Team:Tsinghua-A/Experiments

Protocols
  • Firstly, we need to acquire the elements.
  • We using PCR, Overlap PCR and Gibson to get the elements such as EBFP2, EBFP2N(1-154), EBFP2C(155-238), EBFP2N(1-154):IntN and IntC:EBFP2(155-238).

  • Secondly, we construct 9 parts.
  • We combine the elements with the vector and transfer the vector into the competent E. coli cells. After that, we select the positive clones and use colony PCR to identify the recombinants. Culturing the positive clones and sequencing the isolated plasmids to make sure that we have get the needed parts.

    • Part 1 TRE-Kozak-EBFP2N(1-154)
    • With this part, the expression of EBFP2N(1-154) could be induced by Dox. (See more details: http://parts.igem.org/Part:BBa_K1918101)


    • Part 2 TRE-Kozak-EBFP2C(155-238)
    • With this part, the expression of EBFP2C(155-238) could be induced by Dox. (See more details: http://parts.igem.org/Part:BBa_K1918102)


    • Part 3 TRE-Kozak-EBFP2N(1-154):IntN
    • With this part, the expression of EBFP2N(1-154):IntN could be induced by Dox. (See more details: http://parts.igem.org/Part:BBa_K1918103)


    • Part 4 TRE-Kozak-IntC:EBFP2C(155-238)
    • With this part, the expression of IntC:EBFP2C(155-238) could be induced by Dox. (See more details: http://parts.igem.org/Part:BBa_K1918104)


    • Part 5 CMV-Kozak-EBFP2C(155-238)
    • With this part, the expression of EBFP2C(155-238) could be all the time. (See more details: http://parts.igem.org/Part:BBa_K1918105)

  • Part 6 CMV-Kozak-EBFP2N(1-154):IntN
  • With this part, the expression of EBFP2N(1-154):IntN could be all the time. (See more details: http://parts.igem.org/Part:BBa_K1918106)


  • Part 7 CMV-Kozak-IntC:EBFP2C(155-238)
  • With this part, the expression of IntC:EBFP2C(155-238) could be all the time. (See more details: http://parts.igem.org/Part:BBa_K1918107)


  • Part 8 CMV-Kozak-EBFP2N(1-154)
  • With this part, the expression of EBFP2N(1-154) could be all the time. (See more details: http://parts.igem.org/Part:BBa_K1918108)


  • Part 9 TRE-Kozak-EBFP2
  • With this part, the expression of EBFP2 could be induced by Dox. (See more details: http://parts.igem.org/Part:BBa_K1918109)

    The left figure and the right figure are the FACS test results of group 1 and group 4. The EBFP-A signals tell us that the split florescent proteins could fuse together and our parts work fine.

  • Lastly, we transfect the plasmids to 289H cells and do the FACS test.
Experiments

    We transfect HEK-293 human cells with our plasmid constructions as described in the form [ref: table]. Different concentrations of Dox are applied to cell culture at the same time.

    Transfected cells are cultured for 48 hours before performing flow cytometry, long enough for protein expression level to achieve steady state. FACS examination measures florescent intensity emitted by each cell, from which we obtain a large sample of florescent protein expression level, tens of thousands of cells for each experiment group.

    Data collected from flow cytometry are later analyzed on computers. We estimated probability density function (p.d.f.) from data using kernel density estimation, a nonparametric statistics method. Given high and low Dox concentration input, cells exhibit different probability distributions, as illustrated in the example below [ref: fig].

    What we have in hand is the conditional distribution $p\left( {Y\left| {X = x} \right.} \right)$ , given a known level of input $x$ . In order to calculate mutual information $I\left( {X;Y} \right) = \iint {p\left( {x,y} \right){{\log }_2}\frac{{p\left( {x,y} \right)}}{{p\left( x \right)p\left( y \right)}}dxdy}$ and estimate channel capacity, which is $C = \sup I\left( {X;Y} \right)$ , we need to find the input distribution $p\left( X \right)$ and joint distribution $p\left( {X,Y} \right)$ that optimizes the equation. $p\left( X \right)$ , however, is not known in the first place. We first randomly pick a stochastic vector as the initial input distribution and then use an optimization algorithm to iterate the function and maximize $I\left( {X;Y} \right)$ . The final result is the channel capacity.