SPEC: Carbohydrates should be studied in the context of the following:
• starch, the role of salivary and pancreatic amylases and of maltase located in the
intestinal epithelium
• disaccharides, sucrase and lactase.
3 Triose
5 Pentose
6 Hexose
• starch, the role of salivary and pancreatic amylases and of maltase located in the
intestinal epithelium
• disaccharides, sucrase and lactase.
- Biological molecules such as carbohydrates and proteins are often polymers and are based on a small number of chemical elements.
- Monosaccharides= basic molecular units (monomers) of which carbohydrates are composed.
- The structure of a-glucose and the linking of a-glucose by glycosidic bonds formed by condensation to form maltose and starch.
- Condensation reaction= forms disaccharides by REMOVING water and forms a glycosidic bond between its constituent monosaccarides.
- Hydrolysis reaction= breaking poly/disaccarides into monosaccarides by ADDING water
- Lactose intolerance. Biochemical tests using Benedict’s reagent for reducing sugars and non-reducing sugars. Iodine/potassium iodide solution for starch.
- The number of Carbon's in a sugar contribute to its nomenclature (its naming). For e.g:
3 Triose
5 Pentose
6 Hexose
Monosaccaride + Monosaccaride -> Disaccaride Reducing? (Polysaccaride) Enzyme needed
Glucose +
Galactose + Fructose + |
Glucose ->
-> Glucose -> Glucose -> |
|
Starch, Cellulose, Glycogen
|
Maltase
Amylose Lactase Sucrase |
Note: The disaccarides have the SAME molecular formula, but their structure is DIFFERENT. This difference affects the FUNCTION it has our bodies.
Reducing Sugars- Benedict's Reagent (Copper sulfate solution)
ALL MONOsaccarides ARE reducing sugars. This means that they will donate an e- to the Cu2+ in the Benedict's solution (that is blue) and 'reduce' the Cu2+ to Cu (Remember OILRIG-Oxidation is Loss, REDUCTION is GAIN of e-). This is as the Cu2+ will gain 2 e- to become overall balanced in charge (2+ + 2 e- = 0 overall charge).
So a monosaccaride, like glucose will reduce Benedict's solution from blue across a spectrum of colour to red (which indicates a more reducing carbohydrate is present).
Polysaccarides like starch are non-reducing and tend to be v.large molecules that are insoluble. Glycogen is used as an E store and Cellulose is used as support.
Benedict's Solution Method: Used to deduce whether a reducing sugar is present
1) Put the unknown saccaride in in a test tube and add Benedict's reagent. If turns RED you have a monosaccaride. No further testing is nessessary. If stays the same colour (BLUE), the sample could be anything, so further testing IS necessary.
2) Put a new batch of the same sample in HCL (hydrochloric acid) and heat in water bath to hydrolyse the bonds.
Add sodium hydrogencarbonate to NEUTRALISE the solution (pH7). The idea here is that if the sample contains disaccarides, the bonds will break forming 2 monosaccarides which will be reducing.
Then when you add some more Benedict's, the solution will turn from BLUE to a RED colour if a monosaccaride is present (if v.reducing, but there is a spectra of colour before red). You can then deduce if this happen that a disaccaride was originally present before you hydrolysed the bonds.
If the solution stays BLUE it is more than likely a polysaccaride is present or even a disaccaride (as not all disaccarides are reducing sugars).
ALL MONOsaccarides ARE reducing sugars. This means that they will donate an e- to the Cu2+ in the Benedict's solution (that is blue) and 'reduce' the Cu2+ to Cu (Remember OILRIG-Oxidation is Loss, REDUCTION is GAIN of e-). This is as the Cu2+ will gain 2 e- to become overall balanced in charge (2+ + 2 e- = 0 overall charge).
So a monosaccaride, like glucose will reduce Benedict's solution from blue across a spectrum of colour to red (which indicates a more reducing carbohydrate is present).
Polysaccarides like starch are non-reducing and tend to be v.large molecules that are insoluble. Glycogen is used as an E store and Cellulose is used as support.
Benedict's Solution Method: Used to deduce whether a reducing sugar is present
1) Put the unknown saccaride in in a test tube and add Benedict's reagent. If turns RED you have a monosaccaride. No further testing is nessessary. If stays the same colour (BLUE), the sample could be anything, so further testing IS necessary.
2) Put a new batch of the same sample in HCL (hydrochloric acid) and heat in water bath to hydrolyse the bonds.
Add sodium hydrogencarbonate to NEUTRALISE the solution (pH7). The idea here is that if the sample contains disaccarides, the bonds will break forming 2 monosaccarides which will be reducing.
Then when you add some more Benedict's, the solution will turn from BLUE to a RED colour if a monosaccaride is present (if v.reducing, but there is a spectra of colour before red). You can then deduce if this happen that a disaccaride was originally present before you hydrolysed the bonds.
If the solution stays BLUE it is more than likely a polysaccaride is present or even a disaccaride (as not all disaccarides are reducing sugars).
Emulsion Test: For the presence of Lipids
- will go cloudy/white if present
Iodine Test: For the presence of Starch
- turns black if present
- will go cloudy/white if present
Iodine Test: For the presence of Starch
- turns black if present