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Name: ______________________

Date: _____________ Mods: ____

advanced placement

Environmental Science

Carolina: 18-1030

Population Growth in Lemna minor

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Objectives

Upon completion of this exercise, you should be able to

* Identify the similarities and differences between exponential and logistic population growth

* Explain how carrying capacity and biotic potential affect population growth

* Determine a habitat's carrying capacity using a logistic population growth curve

* Estimate a population's biotic potential from population growth data

Discussion

A population is a group of individuals of the same species living in the same area at the same time. In stable environments, the number of individuals in a population remains nearly constant. In rapidly changing environments, however, many individuals may die within a short period of time, leaving only a few survivors to perpetuate the species. In either case, the survival of the population depends on its ability to increase in number.

Models of population growth help scientists understand how a population's size changes over time. These models have applications in microbiology, wildlife management, pest management, agricultural productivity, and toxicology. Consider two simple models of population growth: exponential and logistic.

Exponential Population Growth

When resources are unlimited, the number of individuals in a population grows exponentially: 1, 10, 100, etc. (see Figure 1).

In exponential growth, the number of individuals increases rapidly and without bound. The population's rate of increase, dN/dt, equals a

constant, r, times the number of individuals, N.

dN/dt = rN

The d means "change" so that the above equation reads, "the change in the number of individuals as time changes equals a constant, r, times the number of individuals." The constant r is the population's intrinsic rate of increase and varies from one environment to another. There exists, at least in theory, an environment that is perfect in all respects for a population and in which it attains its maximum rate of increase. This rate is the population's biotic potential, rm. Most environments, however, limit growth and the population's intrinsic rate of increase is necessarily less than the biotic potential.

Integrating the above formula yields the following equation:

Nt =N0ert

where Nt is the number of individuals at time t, n0 is the number of individuals at time 0, e is Euler's constant, and t is the length of time from time 0 to time t. With this equation, it is possible to calculate the number of individuals in the future, the number of individuals in the past, or a population's intrinsic rate of increase. In this experiment, you calculate the maximum value of r in each growing medium and estimate the biotic potential of Lemna minor.

For example, suppose there were three fronds on day 0 and six fronds on day 3. What is the population's intrinsic rate of increase over this time period? Use the formula Nt = N0ert, where Nt equals six, N0 equals three, and t equals three.

Nt = N0ert

6 = 3e3r 2=e3r

ln 2 = ln e3r Take the natural log (ln) of both sides.

ln 2 = 3r Use the identity In ex = x.

r = (ln 2)/3

r = 0.231 per day

Note that r has the same units of time as t and the quantity rt is unitless. The frond population's intrinsic rate of increase over this time period is 0.231 per day.

Exponential population growth does not continue indefinitely. If it did, one population would quickly cover the surface of the earth (see Pre-Lab Question 2). Growth is limited by the availability of resources, such as light, nutrients, and space. The abundance of resources determines the environment's carrying capacity, K, the number of individuals of a species the environment can support indefinitely.

Logistic Population Growth

As a population approaches its environment's carrying capacity, population growth slows to almost zero. Although individuals in the population may die and new individuals be born, the number of individuals in the population remains nearly constant. This pattern is called logistic population growth and is shown in the following figure.

Notice that the population increases exponentially at first, when resources are in such supply they are effectively limitless. When the number of individuals equals half the carrying capacity, the rate of increase is at a maximum (see Optional Activity 8). Once this number is attained, the rate of increase slows, and the graph approaches a horizontal line whose y value is the carrying capacity. This type of curve is called an S curve, or sigmoidal (after the Greek name for the letter s) curve for its shape.

The rate equation that describes logistic population growth is similar to the exponential rate equation except a new term is added. This term represents environmental resistance as the population approaches the environment's carrying capacity. In this experiment, you determine each growing medium's carrying capacity for Lemna minor.

dN/dt = rN [(K-N)/K]

Lemna minor

Lemna minor is a member of the family Lemnaceae (duckweed family). This tiny organism is ideal for population growth experiments because it reproduces quickly, requires minimal space to grow, and requires no maintenance. The free-floating, freshwater plant consists of a green elliptical frond with one root and is found in still waters, from temperate to tropical zones. The plant reproduces by vegetative budding, although it may flower. On average, fronds live for four to five weeks.

Due to its rapid growth rate, duckweed finds wide use in governmental and commercial applications. For example, the US EPA requires companies that make pesticides to determine whether their chemicals affect the growth of aquatic plants. Many companies use duckweed as a test plant. In the test, the pesticide is applied to duckweed's growing medium and any effects on the duckweed's growth rate are taken as a measure of the pesticide's toxicity.

Some companies use duckweed to remove nitrogen and phosphorus from their wastewater. Nitrogen and phosphorus are plant nutrients that in high concentrations (as in wastewater) promote rapid plant growth. If wastewater were released into the environment untreated, new plant growth would clog waterways and cause eutrophication. To remove nutrients from the wastewater, Lemna plants are grown in it. As the plants grow, they naturally take up nitrogen and phosphorus from the wastewater. When the duckweed plants die, they are harvested, composted, and used as mulch. The treated wastewater continues to the next stage of purification.

The high nutrient content of duckweed also makes it a viable alternative to existing livestock feed. Duckweed can even offer advantages in some cases. For example, poultry' feed requires expensive supplemental pigments that occur naturally in duckweed. Although duckweed is a freshwater plant, its range has extended beyond earth's terrestrial limits. In 1982, the space shuttle orbiter Columbia carried an experiment to study the effects of microgravity on duckweed growth.

Lemna minor grows best in environments that most closely approximate its natural habitat. Although the plant can adapt to extreme conditions, duckweed prefers a pH range of 6.5 to 8.5, low salinity, temperatures around 25° C, and plenty of light. In the experiment, you vary the pH from 5 to 9, the salinity level from 1 g of NaCl/L to 4, and the amount of light the fronds receive by placing one to four screens over the tops of the tubes. You also grow fronds in spring water that contains excess nitrate arid phosphate and, as a control, in pure spring water.

Data Collection

You will count the number of fronds in the tubes and record the number and the count date in the data table that you create as part of your pre-lab assignment. To estimate the biotic potential of Lemna minor, you must identify the time period over which the growth rate is at a maximum. This occurs when the number of new fronds per day per frond is at a maximum.

Suppose you collect the following data. On the first day of the experiment, Day 0, you place 3 fronds in the test tube. The next day, the number of fronds is still 3. Two days later, there are 5 fronds. Day 4, there are 6 fronds. First, complete the column "Change in the Number of Fronds since the Previous Count." On Day 0, there is no change because there was no previous count. On Day 1, the previous count was made on Day 0. The number of fronds, 3, did not change and a 0 is recorded in the column. On Day 3, the number of fronds increased to 5 from 3, so a 2 is recorded in the column, and similarly for Day 4. Next, complete the column, "Change in the Number of Fronds per Day." This value equals the value from column three, the number of new fronds in that day's count, divided by the number of days since the previous count. On Day 3, there were 2 new fronds since the previous count on Day 1. Thus, the change in the number of fronds per day equals 2 divided by 2, or 1. The last column, the number of new fronds produced per day per frond, equals the number in the "Change in the Number of Fronds per Day" column divided by the number of fronds from the previous count. For example, on Day 3, the change in the number of fronds per day was 1, and there were 3 fronds on the previous count, Day 1. Thus, the change in the number of fronds per day per frond is 1 divided by 3, or 'A.

Table I. Sample Experimental Population Growth Data

|Day Number |Number of Fronds |Change in die Number of|Change in the Number |Change in the Number |

| | |Fronds since the |of" Fronds per Day |of' Fronds per Day per |

| | |Previous Count | |Frond |

|0 |3 |NA |NA |NA |

|1 |3 |0 |0 |0 |

|3 |5 |2 |1 |0.33 |

|4 |6 |1 |1 |0.20 |

Experimental Determination of r and K

Use the maximum figure from the "Change in the Number of Fronds per Day per Frond" data, to begin calculating the population's intrinsic rate of increase. (If there are two maxima, choose the one that occurs first in time.) In the example above, the maximum occurred on Day 3. Use the exponential growth formula

Nt = N0ert

where Nt is the number of fronds on Day 3, N0 is the number of fronds on Day 1, e is Euler's constant, and t is the length of the time period, 2 days.

Nt = N0ert

5= 3e2r

5/3 = e2r

ln (5/3) = ln e2r Take the natural log (In) of both sides.

ln (5/3) = 2r Use the identity In ex = x

r = 1/2 ln (5/3)

r = 0.255 per day

To determine each tube's carrying capacity, students graph the "Number of Fronds" against the "Day Number" to obtain a logistic population growth curve. Over time, the number of fronds reaches a constant value. This number is the tube's carrying capacity.

**Pre-Lab Questions – answer on page 7!

Procedure

Divide into six groups of four and number each group: 1, 2, 3, 4, 5, or 6.

1. Follow the directions for your group number. Let each member of the group prepare a test tube.

Group 1. Obtain four test tubes and place in a flask or beaker for support. Measure 50 mL of spring water in a graduated cylinder and pour into each test tube. Using a spatula, carefully transfer one colony of three fronds to each test tube. Cover the test tubes with plastic wrap and secure with a rubber band. Poke five small holes in each test tube's plastic wrap. Label each test tube "Spring Water" and number 1 through 4 using an indelible marker. Place the test tubes in a rack.

Group 2. Obtain four test tubes and place in a flask or beaker for support. Pipet 1 mL of saline solution into a test tube and add 49 mL of spring water. Label the test tube "Salinity, 1 g/L." Pipet 2 mL of saline solution into a test tube and add 48 mL of spring water. Label the test tube "Salinity, 2 g/L." In the same way, prepare the "Salinity, 3 g/L' and "Salinity, 4 g/L' test tubes. Using a spatula, carefully transfer one colony of three fronds to each test tube. Cover the test tubes with plastic wrap and secure with a rubber band. Poke five small holes in each tube's plastic wrap. Place the tubes in the test tube rack.

Group 3. Obtain four test tubes and place in a flask or beaker for support. Measure 10 mL of buffer, pH 5, and pour into a test tube. Add 40 mL of spring water and label the tube "pH 5." Measure 10 mL of buffer, pH 6, and pour into a test tube. Add 40 mL of spring water and label the tube "pH 6." In the same way, prepare the "pH 7" and "pH 8" test tubes. Using a spatula, carefully transfer one colony of three fronds to each test tube. Cover the test tubes with plastic wrap and secure with a rubber band. Poke five small holes in each tube's plastic wrap. Place the tubes in the test tube rack.

Group 4. Obtain four test tubes and place in a flask or beaker for support. Measure 50 mL of spring water in a graduated cylinder and pour into each test tube. Using a spatula, carefully transfer one colony of three fronds to each test tube. Cover the test tubes with plastic wrap and secure with a rubber band. Poke five small holes in each tube's plastic wrap. Label the four tubes of spring water, "1 Screen," "2 Screens," "3 Screens," and "4 Screens." Cut a square screen (3x3 inches), cover the top of the tube labeled "1 Screen," and secure with a rubber band. Cut a screen 3x6 inches and fold it in half to make two squares. Use it to cover the top of the tube labeled "2 Screens," and secure with a rubber band. In the same way, place three screens and four screens over the remaining tubes. Place the tubes in the test tube rack.

Group 5. Obtain four test tubes and place in a flask or beaker for support. Measure 50 mL of spring water in a graduated cylinder and pour into a test tube. Label the test tube "Spring Water, 9 A." To an empty test tube, pipet 1 mL of phosphate solution and add 49 mL of spring water. Label the tube “Excess Phosphate A”. To another empty test tube, pipet 1 mL of excess nitrate solution and add 49 mL of spring water. Label the tube “Excess Nitrate A”. Measure 10 mL of buffer, pH 9, and pour into a test tube. Add 40 mL of spring water and label the tube "pH 9 A." Using a spatula, carefully transfer one colony of three fronds to each test tube. Place the tubes in the large test tube rack.

Group 6. Obtain four test tubes and place in a flask or beaker for support. Measure 50 mL of spring water in a graduated cylinder and pour into a test tube. Label the test tube "Spring Water, 9 B." To an empty test tube, pipet 1 mL of phosphate solution and add 49 mL of spring water. Label the tube “Excess Phosphate B”. To another empty test tube, pipet 1 mL of excess nitrate solution and add 49 mL of spring water. Label the tube “Excess Nitrate B”. Measure 10 mL of buffer, pH 9, and pour into a test tube. Add 40 mL of spring water and label the tube "pH 9 B." Using a spatula, carefully transfer one colony of three fronds to each test tube. Place the tubes in the large test tube rack.

3. Once all the test tubes are placed in the test tube rack, place it under the light source and turn on the light.

4. On days your instructor designates, count the number of fronds in your test tube and record the data in your table. Count only live (green) fronds. Fronds that are totally yellow, clear, black, or white are dead. The experiment may continue for 3 to 4 weeks.

5. At the end of the experiment, complete your data table.

Name: _____________________________________________

Date: _____________________________ Mods: ___________

Lemna minor Lab – to turn in!

Pre-Lab Questions

1. Compare and contrast exponential and logistic population growth. Include the respective rate equations.

2. The surface area of the earth is approximately 5.1 x 108 km2 How many days would a colony of three fronds of Lemna minor require before they covered the earth's surface? Assume the surface area of one frond is

1.5 x 10-5 km2 and r is 0.288 per day.

Data Analysis - Graphing

Graph the curve "Number of Fronds versus Time" for your sample in Excel. Attach your graph separately.

2. From your graph, determine the carrying capacity of your test tube environment. Draw a horizontal line to indicate the value of the carrying capacity and label the line with a "K =."

3. Identify the two days that bound the time period over which the "Change in the Number of Fronds per Day per Frond" was at a maximum. Calculate the observed biotic potential, r, of the fronds in your test tube using the equation, Nt = N0ert where Nt is the number of individuals on the later day, N0 is the number of individuals on the earlier day, e is Euler's constant, and t is the length of the time period in days.

Laboratory Questions

1. Is there a relationship between the number of screens on a tube and its carrying capacity? Is there a relationship between the number of screens on a tube and its fronds' intrinsic rate of increase?

2. What assumptions were made about the plants in this experiment?

3. What are possible sources of error in this experiment?

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Time

Figure 1. Exponential Growth Curve

Time

Figure 2. Graph of Logistic Growth

Figure 3. Lemna minor

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