Who cares for the kids?
Male deserts
Male stays
Female
deserts
Offspring fitness not much
improved with even 1 parent,
or BOTH parents can increase
number of offspring by a large
amount if they leave the
current ones
1 parent essential to raise
offspring, but almost as
good as two. Female can
produce more eggs if she
deserts than male can
fertilize if he deserts
Female
stays
1 parent essential to raise
offspring, but almost as good
as two. Male can fertilize more
eggs by deserting than the
female can increase her egg
numbers if she deserts
2 parents are much better
than one, or NEITHER can
increase RS much by
deserting (cf. mateenforced monogamy
example in text).
How many offspring to rear?
• Lack’s hypothesis – parents lay just enough
eggs to maximize total number of surviving
offspring.
• Tradeoff between offspring size and number
– you can make a lot of offspring, but they
will tend to be small and survive poorly, but
if you make them too big, you could
probably have made a larger number of
fairly healthy offspring with the same
resources.
Chick
survival
Lack’s hypothesis
Number of
surviving chicks
Number of eggs
N*
Number
of chicks
What KIND of offspring to rear
• How many males vs. females? The answer is ½ of
each, on average. Why? One half of all genes in
generation x+1 come from males in gen. X, and ½
from females in gen. X. Imagine there were K times
as many males as females in gen. X. Then on
average it MUST be true that each male has 1/K the
RS of the average female. So a family that produces
4 females will have K times the RS of a family with
4 males, thus selecting for female-biased families.
This bias will occur until the sex ratio in the
population is 1:1, at which point the average RS per
male and per female MUST be the same.
Sex vs. number of offspring
• What if males are much more costly to produce
than females (or vice-versa)? Fisher’s answer
is that total INVESTMENT in each sex should
be equal. So if males each cost 2x as much to
raise as females, then you should produce only
½ as many males. In primates, in fact, birth sex
ratios are biased toward males in many species
in which females stay at home and males
disperse, because the females eventually
compete with their mothers for food, thus
leading to a higher cost overall.
Biased sex investment
• The above rules hold “all other things being equal”. If
a female KNOWS that she is in better or worse
condition than average, she may favor one sex or the
other. She may do this pre- or post-natally. This can
occur even in primates (including man) that have only
one offspring at a time. The rule is to favor the sex that
will benefit most from the extra resources in its future
RS. In species in which males benefit especially from
being bigger than other males, a female in very good
condition should favor males, but in poor condition,
she should favor females. Case of wood rats.
Parent-offspring conflict
• The interests of parent and offspring are not
identical.
• Parent is related to itself by 1.0 (identity)
but to its offspring by 0.5; likewise for
offspring
• Offspring will be selected to demand more
resources from parent than parent is willing
to give – why? More effort in raising
offspring often comes at cost of higher
parental mortality
Fitness costs
or benefits
benefits
Cost to parent
(Indirect) cost
to offspring =
½ cost to parent
P*
O*
Resources invested in offspring
Hamilton’s rule
• How can ‘selfish’ NS favor individuals that
give resources and care to others, even to
offspring?
• Answer seems obvious – offspring carry the
parents’ genes, but this principle holds true
even for more distant relatives as well.
• More formally: rb> c, where b=benefit to
receiver, c= cost to donor, r=‘relatedness’ =
chance that 2 individuals share same allele
by common descent (how to calculate?)
GP1
GP2
R2
P1
P2
R1
P3
A
B
r(AB) = (A P1 B) +
(A P2 B) = ½ * ½ + ½ * ½ = ½
S1
C
E
F
D
Rule is: draw or write ALL pathways that connect two individuals to
ANY common ancestor that is NOT already counted (in this example
do not connect A and B through GP1 or GP2, as you must pass via
P2). Then count the number of links in each pathway, and call this L.
Then, r is the sum of (1/2) L across all pathways. For r(B P3), the
pathways are (B P2 GP1 P3) and (B P2 GP2 P3); r =1/4