Carbohydrates

Question 1

Significance of Monosaccharide Structure

Answer 1

1. Readily soluble and easily transported in water
2. Free C=O group gives them reducing ability
3. Pentose and hexose acts as stable building blocks (monomers) for larger molecules
4. Exhibit alpha and beta isomerism to increase diversity of monosaccharides

Question 2

Things to take note for Drawing

Answer 2

1. 2 monomers and disaccharide drawn in correct orientation
   alpha is down beta is up
2. Glycosidic bond alpha (1,4) or (1,6) labelled
3. Condensation reaction + H2O lost

Question 3

Benedict’s Test

Answer 3

1. Place 2 cm3 of test solution in test tube
2. Add equal volume of Benedict’s reagent
3. Heat by immersing tube in boiling water for 3-4 mins

Partially quantitative test —> brick red ppt to remains blue

Sucrose is non-reducing sugar because C=O bond used in glycosidic bond and not free for reducing

Question 4

Acid Hydrolysis

Answer 4

1. Negative test first
2. Boil equal volumes of new sample w dilute HCl for abt 1 min to hydrolysis into monosaccharide
3. Cool contents, neutralise w NaOH and carry out Benedict’s test again

Question 5

Amylose vs Amylopectin vs Glycogen

Answer 5

1. Starch is 20% amylose 80% amylopectin
2. Amylose forms unbranded polymers while amylopectin forms branched polymers; glycogen is even more extensively branched than amylopectin
3. Amylose has only a (1,4) glycosidic bonds, while amylopectin has a (1,4) glycosidic bonds within branch and a (1,6) bonds at branch points

Starch test involves triiodide ion fitting into amylase helix to form blue-black starch-iodine complex

Question 6

Structure + Function for Storage Molecules

Answer 6

1. All are made of many glucose residues, thus:
   a) Large amounts of energy readily available when hydrolysed
   b) Large macromolecule insoluble in water, no effect on water potential
2. All are comprised of helices, with hydrogen bonds forming between hydroxyl groups that project into core of helix, allowing for intramolecular H bonding
   a) Pack more glucose units per unit volume, compacting molecule
   b) Relative fewer hydroxyl groups available for hydrogen bonding with water, making it insoluble
3. Amylopectin and glycogen are extensively branched
   a) Multiple branch ends allow for greater hydrolysis of glycosidic bonds, increasing energy generation per unit time

Question 7

Cellulose Structure + Function

Answer 7

1. Alternate B-glucose monomers, linked via B(1,4) glycosidic bonds, inverted 180 wrt each other resulting in long straight chains with hydroxyl chains projecting out in both directions
2. Parallel cellulose chains held together by hydrogen bonds to form cross linked microfibrils. Microfibrils also have relatively fewer OH groups available for H bonding, and are thus insoluble in water
   Link to Function for 1 and 2: Thus giving cellulose high tensile strength
3. Meshwork of cross-linked microfibrils forms cellulose cell wall
   Link to Function: Thus allowing for close packing that gives high tensile strength but also freely permeable/porous cell wall. Strong and rigid structure protected from osmotic stress

Cellulose synthesised by cellulose synthases embedded in plasma membrane
Chitin and NAG/NAM have similar structures to cellulose (application)