1. Compare the amount of vitamin C per gram of vegetable with the amount per ml
ID: 58828 • Letter: 1
Question
1. Compare the amount of vitamin C per gram of vegetable with the amount per ml of fresh squeezed fruit juice. Estimate how much of the vegetable (used in your experiment) you would have to eat to equal the vitamin C content of 5 ml of fruit juice.
vegetable used = 1.0851 g
0.5 ml of juice used
2. Use a literature source to find typical vitamin C values for the food that you analyzed. Compare these values with what you obtained.
3. Why should the instrument be zeroed using the phosphate-citrate buffer as opposed to using a solution containing DCPIP?
4. If a fruit or vegetable sample contained a molecule that absorbed light close to the max for DCPIP, why is it important to perform an analysis of the sample without DCPIP present?
Explanation / Answer
1&2. For example, if you assume red bell pepper as your experimental vegetable, that contains 1.3mg Vit-C per gram.
If you use orange juice as your experimental interest, that contains 50mg of Vit-C per 100ml, then per 1ml it is 0.5mg. Now, per 5ml o juice, there is 2.5mg of Vit-C.
So that you need to eat 1.92g of vegetable to get 2.5mg of Vit-C, which is present in 5ml of orange juice.
Or else, if we use your values that 1.0851g vegetable would have 1.41063mg Vit-C and 0.5ml orange juice contains 0.25mg of Vit-C. It suggests that your 0.5ml juice has 5.64 times less Vit-C compare to 1.0851g vegetable containing Vit-C. So that you need to eat 0.192g of vegetable to compensate 0.5ml orange juice containing Vit-C.
3. DCPIP is useful in measuring the Vit-C content of foods and juices. While measuring Vit-C with DCPIP, it is used to oppose phosphate-citrate buffer since, buffer itself has citrate that interfere with the actual content of Vit-C in your experimental juice. So that, we need to normalize that buffer containing citrate value and make the setup zero before testing your test sample.
4. We cannot determine the actual Vit-C content if some compound has close absorption max for DCPIP. Its actual O.D can be interfered by that compound, while Vit-C testing. So, it would be better to analyse without DCPIP.
In oxidized state, DCPIP is in blue with a maximal absorption at 600 nm; when reduced, DCPIP is colorless. The principle of this method is a titration of Vit-C with DCPIP changing the colour from blue to colourless.
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