1-Biological Molecules

Question 1

Metabolism

Answer 1

All biological processes that take place within living organisms.

Question 2

Monosaccharides

Answer 2

Sweet tasting soluble substances with the general formula “(CH2O)n” where n can be any number from 3 to 7.
E.g. Glucose is C6H12O6 and is a hexose (6-carbon) sugar.
2 arrangements of alpha and beta glucose.
Alpha glucose has H above carbon-1.
Beta glucose has H below carbon-1.

Question 3

Test for reducing sugars (Benedict’s Test)

Answer 3

For all monosaccharides and some disaccharides.

1. Heat the sample with Benedict’s solution (copper(ii)sulfate).
2. If the the blue solution turns orange-brown due to a red precipitate (copper(i)oxide) forming then a reducing sugar is present.

Question 4

Condensation reaction of disaccharides

Answer 4

2 monosaccharides join together by a water molecule being removed. The bond formed is called a glycosidic bond.

Question 5

1) Glucose + Glucose ——>
2) Glucose + Fructose ——>
3) Glucose + Lactose ——>

Answer 5

1) Maltose
2) Sucrose
3) Lactose

Question 6

Hydrolysis

Answer 6

When water is added to a disaccharide under suitable conditions the glycosidic bond is broken and water molecule is added to form 2 monosaccharides.

Question 7

Test for non-reducing sugars

Answer 7

1) Make sure sample is in liquid form.
2) Add 2cm3 of sample to 2cm3 Benedict’s reagent in a test tube.
3) Place test tube in boiling water for 5 minutes. If solution doesn’t change colour then reducing sugar isn’t present.
4) Add another 2cm3 of sample to 2cm3 dilute HCl and boil to hydrolyse the disaccharide into its respective monosaccharides.
5) Add sodium hydrogencarbonate to neutralise the HCl.
6) Retest and if a orange-brown solution is formed now then a non-reducing sugar was in the original sample.

Question 8

Polysaccharides

Answer 8

Very large, insoluble molecules formed by condensation reactions forming glycosidic bonds between monosaccharides. This feature makes them suitable for storage, e.g. starch. However the polysaccharide cellulose is used to give structural support for plants instead of being used for storage.

Question 9

Test for starch

Answer 9

1) Add 2cm3 of sample into a test tube.
2) Add 2 drops of iodine solution and shake/stir.
If solution goes blue-black then starch is present.

Question 10

Structure of starch

Answer 10

Chains of alpha-glucose monosaccharides linked by condensation reactions formed by condensation reactions. Chains can be branched or unbranched. The unbranched chain is wound into a tight coil to make the molecule very compact.

Question 11

Function of starch

Answer 11

1) Insoluble so water isn’t drawn into cells by osmosis.
2) Large and insoluble so doesn’t diffuse out of cells.
3) Compact so a lot can be stored in a small space.
4) When hydrolysed forms alpha-glucose which is easily transported and readily used in respiration.
5) Branched variant has many ends which can each be acted on by enzymes at the same time so more glucose monomers are released very rapidly.

Question 12

Structure of glycogen

Answer 12

Similar structure to starch but has shorter chains and is more highly branched. Only in bacteria and animal cells, not plant cells.

Question 13

Function of glycogen

Answer 13

1) Insoluble so doesn’t draw water on by osmosis.
2) Insoluble so doesn’t diffuse out of cells.
3) Compact so a lot can be stored in small places.
4) Highly branched so is broken down by enzymes more rapidly to form glucose monomers used in respiration.

Question 14

Structure of cellulose

Answer 14

Straight unbranched chains of beta-glucose which run parallel to each other and allow hydrogen bonds to form between them. Cellulose molecules are grouped together to form microfibrils which are arranged in parallel groups called fibres.

Question 15

Function of cellulose

Answer 15

Provides support and rigidity

Question 16

Lipids

Answer 16

Contains C, H, O
Insoluble in water
Soluble in organic solvents (alcohols/acetone)

Question 17

Role of lipids

Answer 17

Source of energy — when oxidised they provide 2x energy of carbohydrates for the same mass and release water.
Waterproofing — since they’re insoluble they are useful for being waterproof.
Insulation — slow conductors of heat wand retain body heat. Also act as electrical insulators.
Protection — acts as a buffer around delicate organs, e.g. kidneys.

Question 18

Triglycerides

Answer 18

Consist of 3 fatty acids joined to a glycerol molecule by ester bonds formed in a condensation reaction. When hydrolysed produces 3 fatty acids and a glycerol.

Question 19

Structure and Properties of triglycerides

Answer 19

Good source of energy due to a high ratio of energy-storing C-H bonds to C atoms.
Low mass:energy ratio so are good storage molecules because less mass carries more energy which is beneficial to animals as it reduces the mass they have to carry around.
Large and non-polar so are insoluble in water so their storage doesn’t affect osmosis in cells or water potential.
High H:O atom ratio so when oxidised provide an important source of water.

Question 20

Phospholipids

Answer 20

Hydrophilic phosphate head which interacts with water (but not fat) joined to 2 hydrophobic fatty acid tails that mix readily with fat but orient themselves away from water. These 2 distinct ends make the molecule polar.

Question 21

Structure and Properties of phospholipids

Answer 21

Polar so form a bilayer within cell surface membranes. As a result a hydrophobic barrier is formed inside and outside the cell.
Can form glycolipids by combining with carbohydrates in the cell surface membrane which are important for cell recognition.

Question 22

Test for lipids

Answer 22

1) Add 2cm3 of sample to test tube and add 5cm3 of ethanol.
2) Shake thoroughly to dissolve sample in solution.
3) Ad 5cm3 of water and shake gently .
4) If a cloudy white emulsion forms then a lipid is present.
5) A control can be conducted by repeating the process with water instead of lipid and solution should remain clear.

Question 23

Amino acids

Answer 23

Basic monomer unit which combine to form the polymer polypeptides. Consist of a carbon atom attached to 4 different chemical groups:

1) —NH2 basic amino group
2) —COOH acidic carboxyl group
3) —H hydrogen atom
4) —R group which is a variety of different chemical groups.

Question 24

Formation of peptide bond

Answer 24

A condensation reaction where the —H of the amino group of an amino acid and the —OH the the carboxyl group of another amino acid is removed to link the 2 amino acids to form a dipeptide. Dipeptide can be broken down into its respective amino acids by hydrolysis.

Question 25

Condensation reaction

Answer 25

Removal of water (H2O) to bind two molecules together

Question 26

Hydrolysis reaction

Answer 26

Addition of water (H2O) to form 2 separate molecules

Question 27

Polymerisation

Answer 27

Series of condensation reactions that binds many amino acid monomers together to form a polypeptide.

Question 28

Primary Structure of Proteins

Answer 28

Formation of polypeptides through polymerisation. The primary structure determines the ultimate shape of protein and its function. Only has peptide bonds.

Question 29

Secondary Structure of Proteins

Answer 29

The polypeptide chain coils into an alpha-helix shape due to hydrogen bonds forming between H of —NH group and O of —C=O group.

Question 30

Tertiary Structure of Proteins

Answer 30

Alpha helixes twist and fold even more to form a specific 3D shape. Maintained by ionic bonds formed between amino and carboxyl groups; disulfide bridges; hydrogen bonds and the previously formed peptide bonds. The protein can now interact and be interacted with by other molecules in a specific way.

Question 31

Quaternary Structure of Proteins

Answer 31

An amalgamation of tertiary structures joined together. Can also include prosthetic (non-protein) groups associated with molecules. Contains H-bonds, ionic bonds, disulfide bridges, and peptide bonds.

Question 32

Test for proteins

Answer 32

1) Add sample to test tube with equal volume of NaOH solution at RTP.
2) Add few drops of very dilute (0.05%) copper (ii) sulfate solution and gently mix.
3) If peptide bonds are present then solution goes from blue to purple and indicates presence of proteins.

Question 33

Enzymes

Answer 33

Organic catalysts that increase the rate of a chemical reaction without themselves being used as reactants. They do this by decreasing the activation energy required for a reaction to take place.

Question 34

Effect of temperature of enzyme action

Answer 34

Increasing temperature increases kinetic energy of molecules so the frequency of successful collisions increases. The optimum temperature is about 37C. At about 45C the active site starts to change shape since some H-bonds are broken so substrates no longer fit correctly. At 60*C the enzyme is denatured and stops working altogether.

Question 35

Denaturation

Answer 35

The permanent change of structure of an enzyme so it can no longer carry out its function.

Question 36

Effect of pH on enzyme action

Answer 36

Change in pH alters charges of amino acid that make up the active site so substrates no longer fit and a enzyme-substrate complex can no longer be formed.

Question 37

Effect of enzyme concentration on rate of reaction

Answer 37

If substrates are in excess then increasing the enzyme concentration increases rate of reaction. If the substrates are the limiting factor then increasing enzyme concentration doesn’t affect the rate of reaction.

Question 38

Effect of substrate concentration on rate of reaction

Answer 38

If enzymes are in excess then increasing substrate concentration will increase rate of reaction. If enzymes are the limiting factor then increasing substrate concentration won’t increase rate of reaction.

Question 39

Competitive Inhibitors

Answer 39

Substances that bind to active site of enzyme and compete with substrates, therein decreasing the rate of reaction.

Question 40

Non-competitive Inhibitors

Answer 40

Substances that bind to a site on an enzyme that isn’t the active site. This changes the shape of the enzyme as well as the shape of the active site so substrates can no longer bind to the enzyme and rate of reaction is decreased.