Showing posts with label RNA polymerase. Show all posts
Showing posts with label RNA polymerase. Show all posts

TRANSCRIPTION IN PROKARYOTES



Transcription is considered as the very first step in gene expression.  In other words, it is an act of making a copy. In genetics, however, it means copying DNA into an RNA sequence. Unlike DNA replication, transcription involves a single strand of DNA.

TRANSCRIPTION UNIT

A transcription unit in DNA consists of the following regions:

1)      A Promoter
2)      The Structural Gene
3)      A Terminator

Note: Like DNA replication, transcription also occurs in 5’ à 3’ direction.



Note: The RNA strand is similar to the coding strand; the only difference is that in place of T, U is added in the RNA strand (T is replaced by U).


There are three basic steps of bacterial transcription

i)                    Initiation
ii)                   Elongation
iii)                 Termination

Promoters: 

  • ·         These are DNA sequences that promote gene expression.
  • ·         They promote the exact location for the initiation of transcription
  • ·         They are located upstream of the site where transcription of a gene actually begins
  • ·         The basees in a promoter are numbered in reference to the transcription start site



The conventional numbering system of promoters

  • ·         Most of the promoter regions are labelled with negative numbers
  • ·         There is no base numbered zero
  • ·         Bases to the right are numbered in a positive direction
  • Initiation of Bacterial Transcription
  • RNA polymerase is the enzyme that catalyzes the synthesis of RNA i.e., a single RNA polymerase is responsible for catalyzing all three RNAs (rRNA, mRNA, and tRNA).
  • In E.coli, the RNA polymerase holoenzyme is composed of
  • ·         Core enzyme (5-subunit) = α2ββ’ÏŽ
  • ·         Sigma factor (1-subunit) = σ
  • Holoenzyme = Core enzyme + Sigma factor
                       = α2ββ’ÏŽ σ =4, 50,000 mol wt


  • ·         The RNA polymerase holoenzyme binds loosely to the DNA away from -35 region
  • ·         It then scans along the DNA until it finds both -35 and -10 upstream regions.
  • ·         A region within the sigma factor that contains a helix-turn-helix structure is involved in a tighter binding to the DNA
  • ·         The binding of the RNA polymerase to the promoter forms a closed complex
  • ·         The open complex is formed whenTATAAT box in the -10 region is unwound
  • ·         A short RNA strand is made within the open complex (the sigma factor is released at this point). And this marks the end of initiation.





Elongation:

·         The core enzyme now slides down the DNA to synthesize an RNA strand. This is known as elongation phase.

Termination

·         It is the end of RNA synthesis. It occurs when the short DNA-RNA hybrid of the open complex is forced to separate. This releases the newly made RNA as well as the RNA polymerase.

E.coli has two different mechanisms of termination.
a) rho-dependent termination = protein based
b) rho-independent termination = RNA based


  • ·         It is controlled by rho-protein, which tracks along behind the polymerase on the growing mRNA chain.
  • ·         Near the end of the gene, the polymerase encounters a run of G-nucleotides on the DNA template and it stalls. As a consequence, the rho-protein collides with the polymerase.
  • ·         The interaction with the rho-protein releases the mRNA from the termination bubble.
  • ·         It is controlled by specific sequence in the DNA template strand.
  • ·         As the RNA polymerase reaches near the end of the gene being transcribed, it encounters a region rich in C-G nucleotides.
  • ·         The mRNA folds back on itself and the complementary C-G nucleotides bind together. The result is a stable hairpin that causes the polymerase to stall as soon as it begins to transcribe a region rich in A-T nucleotides.
  • ·         The complementary U-A region of the mRNA transcript forms only a weak interaction with the template DNA. This coupled with the stalled polymerase induces enough instability for the core enzyme to break away the new mRNA transcript.
  • ·         Upon termination, the process of transcription is complete.

Note: The bacterial RNA does not require any processing to become active.
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DNA Replication Question Discussion

There are 15 questions on DNA replication. Some questions have been asked in previous medical entrance examinations while others are not. Please go through all these questions and find the right answer. We'll not provide answer key for the same, but on request, we'll provide the same.

Q1. During DNA replication, torques are relieved by enzymes
(a)    DNA ligase
(b)   DNA polymerase
(c)    DNA helicase
(d)   DNA gyrase
(C.O.M.E.D – K’s 2011) 
  
Q2. 1.7m double helical DNA will have base pairs
(a)    3.4 x 109
(b)   5x 109
(c)    1.7x 109
(d)   1.7x 105
(AMU 2011)


Q3. First step in DNA replication is:
(a)    Synthesis of Okazaki segments
(b)   Formation of DNA templates
(c)    Synthesis of RNA primers
(d)   Activity of ssBP
(M.H.T.CET 2009)

Q4. Okazaki fragments are formed in the direction of
(a)    3’-->  5’
(b)   5’-->  3’
(c)    5’-->  5’
(d)   3’ --> 3’
(W.B 2009)


Q5. In DNA molecule
(a)    Proportion of A to A varies with the organism
(b)   There are two antiparallel strands running in 5’--> 3’ and 3’ --> 5’
(c)    Amount of purine and pyrimidine nucleotides is not always equal
(d)   There are two parallel strands running in 5’--> 3’ direction
(C.B.S.E 2009)


Q6. RNA primer is used in
(a)    Translation
(b)   Replication
(c)    Conjugation
(d)   Transformation
(D.P.M.T 2008)
 
Q7.  DNA replication is semi-conservative as
(a)    Only non-parent strand acts as template
(b)   Both strands of new molecule are synthesized de novo
(c)    One of the strands in each new molecule is parental and the other one is new
(d)   Daughter strands are dispersive
(   (C.O.M.E.D- K’s 2008) 


      Q8. Okazaki fragments during DNA replication
(a)    Polymerize in 3’à 5’ direction and form replication fork
(b)   Prove semiconservative DNA replication
(c)    Polymerize in 5’à 3’ direction and explain 3’à 5’ DNA replication
(d)   Cause transcription
(C.B.S.E 2007)


Q9. DNA replication requires
(a)    DNA polymerase
(b)   DNA ligase
(c)    RNA polymerase
(d)   All of the above
(M.P.P.M.T 2007)


Q10. C-value paradox is
(a)    Diploid DNA content
(b)   Haploid DNA content
(c)    Variation in C-value
(d)   Constancy of C-value
(C.P.M.T 2007)


Q11. Which of the enzymes of DNA replication require ATP
(a)    Helicase
(b)   Ligase
(c)    DNA polymerase
(d)   All of the above
(
Q11. How many OriC are found in E.coli
(a)    1
(b)   2
(c)    3
(d)   4
(
Q12. DNA replication in Bacteria is:
(a)    Unidirectional
(b)   Bidirectional
(c)    Multidirectional
(d)   None

Q13. Which of the following enzyme is absent in DNA Pol II complex of Bacteria
(a)    3’à 5’ exonuclease
(b)   5’à 3’ exonuclease
(c)    Both
(d)   None

Q14. How many base pairs are present in OriC region
(a)    250 bp
(b)   100 bp
(c)    300 bp
(d)   400 bp

Q15. Dna B protein is also called as:
(a)    Ligase
(b)   Protease
(c)    Helicase
(d)   Gyrase
 

 
 

 




 
 
 

 









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