# Differential Equations

QuestionLactose (C12 H22 O11 ) is a sugar found in milk. Enzymes are large proteins that enable biochemical reactions which would otherwise not occur. Lactase is the enzyme responsible for breaking down lactose. The molecular structure of lactose has two rings of 6 carbons atoms each. The lactase enzyme has 4092 amino acids, folded into a speci?c shape that helps cleave the lactose molecule. Lactose breaks into two smaller sugars called glucose and galactose, each of which has one ring of 6 carbon atoms. The process is summarized pictorially below. Glucose and galactose have the same chemical formula (C6 H12 O6 ) but atoms around their carbon rings are twisted in slightly di?erent ways.
Lactose-intolerance is a condition su?ered by adult humans unable to produce the enzyme lactase.
Lactose

Free Lactase Enzyme
?
Glucose

Galactose

Lactose
bound to
Lactase Enzyme
?

Free Lactase Enzyme

Let S(t) be the amount of the starting chemical (lactose); in biochemistry this is called the
substrate. Let P (t) be the amount of product (either glucose or galactose; they are produced in equal
amounts).
Let E(t) be the amount of free enzyme.
Let C(t) be the amount of the intermediate complex (enzyme+substrate bound together).
There are three important reactions:
— The enzyme attaches itself to the substrate at a rate ? S E,
— The reverse of the above, where the enzyme detaches from the substrate at a rate ? C,
— The intermediate complex breaks into the free enzyme and ?nal products at a rate ? C.
In standard biochemistry notation ?, ? and ? are denoted k1 , k?1 and k2 .

The relevant system of nonlinear ordinary di?erential equations is thus
dS
dt
dE
dt
dC
dt
dP
dt

= ?? S E + ? C
= ?? S E + ? C + ? C

# = +? S E ? ? C ? ? C

+?C

where S(t) + C(t) + P (t) = M and E(t) + C(t) = N , and ?, ? and ? are all positive.

1. Explain both mathematically and chemically why N and M are constants.
2. Let x = S/M and y = C/N be dimensionless variables. What is the allowed range of
values of x and y?
3. Show that the ODEs are equivalent to
dx
= xy ? x + ?y
d?
dy
= ?(x ? xy ? ?y)
d?
and give expressions for the dimensionless time ? and the dimensionless parameters ?,
? and ?.
4. Classify the steady state at (0, 0) using linear stability analysis. Remember that in the
original variables ?, ? and ? are all positive.
5. Draw a phase diagram for the case ? = 1 , ? = 1 and ? = 1 on the upper grid provided.
9
Plot the nullclines as accurately as possible and then sketch some representative trajectories (include at least those trajectories starting in the corners of the phase diagram).
6. In a realistic biochemical system, the concentration of enzyme needed is much less than
the concentration of the substrate: explain why this corresponds to ?
7. Describe
in words how this much larger value of ? would change the shape of the trajectories.
Draw a new phase diagram for the same parameters as in Q5, but with ?
1 in the
lower grid provided.
8. If the initial amounts of substrate and enzyme are S(0) = M and E(0) = N , use the
trajectories on the phase diagram constructed in Q6 to help sketch S(t) and E(t) for
t > 0. In particular, show and explain in words the expected behaviour for small and
large values of t.

The grids are provided in this assignment because too many students keep
drawing tiny and messy diagrams. Mathematics is not just about calculations it is also about good visualisation (with appropriately sized and neat
sketches) and communication in logical and complete English sentences.

The model presented here was ?rst proposed by Michaelis and Menten in 1913. They realised that the assumption that the amount of substrate is much larger than the amount
of enzyme, is equivalent to assuming that the concentration of the intermediate complex
does not change rapidly on the time-scale of product formation. Thus, they were also
able to ?nd an asymptotic solution in the limit ? ? ? and their approximate formula
(but not always its rigorous derivation) is now a famous and standard result shown to
all biochemistry students.