4a: Biomolecules in action

Question 1

List the largest 3 components of bacterial cell and what percentage they make up.

Answer 1

Water - 70% Proteins - 16.5% Nucleic acids - 7.1%

Question 2

What are the 2 roles of DNA?

Answer 2

Replication and transcription/translation to form mRNA to transfer code to make proteins.

Question 3

What are proteins made from and name some tasks they do.

Answer 3

Proteins are a linear chain of amino acids (20 to choose from). They are involved in chemical reactivity, signalling, structure and mechanical work.

Question 4

Describe the difference between polymers and biopolymers.

Answer 4

Polymers (and carbohydrates) are limited by simple sequences and variable lengths. Proteins and nucleic acids are precise in composition and length.

Question 5

How are the properties of biopolymers controlled?

Answer 5

They have one or very few stable conformations which controls its properties.

Question 6

Explain the difference between changes in configuration and conformation.

Answer 6

Changes in configuration requires the making and breaking of bonds. Changes in conformation requires just the rotation of bonds.

Question 7

Describe the peptide chain rotations.

Answer 7

There is 3 bonds, however the amide bond is conjugated so it is planar and fixed. The other 2 bonds can have many angles but have few stable conformations.

Question 8

Describe a Ramachandran plot.

Answer 8

A 2D graph plotting the rotations of the 2 rotatable bonds in a peptide chain from -180° to 180°. The shaded area represents the unstable areas of conformation and the white areas represent the stable areas.

Question 9

How is the Ramachandran plot for proline different from other amino acids and what use does this give it for peptides?

Answer 9

It has a very limited plot with almost no stability in the alpha helix region so it is useful for sharp turns in tertiary structures. It is a ‘helix breaker’.

Question 10

Why isn’t a left handed helix stable?

Answer 10

Steric clashes between H atoms on beta C and the carbonyl.

Question 11

Describe the different types of helices.

Answer 11

The alpha helix has 4 residues per twist and optimises all the weak interactions.

The 3₁₀ helix has 3 residues per twist and is more unstable as the hydrogen bonds are bent.

The π has a 5 residue twist and straight hydrogen bonds but its VdW forces are much weaker due to the larger hole in the middle meaning it isn’t more favoured.

Question 12

What are the different types of beta sheets?

Answer 12

Parallel or antiparallel, however they can be assumed to have more conformations if the peptide bond is assumed to rotate by a few degrees.

Question 13

What are the dominant weak forces for secondary structures?

Answer 13

Van der Waals forces and hydrogen bonds.

Question 14

Describe the structure of the beta sheet in terms of ideallity and realism.

Answer 14

An ideal beta sheet would be straight and untwisted, however with large side chains this induces steric clashes. With small side chains such as Gly, Ala and Ser the ideal form is possible. Silk for example is predominantly these small side chain amino acids.

Normal beta sheets have a roughly 25° left-handed twist to them for both parallel and antiparallel forms. In a sheet of 4 strands the last strand is roughly 90° from the first. This makes and excellent core for globular proteins.

Question 15

Give a major interaction difference between beta sheets and alpha helices and the effects this can have on cells.

Answer 15

Alpha helices can only interact with local residues to it whereas beta sheets can interact with residues much further away. However extensive beta sheets can cause cells to misbehave and can cause age related diseases such as Alzheimers and Parkinsons.

Question 16

What are the 2 most common angles for φ and ψ?

Answer 16

Alpha helix - (57, 47)

Beta sheets - (120, 120)

Question 17

What 5 weak interactions affect tertiary structures?

Answer 17

VdW, H-bonding, ionic interactions (salt bridges), disulfide bridges (extracellular) and hydrophobic interactions (in folded proteins).

Question 18

Why is the entropy change for protein folding favourable?

Answer 18

It releases the water ‘cages’ which form around the hydrophobic side chains of the peptide.

Question 19

How does protein folding maximise interactions by folding?

Answer 19

The hydrophobic interactions are buried in the centre of the protein whilst the hydrophilic residues are left to interact with a water layer around the protein. The folding also places ionic interactions and H bonds at the surface of the protein making folding a favourable process.

Question 20

How fast to proteins fold?

Answer 20

Between 1 μs and a few minutes.

Question 21

How can proteins increase the functionality of its amino acids?

Answer 21

By linking up other amino acids the properties of a group can be inhanced, such as a serine becoming much more nucleophilic.

Question 22

How much more ATP is produced from glucose when O₂ is present?

Answer 22

18 times more.

Question 23

What is the major difference in functions between myoglobin and hemoglobin?

Answer 23

Myoglobin stores oxygen in tissues, hemoglobin transports oxygen as well as CO₂ and H⁺

Question 24

What is the structure of myoglobin?

Answer 24

8 alpha helices with a heme group residing in the centre.

Question 25

How does myoglobin control the entry and exit of oxygen?

Answer 25

It has a dynamic structure so it can control when oxygen is allowed in and out.

Question 26

What are the differences between free heme and heme in myoglobin?

Answer 26

Free heme is not soluable in water but is in the cavity in myoglobin. The polypeptide is analogous to a organic solvent.

Iron must be in 2+ to bind reversibly to oxygen, free heme gets oxidised to 3+.

Free heme will dimerise to oxygen making the binding non-reversible.

While free heme and myoglobin will favourably bind to CO over oxygen, it is 1000 times less favourable for myoglobin than it is for free heme.

Question 27

How does myoglobin prioritise bonding to oxygen over CO?

Answer 27

Myoglobin stablises the polarity of the oxygen and sterically blocks CO.

Question 28

What is the composition of hemoglobin?

Answer 28

4 subunits (2 alpha, 2 beta) which are similar to myoglobin except they favourably form tetramers. Each subunit has its own heme group.

Question 29

Describe how hemoglobin’s oxgyen affinity changes from myoglobin’s? Explain why this occurs.

Answer 29

Myoglobin has stronger binding of oxygen than Hb so Hb drops off oxygen in the tissues. Sucessive binding of oxygen to Hb increases its affinity to more oxygen. Hb has 2 states, tense ‘T’ and relaxed ‘R’ and can transition between them. In the relaxed state oxygen binding is 100x more favoured but when no oxygen is coordinated to Hb the T state is favoured. As oxygen coordinates to Hb the R state becomes more favoured by 100x each time and therefore the oxygen affinity for successive coordinations increases.

In summary, for each oxygen added, Hb becomes 100x more favoured to the R state.

Question 30

How does the iron in the heme group change when it is coordinated with oxygen?

Answer 30

Fe changes from a high spin, 5 coordinate species to a low spin, 6 coordinate species. It also moves into the plane of the heme which moves a helix and rotates the dimers by 15°. This promotes the R state.

Question 31

What 2 things promote the T state?

Answer 31

Ionic interactions and low pH, acidic conditions promote oxygen release as the pH of the lungs is higher than in the tissues so the T state gains an ionic interaction by gaining a proton.

Question 32

What is sickle cell anaemia and why is it sometimes selected for?

Answer 32

It causes Hb to be mutated and form fibres which can’t carry oxygen. However carriers of the sickle cell anaemia gene are immune to malaria.

Question 33

How does the drug chloroquinine work and how has it been adapted against?

Answer 33

Malaria breaks down Hb for nutrients, the haem group is poisoness so the virus collects them and crystallises them together. Chloroquinine blocks the collection of the haem groups together. The chloroquinine collects in the digestion vacuole as it is protonated when it enters so it cannot return out through the vacuole wall. The virus has evolved a pump to remove the chloroquinine so the drug is no longer effective.

Question 34

How does the enzyme TIM increase the amount of products of glycolysis? Describe its mechanism.

Answer 34

It catalyses the reaction from DHAP to GAP which are 2 molecules in equlibrium. TIM labilises the C-2 hydrogen and the pro-R hydrogen on DHAP. Dehydrogenases remove the products so the equlibrium isn’t left to run.

Question 35

Name and briefly describe the key experiments to working out the rate of the TIM reaction.

Answer 35

Isotope transfer: Tritium labelled DHAP produces GAP with only 3-6% keeping the tritium. Both molecules don’t exchange so the intermediate must.

Isotope discrimination 1: Product has the same activity as solvent with tritated solvent. RDS must come after the proton transfer in the forward reaction, 4 > 3.

Isotope discrimination 2: Running the reaction in reverse shows that tritium is disfavoured in the product hence barrier 2 > 1 and 3.

Exchange vs conversion: In reverse, the reaction is 3x more likely to go to products than reactants, hence 4 > 2. Note that running the reaction forwards has the opposite effect but this can be attributed to the tritium being hard to return with.

Question 36

How fast does TIM increase the rate of the reaction it catalyses?

Answer 36

10¹⁰ times which is its maximum speed due to the equlibrium limiting its reaction speed and the randomness of the reactant and catalysts encounters.

Question 37

What are the 2 types of nucleic bases?

Answer 37

Purine and pyrimidine.

Question 38

What kind of coil is DNA?

Answer 38

A right handed, double helix.

Question 39

What is; the diameter of the helix, the distance between adjacent base pairs and the number of base pairs in a rotation?

Answer 39

Diameter is 20 Å

Base pairs are 3.4 Å apart

A rotation as 10 base pairs

Question 40

Why does DNA have a major and minor groove?

Answer 40

The sugar bonds are directly opposite.

Question 41

How can probing DNA be done? How do you make DNA more heat resistant?

Answer 41

When DNA is heated it becomes single stranded. Small chains of bases can be introduced with labels to probe the DNA. The more GC pairings there is, the more heat resistant the DNA is. This is because GC has 3 hydrogen bonds where AT only has 2.

Question 42

How is DNA replicated?

Answer 42

DNA polymerase forms a new chain of bases, up to 1000 per second, 5’ to 3’. Mistakes are made but these mutations are checked for later.

Question 43

How does DNA fit into cells?

Answer 43

DNA has different topoisomers, relaxed and supercoiled. To convert from supercoiled to relaxed DNA topoisomerases (flat, planar enzymes) break up the interactions and spread the DNA out.

Question 44

How is DNA transcribed? What causes a gene to be expressed generally and in E.coli?

Answer 44

RNApol unzips the DNA and takes a copy of one of the chains. This mRNA transfers the code and the protein is made from the code on it.

The level of expression is only dependant on how well the promoter attracts the RNApol to transcribe at the genes site.

In E.coli the RNA promotors are 2 sequences of 6 base pairs, 35 and 10 bases upstream from the transcription start site. The closer a gene has to the 6 base pairs at this promotor site, the more it will be expressed.

Question 45

How is lactose used in E.coli? How is its use controlled?

Answer 45

Lactose is a useful sugar but it is not preferred over glucose. β galactosidase is the enzyme that cleaves lactose’s galactose-glucose bond and is only present at around 5 copies per cell when it isn’t needed however when its needed it increases to around 5000 copies per cell.

The lac operon contains all the genes to control the use of lactose. LacI, the repressor is a seperate gene that produces a protein that binds to DNA and blocks RNApol so it can’t transcribe the enzyme.

As [glucose] decreases, [cAMP] increases. cAMP binds to CRP which blocks lacI but gives room to RNApol. LacI is bound by allolactose which removes it from blocking the gene. This all gives RNApol enough room to bind and transcribe the lac operon.