(With help from Shenzhen_SFLS)
We have acknowledged that one Ab with wAg + one Ag coming into one Ab with Ag and one wAg is irreversible and the amount of this kind of changes is related to the amount of Ab with wAg(x2) , the amount of Ag (x1) and time.
According to Volterra Model,we can know that ‘dx1/dt=dx2/dt=-k*x1*x2’.
k is a regular value which is related to affinity(f) between one Ab with wAg and one Ag which is defined as p,so we can attain an another equation:’k=pf’.
According to the first equation above,we can know that ‘x2=x1+m’.
We assume that the inchoate value of x1 is x0.
Difference between revisions of "Team:SZMS 15 Shenzhen/Collaborations"
| Line 2: | Line 2: | ||
<head> | <head> | ||
<meta charset="UTF-8"> | <meta charset="UTF-8"> | ||
| − | |||
<link href="https://2015.igem.org/Template:SZMS_15_Shenzhen/indexsccss?action=raw&ctype=text/css" rel="stylesheet" type="text/css"> | <link href="https://2015.igem.org/Template:SZMS_15_Shenzhen/indexsccss?action=raw&ctype=text/css" rel="stylesheet" type="text/css"> | ||
<title>Title</title> | <title>Title</title> | ||
Revision as of 03:23, 18 September 2015
Collabotations
‘dx1/dt=dx2/dt=-k*x1*x2’
’k=pf’
‘x2=x1+m’
‘x1(0)=x0’
And the program code of matlab is:
dsolve('Dx1=-p*f*x1*(x1+m)','x1(0)=x0','t')
ans =
m/(exp(m*(log((m + x0)/x0)/m + f*p*t)) - 1)
x1= m/(exp(m*(log((m + x0)/x0)/m + f*p*t)) - 1)
We can define the extend of combination as E.
‘E=(x0-x1)/x1’ which is simplified as ‘E=1- m/((exp(m*(log((m + x0)/x0)/m + f*p*t)) - 1)*x0)’
’k=pf’
‘x2=x1+m’
‘x1(0)=x0’
And the program code of matlab is:
dsolve('Dx1=-p*f*x1*(x1+m)','x1(0)=x0','t')
ans =
m/(exp(m*(log((m + x0)/x0)/m + f*p*t)) - 1)
x1= m/(exp(m*(log((m + x0)/x0)/m + f*p*t)) - 1)
We can define the extend of combination as E.
‘E=(x0-x1)/x1’ which is simplified as ‘E=1- m/((exp(m*(log((m + x0)/x0)/m + f*p*t)) - 1)*x0)’