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BACTERIAL NUTRITION AND GROWTH

By: Mary Ylane S. Lee

• WHAT FACTORS AFFECT THE GROWTH OF BACTERIA?

A. PHYSICAL OR ENVIRONMENTAL FACTORS

B. NUTRIENT FACTORS OR REQUIREMENTS

• PHYSICAL REQUIREMENTS or ENVIRONMENTAL FACTORS

- Temperature

- Gas (Oxygen and CO2 requirement)

- Osmotic Pressure

TEMPERATURE

Temperature

How does temperature affect growth rate of bacteria?

CARDINAL TEMPERATURE

• MINIMUM GROWTH TEMPERATURE

• OPTIMUM GROWTH TEMPERATURE

• MAXIMUM GROWTH TEMPERATURE

Sternothermal- species that have small growth temperature range.

Eurythermal- species which grow over a wide range of temperature.

Categories based temperatures:

PSYCHROPHILES- Grow well at 0°C and have an optimum growth

temperature of 15°C or lower and a maximum is around 20°C .

FACULTATIVE PSYCHROPHILES-grow at 0°C, have an optima between 20 to 30°CAnd maxima of about 35°CMESOPHILES- Growth optima is around 20 to 45 °C, a

temperature minimum of 15°C to 20°C and a maximum of about 45°C or lower

THERMOPHILES

- Can grow at temperatures of 55°C or higher. Minimum is usually 45°C and have an optima between 55 and 65°C .

HYPERTHERMOPHILES

- Optima is between 80°C and about 110°C. They usually do not grow well below 55°C

• Oxygen Requirements:

How does oxygen affect optimal growth?

-microorganisms fall into several groups with respect to the effect of oxygen

on their growth and metabolism:

may be Aerobic and Anaerobic

1. Obligate Aerobes- completely dependent on atmospheric oxygen for growth.

2. Facultative anaerobes- do not require oxygen for growth but do grow better in its presence.

3. Aerotolerant anaerobes- simply ignore oxygen and grow equally well, whether it is present or not.

4. Strict or obligate anaerobes – do not tolerate oxygen at all and die in its presence.

5. Microaerophiles- damaged by normal atmospheric level of oxygen and require level below the range of 2 to 10% for growth.

GROWTH LOCATION BASED ON GASEOUS REQUIREMENT

obligate aerobes

Facultative anaerobes

Obligate anaerobes

Aerotolerant anaerobes

Microaerophiles

• pH effects:

The majority of organisms live or grow in habitats between pH 6 and 8 because strong acids and bases can behighly damaging to enzymes and other cellular substances.

Acidophiles

- Optimum between pH 0 and 5.5

Neutrophiles

- Between 5.5 and 8

Alkalophiles

- Range of 8.5 to 11.5

Osmotic Pressure/ Water Activity

Osmotic Pressure:• It is the force with which a solvent moves

from a solution of lower solute concentration to a solution of higher solute concentration

Hypertonic Hypotonic

• water must be available for metabolism and growth (80-90% of cell mass)

water activity aw: represents the mole fraction of the total water molecules that are available:

• aw = p = vapor pressure of solution

p0 vapor pressure of water w

measured by its relationship to relative humidity (aw x 100 = RH)

• What is the minimum amount of water allowing growth?lowest water activity allowing growth:most bacteria 0.91most yeasts 0.88most molds 0.80halophilic bacteria 0.75xerophilic fungi 0.65osmophilic yeasts 0.60

Most bacteria require an isotonic or hypotonic environment for optimum growth.

Osmotolerant- grow over wide ranges of water activity or osmotic pressure.

Halophiles- require high concentration of sodium chloride to grow.

NUTRITIONAL REQUIREMENTS

• Essential Nutrients: • Are any molecular or elemental form of nutrient that is

required by an organism.

• Two categories of essential nutrients; macro-nutrients and micro-nutrients.– Macro-nutrients are needed in larger amounts.

• Used to help with cell structure and the cell's metabolism.• Examples are proteins, and carbohydrates.

– Micro-nutrients or trace elements are needed in a lot smaller amount.

– They help enzyme function and help to maintain protein structure.

– They include elements such as zinc, manganese, and nickel.

ELEMENT CELL FUNCTION

C backbone of organic cell components, energy

Hwater, organic components, pH, hydrogen bonds, re-

O water, organic components, respiration

N amino acids, nucleotides, coenzymes, ATP

S amino acids, coenzymes, enzymes

P nucleic acids, phospholipids, coenzymes, ATP

Na, K, Ca, Cl, Mg, Mn

Trace elements: transport, ionic balance, cofactors (e- donor/acceptors)

• THE SOURCE OF COMMON ESSENTIAL NUTRIENTS ARE: CHNOPS

• Carbon• Hydrogen• Nitrogen• Oxygen• Phosphorous• Sulfur

• Carbon Content Nutrients:Carbon content is another way that nutrients are categorized. Sources of nutrients is extremely varied and some microbes will obtain their nutrients entirely from inorganic sources and others require a combination of organic and inorganic sources.

• Organic nutrients:– Contain at least some combination of

carbon and hydrogen atoms.– Natural organic molecules are usually products of

livings things.– Simple to large polymers.

• Inorganic Nutrients:– An element or simple molecule that contains elements

other than carbon and hydrogen.– Natural reservoirs are mineral deposits in the crust of

the earth, bodies of water and the atmosphere.• EX: metals and their salts (magnesium, sulfate,

ferric nitrate, sodium phosphate).• EX: Gases (oxygen, carbon dioxide) and water

According to Energy Source

• Energy Source– Phototroph

• Uses light as an energy source

– Chemotroph– Uses energy from the oxidation of reduced

chemical compounds

According to Electron Source

• Electron (Reduction potential) Source– Organotroph

• Uses reduced organic compounds as a source for reduction potential

– Lithotroph• Uses reduced inorganic compounds as a source for

reduction potential

• Carbon SourcesEven though there is a distinction between the types of carbon cells can absorb as nutrients. It is important to know that the majority of carbon involved in the structure and metabolism of all cells are organic

• Prokaryotes can be divided into various physiological groups based on howthey derive energy and assimilate carbon

According to Carbon Source

• Carbon source– Autotroph

• Can use CO2 as a sole carbon source

(Carbon fixation)

– Heterotroph• Requires an organic carbon source; cannot use CO

as a carbon source

Can also be classified as:

Group Carbon from Energy by

Chemolithotrophs CO2 oxidation of inorganic

compounds

Photolithotrophs CO2 light

Chemoheterotrophs organic compounds oxidation of inorganic

compounds

Photoheterotrophs organic compounds light

Representative microorganisms:• Chemolithotrophs (chemolithotrophic autotroph):- sulfur oxidizing bacteria- nitrifying bacteria• Photolithotrophs (photolithotrophic autotroph)- algae- cyanobacteria- purple and green sulfur bacteria• Chemoheterotroph (chemoorganotrophic heterotroph)- nonphotosynthetic bacteria- fungi- protozoa• Photoheterotrophs (photoorganotrophic heterotroph)- purple and green non-sulfur bacteria

• Nitrogen source– Organic nitrogen

• Primarily from the catabolism of amino acids

– Oxidized forms of inorganic nitrogen• Nitrate (NO

32-) and nitrite (NO

– Reduced inorganic nitrogen• Ammonium (NH

– Dissolved nitrogen gas (N2) (Nitrogen fixation)

• Phosphate source– Organic phosphate

– Inorganic phosphate (H2PO4- and HPO4

• Sulfur source– Organic sulfur– Oxidized inorganic sulfur

• Sulfate (SO42-)

– Reduced inorganic sulfur• Sulfide (S2- or H2S)

– Elemental sulfur (So)

Nutrient Transport Processes

• Simple Diffusion– Movement of substances directly across a

phospholipid bilayer, with no need for a transport protein

– Movement from high low concentration– No energy expenditure (e.g. ATP) from cell– Small uncharged molecules may be

transported via this process, e.g. H2O, O

• Facilitated Diffusion– Movement of substances across a membrane

with the assistance of a transport protein– Movement from high low concentration– No energy expenditure (e.g. ATP) from cell– Two mechanisms: Channel & Carrier Proteins

• Active Transport– Movement of substances across a membrane

with the assistance of a transport protein– Movement from low high concentration– Energy expenditure (e.g. ATP or ion gradients)

from cell– Active transport pumps are usually carrier

proteins

• Active Transport (cont.)– Active transport systems in bacteria

• ATP-binding cassette transporters (ABC transporters): The target binds to a soluble cassette protein (in periplasm of gram-negative bacterium, or located bound to outer leaflet of plasma membrane in gram-positive bacterium). The target-cassette complex then binds to an integral membrane ATPase pump that transports the target across the plasma membrane.

• Cotransport systems: Transport of one substance from a low high concentration as another substance is simultaneously transported from high low.

For example: lactose permease in E. coli: As hydrogen ions are moved from a high concentration outside low concentration inside, lactose is moved from a low concentration outside high concentration inside

• Group translocation system: A molecule is transported while being chemically modified.

For example: phosphoenolpyruvate: sugar phosphotransferase systems (PTS)

PEP + sugar (outside) pyruvate + sugar-phosphate (inside)

• Bacterial Growth

- Growth is an orderly increase in the quantityof cellular constituents.

- Increase in cell numbers

- Bacteria grow by binary fission.

BINARY FISSION

PHASES OF BACTERIAL GROWTH

Lag phase - synthesis of new components or repair

– there is little or no change in the number of cells, but metabolic activity is high.

– DNA and enzyme synthesis occurs; may last from 1 hour to several days.

Log phase - reproduction at maximum rate (shortest generation time)

– the bacteria multiply at the fastest rate possible under the conditions provided.

Stationary phase - no net increase, balance between cell division, cell "death",

- there is an equilibrium between cell division and death.

Death phase- the number of deaths exceeds the number of new cells formed.

GENERATION TIME

The time required for a cell to divide or a population to double is known as the generation time.

• quite short; about 20 –60 minutes under

optimal condition

Most bacteria have a doubling time of 1-3 hours, although some may be greater than 24 hours.

• 5 –10 hours for most pathogenic bacteria

GENERATION TIME

• . . . is defined as the relationship between the number of bacteria in a population at a given time (Nt), the original number of bacterial cells in the population (No), and the number of divisions those bacteria have undergone during that time (n) can be expressed by the following equation:

Nt = No x 2n

The bacterial growth curve:

• Lag phase

• Exponential phase

• Stationary phase

• Death phase

During the exponential phase, the number of cells doubles every generation (1, 2, 4, 8, 16...) and the number of cells in the culture (Nf) at any particular time is given by:

Nf = (Ni)2n

where:

Ni is the initial size of the population

n is the number of generations that have elapsed.

This equation is usually written in the form log10 for mathematical convenience:

log10 Nf = log10 Ni + 0.301n(log10 2 = 0.301)

• From this equation it is possible to determine n (generations elapsed) by knowing the numbers of cells at the beginning and end of a particular incubation period:

n = (log10Nf - log10Ni) / 0.301

• This is useful because it enables calculation of the generation time g (the time between the beginning and end of the incubation period:

g = t/n

EXAMPLE:

• 100 bacteria present at time 0

if generation time is 2 hours,

what is the cell mass after 8 hours?

Cell Mass= No x 2n

= 100 x 24

= 1,600 bacteria

Sample Problem 1

Bacillus cereus divides every 30 minutes. You inoculate a culture with exactly 100 bacterial cells. After 3 hours, how many bacteria are present?

Solution

In 3 hours, B. cereus will divide 6 times. Therefore, n = 6. 26 = 64 or 2x2x2x2x2x2

100 x 64 = 6,400 cells

Using the same example, let’s say you have determined that your sample contains 6,400 bacterial cells. You know that it incubated 3 hours. How many generations have occurred?

• Number of generations = (log cells at end of incubation ) - (log cells at beginning of incubation) / 0.301

n = (log10Nf - log10Ni) / 0.301

• Therefore, (log 6400) - (log 100) / 0.301 = (3.81 - 2) / 0.301 = 6 generations

• To calculate the generation time for a population: 60 min x hours / number of generations

g = t/n

In this example:

• 60 min x 3 hours / 6 generations = 30 minutes per generation

• Practice Problems • 1. You perform a serial dilution and determine that the original

number of cells in your sample was 12, 000. How many bacteria will be present in 12 hours if the generation time is 15 minutes (assume unlimited food and clean environment)?

• 2. You determine that a coconut cream pie has 3 million (3 x

106) Staph. aureus cells in it. You estimate that the food preparer did not wash his hands and probably inoculated the cream with 500 Staph. aureus. He also forgot to refrigerate it. If the pie was made 6 hours ago, how many generations have occurred? How long is each generation?

• • 3. Using the generation time from problem 2, how many

bacteria would be present after 8 hours at room temperature?

• 4. Let’s say that flesh eating Strep. pyogenes divides every 10 minutes at body temperature. You fall down and scrape your knee and get infected with 5 Strep. pyogenes cells. After 4 hours, without medical intervention, how many bacteria will be ravaging your body?

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