Tema 10A: Fermentations
Transcript of Tema 10A: Fermentations
Tema 10A: Fermentations
� Chapter 14 and Chapter 8 � Pages 383 - 402
Lactic Acid Bacteria
� Characteristics: Gram positive, carbohydrate users, proteolysis rare, nonmotile, non-spore forming
� Strict fermentors, � unable to synthesize cytochromes unless heme is
added.� catalase negative� oxidase negative
� Nutritionally fastidious� All make lactic acid (lactate) as predominant
end product
Lactic Acid Bacteria
� Types of fermentation
� Homofermentative: glucose to 2 lactic acids, 85-95% of glucose carbon in lactate
� Heterofermentative: glucose to 1 lactate, 1 ethanol, and 1 carbon dioxide, only 50% or less of glucose carbon in lactate.
� Types of products will define the pathway used and ATP made.
Lactic Acid Bacteria
� Types of organisms� Streptococcus: homofermentative
� Leuconostoc: heterofermentative
� Pediococcus homofermentative
� Lactobacillus; heterofermentative or homofermentative.
Lactic Acid Bacteria
� Streptococcus species:� Group A, flesh eating, toxin (superantigens and
pyrogenic)� Pneumonococci: polysaccharides
� Enterococcus: gut dwellers
� Lactococcus� natural fermentations� Lactic acid production: lowers pH, preserves and
precipitates proteins
Lactic acid bacteria� Homofermentative pathway� Uses Glycolytic pathway to make 2 pyruvates
from glucose� Overview:
� Activation-use 2 ATP� Make ß-carbonyl� C-C bond cleavage� Oxidation/reduction� Substrate-level phosphorylation
Glucose
Glucose-6-P
Homofermentative Pathway in Streptococci
ADPATP
Used 2 ATPMade 4 ATPNet ATP yield=2 ATP/glucose
Fructose-6-P
ADPATP
Fructose-1,6 bis P
Dihydroxyacetone-P 2 Glyceraldehyde-3-P
2 1,3-bisphosphoglycerate
2 3-phophoglycerate
2 2-phosphoglycerate
2 phosphoenolpyruvate
2 pyruvate
2 ADP
2 ATP
2 ADP2 ATP
2 NAD +2 NADH
2 H 2 O
Dihydroxyacetone-Pconverted to glyceraldehyde-3-PPathway shows 2 G-3-P's afterthis step.
Reoxidation of NADH
2 pyruvate2 NADH
2 NAD +2 lactate
Lactate dehydrogenase
ATP
ADP
CH 2
OH
OH OH OH
H HH
H
H
OC C C C CHO
Glycolytic Pathway for Glucose Metabolism
Glucose
Hexokinase or PTS systemATP
ADP
+
Glucose-6-P
Fructose-6-P
Fructose-1,6-bisP
Glyceraldehyde-3-P Dihydroxyacetone-P
G6P isomerase
Phosphofructo-kinase
Fructose-1,6-bis Paldolase
Triose isomeraseNow have 2 G3P's
CH 2
OH
OH OH OH
H HH
H
H
OC C C C C=3OP O
=3 OP O
H
CH 2
OH
OH OH
H H
H O
C C C C CH 2OH
=3OP O
H
CH 2
OH
OH OH
H H
H O
C C C C CH 2 PO 3=
CH 2
OH O
H H
C C=3 OP O
OH
H O
C C CH 2 PO 3=H O
ATP
ADP
Glycolytic Pathway for Glucose Metabolism
Glyceraldehyde-3-P(Metabolism of only one G3P is shown)
=3OP O CH 2
OH O
H H
C C
CH 2
OH
OH
C C S-Enz=3 OP O
CH 2
OH
OH
C C=3OP O O-PO 3
=
CH 2
OH
OH
C C=3 OP O OH
O-PO 3=
HO-CH 2
OH
C C OH
O-PO 3=
CH 2
O
C C OH
O
CH 3
O
C C OH
{ }
NAD +
NADH
PO 4=
H 2 O
ADP
ATP
1,3-bisphospho-glycerate
3-phosphoglycerate
2-phosphoglycerate
phosphoenolpyruvate
pyruvate
Triosephosphatedehydrogenase
Phosphoglyceratekinase
Phosphoglyceratemutase
Enolase
Pyruvatekinase
From the 3 and 4 carbonsof glucoseFrom the 1 and 6 carbons
of glucose
Mechanism of the isomeration reaction
H2CO PO3=
C
HCOH
HOCH
HCOH
H OH
OHC
O
H2CO PO3=
C
HCOH
HOCH
HCOH
H
H
C OH
H2CO PO3=
C
HCOH
HOCH
HCOH
O
HC OH
H
Glucose-6-P cis-enolateFructose-6-P
Isomerization Reaction:Creates an electron attracting keto group at the # 2 carbon
• H dissociates from C2• 2 electrons shift
to form cis enediol• H from hydroxyl group
dissociates• 2 electrons shift
to form keto group.• Forces electrons in
enol bond to shift to C1.
C-C bond cleavage: Aldolase Reaction
H dissociates from C4; 2 electrons shift to form cis enediolH from hydroxyl group (C4) dissociates2 electrons shift to form keto group.Forces electrons in enol bond to shift to C1.
H 2CO PO3=
C O
H 2CO PO 3=
HCOH
HOCH
HC HO
CHOH
C O-
H 2CO PO 3=
H 2CO PO3=
HCOH
CH O
+
C O
H 2CO PO 3=
C
H
OH
Mechanism of the aldolase reaction
Carbonyl betato Carbon with O
Enol formation
Dihydroxyacetone-P
Glyceraldehyde-3-P
In conclusion Streptococcus
� Uses glycolysis to degrade glucose to 2 pyruvates
� NADH’s made in pathway are reoxidized by reducing pyruvate to lactate
� NADH is key cofactor in oxidation reduction reactions
� ATP made solely by substrate level phosphorylation.
Heterofermentative Pathway� 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:� 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
phosphotransacetylase
acetaldehide dehydrogenase
alcohol dehydrogenase
PGALD dehydrogenase
PGA kinase
enolase
PGA mutase
pyruvate kinase
lactate dehydrogenase
R5P epimerase
G6P dehydrogenase
6PG dehydrogenase
phosphoketolase
hexokinase
Glucose
Glucose-6-PADPATP
[6-P-Gluconolactone]
6-P-gluconate
Ribulose-5-P
CO2
H2O
NADPH
NADP+
NADPHNADP+
Xylulose-5-P
Lactate
NADH NAD+
Acetyl-P
Acetaldehyde
Ethanol
Acetate
NADP+
NADP+
NADPH
NADPH
ATP
ADP
ATP yieldused 1 ATPmade 2 ATPnet yield= 1 ATP
Additional oxidation/reduction reactionsdecrease potential ATP yield.
Small amount
What would happen if the organism coulddivert electrons away from ethanol production?
Acetyl-CoA
CoA
P i
Glyceraldehyde-3-P
1,3-bisphosphoglycerate
3-phophoglycerate
2-phosphoglycerate
phosphoenolpyruvate
pyruvate
ADP
ATP
ADPATP
NAD+NADH
H2O
Heterofermentative Pathway in Leuconostoc sp.
G6P dehydrogenase
6PG dehydrogenase
6PG dehydrogenase
CO2
H 2 CO
HCOH
PO3=
HCOH
HOCH
HC OH
HC
O O
H 2 CO
HCOH
PO3=
HCOH
HOCH
HC
HC
O COOH
H 2 CO
HCOH
PO3=
HCOH
HOCH
HCOH
O
COOH
H 2 CO
HCOH
PO3=
HCOH
HCOH
CO
H 2 CO
HCOH
PO3=
H 2 COH
HCOH
C
H2O
NADP+
NADPH
NADPHNADP+
Glucose-6-P 6-P-glucono-lactone
(enzyme-bound)
6-P-gluconate
3-keto-6-P-gluconate
(enzyme-bound)Ribulose-5-P
Mechanism of beta-decarboxylation
CO 2
H +
C
C
C
O
O
O -
C
C
O -
H
OC
C
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.
1 23
O
H 2CO
HOCH
PO3=
H 2COH
HCOH
C
H 2CO
HC
PO3=
HCOH
O
O-PO3=
O
CH 3
C
O-PO3=
+
Phosphoketolase Reaction
Xylulose-5-P
Acetyl-Phosphate
Glyceraldehyde-3-P
E-TPP- CH CH 2
OHOH
E-TPP C CH 2
OH
O-PO3=
Acetyl-Phosphate
Enzyme contains Thiamin pyrophosphate (TPP) as cofactor the function here is transketolation.
Conclusion� 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
� 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: � mass is conserved
� must account for all of the carbon and electrons originally present in the substrate.
Fermentation analysis� 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 withinoculum.
Take time zero and time final samples and measure Glucose and product formation.
Example: Leuconostoc brevis
Compd. Amount (mmol)
# of C’s mmol of C
Glucose 100 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 amountdetected 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 areadded back to a compound derived from glucose.
Thus, the ratio of oxidized products to reduced productsmust equal 1.
Since glucose (C6H12O6) has 2 H’s for every O, productswith 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 forevery 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)
OR value
mmol(ox)
mmol(red)
Glucose 100 0 - -
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 determinedthis is multiplied by the amount detected (see Table)
• Calculate the O/R ratioOR ratio = |178.6/(-171.8)+(-6.8)|
OR ratio = 178.6/178.6 = 1.0
Ratios close to 1 mean all of the electronshave 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 mmolesC2 = 85.9 mmoles + 7.3 mmoles = 93.2 mmolesC1 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
� 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?
� 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.
� Some lactic acid bacteria possess enzymes that reoxidize NADH (and NADPH) by reducing oxygen to water (Dolin’s enzymes)
OxidaseNAD(P)H + H+ + O2 --> H2O2 + NAD(P)+
PeroxidaseNAD(P)H + H+ + H2O2 --> 2 H2O + NAD(P)+
What happens when oxygen is present?
� 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 ATPAcetate kinase
acetyl-P + ADP --> acetate + ATPFor every acetate made, one ATP is made by substrate-level phosphorylation by this reaction.
When Dolin’s enzymes and oxygen are present, 1) acetyl-P goes to acetate and ATP rather than to ethanol, and2) pyruvate is metabolized to acetate and CO2 rather than to lactate.
Pyruvate Acetyl-PAcetyl-CoA
CO2
Acetate
ATPADPNAD+ NADH CoAPi
Streptococcus sp. and Dolin’s enzymes
Glucose
2 Pyruvate
2 Acetate
2 Lactate
2 CO2
2 ATP net
2 ADP
2 NAD+
2 NADH
2 NADH
2 NAD+
No O2 or Dolin' s enzymes
Glucose
2 Pyruvate
2 ATP net
2 NAD+
2 NADH
2 NADH
2 NAD+
With O2 and Dolin's enzymes
2 Acetyl-CoA
2 Pi
2 CoA
2 Acetyl-P
2 CoA
2 ATP
2 O2
4 H2O 4 NAD+
4 NADH
Net of 4 ATP
Summary
� If there is an alternate electron acceptor, less lactate, more acetate, CO2, and ATP
Bifidobacterium sp.� 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.
Bonjoch, X., E. Ballesté, and A. R. Blanch. (2004) Multiplex PCR with 16S rRNA Gene-Targeted Primers of Bifidobacterium spp. To Identify Sources of Fecal Pollution. AEM V70(5):4054-60.
Matsuki, Takahiro, Koichi Watanabe, Ryuichiro Tanaka, and Hiroshi Oyaizu, 1998, "Rapid Identification of Human Intestinal Bifidobacteria by 16S rRNA-Targeted Species- and Group-Specific Primers," FEMS Microbiology Letters, Vol. 617, pp. 113-121.
Outline of pathway� Activation: 2 glucose to 2 fructose-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.
Uses these enzymes to interconvert hexoses and pentoses.
Phosphoketolase : C6 -> C4 + acetyl-P
Transaldolase: C6 + C4 -> C7 + C3
Transketolase:C7 + C3 -> C5 + C5
Phosphoketolase: C5 + C5 -> 2 C3 + 2 acetyl-P
Net Result: 2 C6 -> 2 C3 + 3 acetyl-P
C
C
C
C
C
C
O
C
C
C
C
CH3
C OO-PO3Pi
Phosphoketolase:C6 (or C5) + Pi -> C4 (or C3) + acetyl-P
C
C
C
C
C
C
O
C
C
C
C
+
C
C
C
O
C
C
C
C
C
C
C
+
Transaldolase:C6 + C4 -> C7 + C3
C
C
C
O
C
C
C
C
C
C
C
+C
C
C
C
C
C
C
C
C
C
+O O
TransketolaseC7 + C3 -> C5 + C5
2 Glucose
Fructose-6-P Fructose-6-P+
2 ATP
2 ADP
Erythrose-4-P
PiPhosphoketolase
Acetyl-P
Sedoheptulose -7-PGlyceraldehyde-3-P
Transaldolase
Xylulose-5-P Ribose-5-P
2 Xylulose-5-P
Ribulose-5-P
2 Glyceraldehyde-3-P
2 1,3-bisphosphoglycerate
2 3-phosphoglycerate
2 2-phosphoglycerate
2 phosphoenolpyruvate
2 pyruvate
2 lactate
2 H2O
2 Acetyl-P
2 NAD+
2 NADH
2 ATP
2 ADP
2 ATP
2 ADP
2 NADH
2 NAD+
3 Acetate
3 Acetyl-P
3 ATP
3 ADP
2 glucose -> 2 lactate + 3 acetate
ATP yield:
(7-2)/2 glucose =
2.5 ATP/ glucose
Chapter 8, Fig 8.11
Summary
� Make acetyl-P by phosphoketolaserather 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.