1. AH Biology: Unit 1 Proteomics and Protein Structure 3
Binding to Ligands

2. Think How is protein production controlled?
Why is it important that protein production is controlled?
Why is protein structure important in relation to its function?
Think about these questions and answer them as you study this section on protein structure.

3. Binding to ligands A ligand is a substance that can bind to a protein.
R groups not involved in protein folding can allow binding to these other molecules.
Binding sites will have complementary shape and chemistry to the ligand.
The ligand can either be a substrate or a molecule that affects the activity of the protein.

4. Nucleosomes
DNA binds to a number of proteins, such as those involved in replication, transcription and the regulatory proteins involved in operons.
As DNA is a long linear molecule and has to be packed into the nucleus. It is first super-coiled using histone proteins.
Positively charged histone proteins bind to the negatively charged sugar–phosphate backbone of DNA in eukaryotes.
DNA is wrapped around histones to form nucleosomes, packing the DNA in chromosomes.

5. Nucleosomes
Nucleosomes animation
DNA replication animation

6. Transcription
Other proteins have binding sites that are specific to particular sequences of double-stranded DNA and when bound to can either stimulate or inhibit initiation of transcription.
lac Operon
Transcription animation

7. Binding changes the conformation of a protein
Enzymes and proteins are three-dimensional and have a specific shape or conformation.
As a ligand binds to a protein binding site, or a substrate binds to an enzyme’s active site, the conformation of the protein changes.
This change in conformation causes a functional change in the protein and may activate or deactivate it.

8. Induced fit
In enzymes, specificity between the active site and substrate is related to induced fit.
When the correct substrate starts to bind, a temporary change in shape of the active site occurs, increasing the binding and interaction with the substrate.
Induced fit

9. Induced fit

10. Activation energy lowered
The chemical environment produced lowers the activation energy required for the reaction.
Once catalysis takes place, the original enzyme conformation is resumed and products are released from the active site.
The enzyme is then free to react with more substrate.

11. Allosteric enzymes
An allosteric enzyme is an enzyme that can have its activity altered by a ligand called a modulator.
In allosteric enzymes, modulators bind at secondary binding sites away from the active site.
The conformation of the enzyme changes and this alters the affinity of the active site for the substrate.

12. Modulators
Negative modulators reduce the enzyme’s affinity for the substrate.
Positive modulators increase enzyme affinity for the substrate.

13. Negative modulators

14. Negative modulators
End product inhibition occurs when the final product of a cascade of enzyme reactions interacts with an allosteric site of the first enzyme in the cascade to inhibit it and thus the production of the end product.
This is an example of negative feedback.
End product inhibition animation

15. Competitive inhibition

16. Competitive inhibition example 1
Ethanol is metabolised in the body to acetaldehyde by oxidation with alcohol dehydrogenase, which is in turn further oxidised to acetic acid by aldehyde oxidase enzymes.
Normally, the second reaction is rapid so acetaldehyde does not accumulate in the body.
A drug called disulfiram (Antabuse) inhibits the aldehyde oxidase, which causes the accumulation of acetaldehyde with subsequent unpleasant side effects of nausea and vomiting.
This drug is sometimes used to help people overcome alcoholism.

17. Competitive inhibition example 2
Methanol poisoning occurs because methanol is oxidised to formaldehyde and formic acid, which attack the optic nerve and cause blindness.
Ethanol is given as an antidote for methanol poisoning because ethanol competitively inhibits the oxidation of methanol.
Ethanol is oxidised in preference to methanol and consequently the oxidation of methanol is slowed down and the toxic by-products do not have a chance to accumulate.
The methanol is then excreted in the urine.

18. Competitive inhibition example 3
Ethylene glycol, if ingested, can be poisonous.
Ethylene glycol is oxidised by the same enzymes used in the previous two examples.
Ethylene glycol → glycolaldehyde → glycolic acid.
Glycolic acid is toxic to the nervous system and kidneys.
Describe how ethanol can be used as an antidote.

19. Non-competitive inhibition
Ligand binding to an allosteric site separate from the active site.
Non-competitive inhibition

20. Enzyme kinetics and inhibition
Michaelis-Menten enzyme kinetics.
In enzyme kinetics the initial rate of the reaction is always measured for analysis.
Vmax = initial rate of reaction when all active sites of enzyme are occupied = maximum rate of reaction.
Km = substrate concentration at half Vmax.

21. Competitive inhibition
Vmax is the maximal initial rate of the reaction when all active sites of a known concentration of enzymes are occupied.
Km = half Vmax
Same Vmax, different Km.

22. Competitive inhibition
If the concentration of inhibitor is less than that of the substrate and the substrate has a higher affinity for the active site, is the enzyme inhibited a lot or a little?
If the concentration of inhibitor is more than that of the substrate is the enzyme inhibited a lot or a little?
If the enzyme is inhibited and we then increase the substrate concentration what happens to the initial rate of reaction?

23. Non-competitive inhibition
Vmax is the maximal initial rate of the reaction when all active sites of a known concentration of enzymes are occupied.
Km = half Vmax
Same Km different Vmax.

24. Competitive inhibition
1/V = 1/ initial rate of reaction.
1/[S] = 1/ substrate concentration.
As it is difficult to calculate Vmax practically for Michaelis–Menten graphs a reciprocal regression (Lineweaver Burk plot) is used to calculate Vmax and Km when determining the type of inhibitor affecting an enzyme.

25. Non-competitive inhibition
1/V = 1/initial rate of reaction.
1/[S] = 1/substrate concentration.
As it is difficult to calculate Vmax practically for Michaelis–Menten graphs a reciprocal regression (Lineweaver Burk plot) is used to calculate Vmax and Km when determining the type of inhibitor affecting an enzyme.

26. Enzyme kinetics and inhibition
1/V = 1/initial rate of reaction.
1/[S] = 1/substrate concentration.
As it is difficult to calculate Vmax practically for Michaelis–Menten graphs a reciprocal regression (Lineweaver Burk plot) is used to calculate Vmax and Km when determining the type of inhibitor affecting an enzyme.

27. Positive modulators
Positive modulators increase the enzyme affinity for the substrate by altering the shape of the active site so that it has a better fit for the substrate.
Positive modulation animation of a steroid on a GABAA receptor linked ion channel.

28. Enzyme kinetics questions

29. Cooperativity in haemoglobin
Binding and release of oxygen in haemoglobin.
Some proteins with quaternary structure show cooperativity in which changes in binding at one subunit alter the affinity of the remaining subunits.
This occurs when oxygen binds to haemoglobin.

30. Cooperativity in hemoglobin
Deoxyhaemoglobin has a relatively low affinity for oxygen.
As one molecule of oxygen binds to one of the four haem groups in a hemoglobin molecule it increases the affinity of the remaining three haem groups to bind oxygen.
Conversely, oxyhaemoglobin increases its ability to loose oxygen as oxygen is released by each successive haem.
This creates the classic sigmoid shape of the oxygen dissociation curve.

31. Cooperativity in haemoglobin
Rightward shifts in the dissociation curve indicate decreased affinity of haemoglobin for oxygen, making it harder to bind the oxygen in the lungs but easier to release oxygen to respiring tissues. A higher partial pressure of oxygen is therefore required in the lungs for uptake of oxygen by deoxyhaemoglobin.
Leftward shifts indicate increased affinity, making it easier to bind oxygen in the lungs but increasing the difficulty to release oxygen to respiring tissues. A lower partial pressure of oxygen is therefore required by respiring tissues to release the oxygen from oxyhaemoglobin.

32. Effects of temperature and pH
Low pH = low affinity.
High temperature = low affinity.
Exercise increases body temperature and produces CO2, acidifying the blood.
This has a corresponding effect on the oxyhaemoglobin dissociation curve.

33. Bohr effect
As cells respire aerobically they increase the concentration of carbon dioxide in their surrounding environment.
Respiration also increases temperature as a result of the heat produced.
Carbon dioxide causes the blood and immediate environment around respiring cells to become acidic as carbonic acid forms.
Both an increase in temperature and a decrease in pH cause a right shift in the oxyhaemoglobin dissociation curve.
This ensures that oxyhaemoglobin releases oxygen to respiring body cells more easily.
As the haemoglobin returns to the lungs the carbon dioxide concentration decreases and as a result the affinity for oxygen increases. This ensures that maximal oxygen uptake can be achieved.

34. Oxygen dissociation curve
Oxygen dissociation review in relation to a patient admitted to hospital.
What sort of conditions affect the ability of red blood cells to transport oxygen?
Under what conditions would haemoglobin struggle to bind oxygen?

35. Red blood cell disorders
Sickle cell anaemia
Thalassaemia

36. High-altitude conditions
High-altitude medicine
High-altitude effects: BBC Horizon, ‘How to Kill a Human Being’

37. Think How is protein production controlled?
Why is it important that protein production is controlled?
Why is protein structure important in relation to its function?
Think about these questions again and answer them as part of a discussion in class based on what you have learned in this section.