COEN 6211 LECTURE 1 2021  

COEN 6211 LECTURE 1 2021   COEN 6211 LECTURE 1 2021   Synthetic Biology Course outline Luc Varin, Ph.D (Biology) Lectures: W 14:45 – 17:30 Office Hours: Tue and Thu 13:00 – 16:00 E-mail: • Office hours: I expect students to contact me by E-mail to schedule a ZOOM session during my office hours to ask their questions related to the course material or their assignments. Course outline • Text book: This course has no text book. All the slides used in the lectures will be made available to you, on-line. Also, all of the research papers used for the latter part of the course will be available to you in PDF format. You are expected to attend every lecture, participate in it, and take your own notes. Synthetic biology Lecture 1 Course outline • Grading 1. Qualifying Exam Withdraw (0)/Pass (10) marks 2. Assignments (2) 20 marks 3. Midterm Exam 15 marks 3. Team Project 25 marks 5. Final Exam 30 marks Lecture outline Introduction to Molecular Biology • DNA structure • Gene structure • Transcription and translation • DNA replication • Gene regulation • Molecular cloning tools Flow of genetic information DNA RNA Transcription by RNA polymerase Translation by ribosomes PROTEIN Chromosome structure 7 DNA structure DNA structure DNA structure Orientation of the strands? DNA structure Nucleic acid hybridization Denaturation = dissociation of the two strands Melting = denaturation Renaturation = reassociation of the two strands Annealing = renaturation Annealing can happen between two complementary DNA strands or between DNA and RNA (hybridization) DNA structure Problem 1 5’-ACCTGCCTGACAACTG-3’ Write the sequence of the complementary strand in a 5’ to 3’ orientation DNA structure Problem 2 5’-GGATGCCT-3’ Will this oligonucleotide anneal with the DNA strand shown below? If your answer is yes, locate the position of annealing 5’-CCTACGGATCTCGGATGCCTTT-3’ DNA structure Problem 2B 5’-ATCCCAGAA-3’ Will this oligonucleotide anneal with the DNA strand shown below? If your answer is yes, locate the position of annealing 5’-CCTAGCTTCTGGGATGCCTTT-3’ DNA replication Requirements • DNA template strand • Primer with a free 3’-OH group • NTPs (ATP, TTP, CTP, GTP) • DNA polymerase G DNA synthesis CATTTTA-5’ 5’-ACTCCGCGTACGTCTAGCCTCCCGTAAAATGC-3’ DNA replication Problem 1 5’- TTCCATTCGGCGA -3’ 3’- AGC -5’ What will be the first nucleotide incorporated by DNA polymerase? Synthetic biology Lecture 1 DNA replication Problem 2 The following diagram of a generalized tetranucleotide will serve as a basis for the question below Given that the DNA strand which served as a template for the synthesis of this tetranucleotide was composed of the bases 5’- CCTG- 3’, fill in the parentheses (in the diagram) with the expected bases. Gene structure Promoter Coding sequence Terminator Gene structure Transcription start site Promoter RNA Terminator Promoter structure The template strand is used as a reference by the RNA polymerase to make the mRNA Promoter strength varies with the “quality” of the -35 and -10 sequences Transcription From: 21 Transcription Compare DNA vs RNA synthesis 22 mRNA structure Start codon (first amino acid of the protein) Stop codon (signal to end protein synthesis) 5’ 3’ Ribosome binding site to initiate translation = untranslated region Translation Translation The genetic code Which amino acid? Transcription regulation +1 -35 Box TTGTCA -10 Box TATAA RNA Core promoter = Binding site for RNA polymerase In this configuration transcription is ON The efficiency of transcription depends on the quality of the promoter Transcription: Activation mechanism +1 bad promoter A = Activator of transcription RNA The efficiency of transcription is dependent on the presence of the activator protein Repression of transcription +1 -35 box -10 box operator R = Repressor In this configuration RNA Polymerase cannot bind transcription is OFF Repression of transcription • The lactose operon of E. coli Active repressor Transcription is OFF Induction of gene expression • The lactose operon of E. coli lacI repressor -35 -10 X operator Transcription is ON Inactive repressor = inducer (lactose) Silencing by antisense RNA siRNA 5’ 3’ X Translation Binding of siRNA causes mRNA degradation Repression by mRNA cleavage 5’ 3’ Ribozymes + aptamer Self-cleaving ribozymes can be inhibited or activated by the aptamer binding Construction of a synthetic gene Modular structure • Construct a promoter • Insert an operator • Insert a ribosome binding site • Select a coding sequence (output) -35 box -10 box Output operator Molecular cloning Outline • Restriction enzymes • DNA ligation • Plasmid cloning • Polymerase chain reaction (PCR) • Restriction endonucleases Recognize and cut specific DNA sequences e.g. AsuII cuts after the first T in the sequence TTCGAA • Restriction endonucleases EcoR1 dimer bound to DNA EcoR1 cleaves the sequence GAATTC between the G and the A • DNA ligase Can link together two DNA strands that have double-strand break (a break in both complementary strands of DNA DNA ligase 5’-phospate group on at least one of the extremities Is required for ligation pUC plasmid EcoR1 EcoR1 EcoR1 1) Cut plasmid DNA and insert with EcoR1 2) De-phosphorylate the cut vector (alkaline phosphatase) 3) Mix de-phosphorylated vector and insert + DNA ligase and ATP 4) Transformation 5) Selection De-phosphorylation of cut vector will reduce the number of false positive EcoR1 BamH1 EcoR1 BamH1 1) Cut plasmid DNA and insert with EcoR1 and BamH1 2) Mix cut vector and insert + DNA ligase and ATP 3) Transformation 4) Selection Digestion of the vector with two restriction endonucleases prevents religation of the plasmid Polymerase chain reaction Technique Applications Sample problem Polymerase Chain Reaction • Objective is to produce a specific DNA sequence in-vitro • Amplification of the target DNA can be done from minute amount of starting material • DNA synthesis is catalyzed by a thermo stable DNA polymerase in presence of two primers Polymerase Chain Reaction Cycle 1 Step 1 Denaturation Incubate reaction mixture at 95 degrees Reaction mixture 5’ 3’ 3’ 5’ 5’ 3’ 3’ 5’ Cycle 1 Step 2 Annealing Incubate reaction mixture at annealing temperature (5 degrees below Tm of primers) 5’ 3’ 3’ 5’ Cycle 1 Step 3 Elongation Incubate reaction mixture at 72 degrees 5’ 3’ 3’ 5’ Cycle 2 Step 1 Denaturation Incubate reaction mixture at 95 degrees 5’ 3’ 3’ 5’ Cycle 2 Step 2 Annealing Incubate reaction mixture at annealing temperature (5 degrees below Tm of primers) 5’ 3’ 3’ 5’ Cycle 2 Step 3 Elongation Incubate reaction mixture at 72 degrees 5’ 3’ 3’ 5’ Cycle 3 Step 1 Denaturation Incubate reaction mixture at 95 degrees 5’ 3’ 3’ 5’ Cycle 3 Step 1 Denaturation Incubate reaction mixture at 95 degrees 5’ 3’ 5’ 3’ Cycle 3 Step 2 Annealing Incubate reaction mixture at annealing temperature (5 degrees below Tm of primers) 5’ 3’ 5’ 5’ Cycle 3 Step 3 Elongation Incubate reaction mixture at 72 degrees 5’ 3’ 5’ Only this product will accumulate 5’ REPEAT FOR A TOTAL OF 30 CYCLES Sample problem 1 AGCTTCTCGCCATTG CGCTCAATTGCGCTA TCGAAGAGCGGTAAC GCGAGTTAACGCGAT A) Design two primers to amplify the DNA fragment between the two arrows B) Design two primers to clone this DNA fragment in the EcoR1 site of pUC18 C) Design two primers to clone this DNA fragment in the EcoRI and HindIII sites of pUC18 EcoRI = GAATTC HindIII = AAGCTT Your supervisor is asking you to clone the coding region of gene X (see below) in the bacterial expression vector pQE30 using the polymerase chain reaction (PCR). Write the sequence of two oligonucleotides that will allow you to clone the coding sequence in the vector. The recombinant protein must be as short as possible. Note: The coding sequence must be in frame with the ATG of the vector. The start and stop codons of gene X are underlined. Coding sequence of gene X GTCGATCAAT ATGGAACATG TTTACTCCAA ACCACCGCAC ACCAATTATG GAAACCAAGC CGGAAAAGAA TTCCGGTGGA GAGCGAAAAA AAAGGATTCC GAATCGTGAA CTGCCAAAAA CATTTTGAAG CCAACGATTC CGACGTCATC CTCGCCACCC TAGCTAAATC AGGCACCACT TGGTTAAAAG CTCTTCTCTT TGCTCTCATT CACCGACACA AGTTCCCAGT TTCTGGCAAG CATCCTCTTC TGAAACAGCA GTAGCAGCGT TTAAAGGGAA GTTTATT Oligo #1 5’ Oligo #2 5’ BamH1 = GGATCC HindIII = AAGCTT RBS 6Xhis ATGAGAGGATCG GGATCCGCATGC AAGCTT RBS 6Xhis ATGAGAGGATCG ACGGATCCGCATGC AAGCTT RT-PCR Can be used for cloning • Restriction enzyme sites can be added to the cDNA of interest • Able to generate sticky ends for ligation into vector of choice • 2 sticky ends permits directional cloning Summary • Genes control cell function • Each gene has a promoter part and a coding part • The promoter can have binding sites for repressors (and or enhancers) • Artificial genes can be designed using modules