More

Question 1

Collagen

Answer 1

Flexible structural protein forming connective tissues
The presence of the many hydrogen bonds within the triple helix structure of collagen results in great tensile strength. This enables collagen to be able to withstand large pulling forces without stretching or breaking
The staggered ends of the collagen molecules within the fibrils provide strength
Collagen is a stable protein due to the high proportion of proline and hydroxyproline amino acids result in more stability as their R groups repel each other
Length of collagen molecules means they take too long to dissolve in water (collagen is therefore insoluble in water)

Question 2

Haemoglobin

Answer 2

Haemoglobin is responsible for binding oxygen in the lungs and transporting the oxygen to the tissue to be used in aerobic metabolic pathways
As oxygen is not very soluble in water and haemoglobin is, oxygen can be carried more efficiently around the body when bound to the haemoglobin
The existence of the iron II ion (Fe2+) in the prosthetic haem group allows oxygen to reversibly bind
None of the amino acids that make up the polypeptide chains in haemoglobin are well suited to binding with oxygen

Question 3

Test for lipids

Answer 3

Take a completely dry and grease-free test tube.
2 To 2cm? of the sample being tested, add 5 cm? of ethanol.
3 Shake the tube thoroughly to dissolve any lipid in the sample.
4 Add 5 cm’ of water and shake gently.
5 A milky-white emulsion indicates the presence of a lipid.
6 As a control, repeat the procedures using water instead of the sample; the final solution should remain clear.

Question 4

List for non reducing sugars

Answer 4

, it must first be ground up in water.
• Add 2cm’ of the food sample to 2 cm’ of Benedict’s reagent in a test tube and filter..
• Place the test tube in a gently boiling water bath for 5 minutes. If the Benedict’s reagent does not change colour then a reducing sugar is not present.
• Add another 2cm’ of the food sample to 2cm’ of dilute hydrochloric acid in a test tube and place the test tube in a gently boiling water bath for five minutes. The dilute hydrochloric acid will hydrolyse any disaccharide present into its constituent monosaccharides.
• Slowly add some sodium hydrogencarbonate solution to the test tube in order to neutralise the hydrochloric acid. (Benedict’s reagent will not work in acidic conditions.) Test with pH paper to check that the solution is alkaline.
• Re-test the resulting solution by heating it with 2 cm? of Benedict’s reagent in a gently boiling water bath for five minutes.
• If a non-reducing sugar was present in the original sample, the Benedict’s reagent will now turn orange-brown. This is due to the reducing sugars that were produced from the hydrolysis of the non-reducing suga

Question 5

Reducing sugars

Answer 5

Benedict’s reagent is un alkaline solution of copper II sulphate when a reducing sugar is heated with Benedict’s reagent it forms an insoluble red precipitate of copper oxide

• Add 2 cm’ of the food sample to be tested to a test tube. If the sample is not already in liquid form, first grind it up in water.
• Add an equal volume of Benedict’s reagent.
• Heat the mixture in a gen tly boiling water bath for five minutes

Question 6

Protein test

Answer 6

Place a sample of the solution to be tested in a test tube and add an equal volume of sodium hydroxide solution at room temperattre
• Add a few drops of very dilute (0.05%) copper(II) sulfate solution and mix gently.
• A purple coloration indicates the presence of peptide bonds and hence a protein. If no protein is present, the solution remains blue.

Question 7

Test for starch

Answer 7

Place 2 cm of the sample being tested into a test tube (or add two drops of the sample into a depression on a spotting tile).
• Add two drops of iodine solution and shake or stir.
• The presence of starch is indicated by a blue-black coloration.

Question 8

Starch suited for function

Answer 8

it is insoluble and therefore doesn’t affect water potential, so water is not drawn into the cells by osmosis
• being large and insoluble, it does not diffuse out of cells
• it is compact, so a lot of it can be stored in a small space
• when hydrolysed it forms a-glucose, which is both easily transported and readily used in respiration
• the branched form has many ends, each of which can be acted on by enzymes simultaneously meaning that glucose monomers are released very rapidly.

Question 9

Glycogen suited

Answer 9

it is insoluble and therefore does not tend to draw water into the cells by osmosis
• being insoluble, it does not diffuse out of cells
• it is compact, so a lot of it can be stored in a small space
• It is more highly branched than starch and so has more ends that can be acted on simultaneously by enzymes. It is therefore more rapidly broken down to form glucose monomers, which are used in respiration.

Question 10

Cellulose suited

Answer 10

cellulose molecules are made up of B-glucose and so form long straight, unbranched chains
• these cellulose molecular chains run parallel to each other and are crossed linked by hydrogen bonds which add collective strength
• these molecules are grouped to form microfibrils which in turn are grouped to form fibres all of which provides yet more strength.