Vector Dose, Transduction, and Functional Copy Number

From Administered Dose to a Cell-Level State

An administered vector dose is not itself a transduced-cell fraction. Between the dose and a cell-level outcome lie at least two distinct questions: how many vector genomes reach the target-cell population, and what fraction of those genomes become functional intracellular copies? Treating these as separate steps prevents a high count of vector genomes from being mistaken for a high fraction of functionally transduced cells.

The model developed here is a deliberately explicit translation from dose to copy-number states. It describes a specified target-cell population at a specified assessment point. It does not describe how vector genomes subsequently clear or how transgene product changes over time.

Definition: Let \(D_{\mathrm{vg}}\) be the administered dose in vector genomes (vg), \(f_{\mathrm{eff}}\) the fraction of administered genomes that become intracellular in the target-cell population, and \(N_T\) the number of target cells. The mean number of intracellular vector genomes per target cell is \(\lambda_T=D_{\mathrm{vg}}f_{\mathrm{eff}}/N_T\). Let \(K_T\) be the total intracellular genome count in a randomly selected target cell. If each intracellular genome independently has probability \(q\) of meeting a specified functional-copy criterion, let \(K_F\) be the number of functional copies and \(\lambda_F=q\lambda_T\).

Here, “functional” is an operational classification: the model’s definition must specify what qualifies as a functional intracellular copy. It does not automatically mean that the copy is integrated, that it will persist, or that a measurable amount of transgene product is already present. Those are distinct biological questions. Likewise, \(D_{\mathrm{vg}}\) counts vector genomes, not necessarily infectious or functionally competent units.

Assumptions Behind the Poisson Allocation Model

A common first model assumes that intracellular genomes are allocated randomly and independently among target cells. If the number of genomes reaching the target population is large and each genome has a small chance of reaching any particular cell, the count per cell can be approximated by a Poisson random variable:

K_T\sim \operatorname{Poisson}(\lambda_T), \qquad P(K_T=k)=e^{-\lambda_T}\frac{\lambda_T^k}{k!}, \quad k=0,1,2,\ldots

The mean and variance of this count are both \(\lambda_T\). To separate total from functional copies, assume each intracellular genome independently qualifies as functional with probability \(q\), where \(0\leq q\leq1\). The functional and nonfunctional counts then have distributions

K_F\sim\operatorname{Poisson}(q\lambda_T), \qquad K_N\sim\operatorname{Poisson}((1-q)\lambda_T),

where \(K_N\) counts intracellular genomes that do not meet the functional criterion. Under these assumptions, \(K_F\) and \(K_N\) are independent, and \(K_T=K_F+K_N\). This distinction matters: a cell can have several total intracellular genomes but no functional copies.

Deriving the Functional-Copy Distribution

The functional-copy distribution follows by averaging over the total genome count. Given \(K_T=m\), the functional count is binomial: \(K_F\mid K_T=m\sim\operatorname{Binomial}(m,q)\). For \(k\geq0\), summing over all possible total counts gives

\begin{aligned} P(K_F=k) &=\sum_{m=k}^{\infty}P(K_T=m)P(K_F=k\mid K_T=m)\\ &=\sum_{m=k}^{\infty}e^{-\lambda_T}\frac{\lambda_T^m}{m!} \binom{m}{k}q^k(1-q)^{m-k}\\ &=e^{-q\lambda_T}\frac{(q\lambda_T)^k}{k!}. \end{aligned}

For the last equality, set \(j=m-k\), use \(\binom{m}{k}/m!=1/[k!(m-k)!]\), and sum the exponential series \(\sum_{j=0}^{\infty}[(1-q)\lambda_T]^j/j!=e^{(1-q)\lambda_T}\). The result is a Poisson distribution with mean \(\lambda_F=q\lambda_T\).

A target cell is functionally transduced if it has at least one functional copy. Therefore, the population fraction predicted to be functionally transduced is

P(K_F\geq1)=1-P(K_F=0)=1-e^{-\lambda_F}.

This is not the same as the fraction with at least one intracellular vector genome, which is \(P(K_T\geq1)=1-e^{-\lambda_T}\). If \(\lambda_F=0\), the functional fraction is zero. The distribution of \(K_F\) remains well-defined, concentrated at zero.

Result: When \(\lambda_F>0\), the functional-copy count among functionally transduced cells has the conditional distribution \(P(K_F=k\mid K_F\geq1)=\dfrac{e^{-\lambda_F}\lambda_F^k/k!}{1-e^{-\lambda_F}}\), for \(k=1,2,\ldots\). Its conditional mean is \(\dfrac{\lambda_F}{1-e^{-\lambda_F}}\). When \(\lambda_F=0\), there are no functionally transduced cells; conditioning on \(K_F\geq1\) is impossible, so both conditional quantities are undefined.

The conditional-mean result follows directly by summing the positive counts. For \(\lambda_F>0\),

E[K_F\mid K_F\geq1] =\frac{\sum_{k=1}^{\infty} kP(K_F=k)}{P(K_F\geq1)} =\frac{\lambda_F e^{-\lambda_F}\sum_{j=0}^{\infty}\lambda_F^j/j!}{1-e^{-\lambda_F}} =\frac{\lambda_F}{1-e^{-\lambda_F}}.

The numerator is \(E[K_F]\), because the zero-count term contributes nothing. The result is a mean number of functional copies per functionally transduced cell, not a mean over all target cells. Over all target cells, including those with zero functional copies, the mean is simply \(\lambda_F\).

Worked Example: Many Total Genomes but Few Functional Copies

Suppose a hypothetical dose is \(D_{\mathrm{vg}}=2.00\times10^{11}\) vg. Assume \(f_{\mathrm{eff}}=0.0200\), \(N_T=1.00\times10^8\) target cells, and \(q=0.0100\). These are illustrative model inputs, not measurements from a real study.

The expected number of genomes entering the target-cell population is \((2.00\times10^{11})(0.0200)=4.00\times10^9\) vg. Thus,

\lambda_T=\frac{4.00\times10^9\ \mathrm{vg}}{1.00\times10^8\ \mathrm{cells}} =40.0\ \mathrm{genomes/cell}, \qquad \lambda_F=(0.0100)(40.0)=0.400\ \mathrm{functional\ copies/cell}.

The predicted fraction with any intracellular genome is \(1-e^{-40.0}\), effectively 1.000. In contrast, the functional fraction is \(1-e^{-0.400}=0.3297\), or about 33.0%. The mean functional-copy count among functionally transduced cells is \(0.400/(1-e^{-0.400})=1.213\) functional copies per cell. The difference arises because only one in 100 intracellular genomes is assumed to qualify as functional.

Comparing Doses and Functional Efficiency

The functional fraction depends on the product \(q\lambda_T\), not on either factor alone. Consequently, different combinations of total genome delivery and per-genome functional probability can predict the same fraction of functionally transduced cells. They need not predict the same total intracellular copy burden.

Worked Example: Same Functional Fraction, Different Total Copy Counts

Consider two hypothetical target-cell populations. Population A has \(\lambda_T=1.00\) total intracellular genomes per cell and \(q=0.800\). Population B has \(\lambda_T=2.00\) genomes per cell and \(q=0.400\).

For A, \(\lambda_F=(1.00)(0.800)=0.800\); for B, \(\lambda_F=(2.00)(0.400)=0.800\). Both therefore have the same predicted functional fraction:

P(K_F\geq1)=1-e^{-0.800}=0.5507.

For each population, the mean among functionally transduced cells is \(0.800/(1-e^{-0.800})=1.453\) functional copies per cell. But the fraction with any intracellular genome differs. For A it is \(1-e^{-1.00}=0.6321\); for B it is \(1-e^{-2.00}=0.8647\). The two populations have identical functional-copy distributions under this model, but B has a larger mean total intracellular genome count and a larger nonfunctional-copy mean: \((1-0.800)(1.00)=0.200\) for A versus \((1-0.400)(2.00)=1.20\) for B. Functional transduction alone therefore does not reveal total vector-genome burden.

Worked Example: Dose Needed for a Target Functional Fraction

Suppose a hypothetical target-cell population contains \(N_T=5.00\times10^7\) cells, \(f_{\mathrm{eff}}=0.0150\), and \(q=0.600\). Find the administered dose predicted to produce a 70.0% functionally transduced fraction under the model.

Set \(1-e^{-\lambda_F}=0.700\). Rearranging gives \(\lambda_F=-\ln(0.300)=1.20397\) functional copies per cell. Since \(\lambda_F=q\lambda_T\), the required total intracellular mean is

\lambda_T=\frac{1.20397}{0.600}=2.00662\ \mathrm{genomes/cell}.

Using \(\lambda_T=D_{\mathrm{vg}}f_{\mathrm{eff}}/N_T\), solve for the administered dose:

D_{\mathrm{vg}}=\frac{\lambda_T N_T}{f_{\mathrm{eff}}} =\frac{(2.00662\ \mathrm{genomes/cell})(5.00\times10^7\ \mathrm{cells})}{0.0150} =6.689\times10^9\ \mathrm{vg}.

This is a model-based dose calculation, conditional on the assumed \(f_{\mathrm{eff}}\), \(q\), and homogeneous Poisson allocation. It is not a dose recommendation. Under the stated assumptions, substituting the calculated dose gives \(\lambda_F=1.20397\), and hence \(1-e^{-1.20397}=0.700\).

Diagnostics, Alternatives, and Limitations

The Poisson model is useful because it connects a mean genome exposure to a complete distribution of cell-level copy counts. Its assumptions should be checked against the experimental design and measurements, not inferred merely because the equations are convenient.

Homogeneous exposure: The model treats target cells as having the same \(\lambda_T\) and \(q\). If cell types, locations, receptor levels, or accessibility differ, a single average can conceal subpopulations. A mixture of cell-specific means generally does not have the same zero fraction as a Poisson distribution evaluated at the overall mean.
Independent allocation: Aggregation, shared delivery barriers, or competition can make genome counts more variable than Poisson counts. A variance much larger than the mean, or systematic discrepancies between observed zero fractions and \(e^{-\lambda_T}\), are warning signs.
Independent functional qualification: The model assumes the probability \(q\) applies independently to each intracellular genome. If functional status depends on cell state or on the number of genomes in that cell, simple thinning may fail.
Measurement definition: A vector-genome assay may count material that is not intracellular or does not meet the functional criterion. The estimate of \(f_{\mathrm{eff}}\) and the definition of \(q\) must match the quantities used by the model.

A finite-allocation alternative is to place a fixed number \(M\) of genomes independently among \(N_T\) equally accessible cells. The per-cell count is then binomial, with probability \(1/N_T\) for each genome. Its zero probability is \((1-1/N_T)^M\), which is approximately \(e^{-M/N_T}\) when \(N_T\) is large and \(1/N_T\) is small. This motivates the Poisson approximation; it does not remove the need to assess accessibility and independence.

When cells differ substantially in exposure or functional probability, a model with cell-specific parameters or a mixture distribution may be more appropriate. When the aim is a direct empirical estimate rather than a mechanistic translation from dose, the observed fraction of cells meeting a prespecified functional criterion may be more informative. Such alternatives have their own data and design requirements; none can be selected solely from the administered dose.

A frequent mistake is to calculate \(1-e^{-\lambda_T}\) and call it the functional transduction fraction. That quantity is the fraction predicted to contain at least one total intracellular genome. The functional fraction is \(1-e^{-q\lambda_T}\) under independent qualification. Another mistake is to treat a mean of, for example, 0.4 functional copies per cell as though every cell has 0.4 of a copy. It is a population mean; individual cells have integer copy counts, including a substantial zero-count group.

Key takeaway: Keep dose, total intracellular genomes, and functional copies distinct. Under homogeneous Poisson allocation with independent functional qualification, \(D_{\mathrm{vg}}\) determines \(\lambda_T\) through target access and cell number, while \(q\lambda_T\) determines the functional-copy distribution and the functionally transduced fraction.

Check Your Understanding

Answer each question using the stated model. Full solutions follow the questions.

If \(\lambda_T=1.50\) genomes per cell and \(q=0.400\), find \(\lambda_F\) and the predicted functional fraction.
For \(\lambda_F=0.500\), calculate the mean functional-copy count among functionally transduced cells. State the units.
A model predicts \(1-e^{-\lambda_T}=0.90\). Can the functional fraction be determined from this alone? Explain what additional quantity is needed.
Why is the conditional mean functional-copy count among transduced cells undefined when \(q=0\), if \(\lambda_T\) is finite?

Solutions. (1) \(\lambda_F=q\lambda_T=(0.400)(1.50)=0.600\) functional copies per cell. The functional fraction is \(1-e^{-0.600}=0.4512\), or 45.1%. (2) The conditional mean is \(0.500/(1-e^{-0.500})=0.500/0.3935=1.271\) functional copies per functionally transduced cell. (3) No. The given result concerns cells with at least one total intracellular genome. The functional fraction requires \(q\), because it is \(1-e^{-q\lambda_T}\). (4) If \(q=0\), then \(\lambda_F=q\lambda_T=0\) and \(P(K_F\geq1)=0\). There are no functionally transduced cells on which to condition, so the conditional distribution and its mean are undefined.