Competition Assays for Evolvability Lines

The 96-well pin tool transfers 3.5 µl.

Serial transfer of 3.5 µl of culture to 900 µl of fresh medium achieves 8.0120 generations per day (assuming the 3.5 µl is transfered with the pin tool and it also takes 3.5 µl away after the transfer). This is a 258.1 x dilution.

0.01% of the carbon compound is used. For glucose, this allows growth to approximately 2 x 10⁸ cells per ml. Other compounds may not support this level of growth.

Materials

• 96-Well Pin Tool that transfers 3.5 µl.

Day 1: Reviving Frozen Stocks

Revive the strains to be tested from the freezer by growing them in 900 ml of the competition medium in deep 96-well plates. If using a mixed population, be sure to use at least 3.5 µl of inoculate from the frozen stock to ensure that the sample is representative of the population.

You will need two plates for each competitoon plate that you are creating, one for reviving each of the strains to be competed separately (for example, one for the evolved strains and one for the ancestor strains). Frozen plates of 606/607 in a checkerboard pattern are available to start the ancestor strains for competitions.

Grow 40-48 hours.

Day 3: Preconditioning Cultures

Prepare a deep 96-well plate with 100 µl of saline in each well using the multichannel repeat pipettor. Transfer 3.5 µl from the overnight in LB to this plate with a 96-well pin tool or multichannel pipettemen. Mix by stirring with the pins or pipetting up and down several times. This makes a 25-fold dilution that has a cell density of approximately 2 x 10⁸ cells/ml, which is roughly the saturating cell density achieved during these long term evolvability experiments.

Transfer 3.5 µl from this dilution plate to a new plate filled with 900 µl per well of the medium to be used in the competition using a 96-well pin tool or multichannel pipettor. This achieves approximately the density experienced by cells after a transfer. Grow exactly 24 hours.

Day 5: Begin Competition

Add 900 µl of DM0 to each well and mix to achieve a 2-fold dilution. (This is necessary since we want the total cell number of both strains we are competing to be approximately at what it would be during a normal day of growth, i.e. we need to achieve 8 generations under the exact conditions of evolution.)

Transfer 3.5 µl of each of the two strains to be competed to 900 µl of the medium to be used in the competition. Immediately make a dilution by taking 3.5 µl of each competitions into 900 µl of saline. Put the competition in the incubator at 37°C.

Plate 100 µl of the dilution on TA plates. This dilution should yield 100-500 cells. These counts give the initial frequencies of the two strains in the competition.

Day 6: Count Initial Plates

Day 7: End Competition

After exactly 24 hours of growth. Transfer 3.5 µl of each competitions into 900 µl of saline twice.Plate 100 µl of the dilution on TA plates. This dilution should yield 100-500 cells. These counts give the initial frequencies of the two strains in the competition.

Count the plates from Day 0.

Day 8: Count Final Plates

Count the plates

Variations

For measuring fitness values more precisely, you can continue to serially dilute for multiple days before plating. This is useful, for example, when showing that the mutation in an Ara+ revertant of an Ara- REL606-based strain is neutral. Beware that evolution can happen during longer competitions depending on how strong the selective pressures are.

Calculating Relative Fitness (W)

The relative fitness (W) of strains A relative to straind B is the ratio of their Malthusian parameters (M_A and M_B) over the course of a representative growth cycle.

N = cell number.
PC = plate count on TA.
DF = dilution factor of all transfers combined.
i and f are the initial and final time points.

M_A = N_A(f) / N_A(i) = PC_A(f) * DF / PC_A(i)

M_B = N_B(f) / N_B(i) = PC_B(f) * DF / PC_B(i)

W = M_A / M_B

Note, that there are problems with this measurement under conditions where: (1) One or both populations are declining in numbers over the course of the competition -- which would lead to negative W values -- or (2) There is a large difference in fitness between the two strains being tests. In these cases it is better to use selection rates (r) to measure fitness as discussed here.

This protocol and discussion are modified from the web pages of Richard Lenski

-- Main.JeffreyBarrick - 17 Sep 2007