Central Dogma of Molecular
Biology
“The central dogma of molecular biology deals with
the detailed residue-by-residue transfer of
sequential information. It states that such
information cannot be transferred back from
protein to either protein or nucleic acid.”
Francis Crick, 1958
… in other words
 Protein information
cannot flow back to
nucleic acids
 Fundamental
framework to
understanding the
transfer of sequence
information between
biopolymers
Presentation Outline
 PART I
 The Basics
 DNA Replication
 Transcription
 PART II
 Translation
 Protein Trafficking & Cell-cell communications
 Conclusion
The Basics: Cell Organization
Prokaryotes
Eukaryotes
The Basics: Structure of DNA
The Basics: Additional Points
 DNA => A T C G, RNA => A U C G
 Almost always read in 5' and 3' direction
 DNA and RNA are dynamic - 2° structure
 Not all DNA is found in chromosomes
 Mitochondria
 Chloroplasts
 Plasmids
 BACs and YACs
 Some extrachromosomal DNA can be useful in Synthetic
Biology
… an example of a plasmid vector
 Gene of interest
 Selective
markers
 Origin of
replication
 Restriction sites
The Basics: Gene Organization
… now to the main course
DNA Replication
 The process of copying double-stranded DNA
molecules
 Semi-conservative replication
 Origin of replication
 Replication Fork
 Proofreading mechanisms
DNA Replication: Prokaryotic
origin of replication
 1 origin of replication; 2
replication forks
DNA Replication: Enzymes
involved
 Initiator proteins (DNApol clamp loader)
 Helicases
 SSBPs (single-stranded binding proteins)
 Topoisomerase I & II
 DNApol I – repair
 DNApol II – cleans up Okazaki fragments
 DNApol III – main polymerase
 DNA primase
 DNA ligase
DNA Replication:
DNA Replication: Proofreading
mechanisms
 DNA is synthesised from dNTPs. Hydrolysis of (two) phosphate
bonds in dNTP drives this reduction in entropy.
- Nucleotide binding error rate =>c.10−4, due to extremely short-lived imino and enol tautomery.
- Lesion rate in DNA => 10-9.
Due to the fact that DNApol has built-in 3’ →5’ exonuclease activity, can chew back
mismatched pairs to a clean 3’end.
Transcription
 Process of copying DNA to RNA
 Differs from DNA synthesis in that only one strand
of DNA, the template strand, is used to make
mRNA
 Does not need a primer to start
 Can involve multiple RNA polymerases
 Divided into 3 stages
 Initiation
 Elongation
 Termination
Transcription: The final product
Transcription: Transcriptional
control
 Different promoters for different sigma factors
… Case study – Lac operon
 For control of lactose metabolism
 Consists of three structural genes, a promoter, a
terminator and an operator
 LacZ codes for a lactose cleavage enzyme
 LacY codes for ß-galactosidase permease
 LacA codes for thiogalactoside transcyclase
 When lactose is unavailable as a carbon source, the
lac operon is not transcribed
 The regulatory response requires the lactose repressor
 The lacI gene encoding repressor lies nearby the lac operon
and it is consitutively (i.e. always) expressed
 In the absence of lactose, the repressor binds very tightly to a
short DNA sequence just downstream of the promoter near the
beginning of lacZ called the lac operator
 Repressor bound to the operator interferes with binding of
RNAP to the promoter, and therefore mRNA encoding LacZ
and LacY is only made at very low levels
 In the presence of lactose, a lactose metabolite called
allolactose binds to the repressor, causing a change in its shape
 The repressor is unable to bind to the operator, allowing
RNAP to transcribe the lac genes and thereby leading to high
levels of the encoded proteins.
End of Part I
 Q & A
 Coffeebreak?!

Central Dogma of Molecular Biology powerpoint

  • 1.
    Central Dogma ofMolecular Biology “The central dogma of molecular biology deals with the detailed residue-by-residue transfer of sequential information. It states that such information cannot be transferred back from protein to either protein or nucleic acid.” Francis Crick, 1958
  • 2.
    … in otherwords  Protein information cannot flow back to nucleic acids  Fundamental framework to understanding the transfer of sequence information between biopolymers
  • 3.
    Presentation Outline  PARTI  The Basics  DNA Replication  Transcription  PART II  Translation  Protein Trafficking & Cell-cell communications  Conclusion
  • 4.
    The Basics: CellOrganization Prokaryotes Eukaryotes
  • 5.
  • 6.
    The Basics: AdditionalPoints  DNA => A T C G, RNA => A U C G  Almost always read in 5' and 3' direction  DNA and RNA are dynamic - 2° structure  Not all DNA is found in chromosomes  Mitochondria  Chloroplasts  Plasmids  BACs and YACs  Some extrachromosomal DNA can be useful in Synthetic Biology
  • 7.
    … an exampleof a plasmid vector  Gene of interest  Selective markers  Origin of replication  Restriction sites
  • 8.
    The Basics: GeneOrganization … now to the main course
  • 9.
    DNA Replication  Theprocess of copying double-stranded DNA molecules  Semi-conservative replication  Origin of replication  Replication Fork  Proofreading mechanisms
  • 10.
    DNA Replication: Prokaryotic originof replication  1 origin of replication; 2 replication forks
  • 11.
    DNA Replication: Enzymes involved Initiator proteins (DNApol clamp loader)  Helicases  SSBPs (single-stranded binding proteins)  Topoisomerase I & II  DNApol I – repair  DNApol II – cleans up Okazaki fragments  DNApol III – main polymerase  DNA primase  DNA ligase
  • 12.
  • 13.
    DNA Replication: Proofreading mechanisms DNA is synthesised from dNTPs. Hydrolysis of (two) phosphate bonds in dNTP drives this reduction in entropy. - Nucleotide binding error rate =>c.10−4, due to extremely short-lived imino and enol tautomery. - Lesion rate in DNA => 10-9. Due to the fact that DNApol has built-in 3’ →5’ exonuclease activity, can chew back mismatched pairs to a clean 3’end.
  • 14.
    Transcription  Process ofcopying DNA to RNA  Differs from DNA synthesis in that only one strand of DNA, the template strand, is used to make mRNA  Does not need a primer to start  Can involve multiple RNA polymerases  Divided into 3 stages  Initiation  Elongation  Termination
  • 19.
  • 20.
    Transcription: Transcriptional control  Differentpromoters for different sigma factors
  • 21.
    … Case study– Lac operon  For control of lactose metabolism  Consists of three structural genes, a promoter, a terminator and an operator  LacZ codes for a lactose cleavage enzyme  LacY codes for ß-galactosidase permease  LacA codes for thiogalactoside transcyclase  When lactose is unavailable as a carbon source, the lac operon is not transcribed
  • 23.
     The regulatoryresponse requires the lactose repressor  The lacI gene encoding repressor lies nearby the lac operon and it is consitutively (i.e. always) expressed  In the absence of lactose, the repressor binds very tightly to a short DNA sequence just downstream of the promoter near the beginning of lacZ called the lac operator  Repressor bound to the operator interferes with binding of RNAP to the promoter, and therefore mRNA encoding LacZ and LacY is only made at very low levels  In the presence of lactose, a lactose metabolite called allolactose binds to the repressor, causing a change in its shape  The repressor is unable to bind to the operator, allowing RNAP to transcribe the lac genes and thereby leading to high levels of the encoded proteins.
  • 24.
    End of PartI  Q & A  Coffeebreak?!