CBSE Class 12 Biology Molecular Basis of Inheritance Worksheet Set A

Very Short Answer Questions

Question. How many base pairs would a DNA segment of length 1.36 nm have?
0.34 ×10-6 ×1.36
4 ×106 bp
Answer. Distance between two base pairs = 0.34 nm or 0.34×10–6 nm Number of base pairs in 1.36 nm DNA segment

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

Question. Why hnRNA is required to undergo splicing? 
Answer. hnRNA undergoes splicing in order to remove introns which are intervening or non-coding sequences and exons are joined to form functional mRNA.

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 bound 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 sequencing of genes, DNA fingerprinting, tracing human history, chromosomal location for disease associated sequences (Any one).

Question. Why are proteins either positively or negatively charged?
Answer. If the proteins are rich in basic amino acids, they are positively charged, and if the proteins are rich in acidic amino acids, they are negatively charged.

Question. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer. Nitrogenous bases—Adenine, Thymine, Uracil and Cytosine.
Nucleosides—Cytidine and Guanosine.

Question. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer. Cytosine = 20%, therefore Guanine = 20%
According to Chargaff’s rule,
A + T = 100 – (G + C)
A + T = 100 – 40. Since both adenine and thymine are in equal amounts,
∴ Thymine = Adenine = 60/2=30%

Question. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer. 

Question. If the sequence of one strand of DNA is written as follows :
5′—ATGCATGCATGCATGCATGCATGCATGC—3′
Write down the sequence of complementary strand in 5′→3′ direction.
Answer. In 3′→5′ direction, 3′—TACGTACGTACGTACGTACGTACGTACG—5′
In 5′→3′ direction, 5′—GCATGCATGCATGCATGCATGCATGCAT—3′

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

Short Answer Questions

Question. Draw a schematic representation of dinucleotide. Label the following:
(i) The components of a nucleotide
(ii) 5′ end
(iii) N-glycosidic linkage
(iv) Phosphodiester linkage. 
Answer. Nucleotide = Ribose sugar + Base + phosphate group.

Question. Describe the structure of a RNA polynucleotide chain having four different types of nucleotides. 
Answer. 

Question. Differentiate between codon and an anticodon.
Answer. 

Codon	Anticodon
The sequence of 3 nitrogen bases on mRNA that codes for a particular amino acid during translation is called codon.	The sequence of 3 nitrogenous bases on tRNA that are complementary to the codon on mRNA for a particular amino acid during translation is called anticodon.

Question. A DNA segment has a total of 1500 nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine bases this DNA segment possesses?
Answer. According to Chargaff’s rule A/G = T/G  = 1
G = C, G = 410, hence C = 410
G + C = 410 + 410
         = 820
So, A + T = 1500 – 820
               = 680
A = T, so T = 680/2 = 340
So, Pyrimidines = C + T
                       = 410 + 340
                       = 750

Question. A DNA segment has a total of 2,000 nucleotides, out 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. Describe the structure of a nucleosome.
Answer. 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 around the positively charged histone proteins to form a structure called nucleosome.
• The nucleosome core is made up of four types of histone proteins—H2A, H2B, H3 and H4 occurring in pairs.

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, would the heat killed strain of Streptococcus have tranformed the R-strain into virulent strain? Explain your Answerwer.
Answer. RNA is more labile and prone to degradation (owing to the presence of 2′–OH group in its ribose).
Hence heat-killed S-strain may not have retained its ability to transform the R-strain.

Question. How do histones acquire 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. A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defend your Answerwer.
Answer. The statement is correct because of degeneracy of codons, mutations at third base of codon,usually doe not result into any change in phenotype. This is called silent mutations but at other times it can lead to loss or formation of malformed protein changing the phenotype.

Question. Explain the dual function of AUG codon. Give the sequence of bases it is transcribed from and its anticodon.
Answer. The dual function of AUG codon:
(a) It codes for amino acid methionine.
(b) It is an initiation codon.
The sequence of bases from which it is transcribed is TAC. Its anticodon is UAC.

Question. Protein synthesis machinery revolves around RNA but in the course 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 double stranded nature and has complementary strands. These further resist changes by evolving a process of repair.

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.

Long Answer Type Questions

Question. Give an account of Hershey and Chase experiment. What did it conclusively prove? If both DNA and proteins contained phosphorus and sulphur do you think the result would have been the same?
Answer : Hershey and Chase conducted experiments on bacteriophage to prove that DNA is the genetic material. (Image 117)
(i) Some bacteriophage virus were grown on a medium that contained radioactive phosphorus ( P) 32 and some in another medium with radioactive sulphur ( S) 35 .
(ii) Viruses grown in the presence of radioactive phosphorus ( P) 32 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 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. If both DNA and proteins contained phosphorus and sulphur, the result might change. In case (i)

Question. During the course of evolution why DNA was choosen over RNA as genetic material. Give reasons by first discussing the desired criteria in a molecule that can act as genetic material and in the light of biochemical differences between DNA and RNA.
Answer : A molecule that can act as a genetic material must fulfil the following
(i) It should be able to generate its replica (replication).
(ii) It should chemically and structurally be stable.
(iii) It should provide the scope for slow changes (mutation) that are required for evolution.
(iv) It should be able to express itself in the form of Mendelian. Biochemical differences between DNA and RNA
(i) Both nucleic acid (DNA and RNA) are able to direct their duplication proteins fails for the first criteria.
(ii) RNA is reactive, it also acts are catalyst, hence DNA is less reactive and structurally more stable than RNA.
(iii) Presence of thymine at the place of uracil also confers additional stability to DNA.

Question. Give an account of post transcriptional modifications of a eukaryotic mRNA. K Thinking Process Post-transcriptional modifications include the modification of the mRNA transcript synthesised by RNA polymerase II (in eukaryotes).
Answer : 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. In eukaryotes, three types of RNA polymerases are found in the nucleus
(i) RNA polymerase I transcribes rRNAs (28 S and 5.8 S). (ii) RNA polymerase II transcribes the precursor of mRNA (called heterogeneous nuclear RNA or hnRNA).
(iii) RNA polymerase III transcribes tRNA, 5 S rRNA and snRNAs (small nuclear RNAs). The hnRNA undergoes two additional processes called capping and tailing. In capping, an unusual nucleotide, methyl guanosine triphosphate is added to the 5¢-end of hnRNA. In tailing, adenylate residues (about 200-300) are added at 3¢-end in a template independent manner. Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing. (Image 119)

Question. Discuss the process of translation in detail.
Answer : There are three-stages of protein synthesis
(i) Initiation Assembly of Ribosomes on mRNA In prokaryotes, initiation requires the large and small ribosome subunits, the mRNA, initiation tRNA and three Initiation Factors (IFs). Activation of Amino Acid Amino acids become activated by binding with aminoacyl tRNA synthetase enzyme in the presence of ATP. Transfer of Amino Acid to tRNA The AA-AMP-enzyme complex formed reacts with specific tRNA to form aminoacyl tRNA complex. AA-AMP-Enzyme complex +tRNA ¾® AA tRNA + AMP + Enzyme. The cap region of mRNA binds to the smaller subunit of ribosome. The ribosome has two sites, A-site and P-site. The smaller subunit first binds the initiator tRNA then and then binds to the larger subunit so, that initiation codon (AUG) lies on the P-site. The initiation tRNA, i.e., methionyl tRNA then binds to the P-site.
(ii) Elongation Another charged aminoacyl tRNA complex binds to the A-site of the ribosome. Peptide bond formation and movement along the mRNA called translocation. A peptide bond is formed between carboxyl group (—COOH) of amino acid at P-site and amino group (—NH) of amino acid at A-site by the enzyme peptidyl transferase. The ribosome slides over mRNA from codon to codon in the 5¢®3¢ direction. According to the sequence of codon, amino acids are attached to one another by peptide bonds and a polypeptide chain is formed.
(iii) Termination When the A-site of ribosome reaches a termination codon which does not code for any amino acid, no charged tRNA binds to the A-site. Dissociation of polypeptide from ribosome takes place which is catalysed by a ‘release factor’. There are three termination codons, i e . ., UGA, UAG and UAA.

Question. Define an operon, giving an example, explain an inducible operon.
Answer : The concept of operon was first proposed in 1961, by Jacob and Monad. An operon is a unit of prokaryotic gene expression which includes coordinately regulated (structural) genes and control elements which are recognised by regulatory gene product. Components of an Operon
(i) Structural gene The fragment of DNA which transcribe mRNA for polypeptide synthesis.
(ii) Promoter The sequence of DNA where RNA polymerase binds and initiates transcription of structural genes is called promoter.
(iii) Operator The sequence of DNA adjacent to promoter where specific repressor protein binds is called operator.
(iv) Regulator gene The gene that codes for the repressor protein that binds to the operator and suppresses its activity as a result of which transcription will be switched off.
(v) Inducer The substrate that prevents the repressor from binding to the operator, is called an inducer. As a result transcription is switched on. It is a chemical of diverse nature like metabolite, hormone substrate, etc. Inducible Operon System An inducible operon system is a regulated unit of genetic material which is switched on in response to the presence of a chemical. e.g., the lactose or lac-operon of E.coli. The lactose operon The lac z, y, a genes are transcribed from a lac transcription unit under the control of a single promoter. They encode enzyme required for the use of lactose as a carbon source. The lac i gene product, the lac repressor, is expressed from a separate transcription unit upstream from the operator. lac operon consists of three structural genes (z, y and a), operator, promoter and a separate regulatory gene. The three structural genes (a, y and a) transcribe a polycistronic mRNA. (Image 121) Gene z codes for b-galactosidase (b-gal) enzyme which breaks lactose into galactose and glucose. Gene y codes for permease, which increases the permeability of the cell to lactose. Gene a codes for enzyme transacetylase, which catalyses the transacetylation of lactose in its active form.
When Lactose is Absent
(i) When lactose is absent, i gene regulates and produces repressor mRNA which translate repression.
(ii) The repressor protein binds to the operator region of the operon and as a result prevents RNA polymerase to bind to the operon.
(iii) The operon is switched off.
When Lactose is Present
(i) Lactose acts as an inducer which binds to the repressor and forms an inactive repressor.
(ii) The repressor fails to bind to the operator region.
(iii) The RNA polymerase binds to the operator and transcript lac mRNA.
(iv) lac mRNA is polycistronic, i.e., produces all three enzymes, b-galactosidase, permease and transacetylase.
(v) The lac operon is switched on.

Question. ‘There is a paternity dispute for a child’. Which technique can solve the problem? Discuss the principle involved.
Answer : DNA fingerprinting is the technique used in solving the paternity dispute for a child. DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. The basis of DNA fingerprinting is DNA polymorphism. Although the DNA from different individuals is more alike than different, there are many regions of the human chromosomes that exhibit a great deal of diversity. Such variable sequences are termed ‘polymorphic’ (meaning many forms). A special type of polymorphism, called VNTR (Variable Number of Tandem Repeats), is composed of repeated copies of a DNA sequence that lie adjacent to one another on the chromosome. Since, polymorphism is the basis of genetic mapping of human.

Question. Give an account of the methods used in sequencing the human genome.
Answer : Sequencing of human genome has made it possible to understand the link between various genes and their functions. If there are any gene defects that express as disorders or that increase the susceptibility of an individual to a disease then specific gene therapies can be worked out Methodologies of human genome sequencing The methods involve two major approaches (i) Expressed Sequence Tags (ESTs) This method focusses on identifying all the genes that are expressed as RNA. (ii) Sequence annotation It is an approach of simply sequencing the whole set of genome that contains all the coding and non-coding sequences, and later assigning different regions in the sequence with functions. For sequencing, first the total DNA from cell is i.e., solated and broken down in relatively small sizes as fragments. There DNA fragments are cloned in suitable host using suitable vectors. When bacteria is used as vector, they are called Bacterial Artificial Chromosomes (BAC) and when yeast is used as vector, they are called Yeast Artificial Chromosomes (YACs). Frederick Sanger developed a principle according to which the fragments of DNA are sequenced by automated DNA sequences. On the basis of overlapping regions on DNA fragments, these sequences are arranged accordingly. For alignment of these sequences, specialised computer-based programmes were developed. Finally, the genetic and physical maps of the genome were constructed by collecting information about certain repetitive DNA sequences and DNA polymorphism, based on endonuclease recognition sites.

Question.List the various markers that are used in DNA fingerprinting.
Answer : Dr. Alec Jeffreys developed the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases. DNA fingerprinting uses short nucleotide repeats called Variable Number Tandem Repeats (VNTRs) as markers. VNTRs vary from person to person and are inherited from one generation to the next. Only closely individuals have similar VNTRs.

Question. Replication was allowed to take place in the presence of radioactive deoxynucleotides precursors in E.coli that was a mutant for DNA ligase. Newly synthesised radioactive DNA was purified and strands were separated by denaturation. These were centrifuged using density gradient centrifugation. Which of the following would be a correct result? (Image 123)
Answer : In above case, as E.coli is a mutant for DNA ligase, it will result in no further joining of Okazaki fragments on lagging strand. This will ultimately result into the formation of both high molecular weight fragments (on leading strands) and low molecular weight fragments (on lagging strand). Hence, only the graph (a) could be the appropriate result after centrifugation.

Question. Name the source of energy for the replication of DNA.
Answer. The sources of energy for the replication of DNA are phosphorylated nucleotides or deoxyribonucleoside triphosphates i.e., deATP, deCTP, deGTP and deTTP.

Question. Why is it not possible for an alien DNA to become part of chromosome anywhere along its length and replicate normally? 
Answer. It is not possible for an alien DNA to become a part of the chromosome anywhere along the length and replicate normally because of absence of origin of replication (ori). It is the sequence where DNA replication starts. This site is also necessary for binding of DNA polymerase to start DNA replication. As this site may not present in all alien DNA molecules hence they cannot replicate normally.

Question. What will happen if DNA replication is not followed by cell division in a eukaryotic cell?
Answer. DNA replication doubles the amount of DNA in a cell and cell division again halves the amount of DNA i.e., maintains the normal amount of DNA in the daughter nuclei. Thus, if DNA replication is not followed by cell division in a eukaryotic cell, then amount of DNA will increase than normal, resulting in abnormal conditions such as polyploidy

Question. Name the enzyme and state its property that is responsible for continuous and discontinuous replication of the two strands of a DNA molecule.
Answer. The enzyme DNA-dependent DNA polymerase catalyse the polymerisation of deoxynucleotides.
It catalyses polymerisation only in 5′ → 3′
direction thereby resulting in continuous replication on DNA strand with 3′ → 5′ polarity and discontinuous replication on strand with 5′ → 3′ polarity

Question. Name the enzyme that joins the small fragment of DNA of a lagging strand during DNA replication. 
Answer. The enzyme DNA ligase joins the small fragments of DNA of a lagging strand during DNA replication.

Question. Name the enzyme involved in the continous replication of DNA strand. Mention polarity of templet strand.
Answer. DNA-dependent DNA polymerase is the main enzyme that using DNA as a templet catalyse polymerisation of deoxynucleotides. Polarity of the templet strand, where replication is continous (i.e.,5′ → 3′ direction) is 3′ → 5′.

Question. Discuss the role, the enzyme DNA ligase plays during DNA replication. 
Answer. DNA ligase is an enzyme that catalyses the repair of a single strand break by formation of covalent phosphodiester bond between adjacent 3′ hydroxyl and 5′ phosphoryl groups in double stranded DNA. Hence, the enzyme helps in sealing gaps in DNA fragments and acts as molecular glue.
During replication of DNA, short replicated fragments/segments of DNA or Okazaki fragments are present on the lagging strand. These fragments are joined to form a continuous strand with the help of enzyme DNA ligase.

Question. State the dual role of deoxyribonucleoside triphosphates during DNA replication.
Answer. Deoxyribonucleoside triphosphates (or phosphorylated nucleotides) i.e., deATP, deCTP, deGTP and deTTP serve dual purpose during DNA replication. They act as substrates for the replication process as well as provide energy for the polymerisation of nucleotides.

Question. Compare the roles of the enzymes DNA polymerase and DNA ligase in the replication.
Answer. Differences between DNA polymerase and DNA ligase are

DNA polymerase	DNA ligase
DNA polymerase catalyse the elongation of a new DNA strand during DNA replication, using an existing DNA strand as template.	DNA replication on one of the strand is discontinuous. The discontinuously synthesised fragments are joined by the enzyme DNA ligase.

Question. Describe the experiment that helped demonstrate the semi-conservative mode of DNA replication.
Answer. The work of Mathew Meselson and Franklin Stahl (1958) on E. coli proved semi-conservative replication of DNA.
They first grew bacteria Escherichia coli in a medium containing heavy isotope 15N for several generations. This led to the incorporation of heavy isotope in all nitrogen-containing compounds including bases. They were able to extract the bacterial DNA and centrifuge it in caesium chloride solution. Depending on the mass of the molecule, the DNA would settle out at a particular point in the tube.
The 15N bacteria were then transferred to a growth medium containing the normal, lighter isotope of nitrogen, 14N, where they reproduced by cell division. Meselson and Stahl found that DNA of the first generation was hybrid or intermediate (15N and 14N). It settled in caesium chloride solution at a level higher than the fully labelled DNA of parent bacteria (15N15N). The second generation of bacteria after 40 minutes, contained two types of DNA, 50% light (14N14N) and 50% intermediate (15N14N). At succeeding generation times, the DNA extracts were found to have a lower proportion of 15N as more 14N was incorporated into the bacterial DNA. This observation is possible only if both strands separate during replication and one strand act as template for synthesis of new strand of DNA having 14N. This was conclusive evidence for the semi-conservative method of DNA replication.

Question. (a) Why did Meselson and Stahl use 14N and 15N isotopes in the sources of nitrogen present in the culture medium in their experiment? Explain.
(b) Write conclusion drawn by the them from the experiment.
Answer. (a) Meselson and Stahl used 14N and 15N isotopes for their experiment because 15N is a heavy isotope of nitrogen and can be separated from 14N by density gradient centrifugation using caesium chloride.
Using two different isotopes of nitrogen helped them to isolate the different generations of E. coli bacteria from each other, e.g., they found that DNA of first generation was hybrid (15N and 14N), it settled in density gradient centrifugation at a level higher than the fully labelled DNA of parent bacteria (15N 15N). Succeding generations were also found to settle toward the surface.
(b) From the experiment they concluded that DNA
replication is semiconservative.

Question. 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 synthesise the light and the heavy DNA molecules in the organism used in the experiment?
(c) How did the scientists 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. 
Answer. (a) NH4Cl (Ammonium chloride) was used as a source of nitrogen by Meselson and Stahl in their experiment.
(b) They synthesised light and heavy DNA molecules in the organim (E. coli bacterium) so as to separate different generations on the basis of density gradient centrifugation.
(c) The DNA molecules from each generations, were tested through density gradient centrifugation using cesium chloride. The heavy DNA molecules settled at the bottom whereas successively light DNA molecules settled at the surface. In this way, the DNA molecules were distinguished.
(d) From the experiment scientist concluded that DNA replication is semi conservative in nature.

Question. Explain the mechanism of DNA replication in bacteria. Why is DNA replication said to be semiconservative? 
Answer. DNA replication is a multistep complex process, which in all living cells (including E. coli), requires a dozen enzymes and protein factors. It begins at a particular spot called origin of replication or ori.
For DNA replication to proceed, the first requirement is to unwind the double helix, so that the two strands are free to act as templates. Separating the two strands of DNA is accomplished by the helicase enzymes that travel along the helix, opening the double helix by breaking H–bonds between nucleotide pairs as they move. Unwinding also creates a coiling tension in front of the moving replication fork. This tension is reduced by topoisomerases. In prokaryotes topoisomerase is replaced by DNA gyrase. Prokaryotes have three major types of DNA synthesising enzyme called DNA polymerase I, II and III. All the three DNA polymerases function in 5′ → 3′ direction only for DNA polymerisation and have 3′ → 5′ exonuclease activity.
To initiate DNA synthesis, a small segment of RNA called RNA primer, complementary to the template DNA, is synthesised by a enzyme primase.
The two strands of DNA run antiparallel to each other. On one strand the DNA synthesis is continuous in 5′ → 3′ direction, on the other strand, DNA is synthesized in small stretches resulting in discontinuous DNA synthesis. This happens in the opposite direction to the first strand but maintains the overall 5′ → 3′ direction as required. Such a process is also referred to as semi-discontinuous replication. The short stretches of DNA, each primed by RNA are called Okazaki fragments.
RNA primer are then removed, and the gap is filled by DNA synthesis. Both steps are performed by DNA polymerase I. The enzyme ligase then seals these fragments.
The strand which supports the continuous DNA synthesis is the leading strand and the one, which is replicated in short stretches is called the lagging strand.
During DNA replication, one strand of the daughter DNA duplex is derived from the parental strand while the other strand is formed a new.
Since the daughter DNA duplex comprises of one parental and one new strand, thus the replication process is said to be semi-conservative.

Question. (a) Why does DNA replication occur in small replication forks and not in its entire length?
(b) Why is DNA replication continuous and discontinuous in a replication fork?
(c) Explain the importance of ‘orgin of replication’ in a replication fork.
Answer. (a) For long DNA molecules, since the two strands of DNA cannot be separated in its entire length (due to very high energy requirement), the replication occurs within a small opening of the DNA helix, referred to as replication fork.
(b) DNA-polymerase can polymerise nucleotides only in 5′ → 3′ direction on 3′ → 5′ strand because it adds them at the 3′ end. Since the two strands of DNA run in antiparallel directions, the two templates provide different ends for replication. Replication over the two templates thus proceeds in opposite directions. One strand with polarity 3′ → 5′ forms its complementary strand continuously because 3′ end of the latter is always open for elongation. It is called leading strand. Replication is dicontinuous on the other template with polarity 5′ → 3′ because only a short segment of DNA strand can be built in 5′ → 3′ direction due to exposure of a small stretch of template at one time.
(c) Origin of replication, ‘ori’ is a sequence present in DNA molecule where replication starts. Enzyme helicase acts over the ori site and unwinds the two strands of DNA to start the process of replication. 

Question. (a) Name the stage in cell cycle where DNA replication occur.
(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) DNA replication occurs during S-phase of the cell cycle.
(b) DNA replication is a multistep complex process, which requires a dozen enzymes and protein factors. It begins at a particular spot called origin of replication or ori.
Separating the two strands of DNA is accomplished by the helicase enzymes that travel along the helix, opening the double helix as they move. Unwinding also creates a coiling tension in front of the moving replication fork, a structure that will be formed when DNA replication begins. This tension is reduced by topoisomerases.
The very important DNA synthesising enzyme is DNA polymerase III. It along with other DNA polymerases (I and II) has the ability to elongate an existing DNA strand but cannot initiate the synthesis.
All the three DNA polymerases function in 5′ → 3′ direction only for DNA polymerisation and have 3 ′ → 5′ exonuclease activity.
To initiate DNA synthesis, a small segment of RNA (10 to 60 nucleotides) called an RNA primer complementary to the template DNA is synthesised by a unique RNA polymerase known as primase. While on the one strand the DNA synthesis is continuous in 5′→ 3′ direction, on the other strand, DNA is synthesised in small stretches resulting in discontinuous DNA synthesis. This happens in the opposite direction to the first strand but maintains the overall 5′ → 3′ direction as required. Such a process is also referred to as semi-discontinuous replication. The short stretches of DNA, each primed by RNA are called Okazaki fragments.
RNA primer are then removed, and the gap is filled by DNA synthesis by DNA polymerase I. The enzyme ligase then seals these fragments.
The strand which supports the continuous DNA synthesis is the leading strand and the one, which is replicated in short stretches is called the lagging strand.
(c) DNA replication is said to be semiconservative since daughter DNA duplex comprise of one parental and one newly synthesised strand.

Question. Describe Meselson and Stahl’s experiment that was carried in 1958 on E. coli. Write the conclusion they arrived at after the experiment.
Answer. The work of Mathew Meselson and Franklin Stahl (1958) on E.coli proved semi-conservative replication of DNA.
They first grew Escherichia coli bacteria in a medium containing heavy isotope of nitrogen (15N) for several generations. This led to the incorporation of heavy isotope in all nitrogen-containing compounds including bases. They were able to extract the bacterial DNA and centrifuge it in caesium chloride solution. Depending on the mass of the molecule, the DNA would settle out at a particular point in the tube (heavy DNA molecule can be distinguished from normal DNA by centrifugation in cesium chloride density gradient).
The 15N bacteria were then transferred to a growth
medium containing the normal, lighter isotope of nitrogen, 14N, where they reproduced by cell division. Extracts of DNA from the first generation offspring were shown to have a lower density, since half the DNA was made up of the original strand containing 15N and the other half was made up of the new strand containing 14N. At succeeding generation times, the DNA extracts were found to have a lower proportion of 15N as more 14N was incorporated into the bacterial DNA. This was conclusive evidence for the semi-conservative method of DNA replication. The conclusion they arrived at after the experiment is that the DNA replication is semiconservative. Semiconservative means that when the double stranded DNA helix was replicated, each of the two double stranded DNA helices of newly synthesised strands consisted of one strand coming from the original helix and one newly synthesised. So, in this way at each replication, one strand of parent DNA is conserved in the daughter while the second is freshly synthesised.

Question. (a) Draw a labelled diagram of a “replicating fork” showing the polarity. Why does DNA replication occur within such ‘forks’?
(b) Name two enzymes involved in the process of DNA replication, along with their properties.
Answer. Due to high energy requirement whole of DNA does not open in one stretch. The point of separation proceed slowly towards both direction. It gives the appearance of Y shaped structure called replication fork.
(b) Two enzymes involved in the process of DNA replication are:
(i) Helicase - causes the unwinding of DNA strand.
(ii) Topoisomerase - releases the tension of DNA strand.

Question. Explain the process of DNA replication with the help of a replicating fork.
Answer. DNA replication occurs during S-phase of cell cycle. It is a multistep complex process which requires over a dozen enzymes and protein factors. It begins at a particular spot called origin of replication of ori.
Replication of DNA is energetically highly expensive. The main enzyme of DNA replication is DNA dependent DNA polymerase.
Deoxyribonucleoside monophosphates occur freely inside the nucleoplasm. They are first phosphorylated and changed to active forms. The phosphorylated nucleotides are deATP (deoxyade- nosine triphosphate), deGTP (deoxyguanosine triphosphate), deCTP (deoxycytidine triphosphate) and deTTP (deoxythymidine triphosphate)
Enzyme helicase (unwindase) acts over the Ori site and unzips (unwids) the two strands of DNA by destroying hydrogen bonds. Unwinding creates tension in the uncoiled part by forming more supercoils. Tension is released by enzymes topoisomerases. With the help of various enzymes both the strands of DNA become open for replication. However, whole of DNA does not open in one stretch due to very high energy requirement. The directions. In each direction, it gives the appearance of Y-shaped structure called replication fork.
RNA primer is a small strand of RNA which is synthesised at the 5′ end of new DNA strand with the help of DNA specific RNA polymerase enzyme called primase.
Prokaryotes have three major types of DNA synthesising enzymes called DNA polymerases III, II and I. In eukaryotes five types of DNA polymerases are found - α, β, γ, δ and ε, but the major three being α, β, and ε. The two separated DNA strands in the replication fork function as templates.
As replication proceeds, new areas of parent DNA duplex unwind and separate so that replication proceeds rapidly from the place of origin towards the other end. RNA primer is removed and the gap filled with complementary nucleotides by means of DNA polymerase I. DNA-polymerase can polymerise nucleotides only in 5′ → 3′ direction on 3′ → 5′ strand because it adds them at the 3′ end. Since the two strands of DNA run in antiparallel directions, the two templates provide different ends for replication. Replication over the two templates thus proceeds in opposite directions. One strand with polarity 3′ → 5′ forms its complementary strand continuously because 3′ end of the latter is always open for elongation. It
is called leading strand. Replication is dicontinuous on the other template with polarity 5′ → 3′ because only a short segment of DNA strand can be built in 5′ → 3′ direction due to exposure of a small stretch of template at one time. Short segments of replicated DNA are called Okazaki fragments.
Okazaki fragments are joined together by means of enzyme, DNA ligase. DNA strand built up of Okazaki fragments is called lagging strand.

Question. State the aim and describe Meselson and Stahl’s experiment. 
Answer. The aim of the experiment of Meselson and Stahl was to prove semiconservative replication of DNA.

Question. (a) What did Meselson and Stahl observe when
(i) they cultured E. coli in a medium containing 15NH4CI for a few generations and centrifuged the content?
(ii) they transferred one such bacterium to the normal medium of NH4CI and cultured for 2 generations?
(b) What did Meselson and Stahl conclude from this experiment? Explain with the help of diagrams.
(c) Which is the first genetic material? Give reasons in support of your answer.
Answer. (a) (i) Meselson and Stahl observed that in the E. coli, the DNA became completely labelled with N15 when cultured in 15NH4Cl and after centrifugation of this culture they observed that fully labelled DNA of the bacteria settled at the lowest level (near bottom) in the test tube.
(ii) After culturing, one such bacterium, in normal medium of NH4Cl for two generations they observed that its density changed and it showed 50% of light DNA (N14) and 50% intermediate (N15N14).
(b) They concluded that DNA replicates semi- conservatively.
(c) RNA is the first genetic material as RNAs were the first biocatalysts and essential life processes such as metabolism splicing and translation evolved around RNA.

Question. Who proposed that DNA replication is semiconservative?
How was it experimentally proved by Meselson and Stahl? 
Answer. Waston and Crick proposed that DNA replication is semi-conservative. 

Question. What is a cistron? 
Answer. Cistron is a segment of DNA consisting of a stretch of base sequences that codes for one polypeptide chain, one transfer RNA (tRNA), ribosomal RNA (rRNA) molecule or performs any other specific function in connection with transcription, including controlling the functioning of other cistrons.

Question. Write the function of RNA polymerase II.
Answer. In eukaryotes, RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hn RNA).

Question. Differentiate between exons and introns.
Answer. Exons are the segments in genes which contain coding nucleotide sequences. These sequences are ultimately translated into polypeptide. Thus, exons carry genetic information. Introns are the segments in genes which contain non-coding nucleotide sequences. These do not form part of mRNA and are removed during the processing of hnRNA.

Question. Which one out of Rho factor and Sigma factor act as initiation factor during transcription in a prokaryote? 
Answer. Sigma (σ) factor acts as initiation factor during transcription in a prokaryote.

Question. Which one of an intron and an exon is the reminiscent of antiquity? 
Answer. Intron is considered to be as the reminiscent of antiquity.

Question. Why hnRNA is required to undergo splicing?
Answer. hnRNA undergo splicing in order to remove introns, which are intervening or non-coding sequences, and exons are joined in a defined order to form functional mRNA.

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

Question. When and at what end does the ‘tailing’ of hnRNA take place? 
Answer. Tailing’ of hnRNA take place during modification of hnRNA into functional mRNA. It takes place at 3′-end.

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

 

Very Short Answer

Question. Who identified DNA as an acidic substance?
Answer. DNA as an acidic substance is identified by Friedrich Meischer.

Question. What is DNA?
Answer. DNA is a long polymer of deoxyribonucleotides.

Question. What are the two types of nucleic acids found in the living systems?
Answer. DNA and RNA are the two types of nucleic acids found in living systems.

Question. How the backbone of a polynucleotide chain is formed?
Answer. The backbone of a polynucleotide chain is formed to sugar and phosphates.

Question. What is the length of the DNA?     
Answer. The length of the DNA is usually defined as number of nucleotides present in it.

Short Answer

Question. Define nucleosome?
Answer. A nucleosome is a section of DNA that is wrapped around a core of proteins. Inside the nucleus, DNA forms a complex with proteins called chromatin, which allows the DNA to be condensed into a smaller volume. The nucleosome is the fundamental subunit of chromatin.

Question. Distinguish between chromatin and chromosome?
Answer. 1. Chromosomes appear during the metaphase and exist in the anaphase of the nuclear division whereas chromatin appears in the interphase of the cell cycle.
2. Chromosomes are condensed into chromatin fibres whereas chromatin is composed of nucleosomes.
3. Chromosomes are thick, compact, ribbon-like structures whereas chromatin fibres are thin, long, uncoiled structures.

Question. What do you mean by DNA helix?
Answer. Helix is the description of the structure of a DNA molecule. A DNA molecule consists of two strands that wind around each other like a twisted ladder. Each strand has a backbone made of alternating groups of sugar (deoxyribose) and phosphate groups.

Question. What do you mean by histones and histone octamer?
Answer. Histones: Histones are highly basic proteins found in eukaryotic cell nuclei that pack and order the DNA into structural units called nucleosomes. Histones are the chief protein components of chromatin, acting as spools around which DNA winds, and playing a role in gene regulation.
Histone Octamer: A histone octamer is the eight protein complex found at the center of a nucleosome core particle. It consists of two copies of each of the four core histone proteins (H2A, H2B, H3 and H4).

Question. How euchromatin is different from heterochromatin?   
Answer. 1. Euchromatin is the uncoiled form of chromatin whereas heterochromatin is a part of chromosome. It is tightly packed.
2. Euchromatin contains a low DNA density compared to heterochromatin whereas heterochromatin contains a high density of DNA.
3. Euchromatin does not exhibit heteropycnosis whereas heterochromatin exhibits heteropycnosis.

Long Answer

Question. State the difference between RNA and DNA?
Answer. 1. DNA stands for deoxyribonucleic acid whereas RNA stands for ribonucleic acid.
2. DNA is a long polymer whereas RNA is shorter than DNA.
3. DNA is mostly found in nucleus and nucleoid whereas RNA is mostly found in the cytoplasm.
4. DNA prefers B-form whereas RNA prefers A-form.
5. DNA is double-stranded and it exhibits a double-helix structure whereas RNA is usually single-strand, sometimes it forms secondary and tertiary structures.
6. DNA is more prone to damage by UV whereas RNA is less prone to damage by UV.
7. DNA carries the genetic information necessary for the development, functioning, and reproduction whereas RNA is mainly involved in protein synthesis; sometimes it regulates the gene expression.

Question. What are the salient features of the Double-helix structure of DNA?
Answer. The salient features of the double-helix structure of DNA are:
1. It is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate and the bases project inside.
2. The two chains have anti-parallel polarity. It means that one chain has thepolarity 5’ → 3’ the other has 3’ → 5’.
3. The bases in two strands are paired through hydrogen bond forming base pairs.Adenine forms two hydrogen bonds with the Thymine from opposite strand and vice versa.
4. The two chains are coiled in a right handed fashion. The pitch of the helix is 3.4 nm and there are roughly 10 bp in each turn. The distance between a bp in a helix is approx. 0.34nm.
5. The plane of one base pair stacks over the other in double helix.

Question. Distinguish between prokaryotes DNA and eukaryotes DNA?
Answer. 1. Prokaryotic DNA is found in the cytoplasm of prokaryotic cells as well as circular plasmids whereas eukaryotic DNA is found in the nucleus of the cell, inside the chloroplast and mitochondria.
2. Prokaryotic DNA consists of one copy of the genome whereas eukaryotic DNA consists of more than one copies of the genome.
3. Prokaryotic DNA is not found inside organelles whereas eukaryotic DNA is found inside chloroplast and mitochondria as well.
4. Prokaryotic DNA is not packed with histones whereas eukaryotic DNA found in the nucleus packed with histones.
5. Prokaryotic DNA replication occurs in the cytoplasm whereas eukaryotic DNA replication occurs in the nucleus.

Question. Explain the structure of polynucleotide chain?
Answer. A sequence of nucleotides joined together. RNA generally consists of one polynucleotide chain, while DNA may consist of one chain (single-stranded DNA), or two chains bonded between the bases in a system of complementary pairing: adenine with thymine, guanine with cytosine (double-stranded DNA). DNA molecules have two polynucleotide chains, held together in a ladder like structure. The sugar phosphate backbones of the two chains run parallel to each other in opposite directions. Each rung of the ladder is a pair of nitrogenous bases, one purine and one pyrimidine extending into the center of the molecule. DNA consists of two polynucleotide chains wound around each other to form a double helix. The two chains are held together by complementary base pairing; that is, specific bonding between A and T bases and between G and C bases on the two strands. Double helix is the description of the structure of a DNA molecule. A DNA molecule consists of two strands that wind around each other like a twisted ladder. Each strand has a backbone made of alternating groups of sugar (deoxyribose) and phosphate groups.

Question. State the difference between nucleosome and chromatin?
Answer. 1. Chromatin is a complex of DNA and proteins that form chromosomes within the nucleus of eukaryotic cells whereas nucleosome is the main structural unit of the
eukaryotic chromatin that consists of a length of DNA coiled around a core of histones.
2. Chromatin forms the chromosomes whereas nucleosomes form chromatin.
3. Chromatin appears as a thread-like, looped structure whereas nucleosome appears as beads on a string.
4. Chromatin is the general term for DNA wrapped around histones whereas nucleosome is the basic repeating, structural unit of chromatin.
5. Chromatin is more condensed than nucleosomes whereas nucleosomes are the least condensed chromosome structures.
6. The diameter of a chromatin fiber is 30 nm whereas the diameter of a nucleosome is 11 nm.

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