Chapter 12: Accessory Protein Model

Inputs (Enter first)

SP[x_] := Simplify[PowerExpand[x]]

Off[General::"spell"]

Off[General::"spell1"]

<< Graphics`Graphics`

Model set up -- see Sup. Mat. 12.2 (Enter second)

Example: Evolution of Sperm and Accessory Protein Production in Males

In recent years, it has been well documented that males of some species transfer many other so-called “accessory proteins” during mating, in addition to sperm. Many of these proteins serve to increase female ovulation, and thereby increase the number of eggs to which that male’s sperm has access (Chapman et al. 2003a, 2003b). If males must produce these proteins from a fixed reproductive budget, R, then there is a tradeoff between sperm and protein production. In particular, if multiple males mate with a female and compete for the fertilization of her eggs, then any given male would gain the highest reproductive success if it produced only sperm, while the other males produce the accessory proteins.  Let’s construct a model to understand the evolution of resource allocations in males: (i) amount of sperm produced per mating, s, and (ii) amount of protein produced per mating, a.

Assume that a gene on the Y chromosome affects resource allocation and that the total number of mates a male will be able to obtain is (R/(s + a + c)).

The next step is to specify the total number of male offspring that a male can expect to sire at each mating. For simplicity we assume that a female has a single mate with probability 1 - u and two mates with probability u.  When she is singly mated, we assume that a male fertilizes all of the female’s eggs.  When she is doubly mated, he fertilizes only a fraction (s1/(s1 + s2)), where the male of interest transferred s1 sperm and the competing male transferred s2 sperm.  This fraction assumes that every sperm has an equal chance of fertilizing an egg. Letting  f[x] denote the number of eggs produced by a female who receives a total of x accessory proteins, the number of male offspring sired per mating by a father allocating s1 and a1 resources to sperm and proteins is:

	(1/2) ((1 - u) f[a1 ]  + 
            u (( s1)/(s1 + s2)) f[a1 + a2])

when the competing male allocates s2 and a2, where the 1/2 reflects an assumption of a 50:50 sex ratio.  The total number of offspring sired by the father over all matings is the above multiplied by the number of mates,  (R/(s1 + a1 + c)).

Next, consider a mutant that allocates sm and am resources to sperm and proteins arising within a resident population that allocates s and a resources to sperm and proteins.  The number of male offspring sired by the mutant relative to the number sired by the resident will be:

λ= (((1/2) ((1 - u) f[am] +  
                    u  (sm/(sm + s)) f[am + a]) 
                (R/(sm + am + c)))/((1/2) ((1 - u) f[a] + 
                    (1/2) u f[2  a]) (R/(s + a + c))));

If λ > 1, then the mutant male sires more sons, causing the mutant Y chromosome to spread through the population.

In the following, let's figure out how sperm and accessory protein allocation should evolve through the successive appearance and fixation of mutations.

Part 1:  Analysizing a specific function for female fecundity f[x]=b (x^k)

To be concrete, let's assume that female fecundity rises as f[x]=b (x^k)when she receives x accessory proteins, where k determines the curvature:

\(\(\(		\)\(f[x_] = b\ x\^k;\)\)\)

The curvature of this function depends on k.  For k = 1, it is linear.  For k < 1, there is diminishing returns (negative curvature), which we might expect as a female runs out of eggs:

DisplayTogether[
Plot[f[a] /. k -> 0.25 /. 
          b -> 100, {a, 0, 2}, PlotRange -> All, 
        PlotStyle -> 
          Dashing[{0.005, 0.015}]], 
Plot[f[a] /. 
            k -> 0.5 /. b -> 100, {a, 0, 2}, 
        PlotRange -> All, 
        PlotStyle -> 
          Dashing[{0.02, 0.015}]], 
Plot[f[a] /. 
            k -> 0.75 /. b -> 100, {a, 0, 2}, 
        PlotRange -> All, 
        PlotStyle -> 
          Dashing[{0.04}]], 
Plot[f[a] /. k -> 1 /. 
          b -> 100, {a, 0, 2}, 
        PlotRange -> All]
]

Question 1:  What is the ESS?

By definition, an evolutionarily stable strategy (ESS) cannot be invaded by any mutant strategy.  If a male allocating s* and a* represents an ESS, then λ cannot rise above 1 when mutants appear that cause s or a to vary from .  This will only be true if:

	(Subscript[(((d λ)/(d sm))), \(sm = \(s\^*\), 
            am = \(a\^*\)\)])=0
	
	\(TraditionalForm\`\((\(d\ λ\)\/\(d\ am\))\)\_\(sm = \(s\^*\), am = \(a\^*\)\) = 0\)

(Note that λ = 1 when the mutation doesn't do anything:

\(\(atESS = {s -> \(s\^*\), sm -> \(s\^*\), a -> \(a\^*\), 
          am -> \(a\^*\)};\)\)

λ/.atESS//Factor

The following command solves the first derivative conditions for the ESS:

subESS=Solve[{
Factor[D[ λ,am]/.atESS]==0,
Factor[D[ λ,sm]/.atESS]==0},\({\(a\^*\), \(s\^*\)}\)]//FullSimplify//Flatten

To be biologically valid, k must be less than one (i.e., there must be diminishing returns in the number of eggs produced as a function of the amount of accessory protein received).

For example:

\(\(\(\(\({s -> \(s\^*\), a -> \(a\^*\)} /. subESS\) /. 
            u -> 1\) /. b -> 1\) /. k -> 0.3\) /. 
      c -> 0.1\)

Plot3D[λ/.\({s -> \(s\^*\), a -> \(a\^*\)}\)/.subESS/.u -> 1/.b -> 1/.k -> 0.3/.c -> 0.1,{am, 0.0, 0.1},{sm, 0.1, 0.2}]

Question 2:  Do populations evolve toward this potential ESS?

When populations converge upon a point over evolutionary time, we call that point "convergence" stable.  To be convergence stable, the eigenvalues of the convergence-stability matrix, C, must have negative real parts, where:

	C = (Subscript[(({{((Subscript[d(((dλ)/(d sm))), sm = s, 
                      am = a])/(d s)), ((Subscript[d(((dλ)/(d sm))), sm = s, 
                      am = a])/(d a))}, {((Subscript[d(((dλ)/(d sm))), sm = s, 
                      am = a])/(d a)), ((Subscript[d(((dλ)/(d am))), sm = s, 
                      am = a])/(d a))}})), \(s = \(s\^*\), a = \(a\^*\)\)])

Cmat={{D[Simplify[D[λ,sm]/.{am -> a, sm -> s}],s],D[Simplify[D[λ,sm]/.{am -> a, sm -> s}],a]},
{D[Simplify[D[λ,am]/.{am -> a, sm -> s}],s],D[Simplify[D[λ,am]/.{am -> a, sm -> s}],a]}}/.atESS/.subESS//Simplify

We can write the eigenvalues of Cmat as:

{(1/2) (Trace - (Sqrt[(Trace^2) + 4 Determinant])), 
        (1/2) (Trace + (Sqrt[(Trace^2) - 4 Determinant]))};

These eigenvalues will have negative real parts if the eigenvalues are complex and Trace is negative or if the eigenvalues are real and Trace is negative and Determinant is positive.

The Trace is always negative:

Cmat[[1, 1]] + Cmat[[2, 2]] // FullSimplify

This is easier to see if rewritten as:

-(((((1 - k))^2) (4 (1 - u) + (2^k) u) 
              (2^(-1 - k)) ((4^k) (u^2) + 
                  2 (1 - u) (4 k (1 - u) + 
                        (2^k) (1 + 
                              k) u)))/((c^2) k u (((2 (1 - u) + (2^k) u))^2)));

    FullSimplify[%/%%]

Furthermore, the determinant of Cmat is always positive:

Det[Cmat] // Factor

This is easier to see if rewritten as:

(((2^(-3 - k)) (((1 - k))^5) (((4 (1 - u) + (2^k) u))^4))/((c^4) k u (((2 (1 - u) + (2^k) u))^3)));

    FullSimplify[%/%%]

Thus, when near to the potential ESS at s* and a* , mutations can spread if they bring the population closer to s* and a*.  This point is thus convergence stable.

Question 3:  Does this potential ESS represent a minimum or a maximum on the λ surface?

In order for λ to be maximized rather than minimized when the mutant strategy is the same as the resident strategy, the eigenvalues of the Hessian matrix, H, must not be positive, where:

	H = (Subscript[(({{(((d^2) λ)/(d (sm^2))), \(\(\(d\^2\) λ\)\/\(dsm\ dam\)\)}, {(((d^2) λ)/(dsm dam)), (((d^2) λ)/(d (am^2)))}})), \(s = \(s\^*\), a = \(a\^*\)\)])

(This is just the two-dimensional equivalent to the condition that (((d^2) y)/((dx^2))) < 0 for the function y to be maximized at a point.)

Hmat={{D[ λ,am,am],D[ λ,am,sm]},
    {D[ λ,am,sm],D[ λ,sm,sm]}}/.atESS/.subESS//Simplify

We can write the eigenvalues of Hmat as:

{(1/2) (Trace - (Sqrt[(Trace^2) + 4 Determinant])), 
        (1/2) (Trace + (Sqrt[(Trace^2) - 4 Determinant]))};

The fact that Hmat is symmetric guarantees that these eigenvalues are real.  To both be negative, we then require that Trace be negative and Determinant be positive.

The trace is guaranteed to be negative whenever k < 1 (as required for the ESS to exist):

traceH = FullSimplify[Hmat[[1, 1]] + Hmat[[2, 2]]]

which can be written as  \(\(-\(\(\((1 - k)\)\^2\ \((\(\((1 - k)\)\ \((8\ \((1 - u)\)\  + 2\^k\ u)\)\)\/k + \(2\^\(1 - k\)\ \((4 \((1 - u)\)\  + 2\^k\ u)\)\^2\)\/u)\)\)\/\(4\ c\^2\ \((2 \((1 - u)\)\  + 2\^k\ u)\)\)\)\)\)to see more easily that the whole term is negative (by drawing the negative sign to the front and rewriting each term in a manner that is easy to see is positive).

On the other hand, the determinant need not be positive:

detH = Factor[Det[matnew]]

For the Determinant to be positive, -(-16 + 16 k + 16 u - (2^(1 + k)) u - 16 k u + 
            3 (2^k) k u) must be positive.

Plot3D[-(-16 + 16 k + 16 u - (2^(1 + k)) u - 16 k u + 
            3 (2^k) k u), {k, 0, 1}, {u, 0, 1}, 
      PlotRange -> {0, 15}, ClipFill -> None]

While we cannot solve -(-16 + 16 k + 16 u - (2^(1 + k)) u - 16 k u + 
            3 (2^k) k u) for k (because of the (2^k) k terms), we can solve it for u:

Solve[-(-16 + 16 k + 16 u - (2^(1 + k)) u - 16 k u + 
                  3 (2^k) k u) == 0, u] // FullSimplify // 
      Flatten

Plot[
      (1/(1 + (((2^(-4 + k)) (2 - 3 k))/(-1 + k)))), {k, 
        0, 1}]

The critical value of u, (1/(1 - (((2^(-4 + k)) (2 - 3 k))/(1 - k)))), is greater than one if k < 2/3, implying that any fraction of double matings allows an ESS to exist when k < 2/3 (strong diminishing returns in the number of eggs that a female produces).  If, however, 2/3 < k < 1, then the ESS will exist only if the number of double matings is rare enough: u < (1/(1 - (((2^(-4 + k)) (2 - 3 k))/(1 - k)))).

When  multiple matings are frequent enough (u > (1/(1 - (((2^(-4 + k)) (2 - 3 k))/(1 - k))))),  the potential ESS represents a minimum rather than a maximum on the λ surface.  In this case, ANY mutant can invade the population at s* and a*.  In this case we see evolutionary branching:  a polymorphism arises that causes the population to consist of some males that invest in more accessory proteins and other males that "cheat" by taking advantage of a female's other mates production of accessory proteins.

IMPLICATION:  Most studies have focused on the average allocation to accessory proteins.  The above suggests that there might well be polymophic strategies, which would be well worth investigating experimentally.

For example:

\(\(\(\(\({s -> \(s\^*\), a -> \(a\^*\)} /. subESS\) /. 
            u -> 1\) /. b -> 1\) /. k -> 0.9\) /. 
      c -> 0.1\)

Plot3D[λ/.\({s -> \(s\^*\), a -> \(a\^*\)}\)/.subESS/.u -> 1/.b -> 1/.k -> 0.9/.c -> 0.1,{am, 0.85, 0.95},{sm, 0.95, 1.05},PlotRange -> {0.9997, 1.0002}]

Part 2:  Generalizing the analysis for arbitrary f[x]

In this section, we do not specify the function f[x] and prove that branching is a general phenomenon of this model when u is large enough.

Clear[f]

Question 1:  For general f[x], when will branching occur?

The first two single derivative conditions give us:

eqn1 = D[λ, am] /. atESS // Simplify

eqn2 = D[λ, sm] /. atESS // Simplify

We cannot solve these equations directly for s* and a* because f[a] is not specified:

\(\(Solve[{eqn1 == 0, eqn2 == 0}, {\(a\^*\), \(s\^*\)}] // 
        Simplify\) // Flatten\)

Instead we use these equations to solve for two of the terms.  Various choices are possible, we tried a few choices before settling on c and f'[a], which led to nice interpretable results:

sub=Solve[{eqn1 == 0, eqn2 == 0},{f'[ \(a\^*\)],c}]//Simplify//Flatten

The second derivative condition for an ESS requires that the Hessian matrix have negative eigenvalues, where:

Hmat = {{D[λ, am, am], 
              D[λ, am, sm]}, {D[λ, am, sm], 
              D[λ, sm, sm]}} /. atESS // Factor

Making our substitution from the first derivative conditions:

Hmatnew = Simplify[Hmat /. sub]

For the eigenvalues to be negative, we need the Trace to be negative and the Determinant to be positive.

Simplify[Det[Hmatnew]]

This can be written as:

DetH=((u (- u ((f'[\(2\ \(a\^*\)\)])^2)-\(2\ f[2\ \(a\^*\)]\ curvature\)))/(\(4\ \(s\^*\)\^2\ \((2\ \((1 - u)\)\ f[\(a\^*\)] + u\ f[2\ \(a\^*\)])\)\^2\)));

where curvature is a measure of the curvature of f[x] at two points, f[a], which is relevant when females are singly mated, and f[2a], which is relevant when females are doubly mated:

curvature=2 (1 - u) f''[\(a\^*\)]+u f''[\(2\ \(a\^*\)\)];

Factor[Det[Hmatnew]/DetH]

For DetH to be positive, the curvature must be negative.  In addition, the frequency of doubly mated females, u, must be sufficiently small:

u < (((- curvature) ((f'[\(2\ \(a\^*\)\)])^2))/(\(2\ f[2\ \(a\^*\)]\)));

If curvature is negative then the trace is guaranteed to be negative too:

Simplify[Hmatnew[[1, 1]] + Hmatnew[[2, 2]]]

which can be written as

\(\(TraceH = \(\(-\ u\)\ f[2\ \(a\^*\)] + 2\ \(s\^*\)\^2\ curvature\)\/\(2\ \(s\^*\)\^2\ \((2\ \((1 - u)\)\ f[\(a\^*\)] + u\ f[2\ \(a\^*\)])\)\);\)\)
FullSimplify[%/%%]

Thus, when the frequency of doubly mated females, u, is sufficiently small that  u < (((- curvature) ((f'[\(2\ \(a\^*\)\)])^2))/(\(2\ f[2\ \(a\^*\)]\))), then there is an ESS and not evolutionary branching.  Conversely, when the frequency of doubly mated females, u, is high and this condition is not satisfied, branching is expected.

As an aside, we can look at the diagonal elements of Hmat to determine whether mutants that affect accessory protein allocation only ((((d^2) λ)/(d ((am)^2)))= Hmat[[1,1]]) of sperm allocation only ((((d^2) λ)/(d (sm^2)))= Hmat[[2,2]]) could invade.  As the following are both negative for negatively curved f[x], we conclude that such mutants cannot invade:

curvatureFUNCTION Simplify[
        Hmat[[1, 1]]/curvature /. sub]

Simplify[Hmat[[2, 2]] /. sub]

Thus, branching, when it occurs, must be driven by mutations that alter both a and s simultaneously (as suggested by the saddle shape of the Figure S12.2.2).

Question 2:  For general f[x], will males evolve towards the potential ESS point?

To be convergence stable,the eigenvalues of the convergence-stability matrix,C,must have negative real parts,where:

Cmat = {{D[
                Simplify[D[λ, am] /. {am -> a, sm -> s}], 
                a], D[Simplify[
                  D[λ, am] /. {am -> a, sm -> s}], 
                s]}, 
{D[
                Simplify[D[λ, sm] /. {am -> a, sm -> s}], 
                a], D[Simplify[
                  D[λ, sm] /. {am -> a, sm -> s}], s]}} /. 
          atESS;

The Trace must be negative in order for Cmat to have eigenvalues with negative real parts:

TraceTemp = 
      Cmat[[1, 1]] + Cmat[[2, 2]] /. sub // FullSimplify

which can be written as:

TraceC=-( \(4\ \((1 - u)\)\ f[\(a\^*\)]\ \((u\ f[2\ \(a\^*\)] + 
                      4\ \((\(s\^*\))\)\^2\ \((\(-curvature2\))\))\)\)+u \(f[2\ \(a\^*\)]\) (\(u\ f[2\ \(a\^*\)]\)+2 \(s\^*\) (u f'[\(2\ \(a\^*\)\)]+\(4\ \(s\^*\)\ \((\(-curvature2\))\)\))))/\((4\ \((2\ \((1 - u)\)\ f[\(a\^*\)] + u\ f[2\ \(a\^*\)])\)\^2\ \((\(s\^*\))\)\^2)\);

Now the relevant measure of curvature of f[x] is

curvature2 = ((1 - u) f''[\(a\^*\)]+u  f''[\(2\ \(a\^*\)\)]);

As long as curvature2 is negative, then Trace is necessarily negative.

Factor[TraceTemp/TraceC]

When the eigenvalues are real, the determinant of Cmat must also be positive for the eigenvalues to be negative.  Evaluating the determinant of Cmat:

DetC = Det[Cmat] /. sub // Simplify

which can be written as:

\(\(\(\((\(-\ curvature2\))\)\ u\ f[2\ \(a\^*\)]\ \((4\ \((1 - u)\)\ f[\(a\^*\)] + u\ f[2\ \(a\^*\)])\)\)\/\(2\ \((2\ \((1 - u)\)\ f[\(a\^*\)] + u\ f[2\ \(a\^*\)])\)\^3\ \((\(s\^*\))\)\^2\);\)\)
Factor[%/%%]

As long as curvature2 is negative, then the Determinant is necessarily positive.

Overall, we can conclude that the ESS point that satisfies the first two derivative conditions will be convergence stable as long as f[x] exhibits negative curvature at points \(\(a\^*\)\) and 2\(\(a\^*\)\).

Question 3:  How does the ESS change as the cost of finding a mate varies?

Evaluating the effects of increasing c on the ESS value of sperm allocation using (12.8.12):

Simplify[((-D[
                      Simplify[
                        D[λ, am] /. {am -> a, sm -> s}], 
                      a] D[Simplify[
                      D[λ, sm] /. {am -> a, sm -> s}], c] + 
                D[Simplify[D[λ, sm] /. {am -> a, sm -> s}], 
                    a] D[Simplify[
                      D[λ, am] /. {am -> a, sm -> s}], 
                    c]) /. atESS)/DetC]

cond = Simplify[% /. sub]

which can be written as:

((u \(f[2\ \(a\^*\)]\) (u f'[\(2\ \(a\^*\)\)]+\(2\ \(s\^*\)\ \((\(-\ curvature2\))\)\)))/(\(2\ \((4\ \((1 - u)\)\ f[\(a\^*\)] + 
              u\ f[2\ \(a\^*\)])\)\ \(s\^*\)\ \((\(-\ curvature2\))\)\)));
Factor[%/%%]

If curvature2 is negative (as required for convergence stability), this condition is necessarily positive, and increasing the cost of finding a mate increases the allocation to sperm at the ESS.

Evaluating the effects of increasing c on the ESS value of accessory protein allocation using (12.8.12):

Simplify[((D[
                    Simplify[
                      D[λ, am] /. {am -> a, sm -> s}], s] 
                  D[Simplify[
                      D[λ, sm] /. {am -> a, sm -> s}], c] - 
                D[Simplify[D[λ, sm] /. {am -> a, sm -> s}], 
                    s] D[Simplify[
                      D[λ, am] /. {am -> a, sm -> s}], 
                    c]) /. atESS)/DetC]

cond2 = Simplify[% /. sub]

which can be written as:

\(\(\(u\^2\ f[2\ \(a\^*\)]\^2\)\/\(4\ \((4\ \((1 - u)\)\ f[\(a\^*\)] + u\ f[2\ \(a\^*\)])\)\ \((\(s\^*\))\)\^2\ \((\(-curvature2\))\)\);\)\)
Factor[%/%%]

If curvature2 is negative (as required for convergence stability), this condition is necessarily positive, and increasing the cost of finding a mate increases the allocation to accessory proteins at the ESS.