Chapter 2 Population growth Last week d N N ( t ) = N (0)e ( b d ) - - PowerPoint PPT Presentation
Theoretical Biology 2016 Chapter 2 Population growth Last week d N N ( t ) = N (0)e ( b d ) t d t = ( b d ) N with solution Consider a time scale of years: b is a birth rate (expected number of offspring per year), d is a death rate
dN dt = (b − d)N with solution N(t) = N(0)e(b−d)t
Consider a time scale of years: b is a birth rate (expected number of offspring per year), d is a death rate (expected chance to die per year), 1/d is the expected life span in years, ln[2]/d is the half life & ln[2]/(b-d) is the doubling time. As each individual is expected to produce b offspring each year over its total generation time of 1/d years, the total fitness amounts to
Figures taken from NRC Handelsblad 11 Dec 2010 (left) and Wikipedia (right).
Question 2.1 from today’s practical
Per capita birth rate expected to decline with density Per capita death rate should increase with density From Campbell
dN dt = (b − d)N with solution N(t) = N(0)e(b−d)t
Let d stay the normal death rate, i.e., 1/d remains the expected life span. Multiply d with a function that increases with N Note that this function should be non-dimensional f(N) = (1 + cN) = (1 + N/k) where k = 1/c
dN dt = (b − d)N with solution N(t) = N(0)e(b−d)t
Let d stay the normal death rate, i.e., 1/d remains the expected life span. Multiply d with a function that increases with N Note that this function should be non-dimensional f(N) = (1 + cN) = (1 + N/k) where k = 1/c
Multiply the parameter d with a non-dimensional function f(N) = (1 + N/k) dN dt = [b − d(1 + N/k)]N The dimension of the parameter k is biomass and the death rate has doubled when N = k. Steady state defines the “carrying capacity”: b − d = dN/k so ¯ N = kb − d d = k(R0 − 1)
Mathematically, one can rewrite both models as the classical “logistic equation”: dN dt = rN(1 − N/K) ,
N(t) = KN(0) N(0) + e−rt(K − N(0)) Here r = b − d is the natural rate of increase, and K is the carrying capacity.
Since this model is of the form N ’= bN - cN2 it can be rewritten into classical logistic growth:
Population size 100 80 60 40 20 200 400 500 600 300 Percentage of juveniles producing lambs
Reproduction of sheep on Hirta Island From: Campbell
From: Smith & Smith: Ecology Salicornia europea Grizzly bear
dN dt = (b − d)N with solution N(t) = N(0)e(b−d)t
Let b stay the normal birth rate at low densities. Multiply b with a function that decreases with N Note that this function should be non-dimensional f(N) = (1 - cN) = (1 - N/k) where k = 1/c
Multiply the b parameter with the non-dimensional function f(N) = (1 − N/k): dN dt = [b(1 − N/k) − d]N The dimension of the parameter k is again biomass, and the birth rate becomes zero when N = k. The steady state, or carrying capacity is b − d = bN k yields ¯ N = k(1 − d/b) = k(1 − 1/R0)
Mathematically, one can rewrite both models as the classical “logistic equation”: dN dt = rN(1 − N/K) ,
N(t) = KN(0) N(0) + e−rt(K − N(0)) Here r = b − d is the natural rate of increase, and K is the carrying capacity.
Since this model is also of the form N ’= bN - cN2 both can be rewritten into classical logistic growth:
allergen
C: free ligand (C > NRT), R: free receptors, RT: total receptors, C1: single bound ligand, C2: double bound ligand: RT = R + C1 + 2C2 How does C2, and hence the growth rate, depend on C? C C2
C C2 C1 R
dN dt = (b − d)N
0 ≤ 2C2 RT ≤ 1
C C2
C C2 C C2 C C2
R C C1 C2
RT = R + C1 + 2C2 , dC1 dt = 2konRC − koffC1 − xonRC1 + 2xoffC2 , dC2 dt = xonRC1 − 2xoffC2 . To study the steady state we set dC2/dt = 0 and add this to dC1/dt: dC1 dt = 0 = 2konRC − koffC1 = 2KRC − C1 , where K = kon/koff and R = RT − C1 − 2C2. Solving this gives C1 = 2CK(RT − 2C2) 1 + 2CK ,
10 0.3 0.15 C2 100 10 0.1 0.01 0.001 0.0001
1
C1 = 2CK(RT − 2C2) 1 + 2CK , which can be substituted into dC2/dt = 0 to solve C2 as a function of C: C2 = 1 + 4CK + 4C2K2 + 4CKRTX − (1 + 2CK)
q
(1 + 2CK)2 + 8CKRTX 8CKX where X = xon/xoff.
C2 C
Thus, the number of crosslinks is a bell- shaped function of the ligand concentration C. Cells grow best at intermediate ligand concentrations
dN dt = bN 2C2 RT − dN .
dt − − dC2 dt = xonRC1 − 2xoffC2 .
Low [ligand] Intermediate [ligand] High [ligand]
Increasing density at carrying capacity increases death rate and/or decreases birth rate. Population increase triggers a negative feedback: stable point. Increasing density at N=0: birth exceeds death rate: unstable From Campbell
dN dt = f(N) = rN(1−N/K)
(a)
N dN/dt = f(N) K
⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥
λ > 0 ↓ λ < 0 ↓ + − dh dt = λh
h(t) = h(0)eλt The stability of a steady state can be expressed as a “Return time” TR = −1 λ ∂Nf(N) = r − 2rN/K which for N = K gives λ = −r and a Return time of 1/r.
Let’s formalize this and plot N’ as a function of N. If increasing N decreases N’=f(N) the steady state is stable. This slope is defined by the derivative of f(N) with respect to N.
dN dt = f(N) = rN(1−N/K)
(a)
N dN/dt = f(N) K
⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥
λ > 0 ↓ λ < 0 ↓ + − dh dt = λh
h(t) = h(0)eλt The stability of a steady state can be expressed as a “Return time” TR = −1 λ ∂Nf(N) = r − 2rN/K which for N = K gives λ = −r and a Return time of 1/r.
∂Nf(N) = r − 2rN/K
¯ N = K → ∂Nf(N) = λ = −r ¯ N = 0 → ∂Nf(N) = λ = r
Population size 100 80 60 40 20 200 400 500 600 300 Percentage of juveniles producing lambs
Reproduction of sheep on Hirta Island From: Campbell
Convenient non-dimensional function: f(x) = x h + x is a saturation function 0 ≤ f(x) ≤ 1 with an initial slope of 1/h and a horizontal assymptote f(x) = 1 for x → ∞. At x = h the function is half maximal, i.e., f(h) = 1/2. The inverse function g(x) = 1 − f(x) = 1 1 + x/h can be used to define declining relationships.
f(x) x h x
g(x)
Hill functions can also be made sigmoid f(x) = x2 h2 + x2 is a sigmoid function 0 ≤ f(x) ≤ 1 with an initial slope of 0 and a horizontal assymptote f(x) = 1 for x → ∞. At x = h the function is half maximal, i.e., f(h) = 1/2. See the Appendix of the reader.
x f(x) h x f(x) h
dN dt = f(N) = rN(1−N/K)
(a)
N dN/dt = f(N) K
⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥
λ > 0 ↓ λ < 0 ↓ + − dh dt = λh
h(t) = h(0)eλt The stability of a steady state can be expressed as a “Return time” TR = −1 λ ∂Nf(N) = r − 2rN/K which for N = K gives λ = −r and a Return time of 1/r.
∂Nf(N) = r − 2rN/K
¯ N = K → ∂Nf(N) = λ = −r ¯ N = 0 → ∂Nf(N) = λ = r
¯ x x
← h →
f(¯ x) f(x) . f(¯ x) + f0h f0 = ∂x f(¯ x)
: f(x) ' f(¯ x) + ∂x f(¯ x) (x ¯ x) = f(¯ x) + f 0h. T = ¯ x.
dN dt = f(N) ' f( ¯ N) + ∂Nf( ¯ N) (N ¯ N) = 0 + λh ,
: f(x) ' f(¯ x) + ∂x f(¯ x) (x ¯ x) = f(¯ x) + f 0h. T = ¯ x.
dN dt = dN dt d ¯ N dt = d(N ¯ N) dt = dh dt , dh dt = λh with solution h(t) = h(0)eλt ,
Indeed if λ< 0, h(t) will approach zero where h = ¯
N − N λ = f 0(N) = ∂Nf(N)
dN dt = f(N) = rN(1−N/K)
(a)
N dN/dt = f(N) K
⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥ ⇥
λ > 0 ↓ λ < 0 ↓ + − dh dt = λh
h(t) = h(0)eλt The stability of a steady state can be expressed as a “Return time” TR = −1 λ ∂Nf(N) = r − 2rN/K which for N = K gives λ = −r and a Return time of 1/r.