# Team:Lethbridge Canada/math

### From 2013hs.igem.org

# Contents:

# The Goal:

By mathematically modeling the maximum rate at which DH5α can produce proteins, the correct ratio of enzyme to protein can be determined. One group of cells is built to maximally express the gene of the oxytocin-neurophysin I protein. With respect to protein levels, the amount of enzyme required to completely cut each oxytocin-neurophysin I can be determined. The aim of this is to optimize the expression process to have exactly the amount of enzyme required and not produce an excess or too little. It is important to note that the model determines the theoretical maximum of protein production, not the experimental value.

# Bacterial Growth:

The purpose of including bacterial growth is so the amount of proteins produced per individual cell can be multiplied by the number of DH5α cells present to calculate the total number of proteins produced in that time interval of 1 minute. In the bacterial growth expression, the two variables are represented as such:

*b*= initial number of DH5α bacteria present_{i}*t*= time (manipulated variable)

We assumed a doubling time of 30 minutes.

# mRNA Production:

## Transcription:

The expression below calculates the amount of mRNA that is produced by an individual cell in the time period of 1 minute. Credits given to the 2009 Beijing iGEM team for the average speed of transcription in *E. coli*.

- 4200nt/min = the maximum rate at which DNA is transcribed in to mRNA in
*E. coli* *l*= gene length in (from RBS to the double terminator)_{tgene}

The resultant number is the number of mRNA strands produced per minute in one cell.

## mRNA Decay:

The stability of mRNA varies, but it will always decay. Half-lives can rage from seconds to hours.

*t*= stagnant time of mRNA between transcription and translation_{s}*t*= half-life_{h}

The above expression will act as a coefficient for how much mRNA will remain after the amount of time that mRNA sits stagnant in the cells between the processes of transcription and translation. However, we have assumed that immediately or very shortly after transcription, the mRNA then undergoes translation. The following logic describes this situation:

Because the exponent of the function is approximately or equal to 0, the entire expression for decay is equal to 1. The numerator of the exponent is essentially 0, so nearly any denominator would give the entire exponent a value of 0 when simplified. Although this logic nullifies the purpose of this term, it is important to note that mRNA decay is existent.

## Transcription and mRNA Decay Working Together:

When combining the previous two expressions, we can determine the amount of mRNA produced by one cell per minute.

# Total mRNA:

Expanding on the previous expression, we get the amount of total amount of mRNA produced by the total number of cells existing at any given time. This is achieved by multiplying the amount of mRNA produced by a single cell by the amount of cells that currently exist.

# Translation:

The following expression models the maximum amount of proteins produced per minute.

- 2400Aa/min = the rate at which DNA is transcribed in to mRNA in
*E. coli* *l*= length of the protein to be expressed (Aa)_{protein}

Credits given to the 2009 Beijing iGEM team for the average speed of translation in *E. coli*.

## Deriving the Translational Constant:

By dividing the expression for translation by the expression for transcription, we can determine the rate at which one strand of mRNA can be translated in to one protein.

Because there are three nucleotides (also called a codon) of mRNA for every one amino acid, the gene length (nt) over the protein length (Aa) will always be equal to 3.

By simplifying this expression, we get the following term that is the rate of translation from mRNA.

We used the RBS BBa_B0034 which has an efficiency index of 1.0.

# Protein Production:

By taking the expression from Total mRNA section and incorporating the Translational Constant section we can determine the total protein production. By multiplying the amount of mRNA produced by the rate at which mRNA is translated in to protein, we can calculate the amount of protein produced. We do this by finding the area under the curve with respect to time.

# Assembling the Equation:

Bacteria donâ€™t double forever. They will reach a maximum population for their environment size. For this reason, to accurately model protein production we need to make a piece-wise function. We let *j* equal the amount of time that it takes for DH5α to reach their maximum population size. Before time *j* we use the expression for start time *i* up to time *j*. Because after time *j*, DH5α no longer double, we replace the time variable with *j* so the graph from this point will be a horizontal line parallel to the time-axis. Time *k* is the end time of the protein production period. The area under the curve from *i* to *k* is the total amount of protein expressed (*n _{p}*).

When plotted, the function creates a graph that looks as follows:

The graph illustrates that at time *i* the protein output is zero and increases until time *j*. This uses the first part of the piece-wise function. From time *j* to time *k* (end time) the graph is constant and deals with the second part of the piece-wise function.

# Values

Variable | Description |
---|---|

b_{i} |
Initial number of E. coli |

i |
Start time (t = 0) |

j |
Time it takes for E. coli to reach its maximum population size |

k |
End time |

l_{gene} |
Length of only gene to be expressed (nt) |

l_{tgene} |
Length of gene from RBS to double terminator (nt) |

l_{protein} |
Length of protein (Aa) |

n_{p} |
Number of proteins produced |

t |
Elapsed time |

t_{h} |
Half-life of mRNA |

t_{s} |
Time mRNA sits in the cell between transcription and translation |

RBS | RBS efficiency (B0034 = 1.0) |

Preprooxyphysin | Value |

l_{gene} |
403bp |

l_{tgene} |
418bp |

NEC I | Value |

l_{gene} |
2277bp |

l_{tgene} |
2289bp |