Yeast Lab: Deletion

Question 1

Goal

Answer 1

• To create a mutation in a specific yeast gene (ADE2) by excising it from the chromosome and inserting HIS3 gene in its place
• also known as “targeted mutagenesis”.
• PCR-mediated gene disruption will be used to create a knockout of the ADE gene in yeast

Question 2

knock out mutation

Answer 2

• One way to investigate function of a particular gene is to prepare a strain with a mutation in a gene eliminating the gene from the chromosome producing a knock out mutation
• This is possible because of the high frequency of homologous recombination in yeast
• The phenotypes of the cells lacking a particular function often yield clues as to the function of the protein

Question 3

PCR-mediated gene disruption

Answer 3

• gene excision
• need sequence upstream and down stream of the gene of interest
• Homologous recombination in yeast requires only short regions of homology
• excising ADE2 gene and replacing it with HIS3 in
  yeast strain BY4742 (MATα, his3∆, leu3, lys2∆0, ura3∆)
  
  • ∆ means gene deletion
• steps
  
  • use PCR to amplify a fragment from plasmid pRS403
    
    • cannot replicate in yeast, only in E. coli
  • fragment has
    
    • 40 nucleotides at each end, specific to upstream and downstream regions of ADE2 gene
    • 20 nucleotides at the 3’ ends of each primer, specific for amplification of selectable marker HIS3 from pRS403
    • Transformants that gained the ability to grow on medium without histidine are selected → ADE replaced with HIS3
    • ADE2 gene is eliminated and results in cells that require adenine to survive
• transformants w/HIS3 gene are identified by their red color when grown on low adenine containing plates
• unsuccesful transformatns (integrated HIS3 gene into another location) have a functioning ADE2 gene and are white
• frequency of targeted gene deletion is generally 1-2%
• PCR analysis is used on red & white colonies to confirm the gene deletion

Question 4

Phusion High-Fidelity DNA Polymerase

Answer 4

• high accuracy
• has proofreading ability due to 3’ to 5’ exonuclease activity that can remove misincorporated nucleotides, unlike Taq DNA Polymerase
  
  • results in high fidelity amplification of the template
  • important for applications where the amplified product needs to be functional.
• an engineered enzyme based on an enzyme from Pyrococcus with the addition of a processivity domain
  
  • incorporates more nucleotides before it falls off the template
  • makes it very fast enzyme
  • requires less time for extension.

Question 5

homologous recombination

Answer 5

• Genetic exchange between identical or nearly identical DNA sequences
• Involves strand breakage, template switching, religation
• Yeast is a good organims for this because
  
  • has high frequency of homologous recombination, allowing efficient gene deletion
  • requires a very short sequence homology for homologous recombination to occur ≈ 40 bp required
  • so ologs would require the 40 bp + marker (HIS3)

Question 6

Designing Oligonucleotide Primers

Answer 6

• Length: about 20 bp
• GC content: about 50%
• Tm: 60ºC
  
  • temperature at which 50% of the primer is annealed to the template
  • Tm = 2(#of A + T) + 4(#G+C)
    
    • multiply by 4 because 3 hydrogen bonds for G+C vs 2 hydrogen bonds for A + T
• Tm Annealing step generally at Tm minus 5-10ºC
• written in 5’ to 3’ direction.
• must bind to opposite strands
  
  • forward oligo will be the exact sequence of the DNA top strand
  • reverse oligo will be the sequence on the bottom strand but written in the 5’ to 3’ direction.

Question 7

COVID19 Testing Techniques

Answer 7

• PCR Test
• SARS-Cov-2 is a RNA virus
• Extract RNA from test sample
• Reverse transcriptase
  
  • synthesizes DNA from RNA
  • DNA template needed for PCR
  • makes cDNA libraries
• Real Time PCR amplification
  
  • of specific genes of virus
  • uses fluorescent detection
  • during reaction can see results

Question 8

Week 11

Preparation of PRS403 plasmid

Answer 8

Week 11

• Preparation of PRS403 plasmid
  
  • E. coli transformed with pRS403 were provided
  • Plasmid is isolated and purify from E. coli
    
    • Pellet cells via microfuge
    • Resuspend in buffer containing RNAse to destroy any RNA present
    • SDS disrupts membrane
    • Alkaline buffer (NaOH)
      
      • lyses cells and denatures DNA and not the plasmid
      • plasmid is circular, double stranded, the strands will not separate
    • K acetate
      
      • Neutralizes solution
      • SDS precipitates and traps chromosomal DNA and proteins
    • Centrifuge
      
      • Pellet = extracellular waste
      • Supernatant = plasmid
    • Column binds DNA in high salt
• DNA concentration of the plasmid was determined by using a Nanodrop
  
  • Max absorbance at 260 nm
  • 260/280 ratio ≈ 1.8 – 2.0
• PCR Amplification of HIS3 Gene from pRS403
  
  • Plasmid DNA is diluted and used as template in a PCR reaction to amplify the HIS3 gene
  • Reaction Mix is created totaling 50 uL
  • Primers used
    
    • Oligo 1F (Forward primer)
    • Oligo 2R (Reverse primer)
    • First 40 nucleotides
      
      • specific to upstream (F primer) and downstream (R reverse) regions of ADE2 gene
      • not homologous to any sequence within pRS403
    • 20 nucleotides at the 3’ ends of each primer, specific for regions of HIS3 from pRS403
    • Process
      
      • First cycle
        
        • primers bind to 20 nucleotide complement region ONLY on pRS403
      • Second cycle
        
        • primers bind to 20 nucleotide complement region ONLY on pRS403
        • primers bind to fragment from 1^st cycle that has 40 + 20 + HIS3 + 20 + 40
      • Method used to add specific sequences to the ends of a PCR fragment

Week 11 - follow up

• Gel analysis to confirm amplification of the fragment 40 + 20 + HIS3 + 20 + 40 from pRS403
  
  • 40 = First 40 nucleotides specific to upstream (F primer) regions of ADE2 gene
  • 20 = 20 nucleotides at the 3’ ends of F primer, specific for regions of HIS3
  • 1184 = HIS 3
  • 20 = 20 nucleotides at the 3’ ends of R primer, specific for regions of HIS3
  • 40 = Last 40 nucleotides specific to downstream (R reverse) regions of ADE2 gene
  • 40 + 20 + 1184 + 20 + 40 = 1304 ≈ size of DNA fragment amplified
• Estimation of amount of PCR product recovered
  
  • 1200 bp marker band is 90 ng
  • If your band is about twice the intensity of the marker it lines up w/(1200 bp), then 2 × 90 = 180 ng on gel
  • 10 x 180 = 1800 ng = 1.8 ug total PCR product recovered
    
    • loaded 5 ul of PCR product on gel from 50 uL of PCR Reaction Mix
    • 50/5 = 10
  • Started with only 4 pg of plasmid DNA and recover ≈ 1.8 ug of a specific target sequence
  • Copy and paste link

https://drive.google.com/file/d/1frR5XNHq2wQyFfIwsrp_3nLQd43WZ-F-/view?usp=sharing

Question 9

Week 12

Transform Yeast Strain

Answer 9

Week 12

• Transform Yeast Strain BY4742
  
  • BY4742 (MATα, his3∆, leu3∆, lys2∆, ura3∆)
    
    • his3 is the only nutritional value important
    • leucine, lysine and uracil added to all medium
  • Yeast strain is prepped and transformation mix is added
    
    • PEG - induces DNA binding to cell surface
    • Lithium cation
      
      • permeabilizes whole yeast cells
      • Cells take up DNA
    • Transformation mix
      
      • has PCR product: fragment of HIS3 Gene from pRS403 ≈ 1304 bp
    • Calf thymus DNA
      
      • Denatured, Single-stranded carrier DNA
      • improves efficiency and uptake of target DNA
  • Cells are treated with heat shock to induce transformation
    
    • yeast mutant strain will be transformed with the PCR product
    • Fragment enters the nucleus and is integrated into the yeast chromosome
    • PCR product can insert at the position of the ADE2 gene (within yeast) that complements the ADE2 sequence (40 bp) of the fragment
• After heat shock they are plated on histidine drop out plates, with low amounts of adenine
  
  • No histidine: To select for transformants whose mutant his3 gene was replaced by the one in the plasmid
    
    • If crossing over at these 40 bp occurs during replication, then the HIS3 gene will partially excise ADE2 gene resulting in ade2/HIS3 mutant
  • Low adenine:
    
    • sustains growth of ADE2 knock-outs, but not enough to turn off the adenine biosynthetic pathway, so the mutants can have either a red or pink color, and are identifiable
    • transformants had their ADE2 gene replaced by HIS3, so they require adenine to survive
  • Leucine, lysine and uracil were also added to the medium, to support growth of BY4742 strain, which had the leu3, lys2 and ura3 genes deleted
  • Considerably more white colonies result from PCR fragment being inserted somewhere else, other than the ADE gene.
  • Most will be white.
• Any new growth from this lab can potentially include two types of mutations:
  • ade2/HIS3 (red) and
  • ????/HIS3 (white, inserted outside of our ADE2 target gene)

Question 10

Week 13

Analysis & Replating

Answer 10

• Typical frequency of targeted gene deletion in yeast is 1-2%.
• Count # of red and white colonies from a representative plate and determine frequency in samples
  
  • Frequency: Red/White
  • Plate 1
    
    • Red: 2
    • White: 68
    • Frequency: 2.9
    • Expected Frequency: 1-2%
  • Plate 2
    
    • Red: 1
    • White: 60
    • Frequency: 1.7
    • Expected Frequency: 1-2%
  • Total
    
    • Red: 3
    • White: 128
    • Frequency: 2.3
    • Expected Frequency: 1-2%
• Look for red/white colonies and plate growing colonies onto new histidine dropout plate containing leucine, lysine, uracil and adenine
  
  • Adenine to try to generate an adenine mutation
  • Colonies that grow have integrated HIS3 from PCR fragment
  • Colonies that are red had the ADE2 gene replaced by HIS3 gene
• We are replating to get more cell mass for the colony PCR to test for disruption of the ADE2 gene next week.

Summary

• All colonies denote transformants whose mutant his3 gene was replaced by the HIS3 gene from the plasmid
• The difference between the red and white colonies is the location in the genome where the HIS3 was incorporated
• For the red colonies, the HIS3 gene was incorporated where the ADE2 was in the genome (replacing it)
• For the white colonies, the HIS3 gene was incorporate elsewhere in the genome, not affecting the existing ADE2 gene

Question 11

Week 14

Colony PCR of red & white yeast colonies to confirm integration of HIS3 gene

Answer 11

• Colony PCR of red & white yeast colonies to confirm integration of HIS3 gene
  
  • No need to purify DNA
  • Convenient way to screen a lot of colonies w/out purifying DNA
  • Done on whole cells
  • Lysed in NaOH
  • Crude extract used for PCR
  • Confirms that the ADE2 gene mutations are not generated by chance, and are indeed caused by the HIS3 recombination event
• Using specific primers
  
  • Not using the 1F and 2R primers previously used to generate the HIS3 gene fragment
  • Using
    
    • 4F and 5R
      
      • w/in Ade2 gene
      • a band produced by these primers
        
        • HIS3 was inserted outside of the Ade2 gene, 330 bp
        • white colony
        • 6F and 7R will not generate a band, 6F and 7F will band, but 7F will be too far away from 6F
    • 6F and 7R
      
      • 6F in the upstream region of Ade2 gene, before the start codon
      • 7R within the HIS3 gene
      • If the HIS3 gene inserted within the Ade2 gene
        
        • you will get a fragment using the 6F and 7R primers, 448 bp
        • 4F and 5R will not generate a band
        • red colony
• 2% gel will be generated for follow up
  
  • For seperation of small fragment

Summary

• Colony PCR was used to confirm if the targeted ADE2 gene deletion was successful.
  
  • done via the use of primers generated to anneal within regions of either entirely within the ADE2 gene (4F and 5R primers)
  • or within both the ADE2 and HIS3 gene (6F and 7R respectively)
• Since the samples were taken from cells grown on histidine dropout plates, it is expected that HIS3 gene was successfully recombined, leading to
  
  • excision of the ADE2 genes in the red colonies
  • or it was successfully recombined outside of the ADE2 gene area in the white colonies

Question 12

Calculations

Answer 12

https://drive.google.com/file/d/1frR5XNHq2wQyFfIwsrp_3nLQd43WZ-F-/view?usp=sharing