Diagnostic test for supervisors in Mathematical Biology
This diagnostic test is intended as an aid to potential new supervisors to help them decide if they would be
suitable as supervisors of Mathematical Biology (MB).
MB is a ﬁrst year course in the Natural Sciences Tripos for students who have studied A level maths (or
equivalent). The majority of students will have gone through the British schooling system and will have this
background. Others will have taken other sorts of qualiﬁcations, for example the International Baccalaureate,
Scottish Highers or the various European equivalents. Essential background is that students have a working
knowledge of calculus. Prior study of statistics is not required.
The course covers1 :
1. Introduction to the Growth and Decline of Populations (ﬁrst three quarters of the Michaelmas term);
2. Physiological Modelling (last quarter of the Michaelmas term);
3. Introduction to Modelling of Interacting Populations (ﬁrst half of the Lent term);
4. Introduction to Statistical Methods (last half of the Lent term);
5. Interacting Populations: Ecological Applications (last half of the Easter term).
This document includes one fully-worked Tripos question from each of these topics with the exception of Sta-
tistical Methods, together with a commentary for new supervisors (in blue). There is also a brief description of
the other topics from those parts of the course which were not tested in the particular Tripos question I selected
(which were all actually from the 2009 examination).
Nik Cunniﬀe (firstname.lastname@example.org)
1 Note that there are actually also six lectures in the Easter term on Matrix Algebra, but as the syllabus for these has not yet
been ﬁnalised, these lectures are omitted from this document. However at the very minimum I would expect supervisors to need
to know how to do Gaussian elimination, to calculate Eigenvalues and vectors and so have an understanding of their geometric
Introduction to the Growth and Decline of Populations
The number of students in a particular Cambridge college, R(t), who are aware of a certain scurrilous rumour,
may be modelled by
= βR(N − R),
where N is the total number of students in the college, t is the time since the start of the academic year, and
β is a positive parameter.
a) Explain the right hand side of the model, paying careful attention to the meaning of the parameter β.
b) Find any equilibria exhibited by the model, and determine their stability, using either a graphical or an
c) Solve the model, determining R as an explicit function of t, assuming 1% of students are aware of the rumour
at the start of the academic year.
d) When is the number of students who are aware of the rumour increasing most rapidly? (You must justify
your answer mathematically).
e) Determine the time taken for 75% of the student body to become aware of the rumour.
Like all rumours, this one becomes less interesting with time. An updated model is proposed:
= βe−γt R(N − R),
where γ is a positive parameter.
f ) Explain the alteration to the model, carefully explaining the meaning of the new parameter γ.
g) Given that no more than 75% of the student body ever become aware of the rumour, assuming again that
1% of students initially know it, prove that γ must satisfy
h) Suggest three reasons why even the updated model would be an unrealistic representation of the spread of
a rumour in an actual Cambridge college.
Solution and commentary
Part (a): That students become able to explain a model that they are given is one of the key aims of this
part of the course. Ideally they will also be able to work in the other direction (i.e. to go from a simpliﬁed
biological description to a diﬀerential equation model), but they ﬁnd this much more challenging. However they
are given additional practice in later parts of the course, and hopefully by the end of the course the majority
are able to make some sort of stab at that as well.
The number of new students who hear the rumour per unit of time depends on both the number who have
already heard it (the R term) and the number who remain to hear it (the N −R term). β is some combination of
how often pairs of students meet each other and how likely they are to pass on the rumour in a given encounter
(this is actually a little complex to interpret “properly”, but any related diﬃculties would probably be glossed
over in lectures, although they could form the basis of a supervision discussion with a bright pair of students).
Part (b): Equilibrium analysis is a key idea of this part of the course. The basic idea is that if one knows
values of R for which dR = 0, one of these values of R will be the ultimate size of the population in the limit of
large t. We decide which equilibrium is reached by examining stability after a small perturbation (note there is
clearly a huge simpliﬁcation to do with local and global stability here; we skirt around this in the course with
carefully chosen examples, although this again could be discussed with the best students).
We prefer to teach a graphical method of stability analysis, as the students ﬁnd this easiest; here dR isdt
plotted as a function of R. Then by examining the sign of dR near to equilibria, whether a small upwards
perturbation of R is opposed ( dR < 0) or promoted ( dR > 0) lets one determine whether the equilibrium in
equation is stable or unstable, respectively. I strongly emphasise this method in lectures, on repeated occasions,
and so students have few problems. I also emphasise that this graph of dR vs R can be used to determine the
qualitative behaviour of a diﬀerential equation system without doing any formal calculations, by using it to plot
a direction ﬁeld, and some students appear to appreciate the utility of this shortcut.
Here setting dR = 0 easily leads to R = 0 and R = N as the two equilibria, as the RHS is already factorised.
The graph of dR vs R is quadratic with negative highest order term, and the equilibrium analysis tells us the
roots are 0 and N . Therefore (for example) the equilibrium at R = N is stable via the following chain of
• Imagine slightly increasing R from R = N ;
• Then from the graph dt < 0;
• This means R would decrease;
• Back to R = N ;
• And so the equilibrium is stable (note I would expect students to check “both sides”, although in set-
ting questions we carefully avoid degenerate cases where the stability would depend on the sign of the
perturbation, which once again could be discussed with very strong students).
The equilibrium at R = 0 is similarly unstable.
Part (c): The key mathematical tool in these lectures is the solution of ordinary diﬀerential equations by
separation of variables. The majority will have brieﬂy encountered this at school, but they are often not very
conﬁdent with the technique. However for others it is very easy. This means that balancing supervisions
between going over the basics and stretching the able during this part of the course can be challenging. Note
that this particular example is probably the most diﬃcult that the students routinely see, as there is a need
to use partial fractions to integrate the right hand side (one could also complete the square on the bottom line
and get arctanh, I think, but most students will not have the necessary background for this). However, because
the logistic growth equation is so central to this part of the course, most students eventually master this set of
manipulations, as they end up practicing it about three or four times. Hopefully the vast majority understand it
rather than parroting it oﬀ without really knowing what they are doing; my experience in supervising suggests
this is the case.
In this case separating variables
= β dt,
R(N − R)
1 1 1
+ = β dt,
N R N −R
log R − log(N − R) = βN t + C,
log = βN t + C,
where C is a constant of integration. However since R = 100 when t = 0, C = log 99 , and so
log = βN t,
R = .
1 + 99e−βN t
Part (d): Often of biological interest is how quickly the population is increasing, and in particular when it
is increasing fastest. Clearly this could be determined by diﬀerentiating the above expression for R(t) twice
wrt t and ﬁnding d R as an explicit function of t. However we prefer to teach a quicker method based on
diﬀerentiating the diﬀerential equation implicitly, which is far easier. Students generally understand this, but
often forget to diﬀerentiate implicitly, leaving out the dR on the right hand side of the ﬁrst line of the solution.
The maximum rate of increase occurs when dt2 = 0, and so using the original diﬀerential equation
d2 R dR
= β (N − 2R) = 0,
which means that R = 2 and so
1 + 99e−βN t = 2,
t = log 99.
Part (e): Conﬁdence with logs is absolutely essential. Certain students ﬁnd this very diﬃcult. If this goes
on for more than a few weeks I encourage them to try EMB. This question also emphasises the need to be able
to go from a verbal description of a problem to the mathematics (which students sometimes ﬁnd challenging).
When 75% of the population have heard the rumour
100 1 + 99e−βN t
and so t = βN log 297.
Part (f ): Again interpretation of models is emphasised.
In the updated model the eﬀective value of β decays exponentially with time, corresponding to the decreasing
level of interest in the rumour. The rate at which interest decays is controlled by γ.
Part (g): Extending the basic models to encompass more complex biology is another feature of the course. The
extensions invariably are chosen to allow analytic progress in solution, and so revolve around changing rates to
be exponentially decaying and/or sinusoidal. Students typically ﬁnd this diﬃcult, perhaps because they are not
conﬁdent enough with algebra to feel happy launching into the extended mathematical manipulations required
below. My approach to this has to emphasise that they are actually doing nothing more complex and that they
should have some belief in their mathematical skills.
(Note if the student is quick enough to work by analogy with part (c) that is ﬁne here) Separating variables
= β e−γt dt,
R(N − R)
1 1 1
+ = β e−γt dt,
N R N −R
log R − log(N − R) = − e + C,
R βN −γt
log = − e + C.
N −R γ
However this means C = γ + log 99 , and so
log = 1 − e−γt ,
N −R γ
R = .
1 + 99 exp − βN (1 − e−γt )
The maximum number of students who will ever hear the rumour is therefore given by the limit as t → ∞ of
R = βN ,
1 + 99e− γ
and so the required condition is
− βN 4
1 + 99e γ
99e− γ > ,
e− γ > ,
e γ < 297,
γ > .
Part (h): Of course all of the models can be criticised, often in exactly the same way. However discussion
the limitations of the models can be interesting in supervision.
Various reasons (be generous here)
• Assumes all students equally likely to mix with each other, irrespective of year and subject etc.
• Assumes students form a totally closed community, and in particular that noone can hear rumour from
• No stochasticity in the model
• Number of students who know the rumour should take only integer values
• Why is an exponential decay in interest taken (apart from to allow analytic solution)?
• Noone ever forgets the rumour
• No allowance for diﬀerent rates of spread in holidays
• etc.etc.etc. (accept anything at all reasonable here)
Other topics covered in this part of the course
The above question illustrates most of the topics covered for one particular model. The models covered in detail
• exponential growth:
• monomolecular growth
= β(κ − y).
• logistic growth (as above)
= βy(1 − ).
• what we call “growth that rises and falls”
= αe−γt − βy.
• the von Bertallnﬀy model of ﬁsh growth
= αy 3 − βy.
• Gompertz growth
= βe−αt y.
Supervisors would expected to talk in depth about these equations, including “derivations” and knowledge
of how they might be used. Mathematical techniques required but not discussed above include integrating
factors and the solution of Bernoulli equations by substitution of an appropriate form which linearises the
equation. Supervisors should know about these too. Finally these lectures involve a lot of graph sketching, and
in particular discussion of how the form of graphs depends upon the values of parameters. This is something
students will not generally have seen before, and which can take a fair bit of supervisory support.
A drug is given in tablet form by mouth. It has been formulated in such a way as to dissolve at a constant rate,
R, in the stomach. The drug is absorbed across the stomach wall in a manner in which the rate of transport
is proportional to the concentration of drug with proportionality constant, k. There is a ﬂow (Q) of liquid into
and out of the stomach such that the volume of the stomach contents does not change.
a) Draw a diagram to indicate the ﬂow of soluble drug into and out of the stomach water and write expressions
for each of the ﬂuxes.
b) Write an expression for the rate of change of concentration in the stomach with time after the tablet arrives
in the stomach up to the point where it has almost all dissolved. Indicate which terms are constants and which
c) Derive an expression for the soluble drug concentration in the stomach as a function of time.
d) Write an expression for the concentration at steady state (Css ).
e) Write an expression for the concentration as a function of time after the tablet has completely dissolved.
f ) Sketch a graph which shows the manner in which the concentration of drug in the stomach changes from the
time the tablet is ﬁrst administered until it has all left the stomach.
Solution and commentary
Part (a): In this part of the course students are taught no new mathematics of note. However the lectures
and examples heavily emphasise and practice setting up mathematical models, which in this part of the course
revolve around the “Input-Output” principle, i.e.:
Increase = Input − Output.
One simple way of correctly setting up models which rely on this guiding principle is to draw a carefully labelled
diagram, and this is the approach we follow. Students typically ﬁnd this part of the course rather diﬃcult, but
careful supervision can help here. This particular example is one of the harder questions we have set on this
topic in recent years.
The arrows are ﬂuxes (see below), which in the lecture notes we invariably label with Jsubscript , for example
Ji for the inﬂux (the other subscripts here refer to “eﬄux”, “secondary” and “additional”, respectively).
Part (b): Invariably the students are considering the rate of change of concentration of a particular substance
in a ﬁxed volume. If the concentration is C, the mass of substance is M and the ﬁxed volume is V , then using
the product rule with one of the functions constant in time, the rate of change of mass is
M =VC ⇒ =V .
So if the arrows on the diagram above correspond to rates of change of mass (i.e. ﬂuxes) then V dC will just
be the appropriate sum of the labelled quantities, taking account of direction. I ﬁnd that students can be best
persuaded to set up the equations correctly if it is emphasised that everything in their equation should be a ﬂux
(and so have units “amount of stuﬀ per unit time”...for example Ji clearly has these units as it is a concentration
multiplied by a volume ﬂow rate).
Just using the input-output principle:
V = Ji − Je + Js − Ja ,
V = 0 − QC + R − kC,
dC R Q+k
= − C
dt V V
Part (c): In the non-spatial situation (see below), unless the ﬂuxes are time-varying, typically the governing
diﬀerential equation will look something like
= α − βC.
As the emphasis in this part of the course is on the modelling, rather than the technical details of the solution,
we encourage students to just substitute into the “standard” solution for this class of (monomolecular) equation
α A −βt
C= − e ,
where A is a constant (ﬁxed from initial conditions). Note this illustrates how diﬀerent techniques of solution are
emphasised in diﬀerent lecture blocks, meaning new supervisors should deﬁnitely read the lecture notes to see
the approach taken by a particular lecturer. Sometimes students ﬁnd it tricky to choose the correct condition.
Here α = V and β = Q+k . Hence by back-substituting into the above
R V A −βt
C= − e .
When t = 0, C = 0, so A = V . Hence
C= 1 − e−βt .
Part (d): Once again equilibrium analysis is important. Here of clear biological signiﬁcance is the steady
state concentration in the stomach. This can be found in two ways...either by taking the limit of the solution
or by ﬁnding the value of C such that the governing diﬀerential equation leads to dC = 0. With bright students
it would be worth discussing whether this value would ever be reached in practice (which of course depends on
how quickly the tablet dissolves compared to the physiological properties of the stomach).
Css = .
Part (e): Here the key is to realise what is being asked for is just a small change to the governing equation,
and an appropriate change to the initial condition. This is one of the sorts of question that almost all students
would actually be able to do very easily, but may not have the conﬁdence to be sure they are correct.
Once the tablet has dissolved, the governing diﬀerential equation becomes
= − C,
which is just the equation of exponential decay. Assuming that steady state had been reached before the tablet
fully dissolved (which you have to do to make this question unambiguous), and measuring time from the time
of dissolution, C = Css when t = 0. Hence for times after that
C = Css e− V t
Part (f ): Most students who have made it this far should have no problem here.
The graph should show a monomolecular increase to the steady state, a period of nothing much going on,
then sometime thereafter an exponential decay back to zero.
Other topics covered in this part of the course
The ﬁrst half of this lecture course is basically as above, progressing in a sequence of graduated steps from
simple washout experiments (i.e. a compartment with a clean inﬂux and a single eﬄux and no secondary ﬂux,
leading to exponential decay), to more complex cascades where the eﬄux from one compartment is the inﬂux
to the next. The basic ideas are as above, though. The main diﬃculty for students is linking the biology to the
“boxes and arrows” diagram, and choosing an appropriate initial condition, although both come to the majority
The second half of the course is similar, but investigates spatial systems. The elaboration is to focus on one
small box (as per above) in a spatial system, in a local application of compartmental analysis, and for which the
ﬂuxes will typically depend on the length scale ∆x. Then by considering the limit as ∆x → 0 and ignoring time
dependence, spatial derivatives appear in the governing equations, and these can be solved to determine the
concentration proﬁle as a function of distance. Students ﬁnd this harder, but once they have been persuaded
that it is really exactly the same set of manipulations as before, they tend to calm down.
Introduction to Modelling of Interacting Populations
a) The diﬀerential equation, with initial conditions,
d2 y dy
−6 + 13y = 5e2x − 26x, y (0) = −1 y(0) = 1
has the complementary function y = e3x (A cos 2x + B sin 2x) where A and B are arbitrary constants.
Find the solution to the diﬀerential equations that also satisﬁes the initial conditions.
b) Locate and classify all the stationary points of the function
f (x, y) = −x2 − 2y 2 + xy 2 .
Solution and commentary
Part (a): This question is a fairly routine application of (the second half of) the algorithm for solving linear
constant coeﬃcient second order ordinary diﬀerential equations. This is an area of the course the students
generally really like, as it is very similar in approach to their school work (i.e. learn an algorithm then use it).
The complementary function is given in the question, so all that remains for the students is to ﬁnd a suitable
particular integral and to ﬁnd the values of A and B according to the initial conditions. I guess this question
is slightly complicated by
• The −26x term on the right hand side, which suggests a trial solution involving x alone, but only the
more well-prepared will realise they need a constant term too.
• The product of functions in the complementary function, which needs care when diﬀerentiating.
andd this will have baﬄed weaker students. Also, one problem that is sometimes seen is that students try to ﬁt
A and B to the given data from the complementary function alone, rather than from the full general solution
after ﬁnding and adding on the particular integral. New supervisors should also note that the entire algorithm
is detailed in the formula book we provide, together with suitable trial functions, and so students do not actually
have to commit anything to memory (although the better ones do end up doing so).
Try the particular integral
y = q e2x + ax + b
and diﬀerentiate it to obtain
= 2q e2x + a,
= 4q e2x .
Now substitute into the diﬀerential equation
4q e2x − 6 2q e2x + a + 13 q e2x + ax + b = 5e2x − 26x,
and match the coeﬃcients on either side:
Coeﬃcient of e2x 4q − 12q + 13q =5
Coeﬃcient of x 13a = −26
Coeﬃcient of x −6a + 13b = 0.
q = 1, a = −2, b = − 13 .
The general solution of the diﬀerential equation is
y = e3x (A cos 2x + B sin 2x) + e2x − 2x − 12
Diﬀerentiate to obtain
= 3e3x (A cos 2x + B sin 2x) + e3x (−2A sin 2x + 2B cos 2x) + 2e2x − 2.
Substitute in the initial conditions, y(0) = 13 , y (0) = −1, to obtain
13 = A+1− 13
−1 = 3A + 2B + 2 − 2
A = 0
−1 = 2B,
which has solutions A = 0, B = −1/2. The requisite solution is therefore
y = − 2 e3x sin 2x + e2x − 2x −
Part (b): The second half of this question is again “pure” maths (in the A level sense). They are taught a
method for classifying stationary points based on looking at a particular function of the second partial derivatives
(which they call A, B and C), and although they have also been shown why it works, few will understand it.
This is a possibility to discuss with the better students. I have found that weaker students sometimes ﬁnd
solving the non-linear simultaneous equations fx = fy = 0 is diﬃcult, and get lost. Here I try to encourage
them to be systematic and if they have time to sketch the curves fx = 0 and fy = 0 to get an idea of how many
solutions to expect by looking for intersections.
The partial derivatives are
fx = −2x + y 2 , fy = −4y + 2xy.
At a stationary point, fx = fy = 0. The equation for fy = 0 gives y = 0 or x = 2. Substituting these in turn
into the equation fx = 0, gives
−4 + y 2 = 0 ⇒ y = ±2.
Hence the three stationary points are, (0, 0), (2, 2) and (2, −2). They can be classiﬁed by using the AC − B 2
as described in lectures:
Stationary points (0, 0) (2, 2) (2, −2)
A = fxx −2 −2 −2 −2
B = fxy 2y 0 4 -4
C = fyy −4 + 2x −4 0 0
AC − B 2 8 −16 −16
Maximum Saddle-point Saddle-point
Other topics covered in this part of the course
In the remainder of the course they are taught how to perform a qualitative analysis of non-linear pairs of
coupled ﬁrst order odes, using the phase plane and linearisation near equilibria. The basic idea is to ﬁnd
equilibria, classify them using a criterion based on the Eigenvalues of the Jacobian matrix (although it is
not phrased in these terms) and then to sketch a phase plane and some typical trajectories. This allows the
qualitative properties of the solution to be investigated. An example of this approach is given in the Ecological
Applications question below. Note that although the reasoning behind all results is derived in full in the lectures,
the vast majority of students just end up classifying equilibria by simple-mindedly working out D11 , D12 , D21
and D22 (where Dij is the value of the diﬀerential of the ith equation wrt the jth variable, evaluated at the
equilibrium), forming the derived quantities T = D11 + D22 and ∆ = D11 D22 − D21 D12 , and then looking up
what this corresponds to, with little idea of what this means. Again this is all in the formula book.
Outline of syllabus for Introduction to Statistical Methods
This section of the course is the only statistical content, and is really about hypothesis testing. Students are
taught the following simple tests:
• Binomial tests
• One sample, two sample and paired t-tests
• One way ANOVA
• Linear regression
and are expected to be able to choose the appropriate test given a biological problem. They are also introduced
to two way ANOVA and to backwards stepwise elimination in the context of regression and model selection.
The students generally ﬁnd this material easy (although sometimes dry), and the lecture notes are excellent (so
potential supervisors who are not too sure of all of the details should not be too discouraged). Here an example
question is given without solution, as the details are fairly mechanical.
A biologist studying a polymorphic lizard species samples 150 lizards at random from her study population.
She records the colour morph of each individual, and whether or not each is infected by a certain nematode
parasite. Her results are summarised in the table below:
Blue Red Yellow
Infected 14 26 28
Uninfected 41 19 22
i) Do these results suggest that some morphs are more or less common than others? State which test you
use to address this question, and show your working.
ii) Do these results suggest that the three colour morphs diﬀer in their susceptibility to infection? State
which test you use to address this question, and show your working.
Interacting Populations: Ecological Applications
A particular model for the interaction between two species X and Y is
= aX − bXY,
= mXY − nY,
where a, b, m and n are positive parameters.
a) Brieﬂy explain the biological basis and assumptions of the model, paying careful attention to the meaning
of the parameters and the type of inter-species interaction that is being represented.
b) Calculate the equilibria exhibited by the model, and determine their stability.
A well-known alteration to this model updates the non-linear terms:
dX bX 2 Y
= aX − ,
dt c + X2
dY mX 2 Y
= − nY,
dt c + X2
In the above, c is another positive parameter.
c) Suggest a plausible biological interpretation of such an extension to the model. You may ﬁnd it helpful to
sketch and compare the functions f (X) and g(X), where
f (X) = bX,
g(X) = ,
c + X2
and to comment on the respective meanings of these functions in the original and extended models, and the
biological phenomena that any asymptote(s) and/or point(s) of inﬂexion may be intended to represent.
Solution and commentary
Once again, explaining models is emphasised. Note that this exact model is fully discussed in the lectures, so
this explaination should not be too much of a challenge. However with good students extensions to the basic
models can be discussed, and indeed this question is actually about one such extension.
Part (a): (7 marks maximum)
One mark for any of the ten points below, to a maximum of seven:
• The interaction is between a predator and its prey
• Identiﬁcation of X as prey and Y as predator
• In the absence of the predator, the prey reproduces exponentially, with per capita growth rate (no need
for this exact terminology) a
• In the absence of the prey, the predator dies (due to lack of food) at rate n
• Populations are linked by a predation term, which causes population of prey to decrease (as they are
being eaten or at least killed) and population of predators to increase (as they are eating and so able to
• Predation term is bilinear (mass action) in both X and Y (enough to say that rate at which prey are
predated upon increases indeﬁnitely with the size of both populations)
• Parameter b is prey eaten per unit predator per unit prey per unit time
• Parameter m relates to how many prey must be eaten per predator for it to be able to reproduce
• Up to one bonus for any particularly good explanation of any of these points or other “interesting”
• Or for any criticism of the model, for example exponential growth of prey is not very realistic, the pair
of populations would not be so isolated in reality, really the interaction would be stochastic, or anything
Part (b): (9 marks maximum)
Here the standard result (that there are predator prey cycles) is derived using the D11 , · · · , D22 method (via
proving the non-trivial equilibrium is a neutral focus or centre) as discused before. Note that of course this
is a simpliﬁcation (that the equilibrium is a centre does not mean there are cycles everywhere) and this could
be discussed with better students. As always, with weaker students these elaborations can be safely ignored in
favour of instilling a secure grasp of the basic result.
One mark for each of the nine points below, all necessary for full credit, and with marks carried forward
from any mistake (i.e. wrong expression for say D11 would lose at most one mark if all subsequent calculations
consistent with mistake):
• any attempt to ﬁnd equilibria by setting dt = dt =0
• correctly ﬁnding equilibrium at (0, 0)
• correctly ﬁnding equilibrium at (n/m, a/b)
• any attempt to assess stability by attempting some sort of partial diﬀerentiation of the diﬀerential equa-
• correct expressions for D11 , D12 , D21 , D22
• any attempt to use T and ∆
• getting T and ∆ correct for both equilibria (accelerated argument using the sign pattern of the Jacobian
is ﬁne and should be given full credit)
• saddle at (0, 0)
• neutral focus/centre at (n/m, a/b)
Part (c): (9 marks maximum)
This is the challenging part of this question. A full sketch of g(X), including ﬁnding the crucial inﬂexion
point, will have defeated many, as although it is not diﬃcult, it does require a signiﬁcant chain of extended
non-guided reasoning. Even those who did manage the sketch are unlikely to have been able to back-relate it
to the biology. However, this question illustrates the sort of reasoning and ﬂexibility we are trying to instill in
the students during this part of the course, and therefore what should be taught in supervision.
One mark for any of the points below, to a maximum of nine, but with correct overall shapes for f (X)
and g(X) and some sort of sensible interpretation of both what is included in each model and what it might
represent biologically necessary for full credit, i.e. for ﬁnal ﬁve marks require some sort version of each of the
ﬁve bold points (intended to discriminate the very best candidates):
• f (X) a straight line through the origin
• g(X) goes through origin
• g(X) has asymptote, tending to b as X increases
• one diﬀerentiation of g(X)
• second diﬀerentiation of g(X) and setting equal to zero
• correct inﬂexion point
• f (X) shows predator’s unbounded response in original model to the number of prey
• this unbounded response in f (X) is unrealistic
• asymptote in g(X) corresponds to the predator’s response saturating at high prey density
• due to (for example) satiation of the predators and/or limited amount that can possibly be eaten per unit
time by a single individual
• inﬂexion in g(X) corresponds to very low amount of predation at low prey density but an increasing
response as density of prey increases
• due to (for example) the predator learning how to more eﬃciently predate upon prey which is more
abundant and/or the predator switching to other food at low densities (note strictly-speaking the latter
not all that consistent with the decay term) and/or spatial aggregation eﬀects in the prey such that a
predator is often unable to ﬁnd any prey at all at very low densities, but at higher densities when they
have a better chance of ﬁnding some sort of prey, they will ﬁnd a lot all at once, and will be able to
eat them all and/or some sort of chemical produced by the prey with a threshold response to it in the
predator and/or... Accept anything at all plausible here.
• any comment that this will aﬀect stability of interaction and/or lead to interesting feedback response
where prey can be controlled by predator but can escape if somehow it’s population gets high enough
and/or mention of Holling (giving due credit to any who have done some extra reading)
Other topics covered in this part of the course
In addition to the Lotka Volterra model for predator prey dynamics (basis of this question) the students are
• Lotka Volterra model for competition;
• SIR model for epidemics;
• SIR model with host demography (i.e. births and deaths).