Kasum Akhmedli: A Model of the Transition from Hermaphroditism to Gonochorism

After publishing the column about the transition from hermaphroditism to separate sexes I (D.Sh.) received a letter from Kasum Akhmedli, a leading researcher at the National Academy of Sciences of Azerbaijan. He sent me his model of this transition and a detailed description of it. I am publishing this letter here (with the author's consent), hoping it will contribute to a better understanding of the evolutionary transition under discussion.

Dear Dmytro!

I decided to model the evolution of sexual reproduction, as you described in the column "The Victory of Stability over Optimality, or Why Hermaphrodites Lose to Males and Females" from 24/01/14, by means of chemical kinetics, and now I want to share the results with you. Although in my models the processes proceed under the law of mass action, and the reproducing forms chosen are extremely simple, nevertheless it seems to me they allow, in a rough approximation, judging the effects of the transition from asexual reproduction of hermaphrodites to sexual reproduction. I have no doubt that you have at your disposal tools for more precise mathematical modeling of evolutionary mechanisms, so do not judge me too harshly for using what I had at hand — the Gepasi program for calculating the kinetics of chemical reactions. The concept of the selfish gene in this approach is that, under conditions of a limited resource of the food substrate (SS), competition arises between colonies consuming this resource, in which the gene that replicates faster survives.

First I will describe the model, and then move on to the results.

Notation

• SS and S - a dimeric substrate (and its easily dimerized monomer) on which the genes of the population replicate and evolve. The amount of substrate is limited (otherwise selection cannot occur).

• HHxx - the genome of a diploid monozygous hermaphrodite of the original population.

• HMxx, HMxy, HFxx, MMxx, MMxy, FMxx, FMxy, FFxx - other diploid genomes (zygotes, replicators) that arose as a result of mutations of the original hermaphrodite and recombinations of the sexual process.

• H, M, F - genes that determine the sexual behavior of a hermaphrodite, male, or female (H supports both oogenesis and spermatogenesis, M supports only spermatogenesis, F supports only oogenesis). Gene H is dominant relative to its mutants M and F. Therefore the combinations HMxx, HMxy, HFxx define hermaphrodites, the combinations MMxy and FMxy define males, and FMxx and FFxx define females. The combination MMxx, despite the absence of a y-chromosome, is also classed as male, since such specimens, at meiosis, in the absence of genes H or F, produce only spermatozoa.

• Hxe, Mxe, and Fxe - haploid egg cells (e - denotes egg).

• Hx, Mx, My and Fx - haploid spermatozoa.

Units of measurement

• The unit of quantity for any sample is taken to be the original initial quantity of replicators with the genome HHxx.

• The unit of time is the average lifespan of the original replicator (individual) HHxx, which remains the same for mutants as well.

Model description

Table 1 gives the scheme describing the evolution of sexual reproduction in hermaphrodites.

Table 1. Scheme of reproduction and evolution processes

N	Process	k→	Notes
	Mutations
1	HHxx → HMxx	Kmu	Male mutation
2	HHxx → HFxx	Kmu	Female mutation
3	HMxx → MMxy	Kmu	Emergence of a male
4	HMxx → FMxx	Kmu	Emergence of a female
	Mitosis
5	SS + HHxx → 2*HHxx	Kmi
6	SS + HMxx → 2*HMxx	Kmi
7	SS + HMxy → 2*HMxy	Kmi
8	SS + HFxx → 2*HFxx	Kmi
9	SS + MMxx → 2*MMxx	Kmi
10	SS + MMxy → 2*MMxy	Kmi
11	SS + FMxx → 2*FMxx	Kmi
12	SS + FMxy → 2*FMxy	Kmi
13	SS + FFxx → 2*FFxx	Kmi
	Meiosis
14	HHxx → 2*Hxe	Kh	Oogenesis
15	HHxx → 2*Hx	Kh	Spermatogenesis
16	HMxx → Hxe+ Mxe	Kh	O
17	HMxx → Hx+ Mx	Kh	S
18	HMxy → Hx + My	Kh	S
19	HFxx → Hxe + Fxe	Kh	O
20	HFxx → Hx + Fx	Kh	S
21	MMxx → 2*Mx	Kh	S
22	MMxy → Mx + My	Ks*Kh	S
23	FMxx → Mxe + Fxe	Ke*Kh	O
24	FMxy → Fx + My	Ks*Kh	S
25	FFxx → 2*Fxe	Ke*Kh	O
	Fertilization
26	Hxe + Hx → HHxx	Kf
27	Hxe + Mx → HMxx	Kf
28	Mxe + Hx → HMxx	Kf
29	Hxe + My → HMxy	Kf
30	Hxe + Fx → HFxx	Kf
31	Fxe + Hx → HFxx	Kf
32	Mxe + Mx → MMxx	Kf
33	Mxe + My → MMxy	Kf
34	Mxe + Fx → FMxx	Kf
35	Fxe + Mx → FMxx	Kf
36	Fxe + My → FMxy	Kf
37	Fxe + Fx → FFxx	Kf
	Death
38	HHxx → SS	Kd	Destruction of zygotes
39	HMxx → SS	Kd
40	HMxy → SS	Kd
41	HFxx → SS	Kd
42	MMxx → SS	Kd
43	MMxy → SS	Kd
44	FMxx → SS	Kd
45	FMxy → SS	Kd
46	FFxx → SS	Kd
47	Hxe → S	Kde	Destruction of oocytes
48	Hx → S	Kds	Destruction of spermatozoa
49	Mxe → S	Kde
50	Mx → S	Kds
51	My → S	Kds
52	Fxe → S	Kde
53	Fx → S	Kds
54	2*S → SS	Kss	Substrate regeneration

The column k→ is the rate constants of the processes. The values of the rate constants are given in Table 2, and the initial conditions - in Table 3.

Table 2. Rate constants

Rate constant		Notes
Kmu	1.00E-04	R.C. of mutation
Kmi	0.1	R.C. of mitosis
Kh	1	R.C. of meiosis in hermaphrodites
Ks	1	Acceleration of spermatogenesis M
Ke	1	Acceleration of oogenesis F
Kf	1.00E+04	R.C. of fertilization
Kd	1	R.C. of replicator destruction
Kde	0.05	R.C. of oocyte destruction
Kds	0.05	R.C. of spermatozoa destruction
Kss	100	R.C. of substrate regeneration

Table 3. Initial conditions

Initial conditions
SS₀	10
HHxx₀	1
All others ₀	0

The fact that the average lifespan of replicators was chosen as the time unit is reflected in that, in processes 36-43, the destruction rate constant for all of them is set to Kd=1. The initial value of the number of hermaphrodites

HHxx

0 = 1 does not mean at all that there is only 1 specimen of the hermaphrodite. It means that whatever the initial number was, a thousand or a million, this number is taken as the unit. This frees us from dependence on a specific number and makes the numbers we deal with easier to perceive. The amount of food resource is limited, i.e. as it is consumed (through growth and reproduction of the replicators) it decreases, and upon the death of zygotes and gametes it is restored, providing a cycling of the substrate. The initial amount of substrate

0 is chosen equal to 10, and since the number of the initial population is taken as 1, the total number of all populations can never exceed 11 units. The initial number of the HHxx population, equal to 1, is, as can be seen from the constants, the steady-state number in the absence of mutations and other replicators besides HHxx. Before the model is run, other replicators are absent from the system. The value Kmu=10^-4 means that mutations occur with an average frequency of 1 mutation per 10000 generations. The value of the mitosis rate constant Kmi is chosen so that, given the chosen initial food content, on average 1 division of HHxx occurs per unit of time (which is equal to the average lifespan). The other rate constants are chosen so as to maintain steady-state gamete quantities at the level of 10-15% of the number of zygotes. Ks and Ke are not rate constants, but coefficients increasing the rate constants of spermatogenesis and oogenesis in M- and F-mutants respectively, compared to the original hermaphrodite, as indicated for processes 22-25. The value of 1 indicated in the table for these coefficients means that there is no acceleration, but it can be introduced if desired.

During fertilization, each act involves 1 egg cell and 1 spermatozoon.

As long as the genome contains the dominant gene H, the replicator (HHxx, HMxx, HFxx) is capable, at meiosis, of both oogenesis and spermatogenesis and is a hermaphrodite. Mutation (1) leads to the emergence of gene M, under whose control (and hence, in the absence of genes H or F in the genome) the capacity for oogenesis is lost. Similarly, mutation (2) leads to the emergence of gene F, under whose control (in the absence of genes H or M in the genome) the capacity for spermatogenesis is lost.

Discussion of modeling results

Model 1

As Model 1, the simplest case is considered, when there are no mutations at all (Kmu=0), and the only replicator in the system is the original hermaphrodite HHxx. Fig. 1 shows graphs of the development of equilibrium in this case.

Fig. 1. Development of the steady state in a population of homozygous hermaphrodites.

H – total content of gene H in the system. Herm – number of HHxx zygotes.

Hxe

and

– number of gametes of both kinds (egg cells and spermatozoa) in the system.

– amount of free substrate monomer.

As can be seen, and as should be expected, equilibrium (more precisely, a steady state) is reached within a few generations. The number of zygotes decreases by a couple of %, as a result of the gametes reaching a steady-state number of 1.4% relative to the zygotes each. The amount of substrate SS (not shown) practically does not change (about 10).

Model 2

Let us now consider what happens when both mutations (M and F) are switched on. We take the coefficients Ks and Ke equal to 1, i.e. spermatogenesis in M-males and oogenesis in F-females proceed at the same rates as in the original hermaphrodite. We should expect that, since in this situation females lose 50% of their fertility, hermaphrodites should outcompete the emerging mutants. However, contrary to these expectations, what happens is what is depicted in Fig. 2.

Fig. 2. Evolution of sex under "male" and "female" mutations in a population of hermaphrodites.

H, M and F – total quantities of the respective genes in the system (accounting for their presence in all kinds of zygotes and gametes). Herm – total number of hermaphrodites: HHxx, HMxx and HFxx. Male – total number of males: HMxy, MMxx, MMxy and FMxy. Fem – total number of females: FMxx and FFxx.

As can be seen from the figure, after several tens of thousands of generations, the number of hermaphrodites, along with the total content of the original gene H, falls to zero, and a population is formed consisting exclusively of males and females, each with half the number compared to the original number of hermaphrodites.

I will add (without piling on more figures) that variations of Ks in either direction between values of 0.5 and 1.5 do not lead to a qualitative change in the picture, only somewhat changing the sex ratio.

Model 3

Let us now consider what happens if only the male M-mutation occurs, or only the female F-mutation. First let us leave the male mutation and turn off the female one (exclude processes 2 and 4). Ks is likewise =1.

Fig. 3. Dynamics of the emergence of the male mutation in a population of homozygous hermaphrodites. Notation – as before.

Here it turns out that if only the male mutation arises, and it is not followed by the female mutation, then the mutation spreads like a disease, and within a few thousand generations the entire population dies out. Increasing Ks also does not help stabilize the population.

Model 4

A similar situation arises in the case where only the "female" mutation occurs. Let us put process 2 back and exclude processes 1 and 3.

Fig. 4. Dynamics of the emergence of the female mutation in a population of homozygous hermaphrodites. Notation – as before.

As can be seen from the figure, the female mutation in the absence of the male one also leads to the complete extinction of the population, though somewhat more slowly. Increasing Ke to 1.5 in this case (see Fig. 5) leads to the hermaphrodite population not dying out, but sharply shrinking (to 0.7) and a very small steady-state number of females remaining - 0.005. The content of genes H and F stabilizes at the level of 1.6 (the original content was 2.0) and 0.14, respectively.

Fig. 5. The conditions are the same as in Fig. 4, with the difference that Ke=1.5.

Conclusions

Contrary to intuitive expectation, if in a population of hermaphrodites the male and female mutations occur synchronously, or with not too great a delay of either one after the other, then, despite the loss of 50% fertility in females and males, the hermaphrodite population over time turns into a population with separate sexes. However, if only one of these mutations occurs, and the other never comes about, this has a detrimental effect on the whole population and reduces its numbers (let us not say it dies out, since the model is crude, and life is a tenacious thing).