Download Electronic supplementary material to: “Simultaneous failure of two

1
Electronic supplementary material to: “Simultaneous failure of two sex-
2
allocation invariants”
3
4
António M. M. Rodrigues1,2,*, Andy Gardner2
5
6
1. Department of Zoology, University of Cambridge, Downing Street, CB2 3EJ
7
Cambridge, United Kingdom.
8
9
2. Wolfson College, Barton Road, Cambridge, CB3 9BB United Kingdom.
10
11
3. School of Biology, University of St Andrews, Dyers Brae, St Andrews KY16 9TH
12
United Kingdom.
13
* Corresponding author, email: [email protected]
14
15
Contents
16
Appendix A. Reproductive success
17
Appendix B. Fitness
18
Appendix C. Stable class-frequency and reproductive value
19
Appendix D. Relatedness
20
Appendix E. Selection gradient
21
Appendix F. Convergence stability
22
Appendix G. Self-knowledge model
23
Appendix H. Supplementary figures (Figure H1 & H2 & H3)
24
References
25
1
26
Appendix A. Reproductive success
27
28
Here, we define the class-specific reproductive success of a focal breeder who
29
allocates a proportion xH of their reproductive resources to sons when they are high-
30
fecundity and who allocates a proportion xL of their reproductive resources to sons
31
when they are of low fecundity. This depends on the quality of the focal breeder (i.e.
32
if it is high- or low-fecundity) and on the quality of the breeder’s successful offspring
33
(i.e. if they are either high- or low-fecundity). For convenience let us define the
34
quantities q(xH,xL), and a(xH,xL) as
35
36
𝑞(𝑥H , 𝑥L ) = ((1−𝑥
1
,
)+(1−𝑥
)(1−𝑠))(1−𝑑)+((1−𝑧
H
L
H )+(1−𝑧L )(1−𝑠))𝑑(1−𝑘)
and
(A1)
37
38
𝑎(𝑥H , 𝑥L ) = (1 − 𝑑)𝑞(𝑥H , 𝑥L ) + 𝑑(1 − 𝑘)𝑞(𝑧H , 𝑧L ),
(A2)
39
40
respectively. The reproductive success of a high-fecundity female is then given by
41
wHfHf = (1-xH)a(xH,xL)(1-ϕ), wHfLf = (1-xH)a(xH,xL)(1-ϕ), wHfHm = (1-xH)a(xH,xL)μ,
42
and wHfLm = (1-xH)a(xH,xL)μ, through her successful offspring that become high-
43
fecundity breeding females, low-fecundity breeding females, high-fecundity breeding
44
males, and low-fecundity breeding males, respectively. The fraction of genes a son (or
45
a daughter) inherits from her mother (or father) is denoted by μ (or ϕ).
46
47
The reproductive success of a low-fecundity female is given by wLfHf = (1-s)(1-
48
xL)a(xH,xL)(1-ϕ), wLfLf = (1-s)(1-xL)a(xH,xL)(1-ϕ), wLfHm = (1-s)(1-xL)a(xH,xL)μ, and
49
wLfLm = (1-s)(1-xL)a(xH,xL)μ, through her successful offspring that become high-
2
50
fecundity breeding females, low-fecundity breeding females, high-fecundity breeding
51
males, and low-fecundity breeding males, respectively.
52
53
The reproductive success of a high-fecundity male is given by wHmHf = (xH/(xH+(1-
54
s)xL))(1-xH+(1-xL)(1-s))a(xH,xL)ϕ, wHmLf = (xH/(xH+(1-s)xL))(1-xH+(1-xL)(1-
55
s))a(xH,xL)ϕ, wHmHm = (xH/(xH+(1-s)xL))(1-xH+(1-xL)(1-s))a(xH,xL)(1-μ), and wHmLm
56
= (xH/(xH+(1-s)xL))(1-xH+(1-xL)(1-s))a(xH,xL)(1-μ), through his successful offspring
57
that become high-fecundity breeding females, low-fecundity breeding females, high-
58
fecundity breeding males, and low-fecundity breeding males, respectively.
59
60
The reproductive success of a low-fecundity male is given by wLmHf = ((1-
61
s)xL/(xH+(1-s)xL))(1-xH+(1-xL)(1-s))a(xH,xL)ϕ, wLmLf = ((1-s)xL/(xH+(1-s)xL))(1-
62
xH+(1-xL)(1-s))a(xH,xL)ϕ, wLmHm = ((1-s)xL/(xH+(1-s)xL))(1-xH+(1-xL)(1-
63
s))a(xH,xL)(1-μ), and wLmLm = ((1-s)xL/(xH+(1-s)xL))(1-xH+(1-xL)(1-s))a(xH,xL)(1-μ),
64
through his successful offspring that become high-fecundity breeding females, low-
65
fecundity breeding females, high-fecundity breeding males, and low-fecundity
66
breeding males, respectively.
67
68
The expressions of the class-specific reproductive success define a fitness matrix A,
69
which is given by
70
71
𝑤Hf→Hf
𝑤Hf→Lf
𝐀 = (𝑤
Hf→Hm
𝑤Hf→Lm
𝑤Lf→Hf
𝑤Lf→Lf
𝑤Lf→Hm
𝑤Lf→Lm
𝑤Hm→Hf
𝑤Hm→Lf
𝑤Hm→Hm
𝑤Hm→Lm
𝑤Lm→Hf
𝑤Lm→Lf
𝑤Lm→Hm ).
𝑤Lm→Lm
72
73
3
(A3)
74
Appendix B. Fitness
75
76
Here we define the fitness of an individual according to its condition. This depends on
77
the contribution of an individual to each class, weighted by the corresponding
78
reproductive values, and divided by the mean reproductive value of the focal class
79
(Taylor & Frank 1996; Frank 1998). For example, the fitness of a high-quality female
80
is given by her contribution in terms of offspring to the different classes (as given by
81
the expressions of reproductive success derived above), weighted by the reproductive
82
value of these classes, divided by the mean reproductive value of a high-quality
83
female. This is given by
84
85
𝑊Hf =
𝑣Hf 𝑤Hf→Hf +𝑣Lf 𝑤Hf→Lf +𝑣Hm 𝑤Hf→Hm +𝑣Lm 𝑤Hf→Lm
𝑣Hf
.
(B1)
86
87
The fitness of a focal individual in each one of the other classes is derived in a similar
88
way. The expected fitness of a random individual in the population is given by the
89
class-specific fitness weighted by the frequency (u) and reproductive value (v) of each
90
class (Taylor & Frank 1996; Frank 1998). This is
91
92
𝑊 = 𝑢fH 𝑣fH 𝑊fH + 𝑢mH 𝑣mH 𝑊mH + 𝑢fL 𝑣fL 𝑊fL + 𝑢mL 𝑣mL 𝑊mL.
(B2)
93
94
Expanding the right hand side of this equation, we show that the expected fitness of a
95
focal individual is given by
96
4
97
𝑊 = 𝑐f ((1 − 𝑥H )𝑘(𝑥H , 𝑥L ) + (1 − 𝑥L )(1 − 𝑠)𝑎(𝑥H , 𝑥L )) + 𝑐m (𝑥
98
𝑥H + (1 − 𝑥L )(1 − 𝑠))𝑎(𝑥H , 𝑥L ) + 𝑥
99
𝑠))𝑎(𝑥H , 𝑥L )),
𝑥L (1−𝑠)
(1
H +𝑥L (1−𝑠)
𝑥H
H +𝑥L (1−𝑠)
(1 −
− 𝑥H + (1 − 𝑥L )(1 −
(B3)
100
101
where cf is the class-reproductive value of females, and cm is the class-reproductive
102
value of males.
103
104
Appendix C. Stable class-frequency and reproductive value
105
106
The frequency and the individual reproductive value of each class can be derived
107
from the matrix A (defined in equation A3; Taylor & Frank 1996). More specifically,
108
the elements of the right-eigenvector corresponding to the leading eigenvalue of
109
matrix A give the stable-class frequency of each class (u), while the elements of the
110
left-eigenvector corresponding to the leading eigenvalue of matrix A give the
111
individual reproductive value of each class (v). The stable-class frequencies are given
112
by uHf = uLf = uHm = uLm = ¼. The class-reproductive values are: (1) cf = μ/(ϕ+μ) for
113
females; and (2) cm = ϕ/(ϕ+μ) for males. The individual reproductive values are then
114
given by: (1) vHf = cf (1-zH) for a high-fecundity female; (2) vLf = cf (1-zL)(1-s) for a
115
low-fecundity female; (3) vLf = cm ( zH/(zH + zL(1-s))(1-zH + (1-zL)(1-s)) for a high-
116
fecundity male; and (4) vLm = cm (zL(1-s)/(zH + zL(1-s))(1-zH + (1-zL)(1-s)) for a low-
117
fecundity male.
118
119
120
5
121
Appendix D. Relatedness
122
123
Here, we define the coefficients of relatedness between interacting individuals. We
124
assume vanishingly small genetic variation. First, we focus on haploid inheritance,
125
then we focus on diploid inheritance, and finally we focus on haplodiploid
126
inheritance. We first define recursion equations that describe the dynamics of the
127
coefficients of consanguinity between successive generations, which we then solve
128
for equilibrium. These coefficients of consanguinity enable us to derive the
129
coefficients of relatedness between interacting individuals (Bulmer 1994).
130
131
Haploidy
132
133
We focus on the coefficient of consanguinity between a juvenile female and a juvenile
134
male before dispersal and before mating, which is denoted by f. The probability that
135
two juveniles are offspring of the high-fecundity mother is (1-zH)/(1-zH+(1-zL)(1-s))
136
times zH/(zH+zL(1-s)), and the probability that two juveniles are offspring of the low-
137
fecundity mother is ((1-zL)(1-s))/(1-zH+(1-zL)(1-s)) times zL(1-s)/(zH+zL(1-s)).With
138
probability ½ both siblings derive the same gene, otherwise with probability ½ the
139
genes are identical with probability f. The probability that two juveniles are not
140
siblings is the probability that the juvenile female is an offspring of the high-fecundity
141
mother and the juvenile male is an offspring of the low-fecundity mother, which is
142
given by (1-zH)/(1-zH+(1-zL)(1-s)) times zL(1-s)/(zH+zL(1-s)), plus the probability that
143
the juvenile female is an offspring of the low-fecundity mother and a juvenile male is
144
an offspring of the high-fecundity mother, which is given by ((1-zL)(1-s))/(1-zH+(1-
145
zL)(1-s)) times zH/(zH+zL(1-s)). If the two juveniles are not siblings then they may
6
146
have identical genes if mothers are both natives of the same patch, which occurs with
147
probability φ = (1-d)2/(1-kd)2, times the probability that they share genes in common,
148
which is given by f. The recursion equation is then given by
149
150
𝑓′ = (1−𝑧
151
(1−𝑧
1−𝑧H
𝑧H
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L (1−𝑠)
1−𝑧H
𝑧L (1−𝑠)
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L
(1−𝑧L )(1−𝑠)
+ 1−z
(1−𝑧L )(1−𝑠)
+ 1−𝑧
(1−𝑠)
𝑧L (1−𝑠)
1
1
) (2 + 2 𝑓) +
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L (1−𝑠)
𝑧H
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L
1
1
1
) 𝜑 (4 𝛾 + 2 𝑓 + 4 𝜂).
(1−𝑠)
(D1)
152
153
Let us now focus on the coefficient of consanguinity between two juvenile females,
154
which is denoted by γ. The probability that two juvenile females are offspring the
155
high-fecundity mother is ((1-zH)/(1-zH+(1-zL)(1-s)))2, and the probability that two
156
juvenile females are offspring of the low-fecundity mother is (((1-zL)(1-s))/(1-zH+(1-
157
zL)(1-s)))2. With probability ½ both siblings derive the same gene, otherwise with
158
probability ½ the genes are identical with probability f. The probability that two
159
juveniles are not siblings is the probability that the juvenile female is an offspring of
160
the high-fecundity mother and the other juvenile female is an offspring of the low-
161
fecundity mother, which is given by two time (1-zH)/(1-zH+(1-zL)(1-s)) times ((1-
162
zL)(1-s))/(1-zH+(1-zL)(1-s)). If the two juveniles are not siblings then they may have
163
identical genes if mothers are both natives of the same patch, which occurs with
164
probability φ, times the probability that they share genes in common. With probability
165
¼ they both derive maternal genes, and thus the probability that they share the same
166
gene is γ. With probability ½ one juvenile female derives a maternal gene, while the
167
other derives a paternal gene, and thus the probability that they share the same gene is
168
f. With probability ¼ they both derive paternal genes, and thus the probability that
169
they share the same gene is η. The recursion equation is then given by
170
7
171
𝛾′ = ((1−𝑧
172
(2 1−𝑧
1−𝑧H
H +(1−𝑧L
2
H +(1−𝑧L
(1−𝑧L )(1−𝑠)
1−𝑧H
H +(1−𝑧L )(1−𝑠)
2
(1−𝑧L )(1−𝑠)
) + (1−𝑧
)(1−𝑠)
1
1−𝑧H +(1−𝑧L )(1−𝑠)
1
1
) ) (2 + 2 𝑓) +
)(1−𝑠)
1
1
) 𝜑 (4 𝛾 + 2 𝑓 + 4 𝜂).
(D2)
173
174
Finally, let us focus on the coefficient of consanguinity between two juvenile males,
175
which is denoted by η. The probability that two juvenile males are offspring of the
176
high-fecundity mother is (zH/(zH+zL(1-s)))2, and the probability that two juvenile
177
males are offspring of the low-fecundity mother is (zL(1-s)/(zH+zL(1-s)))2. With
178
probability ½ both siblings derive the same gene, otherwise with probability ½ the
179
genes are identical with probability f. The probability that two juveniles are not
180
siblings is the probability that one is an offspring of the high-fecundity mother and the
181
other is an offspring of the low-fecundity mother, which is given by two time
182
zH/(zH+zL(1-s)) times zL(1-s)/(zH+zL(1-s)). If the two juveniles are not siblings then
183
they may have identical genes if mothers are both natives of the same patch, which
184
occurs with probability φ, times the probability that they share genes in common.
185
With probability ¼ they both derive maternal genes, and thus the probability that they
186
share the same gene is γ. With probability ½ one juvenile female derives a maternal
187
gene, while the other derives a paternal gene, and thus the probability that they share
188
the same gene is f. With probability ¼ they both derive paternal genes, and thus the
189
probability that they share the same gene is η. The recursion equation is then given by
190
191
𝜂′ = ((𝑧
192
1
2
𝑧H
H +𝑧L
2
) + (𝑧
(1−𝑠)
𝑧L (1−𝑠)
H +𝑧L
2
1
1
) ) (2 + 2 𝑓) + (2 𝑧
(1−𝑠)
1
𝑓 + 4 𝜂).
𝑧H
𝑧L (1−𝑠)
1
) 𝜑 (4 𝛾 +
H +𝑧L (1−𝑠) 𝑧H +𝑧L (1−𝑠)
(D3)
193
8
194
We can find the coefficient of consanguinity (f) by solving these recursion equations
195
for equilibrium (i.e. by setting 𝑓′ = 𝑓, 𝛾′ = 𝛾, and 𝜂′ = 𝜂). The coefficient of
196
consanguinity between a mother and herself is p = 1. The coefficient of consanguinity
197
between a mother and her daughter or son is pD = pS = ½p+ ½f. The coefficient of
198
consanguinity between a mother and a daughter or son of the other mother is pF = pM
199
= φ(½γ+½f). The relatedness between a mother and her daughters or sons is rD = rS =
200
pD / p. The relatedness between a mother and a daughter or son of the other mother is
201
rF = rM = pF / p.
202
203
Diploidy
204
205
We first focus on the coefficient of consanguinity between a juvenile female and a
206
juvenile male before mating takes place. The probability that two juveniles are
207
offspring of the high-fecundity mother is (1-zH)/(1-zH+(1-zL)(1-s)) times zH/(zH+zL(1-
208
s)), and the probability that two juveniles are offspring of the low-fecundity mother is
209
((1-zL)(1-s))/(1-zH+(1-zL)(1-s)) times zL(1-s)/(zH+zL(1-s)).With probability ½ both
210
siblings derive a maternal gene (or a paternal gene), in which case they are copies of
211
the same gene with probability ½, otherwise they are identical with probability f. With
212
probability ½ one sibling derives a maternal gene and the other sibling derives a
213
paternal gene, in which case they are identical with probability f. The probability that
214
two juveniles are not siblings is the probability that the juvenile female is an offspring
215
the high-fecundity mother and the juvenile male is an offspring of the low-fecundity
216
mother, which is given by (1-zH)/(1-zH+(1-zL)(1-s)) times zL(1-s)/(zH+zL(1-s)), plus the
217
probability that the juvenile female is an offspring of the low-fecundity mother and
218
the juvenile male is an offspring of the high-fecundity mother, which is given by ((1-
9
219
zL)(1-s))/(1-zH+(1-zL)(1-s)) times zH/(zH+zL(1-s)). If the two juveniles are not siblings
220
then they may have identical genes if mothers are both natives of same the patch,
221
which occurs with probability φ. If both mothers are natives, then: with probability ¼
222
both genes are maternally derived, in which case they are identical with probability γ;
223
with probability ½ one gene in maternally derived while the other is paternally
224
derived, in which case they are identical with probability f; and finally with
225
probability ¼ both genes are paternally derived, in which case they are identical with
226
probability η. The recursion equation is then given by
227
228
𝑓′ = (1−𝑧
229
(1−𝑧
1−𝑧H
𝑧H
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L (1−𝑠)
1−𝑧H
𝑧L (1−𝑠)
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L (1−𝑠)
(1−𝑧L )(1−𝑠)
+ 1−𝑧
(1−𝑧L )(1−𝑠)
+ 1−𝑧
𝑧L (1−𝑠)
1 1
1
1
) (2 (2 + 2 𝑓) + 2 𝑓) +
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L (1−𝑠)
𝑧H
1
1
1
) 𝜑 (4 𝛾 + 2 𝑓 + 4 𝜂).
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L (1−𝑠)
(D4)
230
231
Let us now focus on the coefficient of consanguinity between two juvenile females.
232
The probability that two juvenile females are offspring of the high-fecundity mother
233
is ((1-zH)/(1-zH+(1-zL)(1-s)))2, and the probability that two juvenile females are
234
offspring of the low-fecundity mother is (((1-zL)(1-s))/(1-zH+(1-zL)(1-s)))2. With
235
probability ¼ they both derive a maternal gene, in which case they are copies of the
236
same gene with probability ½, otherwise they are identical with probability f. With
237
probability ½ one the juvenile female derives a paternal gene, while the other derives
238
a maternal gene, in which case they are identical with probability f. With probability
239
¼ they both derive a paternal gene, in which case they are copies of the same gene
240
with probability ½, otherwise they are identical with probability f. The probability that
241
two juveniles are not siblings is the probability that one juvenile female is an
242
offspring of the high-fecundity mother and the other juvenile female is an offspring of
243
the low-fecundity mother, which is given by two time (1-zH)/(1-zH+(1-zL)(1-s)) times
10
244
((1-zL)(1-s))/(1-zH+(1-zL)(1-s)). If the two juveniles are not siblings then they may
245
have identical genes if mothers are both natives of the same patch, which occurs with
246
probability φ, times the probability that they share genes in common. With probability
247
¼ they both derive maternal genes, and thus the probability that they share the same
248
gene is γ. With probability ½ one juvenile female derives a maternal gene, while the
249
other derives a paternal gene, and thus the probability that they share the same gene is
250
f. With probability ¼ they both derive paternal genes, and thus the probability that
251
they share the same gene is η. The recursion equation is then given by
252
253
𝛾′ = ((1−𝑧
254
(2 1−𝑧
1−𝑧H
H +(1−𝑧L
2
1−𝑧H
H +(1−𝑧L )(1−𝑠)
2
(1−𝑧L )(1−𝑠)
) + (1−𝑧
)(1−𝑠)
H +(1−𝑧L
(1−𝑧L )(1−𝑠)
1−𝑧H +(1−𝑧L )(1−𝑠)
1 1
1
1
) ) (2 (2 + 2 𝑓) + 2 𝑓) +
)(1−𝑠)
1
1
1
) 𝜑 (4 𝛾 + 2 𝑓 + 4 𝜂).
(D5)
255
256
Finally, let us focus on the coefficient of consanguinity between two juvenile males.
257
The probability that two juvenile males are offspring of the high-fecundity mother is
258
(zH/(zH+zL(1-s)))2, and the probability that two juvenile males are offspring of the low-
259
fecundity mother is (zL(1-s)/(zH+zL(1-s)))2. With probability ¼ they both derive a
260
maternal gene, in which case they are copies of the same gene with probability ½,
261
otherwise they are identical with probability f. With probability ½ one the juvenile
262
female derives a paternal gene, while the other derives a maternal gene, in which case
263
they are identical with probability f. With probability ¼ they both derive a paternal
264
gene, in which case they are copies of the same gene with probability ½, otherwise
265
they are identical with probability f. The probability that two juveniles are not siblings
266
is the probability that one is an offspring of the high-fecundity mother and the other is
267
an offspring of the low-fecundity mother, which is given by two time zH/(zH+zL(1-s))
11
268
times zL(1-s)/(zH+zL(1-s)). If the two juveniles are not siblings then they may have
269
identical genes if mothers are both natives of the same patch, which occurs with
270
probability φ, times the probability that they share genes in common. With probability
271
¼ they both derive maternal genes, and thus the probability that they share the same
272
gene is γ. With probability ½ one juvenile female derives a maternal gene, while the
273
other derives a paternal gene, and thus the probability that they share the same gene is
274
f. With probability ¼ they both derive paternal genes, and thus the probability that
275
they share the same gene is η. The recursion equation is then given by
276
277
𝜂′ = ((𝑧
278
(2 𝑧
𝑧H
H +𝑧L
2
) + (𝑧
(1−𝑠)
𝑧H
𝑧L (1−𝑠)
H +𝑧L (1−𝑠) 𝑧H +𝑧L (1−𝑠)
2
𝑧L (1−𝑠)
H +𝑧L
1
1 1
1
1
) ) (2 (2 + 2 𝑓) + 2 𝑓) +
(1−𝑠)
1
1
) 𝜑 (4 𝛾 + 2 𝑓 4 𝜂).
(D6)
279
280
We can find the coefficient of consanguinity f by solving these recursion equations for
281
equilibrium (i.e. by setting 𝑓′ = 𝑓, 𝛾′ = 𝛾, and 𝜂′ = 𝜂). The coefficient of
282
consanguinity between a mother and herself is p = ½ + ½f. The coefficient of
283
consanguinity between a mother and her daughter or son is pD = pS = ½p+ ½f. The
284
coefficient of consanguinity between a mother and a daughter or son of the other
285
mother is pF = pM = φ(½γ+½f). The relatedness between a mother and her daughters
286
or sons is rD = rS = pD / p. The relatedness between a mother and a daughter or son of
287
the other mother is rF = rM = pF / p.
288
289
290
Haplodiploidy
291
12
292
We first focus on the coefficient of consanguinity between two mating partners, this is
293
the coefficient of consanguinity between opposite-sex juveniles in a focal patch
294
before dispersal. The probability that two opposite-sex juveniles are offspring of to
295
the high-fecundity mother is (1-zH)/(1-zH+(1-zL)(1-s)) times zH/(zH+zL(1-s)), and the
296
probability that two juveniles are offspring of the low-fecundity mother is ((1-zL)(1-
297
s))/(1-zH+(1-zL)(1-s)) times zL(1-s)/(zH+zL(1-s)). The juvenile male derives his gene
298
from the mother. With probability ½ juvenile female also derives a maternal gene, in
299
which case they are copies of the same gene with probability ½, otherwise they are
300
identical with probability f. With probability ½ one the juvenile female derives a
301
paternal gene, in which case they are identical with probability f. The probability that
302
two juveniles are not siblings is the probability that the juvenile female is an offspring
303
of the high-fecundity mother and the juvenile male is an offspring of the low-
304
fecundity mother, which is given by (1-zH)/(1-zH+(1-zL)(1-s)) times zL(1-s)/(zH+zL(1-
305
s)), plus the probability that the juvenile female is offspring of the low-fecundity
306
mother and the juvenile male is an offspring of the high-fecundity mother, which is
307
given by ((1-zL)(1-s))/(1-zH+(1-zL)(1-s)) times zH/(zH+zL(1-s)). If the two juveniles are
308
not siblings then they may have identical genes if mothers are both natives of the
309
same patch, which occurs with probability φ, times the probability that they share
310
genes in common. With probability ½ the juvenile female gene is maternally derived,
311
and thus the probability that she has the same gene than the juvenile male is γ. With
312
probability ½ the juvenile female gene is paternally derived, and thus the probability
313
that she has the same gene than the juvenile male is f. The recursion equation is then
314
given by
315
𝑓′ = (1−𝑧
316
(1−𝑧
𝑧H
)(1−𝑠)
𝑧H +𝑧L (1−𝑠)
H +(1−𝑧L
1−𝑧H
1−𝑧H
𝑧L (1−𝑠)
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L
(1−𝑧L )(1−𝑠)
+ 1−𝑧
𝑧L (1−𝑠)
1 1
)
(
(
)(1−𝑠)
𝑧H +𝑧L (1−𝑠)
2 2
H +(1−𝑧L
(1−𝑧L )(1−𝑠)
+ 1−𝑧
(1−𝑠)
𝑧H
1
) 𝜑 (2 𝛾 + 2 𝑓).
H +(1−𝑧L )(1−𝑠) 𝑧H +𝑧L (1−𝑠)
13
1
1
1
+ 2 𝑓) + 2 𝑓) +
(D7)
317
318
Let us now focus on the coefficient of consanguinity between two juvenile females.
319
The probability that two juvenile females are offspring of the high-fecundity mother
320
is ((1-zH)/(1-zH+(1-zL)(1-s)))2, and the probability that two juvenile females are
321
offspring of the low-fecundity mother is (((1-zL)(1-s))/(1-zH+(1-zL)(1-s)))2. With
322
probability ¼ they both derive a maternal gene, in which case they are copies of the
323
same gene with probability ½, otherwise they are identical with probability f. With
324
probability ½ one the juvenile female derives a paternal gene, while the other derives
325
a maternal gene, in which case they are identical with probability f. With probability
326
¼ they both derive a paternal gene, in which case they are copies of the same gene.
327
The probability that two juveniles are not siblings is the probability that one juvenile
328
female is offspring of the high-fecundity mother and the other juvenile female is
329
offspring of the low-fecundity mother, which is given by two time (1-zH)/(1-zH+(1-
330
zL)(1-s)) times ((1-zL)(1-s))/(1-zH+(1-zL)(1-s)). If the two juveniles are not siblings
331
then they may have identical genes if mothers are both natives to the same patch,
332
which occurs with probability φ, times the probability that they share genes in
333
common. With probability ¼ they both derive maternal genes, and thus the
334
probability that they share the same gene is γ. With probability ½ one juvenile female
335
derives a maternal gene, while the other derives a paternal gene, and thus the
336
probability that they share the same gene is f. With probability ¼ they both derive
337
paternal genes, and thus the probability that they share the same gene is η. The
338
recursion equation is then given by
339
14
340
𝛾′ = ((1−𝑧
341
(2 1−𝑧
1−𝑧H
H +(1−𝑧L
2
H +(1−𝑧L
(1−𝑧L )(1−𝑠)
1−𝑧H
H +(1−𝑧L )(1−𝑠)
2
(1−𝑧L )(1−𝑠)
) + (1−𝑧
)(1−𝑠)
1
1−𝑧H +(1−𝑧L )(1−𝑠)
1 1
1
1
1
) ) (4 (2 + 2 𝑓) + 2 𝑓 + 4) +
)(1−𝑠)
1
1
) 𝜑 (4 𝛾 + 2 𝑓 + 4 𝜂).
(D8)
342
343
Finally, let us focus on the coefficient of consanguinity between two juvenile males.
344
The probability that two juvenile males are offspring of the high-fecundity mother is
345
(zH/(zH+zL(1-s)))2, and the probability that two juvenile males are offspring of the low-
346
fecundity mother is (zL(1-s)/(zH+zL(1-s)))2. With probability ½ they both derive copies
347
of the same gene with probability ½, otherwise with probability ½ they are identical
348
with probability f. The probability that two juveniles are not siblings is the probability
349
that one is an offspring of the high-fecundity mother and the other is an offspring of
350
the low-fecundity mother, which is given by two time zH/(zH+zL(1-s)) times zL(1-
351
s)/(zH+zL(1-s)). If the two juveniles are not siblings then they may have identical
352
genes if mothers are both natives of the same patch, which occurs with probability φ,
353
times the probability that they share genes in common, which is given by γ. The
354
recursion equation is then given by
355
356
𝜂′ = ((𝑧
𝑧H
H +𝑧L
2
) + (𝑧
(1−𝑠)
zL (1−𝑠)
H +𝑧L
2
1
1
) ) (2 + 2 𝑓) + (2 𝑧
(1−𝑠)
𝑧H
𝑧L (1−𝑠)
) 𝜑𝛾. (D9)
H +𝑧L (1−𝑠) 𝑧H +𝑧L (1−𝑠)
357
358
We can find these three coefficients of consanguinity by solving these three recursion
359
equations for equilibrium (i.e. by setting 𝑓′ = 𝑓, 𝛾′ = 𝛾, and 𝜂′ = 𝜂). The coefficient
360
of consanguinity between a mother and herself is p = ½ + ½f. The coefficient of
361
consanguinity between a mother and her daughter is pD = ½p+ ½f. The coefficient of
362
consanguinity between a mother and her son is pS = ½ + ½f. The coefficient of
363
consanguinity between a mother and the daughter of the other mother is pF = φ(½γ +
15
364
½f ). The coefficient of consanguinity between a mother and the son of the other
365
mother is pM = φγ. The relatedness between a mother and her daughters is rD = pD / p,
366
and the relatedness between a mother and her sons is rS = pS / p. The relatedness
367
between a mother and a daughter the other mother is rF = pF / p, the relatedness
368
between a mother and a son the other mother is rM = pM / p.
369
370
Appendix E. Selection gradient
371
372
The selection gradient for the sex ratio expressed conditionally on the mother’s
373
fecundity is given by the slope of her fitness W on her breeding value for the sex ratio
374
(Taylor & Frank 1996; Frank 1998). The breeding value of a high-fecundity mother is
375
denoted by gfH, while the breeding value of a low-fecundity mother is denoted by gfL.
376
The selection gradients are given by
377
378
379
380
d𝑊
𝜕
d𝑔fH
H
𝜕
𝑐m (𝜕𝑥 (𝑥
H
𝜕
(𝑥
𝜕𝑥H
𝜕
= 𝑐f (𝜕𝑥 ((1 − 𝑥H )𝑎(𝑥H , 𝑥L ))𝑟MD + 𝜕𝑥 ((1 − 𝑥L )(1 − 𝑠)𝑎(𝑥H , 𝑥L ))𝑟MF ) +
H
𝑥H
H +𝑥L (1−𝑠)
(1 − 𝑥H + (1 − 𝑥L )(1 − 𝑠))𝑎(𝑥H , 𝑥L )) 𝑟MS +
𝑥L (1−𝑠)
(1
H +𝑥L (1−𝑠)
− 𝑥H + (1 − 𝑥L )(1 − 𝑠))𝑎(𝑥H , 𝑥L )) 𝑟MM ), and
𝜕
𝜕
(E1)
381
382
383
384
d𝑊
d𝑔fL
= 𝑐f (𝜕𝑥 ((1 − 𝑥H )𝑎(𝑥H , 𝑥L ))𝑟MF + 𝜕𝑥 ((1 − 𝑥L )(1 − 𝑠)𝑎(𝑥H , 𝑥L ))𝑟MD ) +
L
𝜕
𝑐m (𝜕𝑥 (𝑥
L
𝜕
𝜕𝑥L
(𝑥
L
𝑥H
H +𝑥L (1−𝑠)
𝑥L (1−𝑠)
H +𝑥L (1−𝑠)
(1 − 𝑥H + (1 − 𝑥L )(1 − 𝑠))𝑎(𝑥H , 𝑥L )) 𝑟MM +
(1 − 𝑥H + (1 − 𝑥L )(1 − 𝑠))𝑎(𝑥H , 𝑥L )) 𝑟MS ),
385
16
(E2)
386
for a high-quality mother, and for a low-quality mother, respectively. If we expand
387
the right-hand side of these equations, we get the left-hand side (LHS) of inequalities
388
(1) and (2) in the main text, which are the conditions for natural selection to favour an
389
increase in the sex allocation strategy. To determine the optimal sex allocation
390
strategy, we set the LHS of inequalities (1) and (2) to zero, and we solve the system of
391
equations for equilibrium.
392
393
Appendix F. Convergence stability
394
395
Here we determine the convergence stability (CS; Christiansen 1991; Eshel 1996;
396
Taylor 1996) of the optimal sex allocation strategies. To determine if a pair of optimal
397
sex allocation strategies is convergence stable we define the matrix:
398
𝜕
399
(
𝜕𝑊
𝜕𝑧L
(𝜕𝑔 |
𝜕
𝜕𝑊
𝜕𝑧L
fL
𝑥L =𝑧L
(𝜕𝑔 |
fH
𝜕
)
)
𝑥H =𝑧H
𝜕𝑊
𝜕𝑧H
(𝜕𝑔 |
𝜕
𝜕𝑊
𝜕𝑧H
fL
(𝜕𝑔 |
fH
)
𝑥L =𝑧L
𝑥H =𝑧H
)
)
|
|
.
(F1)
∗
𝑧H =𝑧H
,𝑧L =𝑧L∗
400
401
The pair of optimal strategies (zH* and zL*) are convergence stable if both eigenvalues
402
of matrix (F1) have negative real parts (Otto and Day 2007). If mothers are obliged to
403
invest a fixed amount into sons, irrespective of their fecundity, then the condition for
404
convergence stability is
405
406
𝜕
𝜕𝑧
𝜕𝑊
(𝜕𝑔 |
f
𝑧H =𝑧L =𝑧
)|
(F2)
< 0,
𝑧=𝑧 ∗
407
17
408
where gf is the breeding value of a random a random mother in the population. We
409
find that both the facultative and the obligate sex allocation strategies are convergence
410
stable.
411
412
Appendix G. Self-knowledge model
413
414
Life-cycle and fitness
415
416
In the main text we outlined a model where all patches have one high-fecundity and
417
one low-fecundity mother. Here we extend this model and instead of considering that
418
all patches have one high-fecundity and one low-fecundity mother, we consider a
419
model where the quality of each female is defined before the breeding season.
420
Specifically, we assume that juvenile females become high-fecundity mothers with
421
probability ρ and become low-fecundity mothers with probability 1-ρ. This means
422
that: (1) the frequency of patches with two high-fecundity mothers is u0 = ρ2; (2) the
423
frequency of patches with one high-fecundity mother and one low-fecundity mother is
424
u1 = 2ρ(1-ρ); and (3) the frequency of patches with two low-fecundity mothers is u2 =
425
(1-ρ)2. This also means that: (1) the frequency of high-fecundity mothers in patches
426
with two high-fecundity mothers is uH0 = ρ2; the frequency of high-fecundity mothers
427
in patches with one high-fecundity mother and one low-fecundity mother is uH1 = ρ(1-
428
ρ); (3) the frequency of low-fecundity mothers in patches with one high-fecundity
429
mother and one low-fecundity mother is uL1 = ρ(1-ρ); and (4) the frequency of low-
430
fecundity mothers in patches with two low-fecundity mothers is uL2 = (1-ρ)2. Note
431
that we now use two indices. First, we denote the quality of the mother by the letters
432
‘H’ (high-fecundity) and ‘L’ (low-fecundity). Second, we denote the condition of the
18
433
patch by the numbers ‘0’ (patches with two high-fecundity mothers), ‘1’ (mixed
434
patches with one high-fecundity mother and one low-fecundity mother), and ‘2’
435
(patches with two low-fecundity mothers).
436
437
To analyse this extended model we follow the steps delineated above. We start by
438
defining the class-specific reproductive success of a focal breeder. This depends on
439
the quality of the focal breeder and on the condition of the patch (i.e. ‘H’ or ‘L’, and
440
‘0’, ‘1’ or ‘2’) and on the quality and condition of the breeder’s successful offspring
441
(i.e. ‘H’ or ‘L’, and ‘0’, ‘1’ or ‘2’). For convenience let us define the following
442
quantities: 1-𝑧̅ = u02(1-zH0)+u1(1-zH1+(1-zL1)(1-s))+ u22(1-zL2)(1-s); q0(xH0,yH0) =
443
1/(((1-xH0)+(1-yH0))(1-d)+(1-𝑧̅)d(1-k)); q1(xH1,xL1) = 1/(((1-xH1)+(1-xH1)(1-s))(1-d)+(1-
444
𝑧̅)d(1-k)); q2(xL2,yL2) = 1/(((1-xL2)+(1-yL2))(1-s)(1-d)+(1-𝑧̅)d(1-k)); 𝑞̅ =
445
u0q0(xH0,yH0)+u1q1(xH1,xL1) u2q2(xL2,yL2); a0(xH0,yH0) = (1-d)q0(xH0,yH0)+d(1-k)𝑞̅;
446
a1(xH1,xL1) = (1-d)q1(xH1,xL1)+d(1-k)𝑞̅; a2(xL2,yL2) = (1-d)q2(xL2,yL2)+d(1-k)𝑞̅ ; where x
447
denotes the sex ratio strategy of the focal individual, y denotes the sex ratio strategy
448
of the group mate, and z denotes the average sex ratio strategy across the population.
449
We can now define the fitness success of each female.
450
451
The fitness of a focal individual depends on its condition. We follow the method
452
outlined in appendix B. The fitness of a individual, according to its condition, is given
453
by: WH0f = cf(1-xH0)a0(xH0,yH0)/vH0f, WH1f = cf(1-xH1)a1(xH1,xL1)/vH1f, WL1f = cf(1-
454
xL1)(1-s)a1(xH1,xL1)/vL1f, WL2f = cf(1-xL2)(1-s)a2(xL2,yL2)/vL2f, WH0m =
455
cm(xH0/(xH0+yH0))(1-xH0+1-yH0)a0(xH0,yH0)/vH0f, WH1m = cm(xH1/(xH1+xL1(1-s)))(1-
456
xH1+(1-xL1)(1-s))a1(xH1,xL1)/vH1f, WL1m = cm(xL1(1-s)/(xH1+xL1(1-s)))(1-xH1+(1-xL1)(1-
457
s))a1(xH1,xL1)/vL1f, WL2m = cm(xL2/(xL2+yL2))(1-xL2+1-yL2)(1-s)a2(xL2,yL2)/vL2f.
19
458
Relatedness
459
460
We follow the approach outlined above to determine the coefficients of relatedness
461
among interacting individuals. We focus on the coefficient of consanguinity between
462
two opposite-sex juveniles in patches with two high-fecundity mothers (denoted by
463
f0), in mixed patches (denoted by f1), and in patches with two low-fecundity mothers
464
(denoted by f2); on the coefficient of consanguinity between two juvenile females in
465
patches with two high-fecundity mothers (denoted by γ0), in mixed patches (denoted
466
by γ1), and in patches with two low-fecundity mothers (denoted by γ2); and on the
467
coefficient of consanguinity between two juvenile males in patches with two high-
468
fecundity mothers (denoted by η0), in mixed patches (denoted by η1), and in patches
469
with two low-fecundity mothers (denoted by η2). As above, we define a recursion
470
equation for each one of these coefficients of consanguinity, and this gives us a
471
system of equations for each type of inheritance that we then solve for equilibrium.
472
473
Haploidy -- Here we focus on haploid inheritance. In patches with two high-fecundity
474
females and in patches with two low-fecundity females the probability that two
475
juveniles sampled at random are siblings is Pf0 = ½. In mixed patches: (1) the
476
probability that two juveniles of the opposite sex are siblings is given by Pf1 = ((1-
477
zH1)/(1-zH1+(1-zL1)(1-s)))(zH1/( zH1+zL1(1-s)))+( (1-zL1)(1-s))/(1-zH1+(1-zL1)(1-s)))(
478
zL1(1-s)/( zH1+zL1(1-s))); (2) the probability that two female juveniles are siblings is
479
given by Pγ1 = ((1-zH1)/(1-zH1+(1-zL1)(1-s)))2+((1-zL1)(1-s))/(1-zH1+(1-zL1)(1-s)))2; and
480
(3) the probability that two male juveniles are siblings is given by Pη1 = (zH1/(
481
zH1+zL1(1-s)))2+( zH1+zL1(1-s)))2. The probability that a female chosen at random after
482
dispersal is native to the patch is given by: (1) h0 = 2(1-zH0)(1-d)q0(zH0,zH0) in patches
20
483
with two high-fecundity mothers; h1 = ((1-zH1)+(1-zL1)(1-s))(1-d)q1(zH1,zL1) in mixed
484
patches; and (3) h2 = 2(1-zL2)(1-d)q2(zL2,zL2) in patches with two low-fecundity
485
mothers. Thus, the probabilities of co-philopatry are given by φ0 = h02, φ1 = h12, and
486
φ2 = h22. The probability that a focal patch had two high-fecundity females in the
487
previous generation is π0 = u0. The probability that a focal patch had a high-fecundity
488
female and a low-fecundity female is π1 = u1. The probability that a focal patch had
489
two low-fecundity females in the previous generation is π2 = u2. The probability that a
490
disperser was born in a patch with two high-fecundity females is α0 = (u02(1-zH0))/(1-
491
𝑧̅). The probability that a disperser was born in a patch with one high-fecundity
492
female and one low-fecundity female is α1 = (u1((1-zH1)+(1-zL1)(1-s)))/(1-𝑧̅). The
493
probability that a disperser was born in a patch with two low-fecundity females is α2 =
494
(u22(1-zL2))/(1-𝑧̅). The recursion equations are given by
495
496
1
1
1
1
1
1
1
1
1
1
𝑓0 ′ = 𝑃f0 (2 + 2 𝜄) + (1 − 𝑃f0 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 + 4 𝜂1 )), (G1)
497
498
𝑓1 ′ = 𝑃f1 (2 + 2 𝜄) + (1 − 𝑃f1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 + 4 𝜂1 )), (G2)
499
1
1
1
1
1
2
2
4
2
4
1
1
1
1
1
500
𝛾1 ′ = 𝑃γ1 ( + 𝜄) + (1 − 𝑃γ1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 ( 𝛾1 + 𝑓1 + 𝜂1 )), and
501
(G3)
502
503
𝜂1 ′ = 𝑃𝜂1 (2 + 2 𝜄) + (1 − 𝑃𝜂1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 + 4 𝜂1 )),(G4)
504
505
in which the coefficient of inbreeding (denoted by ι) is given by
506
21
507
𝜄 = (𝜋0 (1 − ℎ0 ) + 𝜋1 (1 − ℎ1 ) + 𝜋2 (1 − ℎ2 ))(𝛼0 𝑓0 + 𝛼1 𝑓1 + 𝛼2 𝑓2 ) + 𝜋0 ℎ0 𝑓0 +
508
𝜋1 ℎ1 𝑓1 + 𝜋2 ℎ2 𝑓0 .
(G5)
509
510
Note that we only need four recursion equations, as the recursion equations for the
511
coefficients of consanguinity γ0, η0, f2, γ2, and η2 are all identical to the recursion
512
equation for the coefficient of consanguinity f0. We can find the coefficients of
513
consanguinity by solving these recursion equations for equilibrium. The coefficient of
514
consanguinity between a mother and herself is p = 1. The coefficient of consanguinity
515
between a mother and her daughter or son is pD = pS = ½pM+ ½ι. The coefficient of
516
consanguinity between a mother and a daughter or son of the other mother is pF = pM
517
= (π0φ0+ π2φ2)f0+π1φ1(½γ1+½f1). The relatedness between a mother and her daughters
518
or sons is rD = rS = pD / p. The relatedness between a mother and a daughter or son of
519
the other mother is rF = rM = pF / p.
520
521
Diploidy -- Here we focus on diploid inheritance. In patches with two high-fecundity
522
females and in patches with two low-fecundity females the probability two juveniles
523
sampled at random are siblings is Pf0 = ½. In mixed patches: (1) the probability that
524
two juveniles of the opposite sex are siblings is given by Pf1 = ((1-zH1)/(1-zH1+(1-
525
zL1)(1-s)))(zH1/( zH1+zL1(1-s)))+( (1-zL1)(1-s))/(1-zH1+(1-zL1)(1-s)))( zL1(1-s)/(
526
zH1+zL1(1-s))); (2) the probability that two female juveniles are siblings is given by Pγ1
527
= ((1-zH1)/(1-zH1+(1-zL1)(1-s)))2+((1-zL1)(1-s))/(1-zH1+(1-zL1)(1-s)))2; and (3) the
528
probability that two male juveniles are siblings is given by Pη1 = (zH1/( zH1+zL1(1-
529
s)))2+( zH1+zL1(1-s)))2. The probability that a female chosen at random after dispersal
530
is native to the patch is given by: (1) h0 = 2(1-zH0)(1-d)q0(zH0,zH0) in patches with two
531
high-fecundity mothers; h1 = ((1-zH1)+(1-zL1)(1-s))(1-d)q1(zH1,zL1) in mixed patches;
22
532
and (3) h2 = 2(1-zL2)(1-d)q2(zL2,zL2) in patches with two low-fecundity mothers. Thus,
533
the probabilities of co-philopatry are given by φ0 = h02, φ1 = h12, and φ2 = h22. The
534
probability that a focal patch had two high-fecundity females in the previous
535
generation is π0 = u0. The probability that a focal patch had a high-fecundity female
536
and a low-fecundity female is π1 = u1. The probability that a focal patch had two low-
537
fecundity females in the previous generation is π2 = u2. The probability that a
538
disperser was born in a patch with two high-fecundity females is α0 = (u02(1-zH0))/(1-
539
𝑧̅). The probability that a disperser was born in a patch with one high-fecundity
540
female and one low-fecundity female is α1 = (u1((1-zH1)+(1-zL1)(1-s)))/(1-𝑧̅). The
541
probability that a disperser was born in a patch with two low-fecundity females is α2 =
542
(u22(1-zL2))/(1-𝑧̅). The recursion equations are given by
543
1 1
1
1
1
1
544
𝑓0 ′ = 𝑃f0 (2 (2 + 2 𝜄) + 2 𝜄) + (1 − 𝑃f0 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 +
545
1
𝜂 )),
4 1
(G6)
546
1 1
1
1
1
1
547
𝑓1 ′ = 𝑃f1 (2 (2 + 2 𝜄) + 2 𝜄) + (1 − 𝑃f1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 +
548
1
𝜂 )),
4 1
(G7)
549
1 1
1
1
1
1
550
𝛾1 ′ = 𝑃γ1 (2 (2 + 2 𝜄) + 2 𝜄) + (1 − 𝑃γ1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 +
551
1
𝜂 )),
4 1
and
(G8)
552
23
1 1
1
1
1
1
553
𝜂1 ′ = 𝑃𝜂1 (2 (2 + 2 𝜄) + 2 𝜄) + (1 − 𝑃𝜂1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝑓0 + 𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 +
554
1
𝜂 )),
4 1
(G9)
555
556
in which the coefficient of inbreeding is given by
557
558
𝜄 = (𝜋0 (1 − ℎ0 ) + 𝜋1 (1 − ℎ1 ) + 𝜋2 (1 − ℎ2 ))(𝛼0 𝑓0 + 𝛼1 𝑓1 + 𝛼2 𝑓2 ) + 𝜋0 ℎ0 𝑓0 +
559
𝜋1 ℎ1 𝑓1 + 𝜋2 ℎ2 𝑓0 .
(G10)
560
561
Note that we only need four recursion equations, as the recursion equations for the
562
coefficients of consanguinity γ0, η0, f2, γ2, and η2 are all identical to the recursion
563
equation for the coefficient of consanguinity f0. We find the coefficients of
564
consanguinity by simultaneously solving these recursion equations for equilibrium.
565
The coefficient of consanguinity between a mother and herself is p = ½+½ι. The
566
coefficient of consanguinity between a mother and her daughter or son is pD = pS =
567
½p+ ½ι. The coefficient of consanguinity between a mother and a daughter or son of
568
the other mother is pF = pM = (π0φ0+ π2φ2)f0+π1φ1(½γ1+½f1). The relatedness between
569
a mother and her daughters or sons is rD = rS = pD / p. The relatedness between a
570
mother and a daughter or son of the other mother is rF = rM = pF / p.
571
572
Haplodiploidy -- Here we focus on haplodipoloid inheritance. In patches with two
573
high-fecundity females and in patches with two low-fecundity females the probability
574
that two juveniles sampled at random are siblings is Pf0 = ½. In mixed patches: (1) the
575
probability that two juveniles of the opposite sex are siblings is given by Pf1 = ((1-
576
zH1)/(1-zH1+(1-zL1)(1-s)))(zH1/( zH1+zL1(1-s)))+( (1-zL1)(1-s))/(1-zH1+(1-zL1)(1-s)))(
24
577
zL1(1-s)/( zH1+zL1(1-s))); (2) the probability that two female juveniles are siblings is
578
given by Pγ1 = ((1-zH1)/(1-zH1+(1-zL1)(1-s)))2+((1-zL1)(1-s))/(1-zH1+(1-zL1)(1-s)))2; and
579
(3) the probability that two male juveniles are siblings is given by Pη1 = (zH1/(
580
zH1+zL1(1-s)))2+( zH1+zL1(1-s)))2. The probability that a female chosen at random after
581
dispersal is native to the patch is given by: (1) h0 = 2(1-zH0)(1-d)q0(zH0,zH0) in patches
582
with two high-fecundity mothers; h1 = ((1-zH1)+(1-zL1)(1-s))(1-d)q1(zH1,zL1) in mixed
583
patches; and (3) h2 = 2(1-zL2)(1-d)q2(zL2,zL2) in patches with two low-fecundity
584
mothers. Thus, the probabilities of co-philopatry are given by φ0 = h02, φ1 = h12, and
585
φ2 = h22. The probability that a focal patch had two high-fecundity females in the
586
previous generation is π0 = u0. The probability that a focal patch had a high-fecundity
587
female and a low-fecundity female is π1 = u1. The probability that a focal patch had
588
two low-fecundity females in the previous generation is π2 = u2. The probability that a
589
disperser was born in a patch with two high-fecundity females is α0 = (u02(1-zH0))/(1-
590
𝑧̅). The probability that a disperser was born in a patch with one high-fecundity
591
female and one low-fecundity females is α1 = (u1((1-zH1)+(1-zL1)(1-s)))/(1-𝑧̅). The
592
probability that a disperser was born in a patch with two low-fecundity females is α2 =
593
(u22(1-zL2))/(1-𝑧̅). The recursion equations are given by
594
1 1
1
1
1
1
2 2
2
2
2
2
1
1
1
595
𝑓0 ′ = 𝑃f0 ( ( + 𝜄) + 𝜄) + (1 − 𝑃f0 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 ) ( 𝛾0 + 𝑓0 ) +
596
𝜋1 𝜑1 (2 𝛾1 + 2 𝑓1 )),
597
(G11)
1
1
598
1 1
1
1
599
𝛾0 ′ = 𝑃f0 (2 (2 + 2 𝜄) + 2 𝜄) + (1 − 𝑃f0 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 ) (4 𝛾0 + 2 𝑓0 + 4 𝜂0 ) +
600
𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 + 4 𝜂1 )),
1
1
1
(G12)
25
601
602
1
1
𝜂0 ′ = 𝑃f0 (2 + 2 𝜄) + (1 − 𝑃f0 )((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝛾0 + 𝜋1 𝜑1 𝛾1 ),
(G13)
603
1 1
1
1
1
1
604
𝑓1 ′ = 𝑃f1 (2 (2 + 2 𝜄) + 2 𝜄) + (1 − 𝑃f1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 ) (2 𝛾0 + 2 𝑓0 ) +
605
𝜋1 𝜑1 (2 𝛾1 + 2 𝑓1 )),
606
(G14)
1
1
607
1 1
1
1
1
1
1
608
𝛾1 ′ = 𝑃γ1 (2 (2 + 2 𝜄) + 2 𝜄) + (1 − 𝑃γ1 ) ((𝜋0 𝜑0 + 𝜋2 𝜑2 ) (4 𝛾0 + 2 𝑓0 + 4 𝜂0 ) +
609
𝜋1 𝜑1 (4 𝛾1 + 2 𝑓1 + 4 𝜂1 )), and
1
1
1
(G15)
610
611
1
1
2
2
𝜂1 ′ = 𝑃𝜂1 ( + 𝜄) + (1 − 𝑃𝜂1 )((𝜋0 𝜑0 + 𝜋2 𝜑2 )𝛾0 + 𝜋1 𝜑1 𝛾1 ),
(G16)
612
613
in which the coefficient of inbreeding is given by
614
615
𝜄 = (𝜋0 (1 − ℎ0 ) + 𝜋1 (1 − ℎ1 ) + 𝜋2 (1 − ℎ2 ))(𝛼0 𝑓0 + 𝛼1 𝑓1 + 𝛼2 𝑓2 ) + 𝜋0 ℎ0 𝑓0 +
616
𝜋1 ℎ1 𝑓1 + 𝜋2 ℎ2 𝑓0 .
(G17)
617
618
Note that we only need six recursion equations, as the recursion equations for the
619
coefficients of consanguinity f2, γ2, and η2 are all identical to the recursion equation
620
for the coefficient of consanguinity f0, γ0, η0, respectively. We find the coefficients of
621
consanguinity by simultaneously solving these recursion equations for equilibrium.
622
The coefficient of consanguinity between a mother and herself is p = ½+½ι. The
623
coefficient of consanguinity between a mother and her daughter or her son is pD =
26
624
½p+ ½ι, whilst the coefficient of consanguinity between a mother and her son is pS =
625
½ + ½ι. The coefficient of consanguinity between a mother and a daughter or son of
626
the other mother is pF = pM = (π0φ0+ π2φ2)(½γ0+½f0)+π1φ1(½γ1+½f1), whilst the
627
coefficient of consanguinity between a mother and a son of the other mother is pF =
628
pM = (π0φ0+ π2φ2)γ0+π1φ1γ1. The relatedness between a mother and her daughters is rD
629
= pD / p, whilst the relatedness between a mother and her son is rS = pS / p. The
630
relatedness between a mother and the daughters of the other mother is rF = pF / p,
631
whilst the relatedness between a mother and the sons of the other mother is rM = pM /
632
p.
633
634
Selection gradient
635
636
The selection gradient for the sex ratio expressed conditionally on the mother’s
637
fecundity is given by the slope of her fitness W on her breeding value g for the sex
638
ratio (Taylor & Frank 1996; Frank 1998). The selection gradients are given by
639
640
𝑑𝑊H0
641
𝑑)𝑞0 ℎ0 (𝑐f (𝑢H0f 𝑟D + 𝑢H0f 𝑟F ) + 𝑐m (𝑢H0m 𝑟S + 𝑢H0m 𝑟M )), and
642
(G18)
𝑑𝑔fH0
= −𝑐f 𝑎0 𝑟D + 𝑐m
1−𝑧H0
𝑧H0
𝑎0 𝑟D − 𝑐m 𝑧
1
H0
(𝑢H0m 𝑟S + 𝑢H0m 𝑟M ) + (1 −
643
644
645
𝑑𝑊H1
𝑑𝑔fH1
= −𝑐f 𝑎1 𝑟D + 𝑐m
1−𝑧H1 +(1−𝑧L1 )(1−𝑠)
𝑎1 𝑟D
𝑧H1 +𝑧L1 (1−𝑠)
− 𝑐m 𝑧
1+(1−𝑠)
(𝑢H1m 𝑟S
H1 +𝑧L1 (1−𝑠)
+
𝑢L1m 𝑟M ) + (1 − 𝑑)𝑞1 ℎ1 (𝑐f (𝑢H1f 𝑟D + 𝑢L1f 𝑟F ) + 𝑐m (𝑢H1m 𝑟S + 𝑢L1m 𝑟M )),
646
27
(G19)
647
648
𝑑𝑊L1
𝑑𝑔fL1
= −𝑐f 𝑎1 𝑟D + 𝑐m
1−𝑧H1 +(1−𝑧L1 )(1−𝑠)
𝑎1 𝑟D
𝑧H1 +𝑧L1 (1−𝑠)
− 𝑐m 𝑧
1+(1−𝑠)
(𝑢H1m 𝑟M
H1 +𝑧L1 (1−𝑠)
+
𝑢L1m 𝑟S ) + (1 − 𝑑)𝑞1 ℎ1 (𝑐f (𝑢H1f 𝑟F + 𝑢L1f 𝑟D ) + 𝑐m (𝑢H1m 𝑟M + 𝑢L1m 𝑟S )), and
(G20)
649
650
𝑑𝑊L2
651
𝑑)𝑞2 ℎ2 (𝑐f (𝑢L2f 𝑟D + 𝑢L2f 𝑟F ) + 𝑐m (𝑢L2m 𝑟S + 𝑢L2m 𝑟M )),
652
(G21)
𝑑𝑔fL2
= −𝑐f 𝑎2 𝑟D + 𝑐m
1−𝑧L2
𝑧L2
1
𝑎2 𝑟D − 𝑐m 𝑧 (𝑢L2m 𝑟S + 𝑢L2m 𝑟M ) + (1 −
L2
653
654
Under self-knowledge females know their own fecundity but not that of their patch
655
mates. Therefore we set zH0 = zH1 = zH, and zL1 = zL2 = zL. The selection gradient of
656
high-fecundity mothers is then given by SH = uH0(dWH0/dgfH0) + uH1(dWH1/dgfH1),
657
while the selection gradient of low-fecundity mothers is given by SL = uL1(dWL1/dgfL1)
658
+ uL2(dWL2/dgfL2). To determine the optimal sex allocation strategy, we set these
659
selection gradients to zero (i.e. SH = 0 and SL = 0), and we solve the system of
660
equations for equilibrium.
661
662
663
664
665
666
667
668
669
670
28
671
672
673
674
675
Appendix H. Supplementary figures
676
677
678
Figure H1 | Local mate competition and local resource competition in viscous
679
populations. The strength of local mate competition (given by the third term in the
680
selection gradients) and the strength of local resource competition (given by the
681
fourth term in the selection gradients) for high- and low-fecundity mothers as a
682
function of the dispersal rate (d) for haploidy and diploidy (panels (a) and (b)) and for
29
683
haplodiploidy (panels (c) and (d)), assuming the parameter values k = 0, s = 0.75, zH =
684
zL = z = 0.1.
685
686
687
688
689
Figure H2 | Facultative sex allocation. (a,b) High-fecundity mothers (blue lines) are
690
favoured to invest relatively more into sons than are low-fecundity mothers (red
691
dashed line) in viscous populations (d < 1), under haploidy and diploidy. (c,d) High-
692
fecundity mothers (blue line) are favoured to invest relatively more into sons than are
693
low-fecundity mothers (red dashed lines) in viscous populations (d < 1), though both
30
694
are favoured to invest less into sons, under haplodiploidy. We arbitrarily set the total
695
number of offspring of a high-fecundity mother to 100. Parameter values: k = 0.
696
697
698
699
700
Figure H3 | Obligate sex allocation. (a,b) If mothers are obliged to invest a fixed
701
amount into sons, irrespective of their fecundity, then their investment into sons is
702
independent of the degree of viscosity, under haploidy and diploidy. (c,d) If mothers
703
are obliged to invest a fixed amount into sons, irrespective of their fecundity, then
704
their investment into sons slightly decreases as populations become increasingly
705
viscous (lower d), under haplodiploidy. As both mothers invest the same proportion
31
706
of resources into males, but differ in the absolute amount of resources they have,
707
high-fecundity mothers give birth to more sons (blue lines) than low-fecundity mother
708
(red dashed lines). We arbitrarily set the total number of offspring of a high-fecundity
709
mother to 100. Parameter values: k = 0.
710
References
711
712
Bulmer, M.G. 1994. Theoretical evolutionary ecology. Sinauer, Sunderland,
713
Massachusetts.
714
715
Christiansen, F. B. 1991. On conditions for evolutionary stability for a continuously
716
varying character. Am. Nat. 138, 37–50.
717
718
Eshel, I. 1996. On the changing concept of evolutionary population stability as a
719
reflection of a changing point of view in the quantitative theory of evolution. J. Math.
720
Biol. 34, 485–510. (doi:10.1007/BF02409747)
721
722
Frank, S. A. 1998. Foundations of social evolution. Princeton Univ. Press, Princeton,
723
NJ.
724
725
Taylor, P. D. 1996 Inclusive fitness arguments in genetic models of behaviour. J Math
726
Biol 34, 654–674. (doi:10.1007/BF02409753)
727
728
Taylor, P. D. & Frank, S. A. 1996 How to make a kin selection model. J. Theor. Biol.
729
180, 27–37. (doi:10.1006/jtbi.1996.0075)
730
32