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As written in my lecture handouts, there two main factors in the Geometric Growth Model of populations:

$R_{0}$ is the expected lifetime reproductive output. This way, for unicellulars, for example, when time between division represents life time, if there is no mortality, $R_{0}$ is calculated as $R_{0}=1*2$, where 1 is 100% and 2 is the amount of daughter cells expected to be produced as a result of the mother cell split. We also define that for this isolated incident, $B$, the amount of births is 2, while the mother cell "dies", which leads to $D=1$.

The second factor is $\lambda$, which is the finite rate of increase, which in other words means that this is average per-capita multiplication factor per one time-step. When we measure population growth in time-steps of 1 lifetime, we can conclude that $\lambda=R_0$. We actually look at $\lambda$ as at $\frac{N_{t+1}}{N_{t}}$, where $N_{t+1}=N_{t}+B-D$(1) and $N_t$ is the amount of individuals in the populations at time-step $t$. This way we define $\lambda$ using $b=\frac{B}{N_{t}}$ and $d=\frac{D}{N_{t}}$: $\lambda=\frac{N_{t+1}}{N_{t}}=\frac{N_{t}}{N_{t}}+\frac{B}{N_{t}}-\frac{D}{N_{t}}=1+b-d$.

All fine when we deal with ideal conditions, where all mother cells divide and there are no mortalities or mutations.

But suppose we're told that only 80% of the mother cells will divide, and the remaining 20% will die without division. In this case: $R_{0}=0.20*0+0.80*2$.

What I'm trying to understand, how will $\lambda$ be affected by this? Does $\lambda$ refers to ideal conditions only, or it depends on the natural situation?

(1) We ignore Immigration and the Emigration at this point.

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2 Answers

up vote 2 down vote accepted

*It's been several years since I've worked with similar equations. The following reply is based on memory, and if anyone has a firmer grasp of the materials, please modify or answer as you see fit.

I would assume λ is under ideal conditions or as an average of whatever species you're working with. Your modified lifetime reproductive output still leaves out many variables which can affect λ. By your own words it leaves out mutations and daughter cells that are dead-on-arrival. There could also be predation or conditions where the Mother-cell produces more than one Daughter-cell, resulting in R = 3 instead of 2 and your subsequent λ also being 3. UCLA Bioengineers have actually observed up to 5 daughter cells in cancer lines.

I'm not sure what you mean by "... or it depends on the natural situation?", but ultimately reality is always more complicated than basic equations usually describe. Either B or D can increase or decrease, depending on the species, environment, and how you define t.

Your equation still produces a population increase, just at a slower rate than your ideal conditions. If you're asking if there's a specific definition for "λ" referenced in your equation, I'm not aware of any specific definition for "λ" - it's not a Constant.

Your title question - does the rate of increase depend on mortality - is ultimately "Yes." If mortality is higher than the birth rate, then you will see a decrease of the population until extinction is reached if there isn't an equilibrium or rebound.

However, most populations change at an Exponential rate - NOT Geometric. To that end, you might be interested in the r/K Selection Theory and how they operate within Exponential equations (which are similar to your Geometric equations). In short, r-Selection is the reproductive strategy of producing extraordinary amounts of offspring with a very, very low survival rate. However, because so many are produced you only need a small overall percentage of survival to maintain or increase populations. The tapeworm I mentioned above is one of the stereotypical examples. K-Selection represents species that invest additional resources into raising a few offspring, usually just a handful that are carefully watched over and raised by the parent(s) until they are capable of reproducing themselves. Humans, elephants, most birds, and larger animals in general are K-selected.

For r-Selection your B and D will both be very high, and for K-Selection both B and D will be low. However, both will result in a positive λ simply because it's all about the ratios.

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Thanks for your answer. I will look into r/K Selection Theory. –  Khaloymes Dec 10 '12 at 7:09
r/K selection is just as relevant for populations with discrete growth. However, the whole concept of r/K selection has also been challenged - for an overview see Reznick D, et al. 2002. r-and K-selection revisited: the role of population regulation in life-history evolution. Ecology, 83 (6): 1509–1520 and this nice blogpost Oikos blog: Zombie ideas in ecology: r and K selection. –  fileunderwater May 29 '13 at 9:23
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The same relationship still holds, and $\lambda$ just describes the finite population growth rate under a particular set of conditions. 'Ideal' conditions (whatever that means) are not implied. Generally you can write: $\lambda = e^r = R_0^{1/G}$, where $G$ is the (average) generation time. However, $r$ (the instantaneous/intrinsic rate of increase) is not really relevant for an organism with discrete reproductive events (i.e. geometric growth)

So for your second example where $R_0$=1.6, assuming that $G$=3 timesteps, $\lambda$= 1.169. However, as hinted above an exponential model is generally a better fit for unicellular organisms.

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