1.  Phenylalanine is an amino acid with the formula C₆H₅CH₂CH(NH₂)COOH.
a. What is the %C in phenylalanine?

b. What is the %N in phenylalanine?

c. What is the C:N ratio of phenylalanine?

Calculate the molecular weight of phenylalanine:

9 C =  108
11 H =  11
2 O =   32
1 N =   14
          165

In 165 grams, 108 g are C and 14 g are N.

1a.

Thus the %C = (108/165) X 100 = 65.5%

1b.

and the %N = (14/165) X 100 = 8.5%

1c.

Thus, the C:N ratio is 108/14 or 7.7:1. Alternatively, 65.5%/8.5% is = 7.7:1.

2.      Assume 100 lbs of tree leaves (C:N ratio 100:1) are incorporated into the surface 6 inches of a 10 ft by 10 ft garden area. What is a reasonable estimate of the amount of N that needs to be added so that N does not limit plant growth due to immobilization?

100 lbs residue
X 0.50 (% decomposition)
50 lbs residue decomposed
X 0.40 (% C in residue)
20 lbs C in residue
X 0.25 (% C into microbe biomass)
5 lbs microbe C
/8 (8:1 C:N ratio of microbe)  =
0.625 lb N needed

Have: 50 lbs residue decomposed = 20 lbs C  (100:1 C:N ratio of residue) = 0.2 lb N in residue*

0.625 - 0.2 = 0.425 lbs N needed to prevent immobilization

*NOTE: This same type of problem may be asked where the %N of the residue is known. With this, one needs to calculate the release of N during decomposition (the N is only released from the residue decomposed and not from the total residue added to soil...for this problem it would be from the 50 lbs of residue decomposed, which is the same as used with a known C:N ratio). In problem #3, the %N of the residue is given. See the calculations involving the lbs of residue decomposed.

3.     Six tons of corn-stalk residue are added per acre (40% C and 0.6% N).

a. What is a reasonable estimate of the amount of microbial C present in the soil in microbial tissue 4-8 weeks after application.

b. What is a reasonable estimate of the amount of CO₂ liberated during this same period.

c. What is a reasonable estimate of the total amount of microbial N at the end of 8 weeks?

d. What is a reasonable estimate of the amount of immobilization of soil N (inorganic N taken from the soil) at the end of 8 weeks.

3a.

12,000 lbs residue
X 0.5 (% decomposition)
6,000 lbs residue decomposed
X 0.4 (% C in residue)
2400 lbs C in residue
X 0.25 (% C into microbe biomass)
600 lbs microbe C
 

3b.

 2400 lbs C in decomposed residue
-600 lbs microbe C
1800 lbs C liberated as CO₂

  3c.     

600 lbs microbe C
/8 (8:1 C:N ratio of microbe)  =
75 lbs N needed by microbes
 

3d.

6000 lbs residue decomposed
X 0.006 (% N in residue)
36 lbs of N in decomposed residue

75 lbs needed - 36 lbs have = 39 lbs immobilized

4. Two tons of sucrose (table sugar) are added per acre. You should assume that 100% will be readily broken down (rather than the expected 50% of a natural plant residue). Also, the maximum incorporation of C into microbial tissue is expected to be much quicker because of this readily available energy source. Activity likely will be maximized within a week after application in a warm, moist soil.

a. Sucrose is C₁₂H₂₂O₁₁. What is the percentage C and N in sucrose? (C=12, H=1, O=16, N=14) b. What is a reasonable estimate of the amount of N required to prevent immobilization from the soil (assume 100% breakdown of the sucrose within a week)?

4a.

C = 144
H =   22
O = 176
       342

% C = 144 /342 = 0.421 (42.1%)
% N = 0% (no N present in sucrose)
 

4b.     

4000 lbs sucrose
X 0.42 (% C in sucrose)
1680 lbs sucrose C
X 0.25 (% C in microbe biomass)
420 lbs microbe C
/8 (8:1 C:N ratio of microbe)  =
52.5 lbs N needed