ G + C = 2000 – 1040
= 960
G = 960/2
= 480
∴ Total number of purines (A + G) = 520 + 480
= 1000.

Question. Following are the features of genetic codes. What does each one indicate?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon. 
Answer. Stop codon does not code for any amino acid and terminates the synthesis of polypeptide chain.
Unambiguous codon: one codon codes for one amino acid only.
Degenerate codon: some amino acid are coded by more than one codon.
Universal codon: Genetic code is same for all organisms (bacteria to humans).

Question. Discuss the role of enzyme DNA ligase plays during DNA replication.
Answer. DNA ligase joins or seals the discontinuous DNA fragments.

Question. Mention the role of ribosomes in peptide bond formation. How does ATP facilitate it?
Answer. There are two sites in the large subunit of the ribosome, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst for the formation of peptide bond 23S rRNA in bacteria is a ribozyme. Amino acids become activated by binding with its tRNA in the presence of aminoacyl tRNA synthetase and ATP.

Question. Describe the structure of a nucleosome. 
Answer. Packaging of DNA in eukaryotes
- Roger Kornberg (1974) reported that chromosome is made up of DNA and protein.
- Later, Beadle and Tatum reported that chromatin fibres look like beads on the string, where beads are repeated units of proteins.
- The proteins associated with DNA are of two types— basic proteins (histones) and acidic non-histone chromosomal (NHC) proteins.
- The negatively charged DNA molecule wraps ar-nd the positively charged histone proteins to form a structure called nucleosome.
- The nucleosome core is made up of f-r types of histone proteins—H2A, H2B, H3 and H4 occurring in pairs.
- 200 bp of DNA helix wrap ar-nd the nucleosome by 1¾ turns, plugged by H1 histone protein.
- Repeating units of nucleosomes form the chromatin in nucleus, which is a thread-like structure.
- The chromatin is packed to form a solenoid structure of 30 nm diametre.
- Further supercoiling forms a looped structure called the chromatin fibre.
- These chromatin fibres further coil and condense at metaphase stage of cell division to form chromosomes. Packaging of chromatin at higher level re-ires NHC proteins.

Question. Draw a schematic diagram of a part of d-ble stranded dinucleotide DNA chain having all the f-r nitrogen-s bases and showing the correct polarity. 
Answer.  

Question. Alth-gh a prokaryotic cell has no defined nucleus, yet DNA is not scattered thr-gh-t the cell. Explain. 
Answer. DNA is negatively charged and has positively charged histone proteins. These help DNA to be held in a place in large loops called nucleoid.

Question. Draw the structure of a tRNA charged with methionine. 
Answer. 

Question. Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase.
Answer. Viruses grown in the medium containing 32P contained radioactive DNA but not radioactive protein because DNA contains phosphorus but proteins do not contain phosphorus. Similarly,viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

Question. Recall the experiment done by Frederick Griffith, Avery, Macleod annd McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material,w-ld the heat killed strain of Streptococcus have transformed the R-strain into virulent strain? Explain y-r answer. 
Answer. RNA is more labile and prone to degradation (owing to the presence of 2′–OH gr-p in its ribose).
Hence heat-killed S-strain may not have retained its ability to transform the R-strain.

Question. How do histones ac-ire positive charge?  
Answer. Histones are rich in the basic amino acid residues lysines and arginines, which carry positive charges in their side chains. Therefore, histones are positively charged.

Question. How are the structural genes inactivated in lac operon in E. coli? Explain. 
Answer. The regulator gene produces repressor which when free, binds to the operator region of the operon and prevents RNA polymerase from transcribing the structural genes.

Question. Draw a labelled schematic diagram of a transcription unit. 
Answer.  

Question. How are the structural genes activated in the lac operon in E. coli? 
Answer. Lactose acts as the inducer that binds with repressor protein that cannot bind to operator and hence frees the operator gene. RNA polymerase freely moves over the structural genes, transcribing lac mRNA, which in turn produces the enzymes responsible for the digestion of lactose.

Question. State the difference between the structural genes in a transcription unit of prokaryotes and eukaryotes.  
Answer. Differences between structural genes of prokaryotes and eukaryotes
Prokaryotes                                                                     Eukaryotes
(i) Polycistronic                                                                 Monocistronic
(ii) No split genes present. The coding sequence            Split genes present. The coding sequence is
is not interrupted.                                                              interrupted to form exon and intron.

Question. (i) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion.  
Answer. (i) George Gamow.
(ii) He proposed that there are four bases and 20 amino acids So, there should be atleast 20 different genetic codes for these 20 amino acids.
The only possible combinations that would meet the requirement is combinations of 3 bases that will give 64 codons.

Question. Protein synthesis machinery revolves ar-nd RNA but in the c-rse of evolution it was replaced by DNA. Justify. 
Answer. Since RNA was unstable and prone to mutations, DNA evolved from RNA with chemical modifications that makes it more stable.
DNA has d-ble stranded nature and has complementary strands. These further resist changes by evolving a process of repair.

Question. Explain the two factors responsible for conferring stability to double helix structure of DNA.
Answer. Factors responsible for conferring stability to double helix structure are presence of hydrogenbonds, the plane of one base pair stacks over the other, complementary presence of thymine in place of uracil.
                                          (B)              (C)
Question. (a) (A)   —    DNA   →    mRNA   →   Protein

Look at the above se-ence and mention the events (A), (B) and (C).
(b) What does Central Dogma state in Molecular Biology? How does it differ in some viruses?
Answer. (a) A—DNA replication, B—Transcription, C—Translation
(b) Central Dogma in Molecular Biology states that information flows in the order
            DNA   →  RNA  →  Proteins
It differs in some viruses as the flow of information is in reverse direction, that is, from DNA to RNA.

Question. Retrovirus do not follow central dogma. Comment. 
Answer. Genetic material of retrovirus is RNA. At the time of synthesis of protein, RNA is reverse transcribed to its complementary DNA first, then transcriped to RNA and proteins. Hence,retrovirus are not known to follow central dogma.

Question. What are satellite DNA in a genome? Explain their role in DNA fingerprinting.
                                          OR
Explain the significance of satellite DNA in DNA fingerprinting technique.  
Answer. A small stretch of DNA sequence that repeats many a time, shows a high degree of polymorphism and forms a bulk of DNA in a genome called satellite DNA.
(i) They do not code for any proteins.
(ii) They form large part of the human genome.
(iii) They show high degree of polymorphism and are specific to each individual.

Question.

Why do you see two different types of replicating strands in the given DNA replication fork?
Explain. Name these strands. 
Answer. The DNA-dependent DNA polymerase catalyses polymerisation only in one direction i.e., 5′→3′.
Therefore, in one strand with polarity 3′→5′ continuous replication takes place whereas the other strand with polarity 5′→3′ carries out discontinuous replication.
The strand with polarity 3′→5′ is called leading strand and the strand with polarity 5′→3′ is called lagging strand.

Question. (a) List the two methodologies which were involved in human genome project. Mention how they were used.
(b) Expand ‘YAC’ and mention what was it used for. 
Answer. The two methodologies involved in human genome project are:
(a) (i) Expressed Sequence Tags: Identifying all the genes that are expressed as RNA
(ii) Sequence Annotation: Sequencing the whole set of genome coding or non coding sequences and later assigning different region with functions.
(b) ‘YAC’ stands for Yeast Artificial Chromosome. It is used as a cloning vectors.

Question. Name indicating their functions, a few additional enzymes, other than DNA polymerase and ligase, that are involved in the replication of DNA with high degree of processivity and accuracy. 
Answer. (i) Helicase: opens the helix
(ii) Topoisomerases: removes the supercoiling of DNA relieves the tension due to unwinding
(iii) Primase: synthesises RNA primer
(iv) Telomerase: to synthesises the DNA of telomeric end of chromosomes.

Question. State the dual role of deoxyribonucleoside triphosphates during DNA replication.
                                          ​​​​​​​                     ​​​​​​​OR
Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerisation.
Answer. (i) Deoxyribonucleoside triphosphates act as substrates for polymerisation.
(ii) These provide energy from its two terminal phosphates for polymerisation reaction.

Question. A template strand is given below. Write down the corresponding coding strand and the mRNA strand that can be formed, along with their polarity.
3′ ATGCATGCATGCATGCATGCATGC 5′ 
Answer. Coding strand: 5′ TACGTACGTACGTACGTACGTACG 3′
mRNA strand: 5′ UACGUACGUACGUACGUACGUACG 3′

Question. The base se-ence of one strand of DNA is TACTAGGAT.
(i) Write the base se-ence of the RNA got after transcription of the given se-ence.
(ii) What is the distance maintained between the two consecutive pairs of bases in the DNA molecule?
(iii) Who contributed the base complementary rule?  
Answer. (i) AUGAUCCUA.
(ii) 3.4 Å or 0.34 nm.
(iii) Chargaff contributed the base complementary rule.

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 recognizes the start signal (promoter), and Rho (\(\rho\)) factor recognizes the termination signal.
In simple words: Sigma helps start transcription and Rho helps end it.

Exam Tip: Always specify sigma factor for initiation and rho factor for termination to secure full marks.

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

Exam Tip: Mention both chromatin packaging and higher-order chromosome structuring.

Question 3. During translation what role is performed by tRNA
Answer: tRNA performs two major roles: (i) It acts as an adapter molecule that reads the genetic code on mRNA through its anticodon, and (ii) it transfers specific amino acids to the site of protein synthesis (ribosome).
In simple words: tRNA brings the correct amino acids by reading the mRNA instructions.

Exam Tip: Mention both the adapter function and amino acid carrier function for completeness.

Question 4. RNA viruses mutate and evolve faster than other viruses. Why?
Answer: RNA is unstable compared to DNA due to the presence of a reactive \(2^\prime\text{-OH}\) group at every nucleotide in RNA. Consequently, RNA viruses mutate at a much faster rate.
In simple words: The reactive chemical group in RNA makes it unstable and prone to frequent mutations.

Exam Tip: Highlight the presence of the \(2^\prime\text{-OH}\) group and chemical instability.

Question 5. Name the parts ‘X’ and ‘Y’ of the transcription unit given below.
Answer: X represents the Template strand, and Y represents the Terminator.
In simple words: X is the strand used for copying, and Y marks the end of transcription.

Question 6. Mention the dual functions of AUG.
Answer: (i) It acts as the initiation codon for protein synthesis in mRNA. (ii) It codes for the amino acid Methionine (met).
In simple words: AUG starts the protein-making process and brings the first amino acid (Methionine).

Exam Tip: Both points (initiator and coding for methionine) are mandatory for a 1-mark question.

Question 7. Write the segment of RNA transcribed from the given DNA -
\(3^\prime\) - A T G C A G T A C G T C G T A - \(5^\prime\) - Template Strand
\(5^\prime\) - T A C G T C A T G C A G C A T - \(3^\prime\) - Coding Strand.
Answer: \(5^\prime\) - U A C G U C A U G C A G C A U - \(3^\prime\)
In simple words: The RNA is complementary to the template DNA strand, with Uracil (U) replacing Thymine (T).

Exam Tip: Always remember to write polarity ( \(5^\prime\) to \(3^\prime\)) and substitute T with U in 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 point to the DNA template, the newly formed RNA, the enzyme, and the termination factor.

Exam Tip: Learn diagrammatic representations of transcription and translation from NCERT.

Question 9. Complete the blanks a, b, c and d on the basis of Frederick Griffith Experiment.
S Strain \(\rightarrow\) inject into mice \(\rightarrow\) (a)
R strain \(\rightarrow\) inject into mice \(\rightarrow\) (b)
S strain (heat killed) \(\rightarrow\) inject into mice \(\rightarrow\) (c)
S strain (heat killed) + R strain (live) \(\rightarrow\) inject into mice \(\rightarrow\) (d)
Answer: (a) Mice die
(b) Mice live
(c) Mice live
(d) Mice die
In simple words: Virulent strains kill mice, non-virulent or heat-killed strains do not, but their combination causes lethal transformation.

Exam Tip: Griffith's experiment proves the phenomenon of bacterial transformation.

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 bind to form double-stranded RNA, preventing translation.
(ii) If a single segment of DNA transcribed two different proteins, the genetic information machinery would become excessively complicated and confusing.
In simple words: Copying both strands would create double-stranded RNA and produce conflicting protein instructions.

Exam Tip: This is a very frequent board exam question; write both points clearly.

Question 11. Mention any two applications of DNA fingerprinting.
Answer: (i) It is used in forensic laboratories to identify criminals.
(ii) It is used to determine biological paternity or maternity in family disputes.
In simple words: It helps catch criminals and solve parentage disputes.

Exam Tip: Mention forensics and paternity testing as the primary applications.

Question 12. State the 4 criteria which a molecule must fulfill to act as a genetic material.
Answer: (i) It must be able to replicate (generate its replica).
(ii) It should be chemically and structurally stable.
(iii) It should provide scope for slow changes (mutations) that are necessary for evolution.
(iv) It must 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: Memorize all four properties as they are standard criteria for genetic materials like DNA.

SA-I (3 MARKS)

Question 13. Give six points of difference between DNA and RNA in their structure/chemistry and function.
Answer:
1. DNA is double-stranded; RNA is single-stranded.
2. DNA contains Thymine as a pyrimidine base; RNA contains Uracil.
3. The pentose sugar in DNA is Deoxyribose; in RNA it is Ribose.
4. DNA is chemically quite stable and not very reactive; RNA is reactive due to the \(2^\prime\text{-OH}\) group.
5. DNA dictates the synthesis of polypeptides (stores genetic code); RNA directly performs protein synthesis.
6. DNA is primarily found in the nucleus; RNA is synthesized in the nucleus and transported into the cytoplasm.
In simple words: DNA stores long-term genetic info safely in double strands, while RNA uses single strands and Uracil to make proteins.

Exam Tip: Present differences in a comparative tabular format or clear numbered points.

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: In eukaryotes, the primary transcript (hnRNA) undergoes three major modifications to become fully functional mRNA:
1. Splicing: Introns (non-coding regions) are removed, and exons (coding regions) are joined together in a defined order.
2. Capping: An unusual nucleotide (methylguanosine triphosphate) is added to the \(5^\prime\) end of hnRNA.
3. Tailing (Polyadenylation): Adenylate residues (200-300 bases) are added at the \(3^\prime\) end.
In simple words: hnRNA is processed by cutting out useless parts (introns) and adding protective caps and tails.

Exam Tip: Define capping (\(5^\prime\)), tailing (\(3^\prime\)), and splicing (intron removal) precisely.

Question 15. Name the three major types of RNAs, specifying the function of each in the synthesis of polypeptide.
Answer:
1. mRNA (Messenger RNA): Determines the sequence of amino acids by carrying the genetic code from DNA.
2. tRNA (Transfer RNA): Recognizes codons on mRNA and transports specific amino acids to the site of protein synthesis.
3. rRNA (Ribosomal RNA): Plays structural and catalytic roles (ribozyme activity) during translation.
In simple words: mRNA carries the message, tRNA brings amino acids, and rRNA builds the ribosome factory.

Exam Tip: Clearly state the unique function of each RNA type in translation.

Question 16. Enlist the goals of Human genome project.
Answer: (i) Identify all 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 other sectors such as industries.
(v) Address ethical, legal, and social issues (ELSI) that may arise from the project.
In simple words: HGP aimed to map all human genes, sequence base pairs, store data, and handle ethical issues.

Exam Tip: Mention gene count estimation (~25,000) and base pair sequencing (~3 billion).

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: UAC
(ii) Codon: AUG
(iii) Enzyme: Aminoacyl-tRNA synthetase
In simple words: Methionine is coded by AUG, matched by UAC, and linked by aminoacyl-tRNA synthetase.

Exam Tip: Remember that codons are on mRNA, and anticodons are on tRNA.

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 clear phases:
1. Initiation: RNA polymerase binds to the promoter site on DNA along with the initiation factor (sigma factor, \(\sigma\)).
2. Elongation: The sigma factor separates, and RNA polymerase uses nucleoside triphosphates as substrates to polymerize RNA in the \(5^\prime \rightarrow 3^\prime\) direction following complementarity rules.
3. Termination: When RNA polymerase reaches the terminator region, it binds with the terminator factor (rho factor, \(\rho\)), causing the newly synthesized RNA and the enzyme to release.
In simple words: The enzyme binds with sigma to start, builds the RNA strand, and stops when it hits rho.

Exam Tip: Mention RNA polymerase, sigma factor, and rho factor in their respective roles.

LA (5 MARKS)

Question 19. What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally?
Answer: Semiconservative replication means that in each new DNA molecule, one strand is parental (original) and the other strand is newly synthesized.
Experimental Proof by Meselson and Stahl (1958):
1. They grew E. coli in a medium containing \({}^{15}\text{NH}_4\text{Cl}\) (heavy isotope of nitrogen) for many generations so that DNA became heavy (\({}^{15}\text{N}\)).
2. These bacteria were then transferred to a medium with normal \({}^{14}\text{NH}_4\text{Cl}\).
3. Samples were extracted at definite time intervals (generation 1 after 20 minutes; generation 2 after 40 minutes) and centrifuged in a Cesium chloride (\(\text{CsCl}\)) density gradient.
4. The DNA extracted after one generation had an intermediate density (hybrid \({}^{15}\text{N}-{}^{14}\text{N}\)).
5. The DNA extracted after two generations consisted of equal amounts of light (\({}^{14}\text{N}-{}^{14}\text{N}\)) and hybrid DNA.
6. This confirmed that DNA replication is semiconservative.
In simple words: Meselson and Stahl used heavy and light nitrogen isotopes to prove that every new DNA molecule keeps one old strand and makes one new strand.

Exam Tip: Mention the cesium chloride density gradient centrifugation and the ratio of light-to-hybrid DNA in generations 1 and 2.

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: Components of Lac Operon:
1. Regulatory gene (\(i\)-gene): Codes for the repressor protein.
2. Promoter site: Where RNA polymerase binds to initiate transcription.
3. Operator site: Acts as a switch that controls access of RNA polymerase to structural genes.
4. Structural genes (\(z, y, a\)):
- Gene \(z\) codes for \(\beta\)-galactosidase.
- Gene \(y\) codes for permease.
- Gene \(a\) codes for transacetylase.
Mechanism of Switching:
- In the absence of Inducer (Lactose): The repressor protein synthesized by the \(i\)-gene binds to the operator region, blocking RNA polymerase from transcribing the structural genes (Operator is turned OFF).
- In the presence of Inducer (Lactose): Lactose acts as an inducer and binds to the repressor protein, inactivating it. The repressor can no longer bind to the operator, allowing RNA polymerase to transcribe the structural genes (Operator is turned ON).
In simple words: Lactose turns the operon ON by removing the blocking repressor protein.

Exam Tip: Clearly draw or describe both states: absence of lactose (switch off) and presence of lactose (switch on).

Question 21. State salient features of genetic code.
Answer: Salient features of the genetic code include:
1. Triplet Code: The codon is a triplet; 61 codons code for amino acids, and 3 codons act as stop codons (UAG, UGA, UAA).
2. Unambiguous and Specific: One codon codes for only one specific amino acid.
3. Degenerate: Some amino acids are coded by more than one codon.
4. Universal: The genetic code is nearly universal across living organisms (e.g., a codon specifies the same amino acid in bacteria and humans).
5. Contiguous and Commaless: The code is read on mRNA in a contiguous fashion without any punctuation marks.
6. Dual Function of AUG: AUG acts as the initiator codon and also codes for Methionine.
In simple words: Genetic codes are read in sets of three, are universal, specific, and contain start/stop signals.

Exam Tip: List at least 5 distinct features with appropriate biological terms like triplet, degenerate, and universal.

Question 22. Describe the process of transcription of mRNA is an eukaryotic cell.
Answer: Transcription in eukaryotes is more complex than in prokaryotes and involves the following steps:
1. Involvement of Three RNA Polymerases:
- RNA Polymerase I transcribes rRNAs.
- RNA Polymerase II transcribes hnRNA (precursor of mRNA).
- RNA Polymerase III transcribes tRNA, \(5\text{S}\) rRNA, and snRNAs.
2. Primary Transcript (hnRNA): Contains both coding regions (exons) and non-coding regions (introns).
3. Post-Transcriptional Processing:
- Splicing: Introns are removed and exons are spliced together.
- Capping: Methylguanosine triphosphate is added to the \(5^\prime\) end.
- Tailing: Poly-A tail (200-300 adenylate residues) is added to the \(3^\prime\) end.
4. Transport: The fully processed hnRNA, now called mRNA, is transported out of the nucleus into the cytoplasm for translation.
In simple words: Eukaryotic transcription uses specialized RNA polymerases and modifies hnRNA through splicing, capping, and tailing.

Exam Tip: Always mention the specific functions of RNA Pol I, II, and III along with post-transcriptional modifications.

Question 23. Describe the various steps involved in the technique of DNA fingerprinting.
Answer: DNA fingerprinting (developed by Alec Jeffreys) involves the following steps:
1. Extraction of DNA: Isolation of DNA from blood, hair follicles, or other tissue samples using high-speed refrigerated centrifuges.
2. Amplification: Making multiple copies of target DNA regions using the Polymerase Chain Reaction (PCR).
3. Restriction Digestion: Cutting DNA into fragments using specialized restriction enzymes.
4. Separation of DNA Fragments: Separating the DNA fragments according to size by agarose gel electrophoresis.
5. Southern Blotting: Transferring the separated DNA sequences from the gel onto a synthetic nitrocellulose or nylon membrane.
6. Hybridization: Incubating the nylon membrane with radioactive DNA probes containing specific VNTR sequences.
7. Autoradiography: Exposing the membrane to X-ray film to detect dark bands where radioactive probes bound to target DNA.
In simple words: DNA fingerprinting isolates, cuts, separates, blots, and exposes DNA to create a unique banding pattern for identification.

Exam Tip: Write the steps in chronological order: Extraction \(\rightarrow\) Amplification \(\rightarrow\) Digestion \(\rightarrow\) Electrophoresis \(\rightarrow\) Blotting \(\rightarrow\) Hybridization \(\rightarrow\) Autoradiography.