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The potential for sexual transmission to compromise control of Ebola virus outbreaks John E. Vinson1, John M. Drake1, Pejman Rohani1,2, Andrew W. Park1,2,+ 1: Odum School of Ecology, University of Georgia, Athens GA, 30602, USA 2: Dept. Infectious Diseases, College of Veterinary Medicine, University of Georgia, Athens GA, 30602, USA +: author for correspondence, [email protected] Supplemental Material 1. Fitting Gamma Distribution Function A recent convenience sample of men who had recovered from Ebola virus infection during the 2013-2015 epidemic [3] suggested that the virus remains present in semen for a considerable time after recovering (Fig. S1). We can fit the data to an exponential distribution through the probability density function (pdf) of a gamma distribution: ηα α−1 −ηt t e Γ(α) When the shape parameter α = 1, the pdf of the gamma distribution is of an exponential distribution with rate parameter η. In a compartmental model, an exponential distribution of the sexually infectious class would only have one 1 compartment with an average infectious period of η months. We used maximum likelihood to estimate. The estimated value when α = 1 is η = 0.103 with negative log-likelihood of 49.315. The data also suggests a gamma distribution with α ≠ 1. A compartmental model with the sexually infectious class described through a gamma distribution would have α pseudo-classes with a pseudo-class transition rate of η. We begin by fitting the pdf of the gamma distribution to the data. The estimated parameters are α = 4.929 and η = 0.711. We require that the number of pseudo-classes to be a discrete integer, we must refit the distribution with α = 5. After refitting, the optimized parameters are α = 5 and η = 0.722 with negative loglikelihood 46.008. Fig. S1: Probability of semen containing Ebola virus as a function of month postrecovery (with 95% binomial confidence intervals), adapted from [3]. The dashed line is a fitted gamma distribution with shape parameter α = 5 and rate parameter η = 0.722 with a mean of 6.925 months. The dotted line is a fitted exponential distribution with rate parameter η = 0.103 with mean 9.709 months (gamma distribution with α = 1). To determine which distribution (exponential or gamma) fits the data best we perform a likelihood ratio test defined as: Λ(x) = L(α = 1, η = 0.103|x) L(α = 5, η = 0.722|x) where x is the data of the convenience sample. From the likelihoods given by these parameter values −2 ∗ log(Λ(x)) = 6.615. We determine that the gamma distribution with α = 5 and η = 0.722 is the best fit distribution (α = 0.05, χ2 = 3.841). 2. Next Generation Method for Calculating the Virus's Reproductive Ratio We begin with defining the probability of dying from infection as ψ. We write the equation for İ: ∑5j=1 Yj ψ γI N 1−ψ ∑5j=1 Yj 1 −ϕI = βC e SI + βS S− γI N 1−ψ İ = βC e−ϕI SI + βS S − γI − We now integrate the equation into our system: Ṡ İ Ẏ1 Ẏ2 Ẏ3 Ẏ4 Ẏ5 Ṙ = −βC e−ϕI SI − βS = βC e−ϕI SI + βS ∑5j=1 Yj ∑5j=1 Yj N S− N = pγI − ηY1 = η(Y1 − Y2 ) = η(Y2 − Y3 ) = η(Y3 − Y4 ) = η(Y4 − Y5 ) = (1 − p)γI + ηY5 S 1 γI 1−ψ We can use the next generation matrix (NGM) method to find the R 0 of the system. Following the next generation matrix method, we identify that the I, Y1 , Y2 , Y3 , Y4 , Y5 classes are of interest. We generate two matrices, the gains (G) (an infection event increases this class) and losses (L) (loss of a current or future infector) containing the gains and loss terms differentiated with respect to each class and evaluated at disease free equilibrium: ∑5j=1 Yj d (β SI + βS S) dI C N ∑5j=1 Yj d (βC SI + βS S) dY1 N ∑5j=1 Yj d (βC SI + βS S) dY2 N G= ∑5j=1 Yj d (βC SI + βS S) dY3 N ∑5j=1 Yj d (βC SI + βS S) dY4 N ∑5j=1 Yj d (βC SI + βS S) [dY5 N d 1 ( γI) dI 1 − ψ d 1 ( γI) dY1 1 − ψ d 1 ( γI) dY2 1 − ψ L= d 1 ( γI) dY3 1 − ψ d 1 ( γI) dY4 1 − ψ d 1 ( γI) [dY5 1 − ψ d (ηY1 − pγI) dI d (ηY1 − pγI) dY1 d (ηY1 − pγI) dY2 d (ηY1 − pγI) dY3 d (ηY1 − pγI) dY4 d (ηY1 − pγI) dY5 d (0) dI d (0) dY1 d (0) dY2 d (0) dY3 d (0) dY4 d (0) dY5 d (ηY2 − ηY1 ) dI d (ηY2 − ηY1 ) dY1 d (ηY2 − ηY1 ) dY2 d (ηY2 − ηY1 ) dY3 d (ηY2 − ηY1 ) dY4 d (ηY2 − ηY1 ) dY5 d (0) dI d (0) dY1 d (0) dY2 d (0) dY3 d (0) dY4 d (0) dY5 d (0) dI d (0) dY1 d (0) dY2 d (0) dY3 d (0) dY4 d (0) dY5 d (ηY3 − ηY2 ) dI d (ηY3 − ηY2 ) dY1 d (ηY3 − ηY2 ) dY2 d (ηY3 − ηY2 ) dY3 d (ηY3 − ηY2 ) dY4 d (ηY3 − ηY2 ) dY5 d (0) dI d (0) dY1 d (0) dY2 d (0) dY3 d (0) dY4 d (0) dY5 d (ηY4 − ηY3 ) dI d (ηY4 − ηY3 ) dY1 d (ηY4 − ηY3 ) dY2 d (ηY4 − ηY3 ) dY3 d (ηY4 − ηY3 ) dY4 d (ηY4 − ηY3 ) dY5 We now have matrices G and L as: βC N βS βS G= βS βS [ βS 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0] 1 γ −pγ 0 0 0 0 1−ψ 0 η −η 0 0 0 L= 0 0 η −η 0 0 0 0 0 η −η 0 0 0 0 0 η −η [ 0 0 0 0 0 η] We now invert the L matrix to get: d (0) dI d (0) dY1 d (0) dY2 d (0) dY3 d (0) dY4 d (0) dY5 ]DFE d (ηY5 − ηY4 ) dI d (ηY5 − ηY4 ) dY1 d (ηY5 − ηY4 ) dY2 d (ηY5 − ηY4 ) dY3 d (ηY5 − ηY4 ) dY4 d (ηY5 − ηY4 ) dY5 ]DFE 1−ψ γ 0 0 −1 L = 0 0 [ 0 p(1 − ψ) p(1 − ψ) p(1 − ψ) p(1 − ψ) p(1 − ψ) η η η η η 1 1 1 1 1 η η η η η 1 1 1 1 0 η η η η 1 1 1 0 0 η η η 1 1 0 0 0 η η 1 0 0 0 0 η ] Now, to get matrix M we multiply matrix G with the inverse of L: M = GL−1 (1 − ψ)βC N p(1 − ψ)βC N p(1 − ψ)βC N p(1 − ψ)βC N p(1 − ψ)βC N p(1 − ψ)βC N γ η η η η η (1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N γ η η η η η (1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N γ η η η η η = (1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N γ η η η η η (1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N γ η η η η η (1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N p(1 − ψ)βS N γ η η η η η [ ] The dominant eigenvalue of this matrix M is the R 0 of the system: R0 = βC N αpβS + ) γ η (1 − ψ)(RC0 + RS0 ) = (1 − ψ)( From the expression, increasing the probability of dying from infections decreases the basic reproductive ratio of the virus. 3. Selecting Phi for 0.25 Attack Ratio We are interested in maintaining RC0 ≈ 2.0 while achieving an attack rate/incidence proportion to 0.25, or final outbreak size of 250 individuals (when N=1000), when sexual transmission of the virus is not possible. Therefore, we seek the value of ϕ that achieves this. We begin by reducing our model to Ṡ İ = −βC e−ϕI SI = βC e−ϕI SI − γI − Ṙ ψ γI 1−ψ = γI From the full model (eq. 2), the sexually infectious classes (Y1 , Y2 , Y3 , Y4 , Y5 ) have been removed. We stochastically simulated outbreaks in the system using Gillespie's direct method [8] to link ϕ with the average attack ratio. Values of ϕ were varied through the plausible range ϕ ∈ {0,0.01, . . . ,0.06}. Other parameters used in the simulations are found in Table 1. Replication was at the scale of 1000 outbreaks per ϕ and each simulation began with one symptomatically infectious individual (I0 = 1), with all other individuals assumed susceptible. Simulations ran until there were no infectious individuals (I = 0). Fig. S2: The average size of outbreaks conditioned on a large outbreak occurring (established by the histogram method outlined in the main text) as a function of ϕ. Dashed lines are the standard deviation of the average outbreak size. The dotted lines illustrate the ϕ value (vertical, 0.054) that achieves an attack ratio of 0.25 (horizontal, 250 individuals of a population of 1000). We define an outbreak by the histogram method outlined in the main text. We found the value of ϕ that achieved an attack ratio of of 0.25 was ϕ = 0.054. 4. Sensitivity to the Parameters in the gamma distribution We are interested in the robustness of the observed results to the number of psuedostages for the sexually infectious class and jointly the transition rate between stages. Since the number of pseudostages must be a finite integer, we iterate over α ∈ (1,2, . . . ,10) and found the optimal η. Values of α = 4 and α = 6 were choosen as these had the next closest likelihood values. α η NLL 1 0.103 49.315 2 0.256 47.321 3 0.411 46.459 4 0.567 46.095 5 0.722 46.008 6 0.876 46.095 7 1.031 46.301 8 1.185 46.593 9 1.339 46.947 10 1.492 47.351 Table S1: Fitting results (negative log likelihood, NLL) for gamma distributions applied to the data on persistence of viral RNA in semen. Shape parameter () was constrained to integer values. Most likely value was used in main text Next most likely values ( are explored herein. Four pseudostages The model was modified to include only four pseudostages (α = 4) with a transition rate of η = 0.5667. All other parameter values are held the same as the original model. Models were implemented as stochastic simulations using Gillespie's direct method [8] to link parameter values with increases in average outbreak size, average duration of the outbreak and the probability of an outbreak (calculated using the methods in the main text). Replication was at the scale of 1000 replicates per parameter set and each simulation began with one symptomatically infectious individual (I0 = 1), with all other individuals assumed susceptible in a population of 1,000 individuals. Simulations ran until there were no infectious individuals (I = Y = 0). Figure S3: Heat-maps of the average outbreak size, average duration of a large outbreak, probability of a large outbreak, and the reproductive ratio (R 0 ) of the virus using the model with four pseudostages from 1000 stochastic simulations for each parameter set. Panels A, B, and C are example outbreaks that can occur with the specified parameter combinations, and with α = 4, η = 0.5667. Six Pseudostages The number of pseudostages was increased to six (α = 6) and the transition rate was set to 0.8764 (η = 0.8764). All other parameter values are held the same as the original model. Models were implemented as stochastic simulations using Gillespie's direct method [8] to link parameter values with increases in average outbreak size, average duration of the outbreak and the probability of an outbreak (calculated using the methods in the main text). Replication was at the scale of 1000 replicates per parameter set and each simulation began with one symptomatically infectious individual (I0 = 1), with all other individuals assumed susceptible in a population of 1,000 individuals. Simulations ran until there were no infectious individuals (I = Y = 0). Figure S4: Heat-maps of the average outbreak size, average duration of a large outbreak, probability of a large outbreak and the reproductive ratio (R 0 ) of the virus using the model with six pseudostages from 1000 stochastic simulations for each parameter set. Panels A, B, and C are example outbreaks that can occur with the specified parameter combinations, and with α = 6, η = 0.8764. Sensitivity to Population Size To examine the robustness of our results with the population size, we now increase the size to 10,000 individuals and use the five pseudostage model. However, the parameter ϕ must also be estimated for the higher population size. Selecting Phi for 0.25 Attack Ratio We are interested in maintaining RC0 ≈ 2.0 while achieving an attack rate/incidence proportion to 0.25, or final outbreak size of 2500 individuals (when N=10,000), when sexual transmission of the virus is not possible. Therefore, we seek the value of ϕ that achieves this. We stochastically simulated outbreaks in the system described by equation 2 of the main text using Gillespie's direct method [8] to link ϕ with the average attack ratio. Values of ϕ were varied through the plausible range ϕ ∈ {0,0.01, . . . ,0.05}. To keep the RC0 at ≈ 2.0, βC was reduced to 0.0004 and all other parameters used in the simulations are as in Table 1. Replication was at the scale of 1000 outbreaks per ϕ and each simulation began with one symptomatically infectious individual (I0 = 1), with all other individuals assumed susceptible. Simulations ran until there were no infectious individuals (I = 0). Fig. S5: The average size of outbreaks conditioned on a large outbreak occurring (established by the histogram method outlined in the main text) as a function of ϕ. Dashed lines are the standard deviation of the average outbreak size. The dotted lines illustrate the ϕ value (vertical, ϕ = 0.028) that achieves an attack ratio of 0.25 (horizontal, 2500 individuals of a population of 10,000). We define an outbreak by the histogram method outlined in the main text. We found the value of ϕ that achieved an attack ratio of of 0.25 was ϕ = 0.028. The optimal ϕ value was used in the five pseudostage model described by eq. (2). Parameter values were set to the values in Table 1, except βC which was set to βC = 0.0004. Models were implemented as stochastic simulations using Gillespie's direct method [8] to link parameter values with increases in average outbreak size, average duration of the outbreak and the probability of an outbreak (calculated using the methods in the main text). Replication was at the scale of 1000 replicates per parameter set and each simulation began with one symptomatically infectious individual (I0 = 1), with all other individuals assumed susceptible in a population of 10,000 individuals. Simulations ran until there were no infectious individuals (I = Y = 0). Figure S6: Heat-maps of the average outbreak size, average duration of a large outbreak, probability of a large outbreak (simulation), and the reproductive ratio (R 0 ) of the virus using the model with five pseudostages with a population size of 10,000 individuals from 1,000 stochastic simulations for each parameter set. 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