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The interval between simple dilutions is a constant value added to the preceding dilution to get the next. How should a factor for serial dilutions be mentioned? As there is no common difference between successive serial dilutions, rather a common ratio.
How should the interval between serial dilutions be expressed? - Biology
A simple dilution is one in which a unit volume of a liquid material of interest is combined with an appropriate volume of a solvent liquid to achieve the desired concentration. The dilution factor is the total number of unit volumes in which your material will be dissolved. The diluted material must then be thoroughly mixed to achieve the true dilution. For example, a 1:5 dilution (verbalize as "1 to 5" dilution) entails combining 1 unit volume of solute (the material to be diluted) + 4 unit volumes of the solvent medium (hence, 1 + 4 = 5 = dilution factor). The dilution factor is frequently expressed using exponents: 1:5 would be 5e-1 1:100 would be 10e-2, and so on.
Example 1: Frozen orange juice concentrate is usually diluted with 4 additional cans of cold water (the dilution solvent) giving a dilution factor of 5, i.e., the orange concentrate represents one unit volume to which you have added 4 more cans (same unit volumes) of water. So the orange concentrate is now distributed through 5 unit volumes. This would be called a 1:5 dilution, and the OJ is now 1/5 as concentrated as it was originally. So, in a simple dilution, add one less unit volume of solvent than the desired dilution factor value.
Example 2: Suppose you must prepare 400 ml of a disinfectant that requires 1:8 dilution from a concentrated stock solution with water. Divide the volume needed by the dilution factor (400 ml / 8 = 50 ml) to determine the unit volume. The dilution is then done as 50 ml concentrated disinfectant + 350 ml water.
2. Serial Dilution
A serial dilution is simply a series of simple dilutions which amplifies the dilution factor quickly beginning with a small initial quantity of material (i.e., bacterial culture, a chemical, orange juice, etc.). The source of dilution material (solute) for each step comes from the diluted material of the previous dilution step. In a serial dilution the total dilution factor at any point is the product of the individual dilution factors in each step leading up to it.
Final dilution factor (DF) = DF 1 * DF 2 * DF 3 etc.
Example: In a typical microbiology exercise the students perform a three step 1:100 serial dilution of a bacterial culture (see figure below) in the process of quantifying the number of viable bacteria in a culture (see figure below). Each step in this example uses a 1 ml total volume. The initial step combines 1 unit volume of bacterial culture (10 ul) with 99 unit volumes of broth (990 ul) = 1:100 dilution. In the second step, one unit volume of the 1:100 dilution is combined with 99 unit volumes of broth now yielding a total dilution of 1:100x100 = 1:10,000 dilution. Repeated again (the third step) the total dilution would be 1:100x10,000 = 1:1,000,000 total dilution. The concentration of bacteria is now one million times less than in the original sample.
3. Making fixed volumes of specific concentrations from liquid reagents:
V 1 C 1 =V 2 C 2 Method
Very often you will need to make a specific volume of known concentration from stock solutions, or perhaps due to limited availability of liquid materials (some chemicals are very expensive and are only sold and used in small quantities, e.g., micrograms), or to limit the amount of chemical waste. The formula below is a quick approach to calculating such dilutions where:
V = volume , C = concentration in whatever units you are working.
(stock solution attributes) V 1 C 1 =V 2 C 2 (new solution attributes)
Example: Suppose you have 3 ml of a stock solution of 100 mg/ml ampicillin (= C 1 ) and you want to make 200 ul (= V 2 ) of solution having 25 mg/ ml (= C 2 ). You need to know what volume ( V 1 ) of the stock to use as part of the 200 ul total volume needed.
V 1 = the volume of stock you will start with. This is your unknown.
C 1 = 100 mg/ ml in the stock solution
V 2 = total volume needed at the new concentration = 200 ul = 0.2 ml
C 2 = the new concentration = 25 mg/ ml
By algebraic rearrangement:
V 1 = ( V 2 x C 2 ) / C 1
V 1 = (0.2 ml x 25 mg/ml) / 100 mg/ml
and after cancelling the units,
V 1 = 0.05 ml, or 50 ul
So, you would take 0.05 ml = 50 ul of stock solution and dilute it with 150 ul of solvent to get the 200 ul of
25 mg/ ml solution needed. Remember that the amount of solvent used is based upon the final volume needed, so you have to subtract the starting volume form the final to calculate it.
4. Moles and Molar solutions (unit = M = moles/L)
Sometimes it may be more efficient to use molarity when expressing chemical concentrations. A mole is defined as exactly 6.023 x 1023 atoms, or molecules, of a substance (this is called Avagadro's number, N). The mass of one mole of an element is its atomic mass (g) and is noted for each element in the periodic table. Molecular weight is the mass (g) of a substance based on the summed atomic masses of the elements in the chemical formula. Formula weight refers to chemicals for which no discrete molecules exist for example, NaCl in solid form is made up of Na+ and Cl- ions, but there are no true molecules of NaCl. The formula weight of 1 mole NaCl would therefore be the sum of 1 atomic mass of each ion. The molecular weight (or FW) is provided as part of the information on the label of a chemical bottle. The number of moles in an arbitrary mass of an element or compound can be calculated as:
number of moles = weight (g)/ atomic (or molecular) weight (g)
Molarity (M) is the unit used to describe the number of moles of an element or compound in one liter (L) of solution (M = moles/L) and is thus a unit of concentration. By this definition, a 1.0 M solution is equivalent to one molecular weight (g/mole) of a compound brought up to 1 liter (1.0 L) volume with solvent (e.g., water) at a fixed temperature (liquids expand and contract with temperature and thus can change molarity).
Example 1: To prepare a liter of a molar solution from a dry reagent
Multiply the molecular weight (or FW) by the desired molarity to determine how many grams of reagent to use:
Suppose a compound’s MW = 194.3 g/mole
to make 0.15 M solution use 194.3 g/mole * 0.15 moles/L = 29.145 g/L
You would dissolve the specified mass of reagent in a fraction of the total volume of solvent (at STP) and then raise the volume to exactly one liter by adding additional solvent and mixing thoroughly.
Example 2: To prepare a specific volume of a specific molar solution from a dry reagent
A chemical has a FW of 180 g/mole and you need 25 ml (0.025 L) of 0.15 M (M = moles/L) solution. How many grams of the chemical are needed to make this solution?
#grams/desired volume (L) = desired molarity (mole/L) * FW (g/mole)
by algrebraic rearrangement,
#grams = desired volume (L) * desired molarity (mole/L) * FW (g/mole) #grams = 0.025 L * 0.15 mole/L * 180 g/mole
after cancelling the units,
#grams = 0.675 g
Solutions Containing Multiple Reagents
Complex solutions such as buffers, salines, fixatives, etc., may be comprised of multiple chemical reagents. In preparation of these solutions, each reagent is dealt with separately in determining how much to use to make the final solution. For each, the volume used in the calculations is the final volume of solution needed.
More examples of worked problems: About.com: Chemistry
5. Percent Solutions (% = parts per hundred or grams/100 ml)
Many reagents are mixed as percent solutions either as weight per volume (w/v) when starting with dry reagents OR volume per volume (v/v) when starting with liquid reagents. When preparing solutions from dry reagents, the same mass of any reagent is used to make a given percent concentration although the molar concentrations would be different.
Weight percent (w/v) = [mass of solute (g)/ volume of solution (ml)] x 100, and,
Volume percent (v/v) = [volume of solute (ml)/ volume of solution (ml)] x 100
For example, a 100 ml of 10% solution of any dry reagent would contain 10 g dry reagent in a final volume of 100 ml. A 10% (v/v) solution would contain 10 ml solute/ 100 ml solution volume.
Example 1: If you want to make 200 ml of 3 % NaCl you would need 0.03 g/ml x 200 ml = 6.0 g NaCl in 200 ml water.
When using liquid reagents the percent concentration is based upon volume per volume , and is similarly calculated as % concentration x volume needed = volume of reagent to use .
Example 2: If you want to make 2 L of 70% ethanol from 100% ethanol you would mix 0.70 ml/ml x 2000 ml = 1400 ml ethanol with 600 ml water.
To convert from % solution to molarity , multiply the % solution by 10 to express the percent solution grams/L, then divide by the formula weight.
Molarity = (grams reagent/100 ml) * 10
Example 1 : Convert a 6.5 % solution of a chemical with FW = 325.6 to molarity,
[(6.5 g/100 ml) * 10] / 325.6 g/mole = [65 g/L] / 325.6g/mole = 0.1996 M
To convert from molarity to percent solution , multiply the molarity by the FW and divide by 10:
% solution = molarity * FW
Example 2: Convert a 0.0045 M solution of a chemical having FW 178.7 to percent solution:
[0.0045 moles/L * 178.7 g/mole] / 10 = 0.08 % solution
6. Concentrated stock solutions - using "X" units
Stock solutions of stable compounds are routinely maintained in labs as more concentrated solutions that can be diluted to working strength when used in typical applications. The usual working concentration is denoted as 1x. A solution 20 times more concentrated would be denoted as 20x and would require a 1:20 dilution to restore the typical working concentration.
Example : A 1x solution of a compound has a molar concentration of 0.05 M for its typical use in a lab procedure. A 20x stock would be prepared at a concentration of 20*0.05 M = 1.0 M. A 30X stock would be 30*0.05 M = 1.5 M.
7. Normality (N): Conversion to Molarity
Normality = n*M where n = number of protons (H+) in a molecule of the acid.
Example : In the formula for concentrated sulfuric (36 N H2SO4), there are two protons, so, its molarity= N/2. So, 36N H2SO4 = 36/2 = 18 M.
The 95 percent confidence intervals in the tables have the following meaning:
Before the tubes are inoculated, the chance is at least 95 percent that the confidence interval associated with the eventual result will enclose the actual concentration.
It is possible to construct many different sets of intervals that satisfy this criterion. This manual uses a modification of the method of de Man (1983). De Man calculated his confidence limits iteratively from the smallest concentrations upward. Because this manual emphasizes pathogens, the intervals have been shifted slightly upward by iterating from the largest concentrations downward.
The confidence intervals of the spreadsheet and the tables associated with this appendix may be different. The MPN Excel spreadsheet uses a normal approximation to the log (MPN) to calculate its confidence intervals. This approximation is similar to a normal approximation discussed in Haldane (1939). This approximation is less computationally intense so more appropriate for a spreadsheet than de Man's confidence intervals.
How should the interval between serial dilutions be expressed? - Biology
We estimate the distribution of serial intervals for 468 confirmed cases of coronavirus disease reported in China as of February 8, 2020. The mean interval was 3.96 days (95% CI 3.53–4.39 days), SD 4.75 days (95% CI 4.46–5.07 days) 12.6% of case reports indicated presymptomatic transmission.
Key aspects of the transmission dynamics of coronavirus disease (COVID-19) remain unclear (1). The serial interval of COVID-19 is defined as the time duration between a primary case-patient (infector) having symptom onset and a secondary case-patient (infectee) having symptom onset (2). The distribution of COVID-19 serial intervals is a critical input for determining the basic reproduction number (R0) and the extent of interventions required to control an epidemic (3).
To obtain reliable estimates of the serial interval, we obtained data on 468 COVID-19 transmission events reported in mainland China outside of Hubei Province during January 21–February 8, 2020. Each report consists of a probable date of symptom onset for both the infector and infectee, as well as the probable locations of infection for both case-patients. The data include only confirmed cases compiled from online reports from 18 provincial centers for disease control and prevention (https://github.com/MeyersLabUTexas/COVID-19).
Figure. Estimated serial interval distribution for coronavirus disease (COVID-19) based on 468 reported transmission events, China, January 21–February 8, 2020. A) All infection events (N = 468) reported across 93 cities of.
Fifty-nine of the 468 reports indicate that the infectee had symptoms earlier than the infector. Thus, presymptomatic transmission might be occurring. Given these negative-valued serial intervals, COVID-19 serial intervals seem to resemble a normal distribution more than the commonly assumed gamma or Weibull distributions (4,5), which are limited to positive values (Appendix). We estimate a mean serial interval for COVID-19 of 3.96 (95% CI 3.53–4.39) days, with an SD of 4.75 (95% CI 4.46–5.07) days, which is considerably lower than reported mean serial intervals of 8.4 days for severe acute respiratory syndrome (5) to 14.6 days (6) for Middle East respiratory syndrome. The mean serial interval is slightly but not significantly longer when the index case is imported (4.06 [95% CI 3.55–4.57] days) versus locally infected (3.66 [95% CI 2.84–4.47] days), but slightly shorter when the secondary transmission occurs within the household (4.03 [95% CI 3.12–4.94] days) versus outside the household (4.56 [95% CI 3.85–5.27] days) (Figure). Combining these findings with published estimates for the early exponential growth rate COVID-19 in Wuhan (7), we estimate an R0 of 1.32 (95% CI 1.16–1.48) (5), which is lower than published estimates that assume a mean serial interval exceeding 7 days (7,8).
These estimates reflect reported symptom onset dates for 752 case-patients from 93 cities in China, who range in age from 1 to 90 years (mean 45.2 years, SD 17.21 years). Recent analyses of putative COVID-19 infector–infectee pairs from several countries have indicated average serial intervals of 4.0 days (95% CI 3.1–4.9 days n = 28 unpub. data, H. Nishiura et al., unpub. data, https://doi.org/10.1101/2020.02.03.20019497), 4.4 days (95% CI 2.9–6.7 days, n = 21 S. Zhao et al., unpub. data, https://doi.org/10.1101/2020.02.21.20026559], and 7.5 days (95% CI 5.3–19, n = 6 8). Whereas none of these studies report negative serial intervals in which the infectee had symptoms before the infector, 12.6% of the serial intervals in our sample were negative.
We note 4 potential sources of bias. First, the data are restricted to online reports of confirmed cases and therefore might be biased toward more severe cases in areas with a high-functioning healthcare and public health infrastructure. The rapid isolation of such case-patients might have prevented longer serial intervals, potentially shifting our estimate downward compared with serial intervals that might be observed in an uncontrolled epidemic. Second, the distribution of serial intervals varies throughout an epidemic the time between successive cases contracts around the epidemic peak (9). A susceptible person is likely to become infected more quickly if they are surrounded by 2 infected persons instead of 1. Because our estimates are based primarily on transmission events reported during the early stages of outbreaks, we do not explicitly account for such compression and interpret the estimates as basic serial intervals at the outset of an epidemic. However, if some of the reported infections occurred amid growing clusters of cases, then our estimates might reflect effective (compressed) serial intervals that would be expected during a period of epidemic growth. Third, the identity of each infector and the timing of symptom onset were presumably based on individual recollection of past events. If recall accuracy is impeded by time or trauma, case-patients might be more likely to attribute infection to recent encounters (short serial intervals) over past encounters (longer serial intervals). In contrast, the reported serial intervals might be biased upward by travel-related delays in transmission from primary case-patients that were infected in Wuhan or another city before returning home. If their infectious period started during travel, then we might be unlikely to observe early transmission events with shorter serial intervals. The mean serial interval is slightly higher for the 218 of 301 unique infectors reported to have imported cases.
Given the heterogeneity in type and reliability of these sources, we caution that our findings should be interpreted as working hypotheses regarding the infectiousness of COVID-19, requiring further validation. The potential implications for COVID-19 control are mixed. Although our lower estimates for R0 suggest easier containment, the large number of reported asymptomatic transmission events is concerning.
Dr. Du is a postdoctoral researcher in the Department of Integrative Biology at the University of Texas at Austin. He develops mathematical models to elucidate the transmission dynamics, surveillance, and control of infectious diseases.
On December 31 2019, the city of Wuhan (Hubei province, China) reported an outbreak of atypical pneumonia caused by a novel coronavirus, later on named SARS-CoV-2 . Viral and epidemiological characteristics of COVID-19 outbreaks such as asymptomatic transmissions [2, 3] and undiagnosed individuals  hastened the spread of the disease, resulting in a global pandemic. In response to this threat, several countries adopted drastic and unprecedented control measures such as national and regional lockdown in which diagnosed or traced individuals were confined in isolation or quarantine . Concurrently, researchers started to estimate key epidemiological quantities such as the incubation period, the serial interval, the generation time and the reproduction number (see e.g. [3, 6–11]). It is important to note that most of these epidemiological determinants were inferred in situations for which different intervention strategies were in place, possibly affecting the resulting estimates. More precisely, control measures such as quarantine or isolation of infectives intervene in reducing the social contacts of such individuals, and consequently in decreasing their probability of transmission after diagnosis. These interventions lead to a delay of the epidemic peak  and to a decrease of the effective reproduction number . Therefore, it is reasonable to think that other epidemic characteristics are also affected by public health interventions.
A key quantity in epidemic modelling is the generation time or generation interval, i.e. the time difference between the infection times of an infector-infectee pair . In particular, the generation time can be used to compute the basic and effective reproduction numbers by means of the Lotka-Euler equation . Often, the generation time is approximated by the serial interval , i.e. the time difference between the onset of symptoms of an infector-infectee pair, and, by doing so, estimates of the basic reproduction number for the COVID-19 epidemic have been computed [10, 11]. Britton and Scalia Tomba  showed that replacing the generation time with the serial interval could lead to an understimation of the (basic) reproduction number. In fact, while they have the same mean, the variance of the former is larger than that of the latter quantity. In addition, Park et al. discuss that if asymptomatic and symptomatic individuals have different generation time distributions, estimation of the reproduction number is biased if performed using only data for the symptomatic population . Non-pharmaceutical interventions such as isolation, which decrease the social interactions of diagnosed individuals, affect the realized generation time distribution of only diagnosed individuals, likely causing the aforementioned bias. Moreover, Park et al. showed how the serial interval can be used to estimate the basic reproduction number . However, in their analysis, they did not consider the effect of control measures on the realized serial and generation intervals. Recently, Ali et al. investigated the relationship between serial interval and non-pharmaceutical interventions, reporting a modification of serial interval distributions as a function of control measures in Chinese data . In particular, they showed that the mean serial interval decreases when control measures are more stringent. Similarly, Sun et al. reported a contraction of the serial and generation intervals when different intervention measures were in place . However, the quantities thereof could contract differently, resulting in a difference between their mean values. In fact, the relationship between serial and generation interval has not yet been explored in the setting of control measures.
In this work we highlight three characteristics of the COVID-19 pandemic dynamic that affect the mean and the variance of the serial and generation intervals. First, the efficacy of control measures, determined by the starting timing of quarantine and isolation relative to exposure and by the magnitude of the contact rate reduction. Second, the presence of an undiagnosed population, of which the individuals escape diagnosis, such that it is not affected by the considered interventions. Third, the relationship between infectiousness, incubation period and time at which interventions take place during the infectious period. After presenting a theoretical description of the effect of control measures on the expected serial and generation intervals, we set up a simulation study to gain insights into the determinants that affect the realized serial and generation intervals in the presence of control measures.
Serial dilutions involve diluting a stock or standard solution multiple times in a row. Typically, the dilution factor remains constant for each dilution, resulting in an exponential decrease in concentration. For example, a ten-fold serial dilution could result in the following concentrations: 1 M, 0.1 M, 0.01 M, 0.001 M, and so on. As is evidenced in this example, the concentration is reduced by a factor of ten in each step. Serial dilutions are used to accurately create extremely diluted solutions, as well as solutions for experiments that require a concentration curve with an exponential or logarithmic scale. Serial dilutions are widely used in experimental sciences, including biochemistry, pharmacology, microbiology, and physics.
Solving Dilution Problems in Solution Chemistry CLEAR & SIMPLE – YouTubeThis video shows how to solve two dilution problems, using the standard dilution formula, M1V1 = M2V2.
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MATERIALS AND METHODS
To examine the theoretical effect of dilution on community inoculum composition, a series of numerical simulations (coded in MATLAB) was performed. Communities were constructed by assigning each of 10 6 individuals a random species identification based on a normal distribution of integers from 1 to 1,000. The mean of this distribution was set at 500 the variance (var) was adjusted in order to create communities of different initial evenness (Fig. (Fig.1A). 1 A). The var levels used were 100, 250, 1,000, and 20,000, and a perfectly even initial community was also generated (1,000 types, each containing 1,000 individuals). Dilution of each of these five initial communities was simulated by randomly selecting 1/10 of the individuals from the array representing the undiluted community. The species identification of each individual in this subset was copied to a second array, which served as the initial community for the next dilution in the series (dilutions extended through 10 𢄥 ). For each community, at each dilution level, richness, evenness, and diversity were calculated. Richness (S) was taken to be the number of organism types (species) in the community. Diversity was expressed as the Shannon-Wiener index (H′) as fol- i lows: H′ = Σ pi ln pi, where i indicates each species or category and pi is the 1 proportion of individuals of each species (37). Evenness (E) was calculated as E = H′/H′max, where H′max = In S (31).
(A) Distribution of individuals among 1,000 types in the initial communities used in simulations. Note that both total abundance and richness were the same in each of the five communities. The mean of each distribution was set at 500, and the variance was altered to simulate communities with a dominant (var = 100, 250, or 1,000) or relatively even (var = 20,000 or even) distribution. (B to D) Simulation results showing how community structure differed in the various initial communities for each serial dilution. The x axis represents the negative exponent of the dilution factor (e.g., 4 corresponds to a 10 𢄤 dilution), and the y axis represents richness (number of types or species) (B), evenness (C), or the calculated value of the Shannon-Wiener diversity index (D).
Batch culture experiments. (i) Microcosm setup.
Raw sewage was collected from the Cape Canaveral Air Station Wastewater Treatment Facility (Kennedy Space Center, Fla.) and used as the inoculum for the microcosm experiments. A single large sample (2 liters) was collected from the equilibration basin and allowed to settle for approximately 2 h to remove large particles. Tenfold serial dilutions (through 10 𢄦 ) of the supernatant were prepared in autoclaved sewage each of these different dilutions then served as an inoculum for a series of batch culture incubations. Before dilution, the concentration of cells in the supernatant, determined by acridine orange direct counting (AODC 15), was 1.8 × 10 6 cells/ml.
The batch cultures were established by adding 1 ml of inoculum to 60 ml of autoclaved sewage in a 125-ml Erlenmeyer flask. Seven treatments (10 0 through 10 𢄦 ) were maintained in this experiment, with three replica communities at each dilution. All flasks were capped with sterile foam plugs to prevent contamination and kept on a shaker table, operated at 150 rpm, to maintain aerobic conditions. Each day, 20 ml of liquid was removed from each flask and replaced with 20 ml of sterile sewage. After 9 days (three retention times), flasks were harvested and analyzed.
(ii) Cultural counts and diversity of colony morphology.
For each flask, a serial dilution of the regrown community was plated, in duplicate, onto both R2A agar (35) and sterile sewage medium (SM prepared by mixing 15 g of agar per liter of sewage supernatant). Plates were incubated at room temperature (ca. 23ଌ), and the number of CFU on R2A agar was determined after 3 days. Growth on SM was evaluated after 6 days.
The diversity of colony morphologies on R2A was compared across the different dilution treatments. For each flask, two plates were selected on each plate, 25 colonies were randomly chosen and colony morphology was described based on size, pigmentation, form, elevation, and surface. Richness (number of distinct colony morphologies), evenness, and diversity (Shannon-Wiener diversity index) were then calculated.
For each flask, a 10 𢄡 dilution of the microbial community was prepared (in sterile water) and inoculated into Biolog GN microplates, which were then incubated at room temperature (ca. 23ଌ). Color formation in each of the 96 wells of each plate was monitored by periodically (every 2 to 4 h) measuring the A590 using a Biotek EL 320 microplate reader. Data were normalized using a blank-corrected average well color development of 0.75 absorbance unit and analyzed using a principal-components analysis (PCA) (8, 10).
(iv) Dilution-extinction analysis of CLPP.
Dilution-extinction analysis was performed on a subset of the regrown communities (one replicate flask from each of the 10 0 , 10 𢄢 , 10 𢄤 , and 10 𢄦 dilution treatments) to determine the relationship between cell density (I) and functional richness (number of positive wells, R) (9). Serial dilutions of the microbial suspensions were inoculated into Biolog GN microplates and incubated at room temperature for 7 days, and A590 was measured. A positive response was defined as any value greater than 0.25 absorbance unit (after correction for the control well), and a hyperbolic model, R = (Rmax × I)/(KI + I), where Rmax equals the maximum (asymptotic level) of R and KI is the value of I when R is one-half of Rmax, was fitted to the data (9).
(v) Molecular analysis of whole-community DNA. (a) DNA extraction and quantification.
At harvest, an approximately 40-ml sample was collected from each flask and the suspended microbial community was concentrated by centrifugation (23,000 × g, 20 min). The cell pellet was resuspended in 200 μl of phosphate-buffered saline and stored at ଌ. Whole-community DNA was extracted using the High Pure PCR Template Preparation Kit (Boehringer Mannheim, Indianapolis, Ind.) and quantified using the PicoGreen double-stranded DNA quantification reagent (Molecular Probes, Eugene, Oreg.).
Amplified fragment length polymorphism (AFLP) was completed using the Perkin-Elmer Microbial Fingerprinting Kit (PE Applied Biosystems, Foster City, Calif.) following the manufacturer's instructions for analysis of individual bacterial strains. Three different pairs of primers, each with a different fluorescent label, were used for selective AFLP amplification: EcoRI-AA (JOE labeled) with MseI-CA, EcoRI-AC (FAM labeled) with MseI-CC, and EcoRI-AT (NED labeled) with MseI-CT. For the complete primer and adapter sequences and an explanation of the primer selection criteria used, see the PE Applied Biosystems AFLP Microbial Fingerprinting Protocol (PE Applied Biosystems, Foster City, Calif.).
Selective amplification products were resolved using an ABI Prism 310 Genetic Analyzer by following the manufacturer's instructions with slight modification for each sample with each primer pair, 1 μl of PCR product was analyzed using a sample injection time of 10 s. Data were analyzed using the Genotyper software (PE Applied Biosystems), and the presence or absence of each peak in each sample was coded as 1 or 0. Such a data matrix was prepared for each primer pair, and the information from the three primer pairs was pooled into a single large data set. The Jaccard coefficient was used to determine distances between samples (relative similarity), and a cluster analysis (unweighted pair group clustering using arithmetic averages and between-groups linkage) was performed. A bootstrapping analysis was then used to assess the significance of each group and subgroup in the cluster analysis (6).
A PCA was also performed on the original pooled data matrix (SPSS 9.0), and plots of the first two principal components (PCs) were made. As PCA is not mathematically appropriate for use with binary data, its application in this study was solely to aid in visualization of the relationships among the samples and not for statistical evaluation. Such an approach has been used several times to compare samples profiled using a variety of similar genetic techniques (6, 7, 47, 48) PCA generally provides the same information (groupings and relative distances among samples) as the above-outlined cluster analysis.
Each T-RFLP (terminal restriction fragment length polymorphism) reaction mixture (50 μl) contained 25 ng of community DNA 10 mM Tris-Cl (pH 8.3) 50 mM KCl 1.5 mM MgCl2 200 μM each dATP, dCTP, dGTP, and dTTP each primer at a concentration of 0.1 μM and 1.25 U of Taq DNA polymerase (21). The bacterial 16S rRNA gene was amplified using two primers: 1392 Reverse (5′ ACGGGCGGTG TGTRC) and 27 Forward (5′ AGAGTTTGATCCTGGCTCAG [labeled with the fluorescent tag 6-FAM]). The PCR program was 94ଌ for 3 min, followed by 30 cycles of 94ଌ for 30 s, 56ଌ for 45 s, and 72ଌ for 2 min, with a final extension at 72ଌ for 3 min. PCR products were purified using the Wizard PCR Preps DNA Purification System (Promega, Madison, Wis.) and eluted in a final volume of 50 μl. Portions (10 μl) of the purified PCR product were then digested with either the HhaI or MspI restriction enzyme, using the manufacturer's recommended reaction buffer and 20 U of enzyme (New England Biolabs, Beverly, Mass.). Digests were incubated at 37ଌ for 4 h.
The lengths of the fluorescently labeled terminal restriction fragments were determined for each sample using the ABI 310 Genetic Analyzer. Three-microliter portions of each digested product were mixed with 24 μl of deionized formamide and 1 μl of GeneScan-1000 size standard (PE Applied Biosystems), denatured at 95ଌ for 5 min, and quickly chilled on ice. Electrophoresis was performed using the same conditions as for AFLP but with a 40-min run time. Data were analyzed using the GeneScan software with a peak height detection of 100. As with the AFLP analysis, the presence or absence of each T-RFLP in each sample was determined and the data from each restriction enzyme were pooled for cluster analysis, bootstrapping analysis, and PCA.
How to Calculate Units of Concentration
Once you have identified the solute and solvent in a solution, you are ready to determine its concentration. Concentration may be expressed several different ways, using percent composition by mass, volume percent, mole fraction, molarity, molality, or normality.
Percent Composition by Mass (%)
This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100.
Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.
20 g NaCl / 100 g solution x 100 = 20% NaCl solution
Volume Percent (% v/v)
Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Volume percent is defined as:
v/v % = [(volume of solute)/(volume of solution)] x 100%
Note that volume percent is relative to the volume of the solution, not the volume of solvent. For example, wine is about 12% v/v ethanol. This means there is 12 ml ethanol for every 100 ml of wine. It is important to realize liquid and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution.
As another example, 70% v/v rubbing alcohol may be prepared by taking 700 ml of isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution (which will not be 300 ml).
Mole Fraction (X)
This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Keep in mind, the sum of all mole fractions in a solution always equals 1.
Example:What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18 molecular weight of glycerol = 92)
90 g water = 90 g x 1 mol / 18 g = 5 mol water
92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol
total mol = 5 + 1 = 6 mol
xwater = 5 mol / 6 mol = 0.833
x glycerol = 1 mol / 6 mol = 0.167
It's a good idea to check your math by making sure the mole fractions add up to 1:
xwater + xglycerol = .833 + 0.167 = 1.000
Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!).
What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution? (The molecular weight of CaCl2 = 110)
11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2
100 mL x 1 L / 1000 mL = 0.10 L
molarity = 0.10 mol / 0.10 L
molarity = 1.0 M
Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water.
Example:What is the molality of a solution of 10 g NaOH in 500 g water? (Molecular weight of NaOH is 40)
10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.05 M / kg
molality = 0.50 m
Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capacity of a given molecule. Normality is the only concentration unit that is reaction dependent.
1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H + ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.
- Grams per Liter (g/L)
This is a simple method of preparing a solution based on grams of solute per liter of solution.
- Formality (F)
A formal solution is expressed regarding formula weight units per liter of solution.
- Parts per Million (ppm) and Parts per Billion (ppb)Used for extremely dilute solutions, these units express the ratio of parts of solute per either 1 million parts of the solution or 1 billion parts of a solution.
A sample of water is found to contain 2 ppm lead. This means that for every million parts, two of them are lead. So, in a one gram sample of water, two-millionths of a gram would be lead. For aqueous solutions, the density of water is assumed to be 1.00 g/ml for these units of concentration.
MICROSOFT OFFICE 2003
- Once the data is highlighted, look on the upper toolbar and click on &ldquoInsert&rdquo, then on &ldquoChart&rdquo. A box with graph options will appear. Select &ldquoXY(Scatter)&rdquo and click on &ldquoNext&rdquo. (Leave the chart sub-type alone: the top chart should be highlighted.)
Your data will appear already plotted on an x-y axis system. Now you will learn how to set-up the axes, write the title of the graph and write the legend. You will also command the computer to draw best-fit lines and calculate the slope of each plot. You can also choose to make some cosmetic touches (changing the size of fonts, colors of the plot etc.&hellip) on your graph before printing.
- Find &ldquoSeries&rdquo towards the top of the graph and click on it. &ldquoSeries1&rdquo should appear highlighted. To the right of this there is an empty space for you to write the NAME of this plot: click on this empty box and type &ldquoAbsorbance vs. Bacterial Numbers&rdquo.
- Click on &ldquoNext&rdquo. Another box will appear with options to modify titles on the graph.
a. Make sure that in the window &ldquoChart Title&rdquo your graph is titled (e.g. Using Turbidemetry to Guesstimate Bacterial Numbers&rdquo).
b. Click on space below Value(X)Axis and type: Absorbance
c. Click on space below Value(Y)Axis and type: Bacterial Numbers (X 106)
d. Click on &ldquoAxes&rdquo (upper left of graph) and make sure that Value(X) Axis and Value(Y) Axis have a &radic .
e. Click on &ldquoGridlines&rdquo and add lines. Click on &ldquoMajor Gridlines&rdquo on the x-axis and y-axis.)
f. Click on &ldquoLegend&rdquo and choose the location of the legend (I chose bottom).
- Click on &ldquoNext&rdquo click on the button titled Place chart as object in sheet 1 (NOT as new sheet). Click on &ldquoFinish&rdquo.
Your graph will appear on the screen as shown below. You can alter its size by clicking the mouse button on the black squares around the graph and dragging the mouse with the mouse button pressed down.
- Now you are going to add the line that goes through the data points (trendline). Click on the graph to make sure that the black squares around the graph are there.
Move the pointer to any of the data points of a particular series (any &Delta, for example). As you do this, information about this data point will appear on the screen. Press the RIGHT mouse button. A box with choices appears. Select and click on &ldquoAdd Trendline&rdquo. Another box will appear. Select Linear (should already be highlighted). Before clicking OK, go to &ldquoOptions&rdquo on the top of the box. Click on &ldquoDisplay equation on chart&rdquo. Click on &ldquoOK&rdquo. The graph will show a best-fit line. On the graph you should see an equation which gives you the value of the slope for the line).
THE GRAPH SHOULD LOOK LIKE THE PICTURE ABOVE, BUT WITH A LINE AND THE SLOPE FORMULA.
What is CFU Calculation?
Now that you have some data, you can do the CFU calculation in the original sample.
1. Take the amount you plated (0.5 mL) and multiply by the dilution factor (0.01) to yield 0.005. You must do this to find the dilution factor which yielded your CFU count. Here, half a milliliter of the 1:100 dilution allowed you to count CFU.
2. Divide the CFU from the dilution (179) by the result from Step 1 (0.005) to yield 35,800 CFU.
This means that the original 1 mL of sample that was diluted contains 35,800 CFU. Another way to put this is to say that the original sample has 35,800 CFU/mL.
CFU is a great tool to find how many bacteria, fungi or any microorganisms there are in a given sample. The plating method described here is particularly helpful for very concentrated samples or organisms that don't grow well in liquid culture.