transcription and translation- an a k. ghosh

7/31/2019 Transcription and Translation- An a K. Ghosh

1/90

Overview of Transcription, Translation andRecombinant DNA Technology

BY

Prof. Ananta K. Ghosh

Department of Biotechnology

Indian Institute of Technology Kharagpur


2/90

Eukaryotic cellswith a nucleus

Nucleus Mitochondria Chloroplast Ribosomes RER SER Golgi body Cytoplasm

Vacuoles

Prokaryotic cellswithout a nucleus

Cytoplasm Ribosomes Nuclear Zone DNA Plasmid Cell Membrane Mesosome Cell Wall

Capsule (or slime layer) Flagellum


3/90

M a c r o m o l e c u l e sProtein

Nucleic acids

olygosaccharides

Lipids

Complex macromolecules


4/90

Nucleic acids

Deoxyribonucleic acid (a polymer of deoxyribonucleotides)

Ribonucleic acid (a polymer of ribonucleotides)

A nucleotide is made up of Sugar, Nitrogenous base andphosphate


5/90

DNA RNA


6/90

DNA RNA


7/90

Nucleotide tri

phosphate


8/90


9/90


10/90

DNA consists of two strands running anti-parallel andforming double hellical structure


11/90


12/90


13/90

DNA vs. RNA

DNA Double Helix

Deoxyribose sugar Adenine pairs with

Thymine (A-T)

Stays in nucleus

RNA Single strand

Ribose sugar

Uracil replacesThymine!

Leaves nucleus todo the work


14/90

The general molecular formula of an amino acid is RCH(NH2)COOH

C

C

R

OO

N

H

H

H

H

carboxylic acidgroup

amine group

Side chain:

R characterisesthe amino acid

Proteins are made up of one or more polypeptide. Eachpolypeptide is a chain of co-valently bonded amino acids

Proteins


15/90

Formation of the peptide bond

Two amino acid molecules;the nature of the R group (R1

and R2) determines the aminoacid

The molecules must be

orientated so that thecarboxylic acid group of onecan react with the amine groupof the other

The peptide bond forms withthe elimination of a watermolecule; it is anotherexample of a condensation

reaction


16/90


17/90

SG

Y

A

V


18/90


19/90

DNA

mRNA

Transcription

The Central Dogma of Molecular Biology

Cell

Polypeptide

(protein)

Translation Ribosome

This describes the flow of information from DNA into RNA (most commonlymRNA) through transcription (copying the same code from one molecule toanother), and then expressing the code into a functional molecule by

translation (converting from a nucleic acid code into an amino acid code).


20/90


21/90

7

Prok aryotic G ene StructureProm oter C D S T erm inator

transcription

Genom ic DNA

m R N A

protein

U T R U T R

translation

Promoter is a DNA sequence usually present upstream of coding regionswhere RNA polymerase binds to initiates transcription.

Gene is the structural

and functional unit ofheridity which carrygenetic informationfrom one generation tonext. In molecular

terms Gene is a part ofchromosomes (DNA)which codes forfunctional RNA orprotein

Gene transcription in prokaryotes

UTR (Untranslated sequences): 5 UTR contains ribosome binding sites

for protein synthesis; 3UTR helps in stability of RNA

CDS: coding sequences for protein synthesis

Terminator: Sequence for ending RNA synthesis


22/90

RNA transcript is complementaryto template strand and identicalto coding strand

Requirement for transcription in prokaryotesGene or DNA to be transcribed, RNA polymerase, rNTPS

and cellular environment

Different genes are transcribed from different strands


23/90

Bacteria has oneRNA polymeraseto synthesize allthree RNA:(mRNA, rRNA,tRNA)

RNA polymerase binds to promoter of a gene to initiate transcription


24/90

A DNA fragment is labeled atone end with 32P (red dot).2. Portions of the sample then aredigested with DNase I in thepresence and absence of RNApolymerase holoenzyme.3. A low concentration of DNase Iis used so that on average each

DNA molecule is cleaved just once(vertical arrows).4. The two samples of DNA thenare separated from protein,denatured to separate the strands,

and electrophoresed. The resultinggel is analyzed by autoradiography,

The set of DNA fragments leftafter DNase I digestion reveals the

promoter

Identification of promoter by DNase 1 footprinting technique


25/90

1 3

P r o m o t e r

P r o m o t e r s s e q u e n c e s c a n v a r y t r e m e n d o u s l y .

R N A p o l y m e r a s e r e c o g n i z e s h u n d r e d s o f

d i f f e r e n t p r o m o t e r s


26/90

Stages of Transcription

Chain Initiation

Chain Elongation

Chain Termination


27/90

Bacterial Transcription Initiation Promoter recognition by RNA polymerase

Formation of Transcription Bubble by unwinding DNAstrands

Addition and Bond creation between rNTPs to start RNAsynthesis

Escape of transcripton apparatus from promoter (promoterclearance)

How RNA Polymerase finds promoter and Initiates


28/90

How RNA Polymerase finds promoter and InitiatesTranscription

Core enzyme has the ability to synthesize RNA on a DNA template butcannot transcription at proper site

Core polymerase has general affinity for DNA and loosely binds atrandom sites in DNA without discriminating promoter and othersequences.

Binding of sigma introduce a major changes in the polymerase and theholoenzyme drastically reduced ability to recognize loose binding sites,

and the enzymes moves along the DNA by directly displaced by anothersequences.

When it reaches the promoter sequences, sigma factor recognizespecifically -35 sequence and binds tightly.

The holoenzyme occupies -50 to +20 regions of DNA and unwinds DNA(17 bp) from -10 regions and adds ribonucleotide (G or A) in the +1 site.

After the synthesis of 6-9 nucleotides long RNA without movement ofenzyme, sigma factor falls off from holoenzyme and the core enzymeenters the elongation process


29/90


30/90


31/90

During ElongationRNA polymeraseunwinds DNA aheadof it, transcribe theregion and rewindsthe DNA at the back

and RNA comes outof the complex.

Transcription occurs

in the TranscriptionBubble at the rate of50 nt/sec.

Elongation continues

till Core enzymesreaches theterminator sequences.


32/90


33/90

When a sequence of DNA is transcribed whose nacent RNA transcriptcontains a series of U residues at the 3 end proceedded by a GC rich selfcomplementary sequences, the complementory sequences base pair with oneanother, forming a stem loop structure.

This stem loop structures interacts with the surface of RNA polymerase

causing it to pause. During this time the rU-dA base pairs at the 3 end ofRNA chain ( which are extremely unstable) melt releasing the RNA from thetranscription complex to terminate transcription

Rho independent transcription termination



34/90



35/90

1) Rho binds a

stretch of GC rich72 nt longsequence ofnacent RNAupstream of the

terminator.

2) Rho acts ashexamer, afterbinding to RNAbreaks ATP by itsATPase activityand with thisenergy movesthrough RNA to

catch DNA-RNAhybrid andUnwinds by itshellicase activityand terminates

transcription.


36/90


37/90

R l ti f t i ti i k t


38/90

Regulation of transcription in prokaryotes

Gene regulation has been well studied in E. coli. Although there are lotof genes are present but they are not all expressed all the time. it isdetermined by the growth status of the cell, metabolic condition etc.

As an example, When a bacterial cell encounters a potential foodsource it will manufacture the enzymes necessary to metabolize thatfood.

In 1959 Jacques Monod and Fracois Jacob looked at the ability of E. coli

cells to digest the sugar lactose

In the presence of the sugar lactose, E. colimakes an enzyme calledbeta galactosidaseto breaks down the sugar lactose so the E. colican

digest it for food but not in the absence of lactose

It is the LAC Z gene in E colithat codes for the enzyme betagalactosidase and this geneis present in lac operon ( cluster of genes

transcribed by same promoter as polycistronic mRNA)


39/90

La ct o se o p e r o n : a r e gu la t o r y ge n e a n d

3 s tu ct u r a l ge n e s , a n d 2 co n t r o l e le m e n t s

lacI

Regulatory gene

lacZ lacY lacA DNA

m-RNA

-Galactosidase

Permease

Transacetylase

Protein

Structural GenesCis-actingelements

PlacI Plac Olac

The LAC operon


40/90

1. When lactose is absent A repressor protein is continuously synthesised. It sits on

a sequence of DNA just in front of the lacoperon, theOperator site

The repressor protein blocks the Promoter site wherethe RNA polymerase settles before it starts transcribing

Regulator

genelac operon

Operator

site

z y aDNA

I O

Repressor

protein

RNA

polymeraseBlocked

2007 Paul Billiet ODWS


41/90

2. When lactose is present A small amount of a sugar allolactose is formed within

the bacterial cell. This fits onto the repressor protein at

another active site (allosteric site) This causes the repressor protein to change its shape (a

conformational change). It can no longer sit on theoperator site. RNA polymerase can now reach itspromoter site

Promotor site

z y aDNA

I O

2007 Paul Billiet ODWS


42/90

Eukaryotic Transcription is Horribly Complicated

Three different polymerases:

RNA polymerase I: synthesizes rRNA in the nucleolus.RNA polymerase II: synthesizes mRNA in the nucleoplasm.RNA polymerase III: synthesizes tRNA, 5S rRNA, small RNAs in the nucleoplasm

All eukaryotic RNA polymerases have 12-16 subunits (aggregates of >500 kD).

Some subunits are common to all three RNA polymerases such as TBP.

Multiple promoter types :TATA Box, Initiator elements, CpG island forpol I), core elements, upstream core elements (

pol I), A box, B Box, C Box for pol III)

Each RNA polymerase recognizes its own promoter

Many proteins (transcription factor) are involved in promoter recognitionby RNA Polymerase

Eukaryotic Transcription


43/90

6

E u k a r y o t i c G e n e S t r u c t u r e5 - P r o m o t e r E x o n 1 I n t r o n 1 E x o n 2 T e r m i n a t o r 3

U T R s p l i c e s p l i c e U T R

t r a n s c r i p t i o n

t r a n s l a t i o n

P o l y A

p r o t e i n

Eukaryote Promoter (Pol II)

Transcription by Polymerase II


44/90

p y y

Three Steps:

Intiation:

Binding of transcription factors and Pol II to promoter,

DNA strand separation and beginning of RNA synthesis.

Elongation:

Continuous Process of RNA synthesis by RNA pol II.

Termination:

Ending of transcription after transcribing a polyA signalsequence.


45/90


46/90

Post transcriptional modification of Pre mRNA


47/90

In eukaryotes, the primary transcript (pre mRNA) must bemodified by:

addition of a 5 cap

addition of a 3 poly-A tail

removal of non-coding sequences (introns- non codingsequence) and joining of coding sequences(Exons) by

splicing through the formation of spliceoome ( with the helpof snRNPs)

Post transcriptional modification of Pre-mRNA

Capping at 5 end of mRNA


48/90

Capping at 5 end of mRNA


49/90

Splicing mechanism


50/90

Splicing mechanism

U-rich small nuclearRNA in association

of proteins calledsnRNP formsspliceosome andhelp in the splicing

process.


51/90


52/90

DNA

Cytoplasm

Nucleus

Eukaryotic Transcription

ExportG

AAAAAA

RNATranscription

GAAAAAA

RNAProcessing

mRNA

Eukaryotic gene regulation


53/90

Eukaryotic gene regulation


54/90

http://www-

class.unl.edu/biochem/gp2/m_biology/ani

mation/gene/gene_a2.html

Translation is the process of decoding a mRNA


55/90

Translation is the process of decoding a mRNA

molecule into a polypeptide chain or protein


56/90

Transcription and translation ineukaryotic cells are separated inspace and time.

Extensive processing of primaryRNA transcripts in eukaryotic cells.

Translation


57/90

Translation or protein synthesis requires theparticipation of multiple types of RNA:

messenger RNA (mRNA) carries the informationfrom DNA that encodes proteins

ribosomal RNA (rRNA) is a structural

component of the ribosometransfer RNA (tRNA) carries amino acids to the

ribosome for translation

The Genetic Code


58/90

The genetic code is the way in which the nucleotide sequence in nucleicacids specifies the amino acid sequence in proteins.

A codon is a set of 3 nucleotides that specifies a particular amino acid.

Therefore, mRNA carries information from DNA in a three letter geneticcode.

A three-letter code is used because there are 20 different amino acids

that are used to make proteins.

If a two-letter code were used there would not be enough codons toselect all 20 amino acids.

That is, there are 4 bases in RNA, so 42 (4x 4)=16; where as 43(4x4x4)=64.


59/90

GENETIC CODE


60/90

There is a total of 64 codons with mRNA, 61 specify aparticular amino acid.

The remaining three codons (UAA, UAG, & UGA) are stop

codons, which signify the end of a polypeptide chain(protein).

This means there are more than one codon for each of the 20amino acids.

Besides selecting the amino acid methionine, the codon AUG

also serves as the initiator codon, which starts thesynthesis of a protein

Deciphering the Genetic code

Marshall Nirenberg Khorana and their collegous deciphered the genetic code by


61/90

Marshall Nirenberg, Khorana and their collegous deciphered the genetic code by

adding homopolymers such as UUU, AAA, CCC, or co-polymers such as ACA,CAA, AAC of synthetic nucleotide triplets to cell extracts containing 20 aminoacyl tRNA which are capable of limited translation

In each extarct one amino acid is radioactively labelled and rest 19 are unlabelledand the reaction mixture are passed through filter. Since ribosomes binds to filter,

if the added trinucleotide caused the labelled aminoacyl trNA to attach to theribosome, then radioactivity would be detected on the filter ( positive test)otherwise label will pass thrrough-a negative test.

mRNA contains codons which code for amino acids.


62/90

tRNA - Transfer RNA.

Each tRNA molecule folds as cloverleaf structure and has 2 important


63/90

Each tRNA molecule folds as cloverleaf structure and has 2 important

sites of attachment.One site, called the anticodon, binds to the codon on the mRNAmolecule.The other site attaches to a particular amino acid.During protein synthesis, the anticodon of a tRNA molecule base pairswith the appropriate mRNA codon.

tRNA Structure


64/90

Aminoacyl tRNA synthetase

There are 20 different anminoacyl tRNA synthetases, one for each amino acid.

Ribosome

A d f 2 b i l d ll


65/90

Are made up of 2 subunits, a large one and a smaller one,each subunit contains ribosomal RNA (rRNA) & proteins.

Protein synthesis starts when the two subunits bind to mRNA.

Translation has 3 Steps, Each Requiring


66/90

Different Supporting Proteins

Initiation

Requires Initiation Factors

Elongation Requires Elongation Factors

Termination

Requires Termination Factor


67/90

Initiation:

1. Binding of initiationfactors to small subunit.

2. Binding of first tRNA andmRNA to small subunit.

3. Binding of large subunit.

Elongation:


68/90

Elongation:

1. Binding of nexttRNA using EFs at

A site.

2. Peptide Bondformation between 2

amino acids.

3. Translocation ofribosome.

E P A

E P A

E P A

E P A

E P A


69/90

Overview of Prokaryotic Translation


70/90

y

Translation - Initiation


71/90

AE

Large

subunitP

Smallsubunit

Translation Initiation

fMet

UAC

GAG...CU-AUG--UUC--CUU--AGU--GGU--AGA--GCU--GUA--UGA- AT GCA...TAAAAAA5

mRNA

3

Translation - Elongation


72/90

AE

Ribosome P

Arg

Aminoacyl tRNA

PheLeu

Met

SerGly

Polypeptide

CCA

Translation Elongation


mRNA

3



73/90

AE

Ribosome P

Phe

Leu

Met

SerGly

Polypeptide

Arg

Aminoacyl tRNA

UCUCCA



mRNA

3



74/90

AE

Ribosome P

CCA

Arg

UCU

PheLeu

Met

Ser

Gly

Polypeptide



mRNA

3



75/90

AERibosome P


Aminoacyl tRNA

Ala

CCA

Arg

UCU

PheLeu

Met

Ser

Gly

Polypeptide


mRNA

3



76/90

AERibosome P


Arg

UCU

PheLeu

Met

Ser

Gly

Polypeptide

CGA

Ala


mRNA

3


77/90


78/90


79/90

http://www-class.unl.edu/biochem/gp2/m_biology/animation/gene/gene_a3.html


80/90

History of recombinant DNA


81/90

technology

Recombinant DNA technology is oneof the recent advances in

biotechnology, which was developedby two scientists named Boyer and

Cohen in 1973.

Basic steps in Recobinant DNA


82/90

1. Isolate the gene

2. Insert it in a host using a vector (plasmid)

3. Produce as many copies of the host as

possible4. Separate and purify the product of the gene

Technology


83/90

DNA recombinanttechnology

Ligase


84/90

Applications of


85/90

Recombinant DNATechnology

Large-scale production of humanproteins by geneticallyengineered bacteria.

Such as : insulin, Growthhormone, Interferons and

Blood clotting factors (VIII & IX)

Production of Human


86/90

Insulin(???)1 ) O b t a i n i n g t h e h u m a n i n s u l i n g e n eHuman insulin gene can be obtained by

making a complementary DNA (cDNA) copyof the messenger RNA (mRNA) for humaninsulin.

2 ) Jo i n i n g t h e h u m a n i n s u l i n g e n e

i l i d ( )


87/90

i n t o a p l a sm i d ( ) v e ct o r

The bacterial plasmids and the cDNA are

mixed together. The human insulin gene(cDNA) is inserted into the plasmid throughcomplementary base pairing at sticky ends.

3 ) I n t r o d u c i n g t h e r e c o m b i n a n t

DN A l id i t b t i


88/90

DN A p lasm id s in t o b act e r i aThe bacteria E.coli is used as the host cell. IfE.coli and the recombinant plasmids are mixed

together in a test-tube.

4 ) Se l ect i n g t h e b a ct er i a w h i ch

h t k t h t


89/90

h a v e t a k e n u p t h e c o r r e c tp iece o f DNAThe bacteria are spread onto nutrient agar. Theagar also contains substances such as an

antibiotic which allows growth of only thetransformed bacteria.


90/90

Thank You

transcription and translation- an a k. ghosh

Documents