The Central Dogma (Francis Crick, 1958)

1. The Central Dogma(Francis Crick, 1958) (Transcription) (Translation) DNA  RNA  Protein (Gene/Genotype)(Phenotype) An informational process between the genetic material (genotype) and the protein (phenotype)

2. Classes of RNA for Transcription and Translation • Informational RNA (intermediate in the process of decoding genes into polypeptides) • Messenger RNA (mRNA) • Functional RNAs (never translated into proteins, serve other roles) • Transfer RNAs (tRNA) • Transport amino acids to mRNA and new protein • Ribosomal RNAs (rRNA) • Combine with an array of proteins to form ribosomes; platform for protein synthesis • Small nuclear RNAs (snRNA) • Take part in the splicing of primary transcripts in eukaryotes • Small cytoplasmic RNAs (scRNA) • Direct protein traffic in eukaryotic cells • Micro RNAs (miRNA) • Inhibits translation and induces degradation of complementary mRNA

3. RNA nucleotide sequences are complementary to DNA molecules New RNA is synthesized 5’ to 3’ and antiparallel to the template DNA template

4. Only one strand of the DNA acts as a template for transcription The template strand can be different for differentgenes But…. For each gene only one strand of DNA serve as a template

5. Single RNA polymerase (Prokaryotes) • Core enzymeHoloenzyme • 2 ,1 and 1 ’ subunits 2 , 1 , 1 ’ subunits plus • σ subunit • Polymerizes RNAFinds initiation sites

6. - 35 bases from initiation of transcription • Recognized by RNA polymerase • - 10 bases from initiation of transcription • Unwinding of DNA double helix begins here

7. Termination RNA polymerase recognizes signals (sequence) for chain termination Releases the RNA and enzyme from the template

9. Simultaneous transcription and translation in prokaryotes Green arrow = E. coli DNA Red arrow = mRNA combined with ribosomes

11. Splicing removes the introns and brings together the coding regions

13. The Central Dogma(Francis Crick, 1958) (Transcription) (Translation) DNA  RNA  Protein (Gene)(Phenotype) An informational process between the genetic material (genotype) and the protein (phenotype

14. Translation (protein synthesis) Peptidyl site: peptidyltransferase attaches amino acid to chain Aminoacyl site: new amino acid brought in Ribosome moves in this direction

15. Cells have adapter molecules called tRNA with a three nucleotide sequence on one end (anticodon) that is complementary to a codon of the genetic code. • There are different transfer RNAs (tRNAs) with anticodons that are complementary to the codons for each of the twenty amino acids. • Each tRNA interacts with an enzyme (aminoacyl-tRNA synthetase) that specifically attaches the amino acid that corresponds to its anticodon. • For example, the tRNA to the right with the anticodon AAG is complementary to the UUC codon in the genetic code (mRNA). That tRNA would carry the amino acid phenylalanine (see genetic code table) and only phenylalanine to the site of protein synthesis. • When a tRNA has its specific amino acid attached it is said to be “charged.”

17. Homopolymer was then added to a test tube containing cell-free translation system, 1 radioactively labeled amino acid and 19 unlabeled amino acids Proteins were isolated and checked for radioactivity Procedure was repeated in 20 tubes, with each tube containing a different radioactively labeled amino acid Only one tube contained radioactively labeled protein; the amino acid that was labeled (phenylalanine) is therefore specified by UUU

18. Next synthesized heteropolymers The artificial RNA sequence would depend upon the ratio of the two or more NDPS added ADP and CDP in a 1 to 5 ratio 1/6 probability of incorporating an A being incorporated 5/6 probability of incorporating a C being incorporated The resulting RNA molecule would be a collection of different codons that are made-up of A and C The numbers of different codons in the RNA molecule is a matter of probability Genetic Code

19. Genetic Code ADP and CDP added in a 1 to 5 ratio AND if codon is a triplet

20. Genetic Code The poly (AC) RNAs produced proteins containing 6 amino acids 2C:1A (11.6%) 1A : 2C (11.6%) 2A : 1C (2.3%) 1A : 2C (11.6%)

21. Using the table below, can you translate this nucleotide sequence? 5’UUCGAUGCCCGGGGUCCUGAAAUUGUUCUAGA 3’ • The first step is to look for the AUG start codon. • Next, group the nucleotides into a reading frame of 3 nucleotides per codons and use the table to find the amino acid that corresponds to each codon. • Stop translating the mRNA when you reach a stop codon. • Is this what you got? Met-Pro-Gly-Val-Leu-Lys-Leu-Phe-Stop

23. Pathways Gene A Gene B Enzyme A Enzyme B Substrate ------- intermediate ------ product Most often the final product of the biochemical pathway is something essential to life, like amino acids, nucleotides, etc.

24. Pathways Gene A Mutant Gene B Enzyme A No Enzyme B Substrate ------- intermediate No Product Can use mutants to work out pathways, and identify which gene catalyzes which step

25.    substrate + = growth - = no growth No. 1 No. 2 No. 3 ornithine citrulline arginine