Representing One Signal as a Function of Another.

A common form found in mathematical models of biochemical clocks is [1a]or [1b]

This states that the production rate s(t) can be broken into two parts, a function of p (also known as the clearance rate of p) and the derivative of p. If p is constant, then dp/dt = 0 and n(p) is constant. We also note that where N(p) is the indefinite integral of n(p), meaning that these two parts are orthogonal (see Theorem 1 below for an alternate proof).

Now, consider oscillations in a more general network whose dynamics may not be initially of this form. Surprisingly, this form still applies, as shown by the following definition and theorem:

The proof of this theorem is given in SI Text given that s(t), r(t), f₁(r), and f₂(r) have a Fourier series representation. This proof also provides a way of calculating f₁ and f₂. This decomposition is possible even if model equations are not available (See Example 1 below).

Assume f₁ and f₂ are known, ideally because we have a mathematical model of the form given by Eqs. 1a and 1b, but otherwise by calculation. What does knowledge of f₁ and f₂ tell us about the relationship between s(t) and r(t)? Some basic properties can be immediately seen. For example, if df₁(r)/dt = 0, then s(t) is a function of r(t). Likewise, if f₂(r) = 0, then s(t) and r(t) are orthogonal.

The rest of this manuscript will derive properties of the systems based on knowledge of f₁ and f₂.

Example 1: Experimental Data on the Mammalian Circadian Clock.

We use data from ref. 17 that continuously measure luciferase reporters of two genes in the mammalian circadian clock (Bmal1 and Per2). The Bmal1 marker is shown in Fig. 1A, where digitized data are shown as well as a best fit that we consider r(t). Fig. 1 B and D show f₂(r) and df₁(r)/dt, respectively, which when added yield the rhythm in the Per2 marker shown in Fig. 1C. Here, we simply illustrate that our methodology can be used on real data. Data from any clock system could have been used and/or other markers of rhythms including more accurate luciferase fusion proteins.

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Fig. 1.

Here, we demonstrate the geometry described in Fig 1 with data presented in the abstract of the HTML version of ref. 17 around day 2. They record markers of the rhythms of two key genes in the circadian clock: Bmal1 (A) and Per2 (C). Both rhythms were well fit by a sinusoid, which is shown. Considering Bmal1 marker as r(t) and Per2 marker as s(t), B and D show f₂(r) and df₁(r)/dt, respectively. Adding them together yields the rhythm of the Per2 marker shown in C.

Inferring One Signal from the Next.

Consider Eq. 1a:

Let ρ(t) ≡ f₂(r(t)). Assume f₁ is a function of f₂ or f₁(r) = q(ρ(t)). Then we have scaling time, we then have and dT/dt ≡ dρ/dq, assuming that dρ/dq ≠ 0, which allows a one-to-one mapping between T and t. Here, an exact solution can be found: which shows that ρ(T) is simply a weighted average of the previous signal s(T). However, this weighting uses T rather than t. From this, the following prediction can be made: As dt/dT lessens, t proceeds slower with respect to T, giving more weight to that part of the signal in determining ρ(T).

Example:

Consider the case where dp/dt = s(t) - n(p(t)) with s(t) = 1 + sin(t) and n(p(t)) = 2p(t)/(0.5 + p(t)). s(t) is shown in Fig. 2A (blue curve) along with n(p(t)) (red curve). Scaling time yields s(T) shown in Fig. 2B (blue curve). One can see that at points where p(t) saturates n(p(t)), s(T) becomes more compressed and includes more of the signal s(t) in a smaller time span. This is then passed through a linear filter with rate constant η = 1.22 yielding ρ(T) (red curve in Fig. 2B). Plotting ρ(T(t)) yields n(p(t)) shown in Fig. 2A.

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Fig. 2.

Inferring the time course of a signal in the feedback loop. (A) Simulations of the system dp/dt = s(t) - n(p(t)) with production rate s_i(t) = 1 + sin(t) (blue curve) and saturating clearance rate n(p(t)) = 2p(t)/(0.5 + p(t)) (red curve). (B) We show (see text) that the above system can be converted to the linear filter, dρ/dT = s(T) - ηρ(T), with η = 1.22 and dT/dt = (1/η)dn/dp. Both s(T) (blue) and ρ(T) (red) are plotted. Over the region 2 < t < 4, n(p(t)) saturates (dn/dp is small), so t proceeds quickly with respect to T, and s(T) is condensed around T = 2.

Comparing Signal Amplitudes.

Again consider signals in the form of Theorem 1 and again with the mean of all signals subtracted off. Because df₁(r)/dt and f₂(r) are orthogonal, we also know that [2]see Fig. 3 for an illustration. This equation tells us that the amplitude of s(t) is greater than the amplitude of df₁(r)/dt or f₂(r).

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Fig. 3.

Geometric representation of the signals within a step of a biochemical feedback loop. We show how a periodic signal s(t) can be represented as s(t) = df₁(r)/dt + f₂(r).

What does this tell us about the relationship between the amplitude of s(t) and the amplitude of r(t)? We then have the following formula: or

Therefore, this ratio of the amplitude of s(t) and the amplitude of r(t) depends on:

1. how either f₁ or f₂ increases the amplitude or decreases the amplitude of r(t) [i.e., ||f₁(r)||²/||r(t)||² and ||f₂(r)||²/||r(t)||²], and

2. ||df₁(r)/dt||²/||f₁(r)||². This can be viewed as a generalized estimate of the frequency of f₁(r).

Let us describe what we mean by a generalized estimate of the frequency of f₁(r). If f₁(r) = cos(wt + ϕ) then this ratio is the exact frequency w. Otherwise, let us assume that f₁(r) has a Fourier series , as in Example 2, this ratio is equal to . Thus, it not only depends on the frequency of f₁(r) but also on the frequencies of the harmonics of f₁(r). As the amplitudes of higher harmonics increase, so does this generalized frequency.

Comparing Signal Shapes.

We next compare the signal shape between s(t) and r(t) using f₁(r) and f₂(r) and the geometry outlined in Fig. 3. To start, we take the inner product of Eq. 1 or Theorem 1 with r(t): where the second equality comes from the fact that df₁(r)/dt and r(t) are orthogonal (Theorem 1).

With the definition of the inner product, we find or [3]where the second equality uses Eq. 2. So how does θ_s,r depend on f₁(r) and f₂(r)? It depends monotonically on θ_f2,r, so as f₂(r) distorts the shape of r(t), it distorts s(t) from r(t). Also, df₁(r)/dt distorts s(t) from r(t). The larger the magnitude of df₁(r)/dt, the greater the difference in shape between s(t) and r(t).

Requirement for Oscillations in a Negative Feedback Loop.

Now, assume we have the general structure of a feedback loop where the signal s(t) in the ith step of the feedback loop is a function of the clearance rate of the previous element in the feedback loop, or that the model has the following form: [4]

For simplicity of notation, we drop the i subscript in f_2,i(r_i) and represent it as f₂(r_i) while noting that this function can be different for different steps. We also assume there are q elements of the loop and that r₀(t) = r_q(t), which gives the structure of a feedback loop. We also sometimes abbreviate s_i-1(f₂(r_i-1)) as s_i-1.

This theorem is proved in SI Text. It is perhaps more interesting when the inequality of Theorem 2 is not satisfied, which then indicates that oscillations can not be present.

So when do oscillations appear in genetic networks? The most common answer found in the biological literature is that negative feedback and delay cause oscillations. However, many genes in the genome are under the control of negative feedback, and these feedbacks always have a delay. This theorem states that something more is needed, and indeed, most genetic networks with negative feedback and delay do not show oscillations. For example, if there are three elements in the network (q = 3) the left-hand side is eight that would require special biological mechanisms to achieve this high gain.

Although similar secant conditions have been presented previously (2, 13, 21–23), this condition involves an equality using the θ_si-1,si. To maximize the chances that a biochemical feedback loop will oscillate, one should aim for θ_si-1,si = π/q, or that each step of the feedback loop will cause the same distortion of the previous signal as in the previous step. Thus, if the steps in the feedback loop are symmetric in the sense that they distort the signals in the same way, then the network will likely oscillate. The following examples illustrate this point.