CEN 551: Biochemical Engineering Instructor: Dr. Christine Kelly
Transcript of CEN 551: Biochemical Engineering Instructor: Dr. Christine Kelly
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CEN 551: Biochemical Engineering
Instructor: Dr. Christine Kelly
Chapter 8
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Schedule• Exam 2. Thursday, March 4, before spring
break.
• Exam 2: Take home on chapter 8 material, in-class on Chapter 9, 10 and 11.
• Take home exam - Genetic engineering.
• In-class - operation of bioreactors, scale-up and control of bioreactors, recovery and purification.
• 3 weeks from Thursday.
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Homework• By Friday at 5:00 pm email
[email protected] a 1 paragraph description of your project topic.
• Chapter 8– Problems 2, 6, 7, and 8.
– Due Thursday, February 19.
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Chapter 8: How cellular information is altered
• Mutation and Selection
• Natural Mechanisms for Gene Transfer and Rearrangement
• Genetically Engineering Cells
• Genomics
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We can alter cells by using mutation or genetic engineering. Mutation is subjecting the cells to stress causing changes in the genetic make-up. Genetic engineering is the purposeful transfer of DNA from one type of organism to another.
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Mutations and Selection• Mutations = mistakes in the genetic code (can
arise from replication and/or damage)• Mutant = organism with a genetic mutation• Wild type = the organism without the genetic
organism• Genotype = genetic construction of an
organism• Phenotype = characteristics expressed by an
organisms.• Expression = usually refers to
transcription+translation+posttranslation processing
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Examples
• Strain A has the tol operon for toluene degradation, and is in a reactor growing on glucose.
• Strain B has the tol operon for toluene degradation, and is in a reactor growing on toluene.
• These strains have the same genotype, but different phenotypes.
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Point mutation: single base changeConsequences – base change may or may not result
in an amino acid change.
If the amino acid is the same as before the mutation there is no consequence.
If the amino acid is different, but not in the region of the active site, there may be no consequences.
If the mutation is in the active site, there may be some enzyme activity consequence.
If the mutation changes the amino acid to a stop codon, the resulting protein will be truncated and probably not active.
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Selection
Selectable mutation: confers upon the mutant an advantage for growth, survival or detection under a set of environmental conditions that the wild type does not have.
ExamplesAntibiotic resistance
Ability to grow on tolueneInability to produce lysine
Ability to produce bioluminescenceAbility to produce more of an enzyme
Inability to grow at higher temperatures
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Serial Dilution Plating
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Natural Mutation Rates
• 10-3-10-9 mutations per cell conversion• 10-6 = 1 mutation/1,000,000 divisions• How do we increase mutation rates?• Why do we want to increase mutation
rates?
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Increase Mutation Rates
Mutagens: chemicals, radiation
Lots of growth (i.e. lots of divisions)
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Why do we want to increase mutations?
We want a cell to develop specific characteristics that are advantages for us.
For example, removing feed back inhibition of lysine to increase lysine production
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Natural Gene Transfer/Rearrangement
1. Transformation: uptake of free DNA by a cell. The cell membrane has to be permeable to DNA.
2. Transduction: DNA is carried into the call in a phage.
3. Conjugation: Cell to cell transfer of DNA. Also called mating.Once the DNA is inside the cell it can remain separate from the chromosome in self replicating plasmid, or integrate into the chromosome. To integrate, the DNA must be complementary to the chromosomal DNA on the ends.
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Mutation and Selection
Using mutation and selection engineers and microbiologists were able to increase penicillin from 0.001 g/L to 50 g/L.
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Genetic Engineering
Using natural mechanisms to purposefully manipulate DNA. The DNA is manipulated outside of the cell, and then sent into the cell.
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Genetic Engineering Tools• Restriction enzymes: enzymes that cut DNA
at specific sequences. Different enzymes will cut at different sequences.
• Gel electrophoresis (Southern Blot): A method to detect what sizes of DNA a sample contains.
• Polymerase chain reaction (PCR): A process used to make many copies of a piece of DNA.
• Plasmid: self replicating, circular piece of DNA that can survive in a cell.
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http://www.accessexcellence.org/AB/GG/nucleic.html
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Enzyme Organism from which derived Target sequence
(cut at *)5' -->3' Ava I Anabaena variabilis C* C/T C G A/G G
Bam HI Bacillus amyloliquefaciens G* G A T C C Bgl II Bacillus globigii A* G A T C T Eco RI Escherichia coli RY 13 G* A A T T C Eco RII Escherichia coli R245 * C C A/T G G Hae III Haemophilus aegyptius G G * C C Hha I Haemophilus haemolyticus G C G * C Hind III Haemophilus inflenzae Rd A* A G C T T Hpa I Haemophilus parainflenzae G T T * A A C Kpn I Klebsiella pneumoniae G G T A C * C Mbo I Moraxella bovis *G A T C Mbo I Moraxella bovis *G A T C Pst I Providencia stuartii C T G C A * G Sma I Serratia marcescens C C C * G G G SstI Streptomyces stanford G A G C T * C Sal I Streptomyces albus G G * T C G A C Taq I Thermophilus aquaticus T * C G A Xma I Xanthamonas malvacearum C * C C G G G
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Gel Electrophoresis
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Polymerase Chain Reaction
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PCR
Primers
DNA polymerase
nucleotides
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Polymerase Chain Reaction (PCR)
PCR allows scientists to extract and analyze bits of microbial DNA from samples, meaning they don’t need to find and grow whole cells. PCR is an essential element in DNA fingerprinting and in the sequencing of genes and entire genomes. Basically, it’s like a technique to photocopy pieces of DNA. In a matter of a few hours, a single DNA sequence can be amplified to millions of copies. PCR lets scientists work with samples containing even very small starting amounts of DNA.
http://www.microbeworld.org/htm/aboutmicro/tools/genetic.htm
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The technique makes use of the DNA repair enzyme polymerase. This enzyme, present in all living things, fixes breaks or mismatched nucleotides in the double-stranded DNA helix. These breaks or mismatches could cause genes to malfunction if left unfixed.
http://www.microbeworld.org/htm/aboutmicro/tools/genetic.htm
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Polymerase uses the intact half of the DNA molecule as a template and attaches the right nucleotides, which circulate constantly in the cell, to the complementary nucleotide at the site of the break. (DNA consists of two strands of nucleotide bases, which are represented as A, G, C, and T. In the laws of DNA base-pairing, A joins with T and G with C.)
http://www.microbeworld.org/htm/aboutmicro/tools/genetic.htm
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Not all polymerases are created equal, however. Many fall apart in high heat. PCR was developed in 1985 following the discovery of an unusual heat-loving bacterium called Thermus aquaticus in a hot spring in Yellowstone National Park. This bacterium’s polymerase, dubbed Taq, does its job of matching and attaching nucleotides even in the high heat generated by the successive “photocopying” cycles required during PCR. Taq made PCR possible.
http://www.microbeworld.org/htm/aboutmicro/tools/genetic.htm
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http://www.microbeworld.org/htm/aboutmicro/tools/genetic.htm
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Plasmids and Cloning
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Movies• 65601 missense mutation• 65701 nonsense mutation• 95301 bacterial transformation• 153301 mutation• 151401 virulence transformation• 156401 heat DNA• 92201 restriction enzyme, recombination• 112601 PCR• 165401 Sequencing
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Example
• I have two organisms: 1. A fast growing yeast that grows well in a fermentor. 2. A fungi that is difficulty to grow.
• The fungi produces an enzyme that may be valuable, but I cannot grow enough fungi to produce enough enzyme to even test the enzyme.
• How can I use the genetic engineering tools to get enough enzyme?
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Genetic Engineering1. PCR the enzyme DNA from the fungi, get a bunch of
the DNA that encodes for the valuable enzyme.2. Find a restriction enzyme that will cut the valuable
enzyme DNA on ether side (but not in the middle).3. Obtain a plasmid that will replicate in the yeast, that
has a site that the same restriction enzyme will cut downstream of a strong promoter.
4. Cut the valuable enzyme DNA and the plasmid with the restriction enzyme.
5. But the valuable enzyme DNA and plasmid together and let them recombine.
6. Get the plasmid into the yeast.7. At all steps, use gel electrophoresis to check and make
sure you have the right DNA.
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White-Rot Fungi• Fungi with mycelium type growth.• Able to degrade lignocellulosic materials using
several enzyme systems (lignin and manganese peroxidases, laccases).
• Expresses and secretes MnP under nitrogen limitation at low concentrations.
• Expresses several degradative enzymes – has been widely studied for bioremediation applications.
• Not suitable for conventional industrial fermentations.
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White-rot Fungi
P. chrysosporium
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Manganese Peroxidase• Glycosylated enzyme that uses H2O2 to
oxidize manganese, which in turn oxidizes lignin.
• White-rot fungi produces a 41-47 kDa MnP under secondary metabolism.
• Native fungal secretion signal directs secretion out of the cell.
• Requires a heme cofactor for ligninolytic activity.
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MnP crystal structure.
(Sundaramoorthy et al., 1994).
glycosylation
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Pichia pastoris• Methylotrophic (methanol as a sole carbon source)
yeast.
• Capable of eukaryotic post translational modifications.
• Higher yields, less expensive, higher cell density, and easier to scale up than mammalian and fungal systems.
• Secretes only small amounts of native proteins.
• Many cloning and expression vectors available.
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Efforts to Increase Production of MnPHomologous expression
• P. chrysosporium primary metabolism: low concentration
Heterologous expression
• Bacteria (E. coli): inactive inclusion bodies
• Insect cells: active enzyme, low concentration (5 mg/L), heme addition, expensive
• Fungal (Aspergillus spp.): active enzyme, higher concentration (100 mg/L), heme addition
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Cloning of mnp1. The white-rot fungi was grown under
nitrogen limitation.2. The total RNA was extracted from the
culture.3. Reverse transcriptase polymerase chain
reaction (RT-PCR) was performed with oligo dT primers to create DNA complementary to the mRNA.
4. PCR was performed with primers specific for the MnP gene.
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tubes
Thermocycler (PCR machine)
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4. PCR was perfomed again with restriction enzyme sites built into the primer sequence. So the ends of the MnP gene would have the correct restriction sites.
5. The PCR product was cut with restriction enzymes.
6. E. coli with the pGAP vector was grown, and then the pGAP vector was isolated from the culture. The pGAP vector was cut with the same restriction enzymes.
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6. The PCR product (a tube full of MnP gene DNA) was mixed with the pGAP vector and the enzyme ligase. The sticky ends of the MnP gene were complementary to the sticky ends of the pGAP vector, and they hybridized. The ligase enzyme will then ligated the nucleotides together. The result was a tube with some pGAP vector (dimers) with no MnP insert, some MnP gene DNA (dimers), and some pGAP vector with an MnP insert.
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7. This mixture of DNA was placed in a cuvette with live P. pastoris cells, and an electric current passed through the solution.
cuvette
electroporator
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8. The electric current caused the membrane to be permeable to DNA. The DNA in the solution went into the P. pastoris cells.
9. Samples from the tube of P. pastoris cells were plated to selective medium with the antibiotic zeocin. The pGAP vector has antibiotic resistance genes. The cells that grew into visible colonies had the pGAP vector insert. These colonies were grown, and tested to check if the cells produced MnP using an enzyme activity method.
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Pichia pGAP vector
constitutive
promoter
affinityantigen
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mnp1 cDNA Sequencing Result(fungal secretion signal, 2 mutations)
TCAGCTCTCAAGGACATCCGCACTCGAATATCGCAATGGCCTTCGGTTCTCTCCTCGCCTTCGTGGCTCTCGCCGCCATAACTCGCGCCGCCCCGACTGCGGAGTCTGCAGTCTGTCCAGACGGTACCCGCGTCACCAACGCGGCGTGCTGCGCTTTCATTCCGCTCGCACAGGATTTGCAAGAGACTCTGTTCCAGGGTGACTGTGGCGAAGATGCCCACGAAGTCATCCGTCTGACCTTCCACGACGCTATTGCAATCTCCCAGAGCCTAGGTCCTCAGGCTGGCGGCGGTGCTGACGGCTCCATGCTGCACTTCCCGACAATCGAGCCCAACTTCTCCGCCAACAGCGGCATCGATGACTCCGTCAACAACTTGCTTCCCTTCATGCAGAAACACGACACCATCAGTGCCGCCGATCTTGTACAGTTCGCCGGTGCGGTCGCGCTGAGCAACTGCCCAGGTGCTCCTCGCCTCGAGTTCATGGCTGGACGTCCGAACACTACCATCCCCGCAGTTGAGGGCCTCATTCCTGAGCCTCAAGACAGCGTCACCAAAATCCTGCAGCGCTTCGAGGACGCCGGCAACTTCTCGCCGTTCGAGGTCGTCTCGCTCCTGGCTTCACACACCGTTGCTCGTGCGGACAAGGTCGACGAGACCATCGATGCTGCGCCCTTCGACTCGACACCCTTCACCTTCGACACCCAGGTGTTCCTCGAGGTCCTGCTCAAGGGCACAGGCTTCCCGGGCTCGAACAACAACACCGGCGAGGTGATGTCGCCGCTCCCACTCGGCAGCGGCAGCGACACGGGCGAGATGCGCCTGCAGTCCGACTTTGCGCTCGCGCGCGACGAGCGCACGGCGTGCTTCTGGCAGTCGTTCGTCAACGAGCAGGAGTTCATGGCGGCGAGCTTCAAGGCCGCGATGGCGAAGCTTGCGATCCTCGGCCACAGCCGCAGCAGCCTCATTGACTGCAGCGACGTCGTCCCCGTCCCGAAGCCCGCCGTCAACAAGCCCGCGACGTTCCCCGCGACGAAGGGCCCCAAGGACCTCGACACGCTCACGTGCAAGGCCCTCAAGTTCCCGACGCTGACCTCCGACCCCGGTGCTACCGAGACCCTCATCCCCCACTGCTCCAACGGCGGCATGTCCTGCCCTGGTGTTCAGTTCGATGGCCCTGCCTAA
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Expression of rMnP1
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Antibody based slot blot detection to identify rMnP protein in the supernatant
of recombinant P. pastoris culture
medium No insert
Fungi Positive control
Recombinant P. pastoris
blank
Recombinant P. pastoris
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Western Blot rMnP1 Detection(various media and concentration methods)
123456
fungiNo insert
Recombinant P. pastoris
45 kDA
210 kDA
27 kDA
90 kDA
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Fungal MnP
Intracellular rMnP
45 kDa
35 kDa
Intracellular Expression of rMnP
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Glycosylation• Three potential N-
glycosylation sites.
• Native MnP only glycosylated at one site as indicated by crystal structure.
Asparagine Serine/threonine
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native MnPrMnPuntreated untreateddegly1 degly2
degly2=EndoH
degly1 = PNGaseF
degly1 degly2
Effect of Deglycosylation on rMnP
150 kDa
45 kDa
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Western Blot Time Course rMnP
fungi Recombinant P. pastoris
time
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Problem 8.3. Isolate a temperature sensitive mutant able
to grow at 30oC but not at 37oC.1. Expose cells to mutagen (uv light,
chemicals).2. Dilution plate the culture to obtain a plate
with widely separated colonies.3. Replicate plate.4. Place one plate at 30oC and the other at
37oC.5. Select the colony that grew at 30oC but not
at 37oC.
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1. Expose cells to mutagen (uv light, chemicals).
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Serial Dilution Plating
2. Dilution plate the culture to obtain a plate with widely separated colonies.
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3. Replicate Plating
Can produce plates with identical strains in the same location.
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30oC 37oC
4. and 5. This strain grew at 30oC, but not at 37oC.
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HW Problem 8.2Obtain a methionine overproducer.• The primary objective is to mutate enzyme r so the
allosteric inhibitor site is no longer functional – in other words, remove the feed back inhibition by methionine.
1. Subject the culture to a mutagen – chemical or radiation.
2. Plate the mutagenized culture on an appropriate medium with a dilution sufficient to allow individual colonies, widely separated.
3. Screen the colonies for methionine over production.
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HW Problem 8.6Given the amino acid sequence of the peptide,
what is the sequence of steps to obtain an E. coli strain expression the peptide.
• Reverse translate the amino acid sequence to obtain a DNA sequence that will encode for the peptide.
• Design end sequences to create a restriction site at each end of the peptide DNA.
• Chemically synthesize the DNA.
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• Design primers for the the synthesized DNA.• PCR using the synthesized DNA as a template.• Cut the PCR product DNA with the restriction
enzyme.• Cut an appropriate vector with the same
restriction enzyme.• Mix the cut PCR product with the cut vector with
the enzyme ligase.• Transform (electroporation or mating) the vector
into the E. coli cell.• Plate the transformed cells on selective medium
– with the antibiotic that the vector carries resistance to.
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• Test the colonies that form for expression of the peptide using protein gel electrophoresis with antibody staining.
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HW Problem 8.7A protein converts colorless substrate to blue
product. We want to produce and E. coli expressing this protein. We have a high-copy plasmid with penicillin-resistance.
• We cannot chemically synthesize the DNA because it is too large (protein not a peptide). We must find donor DNA.
• We can screen many organisms for the ability to turn the colorless substrate the blue. We will use on that can as the source DNA to encode for the protein.
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• We can design primers based on the beginning and ending of the protein amino acid sequence. We will need a set of primers because the redundancy of codons that encode for a particular amino acid.
• We can PCR the genomic DNA of or source strain using different combinations of the primers.
• Run the PCR product on a gel and search for the size band the same as the DNA that encodes for the protein. That set of primers was correct for the ending sequence.
• Design new primers to incorporate restriction sites at the ends of the gene and PCR the source DNA.
• Cut both the vector DNA and the PRC product with the appropriate restriction enzyme.
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• Combine the cut vector, the cut DNA and the enzyme ligase to obtain a vector with the gene insert.
• The vector will have either an inducible or constitutive promoter.
• Transform the vector with the insert into the E. coli. And plate the transformed E. coli onto penicillin medium with the colorless substrate.
• Select the colonies that appear and are blue.
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HW Problem 8.8
• Similar problem 8.6. The vector should have an inducible or constitutive promoter. The vector or the gene should have transcription and translation stop codons.
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Clarification, Review, and Take Home Exam
• Assume we have the DNA sequence of a protein. This is the sequence of the ‘coding’ strand.
• The non-coding strand is complementary to the coding strand.
• The mRNA is synthesized from the non-coding strand – making the sequence of the mRNA the same as the sequence for the DNA coding strand (except for the Us).
• The translation start on the mRNA is AUG, so the translation start on the coding DNA strand is ATG. There needs to be some mRNA transcribed before the translation start for the mRNA to bind to the ribosome.
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• The translation stop codons are the nonsense codons indicated in table 4.1 in your text. The most common is UAG.
• The transcription start and stop sites are not as simple as a single codon. They are defined by the promoter and terminator regions which are typically on the vector already. It is OK if the mRNA is longer than the corresponding protein – it has to be longer at the front end. The extra sequences are not translated.
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mRNA
Transcription start
Transcription stop
Translation start
Translation stop
Encodes for protein
DNA
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Take Home Exam
• We would like to produce an enzyme in a recombinant host because the native host is not suitable for bioreactor fermentation.
• The enzyme is to be used for a transformation in commodity scale food process.
• Suggest the steps for creating the recombinant strain.
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• Select the recombinant host – E. coli, yeast, or CHO cells. Briefly explain why you selected the host.
• We know the coding strand DNA sequence. The flanking sequences, including the start and stop are indicated below.
• AGATCTCTGGUUATGGAATTC..AAGCTTCGAUAGGATTAGATCT
• If you use PCR give the sequence of the primers you will use.
• If you use restriction enzymes, name which restriction enzyme you will use.
• Use diagrams with sequence information whenever possible explaining your steps.
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Assume you have a vector for each type of host with the indicated elements.
Origin of replication
Strong promoter
EcoRI BglII HindIII SmaI Transcription
terminator
Promoter and antibiotic
resistance gene