Experiment-31-34 Enzymes – Integrated Science : Book Guide

This Document Contains Experiments 31 to 34 Name____________________________________________________Section________________Date___________ Experiment 31: Enzymes Invitation to Inquiry Enzymes are always a part of all living things; even your foods. For this investigation you will need a package of gelatin dessert, 4 clear glass containers, a carrot, some fresh pineapple, and minimarshmallows. Mix the gelatin according to directions and pour into 4 containers. 1. To the first container do not add anything. It will be your control. 2. Into the second container mix a quantity of shredded carrot. 3. Into the third container mix a quantity of crushed fresh pineapple. 4. Into the fourth container mix a quantity of mini-marshmallows. 5. Place all four containers in the refrigerator for the time specified on the gelatin dessert package. 6. When the proper time has passed, remove the four containers and examine their contents. 7. Describe what you see. 8. Write an explanation of what you think might be going on here. Keep in mind that all living things carry out their metabolic reactions using enzymes. These enzymes are released when cells are damaged or destroyed, such as in cutting. Also recall that gelatin is a protein that is converted from a sol to gel state when it is cooled. Background Every living organism carries out a large number of chemical reactions. It is essential for the life of the organism that these reactions occur at an extremely rapid rate and at a safe temperature. All organisms contain enzymes, which are protein molecules that speed up the rate of chemical reactions without increasing the temperature. All chemical reactions require an initial input of energy to get them started. This is called activation energy. Enzymes do not start a reaction; they merely speed up the reaction already in progress by reducing the need for large amounts of activation energy. Life on Earth would not be possible without this increased rate of reaction. Each reaction in a cell requires a specific enzyme to allow the reaction to proceed at the proper rate. Because there are hundreds of different reactions necessary in the life of the cell, hundreds of different enzymes are present in the cell. For an enzyme to work in a reaction, it must fit with its substrate (the molecule that will be altered). Each type of enzyme has a specific physical shape that fits the physical shape of its substrate. When an enzyme reacts with a substrate, the two molecules physically combine to form an enzyme-substrate complex. The substrate is changed into the new end product, but the enzyme is not changed by the reaction. The number of times one molecule of enzyme can react with a substrate in a period of time is known as the turnover number (example: 500,000 per second). This means that one enzyme molecule reacts with 500,000 substrate molecules in a second. If a molecule has a shape that is almost the same as that of the normal substrate, this nonsubstrate molecule might bind to the enzyme. When the enzyme is attached to this molecule, the enzyme is not free to combine with the normal substrate molecule. The nonsubstrate molecule attached to the enzyme is called a competitive inhibitor. An inhibitor slows down the normal turnover number of an enzyme because it does not allow the normal substrate to have access to the enzyme. A general equation for an enzymatic reaction is shown in Figure 31.1. The specific reaction for today’s exercise is: + Oxygen + Enzyme+ Water + Enzyme + O2 + Tyrosinase + H2O + Tyrosinase + O2 + Tyrosinase + H2O + Tyrosinase Figure 31.1 In this exercise you are asked to determine how various factors influence the turnover number of an enzyme. The substrate is a clear, colorless molecule known as pyrocatechol. The enzyme is one naturally found in potatoes called tyrosinase. (You may use ground-up potatoes as an inexpensive source of the enzyme.) A reaction between this enzyme and the substrate results in the formation of an end product having a yellowish-brown color. The appearance of this color indicates that a reaction between the enzyme and the substrate has occurred. The degree of color change corresponds to the amount of end product produced. When performing this experiment, use the following scale to rank the degree of color change: Note: Potatoes are an inexpensive source of the enzyme tyrosinase. The enzyme is obtained by grinding a potato in water and straining the blended potato through a strainer. If you are using potatoes as a source of enzyme you will have best results if you do the following: 1. Grind up the potato just before you need it. 2. Use a chilled potato from a refrigerator, and ice cold water to blend the potato. 3. After filtering the ground-up potato, store the liquid containing the enzyme in a small container in a beaker of ice. If you use potato juice as a source of enzyme you may see a pink color develop after a few minutes. This color change is not related to the reaction we are using in this exercise. Ignore any pink color that develops and record only the intensity of the yellow color that develops. Your instructor may use a pure solution of the enzyme tyrosinase. If this is used, no pink color appears. During this lab exercise, you will observe the 1. normal reaction of the enzyme, value of a control in an experiment, and enzyme specificity. 2. influence of temperature on the turnover number. 3. influence of concentration of the enzyme on the turnover number. 4. influence of concentration of the substrate on the turnover number. 5. influence of pH on the turnover number. 6. influence of inhibitors on the turnover number. When everyone has completed this exercise, your instructor may wish to discuss the results obtained by the class. Procedure Throughout this exercise it is important to clearly label all test tubes so that you can identify them. Control This portion of the exercise demonstrates the normal reaction that occurs between the enzyme (tyrosinase) and the substrate (pyrocatechol) used in today’s exercise. You can also determine the value of a control and whether an enzyme is substrate specific. Note: All glassware must be clean for this experiment to work properly. 1. Put 10 mL of distilled water into a test tube and add 10 drops of enzyme. To this mixture add 10 drops of substrate. Mix the contents of the tube by holding the tube in one hand near the top and gently tapping the base of the tube with your other hand. Do not cover the opening of the test tube with your thumb! 2. Put 10 mLof distilled water into a second test tube. Add 10 drops of substrate and mix. This tube contains no enzyme. 3. Put 10 mL of distilled water into a third test tube. Add 10 drops of enzyme and 10 drops of sucrose (the wrong substrate). Mix this tube. 4. Observe each tube after 5 minutes and note any changes in color. Record the color in Table 31.1. The color change in the first test tube indicates what normally occurs in this particular enzymesubstrate reaction. A control, a basis of comparison for a reaction, is an essential part of any experiment. This first test tube serves as a control during the remainder of this exercise. You can compare experimental tubes with this one to determine whether a reaction has occurred. a. What is the purpose of the second tube containing only water and substrate? b. What can you learn from the third tube containing water, enzyme, and sucrose? Table 31.1 Enzyyme Reactions Tube Contents of tube Original color Color after 5 minutes Rank a Water, 10 drops of enzyme, 10 drops of pyrocatechol Clear Yellow +4 b Water, 10 drops of pyrocatechol Clear Clear 0 c Water, 10 drops of enzyme, 10 drops of sucrose Clear Clear 0 Temperature In this portion of the exercise, you will examine the effect of different temperatures on the activity of the enzyme. You need to allow the water in the test tubes to come to a specific temperature before you add the substrate and enzyme. After these are mixed with the water, they need to stay at that specific temperature for 5 minutes to see if the temperature influences the enzyme’s activity. 1. Take five test tubes and label them a, b, c, d, and e. Fill each with 10 mL of distilled water. Add 10 drops of enzyme to each tube. 2. Place tube a in an ice water mixture in a water bath for 5 minutes to allow it to cool. Similarly, place tubes b through e in water baths of 20°, 40°, 60° and 100°C, respectively. These tubes are to remain in their respective water baths until they reach the designated temperature. 3. After allowing the tubes to come to their appropriate temperature, add 10 drops of substrate to each tube. Mix the contents and immediately return the tubes to their appropriate water baths for an additional 5 minutes. 4. After 5 minutes, remove the test tubes from the water baths and observe the color of the tubes. Record the color intensity of each tube in Table 31.2 and rank them from darkest to lightest. Graph your results on the grid provided. Concentration of Enzyme In this experiment you will examine the effect of altering the number of enzyme molecules on the number of product molecules produced. 1. Take three test tubes and label them a, b, and c. Fill each with 10 mL of distilled water. Place them in a test tube rack. 2. Add three drops of enzyme to tube a, nine drops to tube b, and 27 drops of enzyme to tube c. 3. Add 10 drops of substrate to each of these tubes. Mix thoroughly and observe after 5 minutes. 4. Record the color intensity of each tube in Table 31.3, and rank them from darkest to lightest. Graph your results on the grid provided. Effect of Enzyme Concentration on Amount of End Product Concentration of Substrate In this experiment you will examine the effect of altering the number of substrate molecules on the number of product molecules produced. 1. Take three test tubes and label them a, b, and c. Fill each with 10 mL of distilled water. Place them in a test tube rack. 2. Add three drops of substrate to tube a, nine drops of to tube b, and 27 drops of substrate to tube c. 3. Add 10 drops of enzyme to each of these tubes. Mix thoroughly and observe after 5 minutes. 4. Record the color intensity of each tube in Table 31.4, and rank them from darkest to lightest. Table 31.4 Concentration of Substrate Tube Contents of tube Original color Color after 5 minutes Rank a Water, 10 drops of enzyme, and 3 drops of substrate Clear Pale Yellow + b Water, 10 drops of enzyme, and 9 drops of substrate Clear Medium Yellow ++ c Water, 10 drops of enzyme, and 27 drops of substrate Clear Dark Yellow/ Golden Brown +++ Effect of Substrate Concentration on Amount of End Product pH pH is a measure of the number of hydrogen ions (H+) present in a solution. Solutions that have many hydrogen ions are called acids and have a low pH. Solutions with few hydrogen ions are called bases and have a high pH. Solutions with a pH = 7 are called neutral solutions. In this experiment you will assess the effect of different pH values on the effectiveness of enzymes. 1. Take five test tubes and label them a, b, c, d, and e. Fill tube a with 10 mL of water that has been adjusted to a pH of 3. Place the tube in a test tube rack. 2. Fill tube b with 10 mL of water that has been adjusted to a pH of 5. Place the tube in a test tube rack. 3. Fill tube c with 10 mL of water that has been adjusted to a pH of 7. Place the tube in a test tube rack. 4. Fill tube d with 10 mL of water that has been adjusted to a pH of 9. Place the tube in a test tube rack. 5. Fill tube e with 10 mL of water that has been adjusted to a pH of 11. Place the tube in a test tube rack. 6. Add 10 drops of enzyme to each tube. 7. Add 10 drops of substrate to each tube. Mix the tubes thoroughly and observe after 5 minutes. 8. Record the color intensity of each tube in Table 31.5, and rank them from darkest to lightest. Graph your results on the grid provided. Table 31.5 Effect of pH Tube Contents of tube Original color Color after 5 minutes Rank a Water adjusted to pH of 3, 10 drops of enzyme, and 10 drops of substrate Clear Clear 0 b Water adjusted to pH of 5, 10 drops of enzyme, and 10 drops of substrate Clear Clear to Pale yellow 0/+ c Water adjusted to pH of 7, 10 drops of enzyme, and 10 drops of substrate Clear Yellow ++ d Water adjusted to pH of 9, 10 drops of enzyme, and 10 drops of substrate Clear Clear or Pale yellow 0/+ e Water adjusted to pH of 11, 10 drops of enzyme, and 10 drops of substrate Clear Clear 0 Inhibitors Inhibitors interfere with the ability of enzymes to interact with enzymes. In this experiment you will identify one of two molecules as being an inhibitor. (The following steps must be performed in the order presented or results will vary.) 1. Take six test tubes and label them a through f. Fill each tube with 10 mL of distilled water. Place them in a test tube rack. 2. Add one drop of phenylthiourea to tube a,10 drops of phenylthiourea to tube b, and 20 drops of phenylthiourea to tube c. 3. Add one drop of tyrosine to tube d, 10 drops of tyrosine to tube e, and 20 drops of tyrosine to tube f. 4. Add 10 drops of enzyme to each of the six tubes. 5. Add 10 drops of substrate to each of the six tubes. Mix the tubes thoroughly and observe after5 minutes. 6. Record the color intensity of each tube in Table 31.6 and rank them from darkest to lightest. Table 31.6 Inhibitors TubeContents of tubeOriginalColor after Rank color5 minutes Water, 1 drop of phenylthiourea, 10 adrops of enzyme, and 10 drops ofClearClear or 0/+ substratePale yellow Water, 10 drops of phenylthiourea, b10 drops of enzyme, and 10 dropsClearClear0 of substrate Water, 20 drops of phenylthiourea, c10 drops of enzyme, and 10 dropsClearClear0 of substrate d Water, 1 drop of tyrosine, 10 drops of enzyme, and 10 drops of substrate Clear Dark Yellow +++ e Water, 10 drops of tyrosine, 10 drops of enzyme, and 10 drops of substrate Clear Dark Yellow +++ f Water, 20 drops of tyrosine, 10 drops of enzyme, and 10 drops of substrate Clear Dark Yellow +++ Results 1. Which of the two substances (phenylthiourea or tyrosine) was the inhibitor in the exercise you just completed? What was your evidence? No end product was produced (yellow color) in the presence of phenylthiourea, therefore, one can conclude that phenylthiourea is an inhibitor of tyrosinase. 2. If you were to look at the three test tubes involved in the enzyme concentration experiment after 24 hours, how would the colors of each compare? Explain. They should all be the same color since they had the same amount of substrate. Those with less enzyme will simply take longer to use up all the substrate. 3. Both high and low temperatures reduced the amount of color (product) produced. However, the cause of the reduction is different in the two cases. Explain what the differences are. High temperatures disrupt the bonds holding proteins in their threedimensional shape, thereby denaturing the enzyme. Low temperatures slow the rate of molecular motion, therefore there are fewer collisions between enzymes and substrates per unit time. 4. Since an enzyme-substrate complex must be formed for an end product to be produced, how might an inhibitor reduce the effectiveness of an enzyme? Competitive inhibitors are usually molecules with enough chemical similarity to the substrate that they can bind to the active site of the enzyme and block the substrate from interacting with the enzyme. 5. What is turnover number? How did you estimate turnover number in this series of experiments? The turnover number is the number of times an enzyme molecule can combine with copies of substrate molecules per minute under optimum conditions. Turnover number was estimated by comparing the color of the various test tubes, since the product was a colored material. 6. Why does increasing the amount of substrate increase the amount of color produced in the test tubes? Increasing the amount of substrate increases the number of possible product molecules that can be produced. 7. At what pH was the enzyme most effective? A level of 7 was generally most effective, but occasionally 5 depending on other factors. At what other pHs did the enzyme work but not as effectively? Generally there was no reaction at pH 3, some at 5, most at 7, none at 9, and none at 11. Explain why changing the pH alters the effectiveness of the enzyme. Changing the pH changes the shape of the enzyme and substrates can not bind as effectively. At extreme pHs the enzyme may be destroyed. 8. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) Name____________________________________________________Section________________Date___________ Experiment 32: Photosynthesis and Respiration Invitation to Inquiry Try this: 1. Get an aquatic plant from a pond or pet store. 2. Place one of the plants in a clear glass container of tap water. 3. Fill another clear glass container with tap water. 4. Place both containers in a well-lighted area such as under a fluorescent light. 5. Observe both carefully. 6. Count the bubbles on the interior glass surface of both containers. Collect and record your data. Do you see a difference in the number of bubbles in the two containers? Hypothesize what the bubbles might be. Why might there be a difference? Background Photosynthesis is a metabolic process that combines carbon dioxide (CO2) and water (H2O) to form sugar (C6H12O6) and oxygen (O2). The process takes place within the chloroplasts of plants and algae. The chloroplasts contain the green pigment, chlorophyll and the enzymes necessary for photosynthesis. The chlorophyll traps light that serves as the energy source that allows the process to take place. A simplified equation for photosynthesis is: 6 H O2 + 6 CO2 →ChloroplastsLight C H O6 12 6 + 6 O2 Inorganic Raw Materials End Products All organisms require energy to sustain themselves. Nearly all organisms, including plants and animals carry on aerobic respiration in which sugar and oxygen react to form carbon dioxide, water, and a source of energy known as ATP (adenosine triphosphate). Mitochondria are cellular structures that contain the enzymes necessary for the many individual steps of aerobic respiration. A simplified equation for aerobic respiration is: C H6 12O6 + 6 O2 →MitochondriaEnzymes 6 H O2 + 6 CO2 + Energy Raw Materials End Products Look closely at the balanced equations for photosynthesis and respiration and notice that the end products of one reaction are the raw materials for the other. Only organisms containing chlorophyll can perform photosynthesis, whereas respiration can take place in virtually every organism. Many plants perform photosynthesis and respiration simultaneously. The P/R ratio (photosynthesis/respiration ratio) compares the rate of photosynthesis to the rate of respiration. Knowing this ratio can help explain what happens in a plant at different times in its life. For instance, the P/R ratio is different for a corn plant during spring, summer, and fall. It is also different for day and night. Animals, on the other hand, can engage only in respiration. Because animals are incapable of converting inorganic raw materials into organic molecules they must obtain energy-rich organic molecules by eating plants or other animals. They also need oxygen to allow them to release energy from organic molecules. Thus, animals are dependent on plants for the two end products of photosynthesis, namely organic molecules (glucose) and oxygen. Water has many gases dissolved in it, including oxygen and carbon dioxide. Aquatic organisms use these gases when they carry on photosynthesis and aerobic respiration and release gases into the water as well. This exercise allows you to make measurements of the rates of photosynthesis and respiration by measuring the amount of oxygen and carbon dioxide present in the water in which the organisms live. This laboratory activity gives you an opportunity to set up an experiment with proper controls, quantitatively test water samples for oxygen and carbon dioxide content, and collect and analyze data. You may also learn that experimental work sometimes yields results that are difficult to interpret. During this exercise you allow organisms to engage in their normal biochemical processes. Evidence that the organisms have carried on photosynthesis or respiration is revealed by sampling the oxygen and carbon dioxide content of the water in which they live. The tests for measuring the oxygen and carbon dioxide content appear on page 261. Follow these test procedures carefully because you will be measuring very small quantities—parts per million (ppm)—of oxygen and carbon dioxide. If your water sample contains 8 ppm of oxygen, it means that there are eight oxygen molecules dissolved in every 1 million molecules of your sample. During this lab exercise, your group will: 1. Determine dissolved oxygen and dissolved carbon dioxide concentration in aged tap water. (The purpose for doing this is to get baseline data as well as to give you practice with these complex tests.) 2. Set up controls of aged water and three experimental situations: plants in light, plants in darkness, and fish. 3. Determine dissolved oxygen and dissolved carbon dioxide concentrations of the controls and the three experimental situations at the end of an hour. 4. Use the data collected to answer questions. Procedure Initial Trial Fill a large beaker from the container labeled aged water. The aged water is simply tap water that has been sitting open to the atmosphere overnight. Therefore, the aged water has the same concentrations of carbon dioxide and oxygen dissolved throughout the container. In addition, it is equilibrated to room temperature. Since everyone in class will use this aged water to set up their experiments, everyone will start with water that contains the same amount of carbon dioxide and oxygen and has the same temperature. Test this aged water for dissolved O2 and CO2. The directions for the dissolved oxygen test and the dissolved carbon dioxide test are found on pages 260 and 262. Record the results in the proper column of Data Table 32.1. Data Table 32.1 Dissolved Oxygen and Carbon Dioxide Results The oxygen and carbon dioxide levels of the aged water willDissolved O2 (ppm) Dissolved CO2 (ppm) vary somewhat depending on the temperature of the water, but(sodium thiosulfate)number of drops of (number of drops of NaOH) all students in any given lab should have results that are Initial aged water similar. The readings for the controls should be the same as in Control in light the aged stock water. If everything is working properly, the water plant in the light should have a higher oxygen Control in dark concentration and a lower carbon dioxide concentration. The Plant in light water plant in the dark should have a lower oxygen concentration and a higher carbon dioxide concentration. Plant in dark Oxygen should also decrease and carbon dioxide increase in Goldfish the fish water, but in greater amounts. Controls Fill four large test tubes with aged water; cork the tubes in such a way that no air is trapped. Label these tubes Controls and place two of them in a test tube rack marked Light. Place the other two control tubes in a test tube rack in the dark. Your instructor will designate this location. At the end of the hour you will use the water from one of each pair of tubes to test for oxygen content and the water in the other tube to test for carbon dioxide content. Experimental Tubes Plant in Light Fill two other large tubes with aged water. Place several healthy, green sprigs of Elodea or other water plants in each tube. There should be plants from the top to the bottom of the test tube, but the plants should not be jammed together in a clump. Cork the tubes without trapping air. Label these tubes Plant in Light and place them in the test tube rack in front of a fluorescent light source for 1 hour. It is best to use fluorescent lights because incandescent lights tend to heat up the water in the tubes and change the amount of gases that can remain dissolved in the water. See Data Table 32.2 on page 264. Plant in Darkness Fill two more large tubes with aged water and Elodea or other water plant; cork tubes. Label these tubes Plant in Dark and place them in the designated dark area for 1 hour. Dissolved Oxygen Test Follow this procedure to determine the dissolved oxygen content of the initial aged water, control containers, and the three sets of experimental containers. 1. Carefully fill (overflow) the small glass-stoppered bottle with the water to be tested. (Do not put plants or fish into this bottle). (Do not use water that has been used for any other test.) Insert the glass stopper and pour off any excess water trapped on the outside of the bottle around the stopper. 2. Remove the stopper. Cut open chemical packets a and b and carefully empty the contents of both into the bottle. Since the quantity of chemicals in the packets is premeasured, it is important that you not spill any of the contents and that you get all the contents into the bottle. Replace the stopper carefully (do not trap any air bubbles) and shake for 30 seconds or until all granules are dissolved. A brown precipitate will form. (If granules still remain after 30 seconds of shaking, proceed to step 3.) 3. Set the bottle on the lab table. After the brown precipitate has settled out about halfway, empty the contents of the large packet (c) into the bottle. Replace th stopper and shake again until the precipitate has completely dissolved. If you have brown particles in the water, vigorous shaking should get rid of them. You should now have a clear, yellowish liquid. 4. Fill the small measuring test tube with the yellow liquid and pour one full tube of the yellow liquid into a clean glass beaker. It is a good idea to rinse any glassware before you use it to be sure that it is clean. 5. While gently swirling the beaker to mix its contents, add sodium thiosulfate solution drop by drop (count the drops) until the yellow color turns clear, like water. The color change is best seen when the beaker is placed over white paper. The number of drops of sodium thiosulfate used equals the parts per million (ppm) of dissolved oxygen. Your result should be between 1 and 20 ppm. If your result is different from this, call it to your instructor’s attention. Your instructor should be able to help you determine what is wrong. 6. When you are finished with the test, pour the liquids into the large waste chemicals container provided and rinse your glassware. Animal Fill two other large tubes with aged water and place a goldfish in each test tube. Cork the tube without trapping air bubbles. Label these tubes Goldfish and place them with the plant in the light for 1 hour. Analysis of Results Often a visual presentation of data helps one see patterns or trends and makes interpretation easier. Use the data from Data Table 32.1 to construct a bar graph in the space provided below. Use two different colors or shades to distinguish between oxygen and carbon dioxide. Dissolved Carbon Dioxide Test Follow this procedure to determine the dissolved carbon dioxide content of the initial aged water, control containers, and the three sets of experimental containers. 1. Use a graduated cylinder to measure 30 mL of water to be tested. (Do not put plants or fish into the bottle). (Do not use water that has been used for any other test.) Pour the 30 mL of water into a small beaker. 2. Add five drops of phenolphthalein to the water. (Phenolphthalein is an acid/base indicator.) 3. While gently swirling the beaker to mix its contents, add sodium hydroxide solution drop by drop (count the drops) until a light pink color appears and stays pink. The change to a pink color is easiest to detect when the beaker is placed over white paper. The number of drops of sodium hydroxide solution used to get the pink color equals the parts per million (ppm) of dissolved CO2 in the water. Your result will be between 1 and 20 ppm. If your results are different from this, call it to your instructor’s attention. Your instructor should be able to help you determine what is wrong. 4. When you are finished with the test, pour the liquids into the large waste chemicals container provided and rinse your glassware. Results Use the equations for photosynthesis and aerobic respiration to help you think about the questions. Remember reactants decrease in concentration as products increase in concentration. Photosynthesis: 6 H O2 + 6 CO2 →ChloroplastsLight C H O6 12 6 + 6 O2 Aerobic Respiration: C H6 12O6 + 6 O2 →MitochondriaEnzymes 6 H O2 + 6 CO2 + Energy(ATP) 1. How does the P/R ratio change from summer to winter for a plant growing in Canada? P/R ratio will be higher in the summer and lower in the winter. 2. What does it mean if the P/R ratio is 2/1? Which of the tubes (Plant in light, Plant in dark, Goldfish) had a P/R ratio similar to 2/1? The plant is carrying on photosynthesis twice as fast as it is carrying on respiration. The plant in light had a P/R similar to 2/1. 3. Why was there less CO2 in the Water Plant in Light tube at the end of the hour than in the control? The plant was using CO2 in the process of photosynthesis. 4. If you could have measured the number of individual water molecules, would the number of water molecules in the tube have differed at the beginning and at the end of the hour for the Water Plant in Light? Why? Yes, the plant would have been using Hphotosynthesis. 2O in the process of 5. Why was there less oxygen in the Water Plant in Dark tube at the end of the hour?The plant was using oxygen while carrying out respiration. No photosynthesis could take place without light. 6. Why was there less oxygen in the Goldfish tube at the end of the hour? The Goldfish used oxygen in aerobic respiration. 7. How do you think the results would have differed if the Goldfish tube had been placed in the dark? There would be no difference since animals respire in the dark or the light. 8. Compare the processes occurring in the plants in the dark with the processes occurring in the Goldfish in the light or in the dark. Both the plant and the Goldfish are performing respiration in the dark. 9. If a fish had been placed in the tube with the plants in the light, what results would you expect? Why? The concentrations of oxygen and carbon dioxide should be similar to the controls. The plants would have been giving off the oxygen because photosynthesis was proceeding more rapidly than plant respiration but the fish was using it up in respiration. The fish would be giving off CO2 from respiration, while the plants would use it up to carry out photosynthesis. 10. If a fish had been placed in the tube with the plants in the dark, what results would you expect? Why? A lower concentration of oxygen and a higher concentration of carbon dioxide could be expected. Both the plants and the fish would be carrying on respiration. 11. If you have $10,000 and you loaned your friend a nickel, how many parts per million of your money have you given to your friend? 5 ppm Name____________________________________________________Section________________Date___________ Experiment 33: The Chemistry and Ecology of Yogurt Production Invitation to Inquiry Fermentation can result in either souring or spoiling. These are two different things! We eat many soured foods such as yogurt, cheese, and pickles. They do not cause illness. However, foods that have spoiled can be very harmful. One method of preventing spoilage is to add salt (NaCl) to the food. 1. Get two kinds of cottage cheese: one with salt and the other without salt. 2. Open each container and mix the cheese curds with your unwashed hands (preferable dirty hands!). 3. Place these containers in a warm place with the covers on top but not sealed. 4. Note the time. 5. Examine the containers every 6 hours and take notes on the changes that take place including such items as: a. Color b. Texture c. Smell 6. Write a summary of what you have observed, and draw conclusions about what took place. Background Yogurt is the product of the fermentation of milk sugar (lactose) by two different species of bacteria, Lactobacillus bulgaricus and Streptococcus thermophilus. Fermentation is a metabolic process in which organic molecules are broken down to simpler compounds by organisms without the use of oxygen. It is an example of anaerobic respiration. Organisms that carry on fermentation obtain energy in the form of ATP molecules. In the process of either aerobic or anaerobic respiration, each step in the process is controlled by an enzyme that converts a substrate, the material acted upon by the enzyme, into an end product. The end product of the first reaction becomes the substrate for the next reaction in the series. The enzymes in the bacteria we are using today convert lactose into lactic acid. Lactic acid and ATP are the final end products of this example of fermentation. A simplified version for fermentation is Glucose  →Enzymes 2 lactic acid + 2 ATP Figure 33.1 provides a more detailed description of what happens during fermentation. Because acids have a sour taste, the result of the activity of these bacteria is a sour-tasting dairy product. The production of the acid is responsible for the change in the consistency of the milk. The greatly lowered pH causes the milk protein to coagulate and become thick and viscous. Streptococcus thermophilus also produces some other compounds that are important in determining the final flavor of the yogurt. Figure 33.1 To provide a consistently palatable product, commercial producers of yogurt must have a thorough understanding of the biology of the bacteria they use. They must be aware that changes in the environment alter the activities of the bacteria. Fresh milk normally contains a mixture of different species of bacteria that eventually ferment, or sour, the milk. Because these bacteria produce unwanted flavors and compete with the desired bacteria, it is desirable to reduce the natural bacterial population as much as possible. This is usually done by heating the milk for a specified period of time. Substances that retard the growth of the desired species of bacteria—inhibitors—are another problem for the yogurt producer. These inhibitors are of two varieties. One type is produced naturally by the cow (it makes sense that cows would have evolved a method of inhibiting fermentation while the milk was still in the cow; otherwise, cows might give yogurt). The second kind of inhibitor is a much greater problem and results from the treatment of cows with antibiotics. If cows are being treated with penicillin or some other antibiotic that slows down the growth of disease-causing bacteria, this antibiotic can get into the milk and prevent the desired bacteria from doing their job of making yogurt. With one carton of milk per student, you will: 1. prepare and inoculate the container of milk with a bacterial culture that will produce yogurt. 2. incubate the mixture at the proper temperature. Procedure 1. Empty a one-half-pint container of milk into a clean beaker and place on a hot plate. 2. Add to the milk 11 g of powdered milk. Stir and heat to 96°C. Do not allow the milk to boil. Stir constantly to avoid burning. As you approach 96°C stir constantly or the milk will boil over. 3. As soon as the milk reaches 96°C use hot pads to remove it from the hot plate. Allow the milk to cool to 46°C. Stirring occasionally will reduce the time needed to cool the milk. 4. Add a teaspoon of starter yogurt culture to your empty milk carton from the bulk culture provided. 5. Pour the cooled milk into the carton; close and staple shut. Label the carton with your name and the date. 6. Place your culture in the designated incubator (39°C) until it coagulates (about 6–8 hours). A high temperature of incubation results in a more sour yogurt; a lower temperature of incubation results in a more viscous product. The 39°C temperature is a compromise between these two temperatures. Your product should be sour and viscous like yogurt. 7. Remove the carton from the incubator following coagulation, and cool to 10°C in a refrigerator. 8. When cooled (any convenient time within a week) note the odor and flavor. Compare this to a commercially prepared carton of yogurt. Flavored Yogurts (Optional) You may wish to add some flavoring to the yogurt. The flavorings are added after the incubation period to prevent their change in taste by the bacteria. Here are some suggestions for flavoring your yogurt: cucumbers and black pepper fruit preserves dry fruit-flavored gelatins nuts honey granola chocolate syrup cereals fresh or frozen fruit Yogurt Cheese Yogurt and cheese are both fermented milk products; however, cheeses have the curd (the solid portion) separated from the whey (the liquid portion). The curd of yogurt is also softer and more tart than most cheeses. To convert yogurt into a cheese, you merely need to allow the whey to separate from the curd so that the curd can form into a more firm product. This is easily done by placing the yogurt into a bag of closely knit cheesecloth and suspending the bag over a container to collect the whey. If you suspend the yogurt and allow the whey to drain away for 24 hours, the cheese that is formed has the consistency of cream cheese. If it is allowed to drain for a longer period, the curd becomes more dense and can be sliced. The whey can be saved. It is sometimes used in place of milk as a beverage or on cereals or as an additive in other recipes. Results 1. Why was the milk heated prior to adding the starter culture? The milk was heated to kill any undesirable bacteria already present in the milk. 2. What active ingredient(s) can be found in the yogurt starter culture? Bacteria, namely Streptococcus thermophilus and Lactobacillus bulgaricus, can be found. 3. What would happen if you added the yogurt culture to the heated milk before it had been allowed to cool? Adding the yogurt-producing organisms to the hot milk would kill them. Fermentation would not occur, therefore, no yogurt would be produced. 4. What things might act as inhibitors in the yogurt-making process? Anything that interferes with the ability of the bacteria to metabolize the milk sugars will inhibit yogurt production. Extraneous bacteria that are more efficient metabolizers may cause different products that may interfere with the taste of the yogurt. 5. Now that you have made yogurt, what do you think is different about the process of making Swiss cheese, cheddar cheese, and yogurt cheese? (See directions for making yogurt cheese on page 267.) The organisms involved and the time of curing influence the type of cheese produced. 6. What would happen if you added the yogurt culture to the cooled milk and immediately refrigerated it? The bacteria would not metabolize as rapidly as they would at incubation temperatures, therefore, not enough fermentation would take place to produce yogurt. In addition, other, more cold-tolerant species of bacteria might use the milk as a food source. 7. Which of the following terms apply to the process of making yogurt? (circle them) Name____________________________________________________Section________________Date___________ Experiment 34: DNA and RNA: Structure and Function Invitation to Inquiry All mammals produce milk as the primary food for their young. Milk contains many important nutrients including: water, fat, protein, and the sugar, lactose. Normally, after weaning, mammals close down the enzyme systems responsible for the digestion of lactose in milk. Various kinds of research suggest that our human ancestors were similar to other mammals and lost their ability to digest milk as they grew older and stopped drinking breast milk. If adult mammals drink milk they are unable to digest the lactose in milk and bacteria in their intestine ferment the lactose resulting in the production of gases that produce symptoms such as distention of the belly, abdominal cramps, diarrhea, and foul-smelling stools. This condition is known as lactose intolerance. However, in many cultures the raising of cattle allows for a source of milk that can be consumed as food by adults and approximately 50 percent of the human population is currently lactose tolerant. This means that they can digest the lactose in milk without experiencing the symptoms of lactose intolerance. Recent research has indicated that lactose tolerance came about as a result of a mutation in the DNA of certain people in Eastern Europe. Diagram a scheme of events which would have started with such a mutation and resulted in the spread of this mutation throughout Europe and the world. Keep in mind that the lactose tolerance trait is primarily centered in the Caucasian population. Background It is to your advantage when doing this exercise to keep your textbook open to the chapter that describes nucleic acids and processes of replication, transcription, and translation. Refer to the information and diagrams in your text as needed. Both of the nucleic acids, deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), are constructed of smaller subunits called nucleotides. The different kinds of nucleotides synthesized in the cell differ from one another in the kinds of sugars (deoxyribose or ribose) and the kind of nitrogen base (adenine, guanine, cytosine, thymine, or uracil) they contain. Cells maintain a supply or pool of these nucleotides for use in the processes of DNA replication and RNA transcription. DNA replication is the series of chemical reactions that result in the formation of two identical double-stranded DNA molecules from one original molecule. This is an assembly process that uses the two strands of the existing DNA molecules as templates upon which new DNA nucleotides are aligned. Transcription involves the formation of a copy of the DNA code in the form of RNA molecules. The actual process of manufacturing the RNA is similar to replication, except that RNA nucleotides are matched to only one side of the DNA double helix. RNA molecules are single-stranded. There are three types of RNA produced; messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). Each of these RNA molecules has a special role to play in the functioning of a cell. The mRNA carries the message from the DNA in the nucleus to the cytoplasm of the cell. The tRNA carries amino acids used for the manufacture of proteins. The rRNA along with some proteins forms ribosomes, structures needed to assemble proteins. Translation involves the cooperation of all the kinds of RNA to bring about the synthesis of specific proteins. During translation the structure of the ribosome promotes complementary base pairing between mRNA and tRNA, leading to the formation of peptide bonds between specific amino acids. Strings of amino acids are known as polypeptides. One or more polypeptides are combined to form essential proteins such as enzymes, antibodies, hemoglobin, collagen found in connective tissue, or myosin found in muscle tissue. This exercise will help you understand the details of replication, transcription, and translation. Figure 34.1 summarizes these central concepts of molecular genetics commonly called the central dogma. Transcription Translation Traits Nucleus Ribosome Figure 34.1 This exercise simulates the processes of DNA replication, RNA transcription, and translation. Remember, real DNA is a three-dimensional double helix. The flat plastic structures used in this exercise cannot be made to exactly duplicate the real molecules involved in these cell processes. Work in pairs and be sure that you discuss with your partner what each stage of the exercise represents. If you are unsure at any time, check with the instructor. During this lab exercise you will: 1. Separate the molecular models into pools. 2. Use the models from the pools to construct DNA nucleotides. 3. Use the DNA nucleotides to construct a model of a single strand of DNA. 4. Use the DNA nucleotides to base-pair with the nucleotides on the single strand, thus forming a double strand of DNA. 5. Replicate the double strand of DNA. 6. Use the models from the pool to construct RNA nucleotides. 7. Use the RNA nucleotides to construct tRNA molecules and attach the appropriate amino acid to these molecules. 8. Transcribe DNA into mRNA. 9. Translate and form a protein model. Procedure (Don’t Take Shortcuts!) Building Double-Stranded DNA 1. Separate the pieces of the molecular model into pools. Each pair of students should get a DNA/RNA model kit. Empty the plastic parts onto the table and separate the various parts into piles (pools). These parts represent the types of molecules that are necessary to construct nucleotides. Table 34.1 lists the parts and the numbers of each that should be in your kit. If you find that you do not have the proper number, or if pieces are broken, check with your instructor. Table 34.1 DNA-RNA Model Kit Part Name Symbol Number needed Number present Deoxyribose sugar D 36 Ribose sugar R 18 Phosphate P 36 Uracil U 5 Thymine T 10 Guanine G 8 Cytosine C 8 Adenine A 10 Hydroxyl group OH 3 Hydrogen atom H 3 Amino acids Leu, His, Gly 1 of each type 2. Construct DNA nucleotides. A DNA nucleotide is composed of a phosphate molecule bonded to a deoxyribose sugar molecule bonded to a nitrogen-containing base such as adenine. Use the models from the pools to construct DNA nucleotides. Adenine A Phosphate Deoxyribose P D a. Assemble 36 separate DNA nucleotides. Each nucleotide should look like one of the diagrams in Figure 34.2. b. Assemble: 10 separate nucleotides using adenine 8 separate nucleotides using guanine 8 separate nucleotides using cytosine 10 separate nucleotides using thymine c. Make sure that all the phosphates are on the left side of the sugar. These 36 nucleotides make up the DNA nucleotide pool of your cell. Figure 34.2 DNA nucleotides. Notice that each nucleotide differs in the type of nitrogen base it contains and not in the sugar or the phosphate. 3. Construct one side of a DNA molecule. Use the DNA nucleotides you have just assembled to construct a model of one side of a DNA helix using precisely the sequence given in Figure 34.3. Attach the nucleotides from the pool, end-to-end, with the deoxyribose to the right and the phosphates to the left. The phosphate of one nucleotide should link with the sugar of the next. This sequence of nucleotides will represent the gene in our simulation. During the process of transcription, use this specific sequence as the template for building the mRNA molecule. Figure 34.3 Single DNA strand 4. Construct a DNA double helix. Use complementary nucleotides to base-pair with the nucleotides on the single strand you just assembled to form a double-stranded DNA molecule. To assemble the second strand properly, follow the rule that the nitrogen base, adenine, always base-pairs with the base, thymine, in DNA and that the nitrogen base, guanine, always base-pairs with the base, cytosine. Slide the appropriate nucleotides from their pools into position so that a ladderlike molecule is formed on the table. Again, link the sugar of one nucleotide to the phosphate of the next nucleotide. The DNA is really a spiral molecule but we will use this ladderlike molecule to represent the double-stranded DNA molecule. Replication: Synthesis of DNA When a cell divides, each of the two new daughter cells must receive the same DNA. To accomplish this, the cell usually has two sets of DNA molecules on chromosomes that may be separated into the two newly forming cells. The process of constructing copies of a double-stranded DNA molecule is called DNA replication. If the genetic information is not replicated and separated equally when cells divide, the new cells will not contain the genetic messages necessary to manufacture the proteins needed by the cell. Replication of DNA is going on within your body at this moment. Whenever a wound is repaired, growth occurs, or old cells are replaced, DNA must first replicate. In other words, before any cell division occurs, the DNA molecules make copies of themselves so that each new cell receives the same genes. The diagrams in Figure 34.4 show a cell in the process of division after the DNA has replicated. Chromatids separate and the identical DNA moves into two Figure 34.4 You are now ready to proceed through the process of DNA replication. The way in which you move the pieces of the model is similar to what actually happens in a cell during the replication process. 1. Separate the two strands of DNA. 2. Use the remaining DNA nucleotides from the pool and match the proper nucleotides from the pool with their partners on the two separated DNA strands (A-T, G-C). When this is completed you should have two double-stranded DNA molecules in front of you on the table. Furthermore, they should be identical to each other and to the original double-stranded molecule. Be sure that you understand the process of replication before you go on to the next part of the exercise—transcription. Transcription: Synthesis of RNA Transcription is the process of synthesizing RNA. Three forms of RNA are produced by transcription: messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). In this exercise, we examine the formation of mRNA only but remember that transfer RNA and ribosomal RNA are formed in much the same way. During transcription, RNA nucleotides base-pair with DNA nucleotides on the gene side of the DNA molecule. To prepare for the production of ribonucleic acids during the process of transcription, we must first synthesize RNA nucleotides. Remember that RNA nucleotides have ribose sugar in place of deoxyribose sugar and the uracil base in place of thymine. 1. Retain one of the two copies of double-stranded DNA you have just made. Break the other DNA molecule down into its smallest subunits (phosphate, sugar, base). Some of these subunits are now needed to assemble RNA nucleotides. (DNA is not disassembled to make RNA in cells, but we need to do this with this model so that we have enough pieces to make the RNA we need.) 2. Construct RNA nucleotides using ribose, phosphate, and the bases adenine, guanine, cytosine, and uracil. You need: 5 separate adenine nucleotides 5 separate uracil nucleotides 4 separate guanine nucleotides 4 separate cytosine nucleotides Be sure that all the phosphates are on the left side of the sugar (Figure 34.5). These 18 nucleotides represent the RNA pool in the cell. 3. Separate the DNA double helix into two strands. 4. Use the RNA nucleotides in the pool and match the proper nucleotide from the RNA nucleotide pool with its partner on the gene side of the DNA only. Be sure that you use the original DNA sequence as found in Figure 34.3 on page 274, as the gene. This side is the genetic code; the other side does not carry genetic information but is important in the replication process. Figure 34.5 5. This newly constructed mRNA molecule should consist of nine nucleotides. The order of the RNA nucleotides in the mRNA you constructed is predetermined by the order of nucleotides along the coding strand (gene) of the DNA molecule. 6. Remove the mRNA molecule from the DNA strand and move the RNA to the side. Now put the two separated strands of DNA back together as they were. You have just simulated the process of transcription. The mRNA molecule that was formed has picked up the code from DNA. The DNA is intact and not damaged. The same DNA code can be used again for the transcription of additional mRNAs if necessary. In fact, it is common for more than one RNA transcript to be produced each time the cell expends energy to unwind and open up the DNA helix. Translation: Synthesis of Proteins During translation, the genetic message, which is coded by the order of bases in the DNA, is translated into the structure of a protein by determining the order of amino acids in the protein. Messenger RNA carries the message from the DNA in the nucleus to the ribosomes in the cytoplasm. Transfer RNA picks up amino acids in the cytoplasm and carries them to particular places on the mRNA called codons. A codon is a linear sequence of three bases on the mRNA. Ribosomal RNA is part of the structure of the ribosome. The mRNA and tRNA come together at the ribosome during the synthesis of proteins from individual amino acids. In an actual cell, the ribosomes are composed of proteins and rRNA. We will not manufacture a model of the ribosome but will use the box the model came in to simulate the ribosome as it moves down the mRNA. For translation to occur in the cytoplasm of the cell, all three forms of RNA are needed. You will use the mRNA just constructed as well as three molecules of tRNA. Remember that tRNA molecules in the cell are formed in the same manner as the mRNA but at different gene sites on the DNA. You will make the tRNA molecules directly from the remaining plastic parts to save time. Also keep in mind that because the plastic pieces are not flexible, these tRNA models are different from the real thing. Our models lack a phosphate at one end and are much shorter than actual tRNA molecules. Actually a transfer RNA molecule is composed of a single strand of about 100 nucleotides that folds back on itself. One end of the tRNA has a coding portion called an anticodon at one end and binds to a specific amino acid at its other end. There are actually 64 possible kinds of tRNA, but we will use only three in this simple simulation. Simulate the process of translation as follows: 1. Assemble the three tRNA models as they appear in Figure 34.6. 2. Attach the appropriate amino acid to each of the three tRNA molecules (Figure 34.7). Makesure that the H and OH units are attached to the amino acids. 3. Place the model’s box on the first set of three nucleotides on the mRNA molecule (CUU) to simulate the presence of a ribosome. Pair the appropriate tRNA to the first three nucleotides (codon) on the mRNA. Since the three bases on the tRNA pair with the three bases (codon) on the mRNA, the bases on the tRNA are often referred to as the anticodon. The process should occur as shown in Figure 34.8. Figure 34.6 Figure 34.7 4. Next, attach the appropriate tRNA to the second mRNA codon by matching the mRNA codon to its complementary tRNA anticodon. Note that the two amino acids are brought together so that the H end of one overlaps the OH of the other. 5. The two amino acids that have been brought adjacent to one another become chemically bonded to each other. Bonds between amino acids are called peptide bonds. To simulate the formation of a peptide bond, remove the H and OH from the adjacent amino acids and bond the two amino acids together. Link the H and OH to form a molecule of water. Because a water molecule is removed as the two amino acids are joined it is often called a dehydration synthesis reaction. 6. Once this bonding has occurred, the ribosome (box) shifts to the next codon on the mRNA and the anticodon of the next tRNA can be aligned with the codon on the mRNA. The first tRNA (GAA) is removed from the site and can bond with another leucine present in the cell cytoplasm. The original leucine molecule stays as part of the growing polypeptide chain. 7. When the third tRNA (CUU) binds to its mRNA codon, a second dehydration synthesis takes place, forming another peptide bond. The completion of this process should result in the synthesis of a short polypeptide composed of the three amino acids; leucine, histidine, and glycine, in that order. You should also be able to identify three separated tRNA molecules ready to function in the transfer of more amino acids, two water molecules from dehydration synthesis reactions, and two peptide bonds holding the amino acids together. Transfer RNA molecules may be used over and over again as amino acid–carrying molecules. A messenger RNA molecule can be used several times to produce the same polypeptide, but then it is broken down into its parts, which also return to the pools. All polypeptides, no matter how long they may be, are constructed in this manner. Each gene is responsible for the synthesis of a particular polypeptide that differs from other polypeptides in the sequence of the amino acids it contains and the length of the sequence. Each piece of DNA that contains information for the building of a particular sequence of amino acids is called a structural gene. Table 34.2 shows the standard flow of genetic information from the DNA gene sequence in the nucleus to mRNA, which travels to the ribosome, where mRNA meets tRNAs, which carry the amino acids. The specific complementary base pairing allows one to work forward or backward along this informational transfer scheme. The genetic dictionary, a list of codons and the specific amino acids they code for, has been determined (Table 34.3). Note that most amino acids have more than one codon and that, in this particular dictionary, both mRNA codons and tRNA anticodons are given. Table 34.3 Genetic Dictionary mRNA tRNA Amino acid codons anticodons Amino acid mRNA codons tRNA anticodons Phenylalanine UUU AAA UUC AAG Tyrosine UAU UAC AUA AUG Leucine UUA AAU Histidine CAU GAU UUG AAC CUU GAA CAC GUG Glutamine CAA GUU CUC GAG CUA GAU CAG GUC Asparagine AAU UUA CUG GAC AAC UUG Isoleucine AUU UAA Lysine AAA UUU AUC UAG AUA UAU AAG UUC Aspartic acid GAU GAC CUA CUG Methionine AUG UAC Valine GUU CAA Glutamic acid GAA CUU GUC CAG GUA CAU GAG CUC Cysteine UGU ACA GUG CAC UGC ACG Serine UCU AGA UCC AGG Tryptophan UGG ACC Arginine CGU GCA UCA AGU CGC GCG UCG AGC CGA GCU AGU UCA CGG GCC AGC UCG AGA AGG UCU UCC Proline CCU GGA CCC GGG Glycine GGU CCA CCA GGU GGC CCG CCG GGC GGA GGG CCU CCC Threonine ACU UGA ACC UGG Terminator UAA ACA UGU UAG ACG UGC UGA Alanine GCU CGA GCC CGG GCA CGU GCG CGC Initiator AUG Results Directions: Use the complete codon dictionary (table 34.3) to answer these questions. 1. Use the base sequence for mRNA to complete the columns on the following table (Table 34.4). (Use the mRNA sequence shown in the table. Do not use the mRNA base pair sequence from the exercise you just completed.) Remember that complementary base pairing is the key. Refer to Table 34.2 if you have problems. Table 34.4 Complementary Base Paring DNA mRNA tRNA Amino acid __________ __________ U __________________ __________ __________ U __________________ __________ __________ U __________________ __________ __________ __________ A __________________ __________ __________ U __________________ __________ __________ C __________________ __________ __________ __________ U __________________ __________ __________ G __________________ __________ __________ U __________________ __________ complement gene 2. Use one word to describe the relationship between the gene sequence and the mRNA sequence in Table 34.2. complementary 3. Describe in a sentence the relationship between the gene sequence and the tRNA sequence in Table 34.4. The gene sequence and the t-RNA sequence are identical except that U replaces T in RNA. 4. What will the sequence of mRNA nucleotides be if the following represents the bases in a DNA molecule of a structural gene? DNA gene A A T G G T C C A C C G C T G mRNA 5. If a structural gene contains 300 DNA nucleotides, how many amino acids will be used in the protein synthesis process? 100 if you discount initiator and terminator sequences. 6. If a protein has 150 amino acids, how many DNA nucleotides would make up the structural gene? 450 if you discount initiator and terminator sequences. 7. A protein has the following amino acid sequence. Construct a DNA nucleotide sequence ofthe structural gene. Phenylalanine—Glycine—Glycine—Alanine—Proline—Valine—Asparagine—Alanine Answers can vary. Help students realize that most amino acids have more than one codon. 8. Compare your DNA sequence from question 7 to that prepared by the others in the lab. Are there any variations? Did such variations occur in your answer to question 4? What is the reason for the difference? Are there any advantages to these variations? What are they? Yes, No. Complementary base pairing is a precise one-to-one process while most amino acids have more than one possible codon. Yes, mistakes could be made in the genetic code, but are never expressed if the codon containing the mutation still codes for the same amino acid. 9. Fill in Table 34.5 10. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) Solution Manual Experiment for Integrated Science Bill W. Tillery, Eldon D. Enger , Frederick C. Ross 9780073512259