Chapter 1 - Biological molecules

Question 1

What nature of biomolecules provides evidence for evolution?

Answer 1

The nature of the nucleic acid code being identical in all living organisms, as well as the general way in which molecules and groups of compounds interact within and between cells

Question 2

What are the major groups of biomolecules and what are their funtion?

Answer 2

Carbohydrates - respiratory substrate with structural application in plamsa membranes and cell walls
Lipids - Phospholipid bilayer of all cells, certain hormones and respiratory substrate
Proteins - Cell structure as well as enzymes, chemical messengers and intrinsic components of the blood
Nucelic Acids - genetic code for the production of proteins
Water - Most common and important aspect of all cells (cytoplasm)

Question 3

Explain the principles of polymers and their examples in biomolecules

Answer 3

Polymers and large repeating units made up of monomers. Notable examples include polysaccharides, all proteins or nucleic acids (DNA and RNA), for which the monomers are monosaccharides, amino acids and nucleotides respecitvely. They are formed via condensation reaction in organisms and therefore are broken down through hydrolysis.

Question 4

What are the most common monosaccharides and what can they form?

Answer 4

Glucose and fructose form sucrose
Glucose and galactose form lactose
Glucose and glucose form maltose
Alpha glucose is the monomer for starch and glycogen
Beta glucose is the monomer for cellulose

Question 5

What is the difference between alpha and beta glucose?

Answer 5

Alpha glucose forms starch and glycogen and has its 2 -OH groups on the same side of the molecule, whereas beta glucose has its -OH groups on opposite sides of the molecule and makes cellulose

Question 6

What is the bond which holds all monosaccharides together?

Answer 6

Glycosidic bonds

Question 7

What is starch?

Answer 7

Starch comes in two different forms, amylopectin and amylose. Amylose almost exclusively contains 1-4 glycosidic bonds so it is very straight and unbranched, allowing it to be held in a very tight helix making it compact and insoluble due to its size. Amylopectin contains slightly more 1-6 glycosidic bonds however so contains more chains, meaning it is less compact but can be hydrolysed to release glucose faster than amylose can be

Question 8

What is glycogen?

Answer 8

Glycogen is the main carbohydrate store found in animals, but is less common than lipids are for respiration. Glycogen contains lots of 1-6 glycosidic bonds meaning it is very branched and not very compact, however it can be hydrolysed from multiple different chains at once, which is useful because animals have much higher metabolic requirements than plants do.

Question 9

What is cellulose?

Answer 9

Cellulose is a polysaccharide made from beta glucose, however to form a glycosidic bond each glucose molecule must rotate 180 degrees, resulting in a perfectly straight chain of glucose monomers. These straight chains can be “woven” together to forms structurally strong fibres. This is because they can form hydrogen bonds between each chain which are individually weak but due to their sheer quantity can become very strong

Question 10

How would you determine a reducing sugar?

Answer 10

You would add bendicts solution to a sample of the sugar and then heat up the solution in a water bath for about 10 minutes. After this, the solution should turn from blue to red, which indicates the presence of reducing sugars.

Question 11

Why is the Benedicts test described as “semi-qualitative”

Answer 11

Because the solution will turn a varying shade between blue and red, such as yellow or green, depending on the amount of reducing sugar present, with brick red indicating large amounts of sugar are present.

Question 12

How would you determine a non reducing sugar?

Answer 12

You would add sulfuric acid to the sample and heat it in a water bath, and then neutralise the solution again and carry out the regular Benedicts test.

Question 13

How are triglycerides formed?

Answer 13

One glycerol molecule (propane-1,2,3-triol) bonds to three fatty acid chains in a series of condensation reactions. This forms three ester bonds

Question 14

What are phospholipids?

Answer 14

Phospholipids are molecules which contain a glycerol molecule, 2 fatty acids molecules and a phosphate group.

Question 15

Why are triglycerides useful?

Answer 15

They can be hydrolysed to release water and due to their nature can release respiratory enzymes quickly, as well as being hydrophobic compact stores of energy, so they can’t leave cells and don’t affect water potentials. They also have some structural benefits, such as with adipose tissue.

Question 16

Why are phospholipids useful?

Answer 16

Due to the hydrophobic side of the fatty acids and the hydrophilic side of the phosphate group they form excellent cell membranes due to the fatty acids strong intermolecular forces thanks to long chain lengths, as well as protecting the cell between having a hydrophobic centre and hydrophilic outside

Question 17

What are amino acids and what are they used for?

Answer 17

Amino acids are the monomer of all proteins and contain a hydrogen atom, amine group and carboxylic acid group bonded to a central carbon which is also bonded to an -R group, or a group unique to all of the different amino acids.

Question 18

How many amino acids are there which are common to biological organisms?

Answer 18

20, 10 of which humans can make

Question 19

How do amino acids bond and what can they make

Answer 19

They bond by removing a hydrogen from the amine group and the -OH from the carboxyl group to release water (condensation reaction) and form a peptide bond. They can either form dipeptides or polypeptides (proteins)

Question 20

What is the primary structure of a protein?

Answer 20

It is the seqeunce of the amino acids present in a protein, determined by the genetic code

Question 21

What is the secondary structure of a protein?

Answer 21

It is the primary folding pattern of the protein, either a beta pleated sheet or a alpha helix. It is determined by the hydrogen bonds which form between the amino acids.

Question 22

What is the tertiary structure of a protein?

Answer 22

The tertiary structure is the folding that the protein undergoes next. It is determined by the hydrogen bonds, ionic bonds and disulfide bridges which form. This is often the stage proteins get to to become functional

Question 23

What is the quaternary structure of a protein?

Answer 23

This is the final structure of a protein, and it is when multiple polypeptides bond together to make a protein, such as haemoglobin.

Question 24

What is the lipid test?

Answer 24

You add a small amount of ethanol and then shake the sample, and then add water and shake again and should see a white precipitate form

Question 25

How do you test for starch?

Answer 25

You add iodine and it should turn blue-black

Question 26

How do you test for a protein?

Answer 26

You add Biuret solution and the solution should turn purple

Question 27

How do enzymes work and what are they?

Answer 27

Enzymes are proteins which usually are in the tertiary structure with an active site specific to each substrate which catalyse reactions. They place stress on the bonds in the substrate as they form forces with the R groups in the active site, making it easier and less energetic to break the bonds.

Question 28

What factors affect enzymes?

Answer 28

Temperature, increased temperature results in more successful collisions and therefore a faster reaction, however too hot and the enzyme will denature and the shape with change. Ph being too low or high will also denature the enzyme.

Question 29

What is the induced-fit model of an enzyme and how is it different to old fashioned models?

Answer 29

It is the idea that enzymes’ active sites and not perfectly molded to the substrate, and only become the correct shape when the protein comes into contact with its substrate, when in the past the “lock and key theory” said that the active site was always the correct site. This allows both competitive and non-competitive inhibitors to affect the enzyme.

Question 30

How do competitive and non-competitive inhibitors work?

Answer 30

Competitive inhibitors will bond to the active site of the enzyme and sometimes don’t unbond, meaning the substrate cannot reach the enzyme. Non-competitive inhibitors however will bond to the allosteric site of the enzyme, not the active site and instead change the bonds in the active site of the enzyme, meaning it won’t fit the substrate anymore.

Question 31

How does the concentration of substrate affect each inhibitor and why?

Answer 31

As concentrations of substrate increase, competitive inhibitor activity decreases at a constant rate because it is being outcompeted for the active site by the substrate as it is more present than the inhibitor. Non-competitive inhibitor action however is unaffected by substrate levels as they bond to a different site entirely.

Question 32

What is end product inhibition?

Answer 32

End product inhibition is when the end product of a chain of different enzymes is able to inhibit the first enzyme from the chain