# Cell cycle analysis

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(→Determination of the C and D periods:) |
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This gives: | This gives: | ||

- | '''<math>a_i=\tau-\frac{log(2-F)}{log2}*\tau</math>''' | + | '''<math>a_i=\tau-\frac{log(2-F)}{log2}*\tau</math>''' |

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+ | which is the same as this (log2 is 1): | ||

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+ | '''<math>a_i=\tau-log(2-F)*\tau</math>''' | ||

If you have for example a generation time τ=84 minutes and the portion of cells with 4 origins is 66% the formula gives: | If you have for example a generation time τ=84 minutes and the portion of cells with 4 origins is 66% the formula gives: | ||

- | '''<math>a_i=84- | + | '''<math>a_i=84-log(2-0.66)*84=48.5</math>''' |

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[[Image:C+D_1.jpg|left|250px]] | [[Image:C+D_1.jpg|left|250px]] | ||

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===Determination of the C and D periods:=== | ===Determination of the C and D periods:=== | ||

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'''<math>\frac{oriC}{terC}=2^{\frac{C}{\tau}}</math>''' | '''<math>\frac{oriC}{terC}=2^{\frac{C}{\tau}}</math>''' | ||

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which gives: | which gives: | ||

- | '''<math>\frac{oriC}{terC} | + | |

+ | '''<math>C=log_2(\frac{oriC}{terC})*{\tau}</math>''' | ||

## Current revision

**Cell cycle analysis of Escherichia coli cells**

C period = the time for a round of chromosome replication

D period = the time between the end of a round of chromosome replication and cell division

## Contents |

### Determination of initiation age (a_{i}) and C+D:

From flow cytometry analysis of cells treated with rifampicin and cephalexin (run-out histogram) the proportions of cells that had not initiated replication at the time of drug action (4-origin-cells, streaked) and cells that had initiated (8-origin-cells) can be estimated.The initiation age (a_{i}) can be found from the theoretical age distribution described by this formula,

where F is the fraction of cells that had not initiated and τ is the generation time, or from the estimated graph of the theoretical age distribution (streaked portion).

This gives:

which is the same as this (log2 is 1):

*a*_{i} = τ − *l**o**g*(2 − *F*) * τ

If you have for example a generation time τ=84 minutes and the portion of cells with 4 origins is 66% the formula gives:

*a*_{i} = 84 − *l**o**g*(2 − 0.66) * 84 = 48.5

The C+D period is estimated from the initiation age (a_{i}), the generation time (τ) and the number of generations spanned per cell cycle.

Example:

4-origin-cells: 23 %

Generation time (τ): 27 min

Initiation age (a_{i}): 5 min

### Determination of the C and D periods:

The C period is found from the *oriC/terC* ratio obtained by Southern blot or qPCR analysis (oriC/ter ratio determination) and the generation time (τ):

which gives:

The D period is found from the C+D and C period:

*D* = (*C* + *D*) − *C*

Example (continues):

C period calculated from the *oriC/terC* ratio: 49 min

D period = (C+D) – C

D period = 76 min – 49 min = 27 min

### The theoretical exponential DNA histogram:

A theoretical exponential DNA histogram can be drawn to check whether the obtained values fit with the experimental data. From the C+D period the DNA content of the cells at different time points in the cell cycle can be calculated.

Example:

The individual values of C and D can be varied

to obtain a shape of the theoretical histogram

that gives the best fit to the experimental histogram.

### Calculation of the average number of replication forks when D=τ:

In the example given above, 23% of the cells contain 4 replication forks (4-origin peak in run-out histogram) and 77% contain 12 replication forks (8-origin peak), hence the average number of replication forks in the cell population will be:

(4 x 0.23) + (12 x 0.77) = 10.2 forks

### Calculation of the average number of replication forks when D≠τ:

Example:

4-origin-cells: 23%

8-origin-cells: 77%

τ = 27 min

a_{i} = 5 min

C = 51 min

D = 25 min

C+D = 76 min

12 forks → 8-origin peak in run-out histogram = 77% of the cells

6 and 4 forks → 4-origin peak in run-out histogram = 23% of the cells

The fraction of cells containing 6 forks: F = 2 - 2^{((τ-at)/τ)} = 2 – 2^{((27-2)/27)} = 0.10

The fraction of cells containing 4 forks: 0.23 – 0.10 = 0.13

The average number of replication forks: (6 x 0.10) + (4 x 0.13) + (12 x 0.77) = 10.4 forks