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Formula for determination of MDA concentration in fish tissue?
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+ Nutritional biochemistry
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Michael Margolies
Formula for determination of MDA concentration in fish tissue?
Dear Femi, That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1) OD of your sample is 0.189 0.189 = 156 x 1 x Concentration Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM 1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml (as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
Dear Femi, That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1) OD of your sample is 0.189 0.189 = 156 x 1 x Concentration Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM 1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml (as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
Dear Prof. Pardha-Saradhi Sir, you are such a wonderful person with a good heart. You have really solve my problem and simplify the whole thing for me. Thanks so much sir.
Dear Prof. Pardha-Saradhi Sir, you are such a wonderful person with a good heart. You have really solve my problem and simplify the whole thing for me. Thanks so much sir.
Dear Femi, Can you let me know the reference of protocol you are following? Authors might have given extinction coefficient of MDA-TBA complex/abduct based on their protocol. If you follow the protocol of Kwon et al., where 0.5% TBA in 20% TCA is used, then you can use the extinction coefficient of malondialdehyde-TBA at 532 nm given in their paper, which is 155 mM-1 cm-1. Extinction coefficient will vary based on reaction conditions (which include the way you are preparing reagents). In case if extinction coefficient is 'X' mM-1 cm-1 then concentration of MDA in a sample that is showing absorbance of 0.189 would be 1/0.189 micromoles/ml [if path length is one cm and in most of the spectrophotometric measurements it is one (unless it is otherwise specified)]. I am sure you are aware that one mM MDA will have one micromole MDA/ml [one mM solution will have one mmole in 1000 ml i.e. 1000 micromoles in 1000 ml (i.e. one micromole/ml)]. If extinction coefficient is not stated in the paper, you need to prepare MDA standard curve.
Dear Femi, Can you let me know the reference of protocol you are following? Authors might have given extinction coefficient of MDA-TBA complex/abduct based on their protocol. If you follow the protocol of Kwon et al., where 0.5% TBA in 20% TCA is used, then you can use the extinction coefficient of malondialdehyde-TBA at 532 nm given in their paper, which is 155 mM-1 cm-1. Extinction coefficient will vary based on reaction conditions (which include the way you are preparing reagents). In case if extinction coefficient is 'X' mM-1 cm-1 then concentration of MDA in a sample that is showing absorbance of 0.189 would be 1/0.189 micromoles/ml [if path length is one cm and in most of the spectrophotometric measurements it is one (unless it is otherwise specified)]. I am sure you are aware that one mM MDA will have one micromole MDA/ml [one mM solution will have one mmole in 1000 ml i.e. 1000 micromoles in 1000 ml (i.e. one micromole/ml)]. If extinction coefficient is not stated in the paper, you need to prepare MDA standard curve.
I am very much thank full to Pro.P.Pardha Saradhi for detail information on MDA calculation. I also got cleared about this problem. Sir thank you very much.
I am very much thank full to Pro.P.Pardha Saradhi for detail information on MDA calculation. I also got cleared about this problem. Sir thank you very much.
Dear Femi, That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1) OD of your sample is 0.189 0.189 = 156 x 1 x Concentration Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM 1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml (as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
Dear Femi, That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1) OD of your sample is 0.189 0.189 = 156 x 1 x Concentration Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM 1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml (as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
Hello sir...its really elaborated answer for the MDA conc calculation.. P.Pardha-Saradhi sir may you please mentioned here when wen got MDA nmol/ml then how it convert in nmol/g or nmol/mg? may please... Thank you
Hello sir...its really elaborated answer for the MDA conc calculation.. P.Pardha-Saradhi sir may you please mentioned here when wen got MDA nmol/ml then how it convert in nmol/g or nmol/mg? may please... Thank you
Dear Prof. Pardha-Saradhi Sir, lipid peroxidation was assayed by measuring malondialdehyde (MDA) formation as described by Sharma and Krishnamurthy (1968). extinction coefficient is 1.56 x 10 5 (raise to power of 5) M-1cm-1.
Dear Prof. Pardha-Saradhi Sir, lipid peroxidation was assayed by measuring malondialdehyde (MDA) formation as described by Sharma and Krishnamurthy (1968). extinction coefficient is 1.56 x 10 5 (raise to power of 5) M-1cm-1.
Dear Femi,
That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1)
OD of your sample is 0.189
0.189 = 156 x 1 x Concentration
Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM
1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml
(as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
Dear Femi,
That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1)
OD of your sample is 0.189
0.189 = 156 x 1 x Concentration
Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM
1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml
(as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
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i am getting confusion in calculating of nmol/ml MDA in to nmol MDA /g tissue. i used okhawa method
i am getting confusion in calculating of nmol/ml MDA in to nmol MDA /g tissue. i used okhawa method
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VOTE
Dear Prof. Pardha-Saradhi
Sir, you are such a wonderful person with a good heart. You have really solve my problem and simplify the whole thing for me. Thanks so much sir.
Dear Prof. Pardha-Saradhi
Sir, you are such a wonderful person with a good heart. You have really solve my problem and simplify the whole thing for me. Thanks so much sir.
More
VOTE
Dear Femi,
Can you let me know the reference of protocol you are following? Authors might have given extinction coefficient of MDA-TBA complex/abduct based on their protocol. If you follow the protocol of Kwon et al., where 0.5% TBA in 20% TCA is used, then you can use the extinction coefficient of malondialdehyde-TBA at 532 nm given in their paper, which is 155 mM-1 cm-1. Extinction coefficient will vary based on reaction conditions (which include the way you are preparing reagents).
In case if extinction coefficient is 'X' mM-1 cm-1 then concentration of MDA in a sample that is showing absorbance of 0.189 would be
1/0.189 micromoles/ml [if path length is one cm and in most of the spectrophotometric measurements it is one (unless it is otherwise specified)].
I am sure you are aware that one mM MDA will have one micromole MDA/ml [one mM solution will have one mmole in 1000 ml i.e. 1000 micromoles in 1000 ml (i.e. one micromole/ml)].
If extinction coefficient is not stated in the paper, you need to prepare MDA standard curve.
Dear Femi,
Can you let me know the reference of protocol you are following? Authors might have given extinction coefficient of MDA-TBA complex/abduct based on their protocol. If you follow the protocol of Kwon et al., where 0.5% TBA in 20% TCA is used, then you can use the extinction coefficient of malondialdehyde-TBA at 532 nm given in their paper, which is 155 mM-1 cm-1. Extinction coefficient will vary based on reaction conditions (which include the way you are preparing reagents).
In case if extinction coefficient is 'X' mM-1 cm-1 then concentration of MDA in a sample that is showing absorbance of 0.189 would be
1/0.189 micromoles/ml [if path length is one cm and in most of the spectrophotometric measurements it is one (unless it is otherwise specified)].
I am sure you are aware that one mM MDA will have one micromole MDA/ml [one mM solution will have one mmole in 1000 ml i.e. 1000 micromoles in 1000 ml (i.e. one micromole/ml)].
If extinction coefficient is not stated in the paper, you need to prepare MDA standard curve.
More
VOTE
I am very much thank full to Pro.P.Pardha Saradhi for detail information on MDA calculation. I also got cleared about this problem. Sir thank you very much.
I am very much thank full to Pro.P.Pardha Saradhi for detail information on MDA calculation. I also got cleared about this problem. Sir thank you very much.
More
VOTE
Dear Feli, You should also devise by your protein concentration if you want your MDA expressed in nmoles/mg of proteins.
Dear Feli, You should also devise by your protein concentration if you want your MDA expressed in nmoles/mg of proteins.
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Dear Femi,
That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1)
OD of your sample is 0.189
0.189 = 156 x 1 x Concentration
Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM
1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml
(as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
Dear Femi,
That means extinction coefficient is 156000 M-1cm-1 (that is156 mM-1cm-1)
OD of your sample is 0.189
0.189 = 156 x 1 x Concentration
Concentration = 0.189/156 = 0.00121154 mM = 1.21154 microM = 1211.54 nM
1211.54 nM (that means you have 1211.54 nmoles in 1000 ml). That means you have 1.21154 nmoles/ml
(as you have taken one ml of two ml solution for your assay , you need to multiply accordingly) (This is based on the information furnished by you along with your original question)
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Dear prof Pardha Saradhi
Sir what is the formula for TBA calucation whether it is
TBA =Absorbance /155*weight of sample
Dear prof Pardha Saradhi
Sir what is the formula for TBA calucation whether it is
TBA =Absorbance /155*weight of sample
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Dear Alok,
you need to elaborate on what sort of confusion you have. If you are in Delhi, you may visit my lab in University of Delhi.
Dear Alok,
you need to elaborate on what sort of confusion you have. If you are in Delhi, you may visit my lab in University of Delhi.
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VOTE
Hello sir...its really elaborated answer for the MDA conc calculation.. P.Pardha-Saradhi sir may you please mentioned here when wen got MDA nmol/ml then how it convert in nmol/g or nmol/mg?
may please...
Thank you
Hello sir...its really elaborated answer for the MDA conc calculation.. P.Pardha-Saradhi sir may you please mentioned here when wen got MDA nmol/ml then how it convert in nmol/g or nmol/mg?
may please...
Thank you
More
VOTE
Dear Prof. Pardha-Saradhi
Sir, lipid peroxidation was assayed by measuring malondialdehyde (MDA) formation as described by Sharma and Krishnamurthy (1968). extinction coefficient is 1.56 x 10 5 (raise to power of 5) M-1cm-1.
Dear Prof. Pardha-Saradhi
Sir, lipid peroxidation was assayed by measuring malondialdehyde (MDA) formation as described by Sharma and Krishnamurthy (1968). extinction coefficient is 1.56 x 10 5 (raise to power of 5) M-1cm-1.
More
VOTE
Dear Femi,
It’s my pleasure. Thanks for your compliments.
GOOD LUCK
Dear Femi,
It’s my pleasure. Thanks for your compliments.
GOOD LUCK
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