APS 301' Cooperation

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APS 301. Co-operation Conflict Professor
Francis L. W. Ratnieks Lecture 2 PSR Model
Possible Conflicts in the Origin of Life
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PSR Equilibrium Model
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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.
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Equilibrium Proportion of PSR

• Start by defining two parameters. Both are
  proportions, so can only take values from 0 to 1.
• f proportion of eggs laid by females which are
  fertilised
• p 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.

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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)
Fertilised (f) Females ( eggs) Unfertilised
(1-f)
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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 Box
A/(Box A Box C Box D) or p fp/(fp 1 -
f) At equilibrium, p p pe, so
that pe fpe/(fpe 1 - f) (2f - 1)/f
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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 fpe/(fpe
1 - f) 1 f/(fpe 1 - f) divide both sides
by pe fpe 1 - f f multiply both sides by
(fpe 1 - f) fpe 2f - 1 move 1 - f to
right hand side pe (2f - 1)/f divide both
sides by f
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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.
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PSR Equilibrium
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.
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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.)
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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.
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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. Lets
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 (-).
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Relaxing Assumptions Population Structure
Mated dispersing females One-foundress patches
Female wasp
host patch

-
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 (-).
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Relaxing Assumptions Population Structure
Mated dispersing females Two-foundress patches
host patch

-

-

-
dispersing females
none
-

-
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?
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Possible Conflicts in the Origin of Life
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Hypercycle Early Origin of Life
Genetic material is now 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.
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2-Unit 4-Unit Hypercycles
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. If a mutant replicator occurs will
it be favoured by natural selection?
A
B
A
B
D
C
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Selection for Beneficial Mutants
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.
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Selection for Beneficial Mutants
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.
growth
growth
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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 todays 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-Smith Szathmary. 1995.
The major transitions in evolution. Cambridge UP.
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Lower Probability of Non-Functional Cell
Consider an early cell with multiple gene copies
and in which cell division does not result in
careful equal division of genes as in todays
cells. Instead, it is a matter of luck where a
gene ends up with a 50 chance of ending in each
daughter cell. For a cell to function properly
it must have at least one copy of each gene.
The following slide shows that with linked
genes there is a lower probability that a
daughter cell will lack one or the other
essential gene. By linking up, genes are less
likely to end up in defective daughter cells. In
other words there may be a mutualistic advantage
for genes to link up, especially when the number
of gene copies per cell is low.
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Lower Probability of Non-Functional Cell
AA BB
unlinked
linked
AB AB
A
A
A BB
AB
AB
AB
B
B
AB
A
AA
BB
AB AB
B
AA B
In this example a cell with two copies of each
gene divides, with chance determining how many
copies of each gene end up in each daughter cell
of a pair. When the genes are linked (right) only
2/6 of the possible daughter cells are defective
(lack one or the other gene). When the genes are
unlinked the proportion is higher, 10/18 (left).
A B
A B
AA B
B
BB
AA
A BB
A
AA BB
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Greater Cell Efficiency
Consider again the situation where two genes, A
and B, are needed for cell function and that the
number of genes per cell is not as tightly
regulated as in todays cells. In addition, one
gene replicates within the cell faster than the
other. If this happens the cell will soon have
an imbalance in gene function and will therefore
grow less and divide more slowly. When the genes
are linked they replicate at the same rate gene
products and function are maintained at a
constant ratio and the cell grows and divides
more rapidly. Again there is a mutualistic
advantage of linking. By linking with other
genes, a gene may end up in more daughter cells.
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Greater Cell Efficiency
Gene A tends to replicate faster than gene B, if
the two genes are not linked (above). As a
result, the cell with linked genes may grow
faster and divide sooner because a better balance
is maintained in the number of gene copies in the
cell (below).
AB
AB
AB
AB