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


� 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.


� Types of organisms� Streptococcus: homofermentative

� Leuconostoc: heterofermentative

� Pediococcus homofermentative

� Lactobacillus; heterofermentative or homofermentative.


� 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 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.

tema 10a: fermentations

Documents