Transcript Ecological speciation model

Heterofermentative Pathway
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Uses part of the pentose phosphate pathway
Only one pyruvate is made
Have a decarboxylation and C-C cleavage to
give a C3 and a C2
Overview:
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Activation-use 1 ATP
Two oxidations done to make ß-carbonyl
C-C bond cleavage
G3P to pyruvate like in Streptococcus
Less ATP because more ox/red reactions
Heterofermentative Pathway in Leuconostoc sp.
Glucose
ATP
ADP
hexokinase
Glucose-6-P
NADP+
G6P dehydrogenase
NADPH
[6-P-Gluconolactone]
H2O
6-P-gluconate
NADP +
CO2
6PG dehydrogenase
NADPH
Ribulose-5-P
R5P epimerase
Xylulose-5-P
Glyceraldehyde-3-P
NAD+
PGALD dehydrogenase
NADH
1,3-bisphosphoglycerate
ADP
PGA kinase
ATP
3-phophoglycerate
phosphoketolase
NADPH
alcohol dehydrogenase
Acetaldehyde
NADP+
acetaldehide dehydrogenase
NADPH
Acetyl-CoA
Pi
phosphotransacetylase
CoA
Acetyl-P
ADP
ATP
Acetate
Small
What would happen if the organism could
divert electrons away from ethanol production? amount
ATP yield
used 1 ATP
made 2 ATP
net yield= 1 ATP
PGA mutase
2-phosphoglycerate
enolase
H2O
phosphoenolpyruvate
ADP
pyruvate kinase
ATP
pyruvate
Ethanol
NADP +
Additional oxidation/reduction reactions
decrease potential ATP yield.
NADH
NAD +
Lactate
lactate dehydrogenase
6-P-gluconate
HC
HC
OH
NADP+
HCOH
HOCH
HCOH
NADPH
G6P dehydrogenase
HC
H 2 CO
PO3=
HOCH
O
HCOH
HCOH
HC
HCOH
H 2 CO
PO3=
6-P-glucono- lactone
Glucose-6-P
HCOH
H2O
HOCH
O
HCOH
COOH
O
PO3=
H 2 CO
NADP +
(enzyme-bound)
6PG dehydrogenase
NADPH
COOH
CO 2
H 2 COH
C
C
O
HCOH
6PG dehydrogenase
O
HCOH
HCOH
HCOH
H 2 CO
HCOH
PO =
3
Ribulose-5-P
H 2 CO
PO3=
3-keto-6-P-gluconate
(enzyme-bound)
Mechanism of beta-decarboxylation
O
1
C
O-
C
C
CO2
2
3
C
C
H+
O-
O
1) Carbonyl accepts electrons from C-C
between the 1 and 2 carbons.
2) Carbon dioxide and an enolate are formed.
3) Re-shifting of the electrons forms the keto
sugar.
H
C
C
O
Phosphoketolase Reaction
CH 3
H 2 COH
C
C
O
+
O-PO 3=
HOCH
O
O-PO 3=
HC
HCOH
H 2 CO
Acetyl-Phosphate
O
HCOH
PO3=
H 2 CO
Glyceraldehyde-3-P
PO =
3
Xylulose-5-P
Enzyme contains Thiamin pyrophosphate (TPP) as cofactor the function here is transketolation.
E-TPP- CH
OH
CH 2
OH
E-TPP
C
OH
CH 2
O-PO3
=
Acetyl-Phosphate
Conclusion
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Heterofermentative organisms use a pathway
with a greater number of redox reactions than
Streptococcus.
Make very oxidized and very reduced
compounds.
More NAD(P)H to be reoxidized constrains
ATP synthesis, high energy intermediate used
as an electron acceptor.
Vitamins: essential portions of cofactors that
organism can not make de novo
Fermentation Analysis
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In order to understand how an organism
makes its energy or what biochemical
pathways are present, one must first
know what the products of metabolism
are.
First Law of Thermodynamics:
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mass is conserved
must account for all of the carbon and
electrons originally present in the
substrate.
Fermentation analysis
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From this, we can then figure out the
pathways and amount of ATP made.
Also, inspection of the products will allow us
to make predictions about the cell’s
metabolism.
Initially, we will look at glucose consumption in rich
medium
 Growth factors from media supply cell carbon
 Most of glucose goes to products, only 5-10%
incorporated into cells.
 In industry, one must also account for cell mass.
Experimental set up
Glucose added and inoculated
Control: inoculated but without glucose;
correct for products made from other
medium components or brought in with
inoculum.
Take time zero and time final samples and measure
Glucose and product formation.
Example: Leuconostoc brevis
Compd.
Glucose
Amount
(mmol)
100
# of C’s
mmol of C
6
600
Lactate
96.2
3
288.6
Glycerol
6.8
3
20.4
Ethanol
85.9
2
171.8
Acetate
7.3
2
14.6
CO2
89.3
1
89.3
Have we detected all of the
products? carbon
Calculate the carbon recovery by multiplying the amount
detected by the number of carbon atoms for each compound,
then sum up all of the carbon in the products.
Carbon in glucose = 6 X 100 mmoles =600 mmoles
Carbon in products = (288.6 + 20.4 + 171.8 + 14.6 +
89.3) mmoles
Carbon in products = 584.7 mmoles
% C recovery = (584.7 mmol/600 mmol) * 100
% C recovery = 97.4%
Have detected all of the
electrons?
In a fermentation, electrons removed from glucose are
added back to a compound derived from glucose.
Thus, the ratio of oxidized products to reduced products
must equal 1.
Since glucose (C6H12O6) has 2 H’s for every O, products
with more than 2 H’s per O have been reduced, and
products with less than 2 H’s per O have been oxidized.
OR value of a compound
• To calculate the OR value of a compound, give a
numerical score of +1 for every O and -1 for
every 2 H’s.
• Examples: Glucose (C6H12O6): 6O is +6, 12 H's is -6, 6-6=0
Lactate (C3H6O3): 3O is +3, 6H's is -3, 3-3=0
Acetate (C2H4O2): 2O is +2, 4H's is -2, 2-2=0
Glycerol (C3H8O3): 3O is +3, 8 H's is -4, 3-4 = -1
Ethanol (C2H6O): 1O is +1, 6 H is -3, 1-3= -2
Carbon dioxide (CO2): 2 O's = +2
Example: Leuconostoc brevis
Compd
Amount
(mmol)
Glucose
100
OR
value
0
mmol
(ox)
-
mmol
(red)
-
Lactate
96.2
0
-
-
Glycerol
6.8
-1
-
-6.8
Ethanol
85.9
-2
-
-171.8
Acetate
7.3
0
-
-
CO2
89.3
2
178.6
-
O/R ratio of the fermentation
• Once the OR value of the compound is determined
this is multiplied by the amount detected (see Table)
• Calculate the O/R ratio
OR ratio = |178.6/(-171.8)+(-6.8)|
OR ratio = 178.6/178.6 = 1.0
Ratios close to 1 mean all of the electrons
have been accounted for.
C1 to C2 ratio
• A common C-C cleavage reaction is C3 --> C1 + C2
usually indicating pyruvate is an intermediate.
• If this occurs in your organism, then expect
a C1/C2 ratio of 1.
C1 = 89.3 mmoles
C2 = 85.9 mmoles + 7.3 mmoles = 93.2 mmoles
C1 to C2 ratio = 89.3 mmoles/ 93.2 mmoles
C1 to C2 ratio = 0.96
Value is close to one so probably have pyruvate cleavage.
Conclusion
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Fermentation balance is the first step in
understanding the metabolism of an organism
Must have C recovery close to 100% and an
O/R ratio close to 1.
C1/C2 ratio indicates pyruvate cleavage
You can use the above information in the lab
to determine what analyses are needed to
complete the balance.
What happens if an alternate electron
acceptor is present in a fermentation?
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Electron flow dictates carbon flow and energy yield
Alternate electron acceptors provide fermentative
bacteria a “choice”
The result will be less lactate and ethanol and
more acetate and ATP are made.
We will study the effect of oxygen on the
metabolism of lactic acid bacteria
De Felipe et al., J. Bacteriol. vol 180, p 3804, 1998
Utilization of oxygen by facultative
lactic acid bacteria.
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Some lactic acid bacteria possess
enzymes that reoxidize NADH (and
NADPH) by reducing oxygen to water
(Dolin’s enzymes)
Oxidase
NAD(P)H + H+ + O2 --> H2O2 + NAD(P)+
Peroxidase
NAD(P)H + H+ + H2O2 --> 2 H2O + NAD(P)+
What happens when oxygen
is present?
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When oxygen and Dolin’s enzymes are
present, NAD(P)H is reoxidized by
reducing oxygen to water rather than
pyruvate to lactate or acetyl-P to
ethanol.
More acetate and ATP, less ethanol and
lactate, are made.
Make more ATP
Acetate kinase
acetyl-P + ADP --> acetate + ATP
For every acetate made, one ATP is made by substratelevel phosphorylation by this reaction.
When Dolin’s enzymes and oxygen are present,
1) acetyl-P goes to acetate and ATP rather than to ethanol, and
2) pyruvate is metabolized to acetate and CO2 rather than to
lactate.
CO2
Pyruvate
NAD+ NADH
Acetyl-CoA
Acetyl-P
Pi
CoA
Acetate
ADP ATP
Streptococcus sp. and Dolin’s
enzymes
No O2 or Dolin's enzymes
With O2 and Dolin's enzymes
Glucose
Glucose
2 NAD+
2 ATP net
2 NADH
2 Pyruvate
2 NADH
2 NAD+
2 Lactate
2 NAD+
2 ATP net
2 NADH
2 Pyruvate
2 CoA
2 NAD+
2 CO2
2 NADH
2 Acetyl-CoA
2 O2
4 NADH
2 Pi
2 CoA
4 H2O
2 Acetyl-P
2 ADP
2 ATP
2 Acetate
Net of 4 ATP
4 NAD+
Summary
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If there is an alternate electron acceptor,
less lactate, more acetate, CO2, and
ATP
Bifidobacterium sp.
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Bifid or 2 lobes
Gram positive, curved rod found in the feces
of breast-fed infants,
Requires many growth factors, including
N-acetylglucosamine
Makes 2 lactate and 3 acetate from 2 glucose
Makes high-energy intermediate by
phosphoketolase reaction rather than ox/red.
Schell, M. A. et. al. (2002). The genome sequence of Bifidobecterium longum reflects its adaptation to the
human gastrointestinal tract. PNAS V99 N22 p14422-14427.
Outline of pathway
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Activation: 2 glucose to 2 glucose-6-P uses 2
ATP
Make 2 G3P and 3 acetyl-P from 2 glucose
by transketolase, transaldolase, and
phsophoketolase reactions
2 G3P to 2 lactate by reactions seen in
Streptococcus
3 acetyl-P to 3 acetate and 3 ATP by acetate
kinase
Synthesis of 2 G-3-P and 3 C2
units from 2 glucose.
Phosphoketolase : C6 -> C4 + acetyl-P
Transaldolase: C6 + C4 -> C7 + C3
Transketolae: C7 + C3 -> C5 + C5
Phosphoketolase: C5 + C5 -> 2 C3 + 2 acetyl-P
Net Result: 2 C6 -> 2 C3 + 3 acetyl-P
Uses these enzymes to interconvert hexoses and pentoses.
C
C
C
CH 3
C
O
C
Pi
C
O
O-PO3
C
C
C
C
C
C
O
C
+
C
C
C
C
C
Phosphoketolase:
C6 (or C5) + Pi -> C4 (or C3) + acetyl-P
Transaldolase:
C6 + C4 -> C7 + C3
C
C
O
C
+
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
O
+
C
C
O
O
C
C
C
C
C
C
C
Transketolase
C7 + C3 -> C5 + C5
+
C
C
C
C
C
C
2 glucoses -> 2 lactate + 3 acetate
2 lactate
2 Glucose
2 ATP
2 NAD+
2 NADH
2 pyruvate
2 ADP
Fructose-6-P + Fructose-6-P
Phosphoketolase
Pi
Acetyl-P
2 ATP
2 ADP
2 phosphoenolpyruvate
Erythrose-4-P
2 H2O
Transaldolase
Sedoheptuloase-7-P
Glyceraldehyde-3-P
Ribulose-5-P
2 Xylulose-5-P
3 ATP
3 Acetate
ATP yield:
2 3-phosphoglycerate
Ribose-5-P
3 ADP
2 2-phosphoglycerate
2 ATP
Xylulose-5-P
3 Acetyl-P
2 Acetyl-P
2 ADP
2 1,3-bisphosphoglycerate
2 NADH
2 NAD+
2 Glyceraldehyde-3-P
(7-2)/2 glucose =
2.5 ATP/ glucose
Summary
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Make acetyl-P by phosphoketolase
rather than by ox/red reaction
Don’t have to use acetyl-P as electron
acceptor
ATP yield higher than other anaerobes
for this reason.
Avoidance of ox/red leads to higher ATP
gain.