Ejaculate investment and attractiveness in the stalk-eyed fly, Diasemopsis meigenii

The phenotype-linked fertility hypothesis proposes that male fertility is advertised via phenotypic signals, explaining female preference for highly sexually ornamented males. An alternative view is that highly attractive males constrain their ejaculate allocation per mating so as to participate in a greater number of matings. Males are also expected to bias their ejaculate allocation to the most fecund females. We test these hypotheses in the African stalk-eyed fly, Diasemopsis meigenii. We ask how male ejaculate allocation strategy is influenced by male eyespan and female size. Despite large eyespan males having larger internal reproductive organs, we found no association between male eyespan and spermatophore size or sperm number, lending no support to the phenotype-linked fertility hypothesis. However, males mated for longer and transferred more sperm to large females. As female size was positively correlated with fecundity, this suggests that males gain a selective advantage by investing more in large females. Given these findings, we consider how female mate preference for large male eyespan can be adaptive despite the lack of obvious direct benefits.

accuracy by avoiding the mis-counting of paler pixels as sperm in regions of the spermatophore where visual inspection revealed that there were none.
The programme returned the percentage of the total spermatophore area that was occupied by sperm pixels. Using our independent measurements of sperm content (see Materials and Methods) we converted these percentages into absolute measurements, using the following equation,

Background
Suppose there are two types of females in the population: a proportion q of normal females, and a remaining proportion 1q of super-fecund females. The normal females have fecundity 1, while the super-fecund females have fecundity 1 + h.

Strategy
A male's strategy is a vector s = (s 1 ,s s ) describing the quantity of resources he will assign to a mating with female types 1 and 2. Since both s 1 and s 2 must be positive real numbers, we can characterise a male's strategy as being t = (s 1 ,x), where x is the (positive real) coefficient such that s 2 = x s 1 .
Males have two key parameters, their quantity of resources allocated to reproduction R, and the resource cost that they must pay to obtain a mating, c. A male's strategy will be conditional upon these parameters, so that it will be a function s[R,c]. The optimal strategy for a male depends upon what the rest of the population is doing.
We denote the mean population strategy as s  (s 1 , ys 1 ) , where y is the (positive real)

Fitness function
The fitness function for a male with resources R, and cost c (hereafter referred to as an (R,c)-male) playing strategy s[R,c] is defined as, where n[s, R,c] is the expected number of matings, and v[s, R,c s ] is the expected success per mating. If we denote the partial derivatives of W with respect to s 1 and to s 2 by W 1 and W 2 respectively, then a strategy s can only be a best reply to a population mean strategy s if, and we also have that for i = 1,2, where n i and v i refer to partial derivatives with respect to i.

Number of matings
We define the expected number of matings by, This is the ratio of the quantity of resources that the male possesses to the average quantity of resources he uses per mating. This is an approximation to the expected number of matings given s, R, and c, but simulations showed it to be a reasonable approximation under a range of reasonable parameter values (results not shown).
We then have, so that n 2 n 1  (1 q) q .

Success per mating
The expected success per mating is then defined as The term in brackets in the summand represents the expected success from mating with a randomly chosen female who also mates with k other males. With a probability q this female is a normal female, and the focal male invests s 1 resources and receives expected success s 1 (s 1  ks 1 ) , which is the proportion of total sperm the female receives that belongs to him. With a probability (1q), the female is superfecund, and the focal male invests s 2 resources and receives expected success ((1 h)s 2 ) (s 2  ks 2 ), which is the proportion of total sperm the female receives that belongs to him multiplied by her fecundity coefficient 1 + h.
We then have that,

Analysis
Suppose there exists an evolutionary stable strategy (ESS) s   (s 1  , s 2  ) for each combination (R,c), and suppose the population is playing this strategy. Then the mean population strategy will be s   (s 1  , s 2  )  (s 1  , y  s 1  ) . Since it is an ESS, this strategy must be a best reply to itself, i.e. for all (R,c) combinations, R,c] . (3) Also, which means, from (2) and (3), and then (1), Since v 1 is a sum of positive decreasing functions of s 1 , it is itself positive and decreasing in s 1 . Equation (4) (R,c), which implies that all males must invest (1 + h) times as much sperm in super-fecund females as they do in normal females.
It can be proven that the ESS strategy exists and is always of this form, and also that it will always increase in c and be independent of R, but these details have been discussed previously in (Tazzyman et al. 2009) and are not necessary here.