Need help solving EXERCISE 1! Please show all work! 3. Carbon in the Biosphere W

Question

Need help solving EXERCISE 1! Please show all work!

3. Carbon in the Biosphere What are the residence times of carbon in conti nental and marine vegetation? M -5.6 x 107g(C) 5 x 1016g(C)lyr Take a guess. Trees live for 50 years or more and constitute a large share of continental vegetation. Phytoplankton, accounting for a large share of marine primary productivity, generally come and go in a se- ries of annual blooms and crashes each lasting less than a year. We might therefore expect the residence time of carban in plants to be on the order of 10-100 years on the land and 0.1-1 year in the oceans T11.2 yr To be more accurate, we must know the stocks and flows of car- bon. The stocks are living biomasses and the flows are net primary productivities. The Appendix provides both of these quantities, in conveniently comparable units -g(C) and g(C)/yr. From this we can Terrestrial biomass obtain 2 x 1015g(C) 2.5 × 1016g(C),r stock of iving continental biomass continental net primary productivity 5.6 × 10 5% 1016 g(C)lyr Terrestrial- 17 Marine biomass Figure I1-2 Stocks and flows of biomass in terrestrial and marine biomasses. = 11.2 and instead, we would have obtained much shorter residence times. Car bon flows relatively quickly through respiratory pathways in organ- isms. This respiratory flow is not included in net primary productiv- ity. Using net primary productivity yields a residence time that bears a closer relation to the lifetimes of typical organisms (see Exercise 2) stock of living marine plants marine net primary productivity 2.5 x 101 g(C)yr 0.08 yr 1 month EXERCISE 1: Suppose that the average residence time igncr- ing respiratory pathways) of carbon in the phytoplankton in a lake is two weeks. Zooplankton in the lake, grazing upon the phytoplanlk ton, consume 40% of the net primary productivity and have an incr- poration efficiency of 25% (i, e., 25% of the phytoplankton biomass they eat is incorporated into zooplankton biomass). In other words the net productivity of the zooplankton is 0.25 × 0.40 or 10% of the npp of the algae. If the average residence time (ignoring respiratoy pathways) of carbon in zooplankton biomass is six months, estimate the ratio of the average biomass of the zooplankton papulaticn to that of the phytoplankton population in the lake. Figure II-3 illustrates the flows in and out of the two stocks of plankton. These relations are illustrated in Figure II-2. The considerable uncer tainty in the stock and flow data used here (see Appendix) makes these residence times approximate Notice that the continental residence time of carbon is considerably shorter than the average lifetime of trees. Even though the woody parts of trees constitute the bulk of living continental biomass, only a part of each tree's annual production adds woody tissue; much of the net primary productivity produces leaves, which have less than a one-year residence time Why did we use net primary productivities in this calculation? Had we used data on gross marine and continental primary productivities EXERCISE 2: In this exercise we explore the relation between residence time of biomass and lifetime of individuals in a steady-state

Explanation / Answer

Exercise 1:

The residence time of zooplanktons = 6 months = 0.5 years

The net primary productivity of zooplanktons = 0.25*0.40 = 0.1

The average biomass of zooplanktons = residence time * net primary productivity = 0.5 * 0.1 = 0.05

Similarly, the residence time of phytoplanktons = 2 weeks = 0.038 years

The net primary productivity of phytoplanktons wrt that of zooplanktons = 1

Therefore, average biomass of phytoplanktons = residence time * net primary productivity = 0.038 * 1 = 0.038

Therefore, the average biomass of zooplankton population to that of the phytoplankton population in the lake = 0.05/0.038 = 1.3

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.