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I 've an essay about wheat sprout extract, there's a graph about protein content in first 4 day of wheat sprout. so under the graph there's this text

enter image description here

Phosporylation by endogenous kinases of the macromolecules contained in 1 g of the aqueous wheat sprouts extract. Results are expressed as cpm/mg of protein±SEM at 4 different days of germination. 0 are unsprouted seeds

so my problem is to know how many mg (milligrams?) is content for 1 g of protein in few words I need how many mg of protein is content in 1g of products, a conversion from cpm/mg to mg ... is possible?

cpm = counts per minute

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closed as unclear what you're asking by anongoodnurse, David, AliceD, WYSIWYG Dec 21 '16 at 8:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is cpm (counts per minute)? It does not seem like a common abbreviation. $\endgroup$ – WYSIWYG Dec 12 '16 at 6:57
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    $\begingroup$ I feel like the question isn't written very clearly, and you would get better answers if you can clarify a bit. Unfortunately I can't quite get what you are trying to accomplish so I'm not sure how to make helpful edits. The graph you are displaying is about phosphorylation levels, not content, so maybe that is part of the confusion? I.e., the graph seems to be already normalized to protein content, so from just this information you know nothing about the total protein content on various days, just the relative levels of phosphorylation. $\endgroup$ – Bryan Krause Dec 14 '16 at 19:03
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    $\begingroup$ Probably the graph is from this essay. However I fail to understand are you asking for the unit conversion 1000 miligram = 1 gram? or something else more complicated about enzyme and reaction-rate from the graph? $\endgroup$ – Always Confused Dec 14 '16 at 20:44
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    $\begingroup$ @LXG But that paper just shows there is more phosphorylation at day 2, nothing about the quantity of proteins. $\endgroup$ – Bryan Krause Dec 14 '16 at 22:10
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    $\begingroup$ So you're saying that with this data is just impossible to find the quantity of proteins for 1g or percentange ? and you're saing that the 4 day I have more proteins content that in the seconds ? thanks $\endgroup$ – LXG Dec 15 '16 at 17:35
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E. Cpm to fmol/mg: Enter the specific radioactivity as cpm/fmol, the number of cpm counted, and the protein content of the sample in mg. The result is the number of binding sites in fmol/mg protein.

here is a guide of conversions.

The measure is a relative measure, and it can't be very precise as a discrete measure without knowing the precise measurement conditions, i.e. is it mixed every 5 minutes or every 1 minute, is it dissolved in 1 liter or 100ml? CPM denotes the reactivity and not the weight. You are more likely to get a Mol/L rating associated with a CPM, and two vials with the same CPM rating can have different Mol/L and Mg/L depending on their reactivity.

So, if your paper gives you a comparative measurement, and if it gives you a specific measurement at some stage you should be able to multiply it and figure out the numbers for the various days.

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  • $\begingroup$ ok i've tried but no luck I can't convert 299,000 cpm/mg in 1g to mg can U help me? thanks maybe this can help : Figure 2 shows the phosphorylation by endogenous kinase of the macromolecules contained in 1 g of the aqueous wheat extract from seeds sprouted for four days: 1day163900±17911 cpm/mg of protein; 2day 299,000±35,834 cpm/mg of protein; 3day 120530±11940 cpm/mg of protein; 4 day $\endgroup$ – LXG Dec 18 '16 at 18:13
  • $\begingroup$ Perhaps try a pharmacy forum perhaps, someone who has done a thesis in applied biochemistry can probably say it better. also the chemistry forum, i will have a go tomorrow if you send a comment, my eyes are blurry at the moment! $\endgroup$ – com.prehensible Dec 20 '16 at 0:50

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