Bio Lab 5
The following figure shows an open container where solute has been added to the water, bringing its solute potential to -3.4 bars. What effect does adding solute to distilled water have on its solute potential? On its water potential?
It lowers the solute potential and lowers the water potential.
Describe the net movement of water when a dialysis bag containing a 0.2 M sucrose solution is placed into distilled water, which contains no solutes.
water will move into the bag
If the mass of each dialysis bag shown in the experimental set-up below is measured before each is immersed in its respective solution, and the mass is measured again 45 minutes later, which bag would you predict will show the greatest increase in mass?
0.2 - 0.1m
Based on these data, what is the molarity of the potato cores? Enter your answer to two decimal places.
0.32M
Which beaker(s) contain(s) a solution that is hypertonic to the bag?
0.4, 0.6, 1.0
Beet cores from two different sources are dropped into distilled water. The water potential for each core is slightly different. Using what you know of water potential, indicate which of the following statements is true.
The beet core in beaker A is at equilibrium with the surrounding water.
Using the values provided, arrange the beakers in the order of the mass inside the dialysis bags after the experiment has run for 45 minutes, beginning with the bag that loses the most mass first. Rank the bags according to their mass after 45 minutes.
#1: Bag: 0.2M, beaker: 1.0M. #2: Bag: 0.2M, beaker: 0.6M. #3: Bag: 0.2M, beaker: 0.4M. #4: Bag: 0.2M, beaker: 0.2M. #5: Bag: 1.0M, beaker: 0.2M
Using varying molarities of sucrose solutions, the molarity of potato cores was determined to be 0.35 M at a temperature of 21∘∘ C. Calculate the solute potential (ψψ S). Enter your answer to two decimal places.
-8.55 bars.
The acquisition of water and nutrients occurs at the cell's surface, and this limits cell size, because water and nutrients must reach all parts of the cell as wastes are expelled. Diffusion occurs at a fixed rate, so if a cell were large in size, there would be very little movement of important molecules to its interior. As a cell increases in size, its volume increases; its surface area increases as well, but the relative increase in volume is much greater than that in surface area. Because of this, the amount of surface area available for the transport of molecules into the cell per unit of volume decreases as volume increases. Thus, the smaller the size of a cell, the more favorable the surface area to volume ratio--the greater the ratio of surface area to volume, the greater the availability of necessary materials to all parts of the cell. In this respect, smaller is better, as the surface area to volume ratio is greater. This may seem a bit confusing, but we will look at some mathematical calculations and an actual activity to reinforce this concept. Let's practice by calculating surface area (SA) and volume (V) for a cube that is 1 cm per side by starting with surface area. 1. The formula for calculating the surface area of a cube is: SA = length × width × number of sides (6) Therefore, the surface area of a cube that is 1 cm on a side is 1 cm × 1 cm × 6 sides = 6 cm2 2. The formula for calculating the volume of a cube is: V = length × width × height Therefore, the volume of a cube that is 1 cm on each side is 1 cm × 1 cm × 1 cm = 1 cm3 3. The surface area to volume ratio is: SA ÷ V For a cube that is 1 cm on a side, SA/V ratio is 6 ÷ 1 = 6. Calculate the SA/V ratio for a cube that is 3 cm on each side.
2
Water potential is a physical property that can be used to predict the direction in which water will flow. Because water potential is affected by both solute concentration and pressure, it generally applies only to cells that have walls. Without a cell wall, pressure would not build up within a cell--the cell would simply burst. The pressure in a plant cell can be compared to closing the nozzle on a hose so that no water sprays out. Does water keep moving into the hose? No--the back pressure stops the flow of water. This is exactly what can happen in a plant cell that has become turgid (swollen). Recall that molecules always move from an area of high concentration to an area where their concentration is lower. This concept remains unchanged as we look at water potential. Water moves from areas of high water potential (high free water concentration) and lower solute concentration to areas of lower water potential (lower free water concentration) and higher solute concentration.Solutes decrease the concentration of free water due to the water molecules clustering around the solute molecules. If water moves into a cell with a high solute concentration (low water potential), a cell wall allows pressure to build up inside the cell to the point no more water can be gained. Which of these two plants' cells have the highest water potential?
A
In the first part of this investigation, you will use agar cubes--acting as artificial cells--to study the relationship between surface area and volume. In a common procedure, agar or gelatin is prepared using an indicator solution (1% phenolphthalein) that will change color as an acid (vinegar) or base (NaOH) diffuses into it. When both phenolphthalein and NaOH are added to the agar, it will turn a deep pink. The chart below indicates a color scale of pH for phenolphthalein. Phenolphthalein color indicator Color pH Acid or base Colorless 0-8.2 Acidic to slightly basic Pink to red 8.2-12.0 Basic In preparing the artificial cells, both phenolphthalein and NaOH are added to the heated agar. It is poured into a mold, cooled, and cut into cubes of varying sizes. To standardize our measurements, we will use cubes that are 1 cm, 2 cm, and 3 cm per side. Record the measurements in a data table, and follow the instructions shown in the figure below. As the cubes soak, calculate the surface area, volume, and SA/V ratio for each cube. Record this in your data table. Notice that the distance a material diffuses into the cube is the same regardless of size of the agar cube, but that the percentage of the total cube volume penetrated goes down drastically as the cubes get larger. This provides visual support for our discussion in Part B, explaining why smaller cells have more favorable SA/V ratios. There are many adaptations that contribute to favorable surface area-to-volume ratios. Consider these: intestinal cells have numerous projections called microvilli to enhance absorption in the gut; the epidermal cells of plant roots have root hairs to aid in water and nutrient uptake; and the bodies of planaria (tiny worms that have no gills or lungs) are flattened and very thin to allow adequate diffusion of oxygen and carbon dioxide. If you were given a cube of modeling clay, how could you shape it to have the greatest surface area-to-volume ratio?
Flatten the cube into a pancake with numerous projections
The following figure shows a plant cell in distilled water. Use the information provided in the figure to calculate water potential for the cell and for the contents of the beaker. Assume that the membrane is permeable only to water, not to solutes. What will occur when this system is at dynamic equilibrium?
The cell will gain water until its pressure potential equals 3 bars.
A core has been removed from a beet and dropped into distilled water. Using the information provided, calculate water potential for both the distilled water and the beet core. Drag the numbers on the left to the appropriate blanks on the right to complete the sentences. Numbers may be used more than once.
The water potential in the beaker is __0__ bars. The water potential in the beet core is __-0.2__ bars. Water will flow __into__ the beet core cells.
Plant cell A has lower water potential than cell B. Plant cell B is isotonic to the fluid in the beaker. Plant cell B is at equilibrium with the fluid in the beaker. Plant cell A is at equilibrium with the fluid in beaker. Plant cell B's water potential increased due to an increase in solutes inside the cell Plant cell A has higher water potential than cell B.
True False True False False False
Molecules are in constant motion and tend to move from regions where they are in higher concentration to regions where they are less concentrated. Diffusion is the net movement of molecules down their concentration gradient, the region along which the density of a chemical substance increases or decreases. Diffusion can occur in gases, in liquids, or through solids. An example of diffusion in gases occurs when a bottle of perfume is opened at the front of a room. Within minutes, people farther and farther from the source can smell the perfume. Watch this animation describing diffusion. Osmosis is a specialized type of diffusion that involves the passive transport of water. In osmosis, water moves through a selectively permeable membrane from a region of higher concentration to a region of lower concentration. The plasma membrane of all cells is selectively permeable, as it allows passage of certain types of molecules while restricting the movement of others. Water movement through the plasma membrane is an example of osmosis. In the parts of this investigation that follow, we will examine factors that affect diffusion, and see how diffusion and osmosis occur in cells. Osmosis is a specialized type of diffusion. Which of these is specific only to osmosis?
Water molecules pass through a semipermeable membrane.