Question

In a salt marsh in Georgia, the quantity of solar radiation reaching the ground during the summer is 7,000,000cal/m2/day. The gross productivity of the producers in the salt marsh (mostly microscopic algae) is 0.0040 g/cm2/day. The average rate of respiration (the respiration requirement) is 35% of the gross primary productivity. Ecologists have found that 1 gram of salt marsh producers (the algae) is equivalent to 2,920 calories. Calculate the net primary productivity of the system, in g/m2/day. (remember, 1 m2 = 10,000 cm2)

Answer 1

Answer:

NPP=26 g/m²/day

Explanation:

Important data:

• 1 m² = 10,000 cm²
• NPP = GPP - Respiration
• NPP= Net primary productivity of a system
• GPP= Gross productivity of the producers

Available Data:

• GPP=0.0040 g/cm²/day
• Respiration rate= 35% of the gross primary productivity

We need to calculate the NPP, which equals GPP - Respiration rate.

The respiration rate is 35% of the GPP. This is:

100% GPP--------------- 0.004 g/cm²/day

35% GPP ----------------X = (35 x 0.004)/100 = 0.0014 g/cm²/day

Now that we know the values of GPP and respiration rate, we can calculate the NPP.

NPP = GPP - Respiration

NPP = 0.004 g/cm²/day - 0.0014 g/cm²/day

NPP = 0.0026 g/cm²/day

Now we also know the NPP, but this value is calculated in g/cm²/day, and we need to calculate it in g/m²/day, so if 1 m² = 10,000 cm², then:

1 cm²---------------- 0.0026 g/cm²/day

10,000cm²----------X = (10,000 x 0.0026)/1 = 26 g/m²/day