all molecule tests

Question 1

benedicts

Answer 1

Add Benedict’s reagent to the test tube containing the sample solution. (Benedict’s reagent is blue in colour because it contains copper (II) sulfate ions)

Heat the test tube by using a water bath. Alternatively, you can also use the water beaker by bringing it to a boil for some minutes to heat the test tube.

RESULTS
In the presence of reducing sugar, you will observe the formation of a coloured precipitate because copper (II) sulfate is reduced to copper (I) oxide which is not soluble in water.
Hence, we can conclude that the positive result is the change in colour from blue (no reducing sugar) to brown brick-red(indicating a higher concentration of reducing sugar).
We can say that the above test is semi-quantitative which means that the degree of a colour change indicates the concentration of reducing sugar present in the sample.

Question 2

iodine

Answer 2

To determine whether the starch is present in a sample or not, add some drops of orange or brown iodine in potassium iodide solution to the sample (Since iodine is insoluble in water, hence it is in potassium iodide solution).

Results
In the presence of starch, the iodide ions in the solution will react with the centre of starch molecules to produce a complex with a peculiar blue-black colour
This test helps to show that starch in a sample is digested by enzymes

Question 3

emulsion

Answer 3

Lipids refer to nonpolar molecules that are insoluble in water but dissolve in organic solvents like ethanol.

Add ethanol to the sample
Shake the mixture and add it to a water test tube

Results
A milky emulsion will form in the presence of lipids. It means that the solution will turn cloudy. The milky colour of the solution will be more prominent if more lipid is present
The solution remains clear in the absence of lipids.

Question 4

biuret,

Answer 4

To conduct this test, treat the liquid solution of a sample with potassium or sodium hydroxide to make the solution alkaline
Add some drops of copper (II) sulfate solution (blue) to the sample (Biuret reagent has copper (II) sulfate and an alkali)

Results
The protein is present if you observe the change in colour from blue to purple/lilac. (The change in colour can be very subtle, so it is recommended to hold the test tube up against the white tile while observing the change n colour)
No protein in the sample is present if no colour change is observed
For this test to be effective, there must be at least two peptide bonds present in any protein molecule. Hence, the test will be negative if the sample contains dipeptides or amino acids.

Question 5

non reducing sugars

Answer 5

Non-reducing sugars do not change the colour of the solution, which is blue, and so we have to break the sugar down to monosaccharides by hydrolysis to prove they’re non-reducing. So, when the Benedick’s test gives a negative result, add dilute hydrochloric acid and put this in a water bath; this will hydrolise the bonds between the disaccharides. Hydrogen carbonate is then added to neutralise the solution as Benedick’s reagent doesn’t work under acidic conditions. Re-do the Benedick’s test- if a non-reducing sugar was present, the solution will now produce a red precipitate, if no no reducing or non-reducing sugars were present, the solution will remain blue.