July 4

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(Analysis)
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|[[Image:mean_cfp_fluorescence_vs_IPTG.jpg|thumb|300px|mean CFP vs IPTG conc]]
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|[[Image:mean yfp vs IPTG conc.jpg|thumb|300px|mean YFP vs IPTG conc]]
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|[[Image:mean yfp fluorescence vs IPTG conc.jpg|thumb|300px|mean YFP vs IPTG conc]]
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|colspan="2" align="center"|<font color="Red">'''Scatter plot:'''</font> Each dot here represents an individual cell.
|colspan="2" align="center"|<font color="Red">'''Scatter plot:'''</font> Each dot here represents an individual cell.
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Revision as of 07:43, 8 July 2007

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Experiments

  • Open Loop experiment [AA]

i) pT.luxI.C was transferred to 50 mL Glu M9 in 250 mL flask with 20 ng/mL of [aTc] and 1 and 0.1 uL/mL of inoculum

ii) K12z1 was inoculated in 5 mL Glu M9 in 50 mL tube with 20 ng/mL of [aTc] and 1 and 0.1 uL/mL of inoculum.

iii) pL.LuxR.Y.pR.C, K12Z1 and pL.Y were Inoculated in LB

iv) The flask with OD=0.263 was filtered and stored at 4 degrees

v) K12Z1 strains in aTc(20,0 ng/mL) were filtered at OD ~0.2 and stored at 4 degrees.

vi) 7 dilutions for pL.LuxR.Y.pR.C were made in 30 mL AI medium:30 mL new 2xM9 [x2 inoculum=.1,.01]

vii) The following negative controls were done:

    • K12z1(20 ng/mL aTc)
    • pL.LuxR.Y.pR.C in 3 mL K12Z1 grown medium:3 mL new 2xM9,500 uM IPTG. However M9 turned out to be contaminated.
    • pL.LuxR.Y.pR.C in new M9,500 uM IPTG [x2 inoculum=.1,.01]

viii) The following positive control was done: pL.Y in new M9 induced with IPTG: 100 uM.


  • Open Loop experiment [FF]

i) pT.luxI.C was transferred to 50 mL Glu M9 in 250 mL flask with [aTc]=0 ng/mL and 1,0.1 uL/mL of inoculum.

ii) K12z1 was inoculated in 5 mL Glu M9 in 50 mL tube with [aTc]=0 ng/mL and inoculum=1,0.1 uL/mL.

iii) pL.LuxR.Y.pR.C, K12z1 and pL.Y were inoculated in LB

iv) Flask with OD=0.257 was filtered and stored at 4 degrees

v) 7 dilutions were made for pL.LuxR.Y.pR.C 30 mL AI medium:30 mL new 2xM9 [x2 inoculum=.1,.01] along with the negative control with K12z1.


  • pT.LuxI.C and K12Z1 inoculated in L/B for open loop expts [Aa],[Bb].


  • To Prove scaling argument: aTc=20 ng/mL, OD=0.257 & 0.015 (0.015/0.257=0.06)
    • A 6:100 dilution of the 1:2 diluted 0.257 sample in 2xM9, and a 1:2 dilution of 0.015 sample was made.
    • For each of these above AI containing media, IPTG dilutions were made for 5,50,500 uM in duplicates of inoculum (.01,.1).
    • The argument will be proved if they show exact same CFP expression.


Microscopy

  • Open loop samples [E] were imaged.
  • pT.LuxI.C,K12z1 of both [AA] and [FF] were Imaged on FACS and microscopy; but both K12Z1 samples had contamination.


Analysis

  • We obtained the following scatter plots (mean log) for the equivalence curve for open loop expt [A]:
mean CFP vs IPTG conc
mean YFP vs IPTG conc
Scatter plot: Each dot here represents an individual cell.