Two weeks ago we performed Lab 3.1 (Carbon Cycling Between Goldfish and Elodea) using Spirogyra and zebra fish. Even though the procedure went smoothly, I did not realize that the manual had failed to provide formulas for properly answering questions 5 and 7. Below are the formulas with explanations:

5) If you divide the numbers in column I (fish) by column I (plants), you will have the reciprocal of the ratio of fish to plants needed to provide a 1:1 ratio of oxygen consumption to oxygen production. Here is the justification:

{mg O2 per kg per hour (fish) ÷ mg O2 per kg per hour (plants)} × R = 1

 R = ratio of fish to plants needed to get 1:1 ratio O2 consumption by fish vs. O2 production by plants
 
For example, in the sample data the fish consume 470 mg O2 per kg per hour, and the plants produce 910 mg O2 per kg per hour:

{470 mg O2 per kg per hour (fish) ÷ 910 mg O2 per kg per hour (plants)} × R = 1, R = 910 / 470 = 1.9 mass ratio of fish to plants.

 
7) First you need to calculate how much oxygen was consumed by the fish with the plants. To do this, you need to assume that the fish with plants were consuming oxygen at the same rate per kg body mass as the fish without plants:

a)  ΔO2 ppm (“fish only” flask) ÷ total kg fish in (“fish only” flask) =  ΔO2 ppm / kg fish

b)  ΔO2 ppm / kg fish ÷ total kg fish (fish in “fish with plants” flask) =  ΔO2 ppm attributed to fish only
 
c)  ΔO2 ppm attributed to fish only +  ΔO2 ppm recorded in “fish with plants” flask =  ΔO2 ppm attributed to plants only

Finally, in order to compare this to column I for the “plants only” flask, you need to calculate the mg O2 per kg per hour:

d) [{ΔO2 ppm attributed to plants only × volume flask (L)} ÷ total kg plants (plants in “fish with plants” flask)] ÷ 0.5 hour = mg O2 per kg per hour produced by plants in “fish with plants” flask


The sample calculation below is based on data provided in the “Instructor’s Guide”:
a) -2.5 ppm ÷ 0.0029 kg = -862 ppm per kg
b) -862 ppm per kg × 0.0030 kg = -2.6 ppm
c) 2.6 ppm + 0.3 ppm = 2.9 ppm
d) [{2.9 ppm × 0.27 L} ÷ 0.0026 kg] ÷ 0.5 hour = 600 mg O2 per kg per hour


This is near the range of the 630-910 mg O2 per kg per hour produced by the plants in the “plants only” flasks during the second half hour.