D

*Mave

mpn2

D2ave

This table may now be extrapolated, which will again produce Equation (2.43).

Now, as shown by the previous derivations, whether it is interpolation or extrapolation, the formulas are the same. The difference is only in the placement of the interpolants.

Example 2.5 Samples of river water are analyzed for total and fecal coliforms using the MPN technique. Some of the results are shown below:

Number of Positive Tubes Out of Five Serial Dilution Lactose BGB EC

What are the presumptive and confirmed MPNs and the fecal coliform MPN? BGB stands for "brilliant green bile broth" and EC stands for the medium for fecal coliform.

Solution: For the presumptive total coliform MPN, the three highest dilutions are 1.0, 0.1, and 0.01 corresponding to the positive readings of 3, 2, and 1. From Table 2.4, for a serial dilution of 10, 1.0, and 0.1, and positive readings of 3, 2, and 1, the MPN = 17. The serial dilution of 1.0, 0.1, and 0.01 diluted the sample 10 times more than the serial dilution of 10, 1.0, and 0.1 in the table. Because the experiment is diluting the sample 10 times more, the MPN from the reading in the table should be multiplied by 10. Thus,

MPN = 17( 10) = 170 organisms/100 mL Ans At 95% confidence range: 70 < MPN < 400 Ans

For the confirmed test, the three highest dilutions are 10, 1.0 and 0.1 corresponding to the positive readings of 5, 3, and 2. From the table,

MPN = 140 organisms/100 mL Ans At 95% confidence range: 60 < MPN < 360 Ans

MPN = 90 fecal coliforms/100 mL Ans At 95% confidence range: 40 < MPN < 250 Ans

Example 2.6 A sample of water is analyzed for the coliform group using three sample portions: 10 mL, 60 mL, and 600 mL. Each of these portions is filtered through five filter membranes using the membrane-filter technique. The results of the colony counts are as follows: 10-mL portions: 6, 7, 5, 8, 6; 60-mL portions: 30, 32, 33, 31, 25; and 500-mL portions: 350, 340, 360, 370, 340. What is the number of coliforms per 100 mL of the sample?

Solution: Using the criterion of 20 to 80 as the most valid count in a membranefilter technique, the results of the 10- and 500-mL portions can be excluded in the calculation.

a ^ , T 30 + 32 + 33 + 31 + 25 Average count for the 60-mL portions = -5-

= 30.2 coliforms per 60 mL 30.2

Example 2.7 Find the MPN for a reading of 2, 2, 1. The experiment is performed on a 10-mL, 1-mL, and 0.1-mL serial dilution.

Solution: The reading cannot be found in Table 2.4. It is, however, between the readings 2, 1, 1 and 3, 1, 1. Therefore,

(Rxave R\ave\ \R\ave R2aveJ

For reading 2, 1, 1, MPN = 9 = MPNj; for reading 3, 1, 1, MPN = 14 = MPN2

R = R = RAD1) + RAD2) + RAD3- = 2(10) + 2(1) + 1(0.1) = 1

R = R„(D1)+R12(D2) + R13(D3) = 2(10)+1(1)+ 1(0.1) = -, 90

R = R21(D1)+R22(D2) + R23(D3- = 3(10) + 1 (1) + 1(0.1) = 2 _

2.3.6 Viruses

A virus is a submicroscopic agent of infectious disease that requires a living cell for its multiplication. The two essential components of a virus are protein and nucleic acid. Whereas normal cells contain both RNA (ribonucleic acid) and DNA (deoxyribonucleic acid), a given virus contains only one, not both. A virus cannot multiply on its own as a normal cell does. It has no metabolic enzymes, uses no nutrients, and produces no energy. It is just a particle of protein and nucleic acid. A viral particle is tightly packed inside a protein coat that protects it. This unit is called a virion.

A study was conducted on the fate of viruses from the discharge of two sewage treatment plants into a Houston, TX, ship channel (Gaudy and Gaudy, 1980). Significant virus concentrations were detected 8 miles downstream from the point of discharge, and poliovirus was recovered in oysters 21 miles into Galveston Bay. This study shows that viruses can survive in the outside environment for a considerable length of time. In fact, it is known that some viruses may survive for periods of 40 days in water and wastewater at 20°C and for periods of 6 days in streams at normal conditions. For this reason, treated sewage which can contain a multitude of viruses should never be used to irrigate crops for food consumption. Use of treated sewage for spray irrigation can be a hazard from viruses due to droplets carried by the wind.

Viruses excreted by humans may become a major public health hazard. For example, studies have shown that some 10,000 to 100,000 infectious particles of viruses are emitted per gram of feces from people infected with hepatitis. Viruses producing diseases in humans are excreted in feces, so it is the responsibility of the environmental engineer to ensure that viruses in wastewater treatment plants are effectively disinfected. This is usually done by chlorination followed by proper disposal of the effluent.

Requiring a host cell, the virus has a very unique way of propagating itself. Upon entrance into a cell, it disappears. The nucleic acid mingles with the contents of the cytoplasm and loses its identity; nonetheless, its presence is evident from its effects on the host. In one mode of propagation, using the genetic code carried by the nucleic acid, it overpowers the mechanism of the cell to reproduce and diverts the cell materials and energy for viral replication. The virus takes over the machinery, commanding the cell to manufacture hundreds of viral prototypes for further infection of other host cells. The viruses, multiplying in extreme numbers, cause the host cell to swell and burst into pieces. Figures 2.6 and 2.7 show the shapes of the various virus particles and a virus infecting a cell, respectively.

In the second type, the cell tolerates the presence of the virus. The host cell continues to grow and multiply carrying the virus in its interior in a latent or noninfective form. The virus also multiplies at a proportionate rate.

2.3.7 Protozoa

Protozoa are single-cell protists one step above the bacteria in the trophic level. They possess the ectoplasm, which is a homogeneous portion of the cytoplasm. The ectoplasm helps form the various organs of locomotion, contraction, and prehension, such as cilia, flagella, pseudopods, and suction tubes.

FIGURE 2.6 Structural shapes of viruses: (a) helical, (b) polyhedral, and (c) combination T-even.

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