Explain. It is valid to conclude that a base were added, the rate of the reaction would slow down because the pressure of oxygen barely increased, meaning the enzymes were not working at a faster rate. 6. Predict what would happen if vinegar (also known as acetic acid) were added to a water solution of hydrogen peroxide and
ECF potassium levels affect resting membrane potential. Hyperkalemia (excessive levels of potassium in the blood) and hypokalemia (abnormally low blood potassium levels) both affect the function of nerves and muscles. • Explain how hyperkalemia will initially affect the resting membrane potential and the generation of an action potential. The resting membrane potential is based on the polarization of the cell. If there are higher levels of potassium, then the resting membrane potential will be less than normal and an action potential will be not likely with the same amount stimulation.
Effect of Surface Area to Volume Ratio on Diffusion Rate of Glucose from Potatoes Abstract: The experiment was conducted to determine if differences in surface area had a significant effect on diffusion rate of glucose from potatoes. Potato samples of three different lengths were collected through a process of coring along the middle of the vegetable. Cores were soaked in distilled water for a period before an o-dianisidine-enzyme solution was added. Glucose diffusion rates were determined by using a spectrometer after a period of incubation of the potato samples. Glucose levels were measured at 0.52, 0.57, and 0.67 showing that an increase in surface area corresponded with a non-significant (p=0.096) increase in the glucose diffusion rate.
(Damon, )Since the movements of substances in passive transport take place from an area of high concentration to an area of lower concentration until both areas reach equilibrium, the water molecules will move from the potato strips to the sugar solution when there is less water concentration in the solution. Therefore, this means that as the sugar solution concentration increases, there will be more sugar molecules in the solution and less water molecules, therefore lowering the water concentration in the solution, causing the water molecules from the potato strips to move from the higher concentration area to the sugar solution with lower concentration. Vise versa, when there is lower concentration of sugar solution, there will be an increase of water concentration in the solution, therefore water molecules will move from the solution to the potato strips where there will be a relatively lower water concentration. IV) Variables: Independent- concentration of sugar solution (mol per dm³) Dependent- change in length of potato strip (cm) Control- initial length of potato strips (2cm); type of solution (sugar); type of potato b) Controlling Variables I) Independent: - The concentration of the sugar solution would be changed by using different volume of solvent to dilute the solute which would be measured by the
The data collected could have been false and not accurate due to the amount of water given to them each day. One pot may have got more water than another, which could affect the way they grew. Another error that could have occurred was the way the seeds were placed in each pot could have been different. If you packed down the soil on top of the seeds too hard, it would have been harder for the plants to germinate therefore affecting the concluded
Investigation into the effect of variation in temperature on the permeability of cell membranes using fresh beetroot. Identification of trends and patterns As the temperature was raised, more pigment leaked out of the beetroot disks, which in turn meant that less light was able to be transmitted through the liquid. For example, at 80°c the percentage light transmission through the liquid was 0%. At 20°c the percentage light transmission through the liquid was 76%. This proving the above statement.
If the area around a cell has a higher water concentration it will gain water by osmosis. This is because the water molecules surrounding the potato cell are able to pass freely across the cell membrane in both directions; therefore more water will enter the potato than will leave it so it will swell up. Furthermore, if the area around the potato’s cells has a lower concentration of water than the cell will loose water by osmosis due to more water diffusing out through the partially permeable membrane than entering the cell, this will lead to a reduction in the size of the cell. When the concentration of water molecules in the potato is near equal to that of the area around it, the rate of osmosis will decline and the cell will stay the same size, as there is an equal amount of water entering the cell than there is leaving it. Below is a drawn diagram showing the process of osmosis.
= 0.14±.01 mL H2o/100g of tissue/min Species A, losing water at 0.24 ±.01 mL H2o/100g of tissue/min, is losing water at a 0.1 mL H2o/100g of tissue/min more than Species B. In other words, Species A is transpiring faster that Species B. b. Species B could have a thicker cuticle than Species A. The cuticle lowers temperature and so the rate of evaporation of the leaf. Because of thicker waxy covering protecting leaf, it would be more difficult for water to escape the leaf, and leading to a decreased rate of transpiration in Species B.
Osmotic pressure; If a plant was placed in a waterlogged area, where the external solute to the cell (being less concentrated (or hypertonic) to the cell vacuole contents) the cell will not continue to take in water via osmosis forever. The cell wall made of cellulose acts as a firm barrier to any more expansion. Once the cell is full of water, it is said to be turgid. This means that the inward
It is expected that the the length of the radical would decrease as the salt solution increases. Hypothesis: I think that when the NaCl solution concentration increases the length of the radical would decrease. The fact is that as the salt solution concentration increases it weakens the plant ability to absorb water/ nutrients in order to grow. Variables: Independent: Salt concentration (0.0%, 0.2%, 0.4%, 0.6% and 0.8%) Dependent: Length of radical (cm) Controlled variables: Controlled? Amount of seeds (mung beans) per petri dish (8 beans) How?