H2 HELP WITH: biomolecules!

8/10/2019 H2 HELP WITH: biomolecules!

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BIOMOLECULES

Carbohydrates- Explain how structure of starch makes it a good storage molecule.

1. Helically coiled

2. Compact structure allows many starch molecules to be stored within a small volume3. Glucose monomers joined by 1,4-glycosidic bonds → easily hydrolysed by most

enzymes

4. Hundreds of glucose residues⇒ many hydrogen atoms storing chemical energy

5. Insoluble in aqueous solution thus does not exert osmotic pressure

6. Anomeric carbon involved in glycosidic bond formation makes it relatively stable and

unreactive

Explain why cellulose / starch is insoluble in water

1. A large proportion of the free hydroxyl groups on starch are involved in

intra-molecular hydrogen bonding and thus cannot interact with water.

2. In cellulose, only the surface of the microfibril is accessible to water, and thus hasrelatively fewer OH groups available for hydrogen bonding with water (since the

hydroxyl groups are involved in inter-molecular hydrogen bonding between cellulose

microfibrils)

3. Hence there are insufficient free hydroxyl groups to enable cellulose to be soluble in

water (or effect solubilisation)

Proteins- Describe an alpha helix

1. Is a secondary structure formed by the regular coiling of a single polypeptide chain

2. Is maintained by hydrogen bonds between the CO and NH groups of every 4th amino

acid residue on the polypeptide backbone3. There are 3.6 amino acid residues in every turn of the alpha-helix

Proteins- Describe structure of beta-pleated sheet

Hydrogen bonds form between the C=O group of a peptide in one strand and the N-H group

of another peptide in the adjacent strand. R-groups project above and below the

beta-pleated sheet.

Proteins- Describe the structure of a protein

Idea: Mention the structure that holds the protein together, then mention the bonds that

maintain the structure and what they are between.

1. Primary structure - peptide bonds between amino acid residues2. Secondary structure - hydrogen bonds between NH and CO groups of polypeptide

backbone: a-helix (4 a.a. residues away) and b-pleated sheet (different part of same

peptide, parallel/antiparallel)

3. Tertiary structure - hydrogen bonds, ionic bonds, disulfide bonds and hydrophobic

interactions between R-groups of amino acid residues

4. Quaternary structure - same 4 bonds^ between individual polypeptide subunits

a. Best example = haemoglobin

Compiled by Hope Chow

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Proteins - Outline how a structure of a named globular protein is related to its specific

function. [6]

1. Name protein: haemoglobin

2. found in red blood cells of vertebrates, transports oxygen in blood3. 4 polypeptide subunits, 2 alpha globin and 2 beta-globin subunits

4. each subunit is arranged so that most of its hydrophilic amino acid side chains are

on the external surface while its hydrophobic amino acid side chains are buried in the

interior

a. makes it soluble in blood , capable of carrying O2 from lungs to tissues and vice

versa

5. each subunit is made of protein globin and a prosthetic component called a haem

group

6. each haem group consists of a porphyrin ring and an iron ion

a. Fe

2+

of haem group binds temporarily to O2 so 1 Hb molecule can carry up to 4O2, forming oxyhaemoglobin

7. 4 subunits held together by intermolecular bonds allows movement that influences

affinity for oxygen→ enabling cooperative binding of oxygen

a. binding of one oxygen molecule to one haemoglobin subunit induces a

structural change in remaining 3 subunits, increasing their affinity for oxygen

b. hesitant loading of first oxygen molecule results in rapid loading of the other

three.

Enzymes- Describe how enzymes take part in chemical reactions

1. Specific active site where substrate complementary in shape and charge binds by the

lock and key model2. Results in change in conformation of enzyme and active site allowing a more precise

fit (induced fit hypothesis), forming enzyme-substrate complex

3. Lowers activation energy by:

a. Increased proximity of reactants next to each other increases chance of

reaction

b. Straining of bonds to be broken increases chance of breakage

c. Reactants held in the correct orientation

Explain why an increase in substrate concentration at low substrate concentrations

increases the initial rate of rxn but an increase at high substrate concentrations doesnot do so.

1. At low substrate concentrations, the active sites of the enzymes are readily available

to catalyse the reaction à substrate concentration is limiting

2. As the substrate concentration increases, the rate of reaction increases as more

active sites are occupied by substrates


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3. At higher substrate concentrations, there is saturation of the active site at any one

time. Thus, enzyme concentration becomes limiting while substrate concentration is

no longer limiting.

Explain how a non-competitive inhibitor inhibits the enzyme

1. (inhibitor) acts as a non-competitive inhibitor which binds to a site other than theactive site

2. Causes a conformational change in the 3D conformation of the active site of NOL

enzyme

3. Prevents it from being complementary in shape and charge to the substrate

4. Hence there is a lower number of ez-sub complexes formed.

Explain how temperature affects the rate of an ez-cat rxn

1. Increase in temperature increases the kinetic energy of the ez and sub molecules

2. Increases frequency of successful collisions between the ez and sub forming the

ez-sub complex

3. Increased number of molecules having sufficient energy to overcome the Ea barrier to form the products

4. Reaction rate increases with temperature up to the optimum temperature

5. It increases by the Q10 which is the temperature coefficient. State formula. At 25^C,

Q10 =2

6. Each enzyme has an optimal temperature that allows the greatest number of

molecular collisions and thus rate of reaction

7. Beyond that temperature, thermal agitation and increase in intramolecular vibrations

causes the disruption of the tertiary structure of the enzyme, breaking hydrogen,

ionic and other weak interactions that stabilises the active conformation, leading to

the denaturation of the enzyme.

Describe the mode of action of an enzyme

1. Enzyme is specific, determined by its active site, which is complementary in shape

and charge to the substrate forming an enzyme-substrate complex (regurgitate this)

2. Active site is complementary to (substrate here), cleaving the (bond name here) bond,

in a (reaction type here) reaction.

3. The enzyme works by induced fit in which the binding of the substrate to the contact

residues on the active site cause conformational changes for a more precise fit

4. Enzyme then lowers the activation energy required for the reaction to take place

5. This is done by distorting the bond between the reactants, causing strain on the bond

/ Presence of acidic and basic R groups on the catalytic residues facilitate catalysis

of the bond.


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H2 HELP WITH: biomolecules!