Microbiology MOOC Newsletters: February 2012

Daily Newsletter February 24, 2012

Daily Challenge: This is the end of the genome to metabolome unit. Today would like each of you to consider the concept of metabolic engineering. Your challenge to day is to sum up the concept of metabolic engineering, and to reflect upon your feelings regarding this type of science. There are no right answers, just a reflection of your personal opinions. Make sure though that you define and describe metabolic engineering.

Administrative Note: Open Assignments
Remember, you still have the Certified Associate in Project Management (CAPM) and the Responsible Conduct in Research (RCR) assignments to complete. Do not wait until the last day to finish these assignments. Together they are worth 140points.

Daily Newsletter February 22, 2012

Special Reminder: From the syllabus, you have to complete the RCR training and the CAPM training. These are online tutorials that will help you as you move forward toward professions or graduate school. Make sure to set aside time to finish these.

Daily Challenge: Here is a project for you to work on.

Some bacteria have the ability to produce large carbon polymers for energy storage that appears as inclusion bodies within the cell. Polyhydroxyalkanoates (PHA) is a commercially important linear polyester that is produced by certain bacteria. PHAs can be used as either a thermoplastic or an elstomeric material, and is biodegradable. Many biodegradable plastics are now made from PHA.

You have been asked to maximize the production of PHA. One pathway for PHA biosynthesis involves the use of acetyl CoA:

It is known that b-ketothiolase is inhibited by unbound CoA. Coenzyme A is unbound when there is free NAD⁺ in the cell. All three proteins in this pathway are constitutively produced.

You have three strains of Bacillus subtilis (Gram +, neutrophile, mesophile, chemoorganoheterotroph) that are capable of PHA biosynthesis:

· GSU14PHA – has a knock-out of the gene for NADH oxidase (the first step in the electron transport chain), is a fermentative anaerobe, and a generation time of 90 minutes.

· GSU92PHA – is a facultative anaerobe with a pyruvate decarboxylase knockout. It has a generation time of 15 minutes. This strain requires that the fermentation conditions change when you reach production level population by going to a low oxygen level.
· GSU03PHA – is a facultative anaerobe that was the first in house successful PHA producer. It has a generation time of 10 minutes. This strain requires that the fermentation conditions change when you reach production level population by going to a low oxygen level, adding excess carbon, with minimal nitrogen and phosphorus (phosphate).

Below you will find metabolism and transcriptional information for the three strains. The organism will begin producing PHA when the population density has reached 10⁹ cells/ml. The first number describes the "efficiency" the enzyme has in turning substrate to product. So if I have 100g of a substrate, and the enzyme is 90% efficient, then I will get 90g of product. The second number is the maximum reaction rate (Vmax) for this reaction.

Enzyme	GSU14PHA	GSU92PHA	GSU03PHA
Hexokinase	90% / 10⁸ M sec^-1	90% / 10⁹ M sec^-1	90% / 10⁸ M sec^-1
Phosphoglucose isomerase	100% / 10⁶M sec^-1	100% / 10⁸ M sec^-1	100% / 10⁷ M sec^-1
phosphofructokinase	100% / 10⁹ M sec^-1	100% / 10⁸ M sec^-1	100% / 10⁷ M sec^-1
aldolase	100% / 10⁸ M sec^-1	100% / 10⁷ M sec^-1	100% / 10⁷ M sec^-1
triose phosphate isomerase	90% / 10⁷ M sec^-1	90% / 10⁶ M sec^-1	90% / 10⁸ M sec^-1
glyceraldehyde phosphate dehydrogenase	60% / 10⁷ M sec^-1	75% / 10⁹ M sec^-1	90% / 10⁵ M sec^-1
phosphogylcerate kinase	80% / 10⁷ M sec^-1	80% / 10⁹ M sec^-1	75% / 10⁶ M sec^-1
phosphoglyceromutase	100% / 10⁶ M sec^-1	100% / 10⁶ M sec^-1	90% / 10⁶ M sec^-1
enolase	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
pyruvate kinase	40% / 10⁸ M sec^-1	60% / 10⁸ M sec^-1	80% / 10⁸ M sec^-1
pyruvate dehydrogenase	90% / 10⁶ M sec^-1	90% / 10⁷ M sec^-1	70% / 10⁵ M sec^-1
citrate synthase	00% / 10⁸ M sec^-1	90%* / 10⁸ M sec^-1	90% / 10⁹ M sec^-1
aconitase (a and b)	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
isocitrate dehydrogenase	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
alpha-ketoglutarate dehydrogenase	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
succinyl CoA synthetase	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
succinate dehydrogenase	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
fumarase	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
malate dehydrogenase	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1	90% / 10⁸ M sec^-1
b-ketothiolase	90% / 10⁸ M sec^-1	10% / 10⁹ M sec^-1	10% / 10⁸ M sec^-1
Acetoacetyl Reductase	100% / 10⁸ M sec^-1	100% / 10⁸ M sec^-1	100% / 10⁹ M sec^-1
PHA Synthase	100% / 10⁸ M sec^-1	100% / 10⁹ M sec^-1	100% / 10⁸ M sec^-1

*In anaerobic conditions (GSU92PHA, citrate synthase goes to 10%, b-ketothiolase goes to 90%)

^TIn anaerobic conditions (GSU03, Pyruvate goes to 100%, citrate synthase goes to 10% and b-ketothiolase goes to 90%).

Answer the following questions and provide a reflection and comment.

1) If your goal is to maximize the production of PHA, which strain works will make the most product?

2) If your goal is to maximize the speed of PHA production, which strain will work the best? (Hint: Look for any rate limiting steps in the reaction).

3) Now, production and speed are interrelated. I could have a strain that may not be most efficient, but works so fast that it makes up the difference. Considering this, which strain makes the most PHA?

4) Assuming you can move the genes for these enzymes between the cells, is there a way with what you have to engineer a cell to have a higher production rate of PHA?