Homeostasis and Osmoregulation

One of the major challanges faced by animals is the maintenance of a constant internal environment, called Homeostasis (Hickman et al. Ch. 30).  The term Homeostasis is a catch all term that refers to the control of many variables, such as temperature, salt, water, glucose, etc.  The benefit of constant conditions comes in that the chemical reactions on which life depends are affected by physico-chemical conditions within the cytoplasm.  If the internal conditions change, isozymes (alternate forms of an enzyme that function optimally under differing conditions) are often necessary to ensure that the reactions will preceed.  Therefore, there is an energetic cost associated with allowing the internal environment to vary.  If you maintain a constant temperature you only need to make one form of the enzyme but if your temperature is going to vary, you will need to make many isozymes.

     The excretory system is one of the systems involved in maintaining homeostasis in the body.  The excretory system is responsible for the removal of nitrogenous wastes produced during the metabolism of proteins.  Along with nitrogenous waste, the excretory system is responsible for the removal of water from the body.  There are three main forms of nitrogenous waste used by animals.  Many aquatic organisms release ammonia, a highly toxic compound produced in protein metabolism.  Ammonia is released by small animals and those that live in water.  Urea is a less toxic form of nitrogenous waste but the animal must expend energy to produce it from ammonia.  Urea is often used by large animals and terrestrial animals.  The fact that it can build up in the body to higher concentrations than possible for ammonia means that these animals can produce a concentrated urine, therefore, conserving water.  Uric acid is the most complex, and energetically costly, form of nitrogenous waste.  It's advantage is that it is only weakly soluble in water and therefore can be removed from the body with minimal water loss.  This form of waste is found in many terrestrial insects and in reptiles and bird.  It is adaptive for reptiles and birds in that it can be stored inside the shell of developing embryos without incurring a large water demand.  So as we can see, differences in nitrogenous waste are important not only due to differences in toxicity, but also due to differences in the amount of water required for their removal.

     The control of water content is one of the greatest challenges facing aquatic animals.  Freshwater species usually have cytoplasm that is hyperosmotic (see Hickman et al. pages 47-49 for a discussion of osmotic pressure) compared to the environment and saltwater species are often hypoosmotic compared to seawater.  This results in the net gain or loss of water to the body due to Osmosis, the diffusion of a solvent (in this case water) across a semipermeable membrane ( in this case the cell membrane).   Freshwater organisms are constantly removing excess water that is entering the body whereas saltwater organisms are trying to reduce water loss to the environment.  Many saltwater species have an osmolarity (solute concentration -moles of solute per liter of water) that is equal to that of sea water (isosomotic solutions).  These organisms do not expend any energy to control internal osmolarity and therefore will conform to the osmolarity of the environment.  These organisms are called osmoconformers (See spider crab in graph below).  Organisms that maintain a constant osmolarity regardless of the external environment are called osmoregulators (See Shore crab in brackish water - salinity below normal seawater).



Why would osmoconformers be more common in marine environments than freshwater or brackish environments?

Think about the variability in the environment.

We collected a species of crayfish (
Orconectes rusticus) in south central Kentucky and noticed that it has a wide distribution, including some brackish water.  Based on the diversity of habitats where it is found we think that it is an osmoregulator.

What is our hypothesis?

What would be an experiment that we could conduct to determine if the crabs are osmoregulators or osmoconformers?  Write down your idea below.  Do this before moving on.






We are going to expose two groups of crayfish to water of differing salinities and examine whether they gain or lose water.  An osmoconformer will gain or lose water when placed into water of a different salinity as osmosis occurs.  If the crayfish is an osmoconformer, it will lose weight when placed in saltwater as water is drawn out of its tissues.  If the crayfish is an osmoregulator, it will be able to maintain a constant water concentration in either freshwater or saltwater.  The weight of an osmoregulator will not change when exposed to water with a different salinity.  As a surragate for water content, we are going to monitor the change in weight of the animals over time.  If, as we expect, the crayfish are osmoregulators, we should not see any change in the weight of the animals in the different solutions.

We will test the null hypothesis that there is no difference in the % change in body mass of
crayfish in fresh water (< 1 ppt) as compared to brackish sea water (15 ppt).

Experimental procedure

1.  Gently remove excess water from an organism by removing it from its individual container and blotting it on a paper towel.

2. Weigh the animal to get a starting mass. 

3. Return the 5 members of the control group to their containers of sea water and the 5 members of the experimental group to their individual brackish water containers.

4.  Every 30 minutes, for the next 60 minutes, determine the current weight of the organims.

Analyses

Calculate the % body mass for each measurement by:

 
% change in body mass  =  (the weight at time X - the starting weight)
                                                                    starting weight

Calculate the average
% change in body mass for the control and experimental groups.

Alternative Procedure - if the services of
crayfish are not available, we will use potato tissue as a surragate.

We will examine the effect of body size and salinity differences on the rate of water loss by exposing two sizes of potato chunks to hyperosmotic solutions of 15 ppt and 30 ppt.  We know that potato cells are not osmoregulators so the cells should lose weight as water moves by osmosis. 

Which solution do you think will draw out the water the quickest?  Why?

Which size potato organism do you think will lose weight the quickest? Why?

procedure

1.  Dip each potato chunk in water and then gently remove excess water by blotting it on a paper towel.

2. Weight the potato to get a starting mass. 

3. Place 5 large potato chunk into containers of sea water, 5 large chunks into brackish water, 5 small chunks into containers of sea water, and 5 small chunks into brackish water.

4.  Every 30 minutes, for the next 60 minutes, determine the current weight of the potato chunk after blotting it dry.

Analyses

Calculate the % body mass for each measurement by:

 
% body mass  =  (the weight at time X - the starting weight)
                                                  starting weight

Compare the % body mass after 60 minutes for the large potato chunks in sea water to that of the large potato chunks in brackish water.  Repeat for small potato chunks.

Compare the % body mass after 60 minutes for the large potato chunks in sea water to that of the small potato chunks in sea water.  Repeat for the potato chunks in brackish water.

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