PURPOSE
The purpose of this experiment was to determine the effect of water
pollution (antifreeze) on the survival rate and growth of algae.
I became interested in this idea when my old fish tank became coated with algae. I was going to use bleach and soap on it, but my mom said that if I put too much in and didn’t wash it out well enough then it would kill my fish. So I wondered if certain types of pollution that people dumped into the sea or oil that came from ships would affect the growth and survival rate of algae.
The information gained from this experiment could help beneficial algae grow more efficiently and keep workers from polluting the water. Also it would help scientist know if oil or other pollutants would affect the growth and survival rate of algae.
My first hypothesis was that as the antifreeze had more time in contact
with the algae, the less the dissolved oxygen level would be.
My second hypothesis was that ethylene glycol would have more effect than the other pollutants and would decrease the dissolved oxygen level most.
I based my second hypothesis on information that I got from a 2005 science project, “Effect of Antifreeze Type and Concentration on Soybean Growth.” The conclusions stated that ethylene glycol affected the growth of soybeans more than propylene glycol.
The constants in this study were:
• Type of dissolved oxygen meter
• Water type
• Amount of water
• Type of container
• Type of pollutant
• Species of algae
The manipulated variable was the antifreeze concentration.
The responding variable was how much of the algae survived.
To measure the responding variable, I used a dissolved oxygen meter.
QUANTITY
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ITEM DESCRIPTION
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4 quarts
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Water
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24
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Eggs
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5 , 2 liter
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Plastic containers
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2-liter
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Algae
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12
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Small jars
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2 quarts
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Antifreeze concentrations
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1
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Dissolved oxygen meter
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1) Obtain all supplies and algae
2) Create Pollutant Mixtures in Decreasing Concentrations
A) Pour 8 ml of pollutant into a graduated cylinder and add 12 ml of distilled water. Mix well. This yields 20 ml of polluted water. Pour exactly 10 ml of this into a disposable cup. One half of the pollutant (4 ml) is in this cup and the other half (4 ml) is still in the graduated cylinder. Mark this cup “4 ml”.
B) Add an additional 10 ml of distilled water to the graduated cylinder (still containing 4 ml of pollutant) to make 20 ml total. Mix well. Pour exactly 10 ml of this “half strength” pollutant into a second disposable cup and label “2 ml” because that is how much of the original pollutant is still in this water.
C) Add an additional 10 ml of distilled water to the graduated cylinder (now containing 2 ml of pollutant) to make 20 ml total. Mix well. Pour exactly 10 ml of this “quarter strength” pollutant into a third disposable cup and label “1 ml” because that is how much of the original pollutant is left.
D) Repeat this process again. Add an additional 10 ml of distilled water to the graduated cylinder (now containing 1 ml of pollutant) to make 20 ml total. Mix well. Pour exactly 10 ml of this into a fourth disposable cup and label “0.5 ml”.
E) Add an additional 10 ml of distilled water to the graduated cylinder (now containing 0.5 ml of pollutant) to make 20 ml total. Mix well. Pour exactly 10 ml of this into a fourth disposable cup and label “0.25 ml”
F) Add an additional 10 ml of distilled water to the graduated cylinder (now containing 0.25 ml of pollutant) to make 20 ml total. Mix well. Pour exactly 10 ml of this into a fourth disposable cup and label “0.125 ml”
3) Add Algae to Pollutant Mixtures
A) Go back and add exactly 40 ml of algae/water to each of the 10 ml samples of polluted water in each of the disposable cups. Here is the math:
i) 40 ml algae + 10 ml polluted water = 50 ml total (including 4 ml of pure pollutant). So 4 parts pollutant in 50 parts liquid = 4/50 = 8%
ii) 40 ml algae + 10 ml polluted water = 50 ml total (including 2 ml of pure pollutant). So 2 parts pollutant in 50 parts liquid = 2/50 = 4%
iii) 40 ml algae + 10 ml polluted water = 50 ml total (including 1 ml of pure pollutant). So 1 part pollutant in 50 parts liquid = 1/50 = 2%
iv) 0.5 part pollutant in 50 parts liquid = 0.5/50 = 1%
v) 0.25 part pollutant in 50 parts liquid = 0.25/50 = 0.5%
vi) 0.125 part pollutant in 50 parts liquid = 0.125/50 = 0.25%
B) Control Group - Algae with NO Pollutant
i) Create control groups with absolutely no pollutant. Use 40 ml algae + 10 ml distilled water to keep the amount of algae in control samples equal to algae in treatment samples.
4) Start Observations
5) This is “TIME ZERO.” Do first dissolved oxygen reading for each group.
6) Repeat step 5 with other concentrations
The original purpose of this experiment was to determine the effect of
water pollution on the survival rate and growth of algae.
The results of the experiment were the algae reacted most to the ethylene by dropping dissolved oxygen.
My first hypothesis was that as the antifreeze had more time in contact
with the algae, the less the dissolved oxygen level would be. The results
indicate that this hypothesis should be accepted, because the amount of
dissolved oxygen generally decreased across time.
My second hypothesis was that ethylene glycol would have more effect than the other pollutants and would decrease the dissolved oxygen level most. The results indicate that this hypothesis should also be accepted, because the amount of dissolved oxygen generally was lowest for ethylene glycol.
After thinking about the results of this experiment, I wonder if I used a stronger pollutant like gasoline or oil would affect the algae in a different way? Also if I used higher concentrations with my antifreeze, if that would have brought the results down more?
If I were to conduct this project again I would of used oil or a stronger pollutant, I would have put in higher concentrations, bought a lot more algae so I could use maybe two or three types of pollutant.
Researched
by ----- Natalie F
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