In a previous paper (Whitehouse, Hancock and Haldane, 1932) it was shown that though the epidermis, so that in this respect the epidermis seems to play the part of semi-permeable membrane. It is presumably the dense stratum lucidum, of the external epidermis that tends to stop diffusion.

Note that a semipermeable membrane is not the same as a selectively permeable membrane. Semipermeable membrane describes a membrane that allows some particles to pass through (by size), whereas the selectively permeable membrane “chooses” what passes through (size is not a factor).

A semipermeable membrane is a layer that only certain molecules can pass through. This makes the phospholipid bilayer an excellent semipermeable membrane that allows cells to keep their contents separated from the environment and other cells. …

Answer. Answer: The real answer is It is a barrier with tiny openings that let some, but not all, materials pass through.

The Dialysis tubing provides a semi-permeable membrane. Only allowing smaller molecules to pass through it. Iodine molecules are small enough to pass freely through the membrane, however starch molecules are complex and too large to pass through the membrane. Thus iodine diffused into the tube with the starch.

4. Why is iodine called an indicator? It is called an indicator because after few minutes of diffusion, it changed the starch from white to a purple like color. The purple color indicates the presence of iodine.

From the results of this experiment, it is obvious that glucose and iodine (potassium iodide) has smaller molecular size than starch. Because starch had larger molecular size, the dialysis tubing was not permeable to it (it didn’t allow it to readily pass through the pores of its membrane).

Iodine molecules easily pass in between the molecules of the plastics and thus stain the pieces all the way through. The plastics, in turn, provide unique environments for the iodine, just like air or water, and make the iodine brownish or pinkish in color.

Using iodine to test for the presence of starch is a common experiment. A solution of iodine (I2) and potassium iodide (KI) in water has a light orange-brown color. If it is added to a sample that contains starch, such as the bread pictured above, the color changes to a deep blue.

Amylose in starch is responsible for the formation of a deep blue color in the presence of iodine. The iodine molecule slips inside of the amylose coil. This makes a linear triiodide ion complex with is soluble that slips into the coil of the starch causing an intense blue-black color.

This is a physical test. A chemical test for starch is to add iodine solution (yellow/brown) and look for a colour change. In the presence of starch, iodine turns a blue/black colour. It is possible to distinguish starch from glucose (and other carbohydrates) using this iodine solution test.

Emulsion test for lipids

1. Place two spatulas of the food sample into a test tube or 1 cm 3 if the sample is liquid.
2. Add 2 cm 3 of ethanol to the tube.
3. Allow the contents to settle.
4. Pour the liquid from the top of the mixture into a test tube half-filled with water.

The natural brown color is due to povidone-iodine, the active ingredient in Betadine Sore Throat Gargle.

Biology Notes for IGCSE 2014

• Food test 1 – Starch test.
• Food test 2 – Benedict’s test for Reducing Sugars.
• Food test 3 – Emulsion (ethanol) test for fat.
• Food test 4 – Biuret test for Proteins.

The emulsion test is a method to determine the presence of lipids using wet chemistry. The procedure is for the sample to be suspended in ethanol, allowing lipids present to dissolve (lipids are soluble in alcohols). The liquid (alcohol with dissolved fat) is then decanted into water.

It is royal blue initially and changes to violet when it reacts with protein. Brown paper bags are used to test for lipids. The paper bag becomes translucent (allows light to pass through) in the presence of lipids.

In lab, we used Benedict’s reagent to test for one particular reducing sugar: glucose. Benedict’s reagent starts out aqua-blue. As it is heated in the presence of reducing sugars, it turns yellow to orange. The “hotter” the final color of the reagent, the higher the concentration of reducing sugar.

For this identification, Benedict’s test and Fehling’s test can be used. The main difference between Benedict’s solution and Fehling’s solution is that Benedict’s solution contains copper(II) citrate whereas Fehling’s solution contains copper(II) tartrate.

Galactose is classified as a monosaccharide, an aldose, a hexose, and is a reducing sugar.

It is a reducing sugar with only one reducing end, no matter how large the glycogen molecule is or how many branches it has (note, however, that the unique reducing end is usually covalently linked to glycogenin and will therefore not be reducing). Each branch ends in a nonreducing sugar residue.

Maltose is a reducing sugar. Thus, its two glucose molecules must be linked in such a way as to leave one anomeric carbon that can open to form an aldehyde group. The bond from the anomeric carbon of the first monosaccharide unit is directed downward, which is why this is known as an α-glycosidic linkage.

Because the aglycone is a hemiacetal, lactose undergoes mutarotation. For the same reason lactose is a reducing sugar. The free aldehyde formed by ring opening can react with Benedict’s solution. Thus, a solution of lactose contains both the α and β anomer at the “reducing end” of the disaccharide.

Sugars that can be oxidised by mild oxidising agents are called reducing sugars. A non-reducing sugar is a sugar that is NOT oxidised by mild oxidising agents. All common monosaccharides are reducing sugars. The disaccharides maltose and lactose are reducing sugars. The disaccharide sucrose is a non-reducing sugar.

Starch does not feature a free aldehyde or keto group. Hence starch is considered to be a non-reducing sugar.

Carbohydrates containing free aldehyde and keto functional group are thus reducing sugars. Example: Glucose, lactose. If the groups are not free, then they do not reduce Tollens reagent and Fehling’s solution and are, therefore, classified as Non-reducing sugars. Example: Sucrose, trehalose. Was this answer helpful?