3. In a field of sugar beets in Georgia the net primary productivity is 120 grams/m^2/day, and the respiration loss is 25%. The insolation energy is 800 calories/cm^2/day, and one gram of beets is
equivalent to 2,000 calories.
a. Find the gross productivity of the sugar beets.
b. Using the NPP, calculate the efficiency of photosynthesis for the sugar beets.

Answer and Explanation:

Available Data:

1) Find the gross productivity of the sugar beets.

We need to calculate the GPP, which equals NPP - Respiration rate. The respiration loss is 25% of the NPP. This is:

100% NPP 120 g/m²/day

25% NPP X = (25 x 120)/100 = 30 g/m²/day

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

GPP = NPP - Respiration

GPP = 120 (g/m²/day) - 30 (g/m²/day)

GPP = 90 g/m²/day

GPP = 0.009 g/cm²/day

2) Using the NPP, calculate the efficiency of photosynthesis for the sugar beets.

% Efficiency of Photosynthesis = (NPP/Insolation Energy) X 100

We know the NPP, but this value is calculated in g/m²/day, while the Insolation energy is in g/cm²/day. We need to calculate the % Efficiency of Photosynthesis in cm. To do that, we must transform the information.

So if 1 m² = 10,000 cm², then:

10,000cm²120 grams/m²/day NPP

1 cm² (1 x 120) / 10,000 = 0.012 grams/cm²/day NPP

We also know that 1 gr beets = 2,000 cal, and that the Insolation energy is 800 calories/cm²/day

So we need to transform Insolation energy from calories/cm²/day to grams/cm²/day, so:

2,000 calories 1 gr beet

800 calories/cm²/dayX = 0.4 gr beet.

% Efficiency of Photosynthesis = (NPP/Insolation Energy) X 100

% Efficiency of Photosynthesis = (0.012 (gr/cm2/day) /0.4 (gr/cm2/day)) X 100

% Efficiency of Photosynthesis = 3 gr/cm2/day  

the major parts of the digestive system:

salivary glands.

pharynx.

esophagus.

stomach.

small intestine.

large intestine.

rectum.

accessory digestive organs: liver, gallbladder, pancreas.