When the d2 receptors are blocked for long periods of time they tend to up regulate. This is what causes tardive dyskinesia.

enter image description here

Why do the newer atypical anti psychotics cause such at a lower rate? Added tables for clarification of receptors I am referring to.


D1  35–118  Antagonist  
D2  3.00–106    Antagonist  
D2L 31–38   Antagonist  
D2S 21–52   Antagonist  
D3  7.8–91  Antagonist  
D4  1.6–50  Antagonist  
D4.2    17–102  Antagonist  
D4.4    21–60   Antagonist  
D5  74–90   Antagonist  


D1  265–1,170   ND  
D2  0.45–3.3    Partial agonist 
D2L 0.34–0.74   Partial agonist 
D3  0.8–9.7 Partial agonist 
D4  44–514  Partial agonist 
D5  95–2,590    ND  

Haldol(typical antipsychotic)

D1 (silent antagonist) — Unknown efficiency[citation needed]
D5 (silent antagonist) — Unknown efficiency[citation needed]
D2 (inverse agonist) — 1.55 nM
D3 (inverse agonist) — 0.74 nM
  • $\begingroup$ If we compare the Haldol to the abilify for example, we see that Haldol is an inverse agonist, which means it turns receptor activity down below baseline. This is more than simply antagonist activity. While abilify is a partial agonist, it can activate the dopamine receptors to some extent, while preventing full activation. Perhaps this is part of the explanation. Haldol really turns down receptor activity, while abilify only partially turns it down. So the cells are less likely to respond to abilify by upregulating dopamine receptor levels. Just a guess, so not an answer. $\endgroup$ – user137 Jul 29 '18 at 23:50
  • $\begingroup$ @user137 that was my guess as well, but then I would assume abilify would have lower TD rates then the other drugs which no literature seems to suggest $\endgroup$ – William Jul 29 '18 at 23:55
  • $\begingroup$ So I'm leaving this as comment for now because it isn't well the perfect source. According to Dr. Peter Breggins book he says they are using lower doses in comparison to the older typical anti psychotics. $\endgroup$ – William Jul 30 '18 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.