Biochemical Tests and Structures - Scholarly

Question 1

Front

Answer 1

Back

Question 2

Outline the Benedict’s test for reducing sugars.

Answer 2

1. Add an equal volume of Benedict’s reagent to the sample to be tested. 2. Heat the mixture in a water bath at 100°C for 5 minutes. 3. Positive result: colour change from blue to green to yellow to orange to brick-red. Precipitate forms.

Question 3

Outline the Benedict’s test for non-reducing sugars.

Answer 3

1. Conduct a reducing sugar test; if the result is negative the reagent remains blue. 2. Hydrolyse non-reducing sugars (e.g. sucrose) into their monomers by adding an equal volume of HCl. 3. Heat in a boiling water bath for 5 minutes. 4. Neutralise the mixture using sodium hydrogen carbonate solution. 5. Proceed with the Benedict’s test as usual.

Question 4

How can the Benedict’s test be made more quantitative?

Answer 4

Either measure the time from immersing the solution in the water bath until the first colour change is produced or create standard solutions of known concentrations and compare colour change to estimate concentration.

Question 5

Outline the Biuret test for proteins.

Answer 5

1. Add an equal volume of sodium hydroxide to a sample at room temperature. 2. Add a few drops of dilute copper II sulfate solution. Swirl to mix. (Steps 1 and 2 make the Biuret reagent.) 3. Positive result: colour change from pale blue to purple. Negative result: solution remains blue.

Question 6

Describe how to test for and measure the presence of starch in a sample.

Answer 6

1. Add iodine solution. 2. Positive result: colour changes from yellow-brown to blue-black.

Question 7

Describe the emulsion test for fats and oils.

Answer 7

1. Add ethanol to the sample and shake. 2. Allow the mixture to settle. 3. Add an equal volume of water. 4. Record any observations.

Question 8

Describe the positive result of an emulsion test.

Answer 8

White cloudy emulsion forms.

Question 9

Define monomer.

Answer 9

A single subunit that is used to build larger polymers.

Question 10

Define polymer.

Answer 10

A large molecule comprising of repeating subunits (monomers) often joined by condensation.

Question 11

Define macromolecule.

Answer 11

A large biological molecule.

Question 12

Define monosaccharide disaccharide polysaccharide.

Answer 12

1. Monosaccharide: A single unit of carbohydrate. 2. Disaccharide: 2 units of carbohydrate joined by condensation held by a glycosidic bond. 3. Polysaccharide: A polymer with monomers of monosaccharides joined together by condensation held by glycosidic bonds.

Question 13

Draw and describe the structure of α-glucose and β-glucose.

Answer 13

Both are hexose monosaccharides (6C) with a ring structure.

Question 14

What is the difference between α-glucose and β-glucose?

Answer 14

The OH group on carbon 1 for α-glucose is below the plane while that for β-glucose is above the plane.

Question 15

What happens in condensation reactions with reference to glycosidic bonds?

Answer 15

A chemical bond forms between two molecules and a molecule of water is produced. H is removed from one molecule and OH from another. When this occurs between carbohydrates it is referred to as a glycosidic bond which is a type of covalent bond.

Question 16

What is meant by the terms reducing and non-reducing sugars?

Answer 16

Reducing sugar: is able to reduce other sugars to form di- or polysaccharides due to free groups which can be condensed. All monosaccharides are reducing sugars. Non-reducing sugars: do not possess a condensable free group and cannot reduce other sugars.

Question 17

State whether the following are reducing or non-reducing sugars: glucose fructose maltose sucrose.

Answer 17

Reducing: Glucose Fructose Maltose. Non-reducing: Sucrose.

Question 18

State how sucrose is formed via the formation of a glycosidic bond.

Answer 18

Glucose and fructose form sucrose via the omission of H2O in a condensation reaction forming a glycosidic bond between them.

Question 19

How are glycosidic bonds broken? What is this known as?

Answer 19

Water is used to break the glycosidic bond. This is known as a hydrolysis reaction.

Question 20

Describe the structures of amylose and amylopectin.

Answer 20

Amylose: 14-glycosidic bonds; unbranched helical structure. Amylopectin: 14 and 16-glycosidic bonds; branched structure.

Question 21

How do the structures of amylose and amylopectin relate to their function?

Answer 21

Amylose and amylopectin act as storage polymers of α-glucose in plant cells. They are insoluble have no osmotic effect on cells and are large thus do not diffuse out of cells. Amylose is compact due to its unbranched helical structure. Amylopectin has many terminal ends for rapid hydrolysis into glucose.

Question 22

Describe the structure and functions of glycogen.

Answer 22

Glycogen is the main storage polymer of α-glucose in animal cells (but also found in plant cells) and consists of 14 and 16-glycosidic bonds. It is more branched than amylopectin has many terminal ends for hydrolysis is insoluble has no osmotic effect does not diffuse out of cells and is compact.

Question 23

Describe the structure and functions of cellulose.

Answer 23

Cellulose is a polymer of β-glucose polysaccharide that gives rigidity to plant cell walls preventing bursting under turgor pressure and holding the stem up. It has 14-glycosidic bonds and consists of a straight-chain unbranched molecule. Alternate glucose molecules are rotated 180 degrees and H-bond crosslinks between parallel strands form microfibrils giving it high tensile strength.

Question 24

What is the difference between saturated and unsaturated fats?

Answer 24

Saturated fats have no C=C bonds and are solid at room temperature due to strong intermolecular forces. Unsaturated fats have one or more C=C bonds and are liquid at room temperature due to weak intermolecular forces.

Question 25

Describe the structure of a triglyceride with reference to how it is formed.

Answer 25

One molecule of glycerol forms ester bonds with three fatty acids (these can be saturated or unsaturated) via condensation reactions.

Question 26

Relate the structure of triglycerides to their functions.

Answer 26

Triglycerides have a high energy-to-mass ratio for energy storage. Their insoluble hydrocarbon chains have no effect on the water potential of cells used for waterproofing. They are slow conductors of heat for thermal insulation less dense than water for buoyancy in aquatic animals and protect organs. They have a high melting point and are very stable molecules resistant to large amounts of force.

Question 27

Describe the structure and function of phospholipids.

Answer 27

Phospholipids are polar molecules with a glycerol backbone attached to 2 hydrophobic fatty acid tails and 1 hydrophilic polar phosphate head. They form a phospholipid bilayer in water and are a component of membranes; the hydrophobic tails allow for control of the movement of water-soluble molecules in and out of cells.

Question 28

Describe the common structure of an amino acid.

Answer 28

An amino acid consists of an amine group (-NH2) a variable side chain (R) a carboxyl group (-COOH) and a hydrogen atom.

Question 29

How are peptide bonds formed?

Answer 29

Peptide bonds are formed when the OH is lost from the carboxyl group and an H from the amine group via condensation forming a peptide bond (-CONH).

Question 30

Describe and name the process by which peptide bonds are broken.

Answer 30

Peptide bonds are broken by hydrolysis. H2O is split into H and OH with OH returning to the carboxyl group and H to the amine.

Question 31

What is the primary structure of a protein?

Answer 31

The primary structure of a protein is the sequence of amino acids in a polypeptide held by peptide bonds.

Question 32

What is secondary structure?

Answer 32

Secondary structure refers to the regular folding of a polypeptide into alpha helices and beta-pleated sheets held by hydrogen bonds.

Question 33

What is tertiary structure?

Answer 33

Tertiary structure is the further coiling of a protein into its functional 3D shape held by hydrogen ionic and disulfide bonds and hydrophobic interactions.

Question 34

What is quaternary structure?

Answer 34

Quaternary structure refers to the folding of 2 or more polypeptides into a 3D shape which may include prosthetic non-protein groups held by hydrogen ionic and disulfide bonds and hydrophobic interactions.

Question 35

Describe the structure of haemoglobin.

Answer 35

Haemoglobin is a globular conjugated protein with a prosthetic group consisting of 2 α chains 2 β chains and 4 prosthetic haem groups. It is water-soluble dissolving in plasma and the Fe2+ in the haem group forms coordinate bonds with O2. Its tertiary structure changes to make it easier for subsequent O2 molecules to bind (cooperative binding).

Question 36

State the main differences between globular and fibrous proteins.

Answer 36

Globular proteins are soluble and involved in physiological processes (e.g. haemoglobin) while fibrous proteins are insoluble and have a structural role (e.g. collagen).

Question 37

State the features of a globular protein with reference to haemoglobin.

Answer 37

Globular proteins are spherical and compact with hydrophilic R groups facing outwards and hydrophobic R groups facing inwards usually making them water-soluble. They are involved in physiological roles (e.g. metabolic processes) such as enzymes like amylase and insulin. Haemoglobin consists of 2 polypeptide chains linked by 2 disulfide bonds. Its compact nature allows it to transport more oxygen per unit of blood.

Question 38

Describe the structure of collagen.

Answer 38

Collagen is a fibrous protein that is insoluble consisting of long strands with high tensile strength. It is made up of 3 polypeptide chains coiled to form a triple helix with every third amino acid being glycine (the smallest amino acid). Collagen molecules lie parallel to form collagen strands held by staggered covalent cross-bridges between lysine residues.

Question 39

How does hydrogen bonding occur in water?

Answer 39

The hydrogen on one water molecule is attracted to the lone pair of the - oxygen on another water molecule.

Question 40

What are the properties of water due to hydrogen bonding?

Answer 40

1. High surface tension. 2. Acts as a solvent for water-soluble molecules serving as a transport medium. 3. High specific heat capacity. 4. High latent heat of vaporization. 5. Higher boiling point.

Question 41

What is the state of ice compared to liquid water in terms of density?

Answer 41

Ice is less dense than liquid water.

Question 42

At what temperature is liquid water typically expected to be at room temperature?

Answer 42

Liquid water is typically expected to be at room temperature.

Question 43

What is the significance of the density of ice compared to liquid water?

Answer 43

The lesser density of ice compared to liquid water is significant because it allows ice to float on water.