Molecular Basis Of Inheritance

In a DNA strand the nucleotides are linked together by phosphodiester bond. It is a linkage between 3’Carbon of one deoxyribose sugar molecule and 5’ Carbon of the other sugar molecule.

Nucleotide: It is composed of a nitrogenous base, pentose sugar ( either deoxyribose or ribose) and phosphate group (PO4).

Nucleoside: It is composed of a nitrogenous base and pentose sugar (either deoxyribose or ribose). But it does not have a phosphate group.

Both deoxyribose and ribose belong to pentose class of sugars.

Due to paring of the purine base (adenine and guanine) and pyrimidine base (cytosine and thymine) by hydrogen bonds the diameter of the DNA remains constant, leading to uniform width throughout the DNA.

The net electric charge on DNA is negative because of presence of phosphate ions in the nucleotides. The net electric charge on histones is positive because of presence of lysine and arginine amino acid.

•A promoter is a region of DNA where RNA polymerase binds to initiate transcription. It is present at the 5’ end of the transcription unit.

•A terminator is a sequence of DNA that causes RNA polymerase to terminate transcription. It is present at the 3’ end of the transcription unit.

•A transcription unit is the sequence between sites of initiation and termination by RNA polymerase; may include more than one gene.

•Upstream identifies sequences in the opposite direction from expression; for example, the bacterial promoter is upstream of the transcription unit, the initiation codon is upstream of the coding region.

•Downstream identifies sequences proceeding farther in the direction of expression; for example, the coding region is downstream of the initiation codon.

Sickle cell anaemia is a genetic disorder in which Valine is substituted instead of Glutamic Acid in the beta Haemoglobin chain.This mutation leads to the change of RBC’S shape from disc shape to sickle like shape leading to decreased oxygen in the body.

a) As it is a genetic disorder it cannot be treated with iron supplements.

b) The substitution of valine instead of Glutamic Acid in the beta Haemoglobin chain shows it is a molecular disease.

c) Individual who has only one sickle cell mutation gene that is from either mother or father and not from both, the person will not inherit the disorder but would get the sickle cell trait .This trait is protective against malaria.

This shows all the above statements stands correct.

a) AUG codon only codes for methionine.

b)AUG codon is an initiation or start codon which starts the transcription of messenger RNA or m-RNA.

c) It codes for methionine in eukaryotes and modified methionine in prokaryotes .

All the statements holds true.

RNA was first genetic material because:

DNA can only store information and protein can only catalyse reactions none of these alone is sufficient to survive but in case of RNA.

a. It can store genetic material like DNA.

b. It can catalyse chemical reactions just like protein.

Hence it shows RNA was the first genetic material.

In mature m-RNA of the eukaryotes exons (region outside the gene) appear but introns (region inside the gene) do not appear because these are removed at the time of RNA splicing.

According to the Human Genome project, chromosome 1 has approximately 2968 genes which is the highest and chromosome Y has approximately 231 genes which is the least number of genes.

Meselson and Stahl gave experimental proof of the semi-conservative nature of the DNA. They didn’t contribute to the double helix model of DNA.

Remove or replace the 3’ OH group in deoxyribose to prevent polymerisation of the nucleotides because 3’OH group is involved in the polymerisation of nucleotides.

It is because DNA dependent DNA polymerase catalyses polymerisation only in one direction (5’ → 3’).So, the new strand to be formed in 3’→5’ is synthesised in short fragments, called Okazaki fragments which are then joined together by DNA ligase enzyme.

In the process of initiation enzyme DNA helicase, DNA polymerase are involved. Elongation is catalysed by RNA polymerase and Termination is catalysed by RNAse H and DNA ligase enzyme.

Gene expression is the transfer of information from a gene to synthesise a functional gene.Prokaryotes control gene expression at the level of transcription as their organisation is simple.

Regulatory proteins in transcription in prokaryotes act as both activator and repressor. Activator increases the transcription of a gene when an external stimulus is applied and Repressors prevent the transcription of a gene when an external stimulus is applied.

Chromosome 1 was found after 20 years of the beginning of the human genome project, which made it the last chromosome to be discovered.

a. m-RNA or Messenger RNA is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.

b. t-RNA or transfer RNA carries amino acids to ribosomes.

c. r-RNA or ribosomal RNA is the constituent component of ribosomes.

According to Chargaff’s rule of base pairing:

1) Adenine is equal to thymine. They are bonded by two hydrogen bonds.

2) Guanine is equal to cytosine. They are bonded by three hydrogen bonds.

3)The ratio of adenine to thymine and Guanine to Cytosine is equal to 1 which proves DNA is a double stranded structure.

4)In the above organism the DNA is not following the Chargaff’s rule hence it shows it is not a double stranded DNA but a single stranded DNA.

In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called c-DNA or complementary DNA.

After 60 minutes or third generation, bacteria has 25 % hybrid (15N 14N) IN 1:3. After 80 minutes or fourth generation bacteria has 12.5% hybrid in 1:7.

Coding Strand :5' - A U G A A U G - 3’

Template strand: 3’ T A C T T A C-5’ ((Adenosine bonds with thymine, and Guanine bonds with Cytosine).

RNA Transcript: 5’ A U G A A U G-3’ (Adenosine bonds with Uracil and Cytosine bond with guanine)

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The base pairing should be complimentary. It is 5’-UAC-3’ . As Adenine bonds with Uracil and Guanine pairs with cytosine.

One end of the t-RNA binds to a specific amino acid which is at the3’ end of the t-RNA.

To initiate translation, the mRNA binds to the smaller ribosomal subunit which is the 40S subunit.

In E. coli, the lac operon gets switched on when lactose is present and it binds to the repressor .

DNA wraps itself around a protein called histone. Histones pack into structural units called nucleosomes.

Euchromatin has transcriptionally active regions of DNA. Heterochromatin has transcriptionally inactive regions of DNA.

1) In bacteria there are three types of DNA polymerases.

2)All the three DNA polymerases can add nucleotides in 5’-3’ direction.

3) DNA III polymerases can proofread the newly formed strand and can sense the wrong base insertions.

4) It deletes wrong base and correct the mistakes by putting the right base in.

5) It cannot correct the substitution of uracil in place of thymine.

6) It can repair the damage done to DNA by UV rays exposure.

1) In DNA both the strands are anti-parallel and complementary.

2) Both the strand acts as template.

3)As DNA synthesis can only take place in 5’-3’ direction. From these two strands only one strand which is 3’-5’ can synthesise the complementary strand in 5’-3’ direction.

4) The left-out strand of the DNA which was 5’-3’ has to be synthesised in opposite direction as short stretches in discontinuous manner.

5) These stretches are also known as ‘Okazaki fragments’ which are later joined together by DNA ligase enzyme.

a) Complementary strand : 5’-T T A C G T C G A T A A T C C-3’

Here, Adenine bonds with thymine & Guanine bonds with cytosine.

b) the mRNA: 5’ A A U G C A G C U A U U G G-3’

In RNA Adenine bonds with uracil instead of thymine & Guanine bonds with cytosine.

Difference in the nucleotide sequence between the individuals is called as DNA polymorphism. This difference can be in a single base pair, insertion or deletion of various base pairs.

Importance of DNA polymorphism:

1. It is important for genetic variation.

2. Used in criminal cases for identification of the culprit.

3. Used in genetic mapping.

1. This abnormal haemoglobin molecule formation occurs because one codon GAG gets replaced by GUG. Which means in the codon GAG adenosine gets replaced by uracil.

2. This change leads to the incorporation of valine in the beta haemoglobin chain instead of Glutamic acid at the 6^th position. This mutation changes the shape of red blood cell from disc shaped to sickle shape that is why it is also known as Sickle cell anaemia.

3. These sickle shaped blood cells have shorter life span than the normal blood cells. Sometimes they also get stuck in the arteries and causes blockage in the blood flow.

4. The following change may cause the death of the individual before attaining maturity.

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. It happens due to uncoordinated regulation of gene expression in the gene sets associated with organ development.

The number of ribonucleoside triphosphates is 10 times the number of deoxy ribonucleoside triphosphates, but only deoxy ribonucleotides are added during the DNA replication because:

1. The enzyme DNA polymerase only recognises deoxyribonucleotides triphosphates.

2. It incorporates the same as deoxy ribonucleotides during DNA replication.

The few enzymes involved in DNA replication other than DNA polymerase and ligase are:

1. Helicases: These are the enzymes which separate the two DNA strands at the replication fork using energy. These enzymes are also called as helix destabilizing enzyme.

2. DNA Clamp: This enzyme is used to promote the replication of the DNA. It binds the DNA polymerase enzyme to the template strand and prevents it from disassociating. It is also known as sliding Clamp.

3. Single Strand Binding Proteins:These enzymes prevent the single strand of DNA to get digested by the nucleases. It also prevents the formation of secondary structure.

The three viruses which have RNA as the genetic material are:

1. Tobacco Mosaic Virus

2. Human- Immuno Deficiency Virus

3. Influenza Virus

1.The experiment was performed in the year 1928 by Frederick Griffith.

2. Griffith used two strains of bacteria Streptococcus Pneumoniae.

3. These were the S-strain (smooth strain) and the R-strain (rough strain) .

4. These strains were then injected into two different mice.

5. He found that only mice injected with S-strain had the disease and was virulent. This was because the S-strain has a polysaccharide covering which protected the strain from mice’s immune system.

6. The R--strain mice did not have the disease and was non-virulent. This was because the R-strain did not have the covering which protected the strain from mice’s immune system.

7. After this, a combination of heat killed S-strain and live R-strain were injected into two different mice.

8. The disease was seen in the mice.

9. Through this Griffith showed that the transformation took place.

10. Some ‘transforming principle’ of the S-strain was picked up by the R-strain which transformed the R-strain to the S-strain and made it virulent. This change in R-strain showed that it was the inherited genetic material.

11. This genetic change was permanent which further showed that DNA was the genetic material.

1. When Griffith performed the experiment. He was unable to conclude the biochemical nature of the inherited genetic material.

2. After which Oswald Avery, Colin McLeod and Maclyn McCarty in 1933-34 performed an experiment to find this genetic material.

3. They purified the biochemicals (DNA, RNA, & Protein) from the heat killed S-strain cells.

4. They noticed that the RNA digesting enzyme (RNases) , Protein digesting enzymes (Proteases) did not affect the transformation of the non-virulent R-strain to the virulent S-strain.

5. This led to the conclusion that DNA was the transforming principle.

Heavy isotope of nitrogen in the Meselson and Stahl’s experiment showed that the DNA is of intermediate density. Which was between N14 and N15.This showed that the hybrid DNA has one strand of N14 and one strand of N15 showing DNA was semi-conservative in nature.

Cistron is the part of the DNA which has the information for an entire polypeptide chain.

Human Genome is the genome of the Homo Sapiens. It consists of the coding regions of the DNA which encodes all the genes. It also consists of all the non-coding regions of the DNA which does not encode for any genes.

Six features of the human genome are:

1.The genome has around 3164.7 million nucleotide bases.

2.99.9 % of the nucleotide bases are same in all humans.

3.The largest gene is Dystrophin having 2.4 million bases.

4.Chromosome 1 has most genes (2968) and Y has the least genes (231).

5.Less than 2% of the genome has the coding sequence for proteins.

6.Only 50% of the total discovered genes have known functions.

During DNA replication, why is it that the entire molecule does not open in one go otherwise the whole molecule would destabilise as it consumes a lot of energy.

2.This is because the opening or unwinding of the DNA creates tension in the molecule. Uncoiled parts start forming supercoils due to the interaction of the exposed nucleotides.

3.To avoid this scenario the helicase enzyme acts on the site of origin (ori site) of the double stranded DNA and opens only a small stretch.

4.With the help of enzymes these exposed strands are copied.

5.After this the site of origin moves in both the directions forming a ‘Y’ shaped structure called Replication Fork.

DNTP’S are Deoxyribonucleotide triphosphate. The two functions that the monomers (d NTPs) play are:

1. They provide energy for polymerisation reaction.

2. They provide deoxyribonucleotides for DNA replication.

Central Dogma is the process of transfer of information within the genes to DNA than RNA and finally to the proteins. Retroviruses do not follow central Dogma because their genetic material is RNA instead of DNA. This RNA is than converted to DNA by the process called Reverse Transcription with the help of enzyme reverse transcriptase. Hence central dogma is not followed.

Distance between two consecutive base pairs: 0.44 nm or 0.44 X 10^-9.

Length of the DNA: 2 X 10 ⁹ X0.44 X 10^-9

Length of DNA = 0.88m

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. This mutation would make the histones acidic and negatively charged in nature and histones would not be able to bind to DNA because both would be negatively charged. Hence, the DNA would not be packed and chromatin would not be formed.

If RNA, instead of DNA was the genetic material, the heat killed strain of Pneumococcus would not have transformed the R-strain into virulent strain. This is because RNA is not thermo stable. Which means that RNA is more prone to degradation by heat because of the presence of 2’OH group in itsribose. As a result of which it would have lost the capability to change the R-strain to virulent strain.

Use of 15N in place of 35S in the original experiment would not give any conclusive result because:

1. 32P is radioactive and 15N is non-radioactive. So, it’s presence won’t be detected.

2. If hypothetically 15N would be radioactive, then it’s presence would be seen both inside the cell as it gets incorporated in the DNA as nitrogenous base and also as a supernatant as it gets incorporated in the amino group of amino acids.

Some amino acids are coded by more than one codon. This phenomenon is called as degeneracy of codon. Hence, from such amino acids multiple nucleotide sequence would be obtained. Forexample, Isoleucine has three codons AUU, AUG and AUA.The following nucleotide sequences are obtained:

a) AUG-AUU

b) AUG-AUG

c) AUG-AUA

All the three sequences code for Met-Ile.

The above statement is correct. It is because if mutation takes place at the third base pair, it does not lead to a phenotypic change. These mutations are known as silent mutations. If the mutations take place at any other base than the third base phenotypic changes may take place. Forexample: Substitution of Valine instead of Glutamic acid in the beta haemoglobin chain of the RBC’S leads to sickle cell anaemia.

Lactose present in the external medium can only enter the bacterium when the bacterium has permease within it. Low level of expression of lac operon is required for permease to form within the bacteria.

The sequencing of human genome opened new windows for treatment of various genetic disorders because:

1. It led to better understanding of genetic disorders.

2. Diagnosis, treatment and prevention of genetic disorders are better understood.

3. Which genes is for which disease is identified.

The total number of genes in humans is far less (< 25,000) than the previous estimate (up to 1,40,000 gene) because of the presence of large portions of repetitive sequencing regions in the human genome. These are the stretches of the DNA sequence that are repeated numerous times hence the non-repetitive gene count is less, if the repetitive portions are not added. These stretches have no coding functions but make it possible to understand chromosome structure, dynamics and evolution.

Advantages:

1. It would lead to better diagnosis, treatment and prevention of genetic disorders.

2. Better understanding of DNA gene sequence would lead to better understanding of biological systems.

3. Health care, agricultural and environment can be understood easily by learning the DNA sequencing of other living organisms.

4. Extensive research on human evolution can be done

5. DNA forensics would also become possible.

Disadvantages:

1.Patenting of the genetic test results can be done. In turn of which gene patenting can also be done.

2.Untreatable genetic disorders might be discovered.

1. VNTR is the Variable Number of Tandem repeats. It is a chromosomal region. In which a small stretch of DNA sequence is repeated numerous times in a single location.

2. These VNTR produce a pattern of bands which is unique for each individual. These VNTR are used to identify criminals in forensic departments.

3. DNA fingerprinting is not an option for bacteriophage as it does not have repetitive sequencing portions like in VNTR’S so a pattern of bands would not be formed. Bacteriophagehas a small genome that has all the coding sequences.

Polymerisation would not take place as the 3’OH group on the sugar is not present to add a nucleotide for forming ester bonds.

Watson and Crick have made information available for developing a model of DNA which are as follow:

1. Chargaff’s rule: They used Chargaff’s rule of base pairing which showed Adenosine forms bond with Thymine (A&T) and Guanine bonds with Cytosine (G & C).

2. Wilkins and Franklin’s rule: This rule was used to know that DNA has a diameter of 20 Å, a regular helix structure with 34Å distance and 10 base pairs of nucleotides in each turn of the spiral.

3. X-ray crystallography is was used to study DNA’S physical structure.

Contribution:

1. They discovered DNA was a double helix structure.

2. They also discovered the complementary base pairing by hydrogen bonds.

(i) Methylated guanosine cap: When translation is initiated methylated guanosine, cap helps to bind mRNA to smaller ribosomal subunit.

(ii) poly-A “tail”: It prolong mRNA’s life.

The alternate splicing of exons is very specific in nature it could be either gender specific, developmental stage specific etc. Which can lead to encoding of various proteins from a single gene. In absence of such splicing there is a requirement of new gene for every protein or isprotein.

Tandems is a chromosomal region in which a small stretch of DNA sequence is repeated numerous times in a single location. It forms a pattern of bands which is unique for each individual and is used in DNA fingerprinting in forensic department.

1. In the year 1952, Alfred Hershey and Martha Chase did an experiment to detect the presence of genetic material. For this experiment they took a virus called ‘Bacteriophage’.

2.The virus was grown in two separate mediums. One medium consisted radioactive phosphorus (P32). The other medium consisted of radioactive Sulphur (S35).

3.Medium containing radioactive Phosphorous had radioactive DNA because DNA is a phosphorus-basedstructure. Medium containing radioactive sulphur had radioactive protein.

4.Now, these viruses were injected into a bacterium called E. coli.

5.For this incorporation of Virus into E. coli, viral coats were removed by agitating them in the blender.

6.E. coli infected with virus had radioactive DNA and no radioactive proteins were present.

7.This showed that ‘DNA’ was the genetic material that got transferred from virus to E. coli and not protein.

CONCLUSION:

This experiment concluded, that DNA is the genetic material that transfers from one cell to another and not protein.

In such case,

a) Radioactive ³⁵ S and ³² p labelled protein capsule would show no radioactivity in the cell, but radioactivity would be detected in the supernatant.

b) Radioactive ³⁵ S and ³² p labelled DNA: Radioactivity is detected in the cells but not in the supernatant.

Such case would lead to non- conclusive experiment results.

The desired criteria in a molecule that can act as genetic material are as follows:

1.It should be chemically and structurally stable.

2.It should be able to replicate, that is replication should take place.

3.Mutation or gradual and slow changes should take place.

4.Should be able to express itself in Mendelian.

5.It must be able to transfer from parent to progeny for further evolution.

Biochemical Differences between DNA and RNA

These biochemical differences between the DNA and RNA makes DNA a desired genetic material.

Post transcriptional modifications are the chemical alterations done to primary transcript of RNA.These modifications convert a gene into functional RNA.

1. In prokaryotes the primary transcript (hn-RNA) has both the non-coding introns inside the gene and the coding exons but they are non-functional.

2. Hence, Splicing takes place. In which introns are removed and exons are joined in definite order.

3. In Eukaryotes (hn-RNA) or primary transcript is absent so, slicing is not needed.

4. In such cases capping and tailing process takes place.

5. In capping Methyl Guanosine Triphosphate nucleotide is added to the 5’ end of hn-RNA known as cap structure.

6. In tailing Adenylate residues are added at the 3’ end of the RNA.

7. After the whole process hn-RNA is processed to form m-RNA or messenger RNA.

8. This m-RNA is transported out of the nucleus for translation.

Translation is the decoding of m-RNA to form amino acid chain with the help of ribosome in the cytosol of the cell.

Molecules Involved In Translation:

1. Amino acids: These are the structural unit of proteins which are joined together to form a protein chain.

2. m-RNA: The m-RNA or messenger RNA has codons, each codon codes for one amino acid.

3. t-RNA: The t-RNA or transfer RNA binds with amino acid carrier and carry the amino acid sequence.

4. Ribosome: These are required for the catalysis of the process.

Steps Involved In Translation:

Initiation:

1. At the 5’ end of the m-RNA the smaller 40S subunit of the ribosome with methionyl-tRNA scans the m-RNA to find the start codon (5’AUG).

2.This codon (AUG) is specific to methionine.

3.Now, the larger 60S subunit of the ribosome binds to the m-RNA.This 60S ribosomal subunit has two t-RNA binding sites.

Site P: This site can hold the peptide chain.

Site A: This site can hold the t-RNA.

Elongation:

1. Now, as the Met-tRNA binds to the P site of the 60S subunit of the RNA. Another aminoacyl- tRNA which has an anticodon complementary to the next codon occupies the site A within the 60S subunit of ribosome.

2. The enzyme peptidyl transferase forms a peptide bond between Methionyl-tRNA and Aminoacyl-tRNA.

3. This makes the t-RNA molecule at the P site to become uncharged and it leaves the ribosome.

4. The ribosome moves along the m-RNA molecule to the next codon which opens the Site A for next aminoacyl-tRNA.

5. The polypeptide chain is built in the direction of N to C terminal.

Termination:

1. Now, as the A site is vacant the stop codon enters the site.

2. None of the t-RNA molecule binds to this codon.

3.The peptide bond of Methionyl-tRNA and Aminoacyl-tRNA become hydrolysed releasing the polypeptide into the cytoplasm.

4. The ribosomal subunit now disassociates.

1. The Operon model was given by François Jacob and Jacques Monad in 1961.

2. Operon is a cluster of gene which have related functions and are involved in the catabolism or degradation of lactose .

3. There are two types of operons, Inducer operon also called Lac Operon (lactose) and Trp Operon also called Repressor operon (Tryptophan operon).

Inducible Operator System:

1. It is an operon also known as Lac Operon found in bacteria E. coli.

2. When E. coli feeds on something it prefers Glucose over lactose.

3. When Glucose is fully taken up by the E. coli, it starts using Lactose.

4. When intake of lactose is started by E. coli, Lac operon gets activated which proves that lactose is an inducer for the Lac Operon.

Structure of Lac Operon:

Working of Lac Operon:

1. In the operon gene ‘i’ undergoes transcription and forms a messenger RNA or m-RNA.

2. This m-RNA undergoes translation to form repressor protein.

3. Now the repressor protein combines together to form repressor tetramer.

4. The promoter site or gene ‘p’ is the binding site for enzyme RNA polymerase.

5. As the repressor tetramer moves towards the operator site to bind to it, lactose present in the E. coli breaks down the tetramer.

6. Due to which the operator gets unblocked and makes it able for enzyme RNA polymerase to move forward from the promoter site.

7. This enables the enzyme RNA polymerase to start transcribing structural gene z,y and a.

8. Gene ‘z’ coding for enzyme p-galactosidase breaks down into glucose and lactose.

9. Gene ‘y’ coding for enzyme permease provides the entry of more lactose in the bacteria E.coli.

10. Gene ‘a’ coding for transacetylase adds the acetyl group to beta galactosidase to activate beta galactosidase.

Conclusion:

It is concluded E. coli can never die because it can survive in both mediums of glucose and lactose.

1. To solve the dispute DNA fingerprinting should be used.

2. In DNA fingerprinting a pattern of bands is formed due to small stretch of DNA sequence which is repeated numerous times called as Variable Number of Tandem Repeats (VNTR’S) , also known as Mini Satellites. These VNTR’s are specific for each individual.

Procedure of DNA Fingerprinting:

1. For DNA fingerprinting, blood sample of the respective person is taken. Now the DNA is extracted from the blood sample.

2. This DNA is cut into smaller fragments by an enzyme called restriction endonuclease enzyme.

3. These DNA fragments are then separated by the process of Gel Electrophoresis. In which electric field is applied to separate the DNA fragments based on size. The shorter ones move faster, then medium and longer ones are separated.

4. The DNA band pattern now formed by Gel electrophoresis is transferred to the Nylon Sheath, this process is also called Southern Bloating.

5. The band is now mixed with radio isotopic solution and is exposed to X-ray radiation.

6.After exposure to X-ray radiation coloured band patterns or VNTR’S are seen which are specific for everyone.

By the method of DNA fingerprinting the paternity dispute can be solved as the VNTR for the father would be specific for an individual.

Principle Involved:

The principle involved behind DNA fingerprinting is the Variable Number of Tandem Repeats (VNTR’S) , also known as Mini Satellites. These are a pattern of band formed due to small stretch of DNA sequence which is repeated numerous times. These VNTR’s are specific for each individual.

1. DNA sequencing is the method used to detect the definite order of bases (Adenosine, Guanine, Thymine and Cytosine) with in the DNA.

2. Methods mainly used are:

A) Maxim Gilbert Method

B) Sanger Method

Maxim Gilbert Method:

In 1977 Maxim Gilbert devised for DNA sequencing.

Procedure of Maxim Gilbert Method:

1. At first the double helix structure of the DNA is denatured into a single stranded DNA by increasing the temperature.

2. The 5’ end is labelled radioactively by Kinase reaction using gamma P ^32.

3. Cleave the DNA strand at specific positions using chemical reactions.

4. For cleavage at specific positions two chemicals are used followed by the addition of piperidene.

a) Dimethyl Sulphate: This chemical only attacks purines (Adenosine and Guanine).

b) Hydrazine: This chemical only attacks pyrimidine (Cytosine and Thymine)

5.In Maxim and Gilbert experiment the chemicals cleaved G, A+G, C and C+T.

6.In four test tubes, four different size of DNA’s are obtained.

7. Now, these DNA’s are separated based on size by gel electrophoresis technique.

8. After this separation the sequence of DNA is obtained.

Advantage:

1.Purified DNA can be read easily.

2.DNA-Protein interactions can be analysed.

3.DNA stretches can also be sequenced.

4.Nucleic acid structure can also be analysed.

Disadvantage

1. This method is not used widely.

2. Toxic and radioactive chemicals are used.

3. The process is complex.

4. Not more than 500 base pair can be analysed.

SANGER METHOD:

1. In Sanger’s method DNA that has to be sequenced is amplified.

2. Now, heat is used to denature the DNA to produce complementary and template strand for DNA sequencing.

3. A primer is than annealed to the 5’ end of the template strand.

4. The primed DNA is then distributed equally among four test tubes.

5. In each of the test tube all four standard deoxynucleotide triphosphate (DNTP’S) which are (dATP, dGTP,dCTP and dTTP) and DNA polymerase are added.

6. To each of the test tube only one of the four dideoxynucleotides (ddATP, ddGTP, ddCTP,ddTTP) are added.

7. The amount of dideoxynucleotide should be 100 times less than the deoxynucleotide so that the fragments can still transcribe.

8. The DNA fragments now formed, are denatured by heating and separated by the process of gel electrophoresis based on size difference.

9. The bands now formed are seen under UV light or X-ray and the sequence can be detected.

Advantages:

1.This the commonly used method.

2.Scientists were able to study the sequencing properly.

3. Various genetic disorders were detected.

Disadvantages:

1. Illegal use of DNA sequence information can be done.

DNA markers are small regions of DNA sequence which are unique to an individual.

The various markers used in DNA fingerprinting are:

1. Restriction Fragment Length Polymorphism (RFLP’S):

It is a non-PCR based approach to identify the DNA sequencing. In this process DNA is digested at specific sites with restriction enzymes. The pattern band of fragments thus formed are separated by Gel electrophoresis and identified under X-ray or UV.

2. Random Amplified Polymorphic DNA (RAPD):

It is a PCR based approach to find out specific DNA sequence. In this process many short primers are formed. From which random DNA segments are amplified using PCR(which means specific DNA sequence information is not required).The band pattern formed from these DNA fragments are separated by Gel electrophoresis and identified under X-ray or UV .

3. Amplified Fragment Length Polymorphism:

In this process DNA is digested with the restriction endonuclease enzyme. Selective amplification of the fragments from a mixture of DNA fragments is done.

4. Southern Bloating:

In this process the fragments of the DNA are transferred to nylon sheath. These fragments are the mixed with radioactive isotope and the band of pattern formed are then visualised under UV or X-ray.

These are the various markers used to identify DNA sequencing.

1. In the above case as replication could take place in the presence of radioactive deoxynucleotides precursors in E. coli that was a mutant for DNA ligase, Okazaki fragments would not be joined on the lagging strand.

2. Hence, on the leading strand high molecular weight fragments would form.

3. On the lagging strand low molecular weight fragments would form.

4. Which concludes option ‘a’ is correct.