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Pigments and Photosynthesis LEARNING OBJECTIVES 1. Apply the procedure of paper chromatography in the separa7on of the photosynthe7c pigments of plants and relate the pigments to their photosynthe7c ability. 2. Discuss how experimenta7on with the Hill reac7on provides evidence of the opera7on of the light reac7ons. 3. Design and implement an experimental procedure to determine factors that affect the rate of photosynthesis in an organism. 4. Analyze and interpret experimental data from the Hill reac7on. INTRODUCTION Photosynthesis is the process by which light energy is converted to chemical bond energy. Autotrophic organisms are responsible for primary produc7on, and photosynthesis is the process by which such produc7on is affected. In terrestrial communi7es, plants are the most significant photosynthe7c organisms. In aqua7c communi7es, phytoplankton contribute the most to primary produc7on. Photosynthesis comprises numerous vital and complex reac7ons that take place in the cells of organisms capable of carrying out this process. The process can be summarized with this chemical equa7on: 6 CO 2 + 12H 2 O + Light Energy C 6 H 12 O 6 + 6O 2 + 6H 2 O The stomata (small openings in the epidermis of the plant) allow CO 2 and H 2 0 to leave the plant and 0 2 to enter the plant. Guard cells surround the stomata and control the opening and closing of the stomata. When guard cells take on water, they pull apart and allow gas exchange to occur and water vapor to escape. See Figure 6. Figure 6. Guard cells (purple) surrounding stomata (green) CELLULAR STRUCTURES Before considering the molecules that are involved in the photosynthe7c reac7ons, let's examine the cellular structures in which these molecules are contained. Most photosynthe7c organisms are eukaryo7c, and the photosynthe7c reac7ons take place in cytoplasmic organelles known as chloroplasts . Although photosynthe7c eukaryotes such as plants and algae are more common, photosynthe7c bacteria are also abundant. Since bacteria are prokaryo7c, their cells do not contain organelles, but they do have membrane layers organized into thylakoids that perform func7ons analogous to chloroplasts.
The chloroplast houses the photosynthe7c process in eukaryo7c cells. In each chloroplast, a double membrane surrounds a fluid-filled compartment known as the stroma . Enzymes, electron carriers, and numerous other molecules essen7al to photosynthesis are dissolved in this fluid. Also within the stroma are abundant thylakoids . Each thylakoid is a 7ny membranous vesicle containing photosynthe7c pigments and electron carriers. These thylakoids are organized in stacks known as grana (singular = granum ). Different photosynthe7c reac7ons take place in the grana and the stroma. See Figure 7. Figure 7. Leaf cross sec7on and chloroplast. haps://en.wikipedia.org/wiki/Chloroplast#/media/File:Figure_08_01_05.png haps://www.pathwayz.org/Tree/Plain/CROSS+SECTION+OF+A+LEAF+%5BBASIC%5D PHOTOSYNTHETIC REACTIONS The total process of photosynthesis comprises two major series of reac7ons. These two series are known as the light reacKons and the dark reacKons (or light-independent reacKons) . This terminology emphasizes the form of energy required to drive these different reac7ons. The light reac7ons require light energy in order to take place. The dark reac7ons can and do take place in the dark-but they also take place in the light. "Dark" is used to emphasize that these reac7ons do not require light energy in order to take place-but they also take place in the light. They do require specific forms of chemical energy that are produced in the light reac7ons. Accordingly, the light reac7ons must begin and produce this appropriate chemical energy before the dark reac7ons will take place. LIGHT REACTIONS AND PHOTOSYNTHETIC PIGMENTS The light reac7ons take place in the membranes of chloroplast thylakoids. These reac7ons are possible because the embedded pigment molecules absorb the energy of sunlight. Each pigment is capable of absorbing specific wavelengths of light energy in the visible spectrum . Photosynthe7c pigments are classified func7onally as primary and accessory . The primary pigment is that molecule that is capable of giving up an excited electron such that light energy is converted to chemical bond energy. In eukaryotes and some bacteria, the single primary pigment is a type of chlorophyll designated as chlorophyll a . In other groups of photosynthe7c bacteria, the primary pigment is a bacteriochlorophyll that performs a similar func7on. When these pigment molecules absorb sufficient energy of the appropriate wavelengths, they release high energy electrons that drive the light reac7ons. There are several accessory pigments in both eukaryo7c and prokaryo7c cells. Whereas chlorophyll a is a green pigment, the accessory pigments may also be green ( chlorophyll b ), or they may appear blue-green, yellow, orange, or brown. The familiar yellow and orange of hues autumn foliage are produced by those accessory pigments known as carotenoids . These pigments absorb wavelengths other than those absorbed by chlorophyll a, and thus increase the organism's total capacity for light absorp7on. You should be able to idenKfy these
pigments in the chromatogram you will he doing today . Through a process known as electron resonance , accessory pigments transfer their absorbed energy to the primary pigment. This transferred energy excites the primary pigment (chlorophyll a ), which then gives up a high energy electron. By both direct absorp7on of light and transferred energy, the chlorophyll a is capable of driving the light reac7ons. As the light reac7ons take place in the thylakoids of the chloroplast, they produce chemical energy in the forms of NADPH and ATP that drive the dark reac7ons in the stroma. See a summary of the light reac7ons below (Figure 8). Figure 8. Light reac7ons. haps://en.wikipedia.org/wiki/Light-dependent_reac7ons DARK REACTIONS AND FORMATION OF GLUCOSE These dark reac7ons are commonly called the Calvin cycle . This name refers to the physiologist Melvin Calvin, who first described these reac7ons in the 1940s and 1950s. The func7on of the Calvin cycle is to produce glucose , the carbohydrate that incorporates energy ini7ally harvested from sunlight. The glucose molecules will be used in many ways in the cells. For example, glucose is the star7ng point for the synthesis of numerous other sugars that are involved in metabolism. Hundreds of glucose molecules may be bonded to one another to form carbohydrate polymers, such as starch, that serve as energy storage. Other polymers of glucose, such as cellulose, are major structural molecules in cells. For every 6 CO 2 molecules used to create a carbohydrate in the Calvin cycle, 18 ATP and 12 NADPH molecules are needed from the light reac7ons. See Figure 9. Figure 9. Calvin cycle. haps://www.khanacademy.org/science/biology/photosynthesis-in-plants/the-calvin-cycle-reac7ons/a/calvin-cycle Please highlight all your answers with a yellow background or use a different color font that is easy to read.
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ACTIVITY 1: PAPER CHROMATOGRAPHY Chromatography is a procedure for separa7ng dissolved substances from one another. A soluKon is made up of a solvent and one or more solutes . In plants, the solvent is water, and there are many kinds of solutes, such as sugars, mineral ions, amino acids, and pigments. We will use chromatography to separate different plant pigments. In chromatography, the solutes are separated from one another according to both their solubility ( or lack of solubility) in different liquids and their adsorpKon to an inert material. As the name suggests, paper (cellulose) is the inert adsorbent used in paper chromatography. The solu7on is applied as a narrow band approximately 2 cm from the lower edge of the strip or sheet of paper. This is iden7fied as the origin of the chromatogram. The chromatographic paper is placed in a glass chromatography jar. This jar contains the solvent system that will be used to separate the solutes in the applied solu7on. The origin should be above the surface of the solvent, or it will dissolve into the solvent in the boaom of the jar. The jar is sealed to maintain an internal atmosphere that is saturated with solvent molecules. The solvent immediately begins to move up the paper. The solvent front can be observed as a wet boundary on the paper. As the solvent front passes through the origin, some of the pigment solutes will be dissolved in this moving solvent system. Depending upon both the degree of solubility of the solute in this solvent and the affinity of the solute for cellulose, the solute will be carried upward a certain distance on the paper. If several solutes in the unknown solu7on have different affini7es and solubili7es, then they will be deposited at different sites on the paper. In this manner, it is possible to separate these uniden7fied solutes from one another. The chloroplasts of most plants contain several different photosynthe7c pigments, one primary pigment (chlorophyll a ) and several accessory pigments (varies according to species). The purpose of the following ac7vity will be to use paper chromatography to separate and iden7fy the photosynthe7c pigments contained within the chloroplasts of a magnolia leaf. CauKon: The 9 ether : 1 acetone solvent used in this procedure is flammable and quite noxious. The chromatography jars should be kept on the side counter or in a fume hood at all 7mes. Do not move the jars. Used solvent should remain in the chromatography jars or properly collected in a waste jar by the lab instructor. PROCEDURE 1. Obtain a sheet of chromatography paper. Hold the paper by its edges so that dirt and oil from your fingers will not get on the paper. Fold the paper lengthwise to form a crease in the paper. This will help reinforce the rigidity of the paper. 2. Obtain a fresh magnolia leaf from the side counter. 3. Place the leaf boaom-side down on the chromatography paper. 4. Place a ruler parallel to the boaom of the chromatography paper on top of the magnolia leaf. One side of the ruler should run parallel 2 cm above the boaom of the paper. 5. Using the edge of a coin or the 7p of blunt forceps, rub 2-3 smooth, con7nuous lines across the leaf such that it leaves a solid green mark on the paper underneath. Press firmly but not too hard-you do not want to tear the leaf. The pigment line must be above the solvent level in the chromatography chamber. 6. Prepare a chromatography chamber with enough 9:1 ether:acetone solvent to completely cover the boaom of the chamber by about 1 cm. This step may have been done for you. 7. Place the pigmented chromatogram into the chamber and close the lid of the chamber. The origin must not be submerged in the solvent. Consult the lab instructor if the solvent level is not correct. Do not move the chamber once the paper is in place. 8. Allow the chromatogram to develop long enough for the solvent front to come within 2-3 cm of the top of the paper. 9. Remove the chromatogram from the chamber and allow it to air dry near an open window or in a fume hood. Replace the lid on the chromatography jar.
10. Once dry, turn off the room lights and shine the UV lamp on your chromatogram. 11. Do not dispose of the solvent in the jar. It should be reused by the next laboratory sec7on. 12. Draw your chromatogram results in the following diagram and iden7fy each of the separated pigments. ACTIVITY 2: LIGHT REACTION MEASUREMENTS THE HILL REACTION In 1937, the English biochemist Robert Hill demonstrated that when isolated chloroplasts were illuminated, ar7ficial electron acceptors were chemically reduced. As recogni7on of his pioneering work, this use of isolated chloroplasts con7nues to be designated as the Hill reacKon . In these oxida7on-reduc7on reac7ons, oxygen gas (0 2 ) also was produced. Hill's work was significant in providing the ini7al descrip7on of the light reac7ons of photosynthesis. Experiments based on this earlier work are useful in developing an understanding of the light reac7ons of photosynthesis. In our experiment, we will isolate chloroplasts from spinach leaves such that we may use these organelles in our study of the Hill reac7on. Aper we have isolated the chloroplasts, we will carefully break open these organelles in order to expose the internal thylakoids. When we illuminate these exposed thylakoids, the chlorophyll a molecules will release excited electrons. By using an ar7ficial electron acceptor, dichlorophenol indole phenol (DPIP), we will be able to detect the ac7vity of the light reac7ons. DPIP intercepts electrons before they are transferred to natural electron carriers in the thylakoid membranes. As the blue DPIP accepts electrons, it is reduced chemically to a colorless form. By monitoring this color change under varying experimental condi7ons, we can study the effects of light quality on the rate of photosynthesis. You will be measuring transmiaance, which will increase over 7me if photosynthesis is occurring. The faster the DPIP changes from blue to colorless, the steeper your upward slope will be. PROCEDURE
1. Expose the spinach to light (light prime) for 15-30 minutes. Leave the spinach in a plas7c bag. Place the bag of spinach in the light of a lamp behind a heat sink. The spinach must look fresh. Limp spinach will not undergo photosynthesis. 2. Loosely fill the blender with the light-primed spinach un7l it is about ¾ full. Pour 0.5 M cold sucrose into the blender cup un7l the blender cup is about ½ full with liquid. 3. Blend the spinach with three short bursts of 10 seconds each. 4. Pour the blended mixture through a double layer of cheesecloth into a large beaker. 5. Fill two screw-top test tubes with the filtered suspension. Cover two of the tubes with foil and label "Thylakoids." Place these on ice. These first five steps may have already been performed for you so that the sampl.es are ready to be used when you come into lab. 6. Refer to Table 1 below to prepare the blank and light treatment. Use a separate glass test tube for each of them. Do NOT add the thylakoid prepara7on to any tube un7l immediately before that tube is to be used in step 8. Table 1. **White light = sunlight or light from a provided lamp. White light includes all the wavelengths of the visible spectrum. 7. Obtain a cuveae for the blank and pour the contents of your blank tube to the cuveae fill line. 8. Place the blank in the spectrophotometer in the holder space labeled "B." The spectrophotometer should already be warmed up and set to 605 nm for you. 9. Take tube 1 and add the 150 μL of thylakoids. Gently swirl the tube to mix. Pour enough of the test tube contents into the cuveae to reach the fill line. Then place the cuveae into the appropriate sample cell in the spectrophotometer. 10. Immediately take the transmiaance reading, following the direc7ons at the spectrophotometer. You will keep using this cuveae and the blank for the en7re seven minutes of the run. Remember to orient the cuveYe in the same direcKon each Kme so that differences in the cuveYe will not affect the readings. 11. Remove the sample cuveae from the spectrophotometer and stand it a distance of 12 inches in front of the appropriate light source (not the heat sink) for 30 seconds. There should be a heat sink between the cuveae and the light source. 12. Following the 30-second exposure to the light, place a small square of foil or Parafilm over the opening of the cuveae and invert the cuveae to mix the contents. Using a Kimwipe, clean the cuveae and then place it back in the spectrophotometer. Record the % transmiaance reading in Table 2. 13. Repeat steps 10-12 for seven minutes for the sample cuveae. Solu7ons Blank Light Phosphate buffer 1.0 mL 1.0 mL dH 2 O 3.0 mL 3.0 mL 0.1% DPIP 1.0 mL Light White ** White ** Thylakoids 150 µL 150 µL
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List three factors that could possibly affect the photosynthesis rate. 1. The distance between the light source and the solu7on. 2. The amount of the substrate in the cuveae. 3. The type of light/color of light used in the experiment. NOW DESIGN YOUR EXPERIMENT Use the space below to write up your experimental ques7on (hypothesis and predic7on) and mini-protocol. Your TA will let you know what types of treatments are available . You will need to run two addi7onal experiments with two different variables. Before beginning your experiment, your group will need to get your TA's approval. Be prepared to explain why you chose these treatments. What is the main idea your experiment is tesKng and what do you expect your results to be? Our experiment tests the rate of photosynthesis of the same solu7on (Table. 1 - light) with yellow and green film against light. Hypothesis: We hypothesize that the yellow light will show the highest rate of photosynthesis because, on the absorbance spectrum, the color yellow absorbs the most light. The green light will also absorb a high rate of photosynthesis, but not as much of the yellow light. Experimental Design: We are going to u7lize the same setup as our first experiment. With the same blank however with 2 new solu7ons, one with yellow pigment and one with green pigment. Our solu7ons will contain 1 mL of phosphate buffer, 3 mL of dH2O, 1 mL of 0.1% DPIP, and 150 uL of thylakoids. We will take a yellow and then a green film and put it in front of a bowl of water that has a white light behind it. From then we will take our solu7on with the absorbed yellow or green light and put it in the spectrometer. We will do this 14 7mes for yellow light and 14 7mes for green light and record our data. 14. Working with one sample/cuveae at a 7me, perform steps 7-12 for your two treatments. Remember to add the thylakoids just before you start a new treatment. Do not dispose of the samples in your cuveYes unKl the end of the experiment. Do the samples in each cuveae look different from when you started the experiment? The cuveaes do not look different. Do the samples in the cuveaes from different treatments look different from each other? They do not look different from each other. Explain why or why not for each sample. What specifically was causing this? And explain why or why not when you are comparing your two treatments. The cuveaes were supposed to become a lighter shade of blue over 7me, however our results didn’t show this. This means that the rate of photosynthesis was not dras7c enough to change its color. While the color didn’t change, our numbers did. 15. Prepare a single graph that presents the results for all of the experimental treatments. Include trendlines and equa7ons for each line on your graph specifying the slope.
I’m sorry for how the graph looks, Connor couldn’t get it emailed. The 7tle says “Photosynthesis Transmitance Based On Yellow and Green Film.” The y-axis says “Spectrophotometer Readings Transmiaance (605nm) and the x-axis says “Seconds.” No film is the blue line, yellow film is the red line and green film is the gray line. 16. Be prepared to report your results to the rest of the class and explain why you think you got those results. Table 2. Transmiaance readings for Hill reac7on experiment. Write your specific treatments into the table. White light Treatment 1: Yellow Film Treatment 2: Green Film Time (sec) Trans (%) Time (sec) Trans (%) Time (sec) Trans (%) 0 19.8 0 19.8 0 20.0 30 20.4 30 21.0 30 20.5 60 21.5 60 21.9 60 21.0 90 22.6 90 22.9 90 21.5 120 23.8 120 23.8 120 22.0 150 25.2 150 25.0 150 22.6 180 26.5 180 26.1 180 23.2 210 27.9 210 27.2 210 23.6 240 29.2 240 28.1 240 24.1 270 30.5 270 29.3 270 24.7 300 32.1 300 30.7 300 25.3 330 33.5 330 31.7 330 25.8 360 34.9 360 32.9 360 26.5 390 36.4 390 34.1 390 27.1 420 37.6 420 34.9 420 27.8
You will need to include this graph and data in your lab handout. You may also need this informaKon to complete an assignment associated with this lab unit. DISCUSSION QUESTIONS 1. One of the pigments in your chromatogram should fluoresce. Which pigment would this be? Why do we see fluorescence with UV light? Chlorophyll B was fluorescent under the UV light. We see fluorescence with UV light because electrons are promoted to an excited state. The pigment reabsorbs the blue-violet range of wavelengths and the electrons fluoresce and red is emiaed. 2. How did the Hill reac7on provide evidence for the light reac7on processes? The Hill reac7on demonstrates that isolated chloroplasts can produce O2 in the absence of CO2. The Hill reac7on provided evidence for the light reac7on processes by having electrons from water being transferred to an electron acceptor and reducing the acceptor. 3. Based on your graph, which treatment was most effec7ve at driving photosynthesis? White light treatment What would account for the differences in your three treatments? How much light each color absorbs Were there any surprising results and can you explain them? White Light: The white light was the most effec7ve at showing the photosynthesis absorbance rate. White light absorbs every single color. This may explain why it was the most effec7ve and had the highest numbers. Treatment 1: Our first treatment was with light with a yellow film over it. It was the second best at showing the photosynthesis absorbance rate. In the absorbance spectrum, yellow absorbs quite a bit of color, however, it doesn’t absorb every color like white does. The yellow film may have also dimmed the light shown on the solu7on. Treatment 2: Our second treatment was with light with a green film over it. It was not as effec7ve as the white light or treatment 1 at showing the photosynthesis absorbance rate. On the absorbance spectrum, it absorbs light, however not as much as yellow or white does. This explains our results. Our hypothesis was incorrect as yellow light did not show the highest rate of photosynthesis. Having no film aka “white light” shows the highest rate of photosynthesis. ACKNOWLEDGEMENTS Labquest Manual, Vernier hap:/ /www2.vernier.com/manuals/labquest2_user_manual.pdf NCSU, Biological Sciences, 2016. BIO 183 Introductory Biology II Laboratory
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