1. Who cares for the kids? Male desertsMale 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. mate- enforced monogamy example in text).

2. 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.

3. Chick survival Number of chicks Number of eggs Number of surviving chicks N* Lack’s hypothesis

4. 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.

5. 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.

6. 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.

7. 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

8. Fitness costs or benefits Resources invested in offspring benefits Cost to parent (Indirect) cost to offspring = ½ cost to parent P*O*

9. 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?)

10. AB P1 P2 GP1 GP2 R1 R2 S1 CD EF r(A  B) = (A  P1  B) + (A  P2  B) = ½ * ½ + ½ * ½ = ½ 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 P3