Lab+2_CellularMorphology_Transport copy copy

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BIO 240L Name: Jirolna Joseph _____________________________ Lab 2a: Microscopy and Cell Morphology Cells are the basic unit of life. In multicellular [eukaryotic] organisms, cells are compartmentalized into structures that have distinct roles related to cell metabolism and survival. These structures are called organelles and facilitate the division of labor within each cell. In addition, different cell types within an organism are responsible for distinct functions. For example, cardiac muscle cells are responsible for contractile events as the heart beats while epithelial cells of the skin serve as a protective barrier. Cells with different functions tend to have different shapes. Cell shape allows a given cell to carry out its functions effectively. For example, nerve cells have many projections called axons and dendrites emanating from their cell bodies. These projections allow neurons to make contact and communicate with many other cells (receive information from these cells and respond by relaying the information to other cells in the nervous system). This shape is distinct from any other cell type. Cells with similar functions are grouped together to form tissues which ultimately organize themselves into organs and organ systems that all work together to produce a functional living organism. 1

BIO 240L Student Name Jirolna Joseph ____________________ Microscopy and Cell Morphology During this lab session you will view different types of cells under the microscope.  If you are unfamiliar with using a microscope, please ask your instructor for assistance. Total magnification is calculated by multiplying the magnification of the ocular lens (always 10X) times the magnification of the objective lenses. Please ask your instructor for help if you are not familiar with these parts of the microscope. 1. What is the total magnification of a sample when the 4X objective is used? ____40x________ (0.25 point) View the following prepared slides of human cells using the 4X, 10X, and 40X objectives only:  Epithelial (cheek)  Connective (blood and bone)  Muscle (skeletal)  Nervous (neurons) View the cells under the microscope and sketch each cell type in the spaces provided below ( at the 400X magnification only ). Include and label any organelles you see: (0.45 point ea) Cheek Skeletal Muscle Blood 2

BIO 240L Bone Neuron Note how each cell and respective tissue type differs from the others in terms of morphology. Their functions are aided by their structures. Cell Morphology Summary Characteristics Cell Type Shape of cell? Yes or No Processes extending from cell? (0.25 point) Yes or No Nucleus present? (0.25 point) Shape of nucleus? Yes or No Multinucleate cell? (0.25 point) Cheek Splatter No Yes N/A No Skeletal Muscle Striated lines like ribs No Yes N/A No Bone Long strand like a fuzzy piece of hair Yes Yes N/A No Neuron Starlike Yes Yes N/A No Red blood cell Circle dots No No N/A No White blood cell Circle dots No Yes N/A No Draw a representative eukaryotic cell and label its major organelles . (1 point) 3

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BIO 240L Lab 2b: Cellular Transport Part I: Tonicity and Membrane Transport Introduction The volume of a cell changes due to osmosis , which is the movement of water through a semi-permeable membrane. How and why does this happen? Molecules collide against each other due to their Brownian movement resulting in the eventual movement of molecules away from each other until all the molecules in the medium (solution, air, water, etc.) are equally spaced from each other. This means that molecules diffuse through a medium until they reach equilibrium. Molecules move from HIGH concentration to LOW concentration of molecules. Semi-permeable membranes such as the plasma membrane of cells allow only certain molecules to go through the membrane. Semi-permeable membranes can select the molecules that go through by the size, bulk, shape, or charge of the particle. This movement across the cell membrane influences cell shape, function and survival. The term “tonicity” is used when comparing the relative solute concentrations of two solutions. If solutions have similar concentrations of solute they are said to be isotonic with respect to each other. If on the other hand a solution has a greater solute concentration compared to another solution, the solution with the higher solute concentration is said to be hypertonic whereas the solution with the lower solute concentration is said to be hypotonic . Note: If there is a low level of solutes (low concentration), relatively speaking, there is a high amount of water, and v.v. Therefore, when the solutes themselves are restricted from diffusing due to size, as is the case when there is a semi-permeable membrane, the water molecules move across the membrane instead, from a solution in which they are abundant (i.e., low solute levels), to a solution in which they are less abundant (higher concentration of solutes). The net result is the same as if the solutes had been allowed to do the moving – equilibrium is achieved. Tonicity is an important factor in biological systems. As such, the cell membrane which is a semi-permeable structure plays a key role in regulating the tonicities of our intra- and extra- cellular fluids. In order for the body to function, the proper balance of solvent and solutes needs to be maintained both inside and outside the cell. This is controlled by the movements of these molecules across the cell membrane. Passive transport (or diffusion) refers to movement that is energetically favorable and therefore does not require energy input (takes place spontaneously). Active transport refers to movement across the cell membrane that requires the input of energy. The purpose of part I of today’s lab is to investigate the effects of osmotic changes (differences in tonicity) on cell volume and shape. My hypothesis about what will happen to the “cells” under the different osmotic conditions is as follows: (be sure to provide a rationale for this hypothesis; 0.20 pt) 4

BIO 240L ____I think the egg in the isotonic solution will stay normal , the one in the hypertonic solution will look shriveled up and smaller and the one in the hypotonic solution will grow bigger in size. __________________________________________________________________________ __________________________________________________________________________ __________________________________________________________ Materials and Methods: We will use deshelled eggs for this experiment. The eggs will be placed in the following solutions in a blind test: -isotonic -hypotonic -hypertonic Make observations of eggs A, B, and C as you compare their cell shape and size to that of a shelled egg: Egg A is round and plump, a bit bigger _________________________(describe shape and size) compared with the shelled egg. It was most likely placed in the hypotonic___________________ solution. (0.2 point) Egg B is shriveled and deflated_________________________(describe shape and size) compared with the shelled egg. It was most likely placed in the hypertonic ___________________ solution. (0.2 point) Egg C is round but smaller_________________________(describe shape and size) compared with the shelled egg. It was most likely placed in the isotonic___________________ solution. (0.2 point) These results confirm that during osmosis, water moves from the (choose one): isotonic / hypotonic / hypertonic solution into the (choose one): isotonic / hypotonic / hypertonic causing its volume to (choose one): increase / decrease / remain the same . (0.6 points total) Part II: Cell Size and Rates of Transport (Diffusion) Overview Size is a limiting factor in terms of rates of transport, i.e., the larger the structure, the slower the rate of diffusion within it. This is because as a structure (cell) gets larger, its surface area to volume ratio shrinks, resulting in less effective transport within the structure and across its surface. As a consequence, organisms must be subdivided into smaller units (cells) the larger 5

BIO 240L they are. If organisms the size of humans, for example, were made of one single, large cell, the rate of transport would be much too slow to support the metabolic needs. To compensate, humans are subdivided into trillions of cells, each of which can effectively facilitate transport within and across each cell at exponentially higher rates. For living things to survive, they must be able to take in nutrients and eliminate wastes. The cell relies on a structure called the cell membrane to perform these functions. The cell membrane is a bilayer, or two-layered structure, that separates the cell from the outside environment. A cell membrane is semi-permeable, which means that it selectively allows things to enter and leave the cell. Some substances, such as water, carbon dioxide, and oxygen, move easily through the cell membrane. Other substances need assistance moving into the cell, and others cannot pass through the membrane at all. One way that substances can move across the cell membrane is by a process known as diffusion . During diffusion, substances move from an area of high concentration to an area of low concentration to maintain an equal concentration on both sides of the cell membrane. This process does not require energy from the cell and will continue until there is no concentration gradient , or unequal distribution of particles. In this lab, you will create enlarged cell models of various sizes using a substance called agar . This gelatinous material is used in laboratories as a medium to support the growth of various organisms. It can also be used in a non-nutritive form as a thickening agent for cooking. The agar that you will use in this lab has been prepared with phenolphthalein , an acid-base indicator. Phenolphthalein is a colorless substance that turns pink in the presence of a base (pH greater than 8.2), such as sodium hydroxide (NaOH). As you soak your cubes in NaOH, you will be able to see diffusion taking place. The purpose of part II of this lab is to investigate the relationship between cell size and transport rates using different-sized agar blocks as simulated cells. My hypothesis about the effect cell size will have on transport (diffusion) rates is as follows: (0.2 point) The bigger the cell size the smaller it is . __________________________________________________________________________ __________________________________________________________________________ Procedure 1. Place the agar that your teacher has prepared and distributed to you in your cardboard tray. Using the plastic knife, cut the agar into 1-cm, 2-cm, and 4-cm cubes. Make sure to measure each side of the agar carefully (length, width, and height) to ensure that each cube is accurate. Answer number 1 on your student data sheet. 2. Calculate the surface area and volume of each cube and record the data in the table below then calculate the surface area to volume ratio: 6

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BIO 240L (0.6 point total for SA/V ratio) Cell Model Surface Area (length X width = cm 2 ) X 6 Volume (length X width X height = cm 3 ) Surface Area to Volume ratio (SA:V) 1 cm 6 1 6:1 2 cm 24 8 24:8 4 cm 96 64 96:64  List the cubes in order of their surface area to volume ratio, from largest to smallest. 1cm, 2cm, 4cm  What happens to the surface area to volume ratio as a cell increases in size? (0.3 point) The surface area to volume ratio decreases as a cell increases.  Do the surface area and volume of a cell increase at the same rate? Explain your answer. (0.45 point) No, because the larger the volume the smaller the surface area.  Do you think molecules will more easily pass in and out of a cell with a large surface area to volume ratio or a small surface area to volume ratio? Explain your answer. (0.4 point) Larger surface area to volume ratio because the molecule is smaller so it will be easier for it to pass. 3. Place the three cubes into the beaker. Try to position them so that they are not touching each other. Pour 100 mL of 0.1-M sodium hydroxide (NaOH) solution into the beaker until the cubes are completely covered with the solution. Record the appropriate start and stop times below. 7

BIO 240L Time cubes placed in NaOH 1:55 Time for diffusion +10 minutes Times cubes removed from NaOH 2:05  What happened to the cubes as you poured the NaOH into the beaker? Why did this change take place? It started turning pink because the ph of the agar was greater or at least 8.2 4. After 10 minutes, remove the cubes from the NaOH solution with the plastic spoon. Place the cubes on your cardboard tray and carefully blot them dry with a paper towel. 5. Watch for any color change in the cubes. Cut each block in half and measure the pale yellow area in the center. Calculate the volume of this yellow region by multiplying length X width X height. Complete the table below: (0.4 point, % of cube that is pink) Cell model Volume of cube (from Table 1) Volume of yellow region of cube (cm 3 ) Volume of pink region of cube (cm 3 ) (vol. cube – vol. yellow) Percentage of cube that is pink ( vol . pink X 100%) vol. cube 1 cm 1 0.008 0.992 99.2% 2 cm 8 3.4 4.6 57.5% 4 cm 64 42.9 21.1 32.9%  Based on your results, which of the cubes above allowed for the highest diffusion rate? (0.25 point) 1cm cube  Are these results consistent with what you predicted? Yes / No (0.15 point) 8

BIO 240L Explain: (0.25 point) ______________________________________________________________________ yes because the inside did get smaller over time ______________________________________________________________________ ______________________________________________________________________  Can you think of one way that you could have increased the rate of diffusion even more (for all cube sizes)? (0.25 point) By pouring in more of the solution and leaving the agar in the solution for a longer time. Lab clean-up: Follow your teacher’s directions for disposing of your agar blocks and the NaOH. Clean your area and return any materials to the proper location. Additional Questions: 1. How many 1-cm cubes would be needed to make a 4-cm cube? Use the data you calculated in Table 1 to explain your answer. (0.25 point) 2. For cellular respiration to occur, oxygen must diffuse into a cell and carbon dioxide must diffuse out. Which cube accurately represents an efficiently respiring cell? Use the data that you have calculated in this activity to support your choice. (0.25 point) 4. Provide at least 2 possible sources of error: (0.4 point) __________________________________________________________________________ __________________________________________________________________________ 5. These are two concepts I learned about membrane transport today: (0.2 point total) 9

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BIO 240L i._________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ ii._________________________________________________________________________ __________________________________________________________________________ __________________________________________________________________________ 10

Lab+2_CellularMorphology_Transport copy copy

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