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).

Transcriptionin Bacteria

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’ώ
• ·         Sigma factor (1-subunit) = σ
• Holoenzyme = Core enzyme + Sigma factor

                       = α2ββ’ώ σ =4, 50,000 mol wt

Stepsof Transcription in Bacteria:

• ·         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

rho-dependent termination:

• ·         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.

Rho-independenttermination

• ·         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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Discussion on Gene Regulation Questions

Q1. Which of the following is true of an Operon?

a)      An operon consists of structural genes only

b)      An operon consists of Strcutural genes and promoter regions

c)       An operon consists of promoter and I gene

d)      None

Explanation:

The correct answer is ‘b’ because an operon is a complete entity that includes promoter and structural genes. Since promoter region is important to bind RNA polymerase, it must be included in an operon to allow structural genes to express themselves.

Q2. The control of gene expression in prokaryotes mostly occurs at which level?

a)      Transcriptional level

b)      Translational level

c)       Replication level

d)      All of the above

Explanation:

The correct answer is ‘a’ because in prokaryotes, transcription and translation occurs simultaneously.  So, if there is no transcription then, no translation as well.

Q3. Control of gene expression in eukaryotes occurs at which level?

a)      Transcriptional level

b)      Translational level

c)       Processing level

d)      All of the above

Explanation:

The correct answer is ‘d’ because in eukaryotes, transcription and translation processes are separated by the nuclear membrane. In other words, transcription and post transcriptional modifications are done in the nucleus and translation occurs in the cytoplasm.

Q4. Which of the following is a constitutive gene?

a)      I gene

b)      Z gene

c)       Y gene

d)      A gene

Explanation:

The correct answer is ‘a’ because I gene is always on. It is also called as housekeeping gene. On the other hand, Z, Y, and Z genes are expressed under certain conditions like the glucose and lactose level.

Q5. In Lac Operon, when the repressor gene binds to the operator, no transcription and translation process occurs. This type of regulation is called:

a)      Positive regulation

b)      Negative regulation

c)       Neutral regulation

d)      None

Explanation:

The correct answer is ‘b’ because the repressor protein, as the name suggests represses the process of transcription and translation. So, it is called negative regulation.

Q6. What happens when the concentration of lactose in the cell increases?

a)      The repressor protein attaches to the lactose and the combined structure binds to the operator to switch off the process of transcription and translation

b)      The repressor protein binds to the operator without binding to the lactose molecules

c)       Lactose molecules inhibit the repressor protein to bind to the operator sequence

d)      The repressor protein attaches to the lactose and the combined structure fails to bind to the operator to switch on the process of transcription and translation

Explanation:

The correct answer is ‘d’ because lactose acts as an inducer. When there is no lactose in the cell, no transcription and translation process occurs. In this situation, repressor protein binds to the operator sequence to switch off the process of transcription and translation. This is because, the cell already has sufficient amount of glucose. However, when the lactose concentration increases, it means, there is low amount of sugar in the cell and lactose needs to be converted to glucose and galactose to get energy. So, lactose induces the repressor protein and the two collectively fails to bind to the operator. So, when repressor is not attached to the operator, the process of transcription and translation occurs smoothly.  

 Q7. Tryptophan in Trp Operon is called as:

a)      Repressor

b)      Co-repressor

c)       Inhibitor

d)      Suppressor

Explanation:

The correct answer is ‘b’ because when the tryptophan concentration in the cell increases due to rapid transcription and translation process, further synthesis is not required. In that situation, tryptophan attaches itself to the repressor protein and both collectively attaches to the operator sequence to switch off the process.  Since tryptophan attaches to the repressor protein to inhibit the process of transcription and translation – it is called co-repressor.

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What Do You Mean By Monocistron And Polycistron

When we define a gene, it is considered as the functional unit of inheritance. So, what is a gene? Well, it is the DNA sequence that codes for a polypeptide. We can thus call a gene as the DNA sequence that codes for tRNA, rRNA, and mRNA.

We know that only mRNA sequence is translated to produce polypeptides. It is therefore, important to give a special name to the DNA sequence that codes for mRNA and eventually proteins. We call it as CISTRON. We can define CISTRON as a DNA sequence that codes for a polypeptide.

In case of EUKARYOTES, we know that a gene codes for a single polypeptide, and therefore, we can call a CISTRON as MONOCISTRONIC in eukaryotes. On the other hand, when we talk about prokaryotes, we know that structural genes (a combination of genes in tandem) code for more than one polypeptide. So, we can call a CISTRON as POLYCISTRONIC in prokaryotes.

It is also important to mention that in eukaryotes, coding DNA sequences are interrupted by non-coding sequences. And we use special terms such as EXON (coding sequence) and INTRON (non-coding sequence).

Note: In prokaryotes, non-coding sequences (INTRONS) are absent.

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Only One DNA Strand Participates in Transcription, Why?

Transcription is considered as the first step in gene expression. It refers to the copying of DNA sequence into an RNA sequence. However, unlike DNA replication, only one strand is transcribed. The question is why both the DNA strands are not copied during transcription.

The DNA strand involved in transcription is called template DNA and the other strand which does not take part in transcription is called coding strand. Now, let's imagine, if both strands participate in transcription i.e., act as a template, they would code for RNA molecule with different sequences. Now, if these RNA strands code for proteins, they would be of different types. This means, one DNA strand would code for different proteins and this would result in complication in terms of the genetic information transfer machinery. Besides, if two RNA molecules are produced simultaneously, they would complement each other and would form a double stranded RNA. The double stranded RNA would eventually resist from being translated into proteins.

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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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