General pharm – Enzyme kinetics

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HY discussion of concepts followed by a quiz at the end

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Vmax

Maximal rate of reaction for a given substrate-enzyme interaction.
Reflects enzyme efficacy (i.e., maximum rate of reaction, or “how high the mountain goes”).
That is: ↑ Vmax = ↑ efficacy; ↓ Vmax = ↓ efficacy.

The substrate concentration at 1/2 Vmax.
Reflects enzyme affinity (i.e., the stronger the enzyme binds to the substrate, the less substrate is needed to attain half Vmax).
That is: ↑ Km = ↓ affinity; ↓ Km = ↑ affinity.

Competitive inhibitor/antagonist

Binds to same site on the enzyme as the substrate itself (i.e., the enzymatic site).
Reversible (almost always).
Shifts curve to the right.
Increases Km –> more substrate is needed to achieve half of max reaction rate because competitive inhibitor is occupying some enzyme binding sites.
Vmax unchanged –> as long as enough substrate is present, max reaction rate can occur.
HY USMLE examples:
- The alpha-1 antagonists phentolamine, terazosin, tamsulosin, prazosin.
- Methotrexate (competitive, reversible inhibitor of dihydrofolate reductase).
- Statins (competitive, reversible inhibitors of HMG-CoA reductase).

Non-competitive inhibitor/antagonist

Binds to a site on the enzyme distinct from the substrate (i.e., an allosteric site).
Irreversible (almost always).
Shifts curve down.
Decreases Vmax –> because it binds to a site on the enzyme distinct from the substrate, no matter how much substrate is added into the system, maximal reaction rate cannot occur.
No change in Km –> because it doesn’t bind to the enzymatic site (i.e., it’s not competing with the substrate), the amount of substrate needed to achieve half of the maximal reaction rate (i.e., Km) won’t be affected.
HY USMLE example:
- The alpha-1 antagonist phenoxybenzamine (used for pheochromocytoma).

Partial agonist

Binds to the same site as the substrate (therefore competitive).
Decreases Vmax –> produces a positive response, but simply does not carry the same efficacy as the substrate itself. Even if more substrate is added to the system to outcompete the partial agonist, because the latter is capable of producing a partial response, the net response is still attenuated. In the case of a competitive antagonist, it produces no response when bound, but once it dislodges from the enzymatic site, the original substrate can reoccupy that site and induce the max response. This is why Vmax is normal with competitive antagonists but decreased with partial agonists.
Can shift the curve left and down.
Km can be decreased, normal, or increased –> in other words, the partial agonist, despite reducing the maximal response (Vmax), actually can have increased affinity for the enzyme compared to the original substrate. The USMLE, if they give you a partial agonist question, will give you decreased Km (i.e., left-shifted curve) in order to emphasize that the agent is indeed a partial agonist.
The net result of giving a partial agonist is therefore similar to giving an antagonist.
HY USMLE examples:
- Clomiphene citrate (estrogen receptor partial agonist at hypothalamus –> hypothalamus thinks there’s decreased estrogen because the binding isn’t as strong as estrogen itself, albeit positive, so GnRH increases).
- Danazol (androgen-receptor partial agonist at the liver –> leads to increased C1-esterase inhibitor production in the treatment of hereditary angioedema; can cause hirsutism).

Now let’s look at two examples:

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Regarding the above findings, which of the following is true?

A) Drug Y is a non-competitive antagonist of Drug X
B) Drug X is a non-competitive antagonist of Drug Y
C) Drug Y is a competitive antagonist of Drug X
D) Drug X is a competitive antagonist of Drug Y
E) Drug Y is a partial agonist of Drug X
F) Drug X is a partial agonist of Drug Y

(The answer is down below in blue)

The main point to observe is that the presence of Drug Y does not decrease the ability of Drug X to achieve its maximum potential effect, even if more Drug X is required to achieve just such.

This means Drug X and Drug Y are competing for the same receptor(s) – i.e., as long as you keep increasing the dose of Drug X, you can eventually overcome and drown out the impact of Drug Y.

Important examples of competitive inhibitors for the USMLE are methotrexate (on dihydrofolate reductase)¹, statins (on HMG-CoA reductase)², and almost all α1-blockers (i.e., tamsulosin³, terazosin⁴, phentolamine⁵, etc.).

If Drug Y were to cause a ↓ in the ability of Drug X to achieve its maximum effect (i.e., a downward shift of the curve), even as the dose of Drug X is continually increased, Drug Y would therefore not be competing for the same receptor as Drug X. This could be due to a negative allosteric effect on Drug X, or it could mean Drug Y binds to Drug X’s receptor at a distinct location and prevents Drug X from binding, while simultaneously not inhibiting or potentiating the receptor itself. In other words, with non-competitive inhibition, no matter how much the dose of Drug X is increased, the impact of Drug Y can never be overcome.

The super-HY non-competitive antagonist you need to know for the USMLE is phenoxybenzamine at α1 receptors.⁶

The answer is C.

Bottom line: Competitive antagonism = right-shifted curve. Non-competitive antagonism = down-shifted curve. Methotrexate, statins, and almost all α1-blockers are competitive inhibitors. Phenoxybenzamine is non-competitive.

1) https://www.ncbi.nlm.nih.gov/pubmed/3122764

2) https://www.ncbi.nlm.nih.gov/pubmed/27313057

3) https://www.ncbi.nlm.nih.gov/pubmed/9117115

4) https://www.ncbi.nlm.nih.gov/pubmed/2457715

5) https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1282628/

6) https://www.sciencedirect.com/science/article/pii/B9780323481106000144

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L-asparaginase is an integral component of treatment of acute lymphoblastic leukemia (ALL) in children. It hydrolyzes L-asparagine to L-aspartate in leukemic cells. A researcher who is investigating this enzyme isolates an actinomycetes strain, Streptomyces sp. NEAE-82, which produces L-asparaginase. He then carries out kinetic studies of the purified enzyme and presents his findings in a report:

Researcher’s report	Km (M)	Vmax (Uml⁻¹min⁻¹)
Estimations prior to investigation:	0.02000	95.00
Findings as per kinetic studies:	0.01007	95.08

Which of the following most accurately describes the researcher’s findings of L-asparaginase in comparison to his prior estimations?

A) ↓ enzyme affinity for substrate; ↑ efficacy
B) ↓ enzyme affinity for substrate; ↓ efficacy
C) ↑ enzyme affinity for substrate; ↑ efficacy
D) ↑ enzyme affinity for substrate; ↓ efficacy

(The answer is down below in blue)

Once again:

Vmax

Reaction velocity (rate of reaction) when the enzyme is fully saturated by substrate, indicating that all the binding sites are being constantly reoccupied.⁷
Reflective of enzyme efficacy (i.e., maximum rate of reaction).⁸
In other words, ↑ Vmax = ↑ efficacy; ↓ Vmax = ↓ efficacy

The substrate concentration at 1/2 Vmax⁹
Inversely reflective of affinity (i.e., if the enzyme binds more strongly to the substrate, then you need less substrate to achieve 1/2 Vmax).⁹
In other words, ↑ Km = ↓ affinity; ↓ Km = ↑ affinity

The true finding for Km was less than the researcher’s prior estimate, meaning the enzyme has greater affinity for substrate than predicted.

The true finding for Vmax was greater than the researcher’s prior estimate, meaning the enzyme achieves greater efficacy than predicted.

(By the way, the researcher’s true findings are actually the real Km and Vmax for L-asparaginase.¹⁰)

The answer is C – ↑ enzyme affinity for substrate; ↑ efficacy.

Bottom line: Km = substrate concentration at 1/2 Vmax. ↑ Km = ↓ affinity; ↓ Km = ↑ affinity. ↑ Vmax = ↑ efficacy; ↓ Vmax = ↓ efficacy.

7) https://www.sciencedirect.com/science/article/pii/B9780123971760000017

8) https://books.google.co.jp/books?id=9-xeDwAAQBAJ&pg

9) https://books.google.co.jp/books?id=MgqI-pfk45QC&pg

10) https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5015098/

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