Download Omtentafråga - Studentportalen

These are possible answers, or references to where you
can find the answers, for the exam in Behavioural
Ecology 2010.03.19 8.15-12.00
Question 1 (Lars)
a)
Imagine a small passerine bird species, like a collared flycatcher, barn swallow or
great tit. Briefly discuss the most important fitness-affecting resource allocation
trade-offs that a male, a female and a newborn offspring face during a breeding
season. Especially point out the main differences between males, females and
offspring (4p)
Males, unlike female have to invest resources into costly pre-mating behaviors, such
as fighting with other males over territories and access to females as well as having a
large secondary sexual ornament. These resource allocations trades of against
resources devoted to the immune system that can affect their subsequent condition
and survival and future growth of ornaments as well as parental care. Like females
males also trade off there investment in parental care against resources devoted to
the immune system that can likewise affect their subsequent condition and survival
and future growth of ornaments and reproduction. Males also can trade off
investment in EPC and finding a second mate, against mate guarding for females to
mate with other males. Females face a trade off between number of eggs laid and
numbers of surviving young and subsequent parental care against resources devoted
to the immune system that can likewise affect their subsequent condition and
survival and future fecundity and reproduction. Females seeking EPC can suffer from
males less willing to help with parental care. Offspring mainly have a trade off
between researches allocated to growth versus the immune system. They also can
allocate resource to begging and competition between siblings in order to get more
food at the cost of spending these resources directly to growth and/or immune
system.
b)
How and why should reproductive effort change with increasing age and when
should an individual make a terminal reproductive investment? (2p)
Reproductive effort should increase with age since residual reproductive value
decreases with age since ageing lower the future prospect of survival and
reproduction with advancing age. That is shift investment from future reproduction
to current reproduction. At an age when the future prospect of survival and
reproduction drops dramatically it would be optimal for an individual to invest all
resource in to the current reproduction at the cost of its own survival, that is do a
terminal investment.
Question 2 (Jan)
The sex ratio among the offspring in ant societies with a monogamous queen (mating
with only one male) is biased towards daughters (3 daughters to each son). In
contrast the sex ratio is even among the offspring in ant societies with a polyandrous
queen (mating with multiple males). Explain how this difference is a test of the role of
haplodiploid inheritance of sex in the evolution of eusociality. (6p)
The idea that eusociality (evolution of social castes and sterility) should be the result
of haplodiploid inheritance is based on that sisters then should be more related to each
other than they are to either their brothers or their offspring. The haplodiploid
inheritance entails that females (mothers) are diploid and produce two kinds of
gametes while males (fathers) are haploid and produce only one kind of gametes.
-
-
At monogamous mating under such haplodiploid inheritance all the daughters’
paternal genes are identical (proportion 0.5 of their genes; all the fathers gametes
are identical) while daughters share only half of their genes inherited from the
mother (proportion 0.25 of their genes; mother produce two kinds of gametes). As a
result sisters share 0.75 of their genes. In contrast their genes in common with
brothers are limited to those genes passed on by the mother, which amounts to a
proportion of 0.25 (half of daughter genome and two kinds of mother gametes
=0.5/2). The high level of relatedness among sisters in a haplodiploid and
monogamous system entails that they would do better (in terms of propagating
genes into next generation) by producing sisters (relatedness 0.75) than by
producing offspring (relatedness 0.5). Furthermore they do better by helping to raise
sisters than to reproduce themselves, which would select for sterility. Given that
female workers control the sex ratio in the colony and the value to them of
producing males and females equal the level of genetic relatedness we would find a
sex ratio of 3 females to 1 male. This is the ratio that has been found in
monogamous ant societies.
A polygynous mating system is a test of the role of haplodiploidy to the extent that
daughters no longer receive identical paternally inherited genes. The level of genetic
relatedness among sisters is therefore reduced to 0.5 and their advantage of
producing sisters to brothers is reduced. Such a reduction in the female to male sex
ratio in parallel to the lower level of relatedness which has been found in
polygynously breeding ants is consistent with that sociality is under control of a
haplodiploid inheritance.
Question 3 (Richard)
You are sitting on a beach and looking at gulls feeding on mussels. You get up and
walk to where the gulls are feeding and you see that there are two kinds of mussels,
one blue and one grey mussel. Luckily for you, you have the book ”Energetic contents
of mussels and handling time of different mussel species by gulls of the genus
Laridae”. It turns out that the blue mussel contains 200 J/individual and takes 10
seconds to open and swallow, while the grey mussel contains 100 J/individual and
takes 20 seconds to handle. After studying this for a while you know that the search
time for the gull to find a grey mussel is 5 seconds while it takes them one minute to
find a blue mussel.
a) If you assume that the gull is behaving according to the optimal diet model,
what can you then predict about the gulls diet? (2P)
b) Next day you see that the gull is taking only one of the two mussel species
despite the fact that both of them exist on the beach. Which mussel is
preferred and how long can the search time maximally be for this mussel so
that the gulls still choose to specialize on this mussel? (2P)
c) Later that evening you return to the beach and find that the tide has changed
the mussel densities so that the search time for the blue mussel is 40 seconds
and the search time for the grey mussel is 5 seconds as before. In addition,
small thick shelled pink mussels have been washes up to shore in enormous
density (search time = 0 seconds). Your book says that the energy content of
the pink mussel is 60 J/individual and that it takes 30 seconds to open and
swallow it. Would it surprise you if the pink mussel would be included in the
diet of the gull? (2P)
For each question; show your calculations and how you came up with your answer!
a)
First calculate E/h for both prey, that is, if the predator have both prey in the hand,
which should he/she choose first? The one with the highest E/h will be the preferred
prey and will always be in the diet. In this case: the blue mussel will have an E/h
equal to 200/10 = 20 J/s and the grey mussel 100/20 = 5 J/s. This means that the blue
mussel will be the preferred prey (this is also part of the answer to question 5b).
Now you have to calculate if the grey mussel should be included in the diet.
Remember that the next best prey (in this case the grey mussel) will be included in the
diet if its E/h is larger or equal to the E/(h+s) for the mussel in the diet (the blue
mussel). Why don’t we calculate with the search time for the next best prey? Imagine
that the predator is out searching for the blue mussel (first ranked) and while
searching for the blue mussel it encounters the grey mussel, that is there is no search
time for the grey mussel. When encountering the grey mussel, should it be included in
the diet?
The E/(h+s) for the blue mussel is then 200/(10+60) which is about 2.8-2.9 J/s. This
means that the grey mussel E/h (=5 J/s) is higher than the E/(h+s) for the blue mussel
and therefore both mussels should be included in the diet.
b)
The blue mussel will be preferred – see calculations for the answer to question 5a.
Now we want to know what the search time maximally can be so that the gull is
taking only the blue mussel. The E/h for the grey mussel is the same as before, that is
5 J/s. Therefore the E/(h+s) for the blue mussel cannot be equal to or larger than 5. So
that means that we have: 200/(10+s) < 5. This means that s < 30 s. In other words, is
search time for the blue mussel is less than 30 seconds then the blue mussel will
always be the only mussel in the gull’s diet.
c)
In this case E/h for the blue mussel is 20 J/s, for the grey it is 5 J/s and for the pink it
is 2 J/s. Therefore, as before the blue will be first ranked, the grey is second in rank
and the pink last in rank. Now calculate if the grey should be in the diet; that is if
E/(h+s) for the blue is less than or equal to E/h for the grey. Now E/(h+s) for the blue
is 200/(10+40) = 4 J/s and E/h for the grey is as before 5 J/s, therefore the grey mussel
should be included.
Now calculate what E/(h+s) is when both the blue and the grey mussel is included in
the diet and compare it to the E/h for the pink mussel. In this case, for every blue
mussel the gull is encountering while searching it will also encounter 8 grey mussels.
Therefore the E/(h+s) for the blue and the grey mussel combined will be (200+8*100)
/ (40+10+8*20+8*5) = 1000/250 = 4 J/s, which is greater than E/h for the pink mussel
(2J/s). Therefore I would be surprised if the pink mussel would be included in the diet
of the gull.
Question 4 (Anders)
a) In the grotesque-finned sea bass you find a genetic correlation between the
male ornament, the grotesque dorsal fin, and female preference for it. You
also know that males don’t contribute directly to female or fry survival. Is this
genetic correlation necessarily a support for a Fisher run-away process behind
the ornament? Explain! (3p)
b) The male’s dorsal fin is regulated by testosterone and may be a handicap if
testosterone reduces immuno-competence. Explain why we call the ornament
a handicap! You now set out in the field to measure fin size, testosterone
levels and parasite loads in males. Which relationships do you expect to find if
the fin is a testosterone-dependent handicap? Explain! Do you think that
these data can show that testosterone reduces immuno-competence? (3p)
In a) I like you to explain that the same genetic correlation may arise from a
handicap process (and that this process is in operation is actually hinted in b!), so the
correlation is not necessarily a support for a Fisher process. In b) the handicap means
that if males want to advertise good health they do so by reducing health through
testosterone – only male in good health can afford this. However, we don’t expect
them to reduce health to below that of males with parasite problems! Thus, in
nature we expect males with a large ornament to have high levels of testosterone,
but at the same time to have few parasites, something that certainly does not show
that T reduces immunocompetence. For that, you need an experiment!
Question 5 (Göran)
a) It is often suggested that the evolution of polyandry is caused by polyandrous
females having offspring with higher genetic quality. This rests on a very
important assumption. Which assumption is this and how would you test it
using the organism of your preference as a model system? (3p)
Answer: see reading assignement paper #20.
b) List and briefly discuss at least six male adaptations to sperm competition; of
which at least three are ”offensive” and three ”defensive”. (3p)
Answer: see lecture handout on postmating sexual selection.
Question 6 (Anna)
Give three examples of forms (or types) of reproductive isolation, briefly explain how
these forms of reproductive isolation may evolve and state whether we should expect
them to be determined by sex-linked genes or not (6p).
I asked for three examples of forms of reproductive isolation, an
explanation of how these isolation form may evolve and whether we
should expect the relevant traits to be determined by sex-linked genes
or not. Here I will list three examples but a full discussion on other
forms of reproductive isolation would of course also lead to full
credit (given that the required information is accurately discussed).
a)
Ecologically based selection against hybrids. The two parental
species have diverged ecologically due to exposed to different
selective regimes during periods in allopatry or because they have been
exposed to strong competition in sympatry and therefore diverged in one
or several ecologically important trait (e.g. the example of beak size
in the finches I showed on the black board). Hybrids typically have an
intermediate phenotype meaning that they are selected against in the
environment of their parental species (they fall between two adaptive
peak). Because ecological traits typically are exposed to the same type
of selection in both sexes we do not expect sex-linked genes to be
relatively more important. (In addition, standing genetic variation is
more important than new mutations in early phases of population
divergence and because sex-chromosomes have a smaller effective
population size ? such genes are more likely to be found on the
autosomes).
b)
Gamete isolation (a form of pre-zygotic isolation). The proteins
involved in gamete recognition are expected to evolve fast due to
exposure of strong selection (i.e. these proteins have important
fitness effects and may be involved in sexually antagonistic armsraces, which means very fast evolution). Fast evolution in turn means
fast potential for divergence between populations (when there is
limited gene flow between them). Population divergence in these
proteins will lead to pre-zygotic isolation through conspecific sperm
precedence. Because these traits have sex-specific fitness effects we
expect the underlying genes to accumulate on the sex-chromosome.
c) Genetic incompatibility (hybrid are not fertile or viable). In
allopatric populations (i.e. populations separated by a geographical
barrier) different mutations will become fixed (due to selection or
drift). When the two populations come into secondary contact and
hybridise some of the genes with different evolutionary histories (i.e.
genes that have not been selected to work well together) are likely to
give rise to problems in hybrids. Haldane?s rule, i.e.
hybrids of the heterozygous sex suffer ?first? during evolution of genetic
incompatibility suggest that sex-linked genes are important. A likely explanation for
Haldane?s rule is that X-linked incompatibility alleles are not masked by dominance
on the homologous chromosome.
Question 7 (Niclas)
a) Name three ecological factors that are commonly associated with the
existence of parental care in animals. (1.5p)
b) A trade-off between mating investment and parental investment is often
assumed and still, in some species (e.g. in some fishes), males both compete
heavily over matings and still they provide all the parental care. Explain how
this may work from an evolutionary point of view. (2p)
c) Which is the most common care-type (i.e. who provides the care) in birds and
why is this care-type so common in this group? (1p)
d) Species 1 has internal fertilization while species 2 has external fertilization.
All else being equal, in which of the species is post-hatching/post birth
paternal care most likely to evolve and what is/are the evolutionary
argument/s for this? (1.5p)
a) High predation levels, high competition over resources, harsh environmental
conditions, high benefit/cost ratio.
b) Parental care itself can be a sexual trait – females sometimes prefer better fathers.
This may cancel out the strong trade-off between care activities and activities linked
to acquiring a mate. Sometimes the cost of care is small and allows for evolution of
care without taking too much resources from other potential traits and through
relatively small evolutionary changes – for instance in species where males have
switched from defending a territory to defending both the territory and the eggs laid
there by a female.
c) Biparental care – both parents are most often required to successfully rear a brood
of offspring in birds since, in most species, the chicks require both protection from
predators and large amounts of food as juveniles.
d) Paternal care is most likely to evolve in the species with external fertilization
because: a) internal fertilization ties the female much closer to the eggs and hence
makes it difficult for the male to even get access to the eggs. In species with internal
fertilisation and egg-laying (like in some fishes), there is also a reasonable time period
between sperm transfer and egg laying which makes it unlikely for males to evolve
care since they are unlikely to even be around the eggs at time of egg laying b) In
many animals, male care is provided only under high certainty of paternity. This
certainty of paternity drops dramatically during internal fertilization since it is so
difficult for males to assess their level of paternity as compared to an external
fertilizer.