Comment On The Use Of Selection Rate Versus Relative Fitness
In most of our research with the long-term lines, we report changes in performance in terms of relative fitness, W. Relative fitness is a dimensionless quantity, which is calculated as the ratio of the growth rate of one strain relative to that of another during their direct competition.
In some cases, however, it is preferable to express performance in terms of selection rates, r, which have units of inverse time. This formulation is useful for (i) certain theoretical purposes (see Lenski et al. 1991); (ii) when one competitor is much less fit than the other (Travisano and Lenski 1996); or (iii) when one or both competitors are declining in abundance, such as during competition assays under starvation conditions or in the presence of an antibiotic (see below).
The following text (extracted from a letter of mine to a colleague) explains the issues in this last context.
Illustrative calculations of W and r in the context of two growing populations
We grow up two strains, A and B, separately to a density of ~4 x 10^{9} cells/ml. We take 0.05 ml of each and add them to 9.9 ml of fresh medium. So the density of each strain at time 0 in the competition experiment is ~2 x 10^{7} cells/ml. Given an appropriate dilution (two times 100-fold) and then plating 0.1 ml, let us say that we find that our sample yields 192 colonies of A and 204 colonies of B. Now we let the mixed populations grow and compete for one day. At the end of this period, the total density has grown back to ~4 x 10^{9} cells/ml. We make an appropriate dilution (now three times 100-fold) and we plate 0.1 ml. We find that our sample yields 319 colonies of A and 107 colonies of B. Here are the relevant calculations.
A(0) |
= Estimated density of A at time 0 = 192 x 10 x 100 x 100 = 1.92 x 10^{7} cells/ml |
B(0) |
= Estimated density of B at time 0 = 204 x 10 x 100 x 100 = 2.04 x 10^{7} cells/ml |
A(1) |
= Estimated density of A at time 1 day = 319 x 10 x 100 x 100 x 100 = 3.19 x 10^{9} cells/ml |
B(1) |
= Estimated density of B at time 1 day = 107 x 10 x 100 x 100 x 100 = 1.07 x 10^{9} cells/ml |
r |
= {ln[A(1)/A(0)] - ln[B(1)/B(0)]}/day = [ln(166.15) - ln(52.45)]/day = (5.113 - 3.960)/day = 1.153 per day |
That is, over the course of one day of competition, the density of A increased by about 1.1 natural-logs more than did the density of B.
By the way, we can define mA = realized Malthusian parameter = ln[A(1)/A(0)]/day for strain A, with mB defined similarly. Hence, the selection rate constant, r, is equal to the difference in the two strains' Malthusian parameters during direct competition.
To convert this selection rate constant into a relative (Darwinian) fitness, W, we would use the ratio of their Malthusian parameters instead of the difference.
W |
= mA/mB = (5.113/day)/(3.960/day) = 1.29 |
That is, during the competition, A increased at a rate about 29% faster than did B. Notice that W is a dimensionless quantity, because the inverse time units cancel in the numerator and in the denominator.
Approximate conversion of r into W
We can also calculate W from r, as follows, but this is only an approximation.
m |
= average Malthusian parameter = ln[N(1)/N(0)]/day ==> where N(1) = A(1) + B(1) and N(0) = A(0) + B(0) = ln[(4.26 x 10^{9})/(3.96 x 10^{7})]/day = 4.678/day |
That is, the total population density increased by about 4.7 natural-logs during the one-day competition experiment.
The following conversion is only an approximation.
W |
= 1 + r/m = 1 + (1.153/day)/(4.678/day) = 1.246 |
If the total population density increases by the same factor as the dilution, d, used to start the competition experiment, then m = ln d. (This might happen if the experiment begins and ends with populations in stationary phase in the same medium.) In our hypothetical example, each population was diluted 200-fold to begin the experiment. So d = 1/(1/200 + 1/200) = 100 and ln d = 4.605/day. Note that this is very close to m, but not usually identical: d is a theoretical expectation, whereas m takes into account the actual data.
I usually prefer to calculate W directly (as above) rather than to use these approximations.
So, we have two measures, r and W, that are clearly related. The former, r, is a difference in Malthusian parameters and has units of inverse time; the latter, W, is a ratio of Malthusian parameters and is a dimensionless quantity. I usually prefer to use W because I find it easier to compare numbers when I don't have to think about the different time units that might be used in different experiments (hours, days, or whatever).
However, W does not make any sense if one or both competitors decline in density during the competition experiment. We don't have to worry about this in most of our experiments, but in your work with antibiotics it is a very common situation.
Illustrative calculation of r when both populations are declining
Let's imagine now another hypothetical where both competitors declined, but not to the same extent. Let's say that the plate counts gave us the following estimates of the initial and final densities for competing strains A and B.
A(0) = 1.54 x 10^{7} cells/ml A(1) = 7.73 x 10^{5} cells/ml B(0) = 2.08 x 10^{7} cells/ml B(1) = 9.55 x 10^{2} cells/ml
So, calculating the Malthusian parameters, we obtain
mA |
= ln[(7.73 x 10^{5})/(1.54 x 10^{7})]/day = -2.992/day |
mB |
= ln[(9.55 x 10^{2})/(2.08 x 10^{7})]/day = -9.989/day |
In other words, A declined by about 3 natural-logs, whereas B declined by almost 10 natural-logs. Expressing this as a selection rate constant,
r |
= (-2.992/day) - (-9.989/day) = 6.997/day |
Thus, A had an advantage of about 7 natural-logs over B.
However, if we express this as a relative fitness, it makes no sense.
W |
= (-2.992/day)/(-9.989/day) = 0.300 |
This W would seem to imply that A was less fit than B, but that is clearly wrong. And if one strain declines, while the other increases, you will obtain negative values for W, which also do not make sense.
So never use W if either or both populations decrease in density, but instead use r.