301. APS Co-operation & Conflict
Professor Francis L. W. Ratnieks Lecture 2 PSR Model & Possible Conflicts in the Origin of Life
PSR Equilibrium Model
Equilibrium Proportion of PSR
PSR can cause a fertilised egg to develop into a haploid male. Does this mean that PSR would take over a population, such that every male carries PSR? Or would it reach some lower level such that some males carry PSR and some do not? We will investigate this using a simple mathematical model to find the equilibrium proportion of PSR-carrying males. That is, a proportion which is constant across generations.
Equilibrium Proportion of PSR
Start by defining two parameters. Both are proportions, so can only take values from 0 to 1. f p proportion of eggs laid by females which are fertilised proportion of PSR+ males in the current generation
Also, we define some special cases of p p* proportion of PSR+ males in the next generation pe equilibrium proportion of PSR+ males From this we can determine what proportion of the males in the next generation carry PSR. PSR+ males arise from fertilised eggs produced by fusion of egg with PSR+ sperm. Males without PSR, PSR-, arise from unfertilised eggs. The next slides show how to determine the equilibrium.
Next Generation Males
The square represents the panmictic union of male and female gametes in the whole N. v. population.
Males (= sperm) PSR+ (p) PSR- (1-p)
A. Males Fertilised (f) Females (= eggs) C. Males Unfertilised (1-f) PSRPSR+
B. Females
D. Males PSR-
Next Generation Males
The proportion that are PSR+ in the next generation, p*, is p* = (Number PSR+ males)/(Number all males)
From the previous slide we can see that this is simply p* p* = = Box A/(Box A + Box C + Box D) or fp/(fp + 1 - f)
At equilibrium, p* = p = pe, so that pe = fpe/(fpe + 1 - f) = (2f - 1)/f
How Did You Do The Last Line?
One feature of most published models is that they do not include every step of the maths. In the previous slide, the last line could have been expanded as follows. As you can see, the maths was no more than basic algebra. pe 1 fpe + 1 - f fpe pe = = = = = fpe/(fpe + 1 - f) f/(fpe + 1 - f) f 2f - 1 (2f - 1)/f
divide both sides by pe
multiply both sides by (fpe + 1 - f) move 1 - f to right hand side divide both sides by f
PSR Equilibrium
The equilibrium, pe = (2f - 1)/f, is the mathematical result. We now have to translate it back into biology. It is simple to plug in some values of f, the proportion of eggs that are fertilized, and see what happens. Half eggs fertilised, f = 0.5. Equilibrium = (1 - 1)/0.5 = 0 All eggs fertilized, f = 1. Equilibrium = (2 - 1)/1 = 1 Quarter eggs fert., f = 0.25. Equilibrium = (0.5 - 1)/0.25 = -2 The last result is impossible as a proportion must be between 0 and 1. What it means is that if only one quarter of the eggs are fertilized, then PSR cannot persist in the population.
PSR Equilibrium
1 0.8
p,
0.6
equilibrium 0.4
0.2 0 0.5
0.6
0.7
0.8
0.9
1
f, proportion of eggs fertilised We can see the result more clearly by plotting the equation pe = (2f - 1)/f. For values of f below 0.5 the equilibrium is zero.
PSR Equilibrium
What our model tells us is that in a panmictic population, PSR can only persist if more than half the eggs are fertilized. Is this likely? Most animals have an even sex ratio, but N. vitripennis typically has female-biased sex ratios. (This is because of "Local Mate Competition", LMC, meaning that males tend to compete with brothers for matings, not with all the males in the population. Where LMC occurs, the ESS sex ratio for a mother is to have less than 50% sons. In addition, there are some other ultraselfish elements in N. vitripennis that manipulate sex ratio towards producing more female offspring.)
Relaxing Assumptions: Population Structure
In modelling we often make simplifying assumptions. One used in this model was to assume that N. vitripennis mate panmictically. But they do not. Females mate before dispersing from their habitat patch with males who were reared in the same patch. Males also cannot fly between patches. How might this affect the equilibrium? It will make it harder for PSR to persist in the population.
Relaxing Assumptions: Population Structure
To see this consider the most extreme case, in which each patch is founded by a single mated female and her offspring mate with each other. There are two types of patches. Those founded by a female mated to a PSR- male, and those founded by a female mated to a PSR+ male. Let's see what happens in each. A patch colonized by a female mated to a PSR+ male will produce only males. A patch colonized by a female mated to a non-PSR male will produce only females mated to PSR- males. All dispersing females will be mated to non-PSR males (-).
Relaxing Assumptions: Population Structure
Mated dispersing females One-foundress patches
host patch Female wasp
+
-
dispersing females
none
-
In a population structured into patches colonized by single mated females, PSR will immediately die out. A patch colonised by a female mated to a PSR male (+) will produce no dispersing females mated to PSR males. All dispersing females will be mated to non-PSR males (-).
Relaxing Assumptions: Population Structure
Mated dispersing females Two-foundress patches
host patch
+ +
+ -
- -
+ When patches are colonized by two females, PSR will not immediately die out. A patch colonized by one female mated to a PSR male (+) and one to a non-PSR male (-) will produce some dispersing females mated to PSR males (+). How many? What about patches colonized by 3 females?
dispersing females
none
Possible Conflicts in the Origin of Life
Hypercycle & Early Origin of Life
Genetic material now is copied very accurately with few mutations thanks to special enzymes (polymerases). But consider an early stage in the evolution of life when RNA was copied without specific enzymes and the error rate was very high. How could more complex life evolve if genetic material could only be copied in very short lengths? One possibility is to have several proto-genes each coding for part of a primitive metabolism. Each replicator is small so that it can be copied despite the high copying error rate. Collectively these different replicators form a self catalysing "hypercycle" in which each replicator catalyses the next in the cycle. However, for the hypercycle to be evolutionary stable it has to be resistant to mutant replicators that do not contribute to the hypercycle. For more complex life to evolve, mutants which help the hypercycle function better should be favoured by natural selection. We will see that this can occur when the hypercycle is contained in box or cell, so that beneficial mutants reap the reward of their contribution to making an improved hypercycle.
Based on Maynard-Smith & Szathmary. 1995. The major transitions in evolution. Cambridge UP.
2-Unit & 4-Unit Hypercycles
A B
In a hypercycle each unit is a replicator. Each catalyses the replication of the next replicator in the cycle and contributes to the overall metabolism.
A
B
If a mutant replicator occurs will it be favoured by natural selection?
D
C
Selection for Beneficial Mutants
A B
A* is a better "target" for the catalytic activity of B, even though it is not a better catalyst for the replication of B. It may be worse. It will be selected to replace A. A** is a better at catalysing the replication of B. But it is not a better target than A. Thus, A** will not be selected to replace A. Thus mutant A**, which makes the hypercycle run better is not favoured but A*, which can make it worse, is.
A*
B
A**
B
Selection for Beneficial Mutants
A* A* A** B
growth
The hypercycles have now been compartmentalised by putting them into little boxes or cells. The cell with A** mutant grows faster, and divides sooner, than that with A* because A** stimulates the replication of B better than does A*. The more cooperative mutant is favoured. The key is that the catalytic advantage of A** is retained locally, benefitting its own cell. This is also another example of how "population structure" affects the balance between coop. and conflict.
B B
growth
A**
B
Possible Origin of Linkage
Genes are normally linked together into chromosomes. However, linked genes are at a potential disadvantage in comparison to unlinked genes in replication, because it takes longer to copy a longer piece of DNA. In today's cells this is not a problem because cell division is tightly regulated. But what about an earlier stage in evolution when cell division and DNA replication may not have been as tightly synchronised as now? Unlinked versions of two genes, A and B, would have been able to replicate more rapidly than the same genes when linked as AB. How, therefore, was linkage selected for? We will consider two possible advantages. First, that linkage could lower the probability of a non-functional cell (lacking an essential gene) following cell division. Second, that a cell may grow more efficiently and thereby divide more quickly if the essential genes divide at the same rate.
Based on Maynard-...