Topic 1 - Biomolecules

Question 1

What is the chemical composition of a bacterial cell?

Include:

• How much that component contributes to the overall weight (%)
• Number of species
• how amazing Ambre is!

Answer 1

Important things to note:

• Ions mentioned in the photo are the main ones
• Large difference in diversity between the macromolecules.

Question 2

Definition of a metabolite?

Answer 2

A substance that is a product of metabolic action or that is involved in a metabolic process.

Question 3

Definition of a cofactor?

Answer 3

They are inorganic or organic chemicals that assist enzymes during catalysis of reactions.

Question 4

What are the main elements involved in living systems?

Answer 4

The main bulk of biological elements:

Ions - Na, Mg, K, Ca, Cl

Main elements organic molecules - C, H, N, O

Extra elements - P, S

Question 5

What are trace elements involved in living systems?

Answer 5

Trace elements used for specialized properties/reactions.

• Transition metals (V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Mo)
• Extra –> Se and I

Question 6

What are the three main types of chemical reactions in living systems?

Answer 6

1. Protonation/deprotonation –> Coming and going of Hydrogen ions
2. Condensation/Hydrolysis –> H₂O coming and going
3. Oxidation/reduction –> e^- / H⁺ coming and going

Note –> Movement of e^- and H⁺ are equivalent.

Question 7

Explain what happens during protonation/deprotonation reactions?

(Acid/Base chemistry)

Examples - Carboxyl/amine group/phosphoric acid

Answer 7

Note -> Acids donates H⁺ / Base accepts H⁺

Question 8

Explain what happens during condensation/hydrolysis reactions?

Examples - Ester/Phosphate ester/Phosphoanhydride

Answer 8

Question 9

Explain what happens during Oxidation/reduction reactions?

Example - Ethane –> ethene

Answer 9

Question 10

What is a phosphodiester bond?

Answer 10

Basically….

R — O — P — O — R

Question 11

How to approach bonding questions with large molecules?

Example –> Practise drawing the deprotonation/condensation of phosphates in ATP.

Answer 11

With large molecules –> find the r-group that isn’t involved in bonding and ignore it.

For example:

Condensation and deprotonation of phosphate groups in ATP –> Ignore the Adenosine and Ribose molecule.

Question 12

Molecules bonding drawing practise.

Reduction of nicotinamide-adenine dinucleotide (NAD⁺) to NADH + H⁺

1. Draw condensation reaction to form nicotinic acid
2. Draw reduction of nicotinic acid.

Answer 12

Question 13

What do the law of thermodynamics help us with?

Answer 13

They determine the conditions under which specific processes can or cannot take place.

Question 14

What is the 1st law of thermodynamics?

Answer 14

1st law of thermodynamics –> States that the total energy of a system and its surroundings is constant.

Hence, the energy content of the universe is constant (cannot be created or destroyed). However, it can take different forms –> Heat a form of kinetic energy or potential energy is released during the occurrence of a process.

Question 15

What is the 2nd law of thermodynamics?

Answer 15

2nd law of thermodynamics –> States that the total entropy of a system plus that of its surroundings always increases.

Sometimes the 2nd law is contradicted –> i.e. plants use CO₂ + other nutrients to form leaves. However, in such cases entropy is decreased locally, whereas, entropy elsewhere increases.

Additionally –> local decreases often results in a heat release –> increases entropy of surroundings.

Question 16

What are the two conditions needed for a chemical reaction?

Answer 16

1. Reaction needs a mechanism
2. Follow the law of thermodynamics.

Question 17

What is chemical potential energy?

Answer 17

Normally during reactions:

Chemical energy —> Heat

This chemical energy is stored as Chemical potential energy.

Chemical potential energy –> Energy stored in the chemical bonds of a substance. It is represented by mu (µ).

Question 18

What is the formula to calculate Gibbs free energy from the chemical potential energy of molecules?

Answer 18

Given the following reaction:

A + B –> C + D

We can say that….

(µ_c + µ_D) - (µ_A + µ_B) = ΔG

Basically…

The change in µ of products minus change in µ of reactants.

Question 19

What is Gibbs free energy?

Answer 19

Gibbs free energy is an accounting tool that keeps track of both the entropy of the system (directly) and the entropy of the surroundings (in terms of the heat released from the system).

Question 20

How can we calculate Delta G from the equilibrium constant?

Answer 20

For a given equilibrium reaction:

A + B <—> C + D

The following equilibrium constant is obtained:

Keq = ((C)(D))/((A)(B))

When the constant and the following equation are used we can calculate Delta G.

ΔG = -RTLn(k)

As we can see, this equation links ΔG and K.

Question 21

What are the different results for the ΔG = -RTLnK equation?

Answer 21

1. K = 1 –> ΔG = 0
2. K > 1 –> ΔG = Negative –> favours products
3. K < 1 –> ΔG = Positive –> favour reactants

Hence….

More negative ΔG is —> more the reaction is favoured –> as more product is formed.

Question 22

How do non-spontaneous reactions in biological systems occur?

Answer 22

Some reactions in biological systems are not spontaneous, hence in order to solve this reactions are coupled in order to form a negative ΔG. ATP helps out a lot as its reaction to form ADP + Pi is spontaneous.

Example –> phosphorylation of glucose.

Question 23

Explain the phosphorylation reaction of glucose?

Glucose –> Glucose-6-P

Answer 23

Glucose phosphorylation

1. Involves condensation reaction at carbon 6
2. Deprotonation of -OH on the phosphate group.

Question 24

Explain how the coupling of ATP and GLC + Pi allow the reaction to proceed.

Answer 24

1. GLC + Pi —> GLC-6-P + H₂O (ΔG = 13 KJ/Mole)
2. ATP + H₂O —> ADP + Pi (ΔG = -29 KJ/mole)

Add both equations and cancel from both sides results in….

3. GLC + ATP —> GLC-6-P + ADP

This overall equation has a resultant ΔG = -16 KJ/Mole –> which makes the overall reaction spontaneous.

Question 25

Explain how the coupling of respiration (glucose combustion) and ATP allow the reaction to proceed.

Answer 25

This is another example showing how the coupling of two reactions allows the overall reaction to be thermodynamically feasible.

1. C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

This reaction has an equilibrium constant equal to 10⁵⁰⁰. This results in a ΔG = - 2870 KJ/mole.

2. ADP + Pi —-> ATP + H₂O

ΔG = + 29 KJ/mole

Hence the net ΔG after both reactions are coupled is -1942 KJ/mole (Very spontaneous). Combustion helps drive the phosphorylation of ATP.

Note –> One mole of glucose oxidised allows for the phosphorylation of 32 ADP molecules.

Question 26

Explain the change in chemical potential energy using an energy diagram.

Include –> Annotations + what happens when an enzyme is introduced.

General graph and Coupling reaction of ATP + GLC –> GLC-6-P graph.

Answer 26

Question 27

What are some important roles of ATP?

Answer 27

ATP plays an important role in…

1. Chemical synthesis
2. Transport
3. Mechanical movement
4. Creating/destroying information (DNA)

Question 28

When is a reaction spontaneous?

Answer 28

A reaction is spontaneous when ΔG is less than or equal to 0.

This is linked to the fact that entropy of the universe only increases when ΔG ≤ 0.

Question 29

Explain the formation of glucose polymers (amylose/glycogen)

Answer 29

Two glucose molecules undergo a condensation reaction between the -OH on carbon 1 and -OH on carbon 4 on another molecule. Water is released as a by-product of the reaction. This form a glycosidic bond.

This process is repeated many times to form a polymer.

Question 30

How to name a long glucose polymer?

Answer 30

(GLCα 1-4 GLC)_n

Basically…

(Molecule/Bond type/Molecule)Number of bonds

Question 31

Explain the formation of glucose polymers (Cellulose).

Answer 31

The same reaction to form α linkage but in this case, beta glucose molecules are used.

This forms —> (GLC β1-4 GLC) _n

The main difference between both isomers is that in alpha the -OH groups are both below the plane of the ring. Whereas, in beta they are above and below the plane of the ring. This influences the geometry of the bond.

Question 32

Explain the impact of alpha and Beta linkages on the shape of the overall polymer.

Answer 32

The different isomers of glucose (Alpha/Beta) have slight differences in the positioning of the -OH groups involved in the glycosidic bond –> this influences the bond geometry and thus the overall shape of the molecule.

1. Amylose –> α glycosidic bond is bent –> hence when repeated over and over again –> Forms helical shape.
2. Cellulose –> β glycosidic bond is straight –> the β linkage makes every successive molecule flip by 180^o relative to the previous residue —> results in an overall linear shape of the molecule.

Question 33

Explain the general metabolism of food (starch) within a living organism.

Answer 33

Food (starch) enters mouth of living organism –> digestive enzyme break down polymer into the monomeric subunits –> absorption into cell –> actions of glucose depend on its current level within the cell.

1. High glucose conc. –> Stored as glycogen (can be broken down in periods of low levels)
2. Low Glucose conc. –> Broken down to for respiration (used to form ATP).

Question 34

Why can humans digest starch but not cellulose?

Answer 34

Potato

(GLC α 1-4 GLC)_n + H₂O —> GLC

Amylose –> can be broken down by amylase which is present in humans.

Cellulose

(GLC β 1-4 GLC)_n + H₂O —> GLC

There is NO enzyme present that can catalyse the following reaction –> humans can’t digest cellulose. However, ruminants have bacteria that can catalyze hydrolysis reaction.

Question 35

What is the conformation of amylose?

Answer 35

Amylose –> helical structure –> forms starch granules

Note –> there are also interchain H-bonds that contribute the helical/curved shape.

Question 36

What is the conformation of cellulose?

Answer 36

Cellulose

In the 2D plane –> Polymers of glucose link together by H-bonds –> form large 2D sheets.

In the 3D plane –> Sheets are held together by van der Waals forces –> packs sheets on top of each other.

Question 37

Using phospholipids as an example, explain the bonds formed in order to form such a molecule.

Answer 37

Phospholipid structure –> alcohol, phosphate, glycerol and fatty acid chains.

All bonds forme between these groups are due to condensation reactions.

Question 38

What are hetero- and homo-polymers? Provide examples.

Answer 38

Heteropolymer - A polymer derived from two or more different (but often similar) types of monomer (DNA -> Bases change)

Homopolymer - A polymer formed from a single type of monomer. ( Cellulose –> only glucose).

Question 39

Explain the bonding that takes place between nucleotides.

Answer 39

Phosphate is used to form a phosphodiester bond between two nucleotides. It binds to carbons 3 on one nucleotide and carbon 5 of the other.

5^I,3^I phosphodiester bond

The DNA polymer is also directional as at one end of the strand you have a 5^I end whereas on the other you have a 3^I end.

Question 40

Explain the condensation reaction between two amino acids.

Answer 40

Condensation reaction between the amino and carboxyl group which releases water as a by-product.

Question 41

What are some general uses of proteins?

Answer 41

1. Enzymes
2. Structure
3. Nutrition
4. Transport
5. Movement
6. Signalling

Question 42

What interactions exist between the two DNA strands?

Answer 42

The two strands link together via H-bonding between the bases. Additionally, the bases are hydrophobic –> bases hide inside while the hydrophilic sugar/phosphate backbone is on the outside.

Question 43

What is the importance of phosphorylating sugars?

Answer 43

The addition of phosphoryl groups is a common modification of sugars.

• Phosphorylation makes sugars anionic –> the negative charge prevents these sugars from spontaneously leaving the cell by crossing lipid-bilayer membranes
• Also prevents them from interacting with transporters of the unmodified sugar.
• phosphorylation creates reactive intermediates that will more readily undergo metabolism.

Question 44

How are oligosaccharides built up?

Answer 44

Oligosaccharides are built by the linkage of two or more monosaccharides by O-glycosidic bond –> in reference to the glycosidic oxygen.

Question 45

Do oligosaccharides have directionality?

Answer 45

• Oligosaccharides have a directionality defined by their reducing and nonreducing ends –> The carbohydrate unit at the reducing end has a free anomeric carbon atom that has reducing activity
• By convention, this end of the oligosaccharide is still called the reducing end even when it is bound to another molecule such as a protein and thus no longer has reducing properties.

Question 46

What are the common disaccharides and how are they formed?

Answer 46

A disaccharide consists of two sugars joined by an O-glycosidic bond.

Sucrose –> Glucose (alpha) and Fructose (Beta)

Maltose –> Two Glucose molecules linked by alpha-1,4-glycosidic linkage.

Lactose –> Galactose and glucose linked by Beta-1,4-glycosidic linkage

Question 47

What are polysaccharides?

Answer 47

They are large polymeric oligosaccharides, formed by the linkage of multiple monosaccharides, and play vital roles in energy storage and in maintaining the structural integrity of an organism.

Question 48

What is the structure of starch?

Answer 48

Two forms of starch

Amylose and Amylopectin

Amylose the unbranched type of starch, consists of glucose residues in alpha-1,4 linkage. Amylopectin the branched form, has about 1 alpha-1,6 linkage per 30 alpha-1,4 linkages,

Question 49

Structure of cellulose? How does the structure impact function?

Answer 49

Cellulose is an unbranched polymer of glucose residue joined by beta-1,4 linkages.

The Beta configuration allows cellulose to form very long, straight chains. Fibrils are formed by parallel chains that interact with one another through hydrogen bonds, generating a rigid, supportive structure. The straight chains formed by b linkages are optimal for the construction of fibers having a high tensile strength.

Question 50

What are the common features of all biological membranes?

Answer 50

1. Membranes are sheetlike structures, only two molecules thick, that form closed boundaries between different compartments.
2. Membranes consist mainly of lipids and proteins. The mass ratio of lipids to proteins ranges from 1:4 to 4:1. They also have lipids linked to proteins
3. Lipids are amphipathic –> lipids spontaneously form closed bimolecular sheets in aqueous media. These lipid bilayers are barriers to the flow of polar molecules.
4. Specific proteins mediate the distinctive functions of membranes.
5. Membranes are noncovalent assemblies –> held together by non-covalent interactions
6. Membranes are asymmetric
7. Membranes are fluid structures
8. Most cell membranes are electrically polarized, such that the inside is negative [typically 60 millivolts (mV)]

Question 51

How to number the carbons on a fatty acid chain? Names of specific carbons?

Answer 51

Fatty acid carbon atoms are numbered starting at the carboxyl terminus.

Note -> Carbon atoms 2 and 3 are often referred to as alpha and beta respectively.

The methyl carbon atom at the distal end of the chain is called the ω carbon atom.

Question 52

How to denote the position of a double bond?

Answer 52

The position of a double bond is represented by the symbol △ followed by a superscript number. For example, cis-△⁹ means that there is a cis double bond between carbon atoms 9 and 10

Question 53

How do the number of double bonds and chain length impact melting point?

Answer 53

Unsaturated fatty acids have lower melting points than do saturated fatty acids of the same length.

• More unsaturation –> lower melting point

Longer chain length means a higher melting point

Overall –> Thus, short chain length and unsaturation increase the fluidity of fatty acids and of their derivatives.