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I am trying to reconcile data that I have found in one publication (Allen 1969) with data that I found in another publication (George 2003) that synthesized this data. The data is root respiration rate, it was originally measured at $27\ ^\circ C$.

Approach

I am trying to convert a rate of oxygen consumed as volume per mass of root per time to carbon dioxide produced as mass per unit mass per time.

In the appendix table, George 2003 reports the range of root respiration rates, converted to $15\ ^\circ C$ and standard units:

$$[11.26, 22.52] \frac{\mathrm{nmol CO}_2}{\mathrm{g}\ \mathrm{s}}$$

In the original publication Allen (1969), root respiration was measured at $27\ ^\circ $C. The values can be found in table 3 and figure 2. The data include a minimum (Group 2 Brunswick, NJ plants) and a maximum (Group 3 Newbery, South Carolina), which I assume are the ones used by George 2003:

$$[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$

Step 1

Transformed George 2003 measurements back to the measurement temperature using a rearrangement of equation 1 from George, the standardized temperature of $15\ ^\circ $C stated in the Georgeh table legend, and Q$_{10} = 2.075$ from George 2003, and the measurement temperature of $27\ ^\circ $C reported by Allen 1969:

$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$

$$[11.26, 22.52] * exp(log(2.075)*(27 - 15)/10)$$

Now we have the values that we would have expected to find in the Allen paper, except that the units need to be converted back to the original:

$$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$

Step 2: convert the units

Required constants:

  • inverse density of $\mathrm{O}_2$ at $27^\circ C$: $\frac{7.69 \times 10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$ first assume that Allen converted to sea level pressure (101 kPa), although maybe they were measured at elevation (Allen may have worked at \~{} 900 kPa near Brevard, NC)
  • molar mass of $\mathrm{O}_2$: $\frac{32\mathrm{g}\ \mathrm{O}_2}{\mathrm{mol}}$
  • treat 10mg, which is in the unit of root mass used by Allen, as a unit of measurement for simplicity

Now convert $$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$ to units of $\frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{root}\ \mathrm{h}}$. The expected result is the original values reported by Allen: $[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$

$$[27.03, 54.07]\ \frac{\mathrm{nmol}\ \mathrm{CO}_2}{\mathrm{g}\ \mathrm{root}\ \mathrm{s}} \times \frac{1\ \mathrm{g}}{100\times10\mathrm{mg}} \times \frac{3600\ \mathrm{s}}{\mathrm{h}} \times \frac{3.2 \times 10^{-8}\ \mathrm{g}\ \mathrm{O}_2}{\mathrm{nmol}\ \mathrm{O}_2}\times \frac{7.69\times10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$$

Result:

$$[23.8, 47.8] \frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$

These are the units reported in the Allen paper, but they appear to be underestimates . Since the ratio of observed:expected values are different, it is not likely that Q$_{10}$ or the atmospheric pressure at time of measurement would explain this error.

Question

Am I doing something wrong?


  • Reference 1: Allen, 1969, Racial variation in physiological characteristics of shortleaf pine roots., Silvics Genetics 18:40-43
  • Reference 2: George et al 2003, Fine-Root Respiration in a Loblolly Pine and Sweetgum Forest Growing in Elevated CO2. New Phytologist, 160:511-522

Footnote 1: The values from reference 2 are adjusted from the $15^\circ C$ reference temperature to the $27^\circ C$ in reference 1 using the Ahhrenius equation, but I am off by an order of magnitude so I do not think that this is relevant:

$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$

$$[26.9, 54.0] = [11.2, 22.5] * exp(log(2.075)*(27 - 15)/10)$$


note: I have been updating the equation based errors pointed out by Mark and rcollyer, but the problem remains

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Parts of this don't make sense. The $1000g/g$ is probably a typo, but how could the final answer have grams divided by kilograms? –  Mark Eichenlaub Apr 1 '11 at 6:41
    
@Mark the answer is grams of CO$_2$ respired per kg root per second. The 1000g/kg is a typo –  David Apr 1 '11 at 14:14
    
where does "root" come from? –  rcollyer Apr 3 '11 at 2:22
    
@rcollyer it is the plant root that is consuming O2 or producing the CO2. I put it in following Marks question above, but I will take it out since it seems to magically appear in the answer, and the answer can be found here. –  David Apr 3 '11 at 15:43

3 Answers 3

As far as I can tell, you went wrong in a couple of places:

  1. instead of converting from $\mathrm{g}$ to $\mathrm{mg}$ your converting to $\mathrm{kg}$ in your first term, and
  2. in your fourth term, the units should be $\mathrm{g\, O_2}$ and $\mathrm{ nmol\, O_2}$, for consistency. This would've let you know that you've inverted your conversion from $\mathrm{ nmol\, CO_2}$ to $\mathrm{ nmol\, O_2}$ in your third term.

When I'm performing unit conversions, I do a couple of things to make my life easier. First, like it is displayed in your question, I write everything on two lines so that I can easily distinguish between the numerator and denominator, but I doubt that is at issue here. Second, like you, I tend to do them all at once, but occasionally, like this case, it may be worthwhile to do it piecemeal. Last, I convert systematically, working from left to right. For instance, in this case, I'd convert from $\mathrm{ nmol\, CO_2}$ to $\mathrm{\mu L\, O_2}$ first, and then add in the conversions from $\mathrm{g}$ to $\mathrm{mg}$ and $\mathrm{s}$ to $\mathrm{h}$ after that.

Edit: in your most recent revision, you've removed you conversion from $\mathrm{CO_2}$ to $\mathrm{O_2}$, so the units you get are

$$\frac{ \mu L\, \mathrm{CO_2}}{10 g h}$$

If I remember correctly, the conversion was a power of 2, so it will not fix the deviation between the two units.

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I didn't notice the answer below explaining the conversion from $\mathrm{CO_2}$ to $\mathrm{O_2}. For consistency sake, you may wish to explicitly include it. Not having read the papers, I'd have agree with Roy on the temperature conversion. –  rcollyer Apr 8 '11 at 19:35

There are a few problems with this formula and calculation. I shall present those that I have found below and hopefully this will help.

First we note that the intended conversion ratios from the data: old = [27.2,56.2]; new = [11.26,22.52] give a conversion factor between new/old = [0.414,0.401]. So perhaps because of rounding errors one is seeking a conversion factor between these two values, of perhaps around 0.41.

This conversion is done in two logically distinct steps: Unit Conversion (U say) - your step 2- and Temperature Conversion (R say) - your step 1. Both of these conversions contain issues.

Step 1 - Temperature Conversion.

The problem here is not so much the formula (as far as I know), but the assumption of $27^0$C. The Allen paper does do plant tests at $27^0$C and also $30^0$C, $35^0$C and $40^0$C. So which value to choose? The main issue for me however is figure 2 in that paper where the data (from table 3) are plotted against mean annual temperature in degrees Fahrenheit ie the true natural ground temperature. This range is $53^0$F to $67^0$F ie [11.6,19.4] Centrigrade range with median at $15.5^0$C. So it is possible that George has taken $15.5^0$C - which gives an R value of 1.04 or even just $15^0$C giving an R value of 1. If that were true it would largely cancel out the error you have found.

Step 2 - Unit Conversion.

There are some problems here however which complicates things.

First the density of $O_2$ would need to change to that at (say) $15^0$C. Now the figure quoted there is not the density but the inverse density (and there is a typo in the microunit). Using the ideal gas law the inverse density would change to 7.37 microunits from your value of 7.69 microunits - a decrease in this term.

Another problem are the errors in the "respiration" formula. This formula does not balance on the Oxygen: LHS=4, RHS=6. Furthermore the LHS chemical is Formaldehyde! So there must be at least a typo here - or perhaps the whole idea of using respiration formulae is wrong here? As this is just a unit conversion it may not rely on the details of plant respiration but just be based on atomic weights (8 mol $CO_2$ = 11 mol $O_2$)??

EDIT I have just seen what is wrong with this chemical formula: it is intended to be an "empirical formula" reducing from the glucose respiration formula (in Wikipedia). So it is 2$O_2$ on the LHS to balance. In that case it is 1 mol $O_2$ = 1 mol $CO_2$.

The George paper does not explain how the $O_2$ to $CO_2$ conversion works and so I dont know how to take this any further, but hopefully the above is helpful.

(Incidentally to "reverse engineer" the formula under such circumstances you might also check the conversion factor implied by the other cited papers in Appendix 1 where it was maybe clearer what was being converted.)

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Thank you for these pointers. I have updated my calculation to account for the 1:1 ratio of O$_2$:CO$_2$, and changed density to inverse density (but the calculation remains the same). The results of the experiment with T at $30,35,40$ are reported as response ratio, e.g. normalized to respiration at T=30 and reported as a percent. So that data can't be converted to a rate as far as I can see (without more assumptions). And it is not likely that he reports measuring a rate at 27C but then adjusts it to 15C without stating it (or the equation used) in the manuscript. –  David Apr 7 '11 at 7:16
    
@David, I agree with the last sentence. The question (which I cannot answer) is whether George just took the results from figure 3 (presented as they were against actual ground temperatures of around 15C) as the basis of doing the conversion. –  Roy Simpson Apr 12 '11 at 9:11

After making the corrections suggested by rcollyer, you were very close. Looking back to the original paper, note that the units were reported in the unexpcted units of $$\frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$

There is a 10 in the denominator, so multiply your answer by 10 and your calculations are pretty close. enter image description here

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