Transcription

 


What enzymes are involved in transcription 

Transcription:

It is the transfer of genetic information from DNA to RNA to the site where it is required for protein synthesis. Both the strands of DNA do not transcribe RNA but only strand called sense strand is used as template for the synthesis of a complementary strand of RNA.Transcription requires DNA_ dependent RNA % polymerases. Eukaryotes have 3 RNA polymerases. 

RNA _ polymerase I_ for r_ RNAs ( 28S, 18S and 5.8S).

RNA polymerase II _for hn RNA, for m_ RNA, SnRNA.

RNA polymerase III _for t_ RNA, 5_ S RNA and sc RNAs(small cytoplasmic RNAs)Different parts of DNA are involved in transcription of various ribonucleotide acids. Prokaryotes have_ only one RNA polymerase which synthesizes all types of RNAs. RNA polymerase of E.coli has a m.wt of about 490,000 and has 6 polypeptides, two of these are identical hence the enzymes contains 5 distinct polypeptides _ ß , ß, ó ,æ and ₩, ó   factor is involved in initiation of transcription and has no catalytic function. Holoenzyme contains 2æ, ß, ß ₩  and ó polypeptides. Ó factor recognizes the start signal or promoter site on the DNA. A factor is required for termination of transcription. A number of other factors are also required for unwinding of DNA duplex, stabilization of unwound DNA strand, base pairing, separation and processing of transcribed RNA.

1):  Activation of Ribonucleotide: Prior to transcription ribonucleotides are activated through phosphorylation. Enzyme phosphylase is required along with energy. Activated ribonucleotides taking part in transcription are ATP, GTP, UTP and CTP.

2): DNA template: On specific signals segments of DNA corresponding to one or more cistrons become derepressed and ready to transcribe. Each DNA segment has a promoter region and a terminator region. Transcription starts at the promoter region and terminates at terminator region.RNA polymerase recognition site and binding site are present at promoter region.Chain opens in the region occupied by TA TA AG nucleotides in most of the prokaryotes. Unwindases and gyrases separate the chain. Termination region has poly A  base sequence or pallindromic sequence ( identical base sequence running  in opposite directions in the two DNA chains). RNA  polymerase binds to the promoter region.The two DNA  strands  uncoil progressively from the site of polymerase binding. One of the two strands  of DNA functions as template for transcription of RNA. It is called master or sense strand.Transcription proceeds  in 5' ➡️3' direction. 

3): Base pairing: Ribonucleotide triphosphates present in the medium come and lie opposite the nitrogen bases of the DNA template. They form complementary pairs_ U opposite A, A opposite T, C opposite G and G opposite C. Two extra phosphates are removed by the activity of pyrophosphatase and energy is released in this process.
   
                                           Pyrophosphatase        Ribonucleotide triphosphate➡️➡️➡️➡️➡️                                                                 Ribonucleotide + PPi + Energy 

4):Chain formation: RNA chain is formed over DNA template by the action of RNA Polymerase on the adjacent Ribonucleotides With the initiation of chain formation,  ó factor of the RNA polymerase separates and the core enzyme moves along the DNA template causing elongation of RNA chain at the rate of 30 nucleotide/ second. As the polymerase reaches the terminator sequence suggesting that 3' _ AAAAAA T_ 5' in the sense strand of DNA is atleast part of transcription terminator sequence. 

5): Separation of RNA: Termination or Rho factor has ATPase activity ( Roberts). It helps in releasing completed RNA chain which is called primary transcript.

6): Duplex formation: The two DNA strands form linkages amongst complementary base pairs after the release of primary transcript. Gyrases, unwindases and helix destabilizing proteins are released. Double helical structure of the DNA is resumed.

7): Post transcription  processing:RNA strand formed after transcription is larger , than the functional RNAs. Post transcription processing is necessary to form functional RNA.
It is of 4 types: (i) cleavage (ii)Splicing 
(iii) Terminal additions.(iv) Nueotide modifications 
(i) Cleavage : Larger RNA precursors are cleaved form smaller RNAs. Primary transcript of r__ RNA is 45 S in eukaryotes.primary _ transcript is cleared by ribonuclease p( RNA enzyme) to form 5_ 7 t_ RNA precursors. 


(ii) Splicing : Eukaryotic transcripts possess extra segments ( introns or intervening  sequences). These are removed by nucleases. Ribozome is a self splicing intron involved in some of these reactions as well as catalysing polymerisation.


(iii) Terminal additions: Additional nucleotides are added to the end of RNAs for specific functions e.g.CCA segments in t_ RNA, cap nucleotides at 5' end of _ m_ RNA or poly _ A segments at 3' end of m_ RNA.


(iv) Nucleotide modifications: These occur commonly in t _ RNA. It is methylation. ( e.g methyl cytosine, methyl guansone), deamina_ tion ( e.g. inosine from adenine), dihydrouracil, pseudouracil etc.

In vitro  synthesis of RNA was first performed by Ochoa ( 1967).
                
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Several Enzymes are involved in the process of transcription:


1): RNA polymerase: The enzyme is responsible for catalysing the synthesis of RNA from a DNA template .It binds to the DNA and unwinds the double helix, allowing the complementary RNA nucleotides to be added.

2): PROMOTER RECOGNITION FACTORS:  These are proteins that bind to specific DNA sequences called promoter regions, which are located upstream of the start site of a gene.They help RNA polymerase in recognizing and binding to the promoter, initiating transcription. 


3): Transcription  Factors: These are proteins that assist in the regulation of transcription. They can  either enhance or inhibit the binding of RNA polymerase to the promoter, thereby controlling the level of gene expression. It produces RNAs by transcription for use in protein synthesis. 


4): Helicases: These enzymes are responsible for unwinding the DNA double helix by breaking  the hydrogen  bonds between  the base pairs, allowing the RNA polymerase to access the template strand. 


5): Topoisomerases: These enzymes help in relieving the torsional stress that build up ahead of the moving RNA polymerase during transcription by breaking and rejoining the DNA strands.



There may be other accessory proteins involved as well, depending on the specific process and the organisms being studied.

Ribonucleic acid or RNA : RNA or ribonucleic acid is a single chain poly_ ribonucleotide which functions as carrier of coded genetic or hereditary information from DNA to Cytoplasm for taking part in translation and enzyme synthesis. It has 70_ 12,000 ribonucleotide joined end to end. Backbone is formed by the alternating phosphate and ribose sugar. Phosphate combines with carbon 5' of its sugar and carbon 3' of next  sugar similar to the arrangement found in DNA steand. Nitrogen bases are attached to  sugars at carbon 1 of the latter .RNA has uracil instead of thymine, rest of the nitrogenous bases are common in DNA and RNA. These bases can be arranged in any sequence which is complementary to the base sequence of DNA _ e.g. ATACTG sequence of DNA will have UAUGAC over RNA. There are 6 Types of RNAs_ r_ RNA, m_ RNA, t_ RNA, genetic ( genomic ), small nuclear and small cytoplasmic. 


TRANSCRIPTION  IN Biology:

The process of copying genetic information from one strand of the DNA into RNA is termed as transcription. The principle of complementarity governs the process of transcription, except the adenosine now forms base pair with uracil instead of thymine.Unlike in the process of replication, which once set in, the  total DNA of an organism gets duplicated, in transcription only a segment of DNA and only one of the strands is copied into RNA. This necessitates defining the boundaries that would demarcate the region and the strand of DNA that would be transcribed. 

Why both the strands are not copied during transcription has the simple?


● First, if both strands act as a template, they would code for RNA molecules with different sequences ( Remember complementarity does not mean identical), and in turn, if they code for proteins, the sequence of amino acids in the proteins would be different. One segment of the DNA would be coding for two different proteins, and this would complicate the genetic information transfer machinery. 

● Secondary, the two RNA molecules if produced simultaneously would be complementary to each other, hence would form a double standard RNA. This would prevent RNA from being translate into protein and the exercise of transcription would become a futile one.



Transcription of Types :

Post transcription processing: RNA strand formed after transcription is larger, than the functional RNAs. Post transcription processing is necessary to form functional RNA.
 It is of 4 types: (i) cleavage (ii)Splicing 
(iii) Terminal additions.(iv) Nueotide modifications 
(i) Cleavage : Larger RNA precursors are cleaved form smaller RNAs. Primary transcript of r__ RNA is 45 S in eukaryotes.primary _ transcript is cleared by ribonuclease p( RNA enzyme) to form 5_ 7 t_ RNA precursors. 


(ii) Splicing : Eukaryotic transcripts possess extra segments ( introns or intervening  sequences). These are removed by nucleases. Ribozome is a self splicing intron involved in some of these reactions as well as catalysing polymerisation.


(iii) Terminal additions: Additional nucleotides are added to the end of RNAs for specific functions e.g.CCA segments in t_ RNA, cap nucleotides at 5' end of _ m_ RNA or poly _ A segments at 3' end of m_ RNA.


(iv) Nucleotide modifications: These occur commonly in t _ RNA. It is methylation. ( e.g methyl cytosine, methyl guansone), deamina_ tion ( e.g. inosine from adenine), dihydrouracil, pseudouracil etc.

In vitro synthesis of RNA was first performed by Ochoa ( 1967).

Transcription Unit:

A transcription unit in DNA is defined primarily by the three regions in the DNA:

(i) A promoter 

(ii) The structural gene

(iii) A terminator

There is a convention in defining the two strands of the DNA in the structural gene of a transcription unit.The two strands have opposite polarity and the NDA_ dependent RNA polymerase also catalyse the polymerisation in only one direction, that is, 5'➡️3, the strands that has the polarity 3'➡️5' acts as a template, and is also referred to as template strand.The other strand which has the polarity ( 5' ➡️3' ) and the sequence same as RNA ( except thymine at the place of uracil), is displaced during transcription. Strangely, this strand ( which does not code for anything) is referred to as coding strand. All the reference point while defining a transcription unit is made with coding strand. To explain the point,a hypothetical sequence from a transcription unit is represented below:

3_ ATGCATGCATGCATGCATGCATGC_5                                                                 Template strand

5_TACGTACGTACGTACGTACGTACG_3  Coding                                                                    Strand 




(i) A promoter gene It acts as initial signal which functions as recognition for RNA polymerase provide the operator gene is switched on. RNA polymerase is bound to the promoter gene. When the operator is functional the polymerase moves over it and it reaches the structural genes to perform transcription. 


(ii) Structural  genes: These genes synthesis m_ RNAs. m_ RNA controls metabolic activity of cytoplasm through translation. An operon has one or more structural genes. Lac_ operon contains  three structural genes ( Z,Y and A) .They transcribed a polycistronic m_ RNA molecule that helps in the synthesis of three enzymes _ ß_ galactosidase, lactose permease, and transacetylase. These 3 enzymes are produced in different molar concentration.

(iii) Terminotor  : The terminator is located towards 3' end ( downstream) of the coding strand and it usually defines the end of the process of transcription. 




The Process of transcription :

1): Activation of Amino Acid

2): initiation 

3): Elongation ( polypeptide chain Formation): 


1): Activation of Amino acid: Amino acids are activated by " activating enzyme called amino acyl t__ RNA synthetase ( Zamencnik and Hoagland).

It requires ATP and Mg++.
                             Mg++
AA+ ATP + Enzy➡️➡️➡️AA~AMP-E+PPi
                                                      Pyrophosphate


This complex reacts with t__ RNA specific for the amino __ acid to form amino  acyl_ tRNA complex. Enzyme and AMP are released. Some times t__ RNA amino acid complex is called  charged t__ RNA 


AA~ AMP-E+t-RNA➡️AA_tRNA+AMP+E

  

2): Initiation:It requires initiation factors. There are three initiation factors in prokaryotes __ IF3, IF2 and IF 1. Eukaryotes have nine initiation factors eIF2, eIF3, eIF1, eIF4 A, eIF4B, eIF4C, eIF5 and eIF6. Out of these IF3 or  eIF2 gets attached to smaller ribosomal subunit in the dissociated state.GTP is also required. m- RNA attaches itself to smaller subunit of ribosome in the region of tunnel. Its cap has nucleotides complementary to the nucleotides present at the 3' end of its r_ RNA.The initiation codon AUG or GUG lies at  p_ site .The initiation factor already present in smaller subunit catalyses the reaction ( eIF2 in eukaryotes and IF3 in prokaryotes).

                                      eIF2
40S subunit + mRNA ---------- 40S- mRNA

Aminoacyl t_ RNA  complex specific for initiation codon ( methionine t  ___ RNA or valine tRNA) reaches P_ site ( D_ site). Anticoden ( e.g. UAC) of met tRNA establish temporary hydrogen bonds with the initiation codon ( e.g.AUG) of m__ RNA.Tjis reaction requires eIF3 in eukaryotes and IF2 in prokaryotes. This step also requires energy supplied by GTP.

                                            eIF3
40S- m-RNA+t- RNA met ----------40S-m-RNA-                                                GTP             tRNAmet

In eukaryotes methionine accepting t_ RNA is charged with non_ formy lated methionine in the cytoplasm and in prokaryotes, plastids & mitochondria it is formulated t_ RNA, t__ RNA  transferring formylated methionine is different from the one that transfers non_formylated methionine. 

The larger subunit of ribosome now combines with 40S- m- RNA- tRNAmet complex to form inact ribosome. It requires initiation factor IF1,in prokaryotes and factors eIF1,eIF4 ( A,B,C) in eukaryotes. The P site of m- RNA -t- RNA complex in enclosed by the intact ribosome and A site remains exposed.
                                                 eIF1,eIF4
40S-m-RNA+t-RNAmet+ 60S------------80S-t-                                                                         RNAmet


3): Elongation ( polypeptide chain formation): An aminoacyl t-RNA complex reaches the A_ site and attaches to m_ RNA codon next to initiation codon with the help of its anticoden. The step requires GTP and elongation factor ( eEFI in eukaryotes and EF_ Ts in prokaryotes). EF_ Tu is most abundant protein of E.coli. A peptide bond is established between the carboxyl group (-COOH) of amino acid attached to t- RNA at P- site and amino group"(-NH2) of aminoacid attached to t_ RNA at A_ site.


Peptidy transferase present in the larger subunit of ribosome catalyses this reaction. In the process, the connection between t_ RNA and the amino acid at the P site breaks. The free t_ RNA  of the P_ site slips away. The A_ site carries peptidyl t_ RNA complex.

Soon after the formation of peptidyl bond and slipping of the freed t_ RNA of P_ site, the ribosome or m_ RNA rotates slightly and this process is called translocation. A translocase ( EF_ G in Prokaryotes and eEF2 in eukaryotes) & energy are required for this .Energy is provided by GTP.After translocation the A_ site codon along with peptidyl_ tRNA complex reaches the P_ site. A new codon is exposed at the A_ site. It attracts a new aminoacyl t_ RNA complex. Translocase (EF_G in Prokaryotes and eEF2in eukaryotes) is replaced by elongation factor ( eEF1, in eukaryotes and EF_ Tu/ EF_ TS in prokaryotes). The process of bond formation and translocation is repeated. Every codon of m_ RNA is exposed to the A site and get decoded through incorporation of amino acids in the peptide chain. Peptide chain elongates and the chan lies in the groove of the larger subunit to protect itself from the cellular enzymes as it is prone to breakdown due to its extended nature. Helix formation begins at a distance.




On being released from the ribosome, proteins form secondary and tertiary structure. When one polypeptide is associated to the other polypeptide, a quaternary structure of the protein is formed The protein gets modified glycosylation and hydrolytic enzymes. These are packed in the vesicles for export or formation of lysosomes, cell wall, enzymes and plasmamembrane.

Transcription Unit and the Gene:

Genes:Genes are linear segment of DNA consisting of a stretch of base sequences that codes for one polypeptide, one  t_ RNA or one r_ RNA molecule. It is called cistron or currently structural gene.The genetic system also contains a number of regulatory genes which control the functioning of structural genes. Genes have many sites or position where mutations can occur. If a single nucleotide is change it form a mutant phenotype e.g., sickle cell anaemia, two defective cistrons may recombine to form a wild type cistron. Genes can be divided into following  types.

1): Constitutive genes: These are called house keeping genes and expressed constantly in a cell. Their produce are required for the normal cellular activities e.g., genes for glycolysis, AT Pase.

2): Non_ constitutive genes ( Luxury genes: These genes are expressed according to the requirement of the body e.g., genes for nitrate reductase. Lactose system in E.coli. These are further divided into inducible and repressible types:

a): Inducible  genes:  These are switched on in response to the presence of a chemical compound called inducer.It is required for the functioning of the produce of gene activity ( e.g.nitrate for nitrate reductase).

(b) Repressible genes: These genes continuously express themselves till a chemical) sometimes and end product) Inhibits or expresses their activity. It is also called feed back repression. 


3): Multigenes : It is a group of similar or nearly similar which meet the requirement of time and tissue specific products e.g., globin gene family E,§ ,ß,y on  chromosomes II œ and § on chromosome 16 .


4): Repeated genes: These genes occur in multiple copies e.g.histron genes, t_ RNA genes, r_ RNA, actin genes. 

 
 5): Single copy genes:   These genes occur in single  copies ( occasionally 2_ 3 times). They form 60 _ 70% of functional genes. These are formed as a result of duplication, mutation, and exon reshuffling between two genes. 

6): Pseudogenes: Though these are homologous to functional genes but do not form functional product due to intervening non_ sense codons,insertions deletions and inactivation of promoter e.g. several of snRNA genes .


7): Processed genes These are eukaryotic genes devoid of introns. These are formed probably due to reverse transcription. These are generally non functional as they do not have promoters. 


8): Split genes Credit for their djscovery goes to Sharp and Roberts ( 1977). Thess have non_ essential regions interspersed with essential or coding parts. These non_ essential parts are called introns, spacer DNA or intervening sequences.Essential parts are called exons. Transcribed intronic regions are removed before RNA passes out in the cytoplasm. The genes are characteristics of eukaryotes. Histone gene and interferon  genes are completely exonic . Thymudylate synthase and  ribonucleotide reductase gene in T4 are also split genes. A neuropeptide is formed in the hypothalamus by the removal of an exon from calcitonin in thyroid transposase.



9): Transposone These are also called jumping genes ( Hedges and Jacob, 1974). These were discovered by Barbara Me Clintock  ( 1951) in maize. These can move from one place in the genome to another They have repetitive DNA at their ends. Enzymes separates the segment from its original by cleaving the repetitive sequence at its ends.



10): Overlapping genes :In ø× 174 genes B,E and K overlap other genes. 



11): Structural genes These have code for chemical substances required for the cells e.g., structural polypeptides, enzymes, transport proteins, proteinaceous hormones, antibodies, non translated RNAs.


12): Regulatory genes They do not produce chemical and control the functioning of structural genes. These genes may be promoters, terminators, operators and repressor producing i or regulator gene. Represser does not take part in cellular activity and regulates the activity of other genes .

Transcription:

Transcription is the process of copying a segment of DNA and RNA .The segments of DNA transcribed into RNA  molecules that can encode proteins are said to produce messenger RNA ( mRNA) .other segments of DNA are copied into RNA molecules called non_ coding  RNAs ( ncRNAs) comprised only 1__ 3% of total RNA samples 

1) Less than 2% of the human genome can be transcribed into mRNA (human genome coding and noncoding DNA) .Noncoding DNA. Greater part of DN in Eukaryotic cells does not code for RNAs. This " extra", or noncoding DNA, seems to have no function. It has two special forms.

(i) Repetition DNA: The noncoding DNA has many base sequences repeated several times. The repeated sequence are collectively called repetitious DNA 

(ii) Jumpimg Genes: Some repetitive DNA sequence are not found at fixed sites in the DNA of different individuals of the same species .such " mobile" DNA segments are often referred to as " jumping genes". The cause mutations and,thus, have a role in evolution. 


2):DNA and RNA are nucleic acids, which use base pairs of nucleotides as a complementary language. During transcription, a DNA sequence is ready by an RNA polymerase, which produces a complementary, antiparallel RNA strand called  a primary transcript.
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