1 ml saliva 37 Blue coloration turned into translucent green. Blue coloration turned to green and finally moderate amount of brick red precipitate suspended in solution. The solution was opaque. 2 10 ml solution B
3 M HCI 37 The blue coloured solution remains unchanged. The blue coloured solution remains unchanged. 3 10 ml solution B
3 M HCI 95 White coloured translucent suspension formed in moderate amount. A very big amount of white suspension formed which is very opaque. Turned back into translucent and remaining solution was blue which is translucent. 4 10 ml solution B
1 ml saliva 95 The blue coloured solution remains unchanged. The blue coloured solution remains unchanged as well. Discussion :
In the experiment, the enzyme amylase was involved. Amylase breaking down the starch suspension into maltose and maltose into glucose when HCL was added into solution B .This is because solution B was hydrolyzed and the H+ ions present break down the bond in between molecules of the solution B. Saliva enzyme works the best at the optimum temperature which at 37°C. Therefore, the solution B in test tube 1 is broken down completely by the saliva enzyme. At temperature 95°C , salivary amylase which was present in saliva content denatured .This is because high temperature would break down the bond holding the tertiary structure and destroyed the 3D structure of salivary amylase enzyme. The active sites of enzyme therefore changed and were no longer complementary to the substrate (carbohydrates).
The Essay on Determine The Macromolecules Present In An Unknown Solution
Every living thing is dependent on large complex molecules, known as macromolecules. The objective of this lab was to correctly identify which macromolecules the unknown solution was comprised of using various substances as experimental controls. There are four major types of biological macromolecules – carbohydrates, lipids, proteins, and nucleic acids – made up of elements such as ...
No simple sugar was produced, hence no reaction between sugar and Benedict’s solution took place caused the solution to remain blue. The product of the experiment conducted in Table 2 is predicted to be maltose and glucose. Both maltose and glucose are carbohydrates. Glucose is a monosaccharide while maltose is disaccharide. The structure of glucose contains single simple sugar unit but the structure of maltose contain two simple sugar unit (also known as two monosaccharides or two glucose units).
Benedict’s test was used to indicate the presence of sugar. Benedict’s Solution which contains the blue copper ions (II) (Cu2+), are hydrolyzed by reducing sugars into red-brown copper (I) ions (Cu+), which is insoluble in solution.
As a result, red-brown precipitate is formed. The results of Benedict’s test and iodine test for solution A is positive and negative respectively, which concludes that solution A is a reducing sugar. For solution B, the results are negative and positive respectively, which concludes in solution B being starch suspension. Therefore, solution B is more complex comparing to solution A. This is because starch is made up of large number of glucose units and has more complex structure. Conclusion : Solution A is a reducing sugar, Solution B is starch suspension.
