# Feed Forward Loop

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For a three node pattern there exitst 13 possible ways to create a directed (with arrows of definite direction) subgraph.

Similar calculations as for negative autoregulation (NAR) show, that one of this three node patterns occurs much more often in the E. coli network that expected at random. This pattern numbered with $5$ is called feed forward loop (FFL). E. coli contains 42 feed forward loops, while the expected number is less than one, thus the FFL is a network motif.

Depending on the sign of the arrows the feed forward loop can be classified into $8$ different types. Each arrow can be positive or negative (zwo possibilities) and there are three arrows this gives $2^3$ possibilities to make a FFL with different signs. This $8$ FFLs are eather coherent (when both incoming arrows of $Z$ have a positive sign) or incoherent (incoming arrows of $Z$ have different signs).

The input function of $Z$ is two-dimensional. In the logic approximation it can eather be an AND or an OR input funciton. It is common to use a symbolic notation like in electronic engineering and boolean logic.

The following four types of Coherent Feed Forward Loops [C#FFL] exist

The following four types of Incoherent Feed Forward Loops [I#FFL] exist

The most abundant FFLs in E. coli are C1FFL and I1FFL, $90\%$ of the E. coli network are of these types.

##  C1FFL with AND input function

We already know the dynamics of simple regulation, which leads to a time course $$Y(t)=Y_{st}(1-e^{-\alpha t}).$$ We use this result to understand the dynamics of the feed forward loop with an AND input function. Let us now study the response to an ON-step. We assume that protein $X$ is present in its inactive form. Suddenly the signal $S_x$ appears and the protein is transformed in its active form $X \rightarrow X^*$. This leads to immediate production of protein $Y$ whose signal $S_y$ we assume present, so that $Y$ is active ($Y^*$).

Since, $Z$ has an AND input function, $Z$ needs both $X^*$ AND $Y^*$ to be present, before it starts producing $Z$, thus $Z$ is only expressed when $Y*$ exceeds a certain threshold $K_yz$ (threshold of Y to produce Z).

delayed response: the C1FFL-AND motif leads to a delayed production of $Z$ in response to the signal $S_x$

The time delay of production of protein $Z$ equals the time $Y$ needs to reach the activation threashold. $$Y_{st}(1-e^{-\alpha \cdot t_K})=K_{yz}\\ e^{-\alpha \cdot t_K}=1-\frac{K_{yz}}{Y_{st}}\\ t_K=\frac{1}{\alpha} \ln \left(\frac{1}{1-\frac{K_{yz}}{Y_{st}}} \right)$$

If the signal $S_x$ is present for a time period smaller than $t_K$, $Y$ will not exceed the threshold $K_{yz}$ and $Z$ will not be produced.

The C1FFL-AND motif acts as filter for signals $S_x$ of a short time period $t < t_K$

An OFF-step occurs, when the signal $S_x$ suddenly disappears. Since, both $X^*$ AND $Y^*$ are needed to activate $Z$ production, the production stops immediately after $X^*$ disappears.

##  C1FFL with OR input function

For an ON-step, where the signal $S_x$ appears, $Y^*$ starts beeing produced as well as $Z^*$, becaus it is enough that eather $X^*$ OR $Y^*$ is present.

The interessting part of this motif is is the OFF-step. When $S_x$ disappears $Y$ immediately stops beeing produced, but $Z$ is still produced until $Y$ reaches the threashold $K_yz$, then also $Z$ stops producing.

delayed response: the C1FFL-OR motif leads to a delay of production stop of $Z$ following an OFF-step of $S_x$

Summary

C1FFLs are sign sensitive delay elements. With a C1FFL-AND element brief fluctuations of ON-pulses can be filtered away, while with C1FFL-OR element brief OFF-pulses can be filtered away.

##  I1FFL with AND input function

Let us now study the incoherent type one feed forward loop with an AND input function. Notice, that $Z$ is transcribed if transcription factor $X^*$ AND NOT $Y^*$ is present. For an ON-step $X^*$ is immediately active and produces $Y$. $Z$ is also produced, since $X^*$ AND NOT $Y^*$ is present. When $Y^*$ exceeds the threshold $K_{yz}$, $Z$ stops beeing produced and decays exponentially to reach the steady state $Z_{st}$.

As $S_x$ appears $Y$ is produced according to simple regulation and existis in its active form as we assume the signal $S_y$ present. $$Y(t)=Y_{st}(1-e^{-\alpha_Y t}).$$ $Z$ is also produced instantaneously and reaches a maximal level at $Z_m=\beta_Z/\alpha_Z$. $$Z(t)=Z_m(1-e^{-\alpha_Z t}).$$ The time span until $Z$ is repressed, equals the time $Y$ needs to reach the repression threashold. $$Y_{st}(1-e^{-\alpha \cdot t_R})=K_{yz}\\ e^{-\alpha \cdot t_R}=1-\frac{K_{yz}}{Y_{st}}\\ t_R=\frac{1}{\alpha} \ln \left(\frac{1}{1-\frac{K_{yz}}{Y_{st}}} \right)$$ After the repression thershold is reached the $Z$ expression level decays to a the steady state $Z_{st}=\beta_Z '/\alpha_Z$ $$Z(t)=Z_{st}+(Z_0-Z_{st})(1-e^{-\alpha t}).$$ We now introduce the new concept of the repression factor $$F=Z_m/Z{st}=\beta_Z/\beta '_Z$$ For a hight repression factor $F$ the steady state $Z_{st}$ concentration is much lower than $Z_m$. For lower and lower $F$ the difference in concentration of $Z_m$ and $Z_{st}$ gets lower and lower. Hence, with high repression factors a pulse can be generated, while for low repression factors the response time can be shortened.

The I1FFL-AND motif can act as a pulse generator or speed the response time

Video Lecture: