Wednesday, November 13, 2013

065 - Hydrogenase does not confer significant benefits to Azotobacter vinelandii growing diazotrophically under conditions of glucose limitation

Remember the uptake hydrogenase? A byproduct of nitrogen-fixation is hydrogen gas, which allegedly would be a waste of energy, except Azotobacter vinelandii has an enzyme that oxidizes it and converts it into ATP to recoup some of the energy cost of nitrogen-fixation. That's the common story, anyway. But in science you can't always rely on common stories, so Kerstin Linkerhägner and Jürgen Oelze decided to test that hypothesis.

Actually Yates and others had already tested to see if hydrogenase-knockout mutants of Azotobacter chroococcum were impaired, and found that wild-type strains outcompeted them, at least in some conditions such as when carbon was limiting in the medium (019). But these results may be questionable because the wild-type could consume both its own hydrogen and that produced by the mutant when they were mixed together. This wouldn't often be the case in the wild though.

So Linkerhägner and Oelze grew A. vinelandii strain DJ (relatively wild-type) and a hydrogenase-knockout strain in carbon-limited chemostats. They tried two different concentrations of oxygen, low and high. In terms of biomass, the biomass levels from DJ at each oxygen concentration over a range of dilution rates matched rather well with the results these authors saw in 066 (two years later), though not perfectly. Interestingly, the mutant strain matched its parent perfectly. As D increased, the amount of biomass they produced increased. Presumably this would've leveled off at the lower oxygen concentration if they had tested higher D values, as in 066, but they stopped.

Their next figure is somewhat conceptually challenging, but I'll try to break it down:
Figure 2, Linkerhägner and Oelze 1995
It shows the inverse of the yield, Y (which is, in this case, grams of biomass produced per mole of substrate [glucose] consumed) on the y-axis, and the inverse of the dilution rate D on the x-axis. The different symbols are the two strains and two oxygen concentrations, but the strains are virtually identical, so all you need to know is that the top line is higher oxygen and the bottom is lower.

So what does this mean? Well, since Y is biomass per glucose, 1/Y must be moles of glucose consumed for each gram of biomass produced. Seems about right; 0.1 moles per gram, i.e. 18 g glucose per gram of biomass.

Since D is the rate at which the cells in the chemostat get diluted by in-flowing fresh medium, 1/D is called the retention time: the amount of time, in hours, for one full volume of culture to be replaced with fresh medium. So if there is 1 liter of culture in the reactor (for example; the paper doesn't clearly say), then 1/D is the time it takes for 1 liter of fresh medium to flow in and 1 liter of culture to be removed at the same rate.

So overall, what this graph is saying is that as the time taken to dilute out the culture increases, the cells consume more glucose to produce the same amount of biomass. In other words, as medium is added and removed more slowly, the cells' use of substrate for growth is less efficient.

What's the point of this graph? Well, you'll notice that the points make linear patterns that converge to a single Y-intersect. This intersect is at a 1/D of 0, meaning that it takes zero time to replace a single volume of culture; so D = infinity. Obviously this is impossible, and no cells could grow fast enough to keep up, so this is just a theoretical maximum. But the Y-value is non-zero; it's about 1/Y = 0.05 moles glucose per gram biomass. When you invert this to get Y = 20 g biomass per mole glucose, this is, in fact, the theoretical maximum yield for these cells in this medium. The most biomass they can produce from 180 g glucose is 20 g. 

The other thing valuable about this graph is the slope of the lines. Since slope is change in y over change in x, the value here is effectively 1/Y over 1/D, or D/Y, which is to say, the sucrose consumed per increase in biomass per hour. Somehow, from this, it's possible to derive the cells' maintenance coefficient, which is related to the minimum amount of substrate that the cells need to persist at all, even without growing. You can't decrease the amount of food you give them below a certain amount, because they will not be able to maintain their basic life support functions, let alone grow and divide.

I'm not sure how to derive the maintenance coefficient from this graph (have to do some more studying), but it's clear that, at the higher level of oxygen, the cells' yield drops more dramatically as the dilution rate slows down than at the lower level of oxygen. So higher oxygen decreases their efficiency of substrate utilization. According to the authors, the cells require 1 mmol glucose to maintain each g of protein at the lower oxygen, and 16 mmol at the higher. Big difference.

Since both strains were identical in the same conditions, neither seemed to have an advantage in theoretical yield or maintenance requirements over the other.

There's more to the paper. They measured the levels of each adenine nucleotide (ATP, ADP, and AMP) in each strain at three different dilution rates, and didn't see any significant differences between the strains.

They also estimated respiratory activities by measuring the oxygen and hydrogen going into and coming out of the reactor.
Figure 3, Linkerhägner and Oelze 1995

For hydrogen, not surprisingly, the wild-type produced much less hydrogen in all conditions than the hydrogenase-knockout strain. Hydrogen increased as D increased (since more glucose was being fed in at higher dilution rates). Oxygen being high or low didn't affect anything in the mutant, but the parent produced about 10x more hydrogen at higher oxygen (though it's hard to see here); probably this is because oxygen inhibits the hydrogenase. This confirms that the mutant is losing hydrogen and the wild-type is not, as expected. (Interestingly, while the widely-accepted figure for hydrogen produced per nitrogen fixed is 1 mole per mole, the mutant strain here consistently gave an amount averaging around 2 moles per mole. Not sure what to make of that.)

Both strains produced the same amounts of fixed nitrogen, increasing with increasing D but independent of dissolved oxygen. So hydrogenase wasn't protecting the nitrogenase from oxygen at all, as some have thought.

Finally, Linkerhägner and Oelze calculated the amount of oxygen consumed as the wild-type oxidized the hydrogen it produced. Not surprisingly, respiration in general was higher at higher dilution rates and also at the higher level of oxygen. The parent and mutant strains' respiration rates were pretty similar; the calculated amount that respiration with hydrogen contributed to total respiration was quite low, around 3% at the lower oxygen and 0.5% at the higher. This is somewhat surprising, because the maintenance requirement is higher at higher oxygen, so the hydrogenase should be more helpful. On the other hand, it wouldn't work as well because of the oxygen.

Since the hydrogenase contributes so little to the cells' respiration, it seems like it doesn't help the cells very much in terms of recouping energy costs of nitrogenase. Maybe the hydrogen respiratory chain is more efficient than the normal one, generating more ATP for the same amount of proton motive force, but the authors cite some papers allegedly suggesting that it isn't.

So here's another example of these authors challenging the paradigm of the time, saying that hydrogenase isn't very helpful for the cells' energy metabolism. Their data seems pretty convincing, but I could be missing something. It'd probably be wise to check out papers that cite this paper and 066, to see what other researchers have to say about these results.

Citation: Linkerhägner, K. & Oelze, J. Hydrogenase does not confer significant benefits to Azotobacter vinelandii growing diazotrophically under conditions of glucose limitation. J. Bacteriol. 177, 6018–6020 (1995).

No comments:

Post a Comment