posted by Boricua on .
The Managing Director of a well-known company on Wall Street thrives on a diet of fruit jam, bread, pasta, and coffee. She exercises intermittently. One day she decides to go to her primary healthcare provider for a routine checkup. The healthcare provider recommends that she take the Benedict's test. Assume that the glucose levels of the patient are high.
• State the results that the test would indicate (specify the color of the solution).
• State the composition and the properties of the ketohexose derived from fruit jam.
• Describe the manner in which ketohexose acts as a reducing sugar in the test.
she had sugar (unknwn sugar, but assume sucrose, dextrose, and fructose mix), and she had starches. What does the Benedect test test for?
the ketohexose in Jam? Your teacher probably wants you to recognize fructose is in berries.
the last question is for you to research.
so would this be my answer?
*** State the results that the test would indicate (specify the color of the solution). ***
Well, Benedict's test detects reducing sugars (such as glucose). Benedict's reagent contains blue copper ions (Cu2+) which are reduced to copper (Cu+) in the presence of reducing sugars. These are precipitated as red copper (I) oxide which is insoluble in water.
The color of the solution would be red.
*** State the composition and the properties of the ketohexose derived from fruit jam. ***
A ketohexose is a ketone-containing hexose (a six-carbon monosaccharide). In other words, it is a monosaccharide which has a 6 carbon backbone with a ketone group on C2.
Since they have three chiral centers, 8 (2^3 =8) different stereoisomers are possible.
*** Describe the manner in which ketohexose acts as a reducing sugar in the test. ***
When the hemi-acetal or ketal hydroxyl group is free, i.e. it is not locked, not linked to another (sugar) molecule, the aldehyde (or keto-) form (i.e. the chain-form) is available for reducing copper (II) ions. When a sugar is oxidized, its carbonyl group (i.e. aldehyde or ketone group) is converted to a carboxyl group.