Figure 2.5. Ray Beverton as a young man, delivering his famous lectures that began post-WWII quantitative fishery science, and at the time of his retirement. Photos courtesy of Kathy Beverton.

Population growth in fluctuating environments and measures of fitness

We now come to one of the most misunderstood topics in evolutionary ecology, although Danny Cohen and Richard Lewontin set it straight many years ago (Cohen 1966, Lewontin and Cohen 1969). I include it here because at my university in fall 2002, there was an exchange at a seminar between a member of the audience and the speaker which showed that neither of them understood either the simplicity or the depth of these ideas.

This section will begin in a deceptively simple way, but by the end we will reach deep and sophisticated concepts. So, to begin imagine a population without age structure for which N(t) is population size in year t and N(0) is known exactly. If the per capita growth rate is l, then the population dynamics are

from which we conclude, of course, that N(t) = lN(0). If the per capita growth rate is less than 1, the population declines, if it is exactly equal to 1 the population is stable, and if it is greater than 1 the population grows. Now let us suppose that the per capita rate of growth varies, first in space and then in time. Because there is no density dependence, the per capita growth rate can also be used as a measure of fitness.

Suppose that in every year, the environment consists of two kinds of habitats. In the poor habitat the per capita growth rate is 11 and in the better habitat it is 12. We assume that the fraction of total habitat that is poor is p so that the fraction of habitat that is good is 1 — p Finally, we will assume that the population is uniformly distributed across the entire habitat. At this point, I am sure that you want to raise various objections such as ''What ifp varies from year to year?'', ''What if individuals can move from poorer to better locations'', etc. To these objections, I simply ask for your patience.

Given these assumptions, in year t the number of individuals experiencing the poor habitat will be ^N(t) and the number of individuals experiencing the better habitat will be (1 — p)N(t). Consequently, the population size next year is

N (t + 1) = (VN (t)+l2(1 — pN (t)) = {pl + (1 — PMN (t) (2.18)

The quantity in curly brackets on the right hand side of this equation is an average. It is the standard kind of average that we are all used to

(think about how your grade point average or a batting average is calculated). If we had n different habitat qualities, instead of just two habitat qualities, and let pt denote the fraction of habitat in which the growth rate is 1t, then it is clear that what goes in the { } on the right hand side of Eq. (2.18) will be ^ni= 1 p-1,-. We call this the arithmetic average. (I am tempted to put ''arithmetic average'' into bold-face or italics, but Strunk and White (1979) tell me that if I need to do so - to remind you that it is important - then I have not done my job.) Our conclusion thus far: if variation occurs over space, then the arithmetic average is the appropriate description of the growth rate.

Let us now assume that per capita growth rate varies over time rather than space. That is, with probability p every individual in the population experiences the poorer growth rate in a particular year and with probability 1 — p every individual experiences the better growth rate. Let us suppose that t is very big; it will be composed of t1 years in which the growth rate was poorer and t2 years in which the growth rate was better. Since there is no density dependence in this model, it does not matter in what order the years happen and we write

If the total time is large, then t1 and t2 should be roughly representative of the fraction of years that are poorer or better respectively. That is, we should expect t1 ~ pt and t2 ~ (1 — p)t. How should you interpret the symbol ~ in the previous sentence? If you are more mathematically inclined, then the law of large numbers allows us to give precise interpretation of what ~ means. If you are less mathematically inclined, this is a case where you can count on your intuition and the world being approximately fair.

Adopting this idea about the good and bad years, Eq. (2.19) becomes

N (t) = 1pt li1—p)'N (0) = [lpl}2—pl\' N (0) (2.20)

The quantity in square brackets on the right hand side of this equation is a different kind of average. It is called the geometric mean (or geometric average) and it weights the good and bad years differently than the arithmetic average does. Perhaps the easiest way to see the differences is to think about the extreme case in which the poorer growth rate is 0. According to the arithmetic average, individuals who find themselves in the better habitat will contribute to next year's population and those who find themselves in the poorer habitat will not. On the other hand, if the fluctuations are temporal, then when a poor year occurs, there is no reproduction for the population as a whole and thus the population is gone.

Suppose that 11 is less than 1 (so that in poor years, the population declines). Show that the condition for the population to increase using the geometric mean is that 12 > ■p/(1 p). Explore this relationship as 11 and p vary by making appropriate graphs. (Do not use three dimensional graphs and recall the advice of the Ecological Detective (Hilborn and Mangel 1997) that you should expect to make 10 times as many graphs for yourself as you would ever show to others.) Compare the results with the corresponding expression making the arithmetic average greater than 1.

If instead of just two kinds of years, we allow n kinds of years, the extension of the square brackets in Eq. (2.20) will be i= i If where the denotes a product (much as denotes a sum, as used above).

Now let us return to Eq. (2.17) for which N(t) = lN(0) and recall that the exponential and logarithm are inverse functions, l = exp(log(l)), which allows us to write N(t) in a different way. In particular we have N(t) = e[log(1)]tN(0), and if we define r = log(l), then we have come back to our old friend from introductory ecology N(t) = ertN(0). That is, if time were continuous, this looks like population growth satisfying dN/dt = rN, in which r is the growth rate. But we can actually learn some new things about fluctuating environments from this old friend, because we know that r = log(l). In Figure 2.6a, I have plotted growth rate as a function of l and I have shown two particular values of l that might correspond to good years and poor years. Note that the line segment joining these two points falls below the curve (such a curve is called concave). This means that the growth rate at the arithmetic average of l is larger than the average value of the growth rates. This phenomenon is called Jensen's inequality.

If we have more than two growth rates, then the expression in square brackets in Eq. (2.20) is replaced by ni=1 If and if we rewrite this in terms of logarithms we see that

J2pt log(li)

From this equation, we conclude that the growth rate in a fluctuating environment is r = in 1 pi log(li ), which is the arithmetic average of the logarithm of the per capita growth rates. We thus conclude that for a fluctuating environment, one either applies the geometric mean directly

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