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 (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 R 0 = b/d
Birth and death rate depend on density Figures taken from NRC Handelsblad 11 Dec 2010 (left) and Wikipedia (right). Question 2.1 from today’s practical
Birth and death rates can depend on density From Campbell Per capita birth rate expected to decline with density Per capita death rate should increase with density
Density dependent death rate d N N ( t ) = N (0)e ( b − d ) t d t = ( b − d ) N with solution 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
Density dependent death rate d N N ( t ) = N (0)e ( b − d ) t d t = ( b − d ) N with solution 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
Density dependent death rate Multiply the parameter d with a non-dimensional function f ( N ) = (1 + N/k ) d N d t = [ 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”: N = kb − d ¯ b − d = dN/k so = k ( R 0 − 1) d
Logistic growth Mathematically, one can rewrite both models as the classical “logistic equation”: Since this model is of the form N ’= bN - cN 2 it can be rewritten into classical logistic growth: d N KN (0) d t = rN (1 − N/K ) , or N ( t ) = N (0) + e − rt ( K − N (0)) Here r = b − d is the natural rate of increase, and K is the carrying capacity.
Density dependent birth rates Percentage of juveniles producing lambs 100 Reproduction of sheep on Hirta Island 80 60 40 20 0 200 300 400 500 600 Population size From: Campbell
Density dependent birth rates Grizzly bear Salicornia europea From: Smith & Smith: Ecology
Density dependent birth rate d N N ( t ) = N (0)e ( b − d ) t d t = ( b − d ) N with solution 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
Density dependent birth rates Multiply the b parameter with the non-dimensional function f ( N ) = (1 − N/k ): d N d t = [ 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 ¯ yields N = k (1 − d/b ) = k (1 − 1 /R 0 ) k
Logistic growth Mathematically, one can rewrite both models as the classical “logistic equation”: Since this model is also of the form N ’= bN - cN 2 both can be rewritten into classical logistic growth: d N KN (0) d t = rN (1 − N/K ) , or N ( t ) = N (0) + e − rt ( K − N (0)) Here r = b − d is the natural rate of increase, and K is the carrying capacity.
Cross linking of receptors activates cells Mast cell degranulation B cells activate and start to divide allergen
Bivalent ligand binding a monovalent receptor 0 ≤ 2 C 2 d N C : free ligand ( C > NR T ), ≤ 1 d t = ( b − d ) N R T R : free receptors, R T : total receptors, d N d t = bN 2 C 2 C 1 : single bound ligand, − dN . C 2 : double bound ligand: R T R T = R + C 1 + 2C 2 C How does C 2 , and hence the C 1 growth rate, depend on C ? C 2 R ? C 2 C
Bivalent ligand binding a monovalent receptor A B C 2 C 2 C C C D C 2 C 2 C C
R T = R + C 1 + 2 C 2 , d C 1 = 2 k on RC − k o ff C 1 − x on RC 1 + 2 x o ff C 2 , d t d C 2 = x on RC 1 − 2 x o ff C 2 . d t To study the steady state we set d C 2 / d t = 0 and add this to d C 1 / d t : d C 1 d t = 0 = 2 k on RC − k o ff C 1 = 2 KRC − C 1 , C C 1 where K = k on /k o ff and R = R T − C 1 − 2 C 2 . C 2 Solving this gives R C 1 = 2 CK ( R T − 2 C 2 ) , 1 + 2 CK
− − d t C 1 = 2 CK ( R T − 2 C 2 ) d C 2 , = x on RC 1 − 2 x o ff C 2 . 1 + 2 CK d t which can be substituted into d C 2 / d t = 0 to solve C 2 as a function of C : 1 + 4 CK + 4 C 2 K 2 + 4 CKR T X − (1 + 2 CK ) q (1 + 2 CK ) 2 + 8 CKR T X C 2 = 8 CKX where X = x on /x o ff . 0.3 Thus, the number of C 2 C2 crosslinks is a bell- shaped function of the ligand concentration C. 0.15 Cells grow best at intermediate ligand concentrations d N d t = bN 2 C 2 C − dN . 0 R T -05 10 0.0001 0.001 0.01 0.1 1 10 100
In retrospect this is intuitive: most crosslinks at intermediate ligand concentrations Low [ligand] High [ligand] Intermediate [ligand]
Stability of steady states From Campbell 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
Stability of steady states d N d h d t = f ( N ) = rN (1 − N/K ) h ( t ) = h (0)e λ t Let’s formalize this and d t = λ h or plot N ’ as a function of N . (a) The stability of a steady state can be expressed as a If increasing N decreases d N/ d t = f ( N ) “Return time” λ > 0 λ < 0 N’=f(N) the steady state ↓ ↓ T R = − 1 is stable. λ This slope is defined by + ∂ N f ( N ) = r − 2 rN/K � ⇥ the derivative of f(N) with � ⇥ � ⇥ � ⇥ which for N = K gives λ = − r respect to N . 0 � ⇥ − � ⇥ � ⇥ 0 K and a Return time of 1 /r . � ⇥ N � ⇥
Stability of steady states d N d h ∂ N f ( N ) = r − 2 rN/K d t = f ( N ) = rN (1 − N/K ) h ( t ) = h (0)e λ t d t = λ h or (a) The stability of a steady state can be expressed as a ¯ d N/ d t = f ( N ) N = K ∂ N f ( N ) = λ = − r “Return time” → λ > 0 λ < 0 ¯ N = 0 ∂ N f ( N ) = λ = r ↓ ↓ T R = − 1 → λ + ∂ N f ( N ) = r − 2 rN/K � ⇥ � ⇥ � ⇥ � ⇥ which for N = K gives λ = − r 0 � ⇥ − � ⇥ � ⇥ 0 K and a Return time of 1 /r . � ⇥ N � ⇥
Modeling non-straight functions Percentage of juveniles producing lambs 100 Reproduction of sheep on Hirta Island 80 60 40 20 0 200 300 400 500 600 Population size From: Campbell
Hill functions f(x) Convenient non-dimensional function: x x h f ( 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 g ( x ) = 1 − f ( x ) = 1 + x/h can be used to define declining relationships. x
Sigmoid Hill functions Hill functions can also be made sigmoid x 2 f ( x ) = h 2 + x 2 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. f ( x ) f ( x ) h h x x
Actually we linearized f(N) at N=K d N d h ∂ N f ( N ) = r − 2 rN/K d t = f ( N ) = rN (1 − N/K ) h ( t ) = h (0)e λ t d t = λ h or (a) The stability of a steady state can be expressed as a ¯ d N/ d t = f ( N ) N = K ∂ N f ( N ) = λ = − r “Return time” → λ > 0 λ < 0 ¯ N = 0 ∂ N f ( N ) = λ = r ↓ ↓ T R = − 1 → λ + ∂ N f ( N ) = r − 2 rN/K � ⇥ � ⇥ � ⇥ � ⇥ which for N = K gives λ = − r 0 � ⇥ − � ⇥ � ⇥ 0 K and a Return time of 1 /r . � ⇥ N � ⇥
Linearization of a function f 0 = ∂ x f (¯ x ) f ( x ) . f (¯ x ) + f 0 h f (¯ x ) x x ¯ ← h → : f ( x ) ' f (¯ x ) + ∂ x f (¯ x ) ( x � ¯ x ) = f (¯ x ) + f 0 h . T = ¯ x .
Stability and Return time : f ( x ) ' f (¯ x ) + ∂ x f (¯ x ) ( x � ¯ x ) = f (¯ x ) + f 0 h . T = ¯ x . d N d t = f ( N ) ' f ( ¯ N ) + ∂ N f ( ¯ N ) ( N � ¯ N ) = 0 + λ h , where h = ¯ N − N λ = f 0 ( N ) = ∂ N f ( N ) d t � d ¯ d t = d( N � ¯ d N d t = d N N N ) = d h d t , d t d h h ( t ) = h (0)e λ t , d t = λ h with solution Indeed if λ < 0, h(t) will approach zero
Stability and Return time d N d h d t = f ( N ) = rN (1 − N/K ) h ( t ) = h (0)e λ t d t = λ h or (a) The stability of a steady state can be expressed as a d N/ d t = f ( N ) “Return time” λ > 0 λ < 0 ↓ ↓ T R = − 1 λ + ∂ N f ( N ) = r − 2 rN/K � ⇥ � ⇥ � ⇥ � ⇥ which for N = K gives λ = − r 0 � ⇥ − � ⇥ � ⇥ 0 K and a Return time of 1 /r . � ⇥ N � ⇥ r -selected vs K selected species
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