No topic Essay Example | Topics and Well Written Essays - 250 words - 8

No topic - Essay Example The leucocytes involved in inflammatory response are basophils, eosinophils, neutrophils and macrophages along with tissue cells called mast cell. The damaged tissues release chemicals like histamine, kinins and prostaglandins that cause widening of blood vessels and increase in blood flow to site of injury. These chemicals initiate the white blood cells or leukocytes to migrate to injured site that is known as chemotaxis. The inflammatory response can be triggered by exogenous or endogenous agents like complement found in the plasma helps in releasing bradykinin and fibrinopeptides that helps antibodies to fight against the foreign substances. Two types of white blood cells are visible near the injured site, neutrophils are first found that help in avoiding the harmful bacteria from spreading while microphages are seen later to help clearing the damaged area of bacteria or dead cells thereby generating new tissue that reside until the injury is healed completely. 3. Proliferation takes up to four weeks or more depending on the severity of injury, in cases of severe injury the area affected may compose of specific tissue cells and other tissue known as granulation tissue which forms the scar tissue if not removed thereby decreasing the functional ability of tissue. 4. The new cells spread in the injured area try to produce a functioning tissue that might take months or years to develop with proper functioning. Stretching is helpful in strengthening the new tissue during remodeling. Organelles are bodies present in the cytoplasm that tend to serve various metabolic activities within the cells while lysosomes are sac like compartments that contain several enzymes helpful in breaking down harmful cell products, waste materials, cell debris and foreign substances in a phagocytic cell example macrophages and neutrophils. Stroma is the supporting framework of an organ composed of

Rate of Glucose Uptake by Yeast Cells Lab Report

Rate of Glucose Uptake by Yeast Cells - Lab Report Example From the equation, it can be concluded that one mole of glucose reacts with two moles Cu2+ to give one mole of Cu2O. Thus, one can weigh the mass of Cu2O formed and relate it to the amount of glucose present in a solution. This method can be used to study the rate of glucose uptake by yeast cells. 2. Start the timer and add 1.00 mL of yeast suspension with a micropipette into one flask containing the glucose solution. Repeat this until 7 replicates are obtained. Please note that this has to be done very fast (within 1 minute if possible). To the remaining flasks, add 1.00 mL of distilled water into each, and label them as blank. 4. After 30 minutes have elapsed, pour the contents of one flask labelled blank sample and one containing the yeast suspension into two separate beakers containing around 150 mL of boiling distilled water. Cool the mixtures to room temperature and transfer them into two separate 250-mL volumetric flasks. Add distilled water to the flasks to make a final volume of 250.00 mL. Shake the flasks well and let them stand until clear supernatants are obtained (the blank solution should not have a sediment if the experiment has been done properly). Label the one obtained from the blank sample as B60 and the other one as Y60 5. Pipette 25.00 mL of supernatant from B60 and pour it in a 100 mL beaker. Add 25 mL of 0.2500 M Benedict's solution (it is in excess) to the beaker containing the supernatant from B60. Heat the resulting mixture to boiling until a red precipitate is formed. Cool this mixture to room temperature. Repeat to get a replicate. 6. Filter the mixture with a Gooch crucible tared with filter paper using the set up shown in Figure 1. Wash the precipitate several times with cold distilled water. Figure 1. Set up for filtration apparatus 7. Use a cloth (this would avoid fingerprints on the surface) to remove the crucible containing the precipitate from the filtration apparatus. 8. Dry the sample to constant mass. Ensure that tongs are used to handle the crucible and that the sample is kept/cooled in a dessicator before weighing. 9. Repeat steps 5-8 with Y60. 10. Repeat steps 4-9 to get data at t =120,180,240,300,360,420 mins and label the samples accordingly. Calculations As per Equation 1, one mole of glucose reacts with two moles of Cu2+ to give one mole of Cu2O. In the above experiment, glucose is the limiting reagent and Cu2+ is in excess. Hence the amount (mole) of Cu2O formed of is directly proportional to the amount of glucose in the solution. Molar mass of Cu2O = (63.55 x 2) + 16 = 143.10 Molar mass of glucose, C6H12O6 = (12.01 x 6) + (1.01 x 12) + (16.00 x 6) = 180.18 No. of moles of Cu2O formed =