Principles of experiments

    The documents cited shows that even after natural weathering, biological decomposition, and acid rain erosion for a time period of time, some stone inscriptions, which are on rock surface containing calcium, remain in good condition because there is a thin layer of naturally formed semitransparent mineralised film that is hydrophilic and compact on such rock surface.

     In some part under the fine film, knife scratches on the stone from over a thousand years ago are still visible indistinctly. According to solid analysis, natural mineralised film is primarily composed of calcium oxalate monohydrate [Ca(COO)2·H2O] whose process of composition is interfered by microorganism.

     There are some advantages of calcium oxalate being a film as follows,

        1) In water, especially in acidulous water, the solubility of calcium oxalate is one order of magnitude smaller than normal calcium carbonate.

        2) Organically oriented calcium oxalate crystal's particles are fine and compact, which prevent the infiltration of toxic and harmful materials.

        3) The system can't be easily broken by external forces in that there is no distinct boundary between calcium carbonate rock and calcium oxalate film.

        So, this research project has chosen E. coli that has been transcribed with lipase to clean some stone relics that are covered with oil; then, we use E. coli, with transcription, that can secrete oxalate acid upon the stone relics. Therefore, there forms calcium oxalate to be protection film in order to preserve lithic relics.

        If we use chemical methods to generate calcium oxalate, we have to immerse the lithic relics in the chemical liquid; however, it is not suitable for the preservation of large-scale outdoor lithic relics. In fact, the natural existing calcium oxalate comes from the reaction of oxalate acid generated by microorganism and calcium compound on the surface of stone. This research project will utilise genetic engineering methods to "graft" the gene that generates calcium oxalate into Escherichia coli (E. coil).

    Then oxalate acid reacts with calcium lithic relics to generate calcium oxalate, which stimulates its natural formation process. Finally, there forms a fine layer to protect the lithic relics.

[Ca(COO)2·H2O]'s crystal

[Ca(COO)2·H2O]'s SEM Picture

Procedures in detail

1.1 Experiment reagents and equipment

    1.1.1 Target genes

        (1) Lipase: lipase is the target gene segment of this experiment, and the gene sequence is synthesised by IDT Inc.

        (2) Oxaloacetate acetylhydrolase, abbreviation OAH, is the target gene segment of these experiments. The main function of OAH is to transfer oxaloacetic acid to oxalate acid; thus the oxalate acid reacts with the calcium ion to form calcium oxalate that preserves the relics. The target gene segment is synthesised by IDT Inc. after found in gene pool.

    1.1.2 Strain and Plasmid This experiment uses plasmid pET-28a, and the E. coli is TOP-10 competent cell, provided by TIANGEN Biotech CO., LTD.

    1.1.3 Enzyme and Reagents This experiment uses restriction enzyme provided by NEB; T4 DNA Ligase, TIANprep Mini Plasmid Kit, and TIANquick Midi Purification Kit are provided by Tiangen Biotech CO., LTD.

    1.1.4 PCR Primer

        (1) Lipase: According to the sequences and the MCS (multiple cloning sites) to the destination vector of the Lipase, XhoI and NcoI are chosen to be the enzyme cutting cites for the inserting enzyme cutting cites. There form the following primers:

    Sense primer: 5’-AAGCAAGCCCATGGAATTTGTAA-3’ (Underlined proportion is Ncoi enzyme cutting cite)

    Anti-sense primer: 5’-TTTTGTTGCTCGAGTTAATTCG-3’ (Underlined proportion is Xhoi enzyme cutting cite)

         (2) OAH: According to the sequence detected by OAH, the primer is synthesised by the Invireogen:

        Sense primer: 5’-ATCATCTTGCATATGGCGTCCACCATTGCTGTT-3’ (Underlined proportion is Ndei enzyme cutting cite)

         Anti-sense primer: 5’-TTTCCTTTTGCGGCCGCTTAG-3’ (Underlined proportion is Noti enzyme cutting cite)

    1.1.5 Substrate Medium

        (1) The experiment contains LB substrate media. Take the example of confecting 200mL; the materials needed are 2g of tryptone, 1g of yeast extract, 4g of sodium chloride, and 200mL of water with 1000× ratio of chloramphenicol.

        (2) The experiment contains LB substrate media. Take the example of confecting 200mL; the materials needed are 2g of tryptone, 1g of yeast extract, 4g of sodium chloride, and 200mL of water with 1000× ratio of Kanamycin Sulfate.

        (3) The experiment contains Kanamycin Sulfate substrate media. Take the example of confecting 400mL; the materials needed are 4g of tryptone, 2g of yeast extract, 4g of sodium chloride, 6g AGAR powder and 1.2mL

Procedures in detail

1.2 Procedures of Experiments

     1.2.1 PCR Proliferation

        Due to high content of G-C base-pairs in target gene segments, GC Buffer is required in PCR proliferation. Refer to Takara's protocols to prepare 200μL of system: 100μL of GC Buffer, 16μL of dntp mixture, 4μL of sense primer, 4μL of anti-sense primer, 4μL of DNA templet, 2μL of Taq, and 70μL of H2O. 60°C-70°C is used for Gradient PCR, the sixth bad is the brightest (Figure 9) thus the temperature of such is best for annealing. In new system, PCR is proceeded in that temperature; finally we get the gene segments with gene linkers.

    1.2.2 Gel electrophoresis, purification

    1.2.3 Digestion

        Put NotI and NdeI restriction endonuclease in gene segment and vector pET-28a respectively.

    1.2.4

         Linking and setting the groups Experimental group:plasmid after digestion + target gene + T4 DNA Ligase + buffer Control group: plasmid after digestion *Experiment conducted under 16°C

    1.2.5

        Use DNA purification kit to extract product after linking; transcribe in TOP-10 competent cell, put on Kanamycin Sulfate substrate media and then culture under 37°C.

    1.2.6

        According to the growth of strains, bacterium plasmid is extracted with digesting with two enzymes above. Recheck the correctness of position and size to assure that it has been linked.

    1.2.7 Sequencing by Sangon Shanghai, China

    1.2.8 Functional Test of Lipase

        (1) Fasten the AGAR Plane: 2.5g of AGAR powder, 6.25mL of tributyrin, 18.75mL 3g/dL of polyvinyl alcohol, 2.5mL 1g/dL Victorian Blue, 225mL of H2O. Heat and resolve the AGAR powder, pure into substrate media, chilling. IPTG solution: weigh 1.2g of IPTG and put it into centrifugal tubes. Add 40mL of H2O, resolve, then to the constant volume of 50mL. Use 0.22μm strainer to filter out the other bacteria, then put into refrigerator of -20°C.

        (2) Transcribe the lipase enzyme plasmid into BL-21 competent cell of E. coli, culture.

        (3) Pick bacteria into small colonies from growing strains on the cultivation board, which has been configured with a medium, Kanamycin, and join the 0.1 mmol/L IPTG solution to carry on the induction, six hours later to take on the culture medium centrifugal supernatant, get with lipase protein nutrient solution.

        (4) From fixed AGAR plate with holes, pick bacteria. Burn alcohol burner against the bottom of AGAR to assure it is stick. Put 30μL into the holes, lie horizontally in 30°C incubator for a while and then observe the change of colour.

Figure 9 Gradient PCR Electrophoresis (red column is the brightest)

Results of Experiments and Analysis

1.3 Results of Experiments and Analysis

     1.3.1 Digestion check

        (1) Use NcoI and XhoI to digest plasmid on lipase after PCR. We know our result is correct because two bright parts of plasmid is in the same position and of the same size.

        (2) Use NdeI and NotI to digest plasmid on OAH after PCR. We know our result is correct because two bright parts of plasmid is in the same position and of the same size. (Figure 10)

    1.3.2 Sequencing

        The results of t sequencing of Lipase and OAH are both correct, for more detail, check appendix 1.

    1.3.3 Functional Test of Lipase

        Let PET-28a-Lipase plasmid be transcribed into BL-21 E. coli (the procedure is the same as that of TOP-10 E. coli), then put on substrate media. Pick E. coli into shake flask for 3 hours. Join IPTG solution (0.1 mmol/L) for abduction for 6 hours. Put into centrifugal machine then supernatant liquid (with lipase) for lipase test. The result shows that the colour has changed besides plasmid with lipase (Figure 11), showing that LipB2 gene is correctly expressed in BL21 E. coli.

         Linking and setting the groups Experimental group:plasmid after digestion + target gene + T4 DNA Ligase + buffer Control group: plasmid after digestion *Experiment conducted under 16°C

1.4 Result Analysis and Prospect

    According to the results of sequencing, actual sequence is identical to designed sequence, which testifies the validity of formulation (appendix 1). In lipase functional test, there is relatively obvious phenomenon under IPTG abduction (Figure 11), which testifies the function of gene.

    Hence, using this reformed lipase can erase the grease cover on relic surface, and we hope to use OAH reconstruct relic surface.

    Meanwhile, recombinant lipase can be widely used in any field for de-greasing, for example, the de-greasing process of steel surface, pre-treatment of high-fat food disposal, etc.

    This research project only recombines the lipase without optimising it according to temperature fluctuation, abduction concentration. We hope our project is helpful in every field.

Figure 10 Digestion result test

Figure 11 LipB2 gene in BL21 E. coli

Mathimatical models

    Since there is no one had done the same experiment as ours before, we assumed the growing pattern is similar as E.coil CVCC249.

    We got the data of this kind of E.coil from a doctoral dissertation,Comparison of Growth Kinetics and Physiological Characteristics of E.coil CVCC249 under both Batch and Continuous Culture by Huaiqiang Zhang, at page34. We used the logistic function data in this form, which is K=1.937 and r=0.550, and we assumed that N0=0.1( since N0 is initial population of E.coil, we can just simply set this data).

    According to logistic function , we can get the bacteria population-time function:(see Figure 1)

        f(t) = 1.937/(1+(1.937-0.1)e^(-0.55t)/0.1)         f(t) = 1.937/(1+18.37*e^(-0.55t))

    After that, we assumed that there will be x% of bacteria can present the property we need. Then we can get population of expressed bacteria:

        fex(t) = x%*1.937/(1+18.37*e^(-0.55t))

    Because we need the oxalic acid-time function, we assume the speed of oxalic acid secretion of per unit of bacteria is v. So the total speed should be:

        V(t) = v*x%*1.937/(1+18.37*e^(-0.55t))

    Since the secretion of oxalic acid should be accumulated, we take the integration of total speed of time. We used WolframAlpha program to find out this integration.

        ∫1.937/(1+18.37*e^(-0.55t)) dt=3.522*log(18.37+e^(0.55t))+C0

    Because the function should began at (0,0) point, we can find out constant C in function.

        0=v*x%*3.522*log(18.37+e^(0.55*0))+C         C=-3.522*log(19.37)*v*x%         C=-10.438*v*x%         Substitute C=-10.438*v*x%

        Soa(t)=v*x%*(3.522*log(18.37+e^(0.55t))-10.438)

    This is oxalic acid-time function.

    We also need to find out lipid-time function. Obviously, it’s more complicated than last function, since the decomposition speed cannot be a constant.

    The fex(t) function should be also available in this part. Starting constructing process, we assumed that initial total lipid is N, lipid that decomposed is n, and the decomposing speed is relate to following factors:

    k: a constant representing efficiency of lipase     (N-n): remain amount of lipid     fex(t): quantity of expressed bacteria

    And there should be a function meaning that decomposing speed is proportional to the remain amount of lipid and quantity of expressed bacteria:

        dn/dt=k*(N-n)*fex(t)

        Solve that differential equation.

        1/(N-n) dn=k*fex (t) dt         ∫1/(N-n) dn=k∫fex (t) dt         -ln(N-n)=k∫f ex(t) dt + C

    Since we already know that ∫f ex(t) dt= x%*(3.522*log(18.37+e^(0.55t))-10.438) in this condition.

        -ln(N-n)=k*x%*(3.522*log(18.37+e^(0.55t))-10.438)         (N-n)=e^(-k*x%*(3.522*log(18.37+e^(0.55t))-10.438))

    So the lipid-time function, Slp(t), were found:

        Slp(t)=(N-n)=e^(-k*x%*(3.522*log(18.37+e^(0.55t))-10.438))

Quantity of Oxalic Acid

Quantity of Bacteria

Figure 1

Notebook

Our origin experiment records

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