If I slid a plant cells with a water potential averaging -.5 MPa Are placed into solution with a water potential of -.3 MPa, which of the following would be the most likely outcome?
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- Water will always move from an area of high water potential to low water potential. If you drop a cell with WP = 0 into the water with WP = -0.9, then what will happen to the cell?A daisy stem is placed in a solution with a water potential of -0.2 bars. After the daisy stem cells have had a chance to reach equilibrium with the surrounding solution, the pressure potential inside the daisy cells is 0.5 bars. At equilibrium, calculate the solute potential of the daisy cells.Consider a plant cell. The value for solute concentration in a plant cell is -0.12 MPa and the turgor pressure is 0.12 MPa. 1- What is the water potential in this plant cell? 2- If this plant cell were placed in a solution with water potential of -0.1 MPa, what would happen to the cell? (Hint: explain where the water goes and what that does to the cell).
- You place a flaccid plant cell (Ψ = -0.7 MPa) into an environment (beaker) of pure (deionized) water (Ψ = 0 MPa). Compare the initial conditions of the flaccid cell and the environment (beaker of pure water) in terms of: A) Water potential (high or low) of the environment B) Solute concentration (high or low) of the environment C) Tonicity (hypertonic, hypotonic, or isotonic) of the environment D) Water potential (high or low) of the cell E) Solute concentration (high or low) of the cell F) Tonicity (hypertonic, hypotonic, or isotonic) of the cell G) Predict the direction of water movement (into the cell, out of the cell, or no net movement) H) Predict the change in turgor pressor of the cell (increase turgor pressure or decrease turgor pressure) I) Predict the fate of the cell (plasmolyzed, turgid, or lysed)If you place a flaccid plant cell with ΨS = -0.4 MPa in pure water, which of the following will occur? A. Water will not enter the cell because the flaccid cell has solutes and low water potential. B. Water enters the cell because the flaccid cell has solutes and low water potential. C. Water enters the cell because the flaccid cell has solutes and high water potential. D. Water will not enter the cell because the flaccid cell has solutes and high water potential.plant cells with an aqueous potential of -600kPa were placed in different aqueous potentials. Determine in which of the following cases after 10 minutes the cells were blocked, started plasmolysis or completely subjected to the latter. Solution A is -400kPa, solution B is - 600kPa, solution C is -900kPa and solution D is distilled water
- Consider a suspension of particles (isoelectric point is at pH 6) in water at pH 2 and a NaCl concentration of 0.001 M. Describe how the strength of repulsion varies with the following changes, assuming all other conditions remain constant. Give a description (more than just increase or decrease) in terms of the effect on the double layer thickness and the zeta potential. (a) Change from 0.001 M NaCl to 0.1 M NaCl, (b) Change from pH = 2 to pH = 5.Consider the transport of K+ ions from a surrounding fluid (where [K + ] = 30 mM) into a cell (where [K + ] = 420 mM) where the membrane electrical potential is -0.15 V. Is this process favorable?Thank you for helping me. I have to questions. In solution B, if we choose two other points on the graph for calculating the slope, wouldnt the values for Km/Vmax be different than 6.1? Why did you choose exactly those points on the graph? Another thing, I understand the method you used to get to Km/Vmax= slope, but how did you then get the value for Vmax to be 10 mM/s??
- You have a semi permeable membrane with a membrane potential of -90mV. You also have two ions that are both permeable to the membrane, Na and Cl. Na has a concentration of 10mM inside the membrane and 120mM outside the membrane. Cl has a concentration of 1.5mM inside the membrane and 77.5mM outside the membrane. Use the nernst equation to calculate the electrochemical equilibrium of both ions, and show in which direction the netflux would be for each ion.Give me the following: Half reactions Standard cell potential/cell potential solution Standard cell potential/ cell potential final answer Standard Gibbs free energy/Gibbs free energy solution Standard Gibbs free energy. Gibbs free energy final answer K/Q solution K/Q final answerWhat is the free energy change for the transport of calcium ions into a cell. The intracellular [Ca2+] is 1 μM and extracellular [Ca2+] is 1 mM. Assume a membrane potential of -100 mV and T=25°C. Give your answer without units to one decimal place.