Central Dogma
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Transcript of Central Dogma
Central Dogma
DNA Replication, Recombination, and Repair
Central Dogma
DNA Replication – process of producing identical copies of original DNA
• strand separation followed by copying of each strand
• fixed by base-pairing rules
DNA replication is bidirectional. involves two replication forks that move in opposite direction
DNA replication requires unwinding of the DNA helix.
expose single-stranded templates
DNA gyrase – acts to overcome torsional stress imposed upon unwinding
helicases – catalyze unwinding of double helix- disrupts H-bonding of the two strands
SSB (single-stranded DNA-binding proteins) – binds to the unwound strands, preventing re-annealing
PrimerRNA primes the synthesis of DNA.
Primase synthesizes short RNA.
DNA replication is semidiscontinuous
DNA polymerase synthesizes the new DNA strand only in a 5’3’ direction. Dilemma: how is 5’ 3’ copied?
The leading strand copies continuously
The lagging strand copies in segments called Okazaki fragments (about 1000 nucleotides at a time) which will then be joined by DNA ligase
DNA Polymerase= enzymes that replicate DNA
All DNA Polymerases share the following:
1.Incoming base selected in the active site (base-complementarity)2.Chain growth 5’ 3’ direction (antiparallel to template)3.Cannot initiate DNA synthesis de novo (requires primer)First DNA Polymerase discovered – E.coli DNA Polymerase I (by Arthur Kornberg and colleagues)
Roger D. Kornberg2006 Nobel Prize in Chemistry
Arthur Kornberg1959 Nobel Prize in Physiology
and Medicine
http://www.nobelprize.org
3’ 5’ exonuclease activity
- removes incorrect nucleotides from the 3’-end of the growing chain (proofreader and editor)- polymerase cannot elongate an improperly base-paired terminus
proofreading mechanisms• Klenow fragment – removes
mismatched nucleotides from the 3’’ end of DNA (exonuclease activity)
• detection of incorrect base- incorrect pairing with the template
(weak H-bonding)- unable to interact with the minor
groove (enzyme stalls)
DNA Ligase = seals the nicks between Okazaki fragments
DNA ligase seals breaks in the double stranded DNA
DNA ligases use an energy source (ATP in eukaryotes and archaea, NAD+ in bacteria) to form a phosphodiester bond between the 3’ hydroxyl group at the end of one DNA chain and 5’-phosphate group at the end of the other.
Eukaryotic DNA Replication Like E. coli, but more complex
Human cell: 6 billion base pairs of DNA to copy
Multiple origins of replication: 1 per 3000-30000 base pairs
E.coli 1 chromosomeHuman 23E.coli circular chromosome; Human linear
DNA Recombination =
recombinases Holliday junction – crosslike structure
natural process of genetic rearrangement
Mutations1. Substitution of base pair
a. transitionb. transversion
2. Deletion of base pair/s
3. Insertion/Addition of base pair/s
DNA replication error rate: 3 bp during copying of 6 billion bp
Macrolesions: Mutations involving changes in large portions of the genome
Agents of Mutations1. Physical Agents
a) UV Lightb) Ionizing Radiation
2. Chemical AgentsSome chemical agents can be
classified further intoa) Alkylatingb) Intercalatingc) Deaminating
3. Viral
UV Light Causes Pyrimidine Dimerization
Replication and gene expression are blocked
Chemical mutagens
• 5-bromouracil and 2-aminopurine can be incorporated into DNA
Deaminating agentsEx: Nitrous acid (HNO2)Converts adenine to hypoxanthine, cytosine to uracil, and
guanine to xanthineCauses A-T to G-C transitions
Alkylating agents
Intercalating agents
AcridinesIntercalate in DNA, leading to insertion or
deletionThe reading frame during translation is changed
DNA Repair
Direct repairPhotolyase cleave pyrimidine dimers
Base excision repairE. coli enzyme AlkA removes modified bases
such as 3-methyladenine (glycosylase activity is present)
Nucleotide excision repairExcision of pyrimidine dimers (need different
enzymes for detection, excision, and repair synthesis)
QUIZ1. Draw the structure of any nitrogenous base of your
picking. (1 pt)
2. What is the difference between the glycosidic bond and the phosphodiester bond? (2 pts)
3. Give the reason why DNA utilizes the deoxyribose while RNA uses the ribose. (2 pts)
4. Enumerate all the enzymes and proteins involved in DNA replication and briefly state their importance/function. A short concise answer will suffice. (4 pts)
5. Give the partner strand of this piece of DNA:5-ACTCATGATTAGCAG-3 (1 pt)
Central DogmaRNA Transcription
Process of Transcription has four stages:
1. Binding of RNA polymerase at promoter sites2. Initiation of polymerization3. Chain elongation4. Chain termination
Transcription (RNA Synthesis)
RNA PolymerasesTemplate (DNA)Activated precursors (NTP)Divalent metal ion (Mg2+ or Mn2+)
Mechanism is similar to DNA Synthesis
Reece R. Analysis of Genes and Genomes.2004. p47.
Limitations of RNAP II:1. It can’t recognize its target promoter and gene.
(BLIND)2. It is unable to regulate mRNA production in
response to developmental and environmental signals. (INSENSITIVE)
Start of TranscriptionPromoter Sites
Where RNA Polymerase can indirectly bind
TATA box – a DNA sequence (5’—TATAA—3’) found in the promoter region of most eukaryotic genes.
Abeles F, et al. Biochemistry. 1992. p391.
Preinitiation Complex (PIC)
Transcription Factors (TF):
Hampsey M. Molecular Genetics of RNAP. Microbiology and Molecular Biology Reviews. 1998. p7.
TFIID binds to TATA; promotes TFIIB binding
TFIIA stabilizes TBP binding
TFIIB promotes TFIIF-pol II binding
TFIIF targets pol II to promoter
TFIIE stimulates TFIIH kinase and ATPase actiivities
TFII H helicase, ATPase, CTD kinase activities
Termination of Transcription
Terminator SequenceEncodes the
termination signalIn E. coli – base
paired hair pin (rich in GC) followed by UUU…
1. Intrinsic termination = termination sites
causes the RNAP to pause
causes the RNA strand to detach from the DNA template
Termination of Transcription
2. Rho termination = Rho protein, ρ
prokaryotes: transcription and translation happen in cytoplasm
eukaryotes: transcription (nucleus); translation (ribosome in cytoplasm)
In eukaryotes, mRNA is modified after transcriptionCapping, methylationPoly-(A) tailsplicing
capping: guanylyl residue
capping and methylation ensure stability of the mRNA template; resistance to exonuclease activity
Eukaryotic genes are split genes: coding regions (exons) and noncoding regions (introns)
Introns & Exons
IntronsIntervening
sequencesExons
Expressed sequences
Splicing
Spliceosome: multicomponent complex of small nuclear ribonucleoproteins (snRNPs)
splicing occurs in the spliceosome!
Central DogmaTranslation: Protein Synthesis
TranslationStarring three types of RNA
1.mRNA
2.tRNA
3.rRNA
Properties of mRNA1. In translation, mRNA is read in groups of bases called “codons”
2. One codon is made up of 3 nucleotides from 5’ to 3’ of mRNA
3. There are 64 possible codons
4. Each codon stands for a specific amino acid, corresponding to the genetic code
5. However, one amino acid has many possible codons. This property is termed degeneracy
6. 3 of the 64 codons are terminator codons, which signal the end of translation
Genetic Code
3 nucleotides (codon) encode an amino acid
The code is nonoverlappingThe code has no punctuation
Synonyms
Different codons, same amino acidMost differ by the last base
XYC & XYU XYG & XYA
Minimizes the deleterious effect of mutation
Encoded sequences. (a) Write the sequence of the mRNA molecule
synthesized from a DNA template strand having the sequence
(b) What amino acid sequence is encoded by the following base sequence of an mRNA molecule? Assume that the reading frame starts at the 5 end.
Practice
A
Answers
(a) 5’ -UAACGGUACGAU-3’ .(b) Met-Pro-Ser-Asp-Trp-Met.
tRNA as Adaptor Molecules
Amino acid attachment site
Template recognition siteAnticodon
Recognizes codon in mRNA
tRNA as Adaptor Molecules
Mechanics of Protein Synthesis All protein synthesis involves three
phases: initiation, elongation, termination Initiation involves binding of mRNA and
initiator aminoacyl-tRNA to small subunit(30S), followed by binding of large subunit (50S) of the ribosome
Elongation: synthesis of all peptide bonds - with tRNAs bound to acceptor (A) and peptidyl (P) sites.
Termination occurs when "stop codon" reached
Translation: InitiationTranslation occurs in the ribosomeProkaryote START
fMet (formylmethionine) bound to initiator tRNA
Recognizes AUG and sometimes GUG (but they also code for Met and Val respectively)
AUG (or GUG) only part of the initiation signal; preceded by a purine-rich sequence
Translation: Initiation
Eukaryote START
AUG nearest the 5’ end is usually the start signal
Elongation
Termination
Stop signals (UAA, UGA, UAG):• recognized by release factors (RFs)• hydrolysis of ester bond between polypeptide and
tRNA
Reference:
Garrett, R. and C. Grisham. Biochemistry. 3rd edition. 2005.
Berg, JM, Tymoczko, JL and L. Stryer. Biochemistry. 5th edition. 2002.