CBSE Class 12 Biology Molecular Basis of Inheritance Worksheet Set E

Long Answer Questions

Question. (a) Write the conclusion drawn by Griffith at the end of his experiment with Streptococcus pneumoniae.
(b) How did O. Avery, C MacLeod and M. McCarty prove that DNA was the genetic material?Explain. 
Answer. (a) At the end of his experiments Griffith concluded that transformation of R strain by the heatkilled S strain indicated the presence of a transforming principle or genetic material. This transforming principle made the R strain virulent.
(b) They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed. They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material.

Question. How did Alfred Hershey and Martha Chase conclusively establish that DNA is the genetic material? Explain.
Answer. • Procedure:
(i) Some bacteriophage virus were grown on a medium that contained radioactive phosphorus (32P) and some in another medium with radioactive sulphur (35S).
(ii) Viruses grown in the presence of radioactive phosphorus (32P) contained radioactive DNA.
(iii) Similar viruses grown in presence of radioactive sulphur (35S) contained radioactive protein.
(iv) Both the radioactive virus types were allowed to infect E. coli separately.
(v) Soon after infection, the bacterial cells were gently agitated in blender to remove viral coats from the bacteria.
(vi) The culture was also centrifuged to separate the viral particle from the bacterial cell.
• Observations and Conclusions:
(i) Only radioactive 32P was found to be associated with the bacterial cell, whereas radioactive 35S was only found in surrounding medium and not in the bacterial cell.
(ii) This indicates that only DNA and not the protein coat entered the bacterial cell.
(iii) This proves that DNA is the genetic material which is passed from virus to bacteria and not protein.

Question. Who proposed that DNA replication is semi-conservative? How was it experimentally proved by Meselson and Stahl?
Answer. Watson and Crick had proposed the semi-conservative scheme for replication of DNA.
For experimental proof, 

Question. (a) Explain what DNA replication refers to.
(b) State the properties of DNA replication model.
(c) List any three enzymes involved in the process along with their functions.
Answer. (a) DNA replication refers to DNA synthesis: Refer to Basic Concepts 10(iii).
(b) DNA replication is :
(i) Semi-conservative,
(ii) Semi-discontinuous,
(iii) Unidirectional
(c) (i) Experimental proof for semi-conservative mode of DNA replication
• Matthew Meselson and Franklin Stahl in 1958 performed experiments on E. coli to prove that DNA replication is semi-conservative.
• They grew E. coli in a medium containing 15NH4Cl (in which 15N is the heavy isotope of nitrogen) for many generations.
• As a result, 15N got incorporated into newly synthesised DNA.
• This heavy DNA can be differentiated from normal DNA by centrifugation in caesium chloride (CsCl) density gradient.
• Then they transferred the cells into a medium with normal 14NH4Cl and took the samples at various definite time intervals as the cells multiplied

Question. (a) Explain the process of DNA replication that occurs in a replication fork in E. coli.
(b) How are translational unit and untranslated regions in mRNA different from each other?
Answer. (a) DNA replication begins at a unique and fixed point called origin of replication or ‘ori’.
Initiation
• The complementary strands of DNA double helix are separated by enzyme, DNA helicase. This is called unwinding of double-stranded DNA.
• The separated strands tend to rewind, therefore these are stabilised by proteins called single strand binding proteins (ssBPs), which bind to the separated strands.
• Unwinding of double-stranded DNA forms a Y-shaped configuration in the DNA duplex, which is called replication fork.
(b) A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide, a stop codon and untranslated regions (UTRs). UTRs are present at both 5′-end and 3′-end of mRNA.

Question. (a) What do ‘Y’and ‘B’ stand for in ‘YAC’ and ‘BAC’ used in Human Genome Project (HGP).
Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the percentage of discovered genes whose functions are known as observed during HGP.
(c) Expand ‘SNPs’ identified by scientists in HGP.
Answer. (a) ‘Y’ stands for Yeast and ‘B’ stands for Bacterial.
‘YAC’ and ‘BAC’ are used as vectors for cloning foreign DNA.
(b) The total human genome percentage is (<) 2% and percentage of discovered genes is (<) 50%.
(c) Single Nucleotide Polymorphism.

Question. Describe how the lac operon operates, both in the presence and absence of an inducer in E.coli.
Answer.

The repressor is synthesised from the i gene. The repressor protein binds to the operator region and prevents RNA polymerase from transcribing the structural genes zya. In the presence of an inducer, the repressor is inactivated by interaction with inducer. This allows RNA polymerase access to promotor and transcription proceeds.

Question. (a) Name the stage in the cell cycle where DNA replication occurs.
(b) Explain the mechanism of DNA replication. Highlight the role of enzymes in the process.
(c) Why is DNA replication said to be semiconservative?
Answer. (a) S-phase/synthetic phase (of interphase).
(b) (ii) Enzymes for DNA replication
• Various enzymes are required as catalysts during DNA replication in living cells.
(a) DNA-dependent DNA polymerase: It catalyses the polymerisation of deoxynucleotides on DNA template at a fast rate. Its average rate of polymerisation is 2000 bp per second.
[It completes process of replication for E.coli within 38 minutes which has only 4.6 × 106.
(c) During DNA replication in the two newly synthesised daughter DNA, one strand is parental (conserved) on the other is newly synthesised. That is why it is called semiconservative.

Question. (a) Explain the process of DNA replication with the help of a schematic diagram.
(b) In which phase of the cell cycle does replication occur in Eukaryotes? What would happen if cell division is not followed after DNA replication? 
Answer. (a) Replication of DNA begins at ori, to form a replication fork. DNA dependent DNA polymerase forms a new strand in 5′ 3′ direction. The replication is continuous on the 3′ 5′ strand whereas it is discontinuous on the 5′ 3′ strand.
The discontinuously synthesised fragments are later joined by the enzyme DNA ligase.
(b) Replication occurs in S phase If cell division is not followed after DNA replication, the cell will undergo polyploidy.

Question. (a) Name two enzymes involved in the process of DNA replication, along with their properties. 
Answer. 
The two strands of DNA cannot be separated in its entire length due to very high energyrequirement. High amount of energy is required to break the hydrogen bonds holding the two strands. Therefore, the replication occurs in small opening of DNA strands called the replication fork.
(a) (i) DNA dependent DNA polymerase: adds nucleotides only in 5 to 3 directions.
(ii) DNA ligase: joins the discontinuously synthesised DNA fragments during replication.

Question. (a) Write the specific features of the genetic code AUG.
(b) Genetic codes can be universal and degenerate. Write about them, giving one example of each.
(c) Explain aminoacylation of the tRNA.
Answer. (a) AUG is the starting codon and codes for methionine.
(b) The genetic code is universal, i.e., a particular codon codes for the same amino acid in all organisms. For example, UUU codes for phenylalanine in all organisms.
Some amino acids are coded by more than one codon, hence the code is degenerate. For example, UUU and UUC both code for phenylalanine.
(c) Amino acids become activated by binding with aminoacyl tRNA synthetase enzyme in the presence of ATP.

These activated amino acids are then linked to their cognate tRNA to form aminoacylated tRNA.

Question. Draw a labelled schematic structure of a transcription unit. Explain the function of each component in the unit in the process of transcription.
Answer. Schematic structure of a transcription unit:

(i) Promoter: It is the binding site for RNA polymerase for initiation of transcription.
(ii) Structural gene: It codes for enzyme or protein for structural functions.
(iii) Terminator: It is the region where transcription ends.

Question. Explain the process of transcription in prokaryotes. How is the process different in eukaryotes?
Answer. • In prokaryotes, the structural gene is polycistronic and continuous.
• In bacteria, the transcription of all the three types of RNA (mRNA, tRNA and rRNA) is catalysed by single DNA-dependent enzyme, called the RNA polymerase.
• All three RNA’s are needed to synthesize a protein in cell. mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNA plays structural and catalytic role durinng translation.
• The transcription is completed in three steps: initiation, elongation and termination.
• Initiation: s (sigma) factor recognises the start signal and promotor region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription. It uses nucleoside triphosphates as substrate and polymerises in a template-dependent fashion following the rule of complementarity.

Question. Explain the process of transcription in eukaryotes. 
Answer. • The structural genes are monocistronic in eukaryotes.
• The process of transcription is similar to that in prokaryotes.
• It takes place in the nucleus.
• Coding gene sequences called exons form the part of mRNA and non-coding sequence called introns are removed during RNA splicing and exons are joined in a defined order.
• In eukaryotes, three types of RNA polymerases are found in the nucleus:
(i) RNA polymerase I transcribes rRNAs (28S, 18S, and 5.8S).
(ii) RNA polymerase II transcribes the precursor of mRNA (called heterogeneous nuclear RNA or hnRNA).
(iii) RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs (small nuclear RNAs). Post-transcriptional modifications
• The primary transcripts are non-functional, containing both the coding region, exon, and non-coding region, intron, in RNA and are called heterogenous RNA or hnRNA.
• The hnRNA undergoes splicing and two additional processes called capping and tailing.
• In capping, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′-end of hnRNA.

Question. (a) Explain the role of DNA dependent RNA polymerase in initiation, elongation and termination during transcription in bacterial cell.
(b) How is transcription a more complex process in eukaryotic cells? Explain.
Answer. (a) The DNA dependent RNA polymerase helps in DNA replication by catalysing the polymerisation in only one direction, i.e., 5′→3′. In bacteria, the RNA polymerase has
co-factors b, b′, α, α′, w and s which catalyse the process. Refer to the above question.
(b) • The structural genes are monocistronic in eukaryotes.
•  The process of transcription is similar to that in prokaryotes.
•  It takes place in the nucleus.
•  Coding gene sequences called exons form the part of mRNA and non-coding sequence called introns are removed during RNA splicing and exons are joined in a defined order.
•  In eukaryotes, three types of RNA polymerases are found in the nucleus:
(i) RNA polymerase I transcribes rRNAs (28S, 18S, and 5.8S).
(ii) RNA polymerase II transcribes the precursor of mRNA (called heterogeneous nuclear RNA or hnRNA).
(iii) RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs (small nuclear RNAs).

Question. Transcription in eukaryotes is more complex process than in prokaryotes. Justify and compare the initiation, elongation and termination in bacterial cells with eukaryotes.
Answer. Transcription is more complex in eukaryotes due to following reasons:
• In prokaryotes only one type of RNA polymerase is involved whereas in eukaryotes three types of RNA polymerases are involved.
• For Description of processing of hnRNA involving-introns/exons/splicing in eukaryotes and for Description of capping and tailing,
• The structural genes are monocistronic in eukaryotes.
• The process of transcription is similar to that in prokaryotes.
• It takes place in the nucleus.
• Coding gene sequences called exons form the part of mRNA and non-coding sequence called introns are removed during RNA splicing and exons are joined in a defined order.
• In eukaryotes, three types of RNA polymerases are found in the nucleus:

Question. How do RNA, tRNA and ribosomes help in the process of translation?
Answer. mRNA provides a template with codons for specific amino acids to be linked to form a polypeptide/protein.
tRNA brings amino acid to the ribosomes reads the genetic code with the help of its anti-codons, initiator tRNA is responsible for starting polypeptide formation in the ribosomes tRNAs are specific for each amino acid.

Ribosomes-(Cellular factories for proteins synthesis) its smaller sub unit binds with mRNA to initiate protein synthesis at the start codon AUG, in its larger sub unit there are two sites present which brings two amino acids close to each other helping them to form peptide bond.
Ribosomes moves from codon to codon along mRNA, amino acids are added one by one to form polypeptide/protein.

Question. (a) Describe the process of synthesis of fully functional mRNA in an eukaryotic cell.
(b) How is this process of mRNA synthesis different from that in prokaryotes?
Answer. (a) • Francis Crick proposed the presence of an adapter molecule which could read the code on one end and on the other end would bind to the specific amino acids.
• However, tRNA was known before the genetic code was postulated and was then called sRNA (soluble RNA). Its role as an adapter molecule was reported later.
Structure
• The secondary structure of tRNA is cloverleaf like but the three-dimensional tertiary structure depicts it as a compact inverted L-shaped molecule.
(b) In prokaryotes, there is a single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. In bacteria, mRNA does not require any processing as it does not have any introns.

Question. Name the major types of RNAs and explain their role in the process of protein synthesis in a prokaryote.
Answer. The three major types of RNAs are—mRNA, tRNA and rRNA.
mRNA: It provides the template for protein synthesis. It also provides site to initiate and terminate the process of protein synthesis tRNA: Its anticodon loop reads the genetic code on mRNA and brings the corresponding amino acid bound to its amino acid binding end on to the mRNA.
rRNA: It forms a structural component of ribosome (23S RNA) and acts as a catalyst for the formation of peptide bond.

Question. Illustration below is a DNA segment, which constitutes a gene:

(i) Name the shaded and unshaded regions of gene.
(ii) Explain how these genes are expressed.
(iii) How is this gene different from prokaryotic gene in its expression? 
Answer. (i) The shaded portions are introns and unshaded portions are exons.
(ii) The primary RNA contains both introns and exons. By the mechanism of splicing, introns are removed and exons are joined to form functional mRNA after capping and tailing. 
• Translation is the process of synthesis of protein from amino acids, sequence and order of amino acids being defined by sequence of bases inm RNA. Amino acids are joined by peptide bonds.
• A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide, a stop codon and untranslated regions (UTRs). UTRs are additional sequences of mRNA that are not translated. They are present at both 5′ end (before start codon) and at 3′ end (after stop codon) for efficient translation process.
(iii) In prokaryotes, the structural gene is continuous and is not differentiated into exons and introns unlike eukaryotes. In prokaryotes, transcription is followed by translation without RNA splicing mechanism.

Question. There is only one possible sequence of amino acids when deduced from a given nucleotide.
But multiple nucleotide sequences can be deduced from a single amino acid sequence. Explain this phenomena.
Answer. Some amino acids are coded by more than one codon (known as degeneracy of codon), hence on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequences will be obtained.
For example, isoleucine has three codons AUU, AUC and AUA. Hence a dipeptide Met–Ile can have any of the following nucleotide sequences:
(i) AUG–AUU
(ii) AUG–AUC
(iii) AUG–AUA

If we deduce amino acid sequences of the above nucleotide sequences, all the three will code for Met–Ile.

Question. Answer the following questions based on Meselson and Stahl’s experiment:
(a) Why did the scientists use 15NH4Cl and 14NH4Cl as sources of nitrogen in the culture medium for growing E. coli?
(b) Name the molecule(s) that 15N got incorporated into.
(c) How did they distinguish between 15N labelled molecules from 14N ones?
(d) Mention the significance of taking the E. coli samples at definite time intervals for observations.
(e) Write the observations made by them from the samples taken at the end of 20 minutes and 40 minutes respectively.
(f) Write the conclusion drawn by them at the end of their experiment. 

Answer. (a) 15N is the heavy isotope of nitrogen and it can be separated from 14N based on the difference in their densities.
(b) 15N was incorporated into newly synthesised DNA.
(c) The two molecules were distinguished by caesium chloride centrifugation in which these two separated into two different bands at different positions based on their densities.
(d) E. coli culture is taken at equal intervals to know the progress of the experiment as generation time of E. coli is 20 minutes.
(e) After 20 minutes the culture had an intermediate density showing a band in the middle tube and after 40 minutes, the culture had equal amounts of hybrid DNA and the light DNA showing two bands, one in the centre and one at the bottom.
(f) They concluded that DNA replicates semi-conservatively.

Question. Where do transcription and translation occur in bacteria and eukaryotes respectively? Explain the complexities in transcription and translation in eukaryotes that are not seen in bacteria.
Answer. Transcription and translation in bacteria occur in the cytoplasm of the cell, whereas in eukaryotes, transcription occurs in the nucleus and translation occurs in the cytoplasm.
Complexities in transcription in eukaryotes
(i) The structural genes are monocistronic and split in eukaryotes.
(ii) The genes of eukaryotic organisms have coding or expressed sequences called exons that form the part of mRNA and non-coding sequences called introns, that do not form part of the mRNA and are removed during RNA splicing.
(iii) In eukaryotes, apart from the RNA polymerase found in the organelles, three types of RNA polymerases are found in the nucleus.
(iv) RNA polymerase I transcribes rRNAs (28S, 18S, and 58S).
(v) RNA polymerase II transcribes the precursor of mRNA (called as heterogeneous nuclear RNA (hnRNA).
(vi) RNA polymerase III helps in transcription of tRNA, 5S rRNA, and snRNAs (small nuclear RNAs).
(vii) The primary transcripts contain both the coding regions called exons and non-coding regions called intron in RNA and are non-functional called hnRNA.
(viii) The hnRNA undergoes two additional processes called capping and tailing.
(ix) In capping, an unusual nucleotide is added to the 5′-end of hnRNA i.e. methyl guanosine triphosphate.
(x) In tailing, about 200-300 adenylate residues are added at 3′-end in a template independent manner.
(xi) Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA called splicing.
Translation in both eukaryotes and prokaryotes is similar.

Question. (a) State the arrangement of different genes that in bacteria is referred to as ‘operon’.
(b) Draw a schematic labelled illustration of lac operon in a ‘switched on’ state.
(c) Describe the role of lactose in lac operon.
Answer. (a) The operon has polycistronic structural genes, i.e., three structural genes adjacent to an operator, a promoter and a regulator.
(b)

(c) Lactose is the inducer that inactivates repressor. This allows RNA polymerase to access promoter and initiate transcription of the structural genes or switch on the operon.

Question. Explain the steps of DNA fingerprinting that will help in processing of the two blood samples A and B picked up from the crime scene.
Answer. Methodology and Technique:
(i) DNA is isolated and extracted from the cell or tissue by centrifugation.
(ii) By the process of polymerase chain reaction (PCR), many copies are produced. This step is called amplification.
(iii) DNA is cut into small fragments by treating with restriction endonucleases.
(iv) DNA fragments are separated by agarose gel electrophoresis.
(v) The separated DNA fragments are visualised under ultraviolet radiation after applying suitable dye.
(vi) The DNA is transferred from electrophoresis plate to nitrocellulose or nylon membrane sheet. This is called Southern blotting.
(vii) VNTR probes are now added which bind to specific nucleotide sequences that are complementary to them. This is called hybridisation.
(viii) The hybridised DNA fragments are detected by autoradiography. They are observed as dark bands on X-ray film.
(ix) These bands being of different sizes, give a characteristic pattern for an individual DNA. It differs from individual to individual except in case of monozygotic (identical) twins.

Please click on below link to download CBSE Class 12 Biology Molecular Basis of Inheritance Worksheet Set E