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Class 12 Biology Chapter 6 Molecular Basis of Inheritance
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Molecular Basis of Inheritance
Chapter – 6
GENETICS AND EVOLUTION
Very Short Answer Type Questions
(A) . Fill in the blanks :
Q.1. In DNA is a long polymer of ___
Ans : Nucleotide.
Q.2. Uracil is present in ____
Ans : RNA.
Q.3. Purine and pyrimidine are collectively known as____
Ans : Nitrogenous base.
Q.4. The distance between two consecutive base pairs is _____
Ans : 0.34 A°.
Q.5. The chromatin that is more densely packed and stains dark is called ____
Ans : Hetero chromatin.
Q.6. DNA replication is termed as _____
Ans : Semiconservative.
Q.7. The DNAM2- dependent DNA polymerases catalyse polymerization in _____ direction.
Ans : 5’→3′.
Q.8. A gene is defined as the ___ unit of inheritance.
Ans : Fundamental.
Q.9. ____ is a segment of DNA coding for a polypeptide.
Ans : Cistron.
Q.10. AUG codes for ____
Ans : Methionine.
Q.11. refers to the process of polymerization of amino acids to form a _____
Ans : Translation Polypeptide.
Q.12. The ____ codes for the repressor in lac operon.
Ans : Gene-i
Q.13. Human genome project gave rise to a new area in biology called ___
Ans : Bioinformaties.
Q.14. Regulation of lac operon by repressor is referred to as _____
Ans : Regulatory gene.
Q.15. ____ arises due to mutation.
Ans : Abnormal variation.
Q.16. _____ and terminator segments flank the structural gene.
Ans : Initiation.
Q.17. An mRNA has some additional sequences that are not translated and are referred as _____
Ans : Termination codon.
Q.18. During transcription RNA polymerase binds to____
Ans : Promoter.
(B). True or False (1 mark) :
Q.1. Central Dogma was proposed Watson.
Ans : False.
Q.2. RNases are RNA digesting enzymes.
Ans : True.
Q.3. DNA duplication is semiconservative.
Ans : True.
Q.4. The process of formation of DNA from RNA is called transcription.
Ans : False.
Q.5. The discontinuously synthesized fragments are joined by the enzyme DNA polymerase.
Ans : True.
(C). Very short answer questions (1 mark) :
Q.1. What is satellite DNA?
Ans : It has been reported that heterochromatin contains certain polygenes to transcribe r RNA, m RNA and t RNA. Its DNA is formed of repeated polynucleotide sequences each being formed of about 300 nucleotides and is called satellite DNA or repetitive or redundant DNA.
Q.2. What is Central Dogma?
Ans : The process of protein synthesis involves one of the central dogma of molecular biology. According to this, genetic information flows from nucleic acid to protein, the flow of information takes place from DNA to RNA (m RNA) and from RNA to protein.
Q.3. Define autoradiography?
Ans : The mechanism of production of an autoradiogram by exposing photographic film to a radioactive substance in close proximity to the film.
Q.4. What is an inducer in lac operon?
Ans : All the genes present on a chromosome are not expressed simultaneously. The cell permits the expression of few genes at a time thus maintaining its economy. Inducible genes are the genes which remain inactive or repressed in a cell and can be activated when a certain substrate is to be metabolised. The phenomenon of action of these genes is called induction and metabolite is called inducer.
A lac operon is defined as several genes operating in random, all controlled by a common structural gene having repressor, promoter and operator. All these genes are essential for metabolism of lactose. Only presence of lactose can switches on the operon for synthesis of enzyme inside the cell and hence it is termed as inducer lac operon.
Q.5. Name the bacteria used by Griffith for his transformation experiment.
Ans : Streptococcus Pneumoniae.
Q.6. Who proved that DNA in chromosomes also replicate semiconservatively?
Ans : Mathew Messelson and Franklin stahl.
Q.7. How many structural genes are present in lac operon?
Ans : Three Lac – z, Lac Y and Lac – A.
Q.8. What is translation?
Ans : Translation refers to a process of polymerization of amino acids to form a polypeptide order and sequence of amino acid are defined by the sequence of base is the mRNA. The amino acids are joined by a bond called peptide bend.
Q.9. What is an intron?
Ans : In some viruses, the DNA sequence coding for polypeptide do not remain continuously but are split into several pieces. These spit genes are called introm.
Q.10. What is an exon?
Ans : DNA codes for mRNA, but complete sequence of DNA is not found in mRNA. The sequences of DNA found in mRNA are known as exon.
Q.11. What are the three major types of RNAS present in bacteria.
Ans : The three types of RNA are : mRNA, tRNA and rRNA.
Q.12. Name the enzyme used to catalyse the polymerization of deoxynucleotides.
Ans : DNA polymerase.
Q.13. Name the enzyme that joins the discontinuous fragments of DNA.
Ans : Ligase.
Q.14. Who proposed the semiconservative DNA replication.
Ans : Methew Messelson and Franklin Stahl.
Q.15. How many bases code for an amino acid.
Ans : Three.
Q.16. Give the site of protein synthesis.
Ans : Ribosome.
Q.17. What is the function of tRNA?
Ans : tRNA carry the amino acids to the ribosome according to the message carried by mRNA.
Q.18. Who proposed the operon model?
Ans : Francis Jacob and Jacques Monod.
Q.19. What is a bacteriophage?
Ans : The virus which kills bacteria.
Q.20. What is a euchromatin?
Ans : Chromatin of nucleus which stain positively with the DNA specific stain indicating the concentration of DNA is called euchromatin.
Q.21. Define chromatin.
Ans : DNA and histone protein together is called chromatin.
Q.22. Give the definition of nucleosome.
Ans : Nucleosome is a structure formed by wrapping of Histone Octomer by negatively changed DNA.
Q.23. Which base triplet code for the amino acid phenylalanine?
Ans : UUU UUC.
Q.24. Name three nonsense codon.
Ans : UAA UAG UGA.
Q.25. What are the hosts for cloning DNA fragments.
Ans : Ribosomes.
Q.26. What are exons?
Ans : DNA codes for in RNA but complete sequence of DNA is not found in m RNA. The sequences of DNA found in m RNA are known as exons. The coding sequences in eukaryotic genes are called as exons.
(D). Short Answer questions (2 marks) :
Q.1. Who coined the term genetic code? What does it mean?
Ans : Marshall W. Nirenberg and J. Heinrich Malhaei coined the term genetic code.
Codon is the triplet of bases that code for a particular amino acid.
Q.2. Discuss the structure of nucleosome.
Ans : Histones are positively charged which organised to form a unit of eight molecular called histone octamer. Negatively charged DNA wrap twice over the october forming a structure called nucleosome.
Q.3. Define genetic material.
Ans : DNA is the genetic material. RNA has been found to be genetic material in TMV, θβ bacteriophage etc. DNA is made of a series of nucleotides which in term form a large number of genes. Genes are the bearer of hereditary characters and chromosome is made up of highly coiled DNA.
Q.4. What are the four features of genetic material?
Ans : The four features of genetic material are :
(i) DNA is a double helical structure.
(ii) Genetic materials have the power of replication to form duplicate of it.
(iii) Stability of genetic material should be existed.
(iv) Genetic material should be able to undergo mutation and such a e change should be stably inherited.
Q.5. How does DNA express its biological information.
Ans : During protein synthesis mRNA copy the information coded in DNA these information to ribosome where protein is synthesized as par the coded information collected from the DNA.
Q.6. Where and when replication does occur?
Ans : DNA replication takes place inside the nucleus to form duplicate copy of parent DNA. Replication occurs when the cell is ready for division so that each daughter cell receives exact copy of parent DNA.
Q.7. Why DNA replication is stated to be semi-discontinuous?
Ans : Waston and Crick suggested a simple mechanism of replication on the basis of its double helical structure. The method of DNA replication is described as semiconservative method because daughter DNA molecule is a hybrid conserving one parental polynucleotide chain and forming the second one.
Q.8. How is the translation of mRNA terminated?
Ans : In the process of formation of poly nucleotide chain in ribosome amino acids are joined end to end and the process continues till three nonsense codons (UUU, UGA, UGA) present in the mRNA are encountered. On reaching the termination codon (nonsense codon) the polypeptide chain is released from ribosome.
Q.9. What is transcription?
Ans : DNA transfers its information to mRNA which moves to the ribosomes to direct protein synthesis. The formation of RNA from DNA template is called transcription. It occurs during the interphase.
Q.10. Name the transcriptional units in DNA.
Ans: Cistron is the unit of transcription. One cistron carries the information of a polypeptide.
Q.11. What are exons?
Ans : DNA codes for mRNA but complete sequence of DNA is not found in mRNA. The sequences of DNA found in mRNA are known as exons. The coding sequences in eukaryotic genes are called as exons.
Q.12. Why hnRNA is required to undergo splicing.
Ans : The eukaryotic genes are.interrupted by non coding DNA segments. The intervening sequences are known as introns. They are absent in RNA. The eukaryotic cells here, transcribes a precursor RNA much longer than mRNA. This is called heterogeneous nuclear RNA (hnRNA). It undergoes shortening by looping out mechanism to relegate unwanted nucleotide sequences. Process is called as RNA splicing. After this molecule undergoes polyadenylation, capping and methylation at 5′ and so that to form the mRNA molecule. The mRNA molecule then moves to cytoplasm as free nucleoprotein complex (informosomes).
Q.13. What is a satellite DNA?
Ans : It has been reported that heterochromatin contains certain polygenes to transcribe rRNA (in the NOR) 5S RNA and tRNA. Its DNA is formed of repeated polynucleotide sequences (about 100 to 100 million times) each being formed about 300 nucleotides and is called satellite DNA or repetitive or redundant DNA. The genes in heterochromatic region perhaps become active for a short period.
Q.14. “The genetic code is degenerate.” Explain.
Ans : The genetic code degenerate. For a particular amino acid more than one codon can be used. One amino acid often has more than one code triplet eg. phenylalanine has two codons, ie uuu and uuc. Argine has six codons CGU, CGC, CGA, CGG AGA, AGG All Codon started with AC (ACU, ACC, ACA, ACG) specify theanine and with CU (CUU, CUC, CUA, CUG) specify leucine. It helps avoid error.
(E). Short Answer Question (3 Marks) :
Q.1. Draw the structure of tRNA.
Ans :
Q.2. Describe the transformation experiment of Griffith. What was his conclusion?
Ans : In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae witnessed a miraculous transformation in the bacteria. During the course of his experiment, a living organism (bacteria) had changed in physical form. When streptococcus pneumoniae bacteria are grown on a culture plate, some produce smooth shiny colonies (s) while others produce rough colonies (R). This is because the S strain bacteria have a mucous coat, while R strain doesn’t. Mice infected with the S-strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia.
S strain → Inject into mice → Mice die
R strain → Inject into mice → Mice alive
Griffith was able to kill bacteria by heating them. He observed that heat killed S strain bacteria injected into mice did not kill them. When he injected a mixture of heat killed and live R bacteria, the mice died. Moreover, he recovered living S-bacteria from. che dead mice.
S strain → Inject into mice → Mice live (heat killed)
S strain
(heat killed)
+ → Injected into mice → Mice die
R strain
(live)
Griffith concluded that the R strain bacteria had some how been transformed by the heat-killed S-strain bacteria. Some transforming principle, transferred from the heat killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material. However, the biochemical nature of genetic material was not defined from his experiments.
Q.3. Write a note on biochemical characterization of transforming principle of Griffith.
Ans : In 1944, sixteen years after Griffith’s experiment, Ostuarald Avery, Colin Macleod and Maclyn Mccarty reported successfully repetition of bacterial transformation, but in vitro. They were able to identify the transforming genetic material. They worked to determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment.
They purified biochemicals (proteins, DNA, RNA etc) from the heat killed S cells to see which one could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed. They also discovered that protein digesting enzymes like Proteases and RNA digesting enzymes like RNA ases did not affect transformation. So the transforming substance was not a protein or RNA. Digestion with DNAse didn’t inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologists were convinced.
Q.4. Write there differences between DNA and RNA.
Ans : The three differences between DNA and RNA are:
| DNA | RNA | ||
| (i) | The sugar is Deoxyribose. | (i) | The sugar is ribose. |
| (ii) | The nitrogenous bases are A, G C and T. | (ii) | The nitrogenous bases are A, G C and U. |
| (iii) | DNA is double stranded structure. | (iii) | RNA is single stranded structure. |
Q.5. Briefly discuss the process of DNA replication.
Ans : While proposing the double helical structure of DNA, Watson and Crick had immediately proposed a scheme for replication of DNA. The scheme suggested that the two strands would separate and act as a template for the synthesis of new complementary strands. After the completion of replication, each DNA molecule would have one parental and one newly synthesised strand. This scheme was termed as semiconservative DNA replication.
Q.6. How did Hershey and Chase proved that DNA is the genetic material?
Ans : Alfred Hershey and Martha Chase in 1952 worked with viruses that infect bacteria called bacteriophages.
The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell. The bacterial cell treats the viral genetic material as if it was its own and subsequently manufactures more virus particles. They worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
They grew some viruses on a medium that contained radioactive phosphorous and some others on medium that contained radioactive sulphur. Viruses grown in the presence of radioactive phosphorous 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.
Radioactive phages were allowed to infect to E. coli bacteria. Then as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.
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 didn’t enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.
Q.7. How is protein synthesis initiated in a cell?
Ans : The process of synthesis of proteins involves one of the central dogma in molecular biology. According to this, genetic information flows from nucleic acids to protein. We know that DNA is stable inherited from parents to offsprings. According to central dogma, the flow of information takes place from DNA to RNA (mRNA) and from RNA to protein. This can be illustrated as follows:
Q.8. Draw labeled schematic sketch of replication fork of DNA.
Ans :
Q.9. Write three goals of Human Genome Project.
Ans : The three goals of Human genome project are:
(a) 95% of gene containing part of human sequence finished to 99.99% accuracy.
(b) To laraliza and estimate all the genes within the human genome.
(c) To construct detailed genetic map of human genome.
Q.10. What is meant by semiconservative of DNA replication?
Ans : According to semi consecutive hypothesis of DNA replication the unzipping of DNA strands takes place and complementary strand of each unzipped strand in formed is such way that the daughter strands are exact replica of the parent DNA strands.
Q.11. What is Frame shift insertion?
Ans : Deletion or insertion of one or a few nucleotides in the DNA molecule results mutation. Therefore, such mutation that results in shifting of reading frame backward or forward by one or more nucleotide is called frame shift mutation.
Consider a statement that is made by following words :
RAM HAS RED CAP
If a letter B is inserted between HAS and RED then the statement would be –
RAM HAS BRE DCA P
If it is followed by word ‘BI’ and ‘BIG’ then the statements would be:
RAM HAS BIR EDC AP
RAM AIAS BIG RED CAP
Same can be repeated by deleting R, E and D one by one as follows:
RAM HAS RED CAP
RAM HAS EDC AP
RAM HAS DCA P
RAM HAS CAP
Q.12. What are the functions of DNA polymerase.
Ans : In DNA replication a set of enzymes are involved. Of these the main ‘enzyme is referred to as DNA polymerase. It catalyses polymerization of DNA nucleotides of a large number in a very short time. The time required to polymerize depend, upon the number of base pair involved. Generally the speed of polymerization is 2000 base pairs per second. A great amount energy is required for polymerization process. Deoxyribonucleotide triphos phate not only provide energy but also acts as substrate.
Due to requirement of high amount energy for separation of two strands of DNA molecules during replication, a small fork called replication fork occurs in the DNA helix. During replication, the DNA-polymerase catalyse polymerization in one direction, 5′-3′ in one strand and in other the direction is 3′-5′. The polymerization in 3′-5′ direction is continuous while in 5′ – 4 3′ direction it is discontinuous. The discontinuous fragments are later joined by enzyme DNA ligase.
The DNA replication does not occur spontaneously or does not occur at any place. The origin of replication at particular site is determined by the requirement of a particular piece of DNA for the purpose of recombination. Generally a vector locates the origin of replication. In eukaryotic cells, DNA replication takes place in S-phase of cell-cycle.
Q.13. What are the functions of mRNA and tRNA.
Ans : (a) Messenger RNA (mRNA) : It acts as messenger to carry the message of the genetic code from the DNA of the nucleus to the site of protein synthesis, the ribosomes. It is also known as nuclear RNA as it is found only in the nucleus. It acts as a template for the translation of the DNA code in the formation of specific protein. It is formed as a complementary strand to one of the two strands of DNA which acts as a template. So, mRNA molecule possesses the same sequence of nitrogenous bases as is found in the template DNA. But at the time of synthesis of MRNA, the thymine of DNA is replaced by uracil.
(b) Transfer RNA : Transfer RNA or TRNA is the second most common RNA of a cell. It is the smallest molecules of RNA. tRNA has great affinity for specific amino acids and ATP. It brings activated amino acids to the proper site on the mRNA template. This RNA is synthesized in the particular region of DNA molecule and has complementary base sequence as in DNA.
In addition to the above three types of RNA, there is another type of RNA which is found in virus and so known as viral RNA or VRNA. This is found in TMV, influenza virus, tobacco necrosis virus.
Q.14. Elucidated the lac operon model of gene expression regulation.
Ans : The control of protein synthesis has been elucidated in E. coli by Francis Jacob and Jacques Monod at the Pasteur Institute in Paris in 1961. It was found that when sugar lactose is added to the cultures of E. coli, it induces the synthesis of three enzymes required for degradation of lactose to glucose and galactose. Enzymes are β galactosidase, permease and transacetylase. Process is referred as induction and enzymes formed are called inducible enzymes. The genes for these three enzymes lie close to each other and are linked.
These genes are called structural genes as they carry necessary information to code for amino acid sequence and thus finalise the structure and function of enzymes (proteins) of the pathway. The lactose or lac operon of E. coli has three structural genes and its tryptophan operon has 55 such genes. These genes are regulated as a unit by a single switch called operator. Promoter gene marks the site at which transcription of mRNA starts and where RNA polymerase enzyme birds. The entire unit is called operon.
The action of structural gene is regulated by operator site with the help of a repressor protein produced by the action of gene “i” called as regulator gene. The genes are expressed or not expressed depends upon whether the operative switch in on or off. With operator switch on, the three genes are transcribed by RNA polymerase into a single stretch of mRNA covering all the three genes. Each gene segment is called cistron and long m RNA covering all the cistrons is known as polycistronic. If lactose is removed from the medium, the enzymes needed for degradation are not produced.
The repressor substance: may combine with operator gene to repress its action in two ways.
| Sl. No. | CONTENTS |
| Chapter 1 | Reproduction in Organisms |
| Chapter 2 | Sexual Reproduction in Flowering Plants |
| Chapter 3 | Human Reproduction |
| Chapter 4 | Reproductive Health |
| Chapter 5 | Principles of Inheritance and Variation |
| Chapter 6 | Molecular Basis of Inheritance |
| Chapter 7 | Evolution |
| Chapter 8 | Human Health and Disease |
| Chapter 9 | Strategies for Enhancement in Food Production |
| Chapter 10 | Microbes in Human Welfare |
| Chapter 11 | Biotechnology: Principles And Processes |
| Chapter 12 | Biotechnology and its Applications |
| Chapter 13 | Organisms and Populations |
| Chapter 14 | Ecosystem |
| Chapter 15 | Biodiversity and Conservation |
| Chapter 16 | Bioresources of Assam |
| Chapter 17 | Environmental Issues |
(I) IAC OPERON (INDUCIBLE OPERON) :
In the operon is generally off, as a result, there is no transcription and thus no formation of proteins (enzymes)
However, it can be turned on if a metabolite is provided to the bacterium from outside. The consists of three structural genes, Lac Z, Lac Y and Lac A. The added metabolite comes in contact with active repressor bound to operator, leads to change in structure of repressor and repressor is removed from operator. Operon gets transcribed and enzymes are produced. The process continues till metabolite (inducer) is consumed. After the consumption of inducer (metabolite) repressor again gets its tertiary form, binds it to the operator and switch off the operon.
In Lac operon lactose when added enters the cells by the action of enzyme permease few molecules of which are usually present in cell. Lactose binds itself to active repressor leading to change in its structure. As a result repressor now fails to bind itself to the operator. Now RNA polymerase starts the process of transcription of operon by binding to promoter site P. All the three enzymes are formed viz. β galactosidase, permease and transacetylase. Finally, all lactose molecules are used up. Now inactive repressor turns active, attaches itself to operator and finally switches off the operon.
(II) REPRESSIBLE OPERON (TRYPTOPHAN OPERON AND ARGININE OPERON)
In this system, operon is usually on so that transcription occurs normally to synthesize enzymes But operon can be switched off due to non- requirement of metabolite.
In tryptophan operon, formation of tryptophan (an amino acid) needs action of five enzymes in succession. The formation of tryptophan is on because regulator gene R forms an inactive repressor called aporepressor which does not attach itself to operator site. As operator site is free of repressor, the operon system remains on and all the five enzymes needed for tryptophan formation are synthesized.
When tryptophan accumulates or added, few molecules of tryptophan act as co-repressor and binds to inactive repressor (repressor-corepressor complex) which turns active and attaches itself to operator, thus switching off the operon.
Q.15. What is DNA finger printing? Mention its application.
Ans : Technique of using DNA fragments, resulting from restriction endonuclease enzyme cleavage to identify particular individuals, is called DNA finger printing.
Introduction: DNA finger printing technique was developed by Alec Jffery (1985, 86) at Leicester University, United Kingdom, Inheritance of
DNA is very stable. Every person has specific pattern of DNA sequence which shows a combination of DNA sequence of both mother and father. Study of DNA finger prints helps in establishment of identity of a person, identification of criminals in case of a murder or rape, and paternity test in case of disputed parentage. This technique helps in identification even when the stains on victims clothes etc. are even several years old.
In India, first test of DNA finger printing was done in June, 1989 for setting a disputed parentage in Madras. Laboratory for DNA finger printing is situated in Hyderabad at the Centre for Cell and Molecular Biology (CCMB). Paternity dispute cases are much more common in India and most of them are referred to CCMB. for DNA evidence.
Southern Blotting technique of DNA finger printing:
VNTR (Variable number of tandem repeats) technique involved Southern blot hybridisation using radiolabeled VNTR as a probe. The steps were:
(a) For DNA finger printing, DNA is isolated from any body cell or (n) even from blood stains, semen stains or hair roots.
(b) DNA sample is first digested with a restriction endonuclease enzyme and the digested sample is then gel electrophoresed.
(c) DNA bonds in the gel are denatured into single strands by treating them with alkali solution.
(d) Gel is laid on the top of a buffer saturated filter paper, placed on a (p) solid support (e.g.glass plate), with its two edges immersed in the buffer.
(e) A sheet of nitrocellulose membrane is placed on top of the gel and a stack of paper towels is placed on the top of this membrane (acts as an absorbent tissue).
(f) A weight of about 0.5 kg is placed on the top of pack of paper towels. Leave this arrangement for a few hours or over night.
The buffer solution is drawn up by the filter paper wick, and passes through the gel to the nitrocellulose membrane and finally to the paper towels. While passing through the gel, the buffer carries with it single stranded DNA molecules, which bind to the nitrocellulose membrane, while the buffer passes through it to the paper towels.
(g) Next day, paper towels are removed and discarded.
(h) Nitrocellulose membrane with single stranded DNA bands is incubated at 80°C for the 2-3 hours to fix the DNA bands permanently on the membrane.
(i) This membrane now has a replica of DNA bands from agarose gel and can be used for hybridization with specific labelled DNA or RNA probes.
(j) The membrane may then be washed to remove any unwound DNA and X-ray film is exposed to the hybridized membrane to get autoradiograph.
Significance: DNA finger printing shows polymorphism of DNA which is used for identification of a .. with much more certainty that has been possible through techniques of blood groups since the number of blood groups available becomes a limitation. The technique has shown the possibility of two persons having same pattern of DNA finger prints is very remote (exception monozygotic i.e. identical twin)
(F). Long Answer Question (5 Marks):
Q.1. Give a brief account of DNA replication.
Ans : DNA replication takes place in the following steps :
(i) Activation of Deoxyribonucleotides : Deoxyribonucleotides such as DAMP, DGMP, DCMP, TMP floating freely in the nucleus serve as the raw materials. All these nucleotides are synthesised from precursor molecules formed from the metabolic products. The nucleotides are then activated by ATP which serves as energy source.
(ii) Uncoiling of Parent DNA Strands : The double stranded DNA helix uncoils to form single stranded DNA by breakage of weak hydrogen bonds. Enzyme helicase help in uncoiling the helix. Topoisomerase, another enzyme may also be responsible for cutting the DNA strand to facilitate uncoiling. As a result of uncoiling. A, T, G C bases become naked at one end. The naked ends project into nucleoplasm. Uncoiling at one end gives DNA strands a Y-shaped structure called replication fork.
(iii) Formation of RNA Primer : For the synthesis of new DNA strand RNA primer is formed in the DNA template. The enzyme primase catalyses the polymerisation of RNA. The RNA primer is removed after the fork is formed. The gap formed by the removal of primer is filled up by nucleotides.
(iv) Base Pairing : Once separation is completed, the nucleotides of separated chains start attracting their nucleotides or precursors from the environment within the cell. The precursor substances present in the nucleoplasm are triphosphates of deoxyribonucleosides such as deoxyadenosine triphosphate, thymidine triphosphate and so on. Each deoxyribose triphosphate splits off into deoxyribonucleoside monophosphate and two phosphate molecules. These deoxyribonucleotides bind with their complementary deoxyribonucleotides of any one of the two separated strands of parent DNA molecule by hydrogen bonds.
Thus deoxyribonucleotides from the nucleoplasm replace each of their complementary bases that are separated from it in the uncoiling process. According to Watson and Crick’s hypothesis only could couple to “T” base and “G” base could couple “C” and so on.
(v) Formation of New DNA strands : Newly formed nucleotides are attached to the nucleotides of the original strand of DNA molecule by hydrogen bonds. Adjacent sugar radicles also unite with one another by phosphodiester bonds. Thus two double helical molecules identical with each other are formed.
In this way the genetic code is faithfully transmitted from old DNA to new DNA. The deoxyribonucleotides are polymerised to form new DNA strand in the 5′-3′ direction. Because the two strands are antiparallel, the new strand forms in opposite direction. The new strand which is formed in continuous stretch in 5′- 3′ direction is called leading strand.
On the other parent strand short DNA segments form in the 5′-3′ direction starting from the RNA primer (in the fork end). This strand is known as okazaki fragment. The short fragments of DNA then join with the help of DNA ligase forming the lagging strand.
(vi) Editing and DNA Repairing : Accurate replication is ensured by DNA polymerase which remove the wrong bases that may be formed. The enzyme nuclease also remove the abnormal region of the DNA caused by mutation and then DNA ligase joins the repaired portion.
(vii) Helix Formation : Daughters and parent strand coil to form a double helix.
Q.2. Describe the structure of DNA with a neat labeled diagram.
Ans : Though different authors had described the structure of DNA differently but the most acceptable and convincing structure of DNA molecule had been illustrated by J.D. Watson and H.C. Crick (1953) by using X-ray diffraction photograph of DNA. The model as illustrated them is known as Watson and Crick’s structural model of DNA. It provided an idea of how a various constituents of DNA are attached to one another.
The model has the structure of a double helix formed bv polynucleotides. The helix has a diameter of about 20 A and each helix has one complete turn every 34A along its length. Each turn consists of a stack of ten nucleotides and so distance between each nucleotide is 3.4 A. In DNA molecule the adjacent deoxyribonucleotides are joined in a strand by phosphodiester bridges or bonds which link the ugar of one nucleotide to the sugar of the adjacent nucleotide.
The molecule consists of two such polynucleotide chains remain twisted around each other to form a helical structure. The sugar and phosphate chain remain on the outside and thus form the back bone of each polynucleotide chain. The purines and pyrimidines remain on the inner side of the helix and face each other.
The two polynucleotide chains are held together by hydrogen bonds. The pairing of the nitrogenous bases of the two strands is very peculiar. The dimension of the purine rings is greater (due to double ring atoms in their molecules) than that of the pyrimidine ring (due to the single ring atoms in their molecules). So, to maintain constant diameter of the two-chains helix molecule of DNA, the purines always pair with pyrimidines with hydrogen bonds.
Because of the shape and chemical nature of the purine and pyrimidine bases, adenine can be attached only to thymine by two hydrogen bonds. Similarly guanine can be attached to cytosine by three hydrogen bonds. Therefore, four combinations of purine pyrimidine such as A-T,T – A, G-C and CG could only be found.
nucleotide of the other chain. Thus whatever the sequence of bases is present in one strand of the DNA molecule, a complementary sequence is present in other strand. Due to this type of base pairing, the two strands of DNA are not identical but complementary to each other. This is called complementary base pairing. The four base pairs may occur in infinite ways and thus provide infinite varieties of DNA molecules.
One end of the DNA strand is called 5′ end and the other end is called 3′ end. This is beçause the deoxyribose attaches itself to a phosphoric acid at 5 carbon position and the same deoxyribose attaches itself to the next phosphoric acid at 3 carbon position. This is called polarity of DNA.
The two nucleotide chains of DNA run in opposite or antiparallel direction. It means that in one chain the 5 carbon position in sugar molecule is in upward direction and in other chain its carbon position is in lower direction. This implies that if in one chain the direction is 5′-3′ and in other the direction is 3′-5′. This is known as antipara- llel direction of DNA molecules.
Q.3. Discuss how the DNA double helix packed to form chromatin in eukaryotes.
Ans : Exceptionally long DNA molecule is packed in such a way that it is 3. accommodated in the nucleus or the cell. The distance between two base pairs in mammalian DNA molecule is 0.34 nm (0.34 x 10⁹m). Now if the total number of base pairs in mammalian DNA is 6.6×10⁹ by the total length of mammalian DNA would be 6.6×10⁹hpx 0.34×10⁹/bp 2.2 meters. Such a long DNA molecule is packed in such a fashion that a large number of DNA molecule depending upon the number of chromosome (say for instance 46 DNA molecule in human cell) is packed within a single nucleus.
DNA is always linear and double stranded. Single DNA duplex makes a complex with basic protein called histone. This nucleoprotein complex is referred to as chromatin which is the unit of genome. The condensed form of chromatin is called chromosome. The nucleus of eukaryotic cells contains few or many chromosomes depending on the species. The size of chromosome varies from species to species. Therefore the size of DNA molecule is also variable.
The largest chromosome of Drosophila however contains a DNA molecule of about 4.0 cm long with molecular weight of 80 x 109 rearly 40 times larger than the DNA of E coli. The DNA of the eukaryotic cells undergoes folding and supercoiling many times to accommodate themselves within the chromosomes. The physical forms of chromosomes also varies according to the cell cycle. During interphase, they are extended or uncoiled. During prophase, chromosomes coil and shorten until they finally reach the metaphase.
During telophase they begin to uncoil and again attain the relaxed condition when the interphase of the next all cycle is reached. Therefore, change of physical form of chromosome during cell cycle possible brings some additional changes of folding and supercoiling pattern of DNA which is present inside the chromosome. It has been calculated that human chromosome No.13 contains a DNA molecule which is about 32000 pm long. This DNA undergoes looping supercoiling form a chromatid about 6 pm long and 0.8 pm in diameter.
Q.4. How Meselson and Stahl proved that DNA replication is semiconservative?
Ans : In 1958 Mathew Meselson and, Franklin Stahl performed the following experiment to show that DNA replicates.
(i) They cultivated E. coli in medium having ¹⁵NH₄ CI as the only nitrogen source and obtained the E. coli population containing “N labelled DNA. This ¹⁵N containing heavy DNA molecule could be distinguished from normal DNA molecule which was not labelled with N₁₅ by centrifugation in cesium chloride density gradient. This proves that the replicates are derived from parental DNA.
(ii) Next, they transferred the E. coli cells containing 15N labelled DNA of the earlier experiment into medium containing less dense isotopic nitrogen and allowed E. coli to multiply several generations. After first generation they could isolate hybrid DNA (¹⁵N to ¹⁴N)
One strand in ¹⁵N and the other is ¹⁴N the cells of the second generation the 50% “N-DNA and 50% hybrid.DNA could be founds. If samples are taken in subsequent generation, the ¹⁴N-DNA will produce true ¹⁴N-DNA but the hybrid DNA will again produce 50% ¹⁴N-DNA and 50% hybrid DNA. It takes 20 minute for DNA to replicate. This experiment is a proof of semi conservative model of DNA replication.
Q.5. Discuss the process of translation in detail.
Ans : During this process, proteins are formed by the ribosomes on the mRNA strand. The genetic information present in mRNA directs the sequential order of the specific amino acids to form a Polypeptide chain’. The main steps of translation are:
(a) activation of amino acids.
(b) attachment of activated amino acids to tRNA.
(c) initiation of protein synthesis.
(d) elongation of the polypeptide chain. and
(e) termination of the chain.
(a) Activation of Amino Acids : The first step in translation is the activation of amino acids. ch of the 20 amino acids of the cytoplasm occurs in inactive or dormant stage and they cannot like part in protein synthesis. The activation is facilitated by ATP. Each of the amino acid is catalysed by its own specific activating enzyme – aminoacyl synthetase. The free amino acids act with aminoacyl synthetase in presence of ATP to form aminoacyl adenylate and phosphopyruvate.
(b) Attachment of Activated Amino Acid to tRNA: The activated amino acids cannot reach he polyribosome to form polypeptide chain as they are intimately bound to enzymes. The Gated amino acid is transferred to its specific tRNA and attached with tRNA molecules. The transfer of the activated amino acid to tRNA is also specific. As a result of attachment of amino acid to a specific tRNA molecule, aminoacyl-tRNA is formed which can be shown as :
The aminoacyl-tRNA complex moves towards the polyribosome to initiate the protein synthesis.
(c) Initiation of Protein Synthesis : In E. coil, the messenger RNA forms the 70 S complex ith F-met tRNA and GTP. The mRNA has the first codon as AUG at the beginning and so it is own as initiation codon. There are three initiators, F1, F2 and F3 which get attached to the 30 S subunit of ribosomes. F1 and F2 factors along with GTP are required for binding of F-met-tRNA the small subunit. The AUG codons are the codes for the methionine. The methionine has ry important role in protein synthesis as it is the amino acid which starts the protein synthesis. !The initiation of protein synthesis takes place through a specific methionyl-tRNA complex known as F-met-tRNA.
The binding of F-met-tRNA with mRNA-30 S subunit requires the Initiation factors (F1 and F2) and GTP. The UAC anticodon of F-met-tRNA binds to AUG codon of mRNA. In every type of protein, formyl methionine occupies the first place in the molecule and when the protein molecule is synthesized completely, formyl methionine detaches from it.
(d) Elongation of the Polypeptide Chain : Elongation of polypeptide chain involves t regular addition of amino acids and relative movements of ribosome and mRNA in the presence of GTP so that a new triplet codon remains available for new aminoacyl tRNA. Elongation the polypeptide chain requires elongation factors which are EF-Tum EF-G and EF-TS in prokaryotic cells and EF-1 and EF-2 in eukaryotic cells. The site where the entry of amino acid takes place is called “A” site or aminoacyl site and the site in which nascent polypeptide chain remains is called “P” site or peptidyl site or exit site of tRNA molecule.
The next step of this process is translocation which comprises the discharge of unloaded tRNA from the P site. All the sequence of events occur very rapidly as the polypeptide chain of about 40 amino acids is produced within 20 seconds.
(e) Termination of Polypeptide Chain : After the polypeptide chain is formed according to e codon of MRNA, the process of termination and release of polypeptide chain occurs. The polypeptide chain elongates until a termination codon on mRNA is reached. The termination of the polypeptide chain is indicated by three special termination triplets in the mRNA. These triplets codons are UAA, UAG and UGA in prokaryotic cell. These are called nonsense codons which do not collaborate with any codons.
Q.6. Discuss the various application of DNA fingerprinting.
Ans : The HGP will enable the scientist to know for the first time how a chromosome looks like and its entire constituents. The knowledge that will emanate from the project will be used for centuries to come. The discovery of the genetic make up of chromosome will surpass any other discoveries of the past. It has more implications than sending on the moon and can be compared to the invention of wheel, that changed the world. Scientists are yet to know full implications of the knowledge that may be generated by the project. At present the scientist are optimistic about the application of the knowledge in the following cases:
(i) At present human beings suffer from a large number of inherited diseases which have no remedy. Disorder in a simple gene can give rise to 3000 different hereditary disorders, many of which cripple millions of healthy and productive lives and others are not so pronounced.
(ii) The cause of the dreaded disease – the cancer may soon be unfolded with the identification of the cancer gene.
(iii) The genomic study will help understand why human beings differ in physical structure, colour, height and why some are resistant to certain diseases and other are not.
(iv) Why human races are different in response to environmental tolerance, food habit, mental apptitude and why some races are more prone or resistant t certain diseases.
(v) The final frontier of genomic study is to understand the mystery that lie in the fertilized egg. How the fertilized egg knows to give rise to so many different types of cells to produce different organs like brain, muscles, vessels, heart, lungs skin, eyes etc. and the genes responsible for such differentiation.
(vi) To utilise the knowledge to improve health of people and to make a healthy society.
Q.7. Write the salient goals of Human genome Project.
Ans : The goals of Human genome projects are :
(i) Genetic Map : 2 to 5 cm resolution map (600-1500 marker)
(ii) Physical Map : 30,000 sTss.
(iii) DNA Sequence : 95% of gene containing part of Human sequence finished to 99.99% accuracy.
(iv) Capacity and Cost of Finished : Sequence 500 Mb/year at <$ 0.25 per finished base.
(v) Human Sequence Variation : 1,00,000 mapped human SNPs.
(vi) Gene Identification : Full length human c DNAs (chromosomal DNA).
(vii) Model Organisms : Complete.
Q.8. Discuss the methodologies involved in Human Genome Project.
Ans : The methods involves two major approaches. One approach focused on identifying all the genes that are expressed as RNA (Referred to as Expressed sequence Tags (ESTS)). The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non coding sequence, and later assigning different regions in the sequence with function (term referred sequencing, the total DNA from a cell is isolated and converted into random Fragments of relatively smaller sizes and cleaned is suitable hast using specialized vectors.
The cleaning resulted into amplification of each piece of DNA fragment so that it subsequently could be sequenced with case. The commonly used hosts were bacteria and yeast. The vectors were could Bacterial Artificial Chromosome (BAC) and Yeast Artificial Chromosome (YAC).
The fragments were sequenced using automated DNA sequencer that worked on the principle of a method developed by Frederick sanger. These sequences were than arranged based on some overlapping regions present in them. This required generation of over lapping fragments for sequencing. Alignment of these sequences was humanly not possible. Therefore specialized computer based programmes were developed. These sequences were subsequently annotated and were assigned to each chromosome. The sequence of chromosome 1 was completed only in May 2006.
Another challenging task was assigning the genetic and physical maps on the genome. This was generated using information on polymorphism of restriction endonuclease recognition sites.
Q.9. Give five salient features of human genome project.
Ans : The salient features of human genome project are:
(i) The Human genome contains 3164.7 million nucleotide bases.
(ii) The average gene consists of 3000 bases, but sizes vary greatly.
(iii) The total number of genes is estimated at 30,000 much lower then the previous estimates of 80,000 to 1,40,000 genes. Almost (99.99%) nucleotide bases are exactly the same in all people.
(iv) The functions of absent 50% discovered genes are unknown.
(v) Less than 2% of the genome codes for proteins.
(vi) Repeated sequences make up very large portion of the human genome.
(vii) Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times.
(viii) Chromosome-I has most genes (2968) and Y has the fewest (231).
(ix) Scientists have identified about 1.4 million locations where single base DNA differences occur in humans.
Q.10. Discuss in detail the lac operon developed by Jacob and Monod.
Ans : Jacob and Monod in 1961 first provided with a model of Lac operon. An operon is defined as several 4 genes situated in tandem, all controlled by a common structural gene having repressor, promoter and operator. The message produced by an operon is polycistronic because the information of all structural genes resides on a single mRNA molecule. The lac operon (lac refers to lactose) consists of one regulatory gene (i gene) and three structural genes (Z, Y and A).
The i gene codes for repressor of the operon. The Z genes codes for beta galactosidase which hydrolyses lactose into galactose and glucose. They Y gene codes for permease which increase permeability of cells to beta-galactosidase. A gene codes for transacetylase whose function is not definitely known. This shows that three genes are required for metabolism of lactose.
The above fact shows that presence of lactose switches on the operon for synthesis of enzyme beta-galactosidase inside cells. It therefore acts as inducer. The permease facilitate entry of lactose inside cell. The i gene synthesises repressor. The repressor protein binds with the operator to prevent transcribing operon by RNA polymerase. The repressor is then inactivated by the inducer (such as lactose). This in turn allows RNA polymerase access to promoter and then transcription starts.
Q.11. How did Griffith explain the transformation of R-strain bacteria to S-strain bacteria.
Ans : Many suspected that DNA may be the genetic material. But this was not conclusive. The experiment of Griffith in 1920 with Streptococcus pneumoniae inspired many others to work in similar line with a view to unfold the mystery regarding the genetic material. S. pneumoniae is known to cause pneumonia. Griffith selected two types of pneumococcus. One type called smooth type or S-type, causes disease (virulent). It had capsule around its body, The other type, called rough type R-type was without capsule and was not virulent (avirulent). He first worked with the virulent and avirulent type by injecting both the types separately in separate mice.
Naturally the virulent bacteria (S-type) killed the mice while the non-virulent bacteria (type) could not. He then killed another sample of the virulent bacteria by heat treatment a injected the same in mice. This time he noticed that the mice survived. He also could not isola any such bacteria from the mice which survived after injection of heat killed virulent form. But he was astonished to find that when the heat-killed virulent form and living avirulent forms we mixed and the mixture was injected, the mice died.
It was a big question, how avirulent bacteria turned to be virulent and killed the mice while the virulent bacteria that were injected along with it were dead. Not only this, the bacteria isolated from dead mice showed that the non-capsulated avirulent type became virulent and capsulated type. It indicated that the non-virulent, non-capsulated bacteria acquired the genetic character of virulent, capsulated bacteria within mice.
Further research showed that such transformation of genetic character takes place even in culture plate when both types are placed together indicating that the mice tissues have no role in transformation of genetic character in bacteria. It was known further that the extract prepared by killing and crushing the virulent form could also convert non-capsulated, avirulent form into capsulated, virulent form. This indicated that something must- be there in the extract which has transmitted the genetic character.
Q.12. Draw a labelled diagram of the double helical structure of a DNA strand.
Ans :
Q.13. Explain the southern blot hybridization technique of DNA fingerprinting.
Ans : To know the technique one should have prior knowledge of DNA structure, replication and base pairing. Forensic science encounter with problem like identification of culprits from the blood stain, hair, skin or any other minute body parts left at the site of the crime. It may also be necessary to ascertain parenthood of one controversial child. In such cases a forensic expert takes the help of southern blot technique. We know that there is a large amount of “junk DNA”- DNA that does not code for protein-in the human genome. Junk DNA is made up of repeated sequences that are called repeats.
Although individuals may have identical genes, there may be different numbers of repeats between these genes. For example, one person may have 7 while another has 12. The more repeats, the longer the junk DNA between genes. One method of DNA fingerprinting- which produces a Southern Blot – begins by taking a DNA sample from something such as skin, saliva, blood, or hair. The DNA is cut into pieces using restriction enzymes. The resulting collection of DNA pieces will consist of some pieces of junk DNA and some genes. The sample DNA pieces and placed into a clear gelatin, where an electric current pushes the DNA pieces through the gel.
Short pieces move farther than long ones, so a piece of DNA that had 7 repeats would move faster than a piece of DNA with 12 repeats. Since DNA has no color more steps must be completed so scientists can “see” particular DNA pieces. The sequences are denatured so only a single strand remains. They are transferred onto a nylon sheet where the strands are permanently fixed. A radioactive probe with a known sequence is then added.
After a radioactive probe of single stranded DNA has been allowed to bond by base pairing with the denatured DNA on the paper an X-ray reveals only the areas where the radioactive probe sits. These are the only things that will show up on the film. This allows researches to identify, in a particular person’s DNA, the occurrence and frequency of the particular genetic pattern contained in the probe. This is then match with the DNA of suspected persons or their close relative to pinpoint the offenders.
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