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10: Carbohydrate Metabolism of Unknowns

  • Page ID
    110866

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    Learning Outcomes

    By the end of this lab (2 lab periods) you will be able to:

    Introduction

    Now that you have isolated your two unknowns, the next step is to work on figuring out which two organisms you were assigned! The first step will be to perform a series of tests that are based on the utilization of carbohydrates. While all bacteria can use glucose as a source of energy, not all bacteria can ferment glucose or other simple carbohydrates. Those that can ferment carbohydrates often produce different fermentation products that can be used to distinguish between organisms.

    The tests that we will be setting up today are:

    Phenol Red Broths

    As we’ve learned in the lecture, many bacteria can undergo fermentation. In fermentation, glucose is first broken down into pyruvate. To regenerate NAD+ for repeated rounds of glycolysis, these bacteria undergo fermentation producing products such as lactic acid and/or ethanol and CO2.

    When bacteria are “fed” sugars other than glucose, the bacteria need to be able to change them into either glucose or into other intermediates that are part of the glycolytic pathway. For a molecule like sucrose, which is a disaccharide composed of glucose and fructose, the enzyme invertase, found in some bacteria, can break the bond between glucose and fructose. The glucose enters glycolysis directly, and the fructose is phosphorylated and enters a few steps down the pathway. This means that the bacteria only need a few extra enzymes in order to use sucrose for food - they don’t need entirely new pathways. However, not all bacteria have all of these “extra” enzymes, so determining which of these carbohydrates can be fermented by your unknown bacteria can be very helpful in determining which one you’ve got.

    Schematic showing how glucose, sucrose, fructose and mannitol relate to glycolysis
    Figure \(\PageIndex{1}\): Glucose, mannitol, sucrose and lactose can be fermented using glycolysis and fermentation pathways with just one or two additional steps (enzymes).

    Each phenol red broth contains ONE of the four sugars (glucose, lactose, sucrose or mannitol) soluble protein and phenol red indicator. Phenol red is a dye that is red between pH 6.8 and 7.4 and yellow below pH 6.8. The indicator will turn pink as the pH increases above 7.4. If acid is produced by fermentation of the sugar, the broths will turn yellow after 24-48 hours of incubation.

    Phenol red broths also contain a small upside down test-tube called a Durham tube. The purpose of the Durham tube is to capture any gas that might be produced by fermentation. Some of our organisms that can ferment sugars produce only acid, but some can produce acid AND gas. You will want to look at the Durham tube carefully after incubation to determine if gas has been produced.

    examples of phenol red broth results
    Figure \(\PageIndex{2}\): Phenol Red Glucose Fermentation Broth Results. Left - no acid produced. Middle - fermentation occurred producing acid. Right - fermentation occurred producing acid and gas. Photo credits - Ron John Vargas, used with permission.

    MR-VP

    The Methyl red (MR) and Voges Proskauer (VP) tests are done by first inoculating one tube of MR-VP media and incubating for 24-48 hours. The resulting culture is then split into 2 tubes and one is tested with methyl red, the other with VP reagents. So we tend to refer to them as “MR-VP”, but in fact, there are two tests here!

    Methyl Red (MR)

    Not all bacteria produce lactic acid and ethanol/CO2 as byproducts of fermentation. Some bacteria are also heterofermentative, meaning they can produce more than one fermentation product! The methyl-red test allows us to identify organisms that produce a variety of acids also called “mixed-acid” fermentation. In this case, pyruvate is also converted to acetic acid, formic acid, and succinic acid. These acids are capable of lowering the pH below 4.4, where the methyl red indicator will turn red. If the acids are not produced, the pH will remain above 4.4 and there will not be a color change.

    The MR pathway and typical test results
    Figure \(\PageIndex{3}\):A. The pathway that produces mixed acids. B. A methyl-red positive result is shown on the right. A MR negative result is shown on the left. Photo credits - I.H., used with permission

    Voges-Proskauer (VP)

    Some organisms produce a different molecule called acetoin as an intermediate during 2,3 butanediol fermentation. The VP reagents (VP Reagent 1 and VP Reagent 2) react with the acetoin and other media components to produce a red color.

    VP test principles and typical test results
    Figure \(\PageIndex{4}\):A. Acetoin is an intermediate in the breakdown of glucose to 2,3-butanediol. B. A VP positive result is shown on the right. A VP negative result is shown on the left. Photo credits - Maithy Nguyen, used with permission

    Citrate Utilization

    You probably recall that citrate is an intermediate in the Krebs Cycle. It is formed when oxaloacetate reacts with Acetyl CoA. It turns out that bacteria can and will use citrate as a source of food if glucose is not available, but only IF the citrate can be brought into the cell. This is accomplished when the organism can make an enzyme called citrate permease.

    Simmons citrate agar contains citrate as the only source of carbon (food) and ammonium phosphate as the only source of nitrogen. For an organism to survive on this medium, it must possess the citrate permease enzyme. When citrate is used, basic byproducts including sodium carbonate are produced, increasing the pH. The organisms also convert the ammonium phosphate in the medium to ammonia and ammonium hydroxide, both of which also increase the pH of the agar. To detect this increase in pH, a dye called Bromothymol blue is also added to the medium. At a pH of 7.6, the medium changes color from green to a brilliant, sapphire blue.

    Schematic of the citrate reaction and typical citrate test results
    Figure \(\PageIndex{5}\): A. When citrate is used as a source of carbon, basic products such as sodium carbonate are produced, raising the pH. B. A citrate-positive result is shown on the left. A citrate-negative result is shown on the right. Photo credits - Ron John Vargas, used with permission.

    Starch Hydrolysis Test

    Think back to early in the semester when we discussed monomers and polymers. We learned that starch is a very large polymeric molecule composed of thousands of tiny glucose molecules linked together by alpha 1,4- and alpha 1,6-glycosidic bonds. Recall that starch is enormous and far too big to cross into a bacterial cell. However, some bacteria secrete enzymes into the environment which can break starch down into glucose molecules. These molecules can easily pass into the cell.

    To determine if an organism produces these enzymes, we grow organisms on a starch plate. Starch molecules can be detected by adding Lugol’s iodine to the plate surface. The iodine will react with starch and turn it a dark blue/black. If bacteria secrete the enzymes that digest starch, then all of the starch around the bacterial growth will be digested and there will be a clear “zone” around the growth. This zone is easy to see by holding the plate up to the light or observing the plate over a lightbox.

    Starch molecules and typical test results for the starch hydrolysis test.
    Figure \(\PageIndex{6}\): A - A small section of a starch molecule showing alpha 1,4 and alpha 1,6 glycosidic linkages. Citation: Starch. (2023, June 4). In Wikipedia. Starch molecule [en.Wikipedia.org] B. Starch hydrolysis detection on starch plates. The organism on the top does not hydrolyze starch. The organism on the bottom does hydrolyze starch as evidenced by the large zone of clearing around the organism. Photo credits - Celeste Towner, used with permission.

    Catalase Test

    You will recall that the enzyme catalase is found in obligate aerobes and facultative anaerobes. It converts hydrogen peroxide, which is an oxidizing agent, into water and oxygen gas. Hydrogen peroxide is often formed as a byproduct of aerobic respiration, and it is quite toxic to cells if it is not broken down.

    Because oxygen gas is produced when catalase breaks down hydrogen peroxide, it is very easy to detect organisms that produce catalase by simply using a 3% hydrogen peroxide solution that can be purchased at the grocery or drug store.

    catalase reaction
    Figure \(\PageIndex{7}\): A positive catalase test showing characteristic bubbling as oxygen gas is produced. No machine-readable author provided. Nase assumed (based on copyright claims), CC BY-SA 3.0, via Wikimedia Commons

    Materials

    Experiment

    Day 1 - Inoculations

    Day 2 - Data Collection

    Schematic for performing the MR-VP tests
    Figure \(\PageIndex{8}\): Schematic for performing the MR-VP tests.

    Data

    Questions

    Your primary goal as you leave lab today is to have a complete and well-documented record of the results you collected today. Here are a few questions that might help you as you begin to identify your organisms.

    Creating a Dichotomous Key

    Dichotomous keys are used to identify plants, animals, and bacteria based on their unique characteristics. The key leads the user to the correct identification by going through a series of choices. It’s a little like playing 20 questions! Each choice either INCLUDES or EXCLUDES the organism as a possible candidate. A dichotomous key always presents you with two choices.

    By completing the many tests we’ve done so far, we’ve answered a lot of questions about our organisms. For example, we’ve asked what shape(s) your organisms are (by gram staining and looking at morphology). We’ve asked if they express citrate permease, and whether or not they ferment a variety of different sugars. This is a good time to use a few of these questions to begin to narrow down which of the unknown organisms you were assigned.

    Table 1: Example Test Result Data

    Test

    Test results for my organism!

    Is it a coccus or bacillus?

    coccus

    Does it reduce nitrate?

    no

    Example dichotomous key
    Figure \(\PageIndex{9}\): Example dichotomous key for the unknown project.

    You can use the “starter” keys below to begin developing your key. Remember to select tests where you had a clear, definitive result. All of the Gram-negative bacteria we are working with are bacillus, but you can separate the Gram-positive bacteria by shape as they are a mixture of coccus and bacillus.

    Gram Positive Two Step Dichotomous Key

    Starter key for your gram positive organism
    Figure \(\PageIndex{10}\): Starter key for your Gram-positive organism.

    Gram Negative Two Step Dichotomous Key

    Screenshot 2023-07-05 at 7.49.45 AM.png
    Figure \(\PageIndex{11}\): Starter key for your Gram-negative organism.

    This page titled 10: Carbohydrate Metabolism of Unknowns is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Sheri Miraglia.

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