Intrinsic and extrinsic noise in gene expression

Status: Done
Confidence: Certain

The question of noise in gene expression has been more and more important recently, both as an interesting question in its own right, and as a necessary piece of information for constructing random processes to describe other aspects of biology.

There is a problem with measuring noise, however: how much of it is actually noise associated with the components of a pathway operating in a thermal bath, and how much is cell to cell variation, the local density of a necessary enzymes, and other things that are extraneous to the pathway?

Enter two papers: "Intrinsic and extrinsic contributions to stochasticity in gene expression" by Swain, Elowitz, and Siggia (PMID: 12237400) and "Stochastic gene expression in a single cell" by Elowitz, Levine, Siggia, and Swain (PMID: 12183631). The first is theoretical, the second a test of the theory in E. coli.

Take an ensemble of cells whose expression XX of some gene we can monitor (say, with a fluorescent protein on the same promoter). We define the noise in XX as


We use the variance instead of the standard deviation because in most cases we are talking about small numbers of things, and this form gives us a good reference point: if XX is Poisson, η2=1\eta^2 = 1.

We want to decompose this into something measuring the stochastic details of our process of interest and something measuring the variation due using this or that particular cell. Let \mathcal{I} and \mathcal{E} be the intrinsic and extrinsic variables respectively which together entirely determine the state of XX. Then (bringing the mean over to the left hand side so we don’t have to carry it through)

η2(t)E[X]\eta^2(t)\cdot E[X]
=var(X)=E[X2]E[X]2= \mbox{var}(X) = E[X^2] - E[X]^2
=E[E[X2|]]E[E[X|]2]+E[E[X|]2]E[X]2= E[E[X^2|\mathcal{E}]] - E[E[X|\mathcal{E}]^2] + E[E[X|\mathcal{E}]^2] - E[X]^2
=E[var(X|)]+E[E[X|]2]E[E[X|]]2= E[\mbox{var}(X|\mathcal{E})] + E[E[X|\mathcal{E}]^2] - E[E[X|\mathcal{E}]]^2
=E[var(X|)]+E[E[X|]2E[E[X|]]2]= E[\mbox{var}(X|\mathcal{E})] + E[E[X|\mathcal{E}]^2 - E[E[X|\mathcal{E}]]^2]
=E[var(X|)]+E[var(E[X|])]= E[\mbox{var}(X|\mathcal{E})] + E[\mbox{var}(E[X|\mathcal{E}])]
=E[var(X|)]+var(E[X|])= E[\mbox{var}(X|\mathcal{E})] + \mbox{var}(E[X|\mathcal{E}])

The first term we can take as a measure of the intrinsic noise: var(X|)\mbox{var}(X|\mathcal{E}) is the variance due to the intrinsic variables, and its expectation if just averaged over the ensemble of different cells (and their corresponding different extrinsic conditions).

The second term is a measure of the extrinsic noise. E[X|]E[X|\mathcal{E}] is the average expression for a given set of extrinsic conditions (i.e., in one cell), and we take its variance across many cells, and this is purely the variation caused by changes in the extrinsic conditions.

The term we added and subtracted in our computation E[E[X|]2]E[E[X|\mathcal{E}]^2] we can experimentally measure: put two different genes which we can assay on the same promoter in a given cell. They will have the same extrinsic conditions, and different intrinsic variables (since they will be in different parts of the heat bath). Let X=1X=1 and X=2X=2 be the different expressions for the two. Then

=dP()(xdP(X=x|))2= \int dP(\mathcal{E}) (\int x dP(X=x|\mathcal{E}))^2
=dP()x1dP(X1=x1|)x2dP(X2=x2|)= \int dP(\mathcal{E}) \int x_1 dP(X_1 = x_1|\mathcal{E}) \int x_2 dP(X_2 = x_2 |\mathcal{E})
=E[E[X1|]E[X2|]]= E[E[X_1|\mathcal{E}] E[X_2|\mathcal{E}] ]

It is important to remember what is actually measurable here. We can measure mean levels of expression in individual cells. Thus η2\eta^2 is measurable, and we are trying to work backwards to the two noise terms. Consider the observable E[E[X1X2|]2]E[ E[X_1 - X_2 |\mathcal{E}]^2 ].

E[E[X1X2|]2]E[ E[X_1 - X_2|\mathcal{E}]^2 ]
=E[E[X1|]2]+E[E[X2|]2]2E[E[X1|]E[X2|]]= E[ E[X_1|\mathcal{E}]^2] + E[E[X_2 | \mathcal{E}]^2] - 2E[E[X_1|\mathcal{E}]E[X_2|\mathcal{E}]]
=E[E[X1|]2]+E[E[X2|]2]2E[E[X|]2]= E[E[X_1|\mathcal{E}]^2] + E[E[X_2|\mathcal{E}]^2] - 2E[E[X|\mathcal{E}]^2]

The first two terms are both direct observables as well, so we can find the last term: E[E[X|]2]=E[E[X1|]2]/2+E[X2|]2E[X1X2|]2E[E[X|\mathcal{E}]^2] = E[E[X_1|\mathcal{E}]^2]/2 + E[X_2|\mathcal{E}]^2 - E[X_1 - X_2|\mathcal{E}]^2. Note that the other term in our extrinsic noise is an observable as well: in its original form, it is E[X]2E[X]^2. We have the whole extrinsic noise now, and the total noise, and their difference is the intrinsic noise. Voila!

the luria-delbruck calculation or the lockless-ranganathan formalism
more miscellany | explore everything