Team:Heidelberg/internal tools
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Team:Heidelberg/internal tools
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From 2014.igem.org
Internal Tools
Number of mutations and mutated sequences
Expected number of mutations in a single sequence: $$p_{m} = \frac{N_{mutations}}{L_{Sequence}} = N_{generations} \cdot r_{mutation} = t_{total} \cdot \Phi \cdot r_{mutation}$$
The expected share of sequences that shows at least one mutation in \(L_{Sequence}\) bp is the probability that \(L_{sequence}\) basepairs stay unchanged when \(\frac{N_{mutations}}{L_{Sequence}}\) mutations are expected: $$p_{M} = \frac{N_{mutated}}{N_{Sequences}} = 1 - p(N_{mutations}=0) = 1 - (1-p_{m})^{L_{Sequence}} $$
With this equation we can also calculate the number of sequences \(N_{Sequences}\) that have to be sequenced in order to find a mutated one with a probability of \(p(N_{mutated} > 0)\). $$ N_{Sequences} = \frac{p(N_{mutated} > 0)}{p_{M}} $$
The probability to find at least one mutated sequence under the given conditions is $$p(N_{mutated}>0) = 1 - (1-p_{M})^{N_{sequences}}$$ which gives $$N_{Sequences} = \frac{ln(1-p(N_{mutated}>0))}{ln(1-p_{M})}$$
Set \(\Phi\) to zero to use the number of generations for the calculation. If \(\Phi\) and the number of generations are given, \(\Phi\) is used.
Consider \(L_{Sequence}\) as the number of basepairs that is expected to be mutated. If half of the sequence you are interested in, is highly conserved choose a lower \(L_{Sequence}\).
\(p_{m} =\) %(bp/bp).
\(N_{mutations} =\) bp per sequence.
The share of sequences that shows at least one mutation in \(L_{Sequence}\) bp is \(p_{M}=\) % of sequences
Diff tool
Marks differences in two strings, ignores newlines.
Comparison:
Glucose Concentratoin
Calculate the ideal glucose concentration in the medium used for either a turbidostat or a single flask.
The glucose concentration in the turbidostat \(c_{G_{T}}\) is increased with the incoming medium with a flow rate of \(\Phi\) and a glucose concentration of \(c_{G_{M}}\). It is decreased by with the medium that leaves the turbidostat with the same flow rate, but a glucose concentration of \(c_{G_{T}}\). Additionally E. coli take up glucose with a concentration of \(c_{E}\) and a rate of \(q\). $$ \frac{\partial c_{G_{T}}(t)}{\partial t} = \Phi \cdot c_{G_{M}} - \Phi \cdot c_{G_{T}} - c_{E} \cdot q $$ In the case of a turbidostat we can assume a dynamic equilibrium: $$ \frac{\partial c_{G_{T}}(t)}{\partial t} = 0 $$ This results in $$ c_{G_{T}} = c_{G_{M}} - \frac{c_{E. coli} \cdot q}{\Phi} $$ $$ \Leftrightarrow c_{G_{M}} (c_{G_{T}}) = c_{G_{T}} + \frac{c_{E} \cdot q}{\Phi} $$
If the concentration of glucose in a flask, \(c_{G_{F}}\) needs to be determined, the functional dependencies are as follows.
As there is no incoming medium, or medium that leaves the flask, the concentration of glucose is only changed by E. coli degrading it. $$ \frac{\partial c_{G_{F}}(t)}{\partial t} = q \cdot c_{E}(t) $$ Exponential growth of the E. coli is assumed, resulting in $$c_{G_{F}}(t) = c_{G_{F}}(t_{0}) - q \cdot c_{E}(t) \cdot t $$ $$ = c_{G_{F}}(t_{0}) -q \cdot \int_{t_{0}}^{t} c_{E}(t_{0}) \cdot exp\left(\frac{t}{t_{E}}\right) dt $$ $$ = c_{G_{F}}(t_{0}) - q \cdot c_{E}(t_{0}) \cdot t_{E} \cdot \left(exp\left(\frac{t}{t_{E}}\right) - exp\left(\frac{t_{0}}{t_{E}}\right)\right) $$ So the glucose starting concentration \(c_{G_{F}}(t_{0})\) needed to get a concentration of \(c_{G_{f}}(t)\) afer a duration of \(t\) is calculated by $$ c_{G_{F}}(t_{0}) = c_{G_{F}}(t) + q \cdot c_{E}(t_{0}) \cdot t_{E} \cdot \left(exp\left(\frac{t}{t_{E}}\right) - exp\left(\frac{t_{0}}{t_{E}}\right)\right) $$
Further calculations for simplification of entering data:
$$
c_{E. coli_{DW}} = c_{E. coli_{OD600}} \cdot 0.36
$$
according to
Because turbidstats are operated at a constant cell density, the flow rate \(Phi\) can be calculated from the generation time \(t_{E}\). $$ \Phi = \frac{ln(2)}{t_{E}} $$
If the E. coli titer in \(g_{DW}/l\) is zero, it is calculated from the OD, else the dryweight value is used. If the glucose concentration in \(mmol/l\) not zero, it is used for the calulation. If the generation time \(t_{E}\) is not zero, it is used to calculate the flow rate \(\Phi\).
