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Roque, Aryana Rose B. and Schuller, Kris Jane Marie

Roque, Aryana Rose B. and Schuller, Kris Jane Marie

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  • 2. CENTRAL DOGMA OF MOLECULARBIOLOGY“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
  • 3. … IN OTHER WORDS  Protein information cannot flow back to nucleic acids  Fundamental framework to understanding the transfer of sequence information between biopolymers
  • 4. THE BASICS: CELL ORGANIZATION ProkaryotesEukaryotes
  • 6. 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
  • 7. … AN EXAMPLE OF A PLASMIDVECTOR  Gene of interest  Selective markers  Origin of replication  Restriction sites
  • 8. THE BASICS: GENE ORGANIZATION … now to the main course
  • 9. DNA REPLICATION The process of copying double-stranded DNA molecules Semi-conservative replication  Origin of replication  Replication Fork Proofreading mechanisms
  • 10. DNA REPLICATION: PROKARYOTICORIGIN OF REPLICATION  1 origin of replication; 2 replication forks
  • 11. DNA REPLICATION: ENZYMESINVOLVED 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
  • 13. DNA REPLICATION: PROOFREADINGMECHANISMS 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 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
  • 16. TRANSCRIPTION: TRANSCRIPTIONAL CONTROL Different promoters for different sigma factors
  • 17. 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.
  • 18. PART IITranslationDefinition: Translation is the process in which the genetic information on a mRNA molecule is made use of to make proteins.
  • 19.  During Translation, a ribosome will attach itself onto the strand of mRNA molecule waiting to be translated. It will cover a single triplet code at a time. The Ribosome has sockets where tRNA molecules can be inserted. The tRNA molecules are linked to a specific amino acids at one one end, and has 3 bases at the other end. The tRNA molecule whose bases are able to pair with the triplet code on mRNA can enter the socket, and release its amino acid before leaving the socket. The ribosome will move on to the next triplet, and another tRNA will be able to enter the socket. The process repeats itself until the end of the mRNA molecule. The amino acids that are released by the tRNA will join together to form a linear chain. The sequence of amino acids is determined by the sequence of triplets on the mRNA molecule.
  • 21. STEPS IN TRANSLATION1. Initiation The small subunit of the ribosome binds to a site "upstream" (on the 5 side) of the start of the message. It proceeds downstream (5 -> 3) until it encounters the start codon AUG. (The region between the mRNA cap and the AUG is known as the 5-untranslated region [5-UTR].) Here it is joined by the large subunit and a special initiator tRNA. The initiator tRNA binds to the P site (shown in pink) on the ribosome. In eukaryotes, initiator tRNA carries methaionine.
  • 22. 2. Elongation An aminoacyl-tRNA (a tRNA covalently bound to its amino acid) able to base pair with the next codon on the mRNA arrives at the A site (green) associated with:  an elongation factor   GTP (the source of the needed energy) The preceding amino acid is covalently linked to the incoming amino acid with a peptidebond (shown in red). The initiator tRNA is released from the P site. The ribosome moves one codon downstream. This shifts the more recently-arrived tRNA, with its attached peptide, to the P site and opens the A site for the arrival of a new aminoacyl-tRNA. This last step is promoted by another protein elongation factor  and the energy of another molecule of GTP. Note: the initiator tRNA is the only member of the tRNA family that can bind directly to the P site. The P site is so-named because, with the exception of initiator tRNA, it binds only to a peptidyl-tRNA molecule; that is, a tRNA with the growing peptide attached. The A site is so-named because it binds only to the incoming aminoacyl- tRNA; that is the tRNA bringing the next amino acid. So, for example, the tRNA that brings Met into the interior of the polypeptide can bind only to the A site.
  • 23.  3. Termination The end of translation occurs when the ribosome reaches one or more STOP codons (UAA, UAG, UGA). (The nucleotides from this point to the poly(A) tail make up the 3- untranslated region [3-UTR] of the mRNA.) There are no tRNA molecules with anticodons for STOP codons. However, protein release factors recognize these codons when they arrive at the A site. Binding of these proteins —along with a molecule of GTP — releases the polypeptide from the ribosome. The ribosome splits into its subunits, which can later be reassembled for another round of protein