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• 产品描述： A-1165442 is a potent, competitive and orally available TRPV1 antagonist with an IC50 of 9 nM for human TRPV1.

• 靶点： IC50: 9 nM (human TRPV1);TRP/TRPVChannel

• 体外研究：
  A-1165442 displays potent, competitive antagonism at recombinant human TRPV1 activated by capsaicin (IC50=9 nM) and incomplete blockade of acid-evoked response (62% block at 30 μM). A-1165442 possesses excellent selectivity (>100-fold) versus other members of the TRP family (TRPA1, TRPM8, TRPV2, TRPV3) and other receptors expressed in peripheral sensory neurons including P2X2/3, Cav2.2, Nav channels, and KCNQ2/3. A-1165442 shows minimal cross-reactivity upon evaluation (10 μM) in a broad screening panel (n=74, CEREP) of cell-surface receptors, ion channels, and enzymes

• 体内研究：
  A-1165442 exhibits excellent pharmacological selectivity, has a favorable pharmacokinetic profile, and demonstrates good efficacy against osteoarthritis pain in rodents. Oral administration of A-1165442 prevents capsaicin-induced nocifensive behaviors in rats, with an ED50 of 9.5 μmol/kg corresponding to plasma concentration of 420 ng/mL (970 nM). A single dose of A-1165442 produces a robust effect on grip force, with an ED50 of 35 μmol/kg measured 1 h postdosing. Repeated dosing of A-1165442 results in an increase in potency relative to acute analgesic efficacy. No significant changes in core body temperature is observed in conscious rats dosed with A-1165442 and this temperature-neutral profile is maintained in conscious dogs

• 参考文献：
  1. Voight EA, et al. Discovery of (R)-1-(7-chloro-2,2-bis(fluoromethyl)chroman-4-yl)-3-(3-methylisoquinolin-5-yl)urea (A-1165442): a temperature-neutral transient receptor potential vanilloid-1 (TRPV1) antagonist with analgesic efficacy. J Med Chem. 2014 Sep 11;57(17):7412-24.

• 溶解性： soluble  in  DMSO

• 保存条件： -20℃

• 配置溶液浓度参考：
  

  	1mg	5mg	10mg
  1 mM	2.316 ml	11.578 ml	23.155 ml
  5 mM	0.463 ml	2.316 ml	4.631 ml
  10 mM	0.232 ml	1.158 ml	2.316 ml
  50 mM	0.046 ml	0.232 ml	0.463 ml

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本计算器可帮助您计算出特定溶液中溶质的质量、溶液浓度和体积之间的关系，公式为：

质量 (mg) = 浓度 (mM) x 体积 (mL) x 分子摩尔量 (g/mol)

• pg ng μg mg g kg =
  fM pM nM μM mM M *
  nL μL mL L *