Friday 11:00-11:50EEMB/ES 171Problem Set #2Due: Wednesday, December 2 2015Problem #1:You are studying decomposition in mixes of forest leaf litter. To do this experiment, you set up litter bags with each of four species of litter separately, and then with mixtures of different litters. The mixtures are made up of equal amounts of the different litters in the bag. Species Litter lignin Litter N kA 45 0.8 -0.40B 25 0.8 -0.6C 25 1.6 -1D 45 1.1 -0.62E 35 1.2 No Data A + B -0.45A + C -0.8A + D -0.49B + C -0.77B + D -0.5C + D -0.86Analyze these data. Start by making a graph of Lignin/N vs. k (analogous to that in the paper by Muratore, Aber, and Melillo). Lignin/ N vs. K 1. What climate regime do you think this experiment is being done in? Why? Hint: use the Muratore, Aber, and Melillo paper.Friday 11:00-11:50I believe that the climate regime this experiment is being done in is an unbroken forest of uneven-aged, well-stocked flora at a low elevation. This conclusion can be made because many factors such as moisture, temperature, and nature of the microorganisms and soil fauna active in decomposition process. Furthermore, this climate contains moisture.2. What do you think the most likely k value for species E is?According to the graph, I believe that the mostly likely k value for E would be -0.75. 3. Come up with a hypothesis for the k value for species B. How would you test this hypothesis?Species B has a relatively low Lignin to N ratio, which is similar to the other species, however the K value for species be is distinctively larger than the others. Species B K value is a less negative than the other values which makes species B K value an outlier in this graph. The larger k value therefore implies a slower decomposition rate, which could be due to under or over saturated pore space or tannins. Pore space in relation to decomposition rate ideally should be 60%. Taking these factors into account I hypothesize that the K value is less negative than predicted because the pore space saturation can be larger or smaller than this level of saturation. Tannins can also affect the K value because they affect nutrient cycling depending on pH and soil nutrient status which ultimately slows decomposition. One possible method to test this hypothesis would be to set up 3 different areas with the same soil and set up a control, soil with water, and soil with reducedmoisture. Once this is set up, we can solve for the K value of each plot and if any of the experimental plots show a more negative k value than we can imply that water is the factor responsible for the fluctuation in k. Also set up soils and check for tannins to seee if this is a factor that can be responsible for the slower decomposition rate.4. Now consider the litter mixtures. Do they behave as simply the average of the litters that make them up? If not, why not? What factors do you hypothesize might cause the behaviors that you observe? How would you test that hypothesis?No, the litter mixtures do not simply behave as the average of the litters that make them up.They do not act as simply the average because there are different factors that can result in the behaviors shown. Some of these factors responsible can be toxins and initial inoculations of microorganisms. In addition to these factors the fluctuation in N and Ligninvalues from the combination of soils, which can change the overall k value. One way to test the effect the combination of soils had on decomposition rate could be by measuring initial N and Lignin values of single soils and N and Lignin values of soils when combined to see if there was a gain or loss of N, lignin, or both when mixed. If this is observed, the conclusion can be made that the changes in Lignin, N, or both are responsible for the change in K (decomposition) values.Problem #2:You are studying the decomposition and mineralization from leaf litter in a forest. The litter has the following characteristics:Friday 11:00-11:50Material % of litter mass C/N ratio Physiological material 20 5 Cellulose 60 No N Lignin 20 No NThree groups of organisms decompose this materialOrganism C/N Ratio Substrate Use Efficiency They eat: Bacteria 5 0.50 Only physiological materialCellulytic fungi 20 0.40 Everything except ligninLignolytic fungi 20 0.25 EverythingAssume: A. Growth rates:a. Bacteria grow faster than fungi and have first crack at any material that they can metabolize (physiological material and dead microbes). b. Cellulytic fungi grow next fastest and will use resources that bacteria don’t.c. Lignolytic fungi grow the slowest of all three groups (i.e. they have last crack at material). B. Microbes will only process a material if they have the nitrogen they need to do so. 1. Given those assumptions, run calculations for the first cycle of decomposition. Calculate:A. How much of each plant carbon pool is decomposed?All plant carbon pool is decomposed.B. How much biomass of each group of microorganisms is produced?Bacteria: Biomass Produced= 10.0 g N Required= 2.0gCellulytic Fungi: Biomass produced= 24.0 g N Required= 1.2 g N Leftover= 2.0 gLignolytic Fungi: Biomass produced= 5.0 g N Required= 0.25 gTotals: Biomass Produced= 39.0 g N Required= 3.45 gC. Is any nitrogen mineralized? If so, how much? Yes, 0.55 g is mineralized. * WORK ATTACHED ON SEPARATE SHEET OF