Team:HIT-Harbin/Application

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<h5 class="hashed"><span>Overview for our application</span></h5>
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<h5 class="hashed"><span>1 lactobacillus: from lactose to lactic acid</span></h5>
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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>
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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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<p>Reaction 1: endocytosis transportation of lactose
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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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<p>Table 1 the definition variables
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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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<p>The unit of concentration is mmol/L.
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L+LZ <=> temp2 : R3(乳糖与半乳糖苷酶结合)
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<p>LS+Lout <=> temp1 : R1(lactose combines with transporting enzyme)
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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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<p>vf = k1*Lout*LS(the reaction rate is proportional to reactants’ concentrations)
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vr = k_3*temp2(逆反应速率与产物成正比)
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<p>vr =k_1*temp1(the reverse reaction rate is proportional to product’s concentration)
 +
<p>temp1 => L+LS : R2(lactose is transported into the cell)
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<p>vf = k2*temp1(the reaction rate is proportional to reactants’ concentrations)
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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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<p>Lout is known as 136mmol/L. We estimate that the time for the overall transportation of lactose into cells is 400s and assume that k1=0.1,k_1=0.05,k2=0.1,LS=5,temp1=0,L=0。Then the simulation result is as below:
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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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<img src="https://static.igem.org/mediawiki/2013/1/1d/0001.png" alt=""/>
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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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<p>Fig 1. Simulation of lactose transportation
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PA => LA : R6(丙酮酸加氢生成乳酸)
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<p>We can get from the simulation that the transportation is completed when time is 300s.
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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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<p>Reaction 2: from lactose to lactic acid
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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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<p>L+LZ <=> temp2 : R3(lactose combines with galactosidase)
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LA=0(为乳酸的浓度)
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<p>vf = k3*L*LZ(the reaction rate is proportional to reactants’ concentrations)
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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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<p>vr = k_3*temp2(the reverse reaction rate is proportional to product’s concentration)
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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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<p>temp2 => Glu+LZ+gala : R4(lactose decomposes into glucose and galactose)
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k5 = 0.1(葡萄糖生成丙酮酸的速率常数)
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<p>vf = k4*temp2(the reaction rate is proportional to reactants’ concentrations)
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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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<p>Glu => 2*PA : R5(glucose decomposes into pyruvic acid, multi-step reaction)
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模拟结果如下:
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<p>vf = k5*Glu(decomposition rate is proportional to glucose concentration)
 +
 
 +
<p>PA => LA : R6(from pyruvic acid to lactic acid)
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<p>vf = rm1*PA/(Km1+PA)(Michealis equation, the reactant is adequate)
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<p>the simulation result is as below:
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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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<p>Fig 2. Simulation of lactic acid produciton
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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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<p>We can get from the simulation that all lactose becomes lactic acid. The final concentration of lactic acid is 272 mmol/L.
 +
 
 +
<p>Then we consider reaction1 and reaction2 together, the simulation result is as below:
   
   
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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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<p>Fig 3. Considering reacton1 and 2 together
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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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<p>We can see that once lactose enters cells, it is consumed, and the transportation of lactose is faster than the production of lactic acid, which ensure that the production of lactic acid is continuing.
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<p>The cells are assumed in normal condition.
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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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<p>Now we are going to analyze the pH switching function f(p)=1-tanh(n*(p-p*)). The curve of f(x)=1-tanh(x) is as below:
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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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<p>Fig 4. Curve of f(x)=1-tanh(x)
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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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<p>When x=1, f(x) is nearly 0, the switch is off. That is, when n=1, the difference between pH in cell and pH threshold should be smaller than 1 that the switch can be on.
 +
 
 +
<p>The curve of f(x)=1-tanh(30*x) is as below:
   
   
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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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<p>Fig 5. Curve of f(x)=1-tanh(30*x)
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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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<p>When x=0.1, f(x) is nearly 0, the switch is off. That is, when n=30, the difference between pH in cell and pH threshold should be smaller than 0.1 that the switch can be on.  
 +
 
 +
<p>Besides, we get the relationship between pH and lactic acid concentration from experiments:
 +
<p>pH=6.32311-0.02168*[LA]+0.0000426042*[LA]^2(0<[LA]<250)
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<p>its curve is as below:
   
   
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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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<p>Fig 5. Relationship between pH and lactic acid concentration
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2 用pH诱导的启动子表达LacR蛋白
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<p>In experiments, the maximum of pH is 6.3, the minimum is 3.6. We assume that the threshold of pH is 5.5.
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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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<h5 class="hashed"><span>2 the expression of LacR through pH-induced promoter</span></h5>
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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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<p>=> LA
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vf = vm-dm2*mRNA2-c*a*LR/(1+a*LR)(转录速率)
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<p>vf = 0
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  <p>=> mRNA1
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<p>vf = v0+v1*f(LA)-dm1*mRNA1
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  => LS : R11(透过酶的生成)
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  <p>=> LR
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vf = ks*mRNA2-ds*LS
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<p>vf = km*mRNA1-dn1*LR
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  => LZ : R12(半乳糖苷酶的生成)
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  <p>=> mRNA2
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vf = kz*mRNA2-dz*LZ
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<p>vf = vm-dm2*mRNA2-c*a*LR/(1+a*LR)  
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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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<p>=> LS
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LR(0) = 10 (LACR初始浓度)
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<p>vf = ks*mRNA2-ds*LS
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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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<p>=> LZ
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mRNA1(0) = 1 (LacR的mRNA的初始浓度)
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<p>vf = kz*mRNA2-dz*LZ
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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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<p>Table 3 the value of variables
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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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<p>the simulation result is as below:
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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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<p>Fig 6. Simulation of pH induction
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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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<p>Form the figure we can see that in 20s (far from 400s) the concentration of transporting enzyme and galactosidase is very small, which means the inhibition resulted from pH induction is transitory for production of lactic acid. In other word, the inhibition can respond to the fast change of pH.
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当n=1时模拟结果如下
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 +
<p>Finally, we put pH induction and lactic acid production together.
 +
<p>When n=1, the simulation is as below:
   
   
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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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<p>Fig 7. Taking pH induction and lactic acid production together (n=1)
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当n=30时,结果如下
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<p>We can see that the concentration of lactic acid doesn’t decline but only slow down (the inhibition). Time to achieve platform phase increases to 800s.
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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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 +
<p>When n=30, the simulation is as below:
   
   
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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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<p>Fig 8. Taking pH induction and lactic acid production together (n=30)
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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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<p>The time to achieve platform phase decreases to 100s.
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Latest revision as of 04:14, 28 September 2013

HIT-Harbin

Application

1 lactobacillus: from lactose to lactic acid

Reaction 1: endocytosis transportation of lactose

Table 1 the definition variables

The unit of concentration is mmol/L.

LS+Lout <=> temp1 : R1(lactose combines with transporting enzyme)

vf = k1*Lout*LS(the reaction rate is proportional to reactants’ concentrations)

vr =k_1*temp1(the reverse reaction rate is proportional to product’s concentration)

temp1 => L+LS : R2(lactose is transported into the cell)

vf = k2*temp1(the reaction rate is proportional to reactants’ concentrations)

Lout is known as 136mmol/L. We estimate that the time for the overall transportation of lactose into cells is 400s and assume that k1=0.1,k_1=0.05,k2=0.1,LS=5,temp1=0,L=0。Then the simulation result is as below:

Fig 1. Simulation of lactose transportation

We can get from the simulation that the transportation is completed when time is 300s.

Reaction 2: from lactose to lactic acid

L+LZ <=> temp2 : R3(lactose combines with galactosidase)

vf = k3*L*LZ(the reaction rate is proportional to reactants’ concentrations)

vr = k_3*temp2(the reverse reaction rate is proportional to product’s concentration)

temp2 => Glu+LZ+gala : R4(lactose decomposes into glucose and galactose)

vf = k4*temp2(the reaction rate is proportional to reactants’ concentrations)

Glu => 2*PA : R5(glucose decomposes into pyruvic acid, multi-step reaction)

vf = k5*Glu(decomposition rate is proportional to glucose concentration)

PA => LA : R6(from pyruvic acid to lactic acid)

vf = rm1*PA/(Km1+PA)(Michealis equation, the reactant is adequate)

the simulation result is as below:  

Fig 2. Simulation of lactic acid produciton

We can get from the simulation that all lactose becomes lactic acid. The final concentration of lactic acid is 272 mmol/L.

Then we consider reaction1 and reaction2 together, the simulation result is as below:

Fig 3. Considering reacton1 and 2 together

We can see that once lactose enters cells, it is consumed, and the transportation of lactose is faster than the production of lactic acid, which ensure that the production of lactic acid is continuing.

The cells are assumed in normal condition.

Now we are going to analyze the pH switching function f(p)=1-tanh(n*(p-p*)). The curve of f(x)=1-tanh(x) is as below:

Fig 4. Curve of f(x)=1-tanh(x)

When x=1, f(x) is nearly 0, the switch is off. That is, when n=1, the difference between pH in cell and pH threshold should be smaller than 1 that the switch can be on.

The curve of f(x)=1-tanh(30*x) is as below:

Fig 5. Curve of f(x)=1-tanh(30*x)

When x=0.1, f(x) is nearly 0, the switch is off. That is, when n=30, the difference between pH in cell and pH threshold should be smaller than 0.1 that the switch can be on.

Besides, we get the relationship between pH and lactic acid concentration from experiments:

pH=6.32311-0.02168*[LA]+0.0000426042*[LA]^2(0<[LA]<250)

its curve is as below:

Fig 5. Relationship between pH and lactic acid concentration

In experiments, the maximum of pH is 6.3, the minimum is 3.6. We assume that the threshold of pH is 5.5.

2 the expression of LacR through pH-induced promoter

=> LA

vf = 0

=> mRNA1

vf = v0+v1*f(LA)-dm1*mRNA1

=> LR

vf = km*mRNA1-dn1*LR

=> mRNA2

vf = vm-dm2*mRNA2-c*a*LR/(1+a*LR)

=> LS

vf = ks*mRNA2-ds*LS

=> LZ

vf = kz*mRNA2-dz*LZ

Table 3 the value of variables

the simulation result is as below:

Fig 6. Simulation of pH induction

Form the figure we can see that in 20s (far from 400s) the concentration of transporting enzyme and galactosidase is very small, which means the inhibition resulted from pH induction is transitory for production of lactic acid. In other word, the inhibition can respond to the fast change of pH.

Finally, we put pH induction and lactic acid production together.

When n=1, the simulation is as below:

Fig 7. Taking pH induction and lactic acid production together (n=1)

We can see that the concentration of lactic acid doesn’t decline but only slow down (the inhibition). Time to achieve platform phase increases to 800s.

When n=30, the simulation is as below:

Fig 8. Taking pH induction and lactic acid production together (n=30)

The time to achieve platform phase decreases to 100s.