SC/BIOL Current Topics in Biophysics TERM TEST ONE

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Page 1 of 1 SC/BIOL 2090.02 Current Topics in Biophysics TERM TEST ONE Name: KEY Student ID: There are three questions. You must complete all three. Ensure that you show your work (that is, equations, calculations and units). Excessive length is not encouraged. QUESTION ONE Many multi-cellular organisms rely on internal pumps to provide nutrients and oxygen throughout their bodies. Gravity can be a counteracting force. As one example, consider a python lying horizontal on the grass versus climbing a tree, head-up. Snake lengths vary, but the heart is usually situated 0.25 of the total length behind the head for tree climbers. For non-tree climbers the heart is situated at about 0.37 of the total length. So, for a 10-meter long tree-climbing snake, the heart is 2.5 meters behind the head. The blood pressure of a treeclimbing snake is about 10.5 kilopascals. Consider what happens when the snake slithers from a horizontal to a vertical orientation (head-up). Quantify and compare two strategies the snake could use when it starts to climb the tree: changing the work per heart beat (volume compression per pump cycle) or changing the heart beat rate. Is one strategy better than the other, or are they equivalent? Explain, showing your calculations. Would it matter whether the vascular system was a closed piping system (without an opening to the outer atmosphere) or open at some location (to allow some pressure equilibration between the vascular system and the outer atmosphere)? Explain. Show your work with clarity. QUESTION TWO What is R (the gas constant)? Why is it available in a bewildering array of values (see constants handout)? Remember that Dr. Lew is not a physicist, and he believes that units are important. QUESTION THREE Acetylcholine is a crucial neurotransmitter in your electrical system. It is released at the pre-synaptic membrane, diffuses across the synaptic cleft and binds to acetylcholine receptor on the post-synaptic membrane, to cause the next neuron to fire electrical impulses. Assuming the diffusion coefficient is 4!10 6 cm 2 s 1, how wide could the synapse gap be if acetylcholine must diffuse across the synaptic cleft in 10 5 sec?

N T = N 0 2 (T/g) g is the generation time N 0 is the number of cells at time T = 0 N T is the number of cells at time T as time increases, t/g = 1, 2, 3..., thus 2 1, 2 2, 2 3, etc. A cube has a surface area of 6 L 2. Its volume is L 3. As long as the shape is constant, the ratio of suraface area to volume will always be (6 L 2 ) / L 3, or 6/L. For a sphere, the surface area is 4 π r 2, and the volume is π r 3 ; the corresponding ratio of surface area to volume is 4/r. Logistic growth curve: N T = K N 0 et g K + N 0 (e T g 1) K is the carrying capacity 2 r L compression = ρ h F cr = E I π 2 2 L eff F cr = E π r4 π 2 4, and F (2 h) 2 cr = ρ π r 2 h Ψ wv = RT % relative humidity ln + ρ V w 100 w gh velocity (meters sec 1 ) 1(kg(f)) 42.4 10 6 (N m 2 ) 9.80665(N) h critical = 436(kg m 3 ) pressure difference (Pascal = 1 kg m 1 sec 1 ) tube radius L k L v = Δp 1 (R 2 r 2 ) l 4 η (area) A 1 = 6 L 2 A k = 6 (k L) 2 A k = 6 k 2 L 2 ( = k 2 A 1 ) (volume) V 1 = L 3 V k = (k L) 3 V k = k 3 L 3 ( = k 3 V 1 ) The scaling coefficient is different for area (k 2 ) and for volume (k 3 ). Heat conduction rates are defined by the relation: P cond = Q / t = k A [(T a - T b ) / L] where P cond is the rate of conduction (transferred heat, Q, divided by time, t); k is the thermal conductivity; T a and T b are the temperatures of the two heat reservoirs a and b; A is the area; and L is the distance. Thermal conductivities of water and air are about 0.6 and 0024 W m 1 K 1, respectively. Thermal radiation is defined by the relation: P rad = σ ε A T 4 where P rad is the rate of radiation; σ is the Stefan-Boltzmann constant (5.6703 10 8 W m 2 K 4 ; ε is the emissivity (varies from 0 to 1, where 1 is for a blackbody radiator); A is the area; and T is the temperature (in Kelvins). The net radiative emission or absorption will depend upon the difference in temperature: P net = σ ε A (T 4 body T4 ambient ) distance (meters) v = Δp 1 R 2 l 4 η density (water = 1 gm cm 3 ) viscosity (water = 0.01 gm cm 1 sec 1 ) distance from center of tube viscosity (water = 0.01 gm cm 1 sec 1, or Pa sec) R e = ρ ν l η J v = Δp π R 4 l 8 η velocity (cm sec 1 ) tube diameter (cm)

J = D c x J = 1 2 Δ2 τ dc dx D = 1 2 Δ2 τ units: moles cm 2 sec 1 a Fick s First Law of Diffusion: The flux is proportional to the concentration gradient c t = J x velocity vector the notation grad ν is sometimes used C(r) = C 0 1 a r r Fick s Second Law of Diffusion: Changes in concentration over time depend upon the flux gradient c t = D 2 c x 2 ν = u x + v y + w z J x = D c x + v x c (cm 2 sec 1 )(moles cm 4 ) with velocity components, u, v, and w, in the three dimensions, x, y, and z. (cm sec 1 )(moles cm 3 ) J V = r2 8 η P x Fick's First law : J r = D C r Fick's Second Law : (steady state) C t = D 1 r 2 C r r2 = 0 r J r (a) = D C 0 4 π a = I D (diffusive current) (mole cm 2 sec 1 ) I m = 4 π a 2 β (metabolic current) C t (cm 2 ) = u C + r C D 2 C r 2 (units of mole sec 1 ) flow velocity concentration concentration gradient Q = Δp l πa 4 8η A (units of mole sec 1 ) P e = 2 a u D µ liquid j = µ * j + RT lna j + V j P + z j FE + m j gh frontal area fluid density D = 1 2 ρ V 2 A C D p o (t) = e λt drag coefficient (shape-dependent) velocity p n = e λt (λt) n n! B C m dυ = 6 π η r υ dt υ(t) = υ 0 e = e µ µ k k! 6 π η r t m

+ + Rotor (n) + mh outside Rotor (n +1) + mh inside + ADP + P i +mh outside gas constant µ = µ º + RT ln(a H + ) + zf Ψ temperature activity of protons ΔG total = n Δµ H + + ΔG ATP = 0 n (RT ln a H + a H + inside outside Δµ H + = RT ln a H + F a H + F = Aυ + Bω N = Cυ + Dω That is, both velocity and rotation contribute to both the force and torque. + ATP + mh inside Faraday constant Voltage [ATP] ΔG ATP = ΔG º ATP + RT ln [ADP][P i ] + FΔΨ) + ΔG o [ATP] ATP + RT ln = 0 [ADP][P i ] inside outside + ΔΨ (units: mv) RT/F is about 25 mv at 20ºC. The work exerted will depend upon the speed of the contraction, and the cross-sectional area of the muscle times its length. Muscle contraction speeds are normally in the range of 3 milliseconds. The initial velocity will equal the impulse force divided by the mass (ν = F impulse /mass). The work done in the leap is proportional to mass and the height of the leap (W mh), while the work of the muscles is proportional to the mass of the muscle (or the whole organism) (W m). It follows then, that the total work is related solely to the height, since the organism s mass cancels out. Thus, the height of the leap is not proportional to the organisms s size, but rather is similar for any organism. D Arcy Thompson describes this as an example of the Principle of Biological Similitude. µ liquid j = µ * j + RT lna j + V j P + z j FE + m j gh a j = γ j c j The activity of water (a j ) is the product of the activity coefficient and the concentration of water RT ln a j = V j Π RT ln a j + V j P + m j gh osmotic pressure Π s = RT j gravitational potential V j = V n j n i,t,p,e,h The partial molal volume of species j is the incremental increase in volume with the addition of species j. For water, it is 18.0 X 10 6 m 3 mol 1. c j The terms inter-relate various properties of water: changes in its activity with the addition of solutes, and the relation to pressure. Van t Hoff relation