Fermentation Analysis

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Bruce Owens
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Fermentation Analysis In order to understand how an organism makes its energy or what biochemical pathways are present, one must first know what the

1 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.

2 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.

3 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.

4 Example: Leuconostoc brevis Compd. Amount (mmol) # of C s mmol of C Glucose Lactate Glycerol Ethanol Acetate CO

5 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 = ( ) mmoles Carbon in products = mmoles % C recovery = (584.7 mmol/600 mmol) * 100 % C recovery = 97.4%

6 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 (C 6 H 12 O 6 ) 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.

7 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 (C 6 H 12 O 6 ): 6O is +6, 12 H's is -6, 6-6=0 Lactate (C 3 H 6 O 3 ): 3O is +3, 6H's is -3, 3-3=0 Acetate (C 2 H 4 O 2 ): 2O is +2, 4H's is -2, 2-2=0 Glycerol (C 3 H 8 O 3 ): 3O is +3, 8 H's is -4, 3-4 = -1 Ethanol (C 2 H 6 O): 1O is +1, 6 H is -3, 1-3= -2 Carbon dioxide (CO 2 ): 2 O's = +2

8 Example: Leuconostoc brevis Compd Amount (mmol) OR value mmol (ox) mmol (red) Glucose Lactate Glycerol Ethanol Acetate CO

9 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.

10 C 1 to C 2 ratio A common C-C cleavage reaction is C 3 --> C 1 + C 2 usually indicating pyruvate is an intermediate. If this occurs in your organism, then expect a C 1 /C 2 ratio of 1. C 1 = 89.3 mmoles C 2 = 85.9 mmoles mmoles = 93.2 mmoles C 1 to C 2 ratio = 89.3 mmoles/ 93.2 mmoles C 1 to C 2 ratio = 0.96 Value is close to one so probably have pyruvate cleavage.

11 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. C 1 /C 2 ratio indicates pyruvate cleavage You can use the above information in the lab to determine what analyses are needed to complete the balance.

12 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; 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

13 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) Oxidase NAD(P)H + H + + O 2 --> H 2 O 2 + NAD(P) + Peroxidase NAD(P)H + H + + H 2 O 2 --> 2 H 2 O + NAD(P) +

14 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.

15 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 CO 2 rather than to lactate. CO 2 Pyruvate Acetyl-CoA Acetyl-P Acetate NAD + NADH CoA P i ADP ATP

16 Streptococcus sp. and Dolin s enzymes No O 2 or Dolin's enzymes Glucose 2 NAD + With O 2 and Dolin's enzymes Glucose 2 NAD + 2 ATP net 2 NADH 2 ATP net 2 NADH 2 Pyruvate 2 NADH 2 Pyruvate 2 CoA 2 NAD + 2 NAD + 2 Lactate 2 CO 2 NADH 2 2 Acetyl-CoA 2 O 2 4 NADH 2 P i 2 CoA 2 Acetyl-P 2 ADP 2 ATP 4 H 2 O 4 NAD + Net of 4 ATP 2 Acetate

17 Summary If there is an alternate electron acceptor, less lactate, more acetate, CO 2, and ATP

Tema 10A: Fermentations. Chapter 14 and Chapter 8 Pages

Tema 10A: Fermentations. Chapter 14 and Chapter 8 Pages Tema 10A: Fermentations hapter 14 and hapter 8 Pages 383-402 Lactic Acid Bacteria haracteristics: Gram positive, carbohydrate users, proteolysis rare, nonmotile, non-spore forming Strict fermentors, unable