# PharmPK Discussion - Calculation of drug concentration in tissues

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• On 1 Dec 2011 at 12:54:28, Hari Kushwaha (harry2002_4u.-a-.yahoo.co.in) sent the message
`Dear all,In a tissue distribution study, rat liver was collected which weighed8.33 g.  0.5 g rat liver was homogenised in 5 ml homogenising solution,0.5 ml of the homogenate was processed, assayed along with calibrationstandards processed similarily and calculated concentration was 74.03ng/ml. How should I proceed to calculate the drug content in ng/g in ratliver?Thank you in anticipation,HariHari Narayan kushwahaSRF, Central Drug Research Institute,Lucknow-226001,India`
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• On 1 Dec 2011 at 16:18:26, (msgcot2.aaa.optimum.net) sent the message
`If you weigh the homogenate taken for assay will be able to calculatethe concentration of drug in ng/g of tissue`
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• On 2 Dec 2011 at 10:08:23, Fabrice Nollevaux (fabrice.nollevaux.-a-.arlenda.com) sent the message
`The following message was posted to: PharmPKDear Hari,Concentration: 74.03 ng in 1 mL of homogenate-> 37.015 ng in 0.5 mL of homogenate-> 370.15 ng in 5 mL of homogenate-> 370.15 ng in 0.5 g of rat liver-> 740.3 ng per g of rat liverDoes this make sense?Cheers,FabriceFabrice Nollevaux,Arlenda SAwww.arlenda.com`
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• On 2 Dec 2011 at 21:43:49, Garima Balwani (garimabalwani.aaa.gmail.com) sent the message
`Dear Hari74.03 ng/ml  means 370.15 ng in 5 ml of homogenate or 0.5 g of ratliver, therefore in 1 g of liver there is 370.15 X 2 =  740.3 ng of drug-Garima BalwaniResearch ScholarDepartment of PharmacyBITS-Pilani`
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• On 2 Dec 2011 at 08:46:04, Christopher Town (town.chris.aaa.sbcglobal.net) sent the message
`Hello Hari,Though the solution is not hard, it is not quite straightforward.  Thecomment by msgcot2 has merit, but beside the weight of the homogenatetaken, you would need the weight of the entire homogenate.  You couldalso use volume measurements for the homogenate and the sample analyzed. You indicate that the tissue was homogenized in 5 mL of homogenizingsolution.  One assumes that you then have 5 mL + 0.5 g of tissue, so thevolume of the homogenate is greater than 5 mL.  So you need to have anestimate of the total volume or mass of the total homogenate.  One mightassume that all tissue has a density of 1 but that in not correct, somass and volume may not be the same.  I suspect that 0.5 g of lungtissue and 0.5 g of muscle will have different volumes.  So to determinethe concentration of drug in a tissue from the information given, youwould use the following equation:((concentration in aliquot/volume of aliquot analyzed)*total volume ofhomogenate)/weight of tissue homogenized=concentration in tissue.Again you could use mass or volume for the size of the sample analyzedand the total homogenate.  Fabrice's estimate of 740.03 ng/g liver islikely pretty close.  However, Fabrice assumes that that total volume ofhomogenate is 5 mL.  I suspect that is is higher.  You would need thatadditional information to accurately calculate the concentration in thetissue.  You may have a further error in that the concentration in thetissue may not be homogeneous.  Your sample may have higher a higher orlower concentration than the total liver.Good Luck,ChrisChristopher Town, PhDHome EMAIL: town.chris.-at-.sbcglobal.net`
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• On 4 Dec 2011 at 00:34:01, Nirat Shah (niratshah.aaa.ymail.com) sent the message
`Dear Hari,          Please go through the below calculation carefully, you will get the tissueconcentration in ng/g appropriately.Consider 0.5 g of liver weight as 1 part. You have added  5ml of homogenising solutionwhich is equivalent to 10 part of tissue weight. Now add both the part to find dilutionfactor i.e., 1+10=11. Calculated concentration was found to be 74.03 ng/ml. Now multiplythis value with dilution factor i.e., 74.03*11=814.33 ng which is present in 0.5 g oftissue. So, 1g of liver tissue contains 814.33/0.5=1628.66 ng/g of liver tissue.I think the above calculation will give appropriate conc. per g of tissue. But to findexact conc in tissue always homogenize the whole tissue not the some part of tissue. Thiswill give exact amount of drug present in the tissueRegards,Nirat P. ShahScientist IIADME Dept.Torrent Research Centre,Gandhinagar-382428`
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• On 5 Dec 2011 at 13:36:29, (dipakmodi.aaa.lupinpharma.com) sent the message
`The following message was posted to: PharmPKDear Hari,I agree with Nirat, but when we are looking for exact concentration oftest article in tissue then we have to apply correction factor of testarticle in tissue residual blood fraction.It has greatly influence the results for the test article whose tissueto plasma distribution ratio is very high or very low in highly perfusedtissues like lung, liver, kidney, etc.Best Regards,Mr. Dipak ModiAssociate ScientistPharmacokinetics and MetabolismNovel Drug Discovery & DevelopmentLupin Research Park46A/47A, Village: Nande, Taluka: MulshiPune-411 042, India`
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• On 6 Dec 2011 at 14:54:28, Christopher Town (town.chris.at.sbcglobal.net) sent the message
`Hello Narat and Hari,I would maintain that the best estimate will be arrived at by measuring the volume or massof the whole homogenate and the corresponding volume or mass of the homogenate sample thatwas analyzed.  In adding 5 mL of a solution to 0.5 g of tissue and homogenizing, it is notlikely that you will obtain exactly 5.5 mL of homogenate.  There are fats, solids andother components in the sample that will likely prevent that.  However, using a factor of11 is going to give a better answer than a factor of 10, because you should obtain avolume closer to 5.5 mL than 5.0 mL.The remaining issue with your answer is the number of significant digits.  I know thatthis has been covered in previous discussions on this Listserve.  However, in multiplyingby 11, a number that you have little precision for, you went from 4 to 5 significantfigures in you answer.  In dividing by 0.5 (which you had a better value but not to 5 or 6significant figures) you went to 6 significant figures I think that when you make yourassumptions, you might want to give your answer as 1.6 X 10^3 ng/g liver tissue.  The ruleof thumb for significant figures (that I was taught) is that you should match the numberof digits in your answer with that from the factor with the lowest number of significantfigures. I think that two would be a reasonable number of significant figures in thiscase. I know that we all want to keep the number of decimal places constant throughout ourcalculations, but there are times, when for accuracy, we should admit that we can'tjustify that number of significant digits.  I would be interested in the comments ofothers on this issue.Best Regards,Christopher Town, PhDIndependent ConsultantHome EMAIL: town.chris.at.sbcglobal.net`
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