LabBench: Photosynthesis

Using the equation S=∑(xi-x¯)2n-1−−−−−−√S=∑(xi-x¯)2n-1 and the information provided in the table below, determine the standard deviation for "older leaves." Have a calculator in hand to do the calculations.

0.8 minutes

How might you make sense of this data and have a value that you can use to compare across different trials? It turns out that the most representative value to use is the median time--the time it takes for 50% of the disks to float, a term called the ET50 (the Effective Time it takes 50% of the disks to float). That way, if a disk pops up immediately or fails to rise for a very long time, the data won't be skewed as it would by using the mean time, where outlying data points would throw off the average. So using the above graph, you can determine the ET50 for this sample set by locating the point on the y-axis for 5 disks (50% of 10 disks). Draw a straight line across to the point halfway between when 4 disks and 6 disks are floating, as shown by the data points. Then draw a line down to the x-axis, which puts the ET50 for this sample set at 7.5 minutes--the time it takes 50% of the disks to float; see the red lines in the graph, below. Examine the following graph with data plotted for 14 leaf disks over 16 minutes. What is the ET50 for the data plotted here?

10 minutes

The mean ET50 for the old leaves was 9.2 minutes with a standard deviation of 0.8 minutes. Between what two values would you expect the ET50 to be 95% of the time?

7.6 minutes and 10.8 minutes

Which of the following describes the best way to graph the data to clearly compare whether older or younger leaves carry out photosynthesis more rapidly?

A bar graph that compares the mean ET50 of the five trials with young leaves on one bar, and the mean ET50 of the five trials with older leaves on another bar.

The floating disk procedure can be used to investigate numerous factors that can affect the rate of photosynthesis. Some possible variable factors are listed below. Predict how each condition would affect the rate of photosynthesis by dragging the different variables to the appropriate column.

Increase rate - Decrease rate 0.2 - 0.01 green portion of a variegated leaf - white portion of variegated leaf 100W bulb - 40W bulb green light - red light 10 cm - 30 cm 22°C - 15°C

For the experiment in Part H, sort each variable according to whether it represents the independent variable, the dependent variable, or a controlled variable. Sort each item to the appropriate bin.

Independent variable: plotted on the x-axis age of the leaf Dependent variable: plotted on the y-axis time it takes five leaf disks to rise Factors held constant: size of leaf disk light intensity sodium bicarbonate concentration species of plant

Not all wavelengths of the visible spectrum are equally absorbed by photosynthetic pigments. Green and yellow wavelengths are reflected, giving leaves their characteristic colors, while the blue and red wavelengths are absorbed. It is the red and blue wavelengths that provide most of the energy that drives photosynthesis. In 1883, Theodor W. Engelmann conducted a classic experiment in which he illuminated a filament of photosynthetic algae with light that passed through a prism, thus exposing different segments of the algae to different wavelengths of light. He then added aerobic, nonphotosynthetic bacteria, and noted that the largest groups of bacteria were found in the areas illuminated by the red and blue light. Using what you know about photosynthesis and cellular respiration, which of the following would best explain the reason for the bacteria congregating in the regions with red and blue light?

More bacteria congregated in the red and blue light regions because the algae in those regions produced the most oxygen.

In the figure below, notice that there are chloroplasts and numerous air spaces in the mesophyll tissues. The green color of leaves is due to the presence of chlorophyll, a pigment that is able to absorb light energy. Chlorophyll is contained in the chloroplasts, the cellular organelles that actively perform photosynthesis. Here, light energy is used to synthesize organic compounds and oxygen is released as a byproduct. The water used in photosynthesis reaches the leaves through the veins, and CO2 enters through the stomata, pores on the underside of the leaf. Oxygen produced in photosynthesis exits through these stomata as well. In the parts of the investigation that follow, we conduct experiments using leaf disk samples to test whether they are undergoing photosynthesis based on whether they float in solution. What gives the leaf disks their buoyancy?

Oxygen accumulating in the air spaces within the mesophyll.

In photosynthesis, energy enters the ecosystem in the form of sunlight and eventually leaves in the form of heat. Plant cells and other producers convert light energy via chloroplasts into chemical energy that is stored in sugars and other organic compounds, releasing oxygen (O2) as a byproduct. This chemical energy is used for movement, growth, reproduction, heat generation, and other cellular processes--including cellular respiration--in producers and organisms that eat producers. All cells, including plant cells, perform cellular respiration, and ATP powers most cellular work. Sort the reactants and products of photosynthesis to the correct bins.

Reactants: light energy H2O CO2 Products: C6H12O6 O2

Which of the following would be the most plausible explanation for the results seen here?

The light source was removed after 20 minutes.

As described in the previous part of this investigation, air spaces in the mesophyll of leaves give them their buoyancy. In the floating leaf disk technique, you will remove the accumulated gases from leaf disks to make them sink when placed in solution. When the disks are exposed to bright light, they will begin to float as oxygen accumulates in the mesophyll during photosynthesis. The following figure shows how to prepare the leaf disks for this investigation. You may wish to copy this image to take to the lab. As shown in the figure, begin by preparing a sodium bicarbonate (baking soda) solution with a drop of dishwashing detergent. The sodium bicarbonate dissociates in water (NaHCO3 + H2O → Na+ + OH− + H2O + CO2) to release carbon dioxide, which is a reactant for photosynthesis. The detergent reduces surface tension and minimizes the possibility of the disks sticking together. Using a one-hole punch, cut 10 or more uniform leaf disks from a firm leaf, such as spinach or English ivy (avoid the veins). Place the disks into a syringe with the sodium bicarbonate solution and create a vacuum, removing the accumulated gas from the air spaces in the disks' mesophyll and replacing it with the sodium bicarbonate solution. Repeat the vacuum procedure as necessary to ensure that all of the leaf disks have sunk to the bottom of the syringe. Once all of the disks have sunk, expose the syringe they're in to bright light, where the chloroplasts will begin to perform photosynthesis. When sufficient oxygen accumulates in the air spaces of the disks' mesophyll, the disks will rise to the surface. The length of time it takes for leaf disks to float again can be used as a measure of the rate of photosynthesis. Using your understanding of photosynthesis, propose the most likely reason the disks have not risen in syringe A.

There is no source of carbon dioxide.

Examine the graph. Do the data suggest that there is a statistically significant difference between the rate of photosynthesis in young leaves versus older leaves?

Yes, because the error bars of the means do not overlap.

The floating disk technique lends itself to independent investigation as it is relatively simple to collect large amounts of data in a fairly short period of time. To build your skill at doing this, consider the following scenario. A student decided to compare the rate of photosynthesis in English ivy leaves that were newly opened to leaves from the same plant that had been on the vine for many weeks. She punched out 10 disks from each leaf type, created a vacuum to remove the gases in the mesophyll, and sank them in 200 mL of a 0.2% sodium bicarbonate solution in plastic cups. The cups were placed four inches from a fluorescent light source. Every minute, the number of disks that had risen in each solution was recorded. This procedure was repeated with fresh disks for a total of five trials. Which metric would provide the most useful information for comparing the trials?

the median time it takes five disks to rise in each trial