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Limiting enzyme: Protein X is an enzyme that acts on a substrate to provide bene

ID: 218538 • Letter: L

Question

Limiting enzyme: Protein X is an enzyme that acts on a substrate to provide benefit to the cell. L is the substrate concentration. Calculate the fitness function f(X,L) assuming a cost that is linear in the protein concentration , c~–?X, and benefit that is a Michaelis-Menten function of the protein concentration , b(L,X) = b0 L X/(X+K), appropriate for cases where enzyme X, rather than its substrate, is limiting.

a. Calculate the optimal enzyme level as a function of L and K.?

b. What is the minimal substrate level Lc required for maintenance of the gene for X by the organism? When is the gene that encodes X lost? Explain.

Explanation / Answer

Protein X acts on substrate L.

Assume that the cost function is linear, i.e. cost= aX ( a is an arbitrary constant ).

The benefit is given as benefit=(b0) * (LX) / ( (X+K)

Fitness function f(X, L)= benefit - cost = [(b0 * L * X) / (X + K) ] - [aX]

(a) At optimal enzyme concentration, d(f(X, L))/dX will be zero ( for any system, slope of tangent at maxima / minima is zero ).

d(f(X, L))/dX = [(b0*L) * ( X+K ) - ( b0*L*X )*(1)] / [ (X+K)^2] - a

0 = [(b0*L) ( X + K - X )] / [(X + K)^2] - a

a = bo*L*K / [(X + K)^2]

[(X + K)^2] = b0*L*K / a

X ( optimal enzyme concentration ) = sqrt [ b0*L*K/a] - K

(b) Maintenance of gene will occur only if the benefit to the cell is greater than the cost of producing protein X.

i.e. the fitness function has to be > 0

i.e. (b0*L*X) / ( X+K ) - aX > 0

(b0)*(L) / ( X+K ) > a

L > a * (X + K) / (b0) ( minimal substrate concentration for gene to persist )

Gene will be lost once the benefit to cell is none or the costs exceed the benefits. This will occur when

L <= (a) * (X + K ) / (b0)

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