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Beyond the use of reasoning and consensus, scientific inquiry involves the testing of proposed explanations involving the use of conventional techniques and procedures and usually requiring considerable ingenuity.
Base your answers to next 2 questions on the diagram below of the field of view of a light compound microscope and on your knowledge of microscopes.
1. In order to center the organism in the field of view, the slide should be moved down and to the right down and to the left up and to the right up and to the left
2. The approximate length of the organism is 500 µm 1,600 µm 50 µm1.6 µm
3. After viewing an organism under low power, a student switches to high power. The student should first adjust the mirror center the organism raise the objective and switch to high power close the diaphragm
4. Using one or more complete sentences, explain why a specimen viewed under the high-power objective of a microscope appears darker than when it is viewed under low power. Base your answers to next 2 questions on the diagram below of a compound light microscope. 5. The letter C represents the mirror the diaphragm the eyepiece the high-power objective
6. Select and name one of the labeled parts, and in one or more complete sentences describe its function. 7. The letter “p” as it normally appears in print is placed on the stage of a compound light microscope. Which best represents the image observed when a student looks through the microscope? p q b d
8. To separate the parts of a cell by differences in density, a biologist would probably use a microdissection instrument an ultracentrifuge a compound light microscope an electron microscope
9. The diagram below represents the field of view of a microscope. What is the approximate diameter, in micrometers, of the cell shown in the field? 50 µm 500 µm 1,000 µm 2,000 µm
Base your answer to the question on the diagram below. 10. How many millimeters long is the organism resting on the metric ruler? The next 3 questions are based on the experiment described below. A test tube was filled with a molasses solution, sealed with a membrane, and inverted into a beaker containing 200 mL of distilled water. A second test tube was filled with a starch solution, sealed with a membrane, and inverted into a beaker containing 200 mL of distilled water. After several hours, the water in each beaker was tested for the presence of molasses and starch.
The diagrams show the setup of the experiment.
Answer each question related to the experiment in one or more complete sentences.
What principle was being tested in the experiment?
11. What reagents were used in the experiment to test for the presence of molasses and starch? Draw one conclusion from this experiment. The next 2 questions are based on the experiment described below. An opaque disk was placed on several leaves of a geranium plant. The remaining leaves of the plant were untreated. After the plant had been exposed to sunlight, a leaf on which a disk had been placed was removed and tested as shown in parts B and C of the diagram below. Answer each question related to the experiment in one or more complete sentences.
What conclusion can be drawn from the result of the experiment? What process was being investigated by the experiment? The next 2 questions are based on the experiment described below. A student added 15 mL of water to each of three test tubes, labeled A, B, and C. A 1-cc piece of raw potato was added to tube B. A 1-cc piece of cooked potato was added to tube C. Five drops of hydrogen peroxide (H2O2) were added to each test tube.
The results are shown in the following diagram.
What conclusion can be drawn from the experiment? Which test tube is the control? Explain the reason for your choice. Base your answers to next 3 questions on the diagram of the measuring device shown below.
What is the name of this measuring device? In one complete sentence describe the procedure that you would follow to read the meniscus. What must a student do to obtain a volume of 85 milliliters of liquid in this measuring device? Add 2.0 mL. Remove 2.0 mL. Add 2.5 mL. Remove 8.7 mL.
Answers Explained 2 Specimens viewed under the microscope appear upside-down, backward, and reversed. WRONG CHOICES EXPLAINED: (1), (3), (4) With any of these choices, the specimen would be moved out of the field of view.
1 The field of view is given as 1.6 mm. 1 mm = 1000 µm. 1.6 mm × 1000 µm = 1600 µm. The diagram shows that three specimens would fit across the field of view. One-third of 1600 µm = 533 µm. Of the choices given, 500 µm (choice 1) is closest to this value. WRONG CHOICES EXPLAINED: (2), (3), (4) Each of these choices is mathematically incorrect.
2 The student should first center the organism. The field of view is smaller under high power; therefore, less of a specimen can be seen. If the organism is not centered, it may fall out of the field of view under high power. WRONG CHOICES EXPLAINED: (1) The mirror is adjusted for maximum light under low power. Because the diameter of the high-power objective is very small, it is impossible to adjust the light under high power. (3) A compound light microscope is parfocal; that is, it is not necessary to lift the high-power objective to focus under high power. The specimen remains in focus when switching from low power to high power. (4) Closing the diaphragm reduces the amount of light entering the objective. Therefore, the specimen would appear very dark and would be difficult to see. The diameter of the high-power objective is smaller than the diameter of the low-power objective. Less light enters through the high-power objective, and therefore the specimen appears darker.
2 The letter C represents the diaphragm. WRONG CHOICES EXPLAINED: (1) The mirror is represented by D. (3), (4) The eyepiece and the high-power objective are not labeled on the diagram. Coarse adjustment (A)—used to focus a specimen under the low-power objective. or Low-power objective (B)—along with the standard eyepiece, magnifies a specimen 100×. or Diaphragm (C)—regulates the amount of light entering the objectives. or Mirror (D)—provides a source of light that illuminates the specimen.
4 The image of a specimen as seen under a microscope is upside-down (d). The right side is on the left side, and the top is on the bottom. WRONG CHOICES EXPLAINED: (1) In this choice (p) there is no change in the way the image of the letter appears. (2) In this choice (q) the image of the letter is reversed in only one direction: The right and left sides are reversed. (3) In this choice (b) the image of the letter is reversed in only one direction: The top and bottom are reversed.
2 The ultracentrifuge is a machine that spins at a very high speed. A test tube of a liquid containing the parts of ruptured cells is placed in the machine. Each cell part has its own density (mass per unit volume). When the machine rotates, the cell parts fall to different levels in the test tube depending on their density. WRONG CHOICES EXPLAINED: (1) A microdissection instrument enables a biologist to remove a cell part from a single living cell. A micromanipulator is an example of such an instrument. (3) A cell is transparent under a light microscope. Its structures cannot be seen unless the cell is stained. A compound light microscope can be used to view, but not to separate, cell parts. (4) An electron microscope uses beams of electrons to view freeze-dried specimens; it cannot be used to separate cell parts for study.
2 Study the information given in the diagram. Notice that the diameter of the circle is 2 mm. Since 1 mm is equal to 1,000 µm, 2 mm are equal to 2,000 µm. In relation to the entire circle, how large is the cell? Is the cell one-half as large or one-fourth as large? Dividing the circle into four parts shows us that the diameter of the cell is about one-quarter the diameter of the circle. Dividing 4 into 2,000 results in 500 µm.
WRONG CHOICES EXPLAINED: (1) 50 µm is too small. The cell is ten times larger than 50. (3) 1,000 µm is too large. The cell is not one-half the diameter. (4) 2,000 µm is the diameter of the circle. The cell is only one-fourth as large. The organism is 26 millimeters long. The principle of diffusion was being tested in the experiment. Benedict’s solution was used to test for the presence of molasses in the beaker. Iodine was used to test for the presence of starch in the beaker. Molasses can diffuse through a membrane. or Starch cannot diffuse through a membrane. No starch was produced in the area covered by the disk. The process of photosynthesis was being investigated. Raw potato contains an enzyme that breaks down hydrogen peroxide. or Cooking a potato destroys the enzyme that breaks down hydrogen peroxide. Test tube A is the control. A control is the part of the experiment that provides the basis of comparison for the variable being tested. The device is known as a graduated cylinder. The meniscus should be read at eye level.
2 To obtain a volume of 85 mL, 2.0 mL must be removed. The graduated cylinder contains 87 mL of liquid.
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