Identifying feasible operating regimes for early T-cell recognition: The speed, energy, accuracy trade-off in kinetic proofreading and adaptive sorting

In the immune system, T cells can quickly discriminate between foreign and self ligands with high accuracy. There is evidence that T-cells achieve this remarkable performance utilizing a network architecture based on a generalization of kinetic proofreading (KPR). KPR-based mechanisms actively consume energy to increase the specificity beyond what is possible in equilibrium. An important theoretical question that arises is to understand the trade-offs and fundamental limits on accuracy, speed, and dissipation (energy consumption) in KPR and its generalization. Here, we revisit this question through numerical simulations where we simultaneously measure the speed, accuracy, and energy consumption of the KPR and adaptive sorting networks for different parameter choices. Our simulations highlight the existence of a “feasible operating regime” in the speed-energy-accuracy plane where T-cells can quickly differentiate between foreign and self ligands at reasonable energy expenditure. We give general arguments for why we expect this feasible operating regime to be a generic property of all KPR-based biochemical networks and discuss implications for our understanding of the T cell receptor circuit.

A schematic of the model we are considering is shown in Fig. 1. As described in the main text, we denote a receptor-ligand complex that has been phosphorlyated n times by X n with X = C for foreign ligands and X = D for self ligands. Furthermore, we denote the maximum number of phosphorylations as N. With this notation, using the law of mass action, we haveĊ Typically, we set: R T = 10 4 , L T f = L T s = 10 4 , κ = 300s −1 , σ = 1s −1 , = 2s −1 , K T = 10 4 , α = 3 × 10 −4 , γ = 10 −3 , τ s = 1s and τ f = 10s. Any deviations from this choice of parameter is explicitly noted.

Accuracy
At steady state, the error rate can be written as In the presence of the kinase feedback K = σK T σ+ (C2+D2) , the set of eqs. (1) are no longer linear and but the steady-state solution can still be found easily using an iterative method.

Energy Consumption
The power dissipation is calculated based on the net flux and the chemical potential difference [1,2]. We define the net flux J α,n , i ∈ [s, f ] at X n X n+1 in the main pathway.
Considering the flux conservation, the power dissipation P i can be written as The total power dissipation is from the contribution of both foreign and self ligands: P = P s + P f .

Role of γ
In KPR, the reversible decay rate is ignored as it has extremely small value. Let γ n,i denote the rate at which a self (i = s) or foreign ligand (i = f ) can directly form a complex at n − th step of the KPR cascade (see Fig. 2 of main text). In such a reaction, the first n − 1 steps of the KPR cascade are bypassed resulting in lower accuracy. There are several natural choices for how to choose γ n,i . One common choice in the literature is to assume that γ n,i is independent of n and given by γ n,i = γ/τ i . However, with this choice never saturates the KPR accuracy bound for an N-step cascade, η min = τ N s /τ N f , especially when N is large (see Fig 2). For this reason, in this work we choose a step-dependent rate, γ n,i = γ n /τ i (i = s, f ), for directly forming a complex C n and D n This functional form is a direct consequence of assuming that there is a constant free energy difference k B T logφ/γb per phosphorylation. Having a large γ will result in a bypassing of the proofreading steps and a high error threshold for any KPR-based circuit.
One choice o γ n /τ i . There are two reasons for this form: 1. the production rate from ligands and receptors to X n+1 should be smaller then the one to X n as one more phosphorylation step is involved. If not, it is hard for the KPR circuit to achieve the theoretical limit, τ N s /τ N s . 2. it is also natural to assume the energy consumption is the same for each phosphorylation step. The free energy difference between nth and n+1th phsphorlyation round can be calculated as: The speed is defined by the mean first passage time(MFPT) for the foreign ligand. Here we mainly follow the procedures in Ref. [3]. The concentration vector is defined as c = [L f , C 0 , C 1 , . . . , C N , p f ]. An final 'dark' state is added because the response is only activated at the end and it can be treated as absorbing markov chain. Added this absorb state, it becomes an irreversible process, which is impossible to calculate the energy consumption. The transfer probability from C N to the 'dark' state is W (irreversible). We set W = 100s −1 , a large value, which means the final step has little effect on MFPT. Without loss of generality, we begin with N = 4 and m = 2, which can be generalized other cases easily. The master equations i.e. eqs. (1) can be rewritten asċ = Ac and But eqs. (1) are not linear. The first order perturbation approximation is adapted and we can linearize (with bar denoting average) to get c =c + δc.
Applying the Laplace transform, δC(s) = The MFPT can be written: which can be calculated numerically. It should be notified that the concentration and probability have the same master equations but a different pre-factor. When choosing the initial condition [1, . . . 0] T , the pre-factor is set to be 1 and δC N (s) solved from eq. (7) is exactly a probability distribution .
It can be observed that a large amount of red points distributes over regimes C and D with η ∼ 100. This is because of γ ∼ 10 and the inverse flux at the final step dominates. In the extreme case: b/φ is very large, C 0 D 0 will occupy most of products and free ligands L s , L f have little concentration.