Sandbox Reserved 1174: Difference between revisions

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=== Fibrosis ===

~~To gain a better understand the role the LPA plays in fibrosis, a study was done with mice who had contracted~~ fibrosis <ref name= "Tager"> PMID:18066075 </ref>. ~~Idiopathic pulmonary~~ fibrosis ~~(IPF) has high rates of mortality~~. ~~Research has been done to study the~~ pathway of ~~the~~ LPA-LPA1 in fibroblast migration ([https://en.wikipedia.org/wiki/Wound_healing Wound Healing]). ~~In the injured lungs,~~ the ~~IPF, fibroblast can be activated. In the lungs genes related to cell migration can be unregulated. The fibroblast migration can be regulated by LPA. The~~ bronchoalveolar lavage (BAL) in mice ~~that~~ had fibrosis ~~was elevated~~. LPA1 plays an active role between lung injury and contracting pulmonary fibrosis. LPA ~~has the ability to lead to~~ a vascular leak after an initial injury ~~which can lead~~ to fibrosis. ~~This study's findings shows that~~ LPA1 is ~~part of~~ a link between lung injury and [http://www.nature.com/nm/journal/v14/n1/fig_tab/nm1685_F4.html pulmonary fibrosis] <ref name= "Tager"/>.

Idiopathic pulmonary fibrosis (IPF) has high rates of mortality.<ref name= "Tager"> PMID:18066075 </ref>. Understanding how LPA can effect fibrosis, is an important factor to finding medication and a cure for this disease. The pathway of LPA-LPA1 is important in mediating fibroblast migration and [https://en.wikipedia.org/wiki/Wound_healing Wound Healing]. Once fibrosis has been contracted LPA levels increase in the bronchoalveolar lavage (BAL) fluid. The study showed that mice lacking LPA1 had protection from mortality and were able to survive fibrosis. LPA1 plays an active role between lung injury and contracting pulmonary fibrosis. The absence of LPA results in a vascular leak after an initial injury, leading to fibrosis. LPA1 is a link between lung injury and [http://www.nature.com/nm/journal/v14/n1/fig_tab/nm1685_F4.html pulmonary fibrosis] <ref name= "Tager"/>.

== Endocannabinoids ==

Sandbox Reserved 1174: Difference between revisions

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@@ Line 53: / Line 53: @@
 === Fibrosis ===
-To gain a better understand the role the LPA plays in fibrosis, a study was done with mice who had contracted fibrosis <ref name= "Tager"> PMID:18066075 </ref>.  Idiopathic pulmonary fibrosis (IPF) has high rates of mortality.  Research has been done to study the pathway of the LPA-LPA<sub>1</sub> in fibroblast migration ([https://en.wikipedia.org/wiki/Wound_healing Wound Healing]).  In the injured lungs, the IPF, fibroblast can be activated.  In the lungs genes related to cell migration can be unregulated.  The fibroblast migration can be regulated by LPA.  The bronchoalveolar lavage (BAL) in mice that had fibrosis was elevated. LPA<sub>1</sub> plays an active role between lung injury and contracting pulmonary fibrosis.  LPA has the ability to lead to a vascular leak after an initial injury which can lead to fibrosis.  This study's findings shows that LPA<sub>1</sub> is part of a link between lung injury and [http://www.nature.com/nm/journal/v14/n1/fig_tab/nm1685_F4.html pulmonary fibrosis] <ref name= "Tager"/>.
+Idiopathic pulmonary fibrosis (IPF) has high rates of mortality.<ref name= "Tager"> PMID:18066075 </ref>.  Understanding how LPA can effect fibrosis, is an important factor to finding medication and a cure for this disease.  The pathway of LPA-LPA<sub>1</sub> is important in mediating fibroblast migration and [https://en.wikipedia.org/wiki/Wound_healing Wound Healing].  Once fibrosis has been contracted LPA levels increase in the bronchoalveolar lavage (BAL) fluid.  The study showed that mice lacking LPA<sub>1</sub> had protection from mortality and were able to survive fibrosis.  LPA<sub>1</sub> plays an active role between lung injury and contracting pulmonary fibrosis.  The absence of LPA results in a vascular leak after an initial injury, leading to fibrosis.  LPA<sub>1</sub> is a link between lung injury and  [http://www.nature.com/nm/journal/v14/n1/fig_tab/nm1685_F4.html pulmonary fibrosis] <ref name= "Tager"/>.
 == Endocannabinoids ==