TLR7/8/9 antagonist 2

Catalog No.: A35333
TLR7/8/9 Antagonist
TLR7/8/9 Antagonist 2 is a potent orally active inhibitor targeting Toll-like receptors 7, 8, and 9. It demonstrates significant inhibitory effects on HEK/hTLR7, HEK/hTLR8, and HEK/hTLR9, with IC50 values of 0.011 μM, 0.029 μM, and 0.052 μM, respectively. This compound exhibits high bioavailability in vivo, making it a valuable tool for investigating auto-inflammatory diseases, including systemic lupus erythematosus and lupus nephritis.
Grouped product items
Size Price Stock Qty
5mg
$855.00
In stock
10mg
$1,400.00
In stock
25mg
$2,795.00
In stock
50mg
$4,100.00
In stock
Bulk Size
Bulk Discount
Free Delivery on orders over $500
Research use only. We do not sell to patients.

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Adooq Products cited in reputable paper
Science VOLUME 380, ISSUE 6663 (2023)
Science VOLUME 369, ISSUE 6510 (2020)
Science VOLUME 356, ISSUE 6336 (2017)
Cell Vol. 185 Issue 23 p4428-4447.e28
Cell Vol 177, Issue 7, p1933-1947.e25
Cell Vol 156, Issue 5, p857-1114
Nature Volume 622 Issue 7982 (2023)
Nature volume 620, pages890-897 (2023)
Nature volume 610, pages540-546 (2022)
Nature volume 588, pages83-88 (2020)
Nature volume 574, pages268-272 (2019)
Nature volume 573, pages539-545 (2019)
Nature volume 567, pages118-122 (2019)
Nature volume 551, pages639-643 (2017)
Nature volume 551, pages247-250 (2017)
Nature volume 548, pages356-360 (2017)
Nature volume 545, pages187-192 (2017)
Biological Activity
DescriptionTLR7/8/9 Antagonist 2 is a potent orally active inhibitor targeting Toll-like receptors 7, 8, and 9. It demonstrates significant inhibitory effects on HEK/hTLR7, HEK/hTLR8, and HEK/hTLR9, with IC50 values of 0.011 μM, 0.029 μM, and 0.052 μM, respectively. This compound exhibits high bioavailability in vivo, making it a valuable tool for investigating auto-inflammatory diseases, including systemic lupus erythematosus and lupus nephritis.
Product Information
Catalog NumA35333
FormulaC23H31N7
Molecular Weight405.54
CAS Number2920729-91-3
SMILESCN1N=CC2=C(C=C(C)N=C21)N(CC3)CCC3C4=NC=C(N5CCNCC5)C=C4C
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