Microbiology MOOC Newsletters

Monday, March 12, 2012

Daily Newsletter March 12, 2012

Administrative Note: Upon the recommendation of a colluege, you are going to try a different peer review service. You will be given an opportunity after this review to comment on which you found more useful and esier to use.

More information about the service will come when the site opens after 5pm tonight. You will have until Wednesday night to upload your papers. You can continue to review until Sunday night (you must review all papers given you to).

Today's Topic: Bacterial Evolution

What do you remember about evolution?
Think for a minute. Could you define evolution? If someone asked you to explain evolution, could you? Today, I want you to do an evolution refresher.

Go through the tutorial sections An Introduction to Evolution and Mechanisms: The Process of Evolution. This tutorial entitled Evolution 101 is part of an effort by Berkeley University to increase awareness and understanding of evolution. It is an excellent resource to refresh your understanding of evolutionary theory and a great place to find out about modern evolutionary research.

I would also recommend reading Problem Concepts in Evolution, for a good discussion of misconceptions and rebuttal against perceived problems with evolutionary theory. The Evolution FAQ from PBS is another excellent resource. A final site of interest is the Index of Creationist Claims, which includes rebuttals to each claim.

Take time to consider evolution. You will need a good foundation for what we will be talking about this week.

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Daily Challenge: During a debate on whether Evolution should be taught in high schools, you have been asked to provide a 10 minute explaination on how biologists view evolution and why the concept of evolution is considered central to biology. NOTE: your doing this from the perspective of a biologist. If you hold religious opinions, you may share them, but only after you explain the scientific theory of evolution. (Remember a theory is a robustly supported hypothesis, and thus it is based on accumulated data, i.e., facts.)

Friday, March 9, 2012

Daily Newsletter March 9, 2012

Daily Challenge: At this stage of the course, you have gone over a number of different features of bacteria, from their genetics to metabolism, and now growth. Today, I want you to consider the following:

A few weeks ago, there was a discussion of an organism that could produce Polyhydroxyalkanoates (PHA). You looked at the metabolism of PHA, and three different strains of Bacillus subtilis that could produce PHA.

From this week, we went over microbial growth. Yesterday's assignment had you looking at concepts of fermentation. With this in mind, consider the following:

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).

The organism will begin producing PHA when the population density has reached 10⁹ cells/ml. The metabolic information can be found here.

NOTE: The question sets below are for your CONSIDERATION. Take notes on the questions, but don't sit and try to answer them fully.

QUESTION Set Alpha:

Describe the general characteristics of B. subtilis, and the specific characteristics of GSU14PHA and GSU92PHA.
Explain the consequences of having an NADH oxidase knock-out. How could the cell maintain a proton motive force without NADH oxidase?
What is the consequence of having a pyruvate decarboxylase knockout?
Describe the typical fermentation pathway (to lactic acid). In describing the purpose of fermentation, explain why there is a need to oxidize NADH + H⁺.
Give a brief account of quorum sensing, and how it affects cell physiology.
Given 100ml of a starter culture with 10⁶ cells/ml, a 40L fermentation take, and a desired population of 10⁹ cells/ml, how long will it take you to get to the production level? Show all of your work. Make sure that you tell how long each strain will take to get to production levels.
Based upon growth to production level and metabolic characteristics, which strain is best suited for production? Explain and justify your answer. You must be able to justify using numbers, and explain what the numbers mean.
Describe the concept of a rate limiting step, and using examples from the next page, show how a rate limiting step can alter the production of a desired end product.
In the introduction, it states that you have to change conditions for GSU92PHA and GSU03PHA before production can begin. Why would you start with one condition (such as to reach production population levels) and then change conditions to start producing PHA?

To improve production, you have a number of genetic alterations that you can perform with in-house plasmids and materials.

Plasmid pPHA12 holds a PHA operon, which contains coding regions for all of the genes needed to convert Acetyl CoA into PHA. In addition, the promoter is changed, having a transcriptional rate of 2.8.
Plasmid pPHA13 holds a Phophoreductase that can use the reducing power of NADH.
Plasmid pAN34 has a restriction site that will knockout NADH oxidase, and holds a phosphoreductase that can be used to oxidize NADH.
Plasmid pPHI32 has restriction sites that allow integration into the chromosome, it holds the PHA operon, but knocks out hexokinase.
Plasmid pPH132 acts as a conversion to the PHA operon, holding mutations of the genes with the following changes in Kinetics: 10⁸M sec^-1, 10⁹M sec^-1, 10⁹M sec^-1.

QUESTION set Beta:

Using the strain selected above, which alterations will maximize production? Explain and justify your answer.
Describe an operon, and discuss the advantages a bacterium has by using an operon instead of the way in which eukaryotes transcribe genes. What advantage does the eukaryote have?
What is a transcriptional rate, and how does it influence the proteome and metabolome?
What is a phosphoreductase, and will it provide an advantage to our B. subtilis system as it tries to make PHA?
What is hexokinase, and what complications can arise with this knockout? How would you have to change your Fermentation process? How would it change bacterial growth?
What is a restriction site, and how is it used in genetic engineering?
What changes will you have to make, if any, to your system to compensate for any genetic alteration?
What changes will you have to make when you up-scale your fermentation to 1,000L? What do you have to consider when moving from 400L to 1,000L?
Describe other genetic alteration you could use that would result in greater production. Be sure to explain your answers.

The Challenge: After considering these questions, reflect on what you have learned. With all the information above, how would you OPTIMIZE this system to produce PHA? Giving a brief description of the organism is important, and the reasons why you selected a specific organism. Provide any genetic modifications you think would be important, and any information as to the culture conditions that would enhance production. This is a thought questions! Take your time and reflect.

For this blog, your word limit is set to 150 words.
You can get up to 8 points for a well considered discussion.
You'll get 5 points for an average disucussion.
You'll get 2 points for just the minimum.

Thursday, March 8, 2012

Daily Newsletter March 8, 2012

Daily Topic: Variations in plant photosynthesis
As discussed in class, there are variations in how plants experience photosynthesis. The C₄ plants have a spacial separation of CO₂ acquisition and utilization. The CAM plants have a temporal seperation of CO₂ acquisition and utilization. In bacteria there are even more variations, as some species can fix carbon through a reverse TCA process.

The image below is one example of a reverse TCA (Kreb's Cycle).

Daily Challenge: Alternative Carbon Fixation routes
Your task today is to look at different ways organisms can fix carbon. Focus your discussion on the C₄ and CAM plants. Discuss how they separate CO₂ acquisition and fixation. Discuss the habitats of plants with these modifications. Discuss why this modification is helpful (hint: discuss photorespiration). Finally, take a moment to discuss why bacteria may have reverse TCA reactions.

Learning objectives: What are you suppose to learn from this exercise? Why are these modifications important? Is it just about knowing different routes of carbon fixation? What is the main goal of learning about different routes of carbon fixation?

Daily Newsletter March 8, 2012

Daily Challenge: Bacterial Growth
You are to work with the bacterial growth equations and your understanding of bacterial growth factors to answer the following questions.

You have been hired to maximize the production of indene by Rhodococcus. Indene is a precursor used in the manufacturing of the AIDS medication Crixivan™. Given conditions of growth are 4% lactose, 3% yeast extract, oxygen saturation of 18%, and a pH of 6.8. You have ten 10ml stored cultures of the organism, with an OD that is equivalent to 10⁵ cells/ml for this organism. Your fermentation vessel holds 15L, and is aerated and agitated to prevent the formation of biofilms (these cells need to be planktonic). The cells have a lag period of 1 hour when first introduced to a new culture from storage, and will enter late log phase when there are 10⁸ cells/ml. From looking at a colleague’s research notes, you find that the bacterium has a generation time of 25 minutes. How long will it take to reach late log phase?
Hint: You are adding stock cultures to a 15L tank. Have you diluted these cultures? What then is your starting culture?

You have been told that you must maintain late log phase for 48 hours in order to produce a sufficient quantity of Indene. How would go about doing this?
Hint: You will need to go back to the textbook and other references to look at fermentation and continuous cultures. This is not about a "right" answer as much as it is about your thought process.

You have been asked to upscale the production to a 10,000L fermentation tank. What complications could arise when upscaling from a 100L fermentation vessel?
Hint: What problems would come up if you had more liquid or a larger volume? Give this some thought. Think about pH effects of metabolizing organisms, temperature differences in a large body of liquid, oxygenation issues.

Someone suggests that you double the lactose and yeast extract concentrations. What effects could this have on your culture (explain)? How would you go about seeing if this was a good or bad idea?

Give a brief description of Rhodococcus. What is a biofilm?

Learning objectives: What are you being asked to do in these questions? Is this about just using the growth equations? How do conditions affect bacterial growth? Is it important to be sensitive to culture environment and conditions when growing bacteria?

Wednesday, March 7, 2012

Daily Newsletter March 7, 2012

Today's Topic: Bacterial Growth Formula

Today we are going to do a little math. Don't get scare, it is simple, but powerful. A critical concept that all microbiologists, and even cell biologists, must deal with is the rate at which cells grow. Even if you never use these equations again, going through them will build a mental picture of rate at which bacterial cells can multiply, and how large populations can become. On a practical side, it is critical as it helps you to understand when your population reaches a point that would be equivalent to a late log stage population; a working culture as it were.

The core bacterial growth formula is N_t = N_o x 2ⁿ

N_o is the original culture, or the 0 generation.
t is time.
n is the number of generations that bacteria goes through
N_t is the final population size (the population at a specific time interval).

This formula is based on the binary fission form of bacterial growth, and describes the log phase of growth.

Thus, it describes the doubling of a bacterial population at every time interval. This time interval is dependent upon the organism, and the environmental conditions.

Using this formula, you can solve more meaningful problems than just the doubling of bacteria. Question: What would be a more meaningful question to ask about bacterial growth?

n = the number of generations a population has undergone.

n = (log N_t – log N_o) / log 2 = (log N_t – log N_o) / 0.301

So if you know the original population size, and the final population size, you can discover the number of generations a bacteria has gone through. Through mathematical manipulation, we can solve for other problems. In the next formula set we look at the generations per hour of a given bacterial culture.

k = the mean growth rate of a bacterial population (generations per hour).

k = n/t = (log N_t – log N_o) / (0.301 x t)

If we know the original and final population size, we can extrapolate n, the number of generations that have occured in the population. If we divide n by the time it took, you will get k, the number of generations that occur per hour. Why would you want to know the number of generations per hour?

An inverse of k allows us to learn how many hours it takes to have one generation. This is also known as a generation time. If you ever read that an organism has a generation time of x minutes, you are looking at the solution to the problem below. As you can see, this provides a powerful tool in understanding an organism.

g = the mean doubling time of a bacterial population (hours per generation).

g = 1/k

Will an organism have the same g in all environmental conditions? Why?

Daily Challenge: Problems
Here are some problems to help you go through these equations. Show your work and answer the associated questions.

Problem: You have been asked to growth, in batch culture, a population of Rhodococcus spp. (the designation ssp represents a generic species; either the species is unknown, unnamed, or in this case unimportant). You are using lactose as a carbon source, and providing a complete array of other essential nutrients by adding Yeast Extract to the nutrient broth. Rhodococcus has a doubling time of 20 minutes in this environment. You start with 100 ml of a 104 culture. Your fermentation vessel is 10L (so it contains 10L of broth, including the starting culture). How long will it take to reach a population of 10^9,?
HINT: What are you looking for? number of generations? generation time? Something else? Which formula will give you the answer?

Question: Using the same system, an assistant attempts to replicate your system, but instead of lactose, they use sucrose as the carbon source. After 72 hours, they have a population of 10⁷. What was the growth rate of this particular species of Rhodococcus in this environment? Was this a more effective environment?

Question: Using the original fermentation equipment, you are asked to grow a population of Rhodococcus spp. to 10^{8,/sup> in 24 hours. Is this possible? Explain.

Learning Objectives:

What would you describe as the learning objectives today? Are you being asked to memorize these equations? Are you being asked to understand these equations? Are you being asked to use these equations? How could you use these equations?}

Tuesday, March 6, 2012

Daily Newsletter March 6, 2012

Today's Topic: Binary Fission
Bacteria do not undergo mitosis or meiosis. They lack a nucleus, thus there is no nuclear division.

When we have an organism that undergoes a replication event followed by cellular division, the process is called binary fission (splitting into two). The assumption is that the two daughter cells will be genetically identical (plasmids that don't replicate can weaken this assumption).

There are some questions you should consider:

What does a cell need to do before it can divide? (Hint: Biomass)
In eukaryotic cells, there is a signal that controls the cell cycle (CDK), are their signals that control bacterial division?
Without a nucleus, how do you move the genophore (DNA molecule) to opposite sides of the cell?
How long does this process take? (Hint: Each species is different)
Is the time to division constant, or is it variable? Why? (Hint: Conditions)

Whenever you come to a topic like cellular division, you need to "think like a cell". Cells ultimately want to survive, and leave healthy daughter cells that survive. You want to increase your population (that is the biological imperative at a cellular level). So how do you do it, and make sure that the daughter cells are fit for survival?

Daily Challenge: Binary Fission
Describe how a bacterium divides into two daughter cells. Look at possible signals, environmental conditions, and even nutritional requirements that may affect the rate of division. See if you can find one bacterium that is considered slow growing and one fast growing. What is different about them?

Monday, March 5, 2012

Daily Newsletter March 5, 2012

Administrative Note: There are two notes today.
1) This is a milestone week. On Thursday you will have your milestone exam.
2) We have passed the midpoint of the semester, and this is when a planned change is to occur. You will notice that each news letter has guiding comments or questions to help you build your own Learning Objectives for this week.

Today's Topic: Vocabulary

Generally your not asked to work on vocabulary as a sole topic, but in this case it is important to become familiar with various terms used to describe optimal conditions for bacterial growth. Your challenge today will involve working through this vocabulary.

Today's Challenge: Vocabulary.
Provide a definition and example for each of these terms:

Defined media
Complex (undefined) media
Broth (liquid) culture
Solid (agar) culture
Aseptic Technique
Pure Culture
Chemostat
Batch culture
Psychrotroph
Psychrophile
Mesophile
Thrmophyle
Hyperthermophile
Acidophile
Alkaliphiles
Halophiles
Osmophiles
Xerophiles
Obligate Aerobes
Facultative Anaerobes
Obligate Anaerobes
Microaerophiles
Aerotolerant anaerobes*
Superoxide
Fastidious
Growth Factors
Trance Minerals

*Aerotolerant Anaerobes: This is an oxygen utilization term that is often down played. There are bacteria that do not require oxygen as a terminal electron acceptor, but they are able to survive in atmospheric oxygen. This is different than the obligate anaerobes which are killed (or inhibited) in atmospheric oxygen. Look to the Streptococci as aerotolerant.

Learning Objectives:
This first newsletter has an easy objective, learn the terms listed above. By writing out a definition and an example, you will have a better grasp on these terms. You will see these terms repeatedly, as they help us to identify growth conditions for bacteria. It would benefit you to learn them.