Estimating Model Parameters
Some examples

Example 1, a monocyclic epidemic: Flax wilt is caused by the fungus Fusarium oxysporum f. sp. lini. Chlamydospores of the fungus persist for several years in the soil, and when flax is planted in an infested field, the young plants are infected through the roots. An extensive soil survey was made of a heavily infested field, and it was found to contain an average of 57 colony-forming units per gram of soil. When a susceptible flax cultivar was planted in this field, the percent of plants showing wilt symptoms increased with time as follows:

      Days After                % Plants
       Planting                 Infected

          10                       18
          20                       56
          30                       82
          40                       91
          50                       96
          60                       98
      

In a plot of disease progress, note how the percent infection asymptotically approaches 100 percent.

[Click on graph]
Graph, Fusarium wilt on flax
Flax wilt disease progress

To estimate the product, QR, we first have to convert percent infection to the proportion, x, and then using the transformation appropriate for the monocyclic model, calculate ln(1/(1-x)).

           t            x          ln(1/(1-x))

          10           .18            0.198
          20           .56            0.821
          30           .82            1.71
          40           .91            2.41
          50           .96            3.22
          60           .98            3.91
      

From a plot of ln(1/(1-x)) versus t, we can fit a straight line to the data points using least squares regression.

[Click on graph]
Graph, Transformed flax wilt
Flax wilt, multiple hit transformation

The slope of the line estimated by the regression equation is 0.076, which is the value of QR. Therefore,

       R = 0.076/57 = 0.0013/CFU/Day.

Example 2, a polycyclic epidemic: Halo blight of beans is caused by the bacterium Pseudomonas syringae pv. phaseolicola. The major source of initial inoculum is infected seeds that when planted give rise to plants with lesions on the primary leaves. Bacteria produced in these lesions are splash dispersed to adjacent healthy plants. New lesions can themselves produce secondary inoculum within about 4-5 days. Under conditions moderately favorable for disease development, the following observations were made of disease progress:

      Days After            % Plants
       Planting              Infected

          10                    1
          20                    4
          30                   15
          40                   31
          50                   65
          60                   88
          70                   94
      

Disease progress shows the sigmoid-shaped curve characteristic of a polycyclic epidemic.

[Click on graph]
Graph, Halo blight on bean
Halo blight disease progress

As with the previous epidemic, we have to convert the percents to proportions (x), but this time the transformation that we use is ln(x/(1-x)).

          t           x           ln(x/(1-x))

         10          .01            -4.60
         20          .04            -3.18
         30          .15            -1.73
         40          .31            -0.80
         50          .65             0.62
         60          .88             1.99
         70          .94             2.75
      

By plotting ln(x/(1-x)), sometimes called the logits of x, versus t, we can fit a straight line to the data points with least squares regression.

[Click on graph]
Graph, Halo blight on bean
Halo blight, logistic transformation

The regression gives us a slope of 0.124/day, which is our estimate of the apparent infection rate, r.

Obviously the more data points that we have, particularly if they are relatively evenly distributed on both sides of 50% infection, the better estimate we will have of the apparent infection rate. However, it is possible to make a rough estimate of the apparent infection rate with just two data points. Let us suppose for a moment that instead of observations every ten days during the epidemic, we only made two observations, one early (day 10) and one late (day 70). How might we estimate the apparent infection rate? In this case we would be using only the first and the final points in the plot of the transformed data above and calculating the slope as the rise over the run:

      r  =  (ln(0.94/(1.0-0.94)) - ln(0.01/1.0-0.01))) / (70 - 10)

         =  (2.75 + 4.60) / 60

         =  0.123/day

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