Team:HIT-Harbin/Application

From 2013.igem.org

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<p>In applications, we plan to enhance Bio-electric Interface, the device designed by 2012 Edinburg iGEM team. If the input is biochemical signal molecules and the output become electrons, B-POM can be coupled with Bio-electric Interface, that ephemeral processes in cells will be precisely measurable by simple electronic methods and analyzable by computer. B-POM is also helpful in yoghurt producing, where the control of pH is inaccurate. When B-POM is transplanted into yoghurt-producing bacteria, let hydrogen ions be the input and lacR be the output, and select the proper B-POM parameters, then the pH will be steady around 5.5. In short, our inspired and reliable device is promising in various fields.</p>
<p>In applications, we plan to enhance Bio-electric Interface, the device designed by 2012 Edinburg iGEM team. If the input is biochemical signal molecules and the output become electrons, B-POM can be coupled with Bio-electric Interface, that ephemeral processes in cells will be precisely measurable by simple electronic methods and analyzable by computer. B-POM is also helpful in yoghurt producing, where the control of pH is inaccurate. When B-POM is transplanted into yoghurt-producing bacteria, let hydrogen ions be the input and lacR be the output, and select the proper B-POM parameters, then the pH will be steady around 5.5. In short, our inspired and reliable device is promising in various fields.</p>
    
    
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1关于乳酸菌利用乳糖产生乳酸
 
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反应1:乳糖由细胞外运到细胞内。
 
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LS+Lout <=> temp1 : R1(乳糖与透过酶结合)
 
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vf = k1*Lout*LS(正反应速率与反应物浓度成正比)
 
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vr =k_1*temp1(逆反应速率与产物成正比)
 
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temp1 => L+LS : R2(乳糖被运到胞内)
 
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vf = k2*temp1(正反应速率与反应物浓度成正比)
 
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其中LS表示透过酶的浓度;Lout表示胞外的乳糖浓度;L表示胞内的乳糖浓度;temp表示与透过酶结合的乳糖的浓度。vf表示正反应速率,vr表示逆反应速率。k1为乳糖与透过酶结合的速率常数,k_1为分解速率常数;k2为乳糖动到胞内的速率常数。
 
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我们已知Lout的浓度为136mmol/L(以下单位都为mmol/L),在这个基础上,我估计乳糖全部运到胞内的时间为400s,由此,我假设k1=0.1,k_1=0.05,k2=0.1,LS=5,temp1=0,L=0。模拟结果如下图
 
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由此我们可以看到在300S时,乳糖已经全都运到胞内了。
 
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反应2:乳糖转变成乳酸
 
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L+LZ <=> temp2 : R3(乳糖与半乳糖苷酶结合)
 
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vf = k3*L*LZ(正反应速率与反应物浓度成正比)
 
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vr = k_3*temp2(逆反应速率与产物成正比)
 
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temp2 => Glu+LZ+gala : R4(乳糖分解生成葡萄糖和半乳糖)
 
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vf = k4*temp2(正反应速率与反应物浓度成正比)
 
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Glu => 2*PA : R5(葡萄糖分解成丙酮酸,此为多步反应)
 
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vf = k5*Glu(假设分解速率与葡萄糖浓度成正比)
 
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PA => LA : R6(丙酮酸加氢生成乳酸)
 
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vf = rm1*PA/(Km1+PA)(假设底物充足,反应速率满足米氏方程)
 
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  各个参数假设如下:
 
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LZ=5(为半乳糖苷酶的浓度)
 
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L=136(为全部运到胞内的乳糖浓度)
 
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Glu=0(为葡萄糖的浓度)
 
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gala=0(为半乳糖的浓度)
 
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PA=0(为丙酮酸的浓度)
 
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LA=0(为乳酸的浓度)
 
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k3 = 0.1(乳糖与半乳糖苷酶结合速率常数)
 
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k_3 = 0.05(复合物分解速率常数)
 
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k4 = 0.1(乳糖分解速率常数)
 
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k5 = 0.1(葡萄糖生成丙酮酸的速率常数)
 
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rm1 = 35(丙酮酸生成乳酸的最大速率)
 
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Km1 = 100(米氏常数)
 
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模拟结果如下:
 
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由此我们可以看到乳糖全都变成乳酸了,平台时LA=272
 
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接下来我们将反应1与反应2结合起来,(原文件见“one.txtbc”)得到如下结果
 
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由此我们可以看出,胞外的乳糖在被运进来的同进也被消耗,并且乳糖的转运速率比乳酸的生成速率快,这使得乳酸能够不断生成。
 
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以上为正常情况下乳糖生成乳酸时的模拟
 
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接下来,我们来分析一下关于pH开关函数:f(p)=1-tanh(n*(p-p*))
 
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f(x)=1-tanh(x)的图像如下图
 
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当x=1时,f(x)约为0,开关为断开,当x=-1时,f(x)约为3,开关为闭合状态。
 
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由此,我们可以看出当n=1时细胞pH比阈值pH小1时pH开关才算完全开  启
 
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f(x)=1-tanh(30*x)图像如下
 
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此时(n=30时)的pH只要小0.1时开关就能完全开启
 
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另外,由实验拟合得的乳酸浓度与pH的关系函数pH=6.32311-0.02168*[LA]+0.0000426042*[LA]^2(0<[LA]<250)其图像如下
 
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反应中pH最高为6.3,最低为3.6,我们假设pH阈值为5.5
 
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2 用pH诱导的启动子表达LacR蛋白
 
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=> LA : R7(假设乳酸为常量272)
 
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vf = 0(因为为常数反应速率为0)
 
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=> mRNA1 : R8(转录生成LacR的mRNA)
 
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vf = v0+v1*f(LA)-dm1*mRNA1(f(x)为pH开关函数,此时f()函数的n=1)
 
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=> LR : R9(生成LacR蛋白)
 
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vf = km*mRNA1-dn1*LR(转录速率)
 
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=> mRNA2 : R10(透过酶与半乳糖苷酶mRNA的转录)
 
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vf = vm-dm2*mRNA2-c*a*LR/(1+a*LR)(转录速率)
 
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=> LS : R11(透过酶的生成)
 
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vf = ks*mRNA2-ds*LS
 
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=> LZ : R12(半乳糖苷酶的生成)
 
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vf = kz*mRNA2-dz*LZ
 
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各个参数含义如下:
 
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LA(0) = 272 (假设全部乳糖转变为乳酸的情况下)
 
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LR(0) = 10 (LACR初始浓度)
 
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LS(0) = 5 (透过酶初始浓度)
 
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LZ(0) = 5 (半乳糖苷酶初始浓度)
 
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mRNA1(0) = 1 (LacR的mRNA的初始浓度)
 
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mRNA2(0) = 1(透过酶与半乳糖苷酶mRNA的初始浓度)
 
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v0 = 1 (LacR的mRNA基础转录速率)
 
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v1 = 30 (LacR的mRNA在pH诱导下增加的速率)
 
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dm1 = 0.5 (LacR的mRNA的降解速率)
 
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km = 1 (LacR的mRNA转录速率常数)
 
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dn1 = 0.5 (LacR的mRNA降解速率常数)
 
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vm = 1 (透过酶与半乳糖苷酶mRNA基础转录速率)
 
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c = 1.2 (透过酶与半乳糖苷酶mRNA转录被抑制的最大程度)
 
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a = 1 (透过酶与半乳糖苷酶mRNA对抑制因子的敏感度)
 
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dm2 = 0.2(透过酶与半乳糖苷酶mRNA降解速率)
 
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ks = 1 (透过酶mRNA翻译速率常数)
 
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ds = 0.5 (透过酶降解速率常数)
 
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kz = 1 (半乳糖苷酶mRNA翻译速率常)
 
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dz = 0.5(半乳糖苷酶降解速率常数)
 
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模拟结果如下
 
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由图可能看出,在20s内(远小于400s)透过酶与半乳糖苷酶的浓度就降到很低了,这说明,pH诱导产生的抑制作用对于产生乳酸来说可以认为是瞬时的,即:抑制作用可能快速响应pH的变化
 
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最后我们将pH诱导部分与产生乳酸部分结合在一起
 
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当n=1时模拟结果如下
 
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由此,我们可以看出,乳酸的产量并没有减小,只是产生的速率变慢了(受到的抑制)到达平台的时间变为800了
 
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当n=30时,结果如下
 
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此时,乳酸的产量达到最大时的时间变成了100s左右。
 
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下面就来分析一下为何抑制作用反而成了加速率作用。
 
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上图为透过酶与半乳糖苷酶浓度的变化,半乳糖苷酶在整个过程中都是受到抑制,而透过酶则是先锐减然后又增加,(假设透过酶与半乳糖苷酶的翻译与降解速率都相同)这使得胞内的乳糖浓度因积累而增加(分解小于运入)。
 
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乳糖的积累,反而导致了转化成乳酸的速率增加,(这里解释一下,乳糖的积累是由于转运速率 与分解速率不相同所致,乳酸产生速率增加是因为胞内乳糖的积累从而使得产生葡萄糖的速率增加,进而使乳酸生成速率变大)从而使得反应到达平台的时间变短。
 
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综上所述,pH诱导产生的抑制作用并不能改变乳酸的产量,只是改就了乳酸的生成速率,如果pH启动子对酸不敏感,则响应pH变化就慢,相当于对乳酸的生成起了缓冲作用,如果pH启动子对酸敏感,则响应pH变化就快,但因为其对透过酶与半乳糖苷酶的抑制作用并不相同,最终导致了生成乳酸速率的加快。
 
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Revision as of 04:00, 28 September 2013

HIT-Harbin

Application

Overview for our application

In applications, we plan to enhance Bio-electric Interface, the device designed by 2012 Edinburg iGEM team. If the input is biochemical signal molecules and the output become electrons, B-POM can be coupled with Bio-electric Interface, that ephemeral processes in cells will be precisely measurable by simple electronic methods and analyzable by computer. B-POM is also helpful in yoghurt producing, where the control of pH is inaccurate. When B-POM is transplanted into yoghurt-producing bacteria, let hydrogen ions be the input and lacR be the output, and select the proper B-POM parameters, then the pH will be steady around 5.5. In short, our inspired and reliable device is promising in various fields.