CBSE Class 12 Biology Molecular Basis Of Inheritance Worksheet Set B

Important Questions for NCERT Class 12 Biology Molecular Basis of Inheritance

Question. A m-RNA also has some additional sequences that are not translated called UTR. The function of UTR is
(a) Charging of t-RNA
(b) Formation of peptide bond
(c) Helps in efficient translation
(d) Helps in translocation

Answer : C

Question. The t-RNA move away from ribosomes after translocation of ribosome in relation to m-RNA, is known as
(a) Acylated t-RNA
(b) Peptidyle t-RNA
(c) Deacylated t-RNA
(d) Charged t-RNA

Answer : C

Question. Which of the following evidence suggests that essential life processes evolved around RNA
(a) RNA used to act as genetic material
(b) RNA can act as catalyst
(c) RNA is highly reactive
(d) Both 1 and 2

Answer : D

Question. Regarding to Meselson and Stahl experiment for semi conservative nature of DNA replication select out the wrong statement
(a) 15N of 15NH4Cl was incorporated in DNA and other compounds
(b) 15N & 14N can be differentiate on the basis of radioactive activity
(c) Heavy and normal DNA molecules could be distinguished by CsCl density gradient centrifugation
(d) 15N used in 15NH4Cl was not a radioactive isotope

Answer : B

Question. In nucleoside which of the following bond exists between sugar and nitrogenaous base
(a) Phosphodiester bond
(b) Hydrogen bond
(c) Phosphoester bond
(d) N-glycosidic bond

Answer : D

Question. By which of the following bond phosphoric acid remain linked with 5' carbon of sugar in one nucleotide
(a) Phosphoester bond
(b) Phosphodiester bond
(c) N-Glycosidic bodn
(d) Hydrogen bond

Answer : A

Question. If E.Coli is allow to grow for 80 minutes in 15NH4Cl medium then what would be the proportion of hybrid and heavy density DNA molecules
(a) 1 : 7
(b) 7 : 1
(c) 14 : 2
(d) 1 : 4

Answer : A

Question. Which of the following cell cycle event is responsible for aneuploidy based chromosomal disorders
(a) Failure of G1 phase
(b) Failure ofDNA replication in s-phase
(c) Failure of segregation/Disjunction
(d) Failure of movement of chromosomes

Answer : C

Question. How much duration of time required for replication of 4.6 x 106pb in E.coli
(a) 83 minutes
(b) 20 minutes
(c) 2 minutes
(d) 3 hrs

Answer : B

Question. What is the rate of polymerisation in E.coli
(a) 20,000 bp per second
(b) 2000 nucleotides per second
(c) 2000 bp per minute
(d) 2000 bp per second

Answer: D

Ques. In an inducible operon, the genes are
(a) usually not expressed unless a signal turns them”on”.
(b) usually expressed unless a signal turns them “off”.
(c) never expressed
(d) always expresser. 

Answer: A

Ques. Select the two correct statements out of the four (i – iv) statements given below about lac operon.
(i) Glucose or galactose may bind with the repressor and inactivate it.
(ii) In the absence of lactose, the repressor binds with the operator region.
(iii) The z-gene codes for permease.
(iv) This was elucidated by Francois Jacob and Jacques Monod.
The correct statements are
(a) (ii) and (iii) (b) (i) and (iii)
(c) (ii) and (iv) (d) (i) and (ii). 

Answer: C

Ques. The lac operon consists of
(a) four regulatory genes only
(b) one regulatory gene and three structural genes
(c) two regulatory genes and two structural genes
(d) three regulatory genes and three structural genes.

Answer: B

Ques. E.coli cells with a mutated z gene of the lac operon cannot grow in medium containing only lactose as the source of energy because
(a) the lac operon is constitutively active in these cells
(b) they cannot synthesise functional beta galactosidase
(c) in the presence of glucose, E.coli cells do not utilise lactose
(d) they cannot transport lactose from the medium into the cell. 

Answer: B

Ques. What does “lac” refer to in what we call the lac operon ?
(a) Lactose
(b) Lactase
(c) Lac insect
(d) The number 1,00,000

Answer: A

Ques. Jacob and Monod studied lactose metabolism in E. coli and proposed operon concept. Operon concept is applicable for
(a) all prokaryotes
(b) all prokaryotes and some eukaryotes
(c) all prokaryotes and all eukaryotes
(d) all prokaryotes and some protozoans. 

Answer: C

Ques. In E. coli, during lactose metabolism repressor binds to
(a) regulator gene
(b) operator gene
(c) structural gene
(d) promoter gene. 

Answer: B

Ques. In negative operon,
(a) co-repressor binds with repressor
(b) co-repressor does not bind with repressor
(c) co-repressor binds with inducer
(d) cAMP have negative effect on lac operon.

Answer: A

Ques. In operon concept, regulator gene functions as
(a) inhibitor
(b) repressor
(c) regulator
(d) all of these. 

Answer: B

Ques. The wild type E.coli cells are growing in normal medium with glucose. They are transferred to a medium containing only lactose as sugar. Which of the following changes take place?
(a) The lac operon is induced.
(b) E.coli cells stop dividing.
(c) The lac operon is repressed.
(d) All operons are induced. 

Answer: A

Ques. The lac operon is an example of
(a) repressible operon
(b) overlapping genes
(c) arabinose operon
(d) inducible operon.

Answer: D

Ques. An environmental agent, which triggers transcription from an operon, is a
(a) depressor
(b) controlling element
(c) regulator
(d) inducer. 

Answer: D

Ques. ‘Lac operon’ in E. coli, is induced by
(a) ‘I’ gene
(b) promoter gene
(c) β-galactosidase
(d) lactose. 

Answer: C

Ques. Expressed Sequence Tags (ESTs) refers to
(a) novel DNA sequences
(b) genes expressed as RNA
(c) polypeptide expression
(d) DNA polymorphism. 

Answer: B

Ques. Identify the correct order of organisation of genetic material from largest to smallest.
(a) Genome, chromosome, gene, nucleotide
(b) Chromosome, genome, nucleotide, gene
(c) Chromosome, gene, genome, nucleotide
(d) Genome, chromosome, nucleotide, gene 

Answer: A

Ques. Satellite DNA is important because it
(a) does not code for proteins and is same in all members of the population
(b) codes for enzymes needed for DNA replication
(c) codes for proteins needed in cell cycle
(d) shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable from parents to children.

Answer: D

Ques. Which of the following is not required for any of the techniques of DNA fingerprinting available at present?
(a) Restriction enzymes
(b) DNA-DNA hybridisation
(c) Polymerase chain reaction
(d) Zinc finger analysis 

Answer: D

Ques. One of the most frequently used techniques in DNA fingerprinting is
(a) VNTR
(b) SSCP
(c) SCAR
(d) AFLP.

Answer: A

Ques. What is it that forms the basis of DNA finger-printing?
(a) The relative proportions of purines and pyrimidines in DNA.
(b) The relative difference in the DNA occurrence in blood, skin and saliva.
(c) The relative amount of DNA in the ridges and grooves of the fingerprints.
(d) Satellite DNA occurring as highly repeated short DNA segments.

Answer: D

Ques. DNA fingerprinting refer to
(a) molecular analysis of profiles of DNA samples
(b) analysis of DNA samples using imprinting devices
(c) techniques used for molecular analysis of different specimens of DNA
(d) techniques used for identification of fingerprints of individuals.

Answer: A

Molecular Basis of Inheritance Questions and Answers

Question. What is DNA polymorphism? What is it important to study it?
Answer : DNA polymorphism refers to the variation in DNA arising through mutation at non-coding sequences.
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 genome, therefore, it forms the basis of DNA fingerprinting too.
The single nucleotide polymorphisms are used in locating diseases and tracing of human history as well as in case of paternity testing.

Question. Based on your understanding of genetic code, explain the formation of any abnormal haemoglobin molecule. What are the known consequences of such a change?
Answer : Due to point mutation in b-globin chain of haemoglobin molecule, glutamic acid (Glu) is replaced by valine (Val) at the sixth position.
Under stress condition erythrocytes lose their circular shape and become sickle-shaped. As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and
thus affect blood supply to different organs

Question. Sometimes cattle or even human beings give birth to their young ones that are having extremely different sets of organs like limbs/position of eye(s) etc. Comment.
Answer : Sometimes cattle or even human beings give birth to their young ones that are having extremely different of organs like limbs/position of eye etc. It happens due to the disturbance in coordinated regulation of expression in sets of genes, which are associated with organ development.

Question. In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxy x 10 ribonucleoside triphosphates, but only deoxy ribonucleotides are added during the DNA replication. Suggest a mechanism.
Answer : DNA polymerase enzyme is highly specific to recognise only deoxy ribonucleoside triphosphates. Therefore, it cannot hold RNA b-nucleotides.

Question. Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.
Answer : The enzymes involved in DNA replication other than DNA polymerase and ligase are listed  below with their functions.
(i) Helicase — Opens the helix
(ii) Topoisomerases — Removes the super coiling of DNA
(iii) Primase — Synthesises RNA primer
(iv) Telomerase — To synthesis the DNA of telomeric end of chromosomes.

Question. Name any three viruses which have RNA as the genetic material.
Answer : In some viruses, RNA is the genetic material.
e.g., Tobacco mosaic viruscs, QB bacteriophage, HIV, influenza virus, etc.

Question. Why are histone positively charged ?
Answer : For wrapping of negatively charged DNA, histone protein is also rich in lysine and arginine carrying positive charge in its chain.

Question. Regulation of lac operon by repressor is referred to as negative.
Answer : The gene that codes for the repressor protein that binds to operator and suppresses its activity, as a result transcription will be switched off.

Question. Differentiate between template strand and coding strand.
Answer : a)Template strand show 3-5 polarity and coding shows 5-3 polarity
       b) template strand takes part in transcription and coding strand does not.

Question. A cistron consist 20 codons . How many amino acids will it code in the polypeptide transcribed?
Answer : 19 – amino acid, because last codon on mRNA will be a terminating codon.

Question. If a double stranded DNA has 20 percent cytosine, calculate the percent of adenine in DNA.
Answer :
 Accoding Chargaff”s rule=A+T=100-(G+C) C=20/hence G=20/ A+T=100-20=60 hence A=60/2=30/

Molecular basis of Inheritance Questions and Answers

Question. Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as the genetic material.
Answer : In Griffith’s experiment, transformation can be defined as a change in the genetic constitution of an organism by picking out up DNA from the environment (from dead organisms). Transformation helps in identification of DNA as a genetic material. When heat was used to kill the virulent bacteria, they died but not their genetic material (DNA). This DNA when picked up by non-virulent bacteria made them capable of causing infection. Since, ability to cause infection could be passed on by these organisms to their progeny, it was concluded that DNA was the material that was inherited.

Question. Who revealed biochemical nature of the transforming principle?
Answer : Oswald, Avery, Colin MacLeod and Maclyn McCarty revealed biochemical nature of the transforming principle. They reported Griffith’s experiment in an in vitro system in order to determine biochemical nature of transforming principle. They reported that DNA from the heat-killed S-type bacteria caused the transformation of non-virulent R-type bacteria into virulent S-type bacteria. They also discovered that proteases and RNase did not affect transformation while DNase inhibited the process. They concluded that DNA is the hereditary material.

Question. During DNA replication, why is it that the entire molecule does not open in one go? Explain replication fork. What are the two functions that the monomers (dNTPs) play?
Answer : While replicating, the entire DNA molecule to keep the whole molecule stabilised does not open in one go because it would be highly expens energetically. Actually unwiding creates tension in the molecule as uncoiled parts. Actually, unwinding creates tension in the molecule as uncoiled parts start forming super coils due to the interaction of exposed nucleotides. Instead, helicase enzyme acts on the double strand at ori site (origin of replication) and a small stretch is unzipped. Immediately, it is held and stabilised by single strand binding proteins. Slowly with the help of enzymes, exposed strands are copied as a point of unwinding moves and ahead in both directions. It gives an appearance of Y-shaped structure which is called replication fork.
The two functions that the monomer units of NTPs play are
(i) They pair up with exposed nucleotides of the template strand and make phosphodiester linkages and release a pyrophosphate. (ii) Hydrolysis of this pyrophosphate by enzyme pyrophosphatase releases energy that will facilitate making hydrogen bonds between free nucleotides and bases of the template strand.

Question. Retroviruses do no follow central dogma. Comment.
Answer : Retroviruses do not follow central dogma of biology (DNA®RNA®Protein) because their genetic material is not DNA. Instead they have RNA that is converted to DNA by the enzyme reverse transcriptase.

Question. In an experiment, DNA is treated with the compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases. From 0.34–0.44 nm calculate the length of DNA double helix (which has 2 × 109 bp) in the presence of saturating of this compound.
Answer : The length of DNA double helix = 2 × 10 × 0 44 × 10− 9 9 . / bp.

Question. What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
Answer : If histones were mutated and made rich in acidic amino acids. They will not be able to serve the purpose of keeping the DNA coiled around them. This is because DNA is negatively charged molecule and histones are positively charged because of basic amino acids. So, they are attracted to each other. If histones become negatively charged, instead of binding, they will rather repel DNA. The packaging of DNA in eukaryotes would not happen. Consequently, the chromatin fibre would not be formed.

Question. What are the functions of
(i) methylated guanine cap?
(ii) poly-A ‘tail’ in a mature on RNA?
Answer : (i) Methylated guanine cap helps in binding of mRNA to smaller ribosomal sub-unit during initiation of translation.
(ii) Poly-A tail provides longevity to mRNA’s life. Tail length and longevity of mRNA are positively correlated.

Question. Do you think that the alternate splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene? If yes, how? If not, why so?
Answer : Functional mRNA of structural genes need not always include all of its exons. This alternate splicing of exons is sex-specific, tissue-specific and even developmental stage-specific. By such alternate splicing of exons, a single gene may encode for several isoproteins and/ or proteins of similar class. In absence of such a kind of splicing, there should have been new genes for every protein/isoprotein. Such an extravagancy has been avoided in natural phenomena by way of alternate splicing.

Question. Comment on the utility of variability in number of tandem repeats during DNA fingerprinting.
Answer : Tandemness in repeats provides many copies of the sequence for finger-printing and variability in nitrogen base sequences present in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

Question. Recall the experiments done by Frederick Griffith, Avery, MacLeod and 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 Pneumococcus have transformed the R-strain into virulent strain? Explain.
Answer : RNA is more liable and prone to degradation (owing to the presence of 2’OH group in its ribose). Hence, heat-killed S-stain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.

Question. What is hnRNA ?
Answer : The precursore RNA transcribed by RNA polymerase that contain both exons and introns.

Question. Write about intron?
Answer : The non-coding sequences in eukaryotic structural gene are called introns.

Question. Differentiate between genetic code and codon.
Answer : Genetic code is the sequence of base triplet in DNA molecule which determined a polyptides where as triplet base on mRNA that code for a partcular amino acid is called codon.

Question. What are UTRs?
Answer : mRNA has some additional sequences that are not translated are called UTRs. UTRs are present at both end.

Question. Distinguish between VNTR and Probe.
Answer : VNTR;- The segment of DNA which shows very high repetitive nucleotide sequences vary from person to person, called VNTR.
PROBE—small segment of DNA which are very specific complimentary to VNTR sequences are called probe.

Question. List the criteria a molecule that can act as genetic material must fulfill. Which one of the criteria are best fulfilled by DNA or by RNA thus making one of them a better genetic material than the other? Explain.
Answer : A molecule that can act as a genetic material must fulfill the following criteria:
(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 Characters’.
In DNA the two strands being complementary if separated by heating come together, when appropriate conditions are provided. Further, 2’-OH group present at every nucleotide in RNA is also now known to be catalytic, hence reactive. Therefore DNA chemically is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids,the DNA is a better genetic material. The presence of thymine at the place of uracil also confers additional stability to DNA.
Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at a faster rate. RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA,however, is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved around RNA.

Question. What background information did Watson and Crick had available with them for developing a model of DNA? What was their own contribution? 
Answer : Watson and Crick had the following informations which helped them to develop a model of DNA:
(i) Chargaff’s Law suggesting A=T and C G.
(ii) Wilkins and Franklin’s X-ray diffraction studies on DNA’s physical structure.
Based on these information, Watson and crick proposed
(i) complementary base-pairing of nitrogenous bases
(ii) semi-conservative mode of replication
(iii) occurrence of mutation through tautomerism.

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. You are repeating the Hershey–Chase experiment and are provided with two isotopes: 32P and 15N (in place of 35S in the original experiment). How do you expect your results to be different? 
Answer : Use of 15N will be inappropriate because method of detection of 35P and 15N is different (32P being a radioactive isotope while 15N is not radioactive but is the heavier isotope of nitrogen).
Even if 15N was radioactive then its presence would have been detected both inside the cell (l5N incorporated as nitrogenous base in DNA) as well as in the supernatant because 15N would also get incorporated in amino group of amino acids in proteins). Hence, the use of 15N would not give any conclusive results.

Question. Answer the following questions based on Hershey and Chases’s experiments:
(a) Name the kind of virus they worked with and why.
(b) Why did they use two types of culture media to grow viruses in? Explain.
(c) What was the need for using a blender and later a centrifuge during their experiments?
(d) State the conclusion drawn by them after the experiments. 
Answer : (a) They worked with bacteriophage because when it attacks a bacteria it only inserts its genetic material in its body.
(b) They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulphur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
(c) Blender was used to agitate the bacteria to remove the viral coats from them. Centrifuge was used to separate virus particle from the bacteria.
(d) Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.

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 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. 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. 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.

Question. (i) DNA polymorphism is the basis of DNA fingerprinting technique. Explain.
(ii) Mention the causes of DNA polymorphism. 
Answer : (i) Allelic sequence variation has traditionally been described as a DNA polymorphism if its frequency is greater than 0.01. Simply, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism. DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual.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 genome, therefore it forms the basis of DNA fingerprinting too.
(ii) The probability of such variations to be observed in non-coding DNA sequences would be higher as mutations in these sequences may not have any immediate effect in an individual’s reproductive ability. These mutations keep on accumulating generation after generation and form one of the basis of variability. There is a variety of different types of polymorphisms ranging from single nucleotide change to very large scale changes. For evolution and speciation, such polymorphisms play very important role.
The single nucleotide polymorphisms are used in locating diseases and tracing of human history.
DNA polymorphisms are due to mutations.

Question. Which methodology is used while sequencing the total DNA from a cell? Explain it in detail.
Answer : Methodologies of HGP:
- For sequencing, the total DNA from cell is first isolated and broken down in relatively small sizes as fragments.
- These 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 (YAC).
- 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.
- These sequences were annotated and were assigned to each chromosome. Sequence of chromosome 1 was completed only in May 2006. It was the last chromosome be sequenced).
- 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.

1 Write the two specific codons that a translational unit of mRNA is flanked by one on either sides.
2 Define nucleosome.
3 State the central dogma of molecular biology.
4 Name the enzyme responsible for synthesis of new DNA strand during DNA replication
5 Distinguish between euchromatin and heterochromatin.
6 State the functions of Ribozyme and release factor in protein synthesis respectively.
7 Differentiate between exons and introns.
8 How do histones acquire positive charge?
9 Both the strands of DNA are not copied during transcription. Give reason.
10 Transcription and translation are coupled in prokaryotes. Comment on the statement.
11 The base sequence in one of the strands of DNA is TAGCATGAT 3
(i) Give the base sequence of its complementary strand.
(ii) How are these base pairs held together in a DNA molecule ?
(iii) Explain the base complementarity rules. Name the scientist who framed this rule.
12 Draw a neat labeled sketch of replicating fork of DNA replication.
13 How would lac operon operate in E.coli growing in a culture medium where lactose is present as source of sugar?
14 a) What are the transcriptional products of RNA polymerase III ? (b)Differentiate between ‘Capping’ and ‘Tailing’. (c) Expand hnRNA.
15 a) Construct a complete transcription unit with promoter and terminator based on the hypothetical template strand given below.