Organic Degradation

Question 1

What is behind the vast majority of the chemical transformations in the carbon cycle?

Answer 1

Catalyzed by microorganisms.

Question 2

What are reservoirs?

Answer 2

Reservoirs are discrete compartments in the cycle that contain C in some characteristic form, simple (e.g., CO2) or complex (e.g., animals), and it is the form that defines the reservoir. Each reservoir contains a certain amount of C, with some reservoirs being larger than others. Thus reservoirs are measured as amounts. (e.g., Petagram (10^15 g) of C).

Question 3

What are fluxes?

Answer 3

Arrows. They represent the movement of C between two reservoirs. Fluxes often, but not always, involve chemical transformations, and a transformation may be the limiting factor for a flux. Thus, microbes are often critical to fluxes. Fluxes are measured as rates. (e.g., Pg of C per year).

Question 4

What is turnover?

Answer 4

A characteristic of a reservoir that is a function of the size of the reservoir and all of the fluxes into and out of the reservoir. Generally, major global reservoirs are constant or only very slowly changing in size, so the fluxes in approximately equal the fluxes out. Turnover is measured as a unit of time (e.g., a), indicating the number of years the average C atom remains in the reservoir. Turnover rates are also sometimes estimated.

Question 5

Describe the sediment and rock carbon reservoir.

Answer 5

The largest C reservoirs. Their fluxes tend to involve slow processes, like weathering of rock. These processes have low specific rates (e.g., g of C weathered per g of rock). However, because these reservoirs are very large, the overall fluxes are substantial (g of rock weathered globally per a). Because the sizes of these reservoirs are large relative to their fluxes, they have very low turnover.

Question 6

Describe the humus and petroleum carbon reservoir.

Answer 6

Large reservoir with low turnover. Humans increased flux of C from petroleum to CO2. Consequences: there is potential positive feedback from the increased flux of C to atmospheric CO2. The resulting warming may accelerate the flux of humus to atmospheric CO2 by melting permafrost and increasing its rate of decomposition, involving microbial respiration. Very substantial flux of petroleum to microorganisms via naturally occurring petroleum seeps.

Question 7

Describe the atmospheric CO2 carbon reservoir.

Answer 7

Small reservoir, high flux, high turnover rate. Historically most CO2 was from microbial decomposition, but now the human contribution is also significant. Atmospheric CO2 has increased 12% in the last 40 years. Fluxes out of the atmospheric CO2 reservoir are mainly from autotrophy–in plants on land and in micro-organisms in the oceans.In the oceans,there is net CO2 fixation, so the oceans are a sink for CO2.

Question 8

Describe wood decomposition.

Answer 8

Major component of the global carbon cycle, because most terrestrial biomass is wood. Wood decomposition affects many other global nutrient cycles (limiting nutrients). This decomposition also has applications in producing biofuels, chemical and bio-based materials. Wood decomposition is a very complex process, because of wood’s chemical complexity and recalcitrance. Indeed, wood has evolved to resist degradation. Many organisms are typically involved in decomposing wood. When a tree dies, its decomposition normally involves an ecological succession, with different fractions of the wood being degraded by different groups of microorganisms. Insects and other animals can also be important by physically breaking up the wood, increasing its surface are available to microorganisms.

Question 9

Describe lignocellulose.

Answer 9

Main structural component of trees. Lignin, cellulose, hemicellulose. Recalcitrant, rate-limiting step of C cycle. Hemicellulose and cellulose are polysaccharides. Ligninis a phenolic polymer surrounding these polysaccharides. Hemicellulose is a polymer of hexoses, pentoses and sugar acids. It interacts with cellulose via hydrogen bonds. Cellulose is a semi-crystalline structure comprised of chains of β(1,4)-linked glucose.

Question 10

What are the three groups of lignocellulose degrading organisms?

Answer 10

Wood-rotting fungi, cytophaga, actinomycetes.

Question 11

Describe wood rotting fungi.

Answer 11

Hyphal growth of these organisms occurs through channels in wood. Exoenzymes, secreted from the tipsof growing hyphae, degrade polymers extracellularly. Lignin degradation is the rate-limiting step in wood decomposition because lignin protects the less recalcitrant cellulose and hemicelluloses from degradation. Wood-rotting fungi do not seem to grow on lignin. Rather,they appear to degrade lignin to access cellulose andhemicelluloses, which they use as growth substrates.`

Question 12

Describe white rot fungi.

Answer 12

Extensively degrade lignin. Their degradation process bleaches the wood, giving rise to the term white rot.

Question 13

Describe brown rot fungi.

Answer 13

Preferentially degrade cellulose, leaving behind brown lignin. Lignin is only partly degraded in order to access the cellulose.

Question 14

Describe cytophaga.

Answer 14

They are long, slender rods with characteristic gliding motility. Cytophaga typically glide over the surface of their substrate, such as plant material, as they consume it.They have extracellular enzymes that are not released into their environment, but rather remain tethered to the cell surface. These enzymes are organized in a complex structure attached to the cell wall. The complex responsible for cellulose-deconstruction in Cytophaga is a cellulosome.

Question 15

Describe actinomycetes.

Answer 15

Actinomycetes are members of the high-GC, gram-positive bacterial phylum and are common in soil (and other) environments. Some members of this group grow on cellulose and hemicelluloses. Maybe degrade ligin, they can grow on lignin depolymerization products. Like fungi, Actinomycetes secrete exoenzymes to degrade polymers.

Question 16

Why is lignin recalcitrant?

Answer 16

High molecular weight, poor water-solubility and a random structure that includes aromatic rings connected by both ether bonds and C-C bonds

Question 17

Describe the lignin degradation mechanism of enzymatic combustion.

Answer 17

Free radical attack (nonspecific), aerobic, same as combustion. enzymatic production of small molecule oxidants, which depolymerize lignin via free radical reactions. These small molecules, or mediators,are able to move via diffusion and penetrate the structure of wood, in order to access and oxidize the lignin, generating aromatic radical cations in it. These radicals result in the breaking of ether and C-C bonds, leading to depolymerization, or “deconstruction”of the lignin. Bond breakage is relatively random, so a wide variety of depolymerization products are generated. Fungi appear to have relatively limited abilities to further degrade these products. However, this process disrupts the structure of wood and releases cellulose and hemicelluloses, which the fungi utilize as growth substrates.

Question 18

Describe the process of degradation of white rot.

Answer 18

White rot fungi such as P. chrysosporium utilize two types of exoenzymes to generate radicals in the lignin: peroxidases and laccases. Fungal lignin-degrading peroxidases include: lignin peroxidase (LiP), that utilizes veratryl alcohol as a mediator; manganese peroxidase (MnP), that utilizes manganese as a mediator; and versatile peroxidase (VP), that can use both and may oxidize lignin directly. For unclear reasons, white rot fungi can have multiple homologs of each exoenzyme (e.g., P. chrysosporium contains ten LiPs and five MnPs). However, these enzymes all function in the same basic way.

Question 19

Describe peroxidases and laccases.

Answer 19

Peroxidases are heme-containing enzymes that utilize peroxide, H2O2, to oxidize mediators. Laccases are copper-containing enzymes that utilize O2 to oxidize mediators. Fungi also possess enzymes that generate the H2O2 for peroxidases.

Question 20

Describe lignin peroxidase (LiP)

Answer 20

LiP exemplifies white rot peroxidases: it is a secreted heme protein and utilizes H2O2 to oxidize mediators such as veratryl alcohol (VA). An oxidase is often additionally secreted to provide the necessary H2O2.
Steps:
1. Peroxidase reaction. Peroxide is reduced to water and LiP is oxidized to (I). In the process, two electrons are lost by LiP to form LiP I.
2. Oxidation of first mediator molecule. The mediator is a small, soluble aromatic compound such as VA. One electron is abstracted from the aromatic nucleus to form a radical, VA+•, and LiP is partly reduced(II). The radical formed on the aromatic ring is highly reactive.
3. Spontaneous free radical chemical reactions ensue as the soluble radical, VA+•, diffuses and encounters high-molecular-weight lignin (or other substrates). In the ensuing reaction, VA+•, is reduced to VA and a radical cation is generated in the lignin. The VA can participate in another peroxidase cycle. The lignin radical is unstable, and undergoes further non-enzymatic reactions (a chain reaction) leading to bond scission.
4. Oxidation of second mediator molecule. LiP(II) catalyzes a second one-electron oxidation of VA, initiating further free radical reactions. The enzyme returns to resting state; LiP, and can react with another molecule of H2O2.

Question 21

Describe brown rot.

Answer 21

Similar to the white rot: the fungi produce highly-reactive, small chemical species that are responsible for lignocellulose degradation at a distance from the fungal hyphae. Produce peroxide that oxidatively activates the system. Moreover, aromatic radical cations are ultimately generated in the lignin, leading to bond breakage. Differences: no peroxidases and laccases, as well as many of the GHs.

Question 22

Describe chelator-mediated Fenton chemistry of brown rot.

Answer 22

In the current model, a process called chelator-mediated Fenton chemistry (CMF) is involved in the initial deconstruction of the lignin and the polysaccharides. In CMF, the fungi secrete various products into their extracellular matrix, including catechols such as variegatic acid(VA), hydroxyquinones (HQ), oxidoreductases and iron reductases (IR). Oxidoreductases produce peroxide. IR and VA reduce ferric iron to ferrous iron. In addition, VA chelates the iron. In the actual Fenton chemistry, ferrous iron reacts with peroxide to form hydroxyl radicals (•OH). Because hydroxyl radicals are so reactive, it is important that they are generated very close to their intended target. The chelation of the iron by VA helps to do this. It has been proposed that the hydroxyl radicals partially degrade the lignin, punching “holes” in it. The hydroxyl radicals then fragment the underlying cellulose and hemicellulose. The poly-saccharide fragments diffuse through the holes in the lignin where they become available to the fungi’s glycosyl hydrolases.

Question 23

Describe the structure of cellulose.

Answer 23

Gives wood its rigidity. It is a large, insoluble, crystalline polymer of glucose subunits. It differs from lignin in having a very regular (crystalline) structure. Although very different from the random structure of lignin, the crystalline structure of cellulose also causes recalcitrance. An important feature of cellulose is the β-1,4 glycoside bond linking monomers. The bond itself is not recalcitrant, but it gives the polymer a linear conformation, permitting the crystalline structure. Starch and glycogen, by contrast, have beta-1,4 glycoside bonds and are not crystalline. H-bonding is also a critical component of the crystalline structure.

Question 24

Describe the enzymes involved in degrading cellulose.

Answer 24

carbohydrate-active enzymes (CAZy). Most enzymes degarding cellulose are cellulases, which hydrolyse glycoside bonds. Since there is only one type of bond in cellulose, all cellulases are beta-1,4 glucanases. Lytic polysaccharide monooxygenases (LPMOs) also contribute to cellulose degradation. PMOs enhance degradation by cellulases by oxidatively cleaving cellulose and disrupting its crystalline structure. PMOs require reductant, which is believed to come from cellobiose dehydrogenation.

Question 25

Describe the mechanism of cellulose degradation.

Answer 25

Endoglucanases attack within the polymers while cellobiohydrolases remove two monomers from the ends of polymers. LPMOs can also attack within polymers and may be able to do so in more crystalline regions than endoglucanases. These enzymes act in concert to degrade the overall structure of cellulose microfibrils. One hypothesis is that the cellulose structure is first attacked at amorphous, non-crystalline regions. Degradation of crystalline regions must occur at the surface, so it may be limited by substrate availability. Process does not require O2 (except LPMOs) and can occur aerobically or anaerobically. The CAZymes are all exoenzymes, occurs extracellularly. Products are cellobiose (dimer) and glucose (monomer). Cellobiose and glucose are then transported into cells where they are readily fed into central metabolism.

Question 26

Describe the cellulosome.

Answer 26

an extracellular enzyme complex that is associated with the cell wall. These cells degrade cellulose when they are in direct contact with it.The enzymes in a cellulosome complex are distinct from the secreted ones, but they must be secreted prior to assembly into the cellulosome. The cellulosome has on the order of 14-18 polypeptides, including exo- and endoglucanases plus xylanases for hemicelluloses. There is also scaffoldin, which is a central protein that binds the substrate and organizes the catalytic peptides.

Question 27

Describe alkane biodegradation.

Answer 27

Relatively recalcitrant. Oxygenases are key enzymes. Following activation with an oxygenase, alkanes are further degraded via β-oxidation, another very broadly important catabolic mechanism.

Question 28

Describe oxygenases.

Answer 28

Incorporate oxygen from O2 gas into organic substrates, which destabilizes the organic molecules and facilitates their further degradation. Dioxygen is not very reactive: oxygenases activate it, thereby harnessing it for degradative processes. Monooxygenases (MOs) incorporate one O atom from O2 into their substrates. By contrast, dioxygenases incorporate both O atoms into their substrates. MOs are sometimes called hydroxylases, because they form alcohols (but, other processes without oxygenases can also hydroxylate substrates). MOs are also sometimes called mixed-function oxidases because they form water in addition to hydroxylating their organic substrate. MOs require an electron donor, such as NAD(P)H. Oxygenase reactions often serve to “activate” molecules, making them susceptible to further degradation. Oxygenases are common in aerobic heterotrophic microorganisms, with some bacteria able to producenumerous (>100) oxygenases.

Question 29

What is terminal oxidation?

Answer 29

1st step. an example of an MO and its key role in activating a substrate for further degradation. This particular example is an alkane monooxygenase. The electron donor for this MO is NADH.

Question 30

What is further oxidation?

Answer 30

2nd step. The alcohol produced by terminal attack is much easier to degrade than the original alkane. Typically, alcohols are oxidized to aldehydes, which are subsequently oxidized to acids. These sequential oxidations can be catalyzed by alcohol and aldehyde dehydrogenases, respectively. This is a common sequence within degradation pathways.

Question 31

What is beta-oxidation?

Answer 31

β-Oxidation is a general mechanism for fatty acid biodegradation. It is also frequently a part of pathways for degradation of complex organic structures. O2 is not required, so β-oxidation is also used by anaerobes. The process starts with activation using ATP and CoASH, which is only required once per fatty acid. Then, there is an oxidative cycle, with cofactors required to accept electrons. Each cycle ends with cleavage at the beta C atom, thus the name, beta oxidation. In each cycle, a C-2 unit, acetyl-CoA is removed. The acetyl-CoA usually goes to TCA cycle. The remaining fatty acyl-CoA undergoes another cycle of β-oxidation, and this is repeated until the entire hydrocarbon is chopped into C-2 units. An extra step is required for fatty acids with an odd number of C atoms. In organisms degrading hydrocarbons, most of the acetyl-CoA will usually be further oxidized via the TCA cycle, yielding CO2 as the product (mineralization) plus reduced cofactors to serve as electron donors for aerobic respiration. A smaller proportion of the acetyl-CoA is typically used as the main building block for anabolism in such organisms.

Question 32

Describe aromatic degradation.

Answer 32

Aromatic rings abundant in lignin and petroleum, relatively recalcitrant. Oxygenases important for aerobic degradation, but degradation can occur anaerobically too

Question 33

Describe convergent pathways of aromatic degradation.

Answer 33

Aromatic compounds tend to be degraded via “upper pathways” which converge on a small number of ring-cleavage pathways, or “lower pathways”. The arrangement resembles a funnel. The upper pathways often yield dihydroxy aromatic rings, called catechols, which are substrates for subsequent ring cleavage. The lower pathways tend to yield TCA intermediates, which feed into central metabolism for both fueling reactions and anabolism. Dioxygenases are critical in both upper and lower pathways.

Question 34

Describe ring hydroxylation of aromatic degradation.

Answer 34

Upper pathways often involve ring hydroxylation, which destabilizes the ring and makes it amenable to cleavage reactions. Ex. Benzene is attacked by a ring-hydroxylating dioxygenase. Ring-hydroxylating dioxygen-ases are a class of oxygenases that hydroxylate aromatic compounds. These enzymes also require a reductant, often NADH. Ring-hydroxylating dioxygenases often produce a cis-dihydrodiol, which is not aromatic. However, the subsequent reaction is usually catalyzed by a dehydrogenase that re-aromatizes the ring, yielding a catechol. Sometimes the dioxygenase displaces (removes) substituents, and sometimes a substituted catechol results.

Question 35

Describe ring cleavage of aromatic degradation.

Answer 35

Catechols are the substrate. A ring-cleaving dioxygenase is the key enzyme. Intradiol dioxygenases cleave the ring between the two hydroxyl substituents. Extradiol dioxygenases cleave the ring at a bond adjacent to the two hydroxyl groups. Again, this is a general class of enzyme. Ring-cleaving dioxygenases require O2 but do not require reductant. As with other dioxygenases, both atoms of oxygen are incorporated into the ring-cleavage product. The productis further degraded to TCA intermediates.

Question 36

Describe how microbial cell factories can valorize lignin.

Answer 36

Lignin is depolymerized to a mixture of aromatic compounds. The types of aromatics depends on the source of the lignin and the depolymerization approach. Uses biological funneling to exploit convergent nature of aromatic compounds. Uses genetic engineering. Ex. the intradiol cleavage of catechol yields muconate, a compound that can be readily transformed to adipic acid, a precursor used in the synthesis of nylon. The deletion of downstream enzymes yields strains that efficiently convert aromatic compounds to muconate.

Question 37

Describe methylotrophy.

Answer 37

Growth on C1 compounds, including methane, methanol, formaldehyde and formic acid. Technically, methylotrophy does not require autotrophy, but methylotrophs have a similar requirement to synthesize cell material from C1 compounds.

Question 38

Describe methanotrophy.

Answer 38

Methanotrophs grow on methane. Most appear to be specialists (only C1), important in aquatic systems, some are endosymbionts in mussels and sponges. These endosymbionts enable growth of their animal host on methane from methane seeps. Methanotrophs typically have complex internal membrane systems,which are the site of methane oxidation. Some of the key enzymes are membrane-associated. Interestingly, membranes of methanotrophs often contain abundant sterols, which is rare in prokaryotes.

Question 39

Describe the pathway of methylotrophy.

Answer 39

1. methane oxidation by methane monooxygenase, makes methanol
2. methanol oxidation to CO2. Many steps, uses dehydrogenases.
   The dehydrogenases in the pathway are coupled to quinone or NAD, and the electrons pass through the ETC, allowing energy conservation.

Question 40

Describe cometabolism.

Answer 40

Cometabolism refers to the degradation of two compounds by the same enzyme(s) whereby that the degradation of the second compound depends on the presence of the first compound. Cometabolism, or fortuitous metabolism, is distinct from simultaneous metabolism, where two compounds are degraded simultaneously by a bacterium by two different pathways. Usually, cometabolism does not support energy metabolism and is not beneficial to the responsible organism.

Question 41

Describe methane monooxygenases and cometabolism.

Answer 41

MMOs provide an example of cometabolism and its relevance to bioremediation. Some MMOs have low substrate specificity and can degrade certain recalcitrant pollutants. For example, MMO can oxidize TCE. Remember, TCE can be aerobically degraded, whereas PCE is not known to be aerobically degraded. More generally, some oxygenases can cometabolize compounds, including other pollutants. For example, some ring-hydroxylating dioxygenases can attack chlorinated aromatic compounds that do not support growth of the organism containing the enzyme.

Question 42

Describe some consequences of cometabolism.

Answer 42

When cometabolized, a pollutant won’t support growth of the population responsible for degradation. Furthermore, cometabolism can lead to toxic by-products such as reactive oxygen species (ROS). Finally, pollutants that are cometabolized are unlikely to induce expression of the necessary enzymes.

Question 43

What are the fundamental metabolic tasks?

Answer 43

Overall objective: Increase biomass
•scavenge nutrients, convert to usable fragments (central metabolites) and energy, synthesize building blocks, polymerize macromolecules, ultimately for growth (biomass synthesis)
•coordinate these processes with each other and other cellular processes (e.g., cell division) to optimize growth

Question 44

Name some relationships between growth rate and composition.

Answer 44

1. Mass increases with growth rate
2. Proportions of components change (RNA increases the most)
   Proteins require a lot of energy to make.

Question 45

What does protein synthesis need to coordinated with?

Answer 45

C and N metabolism, energy metabolism (ATP and GTP)

Question 46

What are fueling reactions?

Answer 46

provide energy (i.e., to meet biosynthetic metabolic needs); transform constituents of the external medium to critical metabolic intermediates.
catabolic: degrade compounds to metabolically useful fragments

Question 47

What are biosynthetic pathways?

Answer 47

Reform fragments from catabolism into building blocks.

anabolic: synonym.

Question 48

What are the two main functions of central metabolism?

Answer 48

1. convert growth substrates to useful energy

2. maintain pools of critical metabolites (i.e., fragments used to make cellular components)

Question 49

What are precursor metabolites?

Answer 49

12 of them.
•used to make all of 75-100 building blocks, coenzymes, and prosthetic groups
•always synthesized regardless of specific fueling reactions employed
•link between catabolism and biosynthesis
•catabolic reactions make them; biosynthetic reactions use them
•central metabolism: series of ~50 reactions that maintain the pools of the PMs

Question 50

What are alpha-keto acids?

Answer 50

pyruvate, oxaloacetate, alpha KG.
They provide carbon skeletons of amino acids.
3 of the PM
The levels of α-ketoacids are an intracellular signal that regulate the balance between C and N catabolism.
• build up of them triggers:
(a) inhibition of C catabolism (and uptake) and;
(b) stimulation of N catabolism (and uptake)

Question 51

What are some regulations of energy metabolism?

Answer 51

At high growth rates, ATP mainly synthesized through glycolysis: enables higher proportion of proteome to be used for ribosomes and metabolic enzymes (i.e., not for oxidative phosphorylation)
•ATP allosterically regulates many enzymes (not all known)
•high [ATP] inhibits PFK–catalyzed FBP production