#BIO156 LAB 4: osmosis and diffusion
#Lab osmosis and diffusion
Answer the following questions:
- List four cell structures that are common to both plant and animal cells. (4 points)
- What structures are unique to plant cells? (2 points) c. What structures are unique to animal cells? (2 points)
- List five structures observed in the cell images and provide the function of each structure. (5 points)
- Structure 1 and function b. Structure 2 and function c. Structure 3 and function d. Structure 4 and function e. Structure 5 and function
- William is observing a single-celled organism under a microscope and notices that it has a nucleus and is covered in small, hair-like structures.
- Provide a probable name for this organism (1 point) b. Explain why William came to this conclusion. (2 points)
- Where in the cell are the chloroplasts located? (5 points) 5. In the Spirogyra cells observed on the virtual microscope, about how many circular green chloroplasts were seen in a single cell? (2 points) 6. What were the percent differences between the volumes of the potatoes in the osmosis experiment for each salt solution? (8 points) a. 0% b. 1.75% c. 3.5% d. 7%
- What extraneous variables might have affected how the results came out in the osmosis experiment? Describe three. (6 points)
- In osmosis, which direction does water move with respect to solute concentration? (2 points)
- Answer the following questions: a. Explain what would happen to a freshwater unicellular organism if it were suddenly released into a saltwater environment. Use the terms isotonic,
#Lab 4 Cell Structure, Osmosis, and Diffusion
Introduction: Connecting Your Learning
The basic building block of life is the cell. Each cell contains several structures, some of which are common to both eukaryotic and prokaryotic cells and some that are unique to specific cell types. This lab will discuss cell structures and how materials are moved in and out of the cell. Specifically, the principles of diffusion and osmosis will be demonstrated by performing a scientific investigation that studies the effect of salt concentration on potato cells.
Focusing Your Learning
In 1662, Robert Hooke investigated the properties of cork when he discovered cells. He named them after small rooms in a monastery because they reminded him of them. Years later, in 1837, Schleiden and Schwann were attributed with developing the cell theory. While their original theory was modified, the fundamental ideas be- hind the theory held true. Three general postulates are included in the cell theory: 1) All organisms are composed of cells. 2) The cell is the unit of life. 3) All cells arise from pre-existing cells.
Because a cell is the basic building block of living things, it is important to become familiar with its characteristics. Several structures comprise a cell. Many of these structures are visible with the use of a standard compound microscope. Below are pictures of the idealized plant and animal cells, illustrating the important structures.
The cell membrane encloses all cells and is responsible for separating the internal environment from the extracellular space (the space between cells). Because other structures within the cell are also surrounded by a membrane, the outer membrane is of- ten called the plasma membrane.
The cell membrane is semi-permeable, allowing certain molecules to enter into the cell freely, while others are prohibited from entering the cell. It is composed of phospholipids, which have a head consisting of a phosphate group and a tail of two fatty acid chains. The phosphate group is attracted to water
(hydrophilic) while the fatty acid chains are repulsed by water (hydrophobic). When in water, the properties of the phospholipids cause them to form two layers: The hydrophobic tails face the inside of the double layer (away from the water), and the hydrophilic heads face out (toward the water). Because two layers are formed, the membrane is made up of a phospholipid bilayer, as seen in the image.
The cell wall surrounds the cell membrane in plant cells, bacteria, and some fungi. In plant cells, the cell wall is composed of cellulose. In bacteria, the wall is made mostly of polypeptides (protein) or polysaccharides (carbohydrates). The cell wall provides support and protection and is responsible for giving plant cells their shape.
Another important structure found only in eukaryotic cells is the nucleus. This structure contains genetic information and is the control center of the cell. Protecting the nucleus is a double-membrane called the nuclear envelope, which, like the plasma membrane, is semi-permeable. It is important to note that although prokaryotes lack a nucleus, they still contain genetic information.
Within the nucleus is the nucleolus. This is the site where ribosomes are formed. Ribosomes function to assemble proteins. Many cells have multiple nucleoli, which contain concentrated areas of DNA and RNA.
Flagella (singular is flagellum) is Latin for a whip. Flagella are whip-like projections of- ten found in prokaryotes, eukaryotic single-celled organisms, and some specific cells (like human sperm). These structures extend beyond the cell membrane and cell wall and are used for locomotion (movement). Although flagella are found in both eukaryotes and prokaryotes, the structure of the flagella is different for each cell type.
Cilia (singular is cilium) are structurally similar to eukaryotic flagella but are smaller and more hair-like. Cilia are found in some eukaryotic organisms. Some cilia are used for locomotion, as in the single-celled paramecium. In other organisms, the cilia act as a filter. Sometimes, cilia are used not to move the cell itself, but to move objects through a cell (akin to a conveyer belt).
Vacuoles are specialized organelles that are responsible for storing starch, water, and pigments. They also act as a repository for metabolic wastes. Some plant cells contain a large, central vacuole, which occupies almost the entire cell. Central vacuoles are responsible for providing support, which is based on the amount of water or pressure against the cell wall. If too much water is lost in the central vacuole, a plant will lose its support and appear to droop.
Centrioles are found in all animal cells and some plant cells. These structures, which occur in pairs, are responsible for the cytoskeleton. The cytoskeleton is composed of microtubules, microfilaments, and intermediate filaments. It is with these long structures that the cytoskeleton provides support, maintains the cell shape, and anchors the organelles. The cytoskeleton is also used for moving structures or products.
Within eukaryotes is an endomembrane system. In this system, the endoplasmic reticulum, which consists of a membrane that forms folds and pockets, connects the nuclear envelope, the Golgi apparatus (or Golgi complex), and cell membranes. This system is often called the factory of the cell because each of the individual organelles contributes to the production and delivery of proteins, lipids, and other molecules.
The nucleus contains the blueprints for proteins. These plans are then passed to the rough endoplasmic reticulum (RER). This structure is composed of several folds of a membrane and is covered with ribosomes (these bumps are why it is called rough endoplasmic reticulum). Once the ribosomes receive the plans, the protein is built. Some proteins will move to the Golgi complex. Other proteins will move to the smooth endoplasmic reticulum (it is called smooth because it lacks ribosomes). These proteins instruct the organelle to build other molecules, such as lipids and carbohydrates. Like some proteins from the RER, some of these molecules will move to the Golgi complex.
The Golgi apparatus is the central post office area of the cell. It receives the products of the rough endoplasmic reticulum and smooth endoplasmic reticulum, packages them, and ships them to their intended destination.
Another structure found only in photosynthetic cells is the chloroplast. This specialized structure belongs to a class of membrane-lined sacs called plastids (like the vacuole). The chloroplast contains pigments and is responsible for creating food through photosynthesis.
Eukaryotic cells contain an organelle called the mitochondria, which is the site of energy production. This structure is often referred to as the powerhouse of the cell. Cellular energy is stored in the form of adenosine triphosphate (ATP).
The ability of a cell to absorb water and nutrients is an important aspect of its survival. Diffusion is the movement of solutes (dissolved molecules) in a solution or matrix from an area of high concentration to an area of lower concentration. Molecules move down the concentration gradient: from an area of high concentration to an area of low concentration. The greater the concentration differential, the faster the rate of diffusion. The size, shape, and composition of the solute also affect the ability of a substance to diffuse. These factors become increasingly important when considering the diffusion of substances across the cell membrane. Diffusion, being a passive process, is quite efficient across small distances. However, as distances become longer, the efficiency of diffusion decreases.
Osmosis is the movement of water across a selectively permeable membrane from an area of lower concentration (of solute) to an area of higher concentration (of solute). Remember that everything in the universe is constantly moving toward a state of equilibrium. Living cells contain a small amount of salt. For example, human cells contain 0.85% NaCl. If the solution outside the cell has this same concentration, the solution is said to be isotonic. Because there is no net difference in solutes between the inside and outside of the cell, there is no net movement of water. Higher concentrations of solutes outside of the cell are termed hypertonic, while lower concentrations are termed hypotonic.
An important concept that affects how well a cell can absorb and pass material through the membrane is the surface-to-volume ratio. This formula for calculating this ratio is:
Surface area ÷ Volume
Because cells constantly interact with their external environments to obtain nutrients and remove wastes, it is critical that they maintain a proper surface-to-volume ratio.
As objects of the same shape increase in size, the surface-to-volume ratio decreases.
For example, suppose there are two cubes. Cube 1 is 1 cm x 1 cm x 1 cm, and Cube 2 is 10 cm x 10 cm x 10 cm. To calculate the surface-to-volume ratio, the formula for determining the surface area (SA) of a cube (length x width x number of sides) and the formula for the volume (V) of a cube (length x width x height) must be known. Once the formulas for calculating the surface area and volume of a cube are known, the surface area to volume ratios can be calculated, as seen below.
CUBE 1 CUBE 2
Surface Area: 1cm x 1cm x 6 sides = 6cm2 10cm x 10cm x 6 sides = 600cm2
Volume: 1cm x 1cm x 1cm = 1cm3 10cm x 10cm x 10cm = 1000cm3
SA/V: 6cm2/1cm3 = 6.0 cm2/cm3 600cm2/1000cm3 = .6cm2/cm3
As shown in the calculations above, the ratio for Cube 2 is significantly smaller than the ratio for Cube 1. The same trend holds true for cells. As a cell gets larger, the SA/V ratio decreases, meaning that it is not as efficient in moving material in and out of the cell. In other words, the size of the cell membrane relative to the contents of the cell decreases as the cell size increases.
An illustration of the importance of maintaining a high surface-to-volume ratio can be found in the human digestive system. Cells in the human digestive system contain villi, which are finger-like projections. Because of their shape, they have a large sur- face area for a small volume.
- Cell Structure and Function a. Label the following idealized plant and animal cells.
- Observing Cell Structures Under a Microscope i. Utilize the Virtual Microscope to view several cell structures. When
using the virtual microscope, complete the following steps in the order provided below. Failure to properly perform the steps in the correct order will result in failure to complete subsequent steps.
i.Click to view optional detailed instructions.
- Drag and drop the desired slide onto the microscope.
iii. Click on the stage clip knob on the left of the microscope stage. iv. Adjust the interpupillary distance. First, click on the title interpupillary
distance. Next, place the pointer on the images and adjust them until the two images are observed as one image.
- Adjust the slide position. Place the pointer on the positioner and adjust the slide so there is a clear view of the specimen.
- Adjust the iris diaphragm until a comfortable light is obtained. vii. Adjust the diopter until a clear image is obtained. Use the line on the slide and move it up or down. viii. Adjust the coarse focus. Use the line and move it up or down until a clear image is obtained.
- Adjust the fine focus. Use the line and move it up or down until a clear image is obtained.
- Adjust the magnification by clicking on the objective numbers on the microscope.
- Using the virtual microscope, view Spirogyra. Identify and draw an image of the chloroplasts. xii. Using the virtual microscope, view the slide of a paramecium. Identify and draw an image showing the cilia. xiii. Using the virtual microscope, view the slide of the Euglena. Identify and draw an image of the flagella.
- Demonstration of Osmosis in a Potato
- Learn to use the caliper.
- Take the Vernier caliper out of the lab kit. Examine the scale on the tool and try to measure the length of an object. Look closely at the scale. The metric scale will be used for measurements in this lab.
- Read the scale by measuring exactly 2 cm (20 mm). Next, measure 4.5 cm (45 mm).
- This caliper is accurate enough to measure to the nearest tenth of a millimeter (measured by the small, scored lines in the window). With the caliper in hand, go to this instructional Web site, which describes how to use a Vernier caliper.
Watch the scale move as in actual measurement.
- Set up the experiment
- Get four 8 oz. cups from the lab kit. Place a piece of tape on each cup or glass. Using a pen or marker, label the tape on each cup with one of the following percentages: 0%, 1.75%, 3.5%, and 7%.
- Using the graduated cylinder, measure out 100 mL of distilled water. Pour the water into the fifth, unlabeled cup.
- Measure out 1.5 level teaspoons of salt and add it to the unlabeled cup containing 100 mL of distilled water. Mix completely. This is the 7% salt solution.
- Using the graduated cylinder, measure out 50 mL of this mixture and pour it from the graduated cylinder into the cup labeled 7%.
- Add distilled water up to the 100 mL mark of the graduated cylinder to make the next dilution. Adding 50 mL distilled water to 50 mL of a 7% solution will result in 100 mL of a 3.5% solution.
- Using the graduated cylinder, measure out 50 mL of the 3.5% solution and pour it from the graduated cylinder into the cup labeled 3.5%.
- Add distilled water up to the 100 mL mark of the solution in the graduated cylinder to make the next dilution. Adding 50 mL of distilled water to the 50 mL of the 3.5% solution will result in 100 mL of a 1.75% solution.
- Using the graduated cylinder, measure out 50 mL of the 1.75% solution and pour it from the graduated cylinder into the cup labeled 1.75%.
- Empty the remaining 1.75% solution down the drain of the sink and rinse out the graduated cylinder with tap water.
- Using a sharp steak or kitchen knife, slice eight pieces of potato exactly 10 mm x 10 mm x 40 mm (1 cm x 1 cm x 4 cm). It is critically important that these potato core pieces are cut as precisely as possible; they need to all start out having the same volume. A single-edge razor blade may work better than a knife.
- Determine the volume of the potato cores. The volume is calculated by multiplying the width x height x length. Therefore, each core starts out with a volume of 4,000 cubic millimeters or 4 cubic centimeters. Measure the cores with both the mm ruler and the calipers. Measuring with the calipers to the nearest millimeter will be good enough for this lab. Create a data table like the one below to record the beginning and ending volumes.
Table 1. Potato core measurements
0% salt solution 1.75% salt solution
3.5% salt solution
7% salt solution
Beginning average volume (cu mm)
Ending average volume (cu mm)
- Place two measured cores into each solution overnight, or for at least 8 hours. That time period is not critical to the results; it can be longer.
- Remove the cores from one of the cups and pat them dry with a paper towel. The solution may now be discarded down the drain of a sink.
- Using the caliper, measure the height, width, and length of the cores, and then determine the volume of each core. Average the measurements for the two cores and record in the data table above. The cores can now be discarded.
- Repeat Steps 11 – 14 three more times: one time for each cup. 3. Illustration of the Importance of Surface-to-Volume Ratios
- Calculate the surface-to-volume ratio of the following potato cubes: 1. CUBE 1: Length, width, and height are all 5 mm 2. CUBE 2: Length, width, and height are all 3 mm
- Effect of cell size on diffusion rate 1. With clean hands, cutting board, and knife, cut the skin off of the pota-
- 2. Using the knife, cut two cubes of potato with dimensions of 1 cm x 1 cm x 1 cm. 3. Using the knife, cut two cubes of potato with dimensions 1.5 cm x 1.5 cm x 1.5 cm. 4. Using the knife, cut two cubes of potato with dimensions of 2 cm x 2 cm x 2 cm. 5. Place distilled water into a cup or glass. Add the vial of food coloring to the water until a dark color is achieved.
- Carefully place the potato cubes in the solution. The cubes must be completely submerged in the water. Let them stand in the solution for 2 to 4 hours.
- After 2 to 4 hours, remove the cubes. Using the knife, cut each cube in half.
- Using the ruler, measure how far the solution has diffused into each potato cube. Record the results. A sample
- data table is included below that may be used to organize and record the results.
- Complete the following calculations to determine the rate of diffusion and record the results.
Rate of Diffusion (cm/min)= Distance of diffusion ÷ time.
Cube Distance of Diffusion Rate of Diffusion
1 cm cubed
1 cm cubed
1.5 cm cubed
1.5 cm cubed
2 cm cubed
2 cm cubed
Average Rate of Diff.
She notices that the grocery store constantly sprays its produce with distilled water. After returning home, she weighs the carrots again and discovers that they weigh only 4.2 lbs. They also no longer seem as crisp and taut. Provide a possible explanation for why the carrots weighed more at the store, based on the information learned in this lab. (5 points)
- People always say that leeches can be removed from the body by pouring salt on them. Based on what the student learned about osmosis, provide an explanation that supports or refutes this information. (5 points)
- What is the rate of diffusion for the potato cubes from the surface-to-volume experiment (procedure 3b)? (6 points)
- Cube 1 b. Cube 2 c. Cube 3
- Assume the potato cubes are cells. Which cube would be most efficient at moving materials into and out of the cube? Briefly explain the answer. (4 points)
- From what was observed in the potato procedure, how do the rate of diffusion and surface-to-volume ratio limit cell size? (5 points)
- One night, Hans decides to cook a hamburger and spaghetti with meatballs. To test ideas of surface-to-volume ratios, he makes a quarter-pound hamburger and a quarter-pound meatball and cooks them at the same temperature. Which food item will cook the fastest and why? (5 points)
- While watching a special on animals, Brianna discovers that hares tend to lose heat through their ears. Based on this and what is known about surface-to-volume ratios, propose an explanation as to why hares that live in hot climates (such as the desert) have large, extended ears. (5 points)
- (Application) How might the information gained from this lab pertaining to cell structures and diffusion be useful to a student employed in a healthcare-related profession? (20 points)