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purple sulfur bacteria carry out photosynthesis in the presence of \(\mathrm{H}_{2} \mathrm{O}\) and \({ }^{14} \mathrm{CO}_{2}\), but only if \(\mathrm{H}_{2} \mathrm{~S}\) is added and \(\mathrm{O}_{2}\) is absent. During photosynthesis, measured by formation of \(\left[{ }^{14} \mathrm{C}\right]\) carbohydrate, the bacteria convert \(\mathrm{H}_{2} \mathrm{~S}\) to elemental sulfur but do not produce \(\mathrm{O}_{2}\). What is the role of the conversion of \(\mathrm{H}_{2} \mathrm{~S}\) to sulfur? Why doesn't photosynthesis produce \(\mathrm{O}_{2}\) in these bacteria?

Short Answer

Expert verified
Purple sulfur bacteria use \(\mathrm{H}_{2} \mathrm{~S}\) as an electron donor, converting it to sulfur, to fix \(\mathrm{CO}_{2}\); no \(\mathrm{O}_{2}\) is produced as water isn't the donor.

Step by step solution

01

Understand Photosynthesis Mechanism

In typical photosynthesis (like in plants), water (\(\mathrm{H}_{2}\mathrm{O}\)) serves as an electron donor, which is split to release oxygen (\(\mathrm{O}_{2}\)). The electrons released are used to convert carbon dioxide (\({ }^{14} \mathrm{CO}_{2}\)) into carbohydrates.
02

Role of Hydrogen Sulfide (\(\mathrm{H}_{2} \mathrm{~S}\))

Purple sulfur bacteria use \(\mathrm{H}_{2} \mathrm{~S}\) instead of water as the electron donor in their photosynthesis process. When \(\mathrm{H}_{2} \mathrm{~S}\) is split, it releases electrons, which are then used to convert \({ }^{14} \mathrm{CO}_{2}\) into \([{ }^{14} \mathrm{C}]\) carbohydrates, while producing elemental sulfur as a byproduct.
03

Examining Sulfur Formation

The conversion of \(\mathrm{H}_{2} \mathrm{~S}\) to sulfur provides the necessary electrons for photosynthesis in these bacteria. This is similar to how water is split to provide electrons in plant photosynthesis, but it avoids oxygen production because sulfur is produced instead.
04

No Oxygen Formation In Purple Sulfur Bacteria

Since \(\mathrm{H}_{2} \mathrm{~S}\) is used instead of water, and the chemical reaction produces sulfur, elemental sulfur is formed instead of molecular \(\mathrm{O}_{2}\). Thus, no oxygen is released during the photosynthetic process in purple sulfur bacteria.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Donor
In photosynthesis, an electron donor provides electrons for energy conversion. Typically, plants use water (\( \mathrm{H}_{2} \mathrm{O} \)) as the electron donor which splits to release oxygen (\( \mathrm{O}_{2} \)). This process converts carbon dioxide into carbohydrates. However, purple sulfur bacteria take a different route.
Instead of water, they use hydrogen sulfide (\( \mathrm{H}_{2} \mathrm{~S} \)) as their electron donor.
  • Hydrogen sulfide's electrons play a key role in facilitating the conversion of carbon dioxide into carbohydrates.
  • This process occurs without producing oxygen.
In these bacteria, hydrogen sulfide's transformation, rather than water splitting, drives photosynthesis. This unique mechanism is critical in environments lacking oxygen, enabling these bacteria to thrive and carry out essential processes.
Hydrogen Sulfide
Hydrogen sulfide (\( \mathrm{H}_{2} \mathrm{~S} \)) serves as a crucial electron donor in the photosynthetic processes of purple sulfur bacteria. It replaces water in typical oxygen-producing photosynthesis.
Here's how it works:
  • When \( \mathrm{H}_{2} \mathrm{~S} \) is split, it releases electrons necessary for photosynthesis.
  • These electrons enable the conversion of \( { }^{14} \mathrm{CO}_{2} \) into \([ { }^{14} \mathrm{C}]\) carbohydrates.
  • This electron transfer doesn’t result in oxygen production.
Thus, in the absence of oxygen, \( \mathrm{H}_{2} \mathrm{~S} \) effectively substitutes water, ensuring that the bacteria can still perform photosynthesis. But instead of oxygen, it produces elemental sulfur as a byproduct.
Elemental Sulfur
When purple sulfur bacteria use hydrogen sulfide for photosynthesis, one of the byproducts is elemental sulfur. This process is different from the one in plants where oxygen gas is produced.
Key points about sulfur formation are:
  • The splitting of \( \mathrm{H}_{2} \mathrm{~S} \) provides the electrons needed for carbohydrate production.
  • The leftover matter from this reaction forms elemental sulfur.
Unlike plants that release oxygen, these bacteria's process culminates with sulfur rather than \( \mathrm{O}_{2} \) gas.
Elemental sulfur remains as solid yellow deposits, often seen in their habitats like sulfur springs. This distinction highlights the adaptive mechanisms of purple sulfur bacteria enabling them to sustain life in anaerobic conditions.

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Most popular questions from this chapter

Wavelengths The rate of photosynthesis in a green plant, measured by \(\mathrm{O}_{2}\) production, is higher when illuminated with light of wavelength \(680 \mathrm{~nm}\) than with light of wavelength 700 \(\mathrm{nm}\). However, illumination by a combination of light of 680 \(\mathrm{nm}\) and \(700 \mathrm{~nm}\) gives a higher rate of photosynthesis than light of either wavelength alone. Explain.

Chemistry of Malic Enzyme: Variation on a Theme Malic enzyme, found in the bundle-sheath cells of \(\mathrm{C}_{4}\) plants, carries out a reaction that has a counterpart in the citric acid cycle. What is the analogous reaction? Explain your choice.

Effect of Venturicidin on Oxygen Evolution Venturicidin is a powerful inhibitor of the chloroplast ATP synthase, interacting with \(\mathrm{CF}_{\mathrm{o}}\) and blocking proton passage through the \(\mathrm{CF}_{0} \mathrm{CF}_{1}\) complex. How would venturicidin affect oxygen evolution in a suspension of well-illuminated chloroplasts? Would your answer change if the experiment were done in the presence of an uncoupling reagent such as 2,4 dinitrophenol (DNP)? Explain.

how an inhibitor of electron passage through pheophytin would affect electron transfer through (a) photosystem II and (b) photosystem I. Explain your reasoning.

Energetics of Phototransduction During photosynthesis, pigment molecules in chloroplasts must absorb eight photons (four by each photosystem) for every \(\mathrm{O}_{2}\) molecule they produce, according to the equation $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NADP}^{+}+8 \text { photons } \rightarrow 2 \mathrm{NADPH}+2 \mathrm{H}^{+}+\mathrm{O}_{2} $$ The \(\Delta G^{\prime \circ}\) for the light-independent production of \(\mathrm{O}_{2}\) is 400 \(\mathrm{kJ} / \mathrm{mol}\). Assuming that these photons have a wavelength of \(700 \mathrm{~nm}\) (red) and that the light absorption and use of light energy are \(100 \%\) efficient, calculate the free- energy change for the process.

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