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Contents

Chapter 1 1

Chapter 2 4

Chapter 3 11

Chapter 4 15

Chapter 5 19

Chapter 6 21

Chapter 7 28

Chapter 8 33

Chapter 9 37

Chapter 10 40

Chapter 11 43

Chapter 12 49

Chapter 13 52

Chapter 14 55

Chapter 1 1. (a) If our DNA contained combinations of three bases instead of four, how many amino acids could be

encoded when a codon contains one, two, or three bases? Number of words = n(letters)m(word length) 31 = 3; 32 = 9; 33 = 27. Can cover the coding of 20 amino acids with three bases and by using a triplet to code for a particular amino acid

(b) Why do you suppose living systems use 4 bases instead of 3 bases in the genetic code? With four bases and more synonyms, the system is more robust. A mutation could wind up coding for the same amino acid, and the same protein would be made. Also stop and start signals can be composed of additional “words.” 2. If a cell is maintaining 200 proteins and the average number of amino acids per protein is 75, what is

the total number of A, G, C, and T bases used to code for the construction of the proteins? 200 prot * 75 AA/prot * 3 bases/AA = 45,000 bases. 3. Match the statement on the left with the best analogous match on the bottom

A car is either a Toyota or a Ford. a

The Naval Tomcat aircraft was based on the F14 prototype. d

Ford water pumps do not work in Toyotas. b

Each musical measure has the same number of beats. e

A taped message in Mission Impossible incinerates after it is read. c

(a) Taxonomy (b) 16S RNA gene in archaea vs. eucarya (c) mRNA lifetime (d) Phylogeny (e) codon triplet (f) universal genetic code (g) Bill Clinton 4. Which compound(s) below would not likely be used by a heterotroph for energy?

(a) carbon dioxide (b) glucose (c) methane (d) fructose (d) reduced iron (Feo) Cannot oxidize carbon dioxide further and Feo is not an organic, C-containing compound. 5. A microbe was found in a fossilized meteor in Antarctica. It appears to have a DNA-like molecule

made of 5 different types of base-like molecules. An analysis of other molecules suggests that 25 different amino acid-like molecules make up something that resembles proteins. If there is a genetic code in use, then what is the minimum number of base molecules per codon? Why is the value a minimum?

52 = 25, so with two bases per codon, can cover minimally. More bases per codon would provide more redundancy/degeneracy as well as start and stop signals.

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6. If the yield for the growth of E. coli on glucose is 0.3 g cell (water-free basis)/g glucose, answer the following:

(a) What will be the mass concentration of E. coli on a water-free basis when provided 5 g

glucose/liter? 5 g gluc/lit * 0.3 g cell dry wt/g gluc = 1.5 g cell dry wt/lit.

(b) What will the total mass of E. coli per liter? 1.5 g cell dry wt * 1 g cell total (hydrated)/0.3 g cell dry wt = 5 g cell total/lit. Note: In any multicomponent system, concentrations can be on different bases just as income can be on a pre-tax or after-tax basis. A good engineer always asks what basis a concentration or other number is on to make sure everyone is talking about the same thing. 7. Assume that a typical protein has 100 amino acids and a “ballpark” molecular weight for an amino acid

is 100 g/mol. How many protein molecules are present per 70 kilograms (i.e., average weight of a human) of hydrated animal cells? If a protein’s typical dimension is 10 Angstroms (1 A = 10-8 cm), could the distance between Pittsburgh and Los Angeles be spanned by aligning the protein molecules end-to-end?

100 mol AA/mol prot * 100 g/mol AA * 1 g prot/g AA = 104 g prot/mol prot. 70,000 g person * 0.3 g solid/g person * 0.6 g prot/g solid * mol prot/104 g prot * 6.0 (1023) molecules/mol = 7.56 (1023) protein molecules. A protein is about 10 Angstroms in size (10-7 cm), which means if all these proteins were laid end to end the total length would be 7.56 (1023) protein molecules * 10-7 cm/prot molecule * m/102 cm * km/103 m = 7.56 (1011) km, which will get one from California and back quite easily. 8. What type of cell on average has more transcription going on, a growing bacterial cell or a stem cell

residing in the bone marrow? Bacterial cell. Stem cells are typically quiescent and undergo division and differentiation only when required for development or repair. 9. Where would one most likely look to isolate a new Archaea--spoiled hamburger or in a boiling hot

spring in Yellowstone National Park? While spoiled hamburger can be pretty nasty, some environments in Yellowstone are more akin to early Earth and present the environmental extremes that Archaea are adapted to. 10. Identify what is not true about the following statement: Bacteria and animal cells use the same genetic

code (i.e., codons) to store protein “recipes” and use the same ribosomal machinery to translate the information into functional proteins.

The genetic code is fairly conserved, so that part is true. If that were not true, bacteria would not be able to produce human or other proteins, and rDNA technology would be not be as advanced as it is. The parts of the ribosomal machinery do, however, exhibit differences. The similarities and differences are now used as the basis for modern taxonomy. 11. Assume a cell possesses 1000 genes, but at any point in time, ten percent are being used (i.e.,

“expressed”). If a typical protein has 100 amino acids, what is the total number of A, G, C, and T bases that encode the cell’s (a) expressed and (b) total genetic repertoire?

(a) 1000 (0.1) genes expressed * 1 prot/gene * 100 aa/prot * 3 bases/aa = 3 104.

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(b) total repertoire = 3 105. 12. A codon in a Martian bacterium contains combinations of four of the five base-like molecules that

makes up what passes for Martian DNA. An average bacterium on Earth has 4000 genes. On Mars life is quite different; hence, a larger repertoire of 6000 genes is needed to provide flexibility. By what factor is the DNA larger in a Martian bacterium compared to an Earthling bacterium? Assume a Martian protein contains about as many amino acids as an Earthling protein.

(a) 1.2 (b) 1.33 (c) 1.67 (d) 2.0 (e) 2.66

Application of basic modern biology. Information is stored in DNA. Each gene codes for a protein. A codon specifies each part (amino acid) of the protein. A codon possesses three bases on Earth. So the Martian DNA is larger than Earth’s by (6000/4000)* (4/3) = 24/12 = 2 = (factor of more genes) (factor by how much larger codons are). 13. A new organism may have been found in the University Center dining hall in the cole slaw. What is

the best thing to do first to characterize the new cell? (a) Sequence the entire genome. (b) Measure mRNA stability. (c) Look for similarity with the 16S RNA encoding stretch of DNA in other known organisms. Modern biology is based on molecular determinism. A key concept is that cells are classified into three groups (Bacteria, Eucarya, and Archaea). They can be distinguished by not having “interchangeable” 16S RNA. If the cell appears to be “new,” then it seems first “best” thing to do is to see which family it belongs to, before getting into more detailed studies. So answer (c) makes the most sense. Another way to view this is that genome sequencing is expensive so eliminate (a). Other data are meaningless if you do not even know what kind of organism it is.

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Chapter 2 1. A typical cylindrical-shaped bacterial cell is 2 µm long and has a radius of 0.5 µm. Assuming that the

yield is 0.3 g cell/g glucose and the density of a hydrated cell is 1.05 g/cm3, how many molecules of glucose are needed to build one cell?

Volume of Cell = πR2L = π (0.5)2 2 = 1.57 µm3. Mass of one cell = 1.57 µm3 * 10-18 m3/µm3 * 106 cm3/m3 *1.05 g/cc * 0.3 g dw/g = 5 10-13 g dw. Mass glucose needed = 5 10-13 g dw *g gluc/0.3 g dw = 16.7 10-13 g glucose. Molecules glucose needed = 1 mol gluc/180 g * 16.7 10-13 g glucose * 6.02 1023 molecules/mol = 5.6 109 molecules of glucose. 2. A life science experiment is to be flown on a space shuttle mission. The experiment entails studying

the effects of microgravity on cell growth kinetics. This experiment is one of many that will be performed, so to manage weight only 1 kg of raw materials can be flown for this particular experiment. What are the optimal amounts of the materials, glucose (C6(H20)6), ammonium chloride (NH4Cl), and phosphate salt (KH2PO4), that can be brought on the mission if one gram of glucose yields 0.3 g cell on a dry weight basis? Optimal means that no excess mass was flown or occupied space.

We want to pick each component such that none is in excess and thus represents mass that was flown for no reason. First find the proportions needed so that none of the N, P, or C sources are in excess. Basis: 1 g cell dry weight. Glucose = 3.33 g. Nitrogen = 1 * 0.14 = 0.14 g N so NH4Cl = 53/14 * 0.14 = 0.53 g. Phosphate = 1 * 0.04 = 0.04 g P so KH2PO4 = 136/31 * 0.04 = 0.175 g. The component masses that add to one kg must also be in this proportion to maximize the cells grown while minimizing the mass of raw materials flown. Basis: 1 kg total of C, N, and P source. x = glucose, y = NH4Cl, and z = KH2PO4. 1 = x + y + z. y/x = 0.53/3.33 = 0.159. z/x = 0.175/3.33 = 0.0526. We now have three independent equations and three unknowns. 1/x = 1 + 0.159 + 0.0526 = 1.21 so x = 0.825. Now having x, we find that y = 0.131 and z = 0.0434. Check 1 =??= 0.825 + 0.131 + 0.0434 = 0.9994; close enough for NASA work. 3. If 50% of ingested iron is not absorbed by the body and 75% of the iron in red blood cells is recycled,

estimate how much iron must be ingested per day to maintain the iron content of the blood. Assume 21.5 mg iron/day is available for recycle.

Md = 21.5

MR = 0.75 * 21.5

Me2 = 0.25* 21.5

Me1 = 0.5 Mi

Mi = ?? Red blood cell inventory

Me1 + Me2 = 0.5 Mi + 0.25*21.5

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Basis: 21.5 mg iron/day available from spent red blood cells (from reading) = Md. Recycle efficiency = 75%; hence, MR = 0.75 * 21.5 and Me2 = 0.25 *21.5. Me1 = 0.5 Mi. There are five variables and we know the values of four; hence, only one equation is needed. An overall mass balance will suffice. Mi = Me1 + Me2. Mi = 0.5 Mi + 0.25 * 21.5. Mi = 10.75 mg iron/day. 4. A 1,000,000 m3 lake has a phosphorous content of 5 mg/m3. A river provides the lake with new water

and nutrients. Rainfall provides nutrient-free water. Another river drains the lake. The local precipitation on the lake surface provides 100,000 m3/year of water. The flow rate of the outfall is 1,000,000 m3/year. What must the phosphorous content of the incoming river water be to maintain a constant phosphorous concentration in the lake?

100,000 m3/yr P conc = 0; pure rain water 900,000 m3/yr 1,000,000 m3/yr Conc = ?? Conc = 5 mg/ m3

1,000,000 m3 5 mg/m3

Mass flow P in = mass flow P out at steady state, and volumetric flow rate of water into the lake will maintain constant lake volume 900,000 x + 100,000 (0) = 1,000,000 * 5 [mg/yr]. x = 5.56 mg/m3. 5. The surface area of a typical human is 1.8 m2. If you shed and replace the outer skin cells every 30

days, estimate how many cans of soda are represented by the carbon content of the lost skin cells. Assume that the typical dimension of a skin cell is 50 µm, the density of a hydrated cell is 1 g/cm3, and a can of soda contains 36 grams of sugar as glucose.

Basis: 1.8 m2 of exterior area of skin cells. Number of skin cells ~ 1.8 m2 /[50*50 10-12 m2] = 7.2 108. Carbon content of a skin cell ~ [50*50*50 10-12 cm3] * 1 g/cm3 *0.3 g dw/g * 0.5 g C/g dw = 1.88 10-8 g C. Cans of soda equiv = total skin cell C/soda C = (7.2 108 * 1.88 10-8 )/(36 *72/180) = 0.94 ~ 1 can of soda. 6. Average daily water inputs and losses for a human at rest are summarized below

Food 1000 g Drink 1200 g Air In 50 g Air Out 400 g Metabolic Water Production 300g Sweat 350 g (evaporated) + 200 g (damp) Urine 1400g Feces 200 g

If you do not drink any water, will you think that you lost weight and if so, how much?

Check the water balance to see if there is a deficit when one does not drink. This problem is also relevant to the now frowned on practice of shunning fluids (and/or wearing a sweat package) prior to a wrestling match in order to “make weight.” It can also explain why you weigh less in the morning.

Water in = 1000 + 50 + 300 = 1350 g. Water out = 400 + 350 + 200 +1400 + 200 = 2550 g.

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You would appear to loose 1200 g (1.2 kg, 2.6 lbs). Indeed this is the basis for why you weigh less in the morning. For eight hours or more you do not drink fluids, yet you still lose water through your lungs and other ways. That first thing in the AM elimination before you get on the scale helps the weight picture as well.

7. Consider the following proposal that has been put to a bioengineer for evaluation. If we put food crops

in a space station, then the carbon dioxide released due to the activity of metabolic processes could be used to produce plant material via photosynthesis. Then, once the process was up and going, the astronauts could eat some of the plants and lower the amount of supplemental food they need to bring with them. That is a steady state inventory of plant mass can be established where new growth from fixing CO2 by photosynthesis is harvested. Develop an operating diagram that relates how the fraction of the total glucose respired that can be derived from plants depends on the efficiency of carbon capture by photosynthesis. Assume that an astronaut exhales on average, 100 ml/min of carbon dioxide. Also assume that plants are 70% by weight water, 50% by weight carbon (water-free basis), and 90% of the plant carbon is equivalent to the carbon in glucose-like carbohydrate (i.e., can be metabolized for energy by respiration). Does this proposal seem feasible? Do you have any ideas about how to improve the process?

Basis 144 lit CO2/d output ~ burn rate. Plants --70 wt% water, 50 wt % C water-free basis, 90% of the carbon in plants equiv to glucose-C.

Assume for now that although the plant carbohydrate could be assimilated by an astronaut, we will just look at the astronaut’s energy needs. The astronaut’s burn rate (CO2 output) will be compared to the burn rate plant carbohydrate could provide.

Process Diagram

144 lit/d * 12 g C/22.4 lit = 77 g C/d; ε = efficiency of C capture

d

ε 77 g C/d ∆Plant Mass ε 77 * 0.9 = ε 69.3 g Carbo-C/d

ε 7.7 g non Carbo-C/d

ε69.3 g Carbo-C/d

F

In general, F [g new C/d] = 77 – ε69.3

Consider a test case. If 50% efficien(0.5*69.3)/77 = 0.45 of total respiratspace station. How much total plant m

Total new plant mass production/d ismaterial/d. Here we eat each day’s inc

Thus, need more than 256 g of plant mplant (kinetics) can increase its mass bwill be more mass associated with theair with a spray mist is used. There

(1 – ε) 77 g C/

F + ε 69.3 = 77

(basis for operating diagram).

t at C capture & use each day’s increment in plant carbo, can supply ion. That sounds pretty good, but look at the whole picture on the aterial needs to be on board?

0.5*77 gC d-1 [g dw/0.5 g C] [g total/0.3 g dw]= 256 g new plant rement to maintain a steady state value of total plant biomass.

ass as a standing crop to support this daily harvest. If a fast growing y 1 percent per day, then will need 25.6 kg total plant biomass. There

plant mass due to soil, water, or a hydroponics setup unless growth in will be mass associated with horticulture supplies and equipment.

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Maintaining and/or finding the spaces for that much plant biomass and supporting gear could prove to be a challenge and has to be compared to total station size, science/technology mission requirement.

8. The doubling time of an infectious microbe is 20 minutes. How long will it take for the cell population to expand 1000-fold?

N = No exp(µt). µ = 0.693/0.33h = 2.1 h-1. N/No = 1000 = exp(µt) = exp(2.1t). 2.1 t = ln (1000) so t = 3.3 hours. They can multiply pretty quickly. If not, then we probably would not get sick. 9. A sink can hold 5 liters of water. When the drain is closed, how long will it take to overflow when the

flow from the faucet is 0.50 liter/min? 5 lit * min/0.5 lit = 10 minutes. 10. A sink can hold 5 liters of water and the drain is closed. You turn on the faucet and then run to answer

the phone. The outflow from the faucet is now 0.50 lit/min. After the faucet runs for 8 minutes while you are on the phone, your roommate notices that the faucet is on. The faucet is turned off by your roommate, but due to a worn washer, the faucet drips at 0.0022 lit/min. You both forget about the sink and go to sleep for eight hours. Will there be water on the floor in the morning when you wake up?

Here, have to make a balance and keep track of total allowed and how much is present at a certain point in time, which is somewhat reminiscent of So = present + used. After eight minutes, 8 min * 0.50 lit/min = 4 lit. So we are 1 lit away from overflow. While dripping overnight, the volume added is 8*60*0.0022 = 1.056 lit. Exceed capacity by 0.056 lit so there will be a small volume of water on the floor. 11. A sink with the drain closed initially holds 5 liters of water (Vo = 5 lit). The drain is opened and water

begins to drain. The rate that water drains from the sink is proportional to the volume remaining in the sink; hence, dV(t)/dt = -αV(t). Derive an equation that shows how the volume remaining depends on the initial volume (Vo) and α.

dV/dt = -αV can be separated as dV/V = -αdt. Integrating both sides using the limits, t = 0, V = Vo and t, V yields V(t) =Vo exp(-αt). 12. A ten-fold cell proliferation in cell mass occurs. Assume that internal control of the kinetics occurs over

the majority of time with Xo = 0.1 g/l; µ = 2 h-1; and Y = 0.5 g cell/g glucose. Derive how S depends on t and then plot S(t) versus time (t) when So is 1.8 g/l and 5 g/l. Discuss how the internal control assumption can cause the long time behavior of S(t) on your plot to differ from reality. What would the long-time section of the curve look like in reality?

Mass Balance on reactant (e.g. glucose; aka “substrate”): So = S(t) + [X(t) – Xo] = initial present = that which remains + that which got used. Y Now for internal control, X(t) = Xo exp (µt). So combining the kinetics and mass balance yields S as a function of time, t. S(t) = So – [Xo exp (µt) – Xo] = So – Xo [exp (µt) – 1]. Y Y Does this answer make sense? If t = 0, S(t) should equal So because no time has been allowed for growth and nutrient consumption to occur.

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Lim t 0, S(t) = So – Xo [1 – 1] = So; makes sense, therefore probably did math OK. Y The general shape of the curves will look like. S(t) 5.0 1.8 time Note: As S becomes small, internal control will likely not be the case. The initial and medium time behavior may be described OK, but long time behavior will be predicted to drop off faster than may actually happen. If we had a kinetic model that described how cell growth rate depends on both high and low concentration, rather than limiting extremes, we could be more accurate at long time. 13. Often a nutrient or energy source is not in excess and the concentration limits the cellular growth rate.

For example, some leukemia cells require the amino acid asparagine, and this amino acid is not prevalent in the blood stream. Consequently, one therapy for leukemia is to inject a patient with the enzyme asparaginase, which catalyzes the removal of the amino group, thereby lowering the availability of asparagine to leukemia cells even further. When external control exists on growth kinetics, estimate by what factor the time will increase for a one thousand-fold expansion of cells when the asparagine concentration in the blood stream is reduced from 10-7 mol/lit to 10-9 mol/lit. Assume that k = 105 h-1 mol–1 lit.

When external control is the case, the concentration of some nutrient will influence the rate. A low value will provide a low rate and increasing concentration will increase rate. So we need to compare how much time is required for the cells to expand in number when (1) no drug is used (S1 = 10-7 mol/lit) to (2) when the drug is used (S2 = 10-9 mol/lit) to gain a sense of the effect. First find out how much time is required for the general case, then use the numbers. Often a general analysis will boil down to something simple, so we do not have to use the numbers and solve the problem for each set of numbers. dN/dt = kSN; for a given case, k and S are constant so let k’ = kS. dN/dt = k’N; this looks like something we have dealt with before: dy/dx = ay. Separating and rearranging into things we have been taught how to integrate ∫dN/N = k’∫dt where the lower limits are t = 0, N = No and upper limits are t and N. ln (N/No) = k’t or t = 1/k ln (N/No) ln(N/No) is the same for both the w/o drug and with drug cases; it equals ln (1000). A fixed amount of expansion is the basis for which the different times needed are being compared. So the factor time increases for a 1000-fold expansion for case 1 vs. case 2 is t2/t1 = [1/k’2]/[1/k’1] = k’1/k’2 = kS1/kS2 = S1/S2 = 10-7 /10-9 = 100-fold more time is required. Answer makes sense. Less amino acid concentration, the slower the bad cells will increase in number and progress the disease. 14. When at rest, a typical person’s ventilation rate is 6000 ml/min (in Standard Temperature and Pressure

conditions or STP). When one exhales, carbon dioxide is emitted at a rate of 200 ml/min (STP). In a 20 m * 15 m * 8 m lecture hall, there are 50 people.

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(a) Estimate the carbon dioxide concentration (mol/lit) in the room after 50 minutes, assuming the concentration was zero at the start of class. Assume that the room is roughly at STP conditions and no one passes out.

Total CO2 output = 200 ml/min-person * 50 people * 50 minutes = 500 liters. CO2 conc = 500 liters * mol/22.4 lit * 1/(20*15*8 m3) * 10-3 m3/lit = 9.3 10-6 mol/lit.

(b) Re-estimate the concentration at the end of class, assuming this time that each dimension of the

room is cut in half. The volume is reduced by a factor of eight so the concentration increases by eight to 7.4 10-5 mol/lit.

(c) Why does the pressure not rise due to all that CO2 release? Because for every mole of CO2 produced, a mole of oxygen was removed by respiration, thereby not adding or subtracting net moles. The water can condense on the cooler windows. It is a reaction in a closed system with no net change in the number of gas molecules if the water condenses and fogs the windows. There is probably a lot of “fog” in the room already.

(d) There are places in Africa where lakes emit CO2. The gas disperses over the surrounding land.

Because CO2 is dense, it tends to sink into the depressions in the surrounding landscape. Small animals fall into these depressions and die from asphyxiation. Larger animals attempt to feed on the smaller animals and they, in turn, perish in the toxic environment. Only animals with suitable physiology and stature can survive the high CO2 concentration that can be established close to the ground, and they prosper from these asphyxiation events by foraging on the victims of asphyxiation. Lakes Nyos and Monoun are particularly infamous for their CO2 releases, which can also kill humans in surrounding communities. For humans, a CO2 concentration that exceeds 4.46 10-3 mol/lit (10 percent) can be lethal. In view of these facts, what do you conclude about the need for ventilation in the lecture hall?

If the air in the room never turns over, eventually there could be a problem. Leaky windows and opening the door between classes will probably help freshen up the air. So the workload in the class will probably kill me before the CO2 in the classroom does. 15. A patient has impaired iron elimination. All iron ingested is absorbed (i.e. Me1 = 0) so feedback

control with respect to absorption in the digestive track is defective due to disease. (a) Show that at steady state, Mi = Md (1 – ε).

Mi = Me2 from overall balance. Md = MR + Me2 = εMR + Me2 from balance at recycle boundary; hence, Me2 = Md(1 – ε). Combining the above two equations yields the result.

(b) If Md still equals 21.5 mg iron/day and the recycle efficiency (ε) is 90 percent, would you advise the patient to continue to ingest 8 mg iron/day?

No. Mi should be 21.5 (0.1) = 2.15 mg/d. 16. Some solution techniques were reviewed/introduced and will be used again. What is the solution to

dy/dt = -ky2 for y = 1 at t = 0. “Solution” means how does y depends on time, t. y = (1 + kt)-1 using separation of variables. 17. A patient has impaired iron absorption. However, 90 percent of the iron in red blood cells is still

recycled and 21.5 mg of iron is available for recycling per day. Generate an operating diagram that depicts how the efficiency of uptake of ingested iron (εU; εU = 1 means all ingested iron is absorbed

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rather than eliminated) depends on the daily mass ingested (Mi). Such a diagram could provide disease management guidance.

Analysis of a mass flow network and looking at the iron problem again. Also a picture (i.e., operating diagram) is often more informative than a pile of numbers. With one glance, the whole situation can be grasped. Me1

εU Mi

Md

Me1 + Me2

MR

Me2

Mi

Red blood cell inventory

εU 1.0

εU = (Mi – Me1)/Mi εU Mi + MR = Md εU Mi + 0.9 (21.5) = 21.5 εU Mi = 0.1(21.5) = 2.15 εU = 2.15/Mi

2.15 Mi

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Chapter 3 1. You are hiking in the mountains. You estimate that you will need at least 6 liters (6 kg) of water a day

to sustain yourself. You have two choices: (1) carry a 1 liter bottle (1 kg) and fill up periodically when water is nearby or (2) carry 6 liters (6 kg). What are the tradeoffs associated with these different solutions? Note that a typical backpacker carries about 18 kg of food, shelter, etc. so the water carried adds to this weight.

(1) is better than (2) because less weight is carried. Weight is important because in case (2) the water carried represents a significant fraction of the total weight carried. Indeed, the trekker may find that carrying the additional weight may result in more than 6 lit/day being required, because the level of exertion increases. (1) is not better than (2) from the standpoint that if water becomes scarce, a problem with supplying the daily need could arise—could have a bad dry spell and become dehydrated and exhausted. 2. You are contemplating the replacement of your current wood stove. It has cracks that admit a lot of air.

Consequently, airflow is high. The high airflow ignites wood quickly and burns it “fast and furious.” The new wood stove has lower airflow. What are the tradeoffs associated with replacing the old wood stove?

The current stove heats up the room quickly, but consumes a lot of wood. The new stove will take longer to heat up the room, so although you will suffer for a while, it will not eat up the woodpile that you risked your life building using a chain saw with a dull chain. (The new stove will also produce more creosote and raise the risk for a flue fire and the burning down of your residence.) 3. A net decrease in energy is required for a process to be spontaneous. If a biological reaction has an

energy change of 10 kcal/mol, how much ATP has to be hydrolyzed to make the reaction spontaneous when coupled to ATP hydrolysis?

One mole of ATP will make the net change= 10 – 7 = 3 kcal/mol, which still is not negative. Two moles will make the net change = 10 – 14 = -4 kcal/mol, which goes “downhill.” 4. A one kg mass falls one meter. When it falls, it compresses a spring. After the spring compresses, it

extends and pushes the mass back up in the air. During compression, 10 percent of the original energy is lost. How much supplemental energy is required to push the mass back up to the original height of one meter?

Potential energy = mgh. Ten percent is lost during compression. So energy available upon release is 0.90 mgh. Thus, need 0.10 mgh extra to add into the mix. 5. A one cubic centimeter volume of cells can produce 0.03 moles of ATP from an energy source such as

glucose. This ATP is used to drive biosynthesis. The concentration of ATP in a cell is also typically 10-3 mol/liter.

(a) If all the ATP produced was not used to drive biosynthesis and other processes, and instead rapidly

broke down to ADP, what would be the temperature change of the cells? 0.03 mol * 7000 cal/mol = 210 cal would be released. This energy would raise the temperature by 210 degrees Centigrade! (The value of 0.03 moles is estimated from 10 g cell dry wt/mol ATP and 1 cc of cells containing 0.30 g cell dwt).

(b) How much energy is being transacted compared to that stored as free ATP? A rough ratio is easiest way to answer. In/out cash flow = 0.03 moles ATP. Balance = 10-6 mol ATP (= 10-

3 mol lit-1* 1 lit/1000 cc). So ratio = 3 10-2/10-6 = 3 104. Quite a lot of funds flowing relative to balance.

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6. A metabolic reaction occurs at a rate of 10-2 mol/h g cell. This means that every hour, within one gram of cells, 10-2 moles of some metabolite is enzymatically transformed to another molecule. The energy released per mole of reacted metabolite is 5 kcal.

(a) How much ATP can be generated from this reaction per gram of cell in one hour, as opposed to the

released energy appearing as heat? 1 g * 10-2 mol g-1 h-1 *5 kcal/mol * 1h = 5 10-2 kcal; so could produce 5 10-2 /7.3 = 6.8 10-3 mole ATP.

(b) If the energy was not captured and stored in ATP, how much would one gram of cells heat up 100

cubic centimeters (100 g) of water in fifty minutes? In two minutes would produce 10-2 mol g-1 h-1 *5 kcal/mol * 1 g * (50/60) h = 4.17 10-2 kcal = 41.7 cal; hence, would heat up 100 g of water by 41.7/100 = 0.417 degrees C. 7. You are contemplating increasing the airflow through your wood stove by installing different exhaust

piping. Currently, you believe that the airflow is too low because the logs burn slowly and the heat released to the room is on the low side of tolerable. In fact, you now put a fair number of logs in the stove in order to achieve a barely tolerable level of heating.

(a) If you increase airflow and load the same amount of wood, what could be an adverse consequence?

The room will overheat and become uncomfortable. Opening the window will cool down the room, but energy will be wasted as will the effort you put into building the wood pile.

(b) What are two control strategies you could use to make the stove with new exhaust piping provide

the desired level of room heating? (1) Manually control how much wood you put in. Add log by log until have right level. Kind of a pain because you will have to get out of your chair a lot and open the stove often. That will make the room smoky and distract you from the card game. (2) Change the exhaust system, but also add a new air control valve (damper) so for a given load, you can control airflow. You may still have to get out of your chair, but at least you are not opening the stove door and dealing with smoke and sparks.

8. Classify the following as continuous or discrete-variable control systems and explain your answer.

(a) Room air conditioner with a constant speed fan. (b) Shivering that becomes more intense as the core body temperature drops. (c) The thermostat in a car’s cooling system that admits water to the radiator when the water

temperature exceeds 220 degrees Fahrenheit.

(a), (b), and (c) are discrete, continuous, and fairly discrete or mixed, respectively. In (a), once the air conditioner is ON, the fan speed and compressor circulation are constant and uninformed about the deviation from the set point. Thus, the response does not increase in proportion to the deviation. The only way to increase the response is for someone to manually alter the fan setting from, for example, low to high. Thus, additional control logic (i.e. a person-based sensor and actuator) is required. In (b), the colder you are (i.e., further the body temperature drops), the more you shiver. Concerning an automobile, here is a deeper explanation and depending on how much you know about cars, the more you can say. The way the problem is worded, the thermostat is closed when T < 220 and open when T > 220. So based on the given info, the only conclusion is discrete. In reality, a thermostat opens at around 180 degree F. The mechanism is a wax melts and its density increases. The increased volume of the wax pushes up a control rod, which opens a path for water to be pumped from the engine ports to the radiator. A thermostat opens continuously over a 10-15 degree F span. Thus, to be exact one can say for T < 180, the radiator cooling system is OFF. Over around 180 –200, the system is variably ON and possibly emulating continuous control. Emulation is the caveat because the partial opening may reflect the wax melting dynamics and a distributed melting

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temperature if the wax is a blend of different materials. Thereafter, for T > ca 200 degrees F, the system is ON. For more information on car parts, consult http://auto.howstuffworks.com/cooling-system8.htm.

9. Figure 3.4 shows, in part, how glucose is degraded, and the reactions involved where ATP formation

can store the energy released by oxidative reactions. The early steps basically entail converting glucose to another sugar, fructose. Glucose and fructose also acquire phosphate groups (glucose-6-phosphate, G6P; fructose-6-phosphate; F6P), which endow them with a negative charge that prevents them from leaking out of the cell. The rest of the steps break the six carbon sugars down to three carbon compounds and then oxidize the glucose-derived fragments.

(a) Will a cell with high biosynthetic demand for ATP and low ATP production rate have a high or

low value of ATP relative to ADP? ATP hydrolysis to ADP will exceed rate of ADP ATP, so ADP/ATP will be higher than when the rates are balanced.

(b) How will a high value of ADP concentration relative to ATP concentration influence the rate of

glucose use? Answer this question in terms of (i) which reaction rates are affected and (ii) what the net affect is on the rate of glucose metabolism.

High ADP will stimulate rate 3 via positive feedback. Because ATP is lower, there will be lessened negative feedback on rates 3 and 10. The stimulatory and lessened restricting effects will combine to increase the rate of glucose use until ATP concentration increases relative to ADP concentration.

(c) How will a high value of ATP concentration relative to ADP concentration affect the rate of

glucose use? Lower ADP will stimulate rate less 3 via positive feedback. Because ATP is higher, there will be increased negative feedback on rates 3 and 10. The lessened stimulatory and increased restricting effects will combine to decrease the rate of glucose use until ATP production decreases and balances use more

10. The autopilot on a boat works as follows. A desired course heading is inputted (e.g., due East). The

global positioning system continually monitors the boat’s position. Over some time span, the boat’s actual trajectory is computed (e.g., NE). If necessary, the boat’s steering is adjusted to reacquire the desired course.

(a) Would a discrete or continuous control system be used?

Continuous would work OK. Depending how far off course the boat is, the more the rudder should be used to correct the trajectory.

(b) What is the signal the controller uses if the desired direction of travel is Eastward?

Error = Actual course – East.

(c) A boat moving at 10 knots is exposed to a 1 knot cross current. The positioning system is accurate to 10 meters. Based on this phenomenon, what are the tradeoffs associated with computing the error signal every 10 seconds or 1 hour? What seems to be a more reasonable time scale based on the tradeoffs?

If exercise control every 10 seconds, the boat does not have long to accumulate off-course error. However, you may be wasting computing power because the rudder corrections will generally be small or zero depending on the accuracy of the GPS. Waiting 1 hour to compute an error and then correct will result in considerable accumulation of off-course deviation. The longer time, however, requires less rapid computing power. A minutes time scale probably is the best compromise.

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(d) For every degree the boat is off course, the rudder can be set to rotate 5 or 20 degrees. Which setting corresponds to a high gain control system and what are some potential disadvantages of using high-gain control for boat steering?

The 20-degree response is high gain. If the response is too great, the boat will over steer in that the corrections could exceed the initial error. The boat will move with a more zigzag course and consume more fuel.

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Chapter 4 1. A swimming pool contains 20 m * 10 m * 2 m of water. Estimate how many people have to stay in the

pool for one hour to raise the water temperature by 1 degree Centigrade if they are not exercising. If an average person has a volume of 0.72 m3, is the answer feasible?

Volume of water = 20*10*2 = 400 m3. Mass of water = 400 m3 * 106 cm3/m3 * 1 g/cm3 = 4 108 g. Heat needed = 4 108 cal = 4 105 kcal. Humans needed: 4 105 kcal = x * 72 kcal h-1 human-1. X = 5556 humans! People volume = 5556 * 0.73 m3 = 4000 m3. No way can squeeze all those people in the pool unless it is a hot tub in California. This result also illustrates the high heat capacity of water. Despite a fair amount of heat released from a person, the effect on the water temperature is nil. Also this explains why despite all the tourists from Quebec in the water at Old Orchard Beach, Maine, each summer, the water never seems to warm up! 2. How much water has to be evaporated per hour to dissipate the basal metabolic heat release rate of 72

kcal/h? 72,000cal/555.56 cal g-1 = 129.6 g. About 1/10 liter would have to be sweated and evaporated to dissipate the heat if other mechanisms did not help out. 3. A 70 kg jogger starts from rest and in 3 seconds a pace of 5 km/h is set. The drag force exerted by the

air, which a runner must over come, is given by Drag Force =FD = CD A ½ρv2

where A, CD, ρ, and v are the runner’s frontal area, drag coefficient, fluid (air) density (29 g/22.4 lit), and speed, respectively. Assume that a runner’s area is in the ballpark of 0.9 m2 and in air at the running speed, CD ~ 1.0. Also recall 1 Newton = 1 kg m s-2; 1 Joule = 1 kg m2 s-2; 1 cal = 4.2 Joules. (a) If the runner is able to convert 50% of available metabolic energy to work, estimate the kcal/h of

metabolic energy used to just overcome air resistance. Need units of kg, m and s. 5 km/h = 5000 m/3600 s = 1.39 m/s. FD = 1.0 * 0.9 m2 *1/2 * 0.029 kg/22.4 lit * 1 lit/103 cm3 *106 cm3/m3 * 1.392 m2/s2 = 1.13 Newtons. Power = FD v = 1.13 * 1.39 = 1.56 J/s. The heat equiv is 1.56 J/s * 3600 s/h * cal/4.2 J * kcal/1000 cal = 1.34 kcal/h. So the runner must burn 2.68 kcal/h. This is pretty low compared to the basal rate. Depends on velocity2 so the value would be higher if the pace were quicker. More energy is used to lift the runner’s mass up and down against gravity due to the gait and limb movement.

(b) Assuming the same efficiency, estimate the kcal/h of metabolic energy needed to get up to the running speed of 5 km/h?

∆Kinetic Energy/∆t = Work/∆t. ∆Kinetic Energy = ½ 70 kg * 1.392 m2/s2 = 67.6 kg m2/s2. ∆Kinetic Energy/∆t = 67.6/3 (J/s) *3600 s/h * cal/4.2 J * kcal/1000 cal = 19.31 kcal/h. Adjusting for efficiency, 38.6 kcal/h of metabolic energy is used. Note there will also be drag force and thus extra work performed during the acceleration. However, the maximum value of the drag contribution is 2.68 kcal/h. Because during acceleration the average velocity will be less than 5 km/h, the integral value of the power consumed to fight drag during acceleration will actually be less that 2.68 kcal/h. So a good estimate is 38.6 + ½ (0 + 2.68) = 40 kcal/h.

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(c) Over a time of one hour, how much extra heat has to be removed from the runner’s body based on the above estimates and assuming a basal rate of 72 kcal/h. What percentage of the basal rate is released?

Normally over one hour, the body releases 72 kcal. The extra metabolic energy involved with the aspects of the run considered is roughly 40 * 3/3600 + 2.68 (1) = 2.7 kcal. ½ of this will be released as heat, 1.35 kcal and the fraction is 1.35/72 * 100 = 1.9 ~ 2%.

(d) Based on your experience with running or observing runners, does this answer make sense? If

not, what may be missing in the energy accounting? Suspect that more heat is really released. When running or watching someone run, the body also moves up and down. This motion represents work against gravity. An example will be provided in Chapter 10 on torso “bobbing” during walking. Other work can be found.

(e) The drag coefficient depends, in part, on speed and the density of air. Assume that at typical running speeds, the drag coefficient curve is flat with respect to speed; hence, CD is constant. So what advice would you give the runner on how to lower their metabolic drain over the run when the running speed is much higher than 5 km/h?

Lower projected area by lowering head and leaning forward somewhat. Some track coaches also advise runners to “ramp-up” when coming off a starting block. A runner’s projected area can also be lowered by wearing tight-fitting clothes. 4. When one mole of ATP is hydrolyzed to ADP, enough energy is liberated to drive the synthesis of 30 g

of cellular material, which includes protein, DNA, etc. ATP, in turn, is obtained from the metabolism of molecules such as glucose. One mole of glucose (MW =180) can yield about 36 mol ATP through substrate- and oxidative-level processes. A can of Coke contains 39 g of sugar. Assuming the sugar in Coke is glucose, how much energy in terms of ATP is in a can of Coke? What total mass (i.e., including water) of cells could be produced based on the energy content of a can of Coke? (Note: Assume the sugar in Coke provides only energy; the carbon to build the mass comes from somewhere else. So consider a can of Coke in terms of energy equivalents).

39 g glucose/can * mol glucose/180 g * 36 mol ATP/mole glucose = 7.8 mol ATP/can. 7.8 mol ATP/can * 30 g cell mass dw/mol ATP * g total/0.3 g dw = 780 g cells/can ~ 0.8 kg tissue/can. 5. In a mechanical system, the potential energy available to do work is ∆U = mg∆h, where ∆h is the

distance a mass is raised above a zero potential energy datum. In a chemical system that has two different solute concentrations, the (Gibbs free) energy that is available to do (chemical) work is

∆G = RT ln [C1/C2],

where R and T are the gas constant (2 cal/mol K) and temperature (Kelvin). C1 and C2 refer to the concentrations (e.g. molarities, M) of a solute on different sides of a membrane.

(a) For a one unit difference in pH across a membrane and at body temperature (37 oC), what is

the energy (in kcal/mol) that is available to do (chemical) work? ∆G = 2 * (273 + 37) ln [10] = 1,427 cal = 1.43 kcal.

(b) This gradient is to be used to drive the reaction ADP + P = ATP. A concentration gradient of

any solute has potential energy. When the solute is charged, a voltage is also established across the membrane, which also adds to the total potential energy. What fraction of the energy needed to drive the reaction is provided by the voltage across the membrane?

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The reaction requires 7.3 kcal to push it. A proton “falling down” a 10-fold concentration gradient can supply 1.43 kcal. The remainder has to come from the electrical work performed: 7.3 – 1.43 = 5.87, which amounts to 100 * 5.87/7.3 = 80%. The contribution required decreases if ADP ATP is coupled to more than one H+ entering the cell via ATPase. If coupled to two H+, then the electrical work drops to 61%. This solution is not totally rigorous, but it points out some of the basic ideas of coupling and membrane processes. 6. Some drugs can effectively produce “holes” in microbial membranes that permit H+ to leak in and out.

Why are such drugs potent antibiotics in that their administration slows microbial growth and thus the course of infection?

The “holes” cause the H+ gradient to collapse. Thus, the cell does not have a reservoir for energy storage and use in coupling. Put another way, oxidative energy is used to drive H+ out. If H+ simply leaks back in, then the energy is wasted. This is akin to having a dam with a leak. Water will never buildup behind it. 7. An interesting discovery was made by the University of Massachusetts researchers, Daniel R. Bond,

Dawn E. Holmes, Leonard M. Tender, and Derek R. Lovley, which was reported in the journal, Science (VOL 295 18 JANUARY 2002). They placed a graphite electrode in the oxygen-deficient sediment of Boston Harbor. A wire connected the deepwater electrode to another electrode near the surface in oxygen-rich water. They observed that a current passed through the wire with enough magnitude to power a calculator or similar electronic device. Notably, current flow through the wire occurred when microbes were present that could metabolize the organic matter in the sediment. Natural resource and bioengineers are quite excited about this finding because the prospect now exists for building biological batteries that can power undersea observatories and other applications. Use a diagram and words to explain how this battery apparently works.

Device

Oxygen-rich water Top electrode Low oxygen water Bottom electrode

e + oxygen = reduced product current flow Organic + microbe = Oxidized + NADH (or other carrier) NADH + electrode = NAD+ + e

Microbes oxidize the organic material and electrons are put on some carrier. There is no oxygen to transfer the electrons to, and somehow a cell-mediated transfer to the bottom electrode occurs. An electron acceptor exists near the surface, oxygen. So current flows through the wire. In a sense, the natural cytochrome chain (Figure 4.3) has been replaced by a man-made surrogate. (Interestingly, polluted harbors may hold the most promise for biological power generation?!) 8. A spring stores potential energy (U) equal to U = ½ kx2, where k is the spring constant and x refers to

the displacement from equilibrium. For a fixed displacement, how does doubling k alter the force available to do work?

A force is provided by the gradient of a potential. This exercise is a reminder from Physics class and an illustration of a nonlinear potential unlike mgh. F = d/dx [½ kx2] = kx. We recover Hooke’s Law. If double k, then double the force available. 9. Which system is capable of exerting a greater force: (a) 10 volts established across a 1 mm air gap or

(b) 1,000 volts established across a 1 m air gap, and why?

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Another simple gradient exercise. Even though 1,000 volts sounds more impressive, (a) wins when gradients are compared. Illustrates that when scale down system size, voltages and other potentials may become smaller, but they can do considerable things because length scales also become smaller.

(a) 10 volt/0.1 cm = 100 volts/cm. (b) 1,000 volts/100 cm = 10 volts/cm.

10. A muscle cell is oxidizing a compound for energy. The oxygen in the tissue is depleted. Briefly state what will happen to the NADH concentration within the cells once the oxygen concentration drops to zero and describe why.

The NADH concentration will rise because a cycle is broken. The NADH will have to be reoxidized and reused as a shuttle via another process (e.g. lactate formation).

S + NAD Sox + NADH H20 02

11. You just ate a 250 Calorie Snickers bar. If it takes about 3 moles of ATP to produce 100 g of protein,

roughly how much protein could be theoretically synthesized using the energy content of the candy bar to drive nonspontaneous biosynthesis?

(a) 1 g (b) 1 kg (c) 3 g (d) 3 kg (e) 5 g

An essential feature of biosystems is that energy is stored and transacted, in part, with ATP. Also energy is required to push reactions that do not want to happen due to entropy and other issues. 250 kcal * (mol ATP/7.3 kcal) * (100 g prot/3 mol ATP) ~ 1 kg. A lot of energy in there.

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Chapter 5 1. The binding of two different ligands, L1 and L2, to a protein is measured. The protein is found to bind

L1 more strongly than L2. In a solution containing protein, L1, and L2, answer the following.

(a) Which will be more prevalent, P-L1 or P-L2, when the total L1 and L2 in the solution are equal?

P-L1 because the protein binds it preferentially over L2.

(b) Will a given protein molecule never bind L2 after a L1 has bound to it?

No, binding is dynamic. So a given protein molecule will bind different L1’s and L2’s.

(c) Will adding more L2 increase the amount of P-L1 present?

No, this is akin to competitive inhibition. L2 still binds, so adding more with distract the protein from binding L1. Not as easy as the first two—so a good check on reading comprehension.

2. A metabolic network is shown below. A mutation results in the first enzyme’s binding site attaining a

shape that fosters increased binding of S4. Discuss what the effect on the regulation of the network will be.

S1 S2 S3 S4

(-)

The sensitivity of how the rate of the first reaction is restricted by S4 level will increase. (More expert types: There could be less throughput if S1 cannot back up and push the reaction. Pushing will require that enzyme-1 is at a low saturation level). 3. In Figure 3.4, ADP is shown to increase the rates of some enzyme-catalyzed reactions while ATP

binding can decrease the rates of others.

(a) For reaction 10 in Figure 3.4, draw diagrams akin to those in Figure 5.3 that show how the enzyme-catalyzed rate is regulated through binding.

PEP ATP

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(b) For reaction 3 in Figure 3.4, assume that the enzyme has two binding sites. One recognizes the substrate, fructose-6-phosphate (F6P). The other can bind either ATP or ADP, but the effect of binding ATP or ADP can differ. With diagrams akin to those in Figure 5.3, show how ATP or ADP binding can alter the rate at which F6P is transformed.

F6P

ATP

F6P

ADP

4. A mutation has occurred in a cell. One of the repressor protein’s binding sites no longer has the right shape to bind an inducer. Discuss what the result will be.

The gene will never be expressed because there is no way to turn it on by removing the repressor. 5. A mutation has occurred in a cell. One of the repressor protein’s binding sites has the right shape to

bind to the repressor region regardless of whether an inducer is present or absent. Discuss what the result will be.

The gene will never be expressed because there is no way to turn it on by removing the repressor. 6. A mutation has occurred in a cell. One of the repressor protein’s binding sites has changed such that the

shape no longer fosters binding to the repressor region, regardless of whether an inducer is present or absent. Discuss what the result will be.

The gene will always be expressed because there is no it can bind to the repressor region. (Yes, a “constitutive mutant”).

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Chapter 6 1. For the reaction, L + P PL, which of the equilibrium constants below will maximize the

concentration of PL when total ligand and protein concentrations are fixed? (a) 10-3 M (b) 10-4 M (c) 103 M-1 (d) 104 M-1 (e) 102 M-1 (f) 10-2 M

The association constants are (c), (d), and (e). The greater the association constant, the more PL tends to be formed. Therefore, the best association constant is (d).

The dissociation constants are (a), (b), and (f). The smaller the value, the lower the tendency for PL to fall apart. The smallest is (b).

(b) and (d) and they are equivalent equilibrium constants. One is just the inverse of the other and thus reflects the direction the reaction is written.

2. The total protein and ligand concentrations are 10-7 M and 10-4 M, respectively. The dissociation

constant is 10-4 M. Which one of the values below equals the fraction of protein that has ligand bound? (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 (e) 0.5 (f) 0.6 (g) 0.7

Because LT >> PT, we can use f = 10-4/(10-4 + 10-4) = 0.5. If instead the total protein and ligand concentrations are 10-7 M and 10-7 M, respectively, without resorting to calculations decide if the fraction of protein that has ligand bound will be smaller, the same, or greater than above, and explain your reasoning. The reaction is P + L = PL. The ligand concentration was decreased 1000-fold. Expect that the reaction will go less to the right and thus f should be less than 0.5. This should be the outcome of the more detailed calculation that will require the solution of a quadratic equation. 3. The total ligand and protein concentrations are 10-4 and 10-6 M, respectively. What is the free protein

concentration when the association constant is 104 M-1? K = 104 = (PL)/(P)(L) LT >> PT thus L ~ LT and f = KLT/(1 + KLT) = (104 10-4)/(1 + 104 10-4) = ½ =0.5; so free = 5 10-7 M. 4. The total ligand and protein concentrations are 10-6 and 10-6 M, respectively. What is the free protein

concentration when the association constant is 106 M-1? K = 106 = (PL)/(P)(L). LT is not >> PT thus L is not ~ LT. Thus K = (f)/[(1 – f)L].

A balance on L yields LT = L + PL = L + fPT so L = LT – fPT. After substituting for L, obtain f2PT – f[PT + LT + K’] + LT = 0. Note K’ is the dissociation constant (10-6 M). Putting in values and canceling out some things yields f2 – 3f + 1 = 0.

The roots are given by the quadratic formula [-(-3) +/- (-32 – 4*1*1)1/2]/(2*1).

Only the f = 0.38 root makes physical sense because 0 < f < 1. Therefore, free protein concentration = (1 – f) PT = 0.62 10-6 M.

5. The association constant for a protein-ligand pair is 10-4 M-1.

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(a) When the total ligand and protein concentrations are 10-3 M and 10-5 M, respectively, what fraction of the protein has ligand bound?

f = L/(K + L) = L/(104 + L). Because (LT) >> (PT), (L) ~ (LT). f = 10-3 /(104 + 10-3) ~ 0; Makes sense—the dissociation constant is huge so PL readily falls apart to P and L. f ~ 0.

(b) When the total ligand and protein concentrations are 10-5 M and 10-5 M, respectively, what fraction of the protein has ligand bound?

Here (LT) = (PT) so (L) need not roughly equal (LT) because total ligand concentration does not numerically overwhelm the total protein concentration. Now the dissociation constant is still huge and LT is even less so expect f ~ 0. Combine mass balance and equilibrium-- f = [LT – f PT]/[104 + LT – f PT). LT = PT = 10-5 M; f = 0.

(c) Redo parts (a) and (b) and let the association constant increase to 104 M-1

The total ligand and protein concentrations are 10-3 M and 10-5 M: f = 10-3 /(10-4 + 10-3) = 1/1.1 = 0.91. The total ligand and protein concentrations are 10-5 M and 10-5 M: f = [10-5 – f 10-5]/[10-4 + 10-5 – f 10-5). f2 –12f + 1 = 0, f = [12 + (144 –4)1/2]/2 = 6 + 5.92 0 < f < 1 so choose root where f = 0.08. 6. In a restaurant, 10 waitstaff people deal with 200 customer persons in a ten hour shift. What is the

turnover frequency for the “waitstaff catalysts”? Turnover numbers are frequencies with the dimension of time-1. The waitstaff are a processing agent and the customers get converted from seated to out the door with full bellies and lighter wallets. 200 people “reacted”/(10 people * 10 h) = 2 h-1. 7. An enzyme’s turnover number and concentration are 5,000 s-1 and 10-3 M, respectively. If the

association constant is 105 M-1, what concentration of substrate will yield a rate equal to 2.5 mol lit-1 s-1. VM = 5000 *10-3 = 5 mol lit-1 s-1. Need to use the dissociation constant instead of association in Michaelis-Menton Equation, 10-5 M. 2.5 = 5.0 S/(S + 10-5). When S = value of dissociation constant, the S/(S + K) term = 0.5 so S = 10-5 M. 8. A drug molecule can bind to the active site on an enzyme. However, unlike the normal substrate, the

enzyme cannot alter the drug molecule. Thus, the drug essentially ties up the enzyme so that less enzyme is available to catalyze the transformation of the usual substrate. The mechanism is

E + S ES; K = [(E)(S)]/(ES) E + D ED; Ki = [(E)(D)]/(ED) ES E + P

(a) Show that when so-called competitive inhibition occurs, where S and D refer to the substrate and drug concentrations, respectively, the rate is

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r = VMS S + K (1 + D/Ki)

Rate, r, will still equal k(ES), but (ES) will be lower because the drug ties up some enzyme and keeps it out of business. Starting with a total balance on enzyme,

(ET) = (E) + (ES) + (ED). Now try to make ET (known) a function of what is sought (ES).

(ET) = K [(ES)/(S)] + (ES) + [(D)/Ki]*(E). (ET) = K [(ES)/(S)] + (ES) + [(D)/Ki]*K [(ES)/(S)]. (ET) {K/(S) + 1 + [K/Ki]* [(D)/(S)]}-1 = (ES).

r = k(ES) = VM (S) (S) + K{1 + [(D)/Ki]}

(b) How would you design experiments and use the information provided by a plot of 1/r vs 1/S to determine the values of VM, K, and Ki?

Set (D) to a fixed level and vary (S). Determine how r depends on the varied (S). For two r-(S) data points or more, can find VM and apparent value of K, which equals K[1 + (D)/Ki]. Either find two parameters numerically or by a double reciprocal plot (1/r vs. 1/S plot); problem concerning double reciprocal plotting given below.

Then set (D) to a different level, repeat r-(S) measurements, and again find VM and the apparent value of K. Now have two values of apparent values of K corresponding to two different drug concentrations:

Value 1 = K[1 + (D1)/Ki] Value 2 = K[1 + (D2)/Ki] So now have two equations for two unknowns, K and Ki.

9. Describe the effect of the following mutations or drug additions.

S1 S2 S3 S4 S5 S6

S7 (-)(-)

(a) The enzyme that catalyzes the transformation of S1 has a mutated regulatory site such that no

binding of S3 occurs. Rates of S3 and S7 formation increase.

(b) The enzyme that catalyzes the transformation of S4 has a mutated regulatory site such that no binding of S6 occurs.

Rate of S6 formation increases and rate of S7 formation decreases.

(c) A drug that binds to the enzyme that catalyzes the transformation of S2 is added.

If VM of enzyme 2 is large compared to others, nothing will happen. If the second enzyme is close to saturation, the throughput might be lowered (overall rate S1 is used).

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10. When enzymes are purchased, a data sheet is usually provided. Practical information on storage and database references (e.g. Enzyme Commission numbering) is provided. Nowhere on the data sheet does the turnover number directly appear! Rather “rubber meets the road” numbers are often given instead. These numbers provide users a sense of how much enzyme mass to use in order to get a specific effect. One common number is how many “units” of activity there are per mass of product. The definition of a unit is also typically provided. Based on a molecular weight of 250,000 for catalase and the unit data provided below in the abstracted data sheet, show that the turnover number is on the order of 104 seconds-1.

Product Name: Catalase from bovine (cow) liver EC Number: 2325771 Storage Temp: below 0°C Unit Definition: One unit will decompose 1.0 µmole of H2O2 per min at pH 7.0 at 25°C based on the H2O2 concentration falling from 10.3 to 9.2 mM when measured by the rate of decrease of light absorption at 240 nm. Activity: 2000-5000 units/mg of protein

(3500 Units/0.001 g ENZ)* (250,000 g ENZ/mol ENZ) * (10-6 mol HOOH decomp/min-unit) * min/60 s

= 14,583 s-1 ~ 104 s-1.

11. If the dissociation constant for catalase-hydrogen peroxide is 10-3 M and 5 mg of enzyme is used per liter, answer the following when the turnover number is 104 s-1.

(a) What is the maximal velocity, VM?

VM = k (ET) = turnover number (or freq) * enzyme concentration.

10,000 mol HOOH eaten * 0.005 g * 1/liter = 2 (10-4) mol HOOH eaten.

mol Enz-second 250,000 g/mol Enz liter-second

(b) Show what the double reciprocal plot (1/r vs. 1/S) looks like when the substrate concentration (hydrogen peroxide) is varied from 10-3 to 10-1 M. What does the intercept and slope equal in terms of VM and K?

Intercept = 1/VM = 5,000. Slope = K/VM = 10-3/2 10-4 = 5. Straight line when plot 1/r versus 1/S with above slope and intercept. This line must pass through the following point. Let S = K = 10-3. Convenient choice we can that when S = K, r = ½ VM.

r = 2 (10-4) S = 2 (10-4) ½ = 10-4 S + K Thus, when 1/S = 103, 1/r = 104 While analysis via a double reciprocal plot could be presented in the text, “discovery” through problem solving might be a reasonable shot at “active learning.”

5 1 slope

1/r 10,000 5,000

10 103 1/S

(c) What appears to be the utility of such a strange plot when one instead plots 1/r versus 1/S data for a

set mass of newly discovered enzyme that has unknown values of k and K?

24

From the reciprocal of the intercept, which yields VM, and the moles of enzyme used in known assay volume (ET), one can determine k because VM = k*ET. From the slope (K/VM), one can then determine K because the value of VM has been found from the intercept value. 12. The enzyme catalase breaks down hydrogen peroxide (HOOH) to oxygen and water according to the

following reaction HOOH H2O + ½ O2.

For fun, a beaker filled with hydrogen peroxide solution and catalase can be used to inflate an object such as a rubber glove with the oxygen released from the reaction. Assume that the total volume (in liters) of the glove is measured every 30 seconds. When 4 mg of enzyme is used, explain how you would use the volume data and initial hydrogen peroxide concentration to determine the turnover number (seconds-1) and dissociation constant. Recall that the ideal gas law relates gas volume to moles of gas, PV = nRT, where P, V, n, R, and T are pressure, volume, moles, a constant, and temperature, respectively. At room temperature (293 K) and atmospheric pressure (1 atm), n (moles) = 4.13 (10-2) V (liters). Your answer should contain both qualitative description (strategy) and “how-to” quantitative analysis (how to process the data). Overall, need to establish how the rate (r) depends on substrate (S) concentration. If link r to S (over each time interval), can then find VM and K from the data. Need two data points (two values of r and S will yield the two unknowns, VM and K, because there will now be two equations and two unknowns), or plot a series of data on a 1/r vs 1/S plot as shown in some of the problems above. From the y intercept, the turnover number can be obtained from VM (obtained from the data) = k (sought) * moles enzyme/vol (known from mass added, beaker fluid volume, and enzyme molecular weight) The rate over each interval can be found as follows. The volume the glove changes, ∆V = V(t + ∆t) – V(t), every 30 seconds can be related to the moles of HOOH degraded over ∆t = 30 seconds. (note: each mole of HOOH degraded yields ½ mole oxygen, so this will have to be in the bookkeeping). Volume change can be related to mole change by the ideal gas law; in general, (∆moles HOOH) is proportional to ∆V. Thus, rate over each 30 second interval, r = ∆moles/t-soln vol ~ ∆V/∆t. The value of S(t) that corresponds to the rate during the ∆t interval can be found from a mass balance. S(t) = So – S reacted = So – α*Gas volume(t), where α is the conversion between volume and moles as provided by the ideal gas law and includes the ½ factor. So by using the ideal gas law and a mass (mole) balance, a table can be created as follows (hypothetical numbers are used to illustrate the data processing): Time (sec)

Total Gas Volume (lit)

Volume Change (lit)

Tot moles O2 made = 2x HOOH used

HOOH left S(t) (Moles/lit)

Rate = r(t) ∆[HOOH]/∆t (mol lit-1 s-1)

0 0 0 0.0550 0 30 0.03 0.03 1.2 10-3 0.0526 8 10-5 60 0.06 0.03 2.5 10-3 0.0500 8 10-5 90 0.08 0.02 3.3 10-3 0.0484 5 10-5 etc The last two columns are the relation between S and r. (One could average values over each ∆t interval to say what an average HOOH concentration and r were to overcome the fact that over wide measurement intervals, S and r can change.) About 20% of the students have been able to do this problem based on some reminders and hints from high school or first-year chemistry. It has also been used as an in-class demonstration where some of the basic calculations are done in advance, but in the backwards direction. Namely, given a turnover number and

25

initial HOOH concentration, estimate how much enzyme needs to be added to inflate the glove within 5 minutes. The minimum time assumes that the enzyme is saturated throughout. Enough HOOH has to be added to produce ½ to 1 liter of gas to inflate a typical lab glove. A safety factor then can be assumed to add a pinch more enzyme if saturation is not expected to occur throughout the inflation period. Thus, a design problem can be posed, solved, and cheaply demonstrated. 13. Below is a branched metabolic pathway that produces two biomolecules, S4 and S5. The disease,

Mellonitis, results from more S4 being produced than S5; hence, the patient has a S5 deficiency relative to S4. Mellonitis results from a mutation in the allosteric enzyme that converts S3 to S5. The mutation results in the binding site having a low association constant for E-S3 formation. Research also shows that the enzyme that converts S3 to S4 is allosteric; binding sites exist for S3 and S4.

S1 S2 S3

(a) How many enzymes make

Six (double-check that students knowthe effects can be “spongy”).

(b) How many regulated enzym

2 (Ditto above)

(c) Will a drug that is a molecuWhy?

Yes. The mimic will bind to the epresumably the drug. The drug binvia an allosteric site and cooperativenecessitate more of the input providincreases a lot due to increased feedproduction might not be that great. So in reality the drug could help anenzyme that uses S4. This would ca Overall, the possibility for more thmetabolic engineering. System mouncertainties.

14. The first enzyme in a pathway i

bind to the enzyme. When ADPcatalyzed rate depends on S conjustify your plot, or mark import

A basic concept is that reactions in with allosteric enzymes via an oftenand ADP binds to the other site. ADthe same. If K stays the same, S = Kto double the turnover number, then

(-)

S4 S5

(-)

up the pathway?

what such a diagram means; otherwise the rest will be tough because

es comprise the pathway.

lar mimic (i.e. analog) of S4 possibly reduce the Mellonitis symptoms?

nzyme that converts S3 to S4 and this enzyme is inhibited by S4 and ding will slow the enzyme down based on the feedback that can occur interactions in the enzyme molecule. Slowing down the enzyme may ed by the upstream enzymes to flow to S5 instead of S4. (note: if S3 back, the effect of increasing the rate of S5 production relative to S4

An elevated value of S3 could overcome increased feedback from S4. d likely not make things worse. Also the mimic could slow down the use S4 to increase and feedback).

an one answer and view illustrates the challenges of drug design and dels can help where inspection can lead to hard-to-see outcomes or

s allosteric and has two sites. Substrate (i.e., reactant S) and ADP can binds, the enzyme’s turnover number increases. Plot how the enzyme-centration when ADP is (i) absent and (ii) present. Add comments that ant features directly on the plot.

biosystems are catalyzed by enzymes and control of rates is often done used mechanism, binding. To apply in this case, S binds to one site P binding increases the turnover number, so only VM increases; K stays where rate = ½ VM is the same. For example, if enough ADP is added

26

W/ADP WO/ADP

rate

S 15. It is desired to bind a nonreactive fluorescent ligand to an enzyme. When the ligand binds to an

enzyme’s active site, the local environment enhances the fluorescence intensity (ligand is brighter). Thus, when the ligand enters the cell, if the enzyme is there and binding occurs, then a fluorescence microscope can be used to visualize cells that possess the enzyme of interest. The enzyme concentration is expected to be around 10-8 mol/lit and the association constant is 108 M-1. What should the total intracellular concentration of the nonreactive fluorescent ligand be in order for 80 percent of the enzyme molecules to be labeled (i.e., have L bound to them)? State your assumptions.

From Chapter 6, analysis depends on how LT compares to PT. Here L ~ K., so use f2PT – f[K + LT + PT] + LT = 0. (6-6) Another way to look at this, the above integrates mass balance and binding together so it is always correct. So if one is not sure, then use it. Also f is given, so it is easy to solve. 0.82 10-8 – 0.8[10-8 + LT + 10-8] + LT = 0; LT =4.8 10-8mol/lit Another way to do this problem is f = L/[K + L] = 0.8 = L/( 10-8 + L); L = 4 10-8. LT = L + PL = 4 10-8 + 0.8 10-8 = 4.8 10-8 mol/lit.

27

Chapter 7 1. A fluorescence-activated cell sorting (FACS) procedure fails to yield cells that bear a particular

receptor from a tissue sample. These cells are important to your tissue engineering work so you need to figure out what the problem is. You suspect that the cells are really there and the fluorescently-labeled probe chosen does not work as well as you expected. Time on a FACS machine is hard to arrange and can be expensive. What is an alternative route you can use to debug this problem and how would you proceed?

Use microscope imaging. Try the label at different concentrations to check if binding occurs and offers significant fluorescence. (May find that either no binding occurs or the binding is so weak that not enough label was used. With this information in-hand, you can either elevate the label concentration used or abandon it in favor of an alternative.) 2. The dissociation constant for a fluorescently-labeled probe is 10-6 M. The probe is to be used to detect

the presence of stem cells in umbilical cord blood and to isolate them by flow cytometry. The total number of cells in the 5 ml blood sample is 108 and it is estimated that 0.1 percent are stem cells. Each stem cell is estimated to possess 100 receptors to which the fluorescent probe will bind.

(a) How much total probe in mol/lit has to be added for fifty percent of the total receptors to have

fluorescent probe bound? Total protein concentration = 108 10-3 102/0.005 lit = 2 109 molecules/lit = 3.3 10-15 mol/lit. f = L/(K + L) = 0.5 = L/(10-6 + L) so L = 10-6 mol/lit. The free ligand concentration is >> the total protein concentration; hence, LT ~ L. LT = 10-6 M.

(b) Which is more likely, fifty percent of the stems cells will be fluorescent or fifty percent of the

receptors on a given stem cell will “glow”? Binding is reversible. Thus, if all the receptors have a similar value of K (i.e., receptors do not interact) it is impossible for one cell to be labeled substantially different from another cell. Rather, the probes will exchange and be uniformly distributed if you wait long enough. (In practice, unequal labeling may occur if the receptor number is small. That is, for a fixed average probability, the Mean Value Theorem suggests that variance will increase as sample size decreases by N½ ). 3. The properties of three different fluorescently-labeled ligands are given below.

(a) What are the tradeoffs (i.e., relative good and bad features of the different ligands)? A is the cheapest, but it has the greatest dissociation constant and the least brightness. Therefore while A is cheap, one may have to use a lot of A for the desired outcome. B is the brightest, so one may not need to use a lot, but B’s dissociation constant is in the middle as is its cost. C costs the most so one may initially dismiss it. However, C has decent brightness and the lowest dissociation constant. Thus, it’s a contest to see if C’s greater brightness means we can use less C than B, and possibly spend the least money.

(b) Which would be the best label to use for a fluorescence-activated cell sorting procedure? Assume that the total cell surface, target protein concentration is 10-14 mol/lit and the product of fraction labeled and fluorescent intensity must be greater than or equal to 1.5 in order to differentiate the desired cells from others.

LABEL DISSOCIATION CONSTANT (M)

FLUORESENCE INTENSITY (MAX = 10)

COST $US/MOL

A 10-6 3 109 B 10-7 9 1010 C 10-8 5 1011

28

This is an engineering decision analysis problem; variations may be good for a test problem. Forces the formation of an objective function akin to that put forth in text when discussing the use of an enzyme. Take a basis of one liter; any consistent basis can be used and would just scale the answer. When f = 0.5, all will provide barely enough or plenty signal. So compare them on the basis of f = 0.5. At 50% labeling, L = K. LABEL AMOUNT (M) FLUORESENCE

INTENSITY (MAX = 10) COST $US

A 10-6 1.5 103 B 10-7 4.5 103 C 10-8 2.5 103 They all cost the same for 50% labeling. Label A will, however, yield the lowest fluorescence. Label B for the same price as A and C will yield the greatest fluorescence, so might as well go with B because there is more bang for same buck. An alternate solution entails figuring out the exact amount of each label needed and the cost magnitude. f = L/(K + L). Find L, which is the minimum amount of label needed (LT > L). LABEL MINIMUM AMOUNT

(M) FLUORESENCE INTENSITY

COST $US

A f = 0.5 when 10-6 1.5 ~ 103 B f = 0.17 when 2 10-8 1.5 2 102 C f = 0.3 when 4 10-9 1.5 4 102 B is again better from the different perspective of same bang for less bucks.

(c) What would C have to cost for it to be as cost effective as B?

Cost cut would have to be ½ (0.5 1011 $/mol) to be as “useful” as B. A free sample of B or C would be even better. 4. Your car has run out of gas. Therefore, its speed declines with distance x as you coast down the

highway. Assume the velocity v (miles/h) depends on distance past the point where you ran out of gas according to

v(x) = 66 – 33 x.

If the gas station is 1.8 miles away, how many minutes will it take to get there? An application of the separation of variables technique. dx/dt = 66 – 33x. Separating variables yields dx/(66 – 33x) = dt. Integrating both sides yields (-1/33) ln [1 – 0.5x] = t; t = 4.2 minutes. 5. What does Equation 7-4b become if 99% conversion is sought? τ = 1/kET [K ln100 + 0.99So]. Quizzes on understanding of what variables mean. 6. If the same amount of enzyme is used, how much longer (e.g., twice as long) does it take for ninety

nine percent of a substrate to be reacted to product as opposed to ninety five percent for the cases below?

(a) The dissociation constant and initial substrate concentration are 10-5 M and 10-3 M, respectively.

τ0.99/τ0.95 = [K ln100 + 0.99So]/ [K ln 20 + 0.95So] ~ 0.99/0.95 = 1.04.

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(b) The dissociation constant and initial substrate concentration are 10-5 M and 10-7 M, respectively.

τ0.99/τ0.95 = [K ln 100 + 0.99So]/ [K ln 20 + 0.95So] ~ ln 100/ln 20 = 1.54.

(c) What is there such a difference in the time factor when only So and not the percent conversion was changed?

In the first case, the enzyme is saturated much of the time, so the concentration dependence on rate is low. The opposite is true in case (b).

7. An enzyme has the following properties: VM = 10-3 mol lit-1 s-1 and K = 10-5 M.

(a) Is the K value an association or dissociation constant and why?

Yes, have to remember stuff from prior chapters. It is a dissociation constant. Can tell by the units. When dissociate ES E + S

K = Prod/Reactants = (E)(S)/(ES)

K’s units are C2/C = concentration units.

(b) How long will it take for the substrate concentration (S) to decrease from 10-5 to 10-6 M?

Enzyme kinetics is the starting point. r = k(ES) = k(ET)(S) = VM(S) = - d(S)/dt. Employing separation of variables yields

(S) + K (S) + K - ∫dt = [K/kET][∫dS/S + ∫dS] . (7-3a)

Integration requites that limits are set; in this case, the limits are

at t = 0, S = So and (7-3b) at t = τ, S = S*. (7-3c)

The general result obtained is:

τ = [1/kET][K ln So/S* + (So – S*)] . (7-4a) Using the numbers provided yields the time required for the substrate concentration to decrease

τ = (1/10-3) [10-5 ln 10 + (10-5 - 10-6)] = 103 *10-5 [ln 10 + 1 – 0.1] = 3.2 10-2 s

(c) If the initial product concentration is 0 M, what is the product concentration after the time found in part (b) has elapsed?

S P; if S disappears, something else (P) appears because of conservation of mass. By mass balance, So = S(t) + P(t) – P(0) so product concentration = 10-5 - 10-6 = 0.9 10-5 M Thus, the same time calculation can be used to answer two questions: 1. How long for a given % of S to disappear? 2. How much time is required for a given amount of product to be made?

(d) By what factor (half, one hundred, etc.) does the time change when the enzyme concentration is doubled?

From general result, time decreases by a factor of two because τ ~ 1/ET.

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(e) By what factor (half, one hundred, etc.) does the turnover number have to change when the enzyme concentration is fixed, yet the reaction time is desired to be reduced by a factor of ten?

Because τ ~ (1/kET), k must increase 10-fold for the time to decrease by a factor of 10. Thus can control time needed in at least two ways:

1. Vary amount of enzyme used per volume of sample. 2. Choose different enzymes (same reaction catalyzed, but differ in turnover number).

8. You are considering the purchase of two different automobiles with a zero interest loan. For the

different scenarios below, pose an objective function and select which auto is better.

(a) Auto A costs $15,000 and the fuel costs are estimated to be $1,100 per year. Auto B costs $25,000 and the fuel costs are estimated to be $1,500 per year. Both autos and the loans will last ten years and have similar insurance and repair costs based on your excellent driving record and ability to fix things yourself.

This is pretty easy and a relatable objective function notion. Objective is MIN (total $/year). Auto A = 1,500 + 1100 = 2600. Auto B = 2500 + 1500 = 4000. Auto A is better. It is even better when you calculate the interest gained on 25,000 – 15,000.

(b) Auto A costs $15,000 and the fuel costs are estimated to be $1,100 per year. This Auto (and simple loan) will last five years and Yugos tend to have zero resale value, but you will buy another Yugo because you think that they are cheap and all that you need. Auto B costs $25,000 and the fuel costs are estimated to be $1,500 per year over the ten-year lifetime of the car and simple loan. Again, there will be similar insurance and repair costs.

Objective is MIN (total $/year). For a ten year horizon, two Auto A’s = 3000 + 1100 = 4100. Auto B = 2500 + 1500 = 4000. Auto B is better (unless possibly you account for interest). Overall, the Yugo has desirable and undesirable properties. Its half-life is short, which is undesirable, but it is also cheap, which is tempting. Thus, the Yugo is akin to a cheap, but unstable enzyme. 9. For an initial analyte concentration equal to 10-4 M, choose the best enzyme, based on the data below,

to use in a sensor that converts the analyte to a detectable product. Assume that converting ninety-five percent of the analyte is acceptable for signal generation. The shelf life of the enzyme-containing sensor component is one year.

Enzyme A: K = 10-5 M, turnover number = 104 s-1, and half-life = 6 months. Enzyme B: K = 10-6 M, turnover number = 105 s-1, and half-life = 3 months. There is good news and bad news with respect to B. B has lower K and higher k, but shorter half-life than A. Use Equation 7-8:

φ = [K ln 20 + 0.95So] / ko exp (-0.693 τS/τH). φΑ = 5 10−8. φΒ = 1.57 10−8.

Enzyme B is better even though its half-life is less. So is low; hence, the lower K and greater k helps in reducing the amount of total enzyme needed. 10. Consider enzyme-based sensor design when So is much less than K.

(a) Show that φ is given approximately by the simplified proportional relationship below for a shelf life of one year.

φ ∼ (Κ/ko) exp (1/τH), where τH has units of years. φ = [K ln 20 + 0.95So] / ko exp (-0.693 τS/τH) = K ln(So/S) ko

-1 exp(0.693/τH).

31

So/S will be high and comparable for two sensors built from different enzymes.

exp(0.693/τH) = exp(0.693) exp(1/τH) = constant exp(1/τH). Thus,

φ ∼ (Κ/ko) exp (1/τH), where half-life has units of years. Such a problem may allow the instructor to see if the students know what the symbols mean, or if plugging and chugging is the case.

(b) Using this relationship may simplify how to decide which enzyme alternative is best under some

circumstances. For example, based on the properties of the two enzymes below, pick the one best for a sensor application when the shelf life is one year.

Enzyme A: K = 10-5 M, turnover number = 104 s-1, and half-life = 6 months. Enzyme B: K = 10-6 M, turnover number = 104 s-1, and half-life = 3 months. Enzyme A will have a ten-fold greater value of (Κ/ko), which is not encouraging for A even through its half-life is pretty good. Thus, unless the exp value of A is ten-fold smaller, Enzyme B will be the better choice.

exp(2) = 7.4 while for B, exp (4) = 54.6. B wins because the value of the exp function does not exceed 74.

32

Chapter 8

1. The yield for growth on methanol is 0.4 g cell/g methanol. If one million pounds of methanol is used per year as a feed for a microbial process that produces protein for animal feed, how many extra pounds of protein can be produced if metabolic engineering increases the cell yield by five percent? What is the extra mass of protein in human equivalents, assuming an average human contains 12.6 kg of protein?

Improved yield = 1.05 * 0.40 = 0.42 g cell/g methanol. Assuming that protein is 60% of cellular dry weight (recall Chapter 1), the increased protein is

(0.42-0.40) g cell/g methanol * 0.60 g protein/g cell * 103 g methanol/2.2 lbs methanol * 106 lbs methanol * = 5.45 106 g protein = 5.45 103 kg protein.

The human equivalent is 5.45 103 kg /12.6 kg = 433 people, which is on the order of ten classes of students.

2. A typical cell yield for the growth of E. coli on glucose is 0.3 g cell dry wt/g glucose when some glucose is lost to carbon by-products such as acetic acid (CH3COOH; MW = 60).

(a) Assuming that the loss entails 40 percent of glucose carbon going to acetate, what fraction of glucose carbon goes to cell mass and carbon dioxide?

Basis: 100 g glucose; 40 g C. Acetate C = 0.4*40 = 16 leaving 40 – 16 = 24 g C for C in cell mass plus CO2. Based on cell yield, cell C = 0.3 (100) 0.5 = 15 g C or 15/40 *100 = 37.5%. Thus, CO2 = 100 – 40 – 37.5 = 22.5%.

(b) If acetate production is stopped by metabolic engineering, and CO2 production per gram of total glucose used remains fixed, what is the cell yield that can be attained and by what factor does it compare to the “wild-type” cell yield? Comment on energetic implications of this estimate.

Assuming that CO2 production stays the same and the acetate C is directed into cell mass will provide a maximal mass or C yield. Note this means that adequate ATP is produced with the same amount of glucose oxidized to CO2. Not likely, but an interesting upper bound calculation. Also assume that cells are 50% C on a dry weight basis.

Basis: 100 g glucose; 40 g C. C going into cell mass = 40 – 9 = 31 g C. Cell mass from 31 g C = 31/0.5 = 62 g cell dry wt. Cell yield MAX on glucose = 62/100 = 0.62 g cell dry wt/g glucose, which is double the wild-type value.

This calculation suggests that the actual yield of metabolically engineered cells may lie between 0.3 and 0.6 g cell/g glucose. What matters is the ATP generating mechanisms and dissipation processes.

3. What might happen to acetate production and cell yield on glucose if instead of deleting pyruvate kinase from a cell, the capacity of the TCA cycle was increased?

Might make less acetate, which is good. However, if excess C is sent to the TCA cycle (i.e., more than needed for glutamate and other syntheses), the C could just get oxidized to CO2 and/or the TCA metabolites could leak out of the cell as their concentration increases. Both effects would tend to not increase the cell yield or potentially offset the positive effect of dropping acetate production.

33

4. Why is developing a tissue engineering strategy for retina replacement potentially easier than that for engineering a replacement for a highly blood-infused organ such as a liver?

The tissue is smaller and has less three-dimensional spatial and cellular organization. Blood vessel generation and connections to the circulatory system are considerably less extensive. 5. When transfected cells from a donor are implanted into a patient, note three hurdles that you anticipate

must be surmounted for the technology to work.

(i) The gene product must be produced at the right level. High enough to have a therapeutic effect, but not too high where side effects can occur.

(ii) If the cells are from another source, immune system rejection must be suppressed. (iii) After the disease is treated (e.g., tumor has regressed), the gene product might have to be

turned off, or at least the level of production altered.

6. The alcohol propanol (CH3-CH2-CH2OH) and the carboxylic acid, acetic acid (CH3-COOH) combine during an esterification reaction. What is the chemical structure of the reaction product?

CH3-(CO)O-CH2-CH2-CH3

7. To follow-up Figure 8.4, lactide and glycolide can be chemically reacted to form polylactide and PGA, respectively. Lactide and Glycolide can also be reacted together to form a polymer that possesses remnants of both monomers joined by an ester linkage. Such a polymer is called a copolymer. Can you think of reasons why copolymers are synthesized for biomaterial and tissue engineering applications?

The copolymer may have stability properties midway between polylactide and PGA. Thus, one can “tune” the absorption/dissolution rate. (Of course copolymerization can be tricky and could lead to amorphous products which may/may not have compromised mechanical properties). 8. A restriction enzyme cuts after the sequence, TTAA. Show how restriction enzyme treatment will alter

the DNA below, which will be done for cloning DNA in metabolic engineering.

GCATACCGTTAAGCGCTAAT GCATACCGTTAA GCGCTAAT CGTATGGC AATTCGCGATTA

Yields two fragments with sticky ends, which can be grafted to other DNA pieces (sealed by ligation) and inserted into a plasmid. 9. The yield of a microbe is increased from 0.3 to 0.6 g cell dry wt/g glucose, because energetic

inefficiencies have been reduced. The metabolically engineered cell has the same kinetics (exponential/internal control rate and external control dependence on glucose) as the unengineered cell. Draw curves below that correspond to the (1) unengineered and (2) engineered cells when growth occurs on glucose in a flask. Assume the initial concentrations of (1) and (2) are equal and each has been provided with the same initial amount (grams/liter) of glucose.

Time

2: Engineered attain twice the final cell concentration because yield is doubled

1: Unengineered

Xo > 0

X = Xoexp(µmt) the same initially

Cell Conc. gm dry wt/lit

34

10. What are two methods used to introduce foreign DNA into bacterial cells? Piggy back on plasmids. An engineered plasmid can be inserted by electroporation. Use competetant cells for taking up and integrating raw DNA.

11. What is the transpose of the matrix below?

1 2 1 transpose is 1 2 3 2 2 2 2 2 3 3 3 3 1 2 3 12. For the matrix A given below, answer the following.

1 1 0 1 1 0

(a) What is AAT? Point out the diagonal and explain why it is symmetric.

1 1 1 0 1 2 1 0 1 1 1 0 = 1 2 diagonal is 2,2. The elements 1 and 1 are equal. 1 0

(b) What are the eigen values, and do they sum to 1?

No, because λ1= 3, λ2 = 1, where the eigen values are found from the solution to: (2 – λ)2 – 1 = 0.

(c) Each element in AAT is reduced by the same factor to rescale the data in the matrix. The factor makes the elements on the diagonal equal to one. What are the eigen values and do they sum to the dimension of AAT?

Yes, because the solution to (1 – λ)2 – 0.25 = 0 yields λ1= 1.5, λ2 = 0.5 and 1.5 + 0.5 = 2 and the matrix is 2 * 2.

13. The correlation matrix A below was generated.

1 1 1 1

(a) What are the eigen values of A? (1 – λ)2 – 1 = 0. λ(λ – 2) = 0; hence λ1 = 2, λ2 = 0.

(b) What is the interpretation of the relative values of the eigen values?

The data say x and y are totally correlated with each other; hence, there is only really one “thing” going on, so one dimension can be used to describe trends.

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14. The correlation matrix below was obtained. Without doing computations, state how many eigen values there are and their values.

1 1 1 1 1 1

1 1 1 There are three eigen values because it is a 3 x 3 square matrix. The data are perfectly correlated so the eigen values are 3, 0, and 0.

15. The correlation matrix below was obtained. The eigen values are 2.34, 0.73, and 0.04. Explain the significance of the relative values of the eigen values.

1 0.9 0.3 M = 0.9 1 0.6 0.3 0.6 1

xx xy xz 1 0.9 0.3 yx yy yz; M = 0.9 1 0.6 zx zy zz 0.3 0.6 1

The x-y correlation is strong and contributes to making one axis in a three dimensional ellipse longer than the other two axes.

36

Chapter 9 1. Air containing a toxic substance is suddenly inhaled and results in the concentration in the lungs

equaling 10 percent by volume. How many breaths are required to reduce the toxic substance concentration to less than one percent by volume and what is the maximum time needed?

If perfect mixing occurs, each breath can reduce the concentration to 100(6000 – 350)/6000 = 94% of its initial value. Thus, the number of breaths, n, is given by 0.1 = 0.94n n = log (0.1)/ln(0.94) = 37. The maximum time is 37/12 = 3.1 minutes. Faster ventilation and a little panic will lower the time. 2. Water, fat, and protein are dominant molecular constituents in the human body. If fat and protein are

about 50 percent by weight carbon, estimate the fraction of carbon in the water-free mass of the human body and compare this value to a typical cell.

Basis: 70 kg human. ~ 12.6 * 2 = 25.2 kg water-free material. 25.2 * 0.5 = 12.6 so % carbon = 100 (12.6/25.2) = 0.5. Of course by inspection, can do this in your head. Protein mass = fat mass and both are 50% by mass carbon; hence, % carbon = 0.5. 3. One type of an artificial heart and battery power supply weighs 4 lbs (1.8 kg). Assume that the mass of

a real heart is roughly proportional to body weight. How does the total mass of the artificial system compare to mass of a real heart in a 70 kg adult and a 20 kg child? What are the quality of life implications for the adult and child after implantation?

Mass real heart in adult = 0.005 * 70 = 0.35 kg. Mass real heart in child = 0.005 * 20 = 0.10 kg. The replacement is more massive than the natural organ (5-fold and 18-fold for adult and child, respectively). Note that the heart mass-to-body mass ratio is actually higher in young children than adults. However, heart chamber volume does increase with age. In children, a higher pulse rate compensates for lower stroke volume, thereby providing adequate oxygen to growing tissues. Thus, the assumption made is conservative. The point is some replacements still strive to achieve what less massive natural organs accomplish. Despite the conservative assumption, there may be challenges with using some replacement technologies in children. The child may be more physically restricted than an adult. For more on organ replacement in children, see the Government Accounting Office Report (Sept. 2001), “Organ Transplants: Allocation Policies Include Special Protections for Children.” http://www.gao.gov./new.items/d01498.pdf . 4. The iron in food is typically broken down into two categories: (i) heme and (ii) nonheme. The heme

form is akin to the form in which iron is present in your blood, and it is the most absorbable form. 100 g of tofu contains 3 mg of nonheme iron and 100 g of beef liver contains 18 mg of iron (British Columbia Ministry of Health File 68d, http://www.healthplanning.gov.bc.ca/hlthfile/hfile68d.html).

(a) Approximately how much tofu or liver would you have to eat daily to meet the FDA recommended

value of iron intake (8 mg/day)? Tofu > 100 * 8/3 > 267 g; greater because tofu has the nonheme form of iron. Liver = 100 * 3/18 ~ 16.7 g.

(b) What possible problems could arise with a diet based exclusively on tofu or liver? All tofu, could be iron deficient if did not consume ~ 300 g per day. Need more bulk intake. All liver, could overload on iron and tend towards hemochromatosis, especially if genetically disposed and you actually liked liver and onions.

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5. Blood flow to the kidneys is typically 1,200 ml/min. Urine flow and glomerular filtration rates are 1.5 ml/min and 125 ml/min, respectively.

(a) What percentage of the blood entering the kidneys is filtered every minute?

Using problem information, the percentage = 125/1200 (100) = 10.4%.

(b) What fraction of the total blood flow is filtered every minute? Using problem information and that from Table 9.1, the percentage = 125/5000 (100) = 2.5% or fraction = 0.025. 6. About how long does it take to filter all the blood in the body if disease causes the glomerular filtration

rate to decrease by a factor of two? What would be an adverse consequence of the lengthened filtration time?

Half the normal rate means that the pace of glomerular filtration is 62.5 ml min-1. If one uses the total blood volume, then 5000 ml/(62.5 ml min-1) = 80 min is the time required, which exceed the healthy value (40 min) by a factor of 2. Kidneys process cell-free fluid, so it is more realistic to use the cell-free volume of blood, which equals 3000 ml. In this case, the answer is 0.6(80) = 48 min versus the healthy value (24 min). Toxic products could accumulate and/or the ability to maintain blood composition could be compromised 7. A drug can bind to blood proteins. The fraction bound is designated by f.

(a) How will the renal clearance of the drug compare to the inulin clearance? The drug’s clearance will be less than 125 ml/min because some of the drug will stick to blood proteins and thus not be present in the glomerular filtrate.

(b) If the glomerular filtration rate is denoted by GFR, how will the renal clearance of the drug (CD) depend on GFR if the drug is not reabsorbed or subject to tubular secretion?

The bound drug will not be filtered. The free drug will pass in the glomerular filtrate. If it is not reabsorbed or secreted, then the clearance will be CD = GFR (1 – f)

8. The renal clearances of four drugs were investigated in clinical trials. Some binding studies have also

been performed to determine if drugs bind to plasma proteins. Some or none of these drugs may be subject to reabsorption or tubular secretion; hence, it was hoped that the clinical studies would shed more light on the drug elimination mechanisms. Data to date indicate that all of the drugs are not metabolized in the body. Based on the data in the table below, indicate what mechanisms may operate in eliminating the drugs from the body by the renal system.

Drug Bind to Plasma Proteins Clearance A NO 125 ml/min B NO 150 ml/min C YES 80 ml/min D YES 125 ml/min

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Drug A does not bind to plasma proteins and its clearance equals the glomerular filtration rate (GFR). Thus, reabsorption and secretion are unlikely unless they occur at the same rate. Drug B does not bind plasma protein, yet its clearance is greater than the GFR. Thus, net tubular secretion is a good bet. Drug C binds to plasma protein and its clearance is less than the GFR. Thus, protein binding perhaps in conjunction with tubular reabsorption occurs. Drug D binds protein yet has a clearance equal to the GFR. One would expect that with protein binding, the clearance should be less than the GFR; hence, active tubular secretion likely occurs. 9. What is the key difference between how protein/peptide and steroidal hormones are produced and

presented to the body? Proteins/peptides are made in advance and packaged in vesicles; they are poised for release. Steroids are not packaged in vesicles and they can bind to blood proteins and travel around the body. 10. In one scenario, some class members vote on when a term paper is due. Based on the due date, other

class members suggest how many pages it should comprise. A second scenario entails the course instructor quelling the discussion and herself setting the date and expected word count. Which scenarios are analogous to antagonistic and permissive control systems?

The class votes on one issue and others add their input leading to a combined decision—permissive. The other is antagonistic (although “persuasiveness” may be a more benign description).

39

Chapter 10 1. The function f(x) = x + 2x2 + x3 describes a velocity profile in a biofluid mechanics problem. It is

interest to simply describe what happens for 1 < x < 3. Show that the first-order Taylor Series approximation (i.e., f(x) ~ f(xo) + f’(xo)[x – xo]) to use in this case is f(x) ~ 21x – 24.

f(x) ~ f(xo) + f’(xo) [xo – x]; when expand around xo = 1, the result is as follows. f(xo) = 1 + 2 + 1 = 4. f’(x) = 1 + 4x + 3x2 so f’(xo) = 1 + 4 + 3 = 8. f(x) ~ 4 + 8(x – 1) = 8x – 4. Some may expand at midpoint, x =2, which actually makes more sense, because this is the midpoint of the interval that one wishes to cover with the approximation. f(xo = 2) = 2 + 8 + 8 = 18. f’(x) = 1 + 4x + 3x2 so f’(2) = 1 + 8 + 12 = 21. f(x) ~ 18 + 21(x – 2) = 21x – 24.

2. The function f(x) = x2 is to be approximated at x = 3. What is the error between the actual value and approximate value when a first-order approximation is used and x = 3.3?

f(xo = 3) = 32 = 9. f’(x) = 2x. f’(xo = 3) = 2*3 = 6. f(x) ~ 9 + 6(x – 3). f(3.3) = 10.9. f(3) ~ 9 + 6(0.3) = 10.8; not too bad of an error. The error is 0.1/10.9 *100 = 0.9%.

3. The relationship between biomechanical power consumption for some task and variables x and y is

P = 10 x2y + 5 y/x.

Show that the optimal performance is given by P* = 12 y.

From inspection, P = y[10 x2 + 5/x]. Thus, P is large as x grows OR as x becomes near zero. The variable y only scales the value of P once x is set. dP/dx = 0 = 20x – 5/x2. x = (5/20) 1/3 = 0.63. Using this value of x in the equation for P yields P = 12y.

4. An adult and a child walk a mile in the same amount of time. L1 (adult) = 1 meter and L2 (child) = 0.5

meter. They optimize stride lengths to minimize how tired they get. By what factor will the child have to take more steps than the adult to complete the mile?

a* ~ L1/2. a2*/a1* = (0.5/1) 1/2 = 0.707. So the child will take 1/0.707 = 1.4 more steps with a shorter stride to cover the same distance. 5. A 185 lb person whose leg length is 1 m walks at a rate of 3 km/h. Assuming that α = 0.1 and β =

0.05, answer the following (a) What is the power consumption in Watts when the stride length is 0.1 m? (b) What is the power consumption in Watts when the stride length is 1 m?

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(c) What is the best stride length and minimum power consumption? PT = αMv3/a + βMgva/L. (10-5) a* = (α/β)1/2 (L/g)1/2 v. (10-6b) PT* = 2(αβ)1/2 (g/L)1/2 Mv2 . (10-7)

(a) M = 185 * 1/ 2.2 = 84 kg and v = 3000 m/3600 s = 0.83 m/s. Power = 0.1*84*0.833/0.1 + 0.05*84*9.8 * 0.83*0.1/1 = 48 + 3.43 = 51.4 watts.

(b) Power = 48/10 + 3.43*10 = 39.1 watts.

(c) Optimal Stride = a* = (0.1/0.05)1/2 (1/9.8)1/2 0.83= 0.375 m. Minimum Power = PT* = 2(0.1∗0.05)1/2 (9.8/1)1/2 84* 0.832 = 25.6 watts; optimal consumes about ½ as much power!! 6. Two hikers are in the midst of the 100 Mile Wilderness section of the Appalachian Trail in Maine.

They got separated when one hiker made a side trip to a water source and got lost (i.e., an adapted scene from A Walk in the Woods). Realizing they are separated, they now seek to hike out and rejoin at an agreed upon emergency rendezvous point. Each hiker has the same amount of food. Lost hiker-1 is tall and lanky while hiker-2 has a more compact build, as delineated below

Hiker-1 M = 160 lb L = 41 in Hiker-2 M = 180 lb L = 36 in

(a) One scenario is that each hikes roughly at the same speed, but to save energy, each hiker

unconsciously optimizes their stride length. Based solely on the power consumption involved with walking to the rendezvous point, which hiker may run out of food first due to their different “biomechanical construction”?

PT* ~L-1/2 M so PT2*/ PT1* = (L1/L2)1/2 M2/M1= (41/36)1/2 180/160 = 1.2. Hiker-2 may run out of food first.

(b) By what factor should hiker-2 adjust his speed to make his power consumption equal to hiker-1?

PT1* = PT2* so this means (1/L2)1/2 M2v2

2 = (1/L1)1/2 M1v12.

(v2/v1)2 = (L2/L1)1/2 M1/M2 = (36/41)1/2 160/180 = 0.83.

So v2 = 0.91 * v1; make sense Hiker-2 needs to slow down by about 10% to conserve energy. Panic is not good when lost. 7. You are backpacking. Your pack weighs 20% of your weight. You desire that your power consumption

be the same as when you walk without a backpack. You also pace yourself so that you do not burn out over a long day. When carrying the pack, by what factor does your walking speed compare to the speed that you would walk without the pack?

If keep power consumption the same and are pacing optimally, PT* = 2(αβ)1/2 (g/L)1/2 Mv2. Let N and B subscripts denote not (N) and carrying (B) a backpack. Thus, MNvN

2 = MBvB2.

MB/MN = 1.2 = vN2/vB

2 So walk 1.2½ times slower = 1.09, which equates to roughly a 10 percent difference. Will be even slower if going uphill to reach a mountain top campsite. 8. Three connected subsystems each possess an efficiency of 85%.

(a) What is the overall efficiency of the entire system? 0.853 = 0.61.

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(b) How much will the overall efficiency increase if each subsystem efficiency is increased to 90%? 0.93 = 0.73 so increase is (0.73 – 0.61)100/0.61 = 19.7%. 9. A battery can provide enough power to a heart assist device to allow for ten hours of mobility. How

much more additional mobility (or safety factor) is obtained when the overall efficiency of converting electrical energy to blood pumping work increases from 90 to 95 percent?

95/90 *10 = 10.56 hours. Thus, the gain is 10.56 –10.0 = 0.56 h = 33 minutes

10. The effect of gravity on the Moon is one-sixth that on the Earth. Consider the prospect that people

actually live on the Moon, and while we are being speculative, assume that the Moon’s gravity is one-ninth that of the Earth’s. Also assume that Moonlings are twice as massive and four times taller than the average Earthling.

(a) For a given walking speed, will an Earthling astronaut or a Moonling do less, equal, or more work

when taking a stroll on the Moon, and by what factor? Be sure to state your assumption(s).

Assume an optimal stride length is used and the Moon people do the same. Therefore, a* = (α/β)1/2(L/g)1/2 v (10-6b) so a* ~ L1/2. Can go two ways. One criterion for equal work is Mr = ar = Lr

1/2. 2/1 = ? = (4/1)1/2 YES. Or can use scaling of PT* if not sure of the criterion. PT* = 2(αβ)1/2 (g/L)1/2 Mv2 and assume that α and β are about the same for Earthlings and Moonlings. Compare an Earthling and Moonling at the same walking speed. PTE*/PTM* = (LM /LE)1/2 (ME/MM) = 2 * ½ = 1 SAME

(b) An average Earthling has a mass of 70 kg. How much extra mass in kg could an Earthling carry on

the Moon and still have the same, minimum value of power consumption as on Earth when not carrying the extra mass (i.e., Earthling + Extra Mass on Moon = Earthling on Earth; Extra Mass = ?)? Be sure to state your assumption(s).

From PT* = 2(αβ)1/2 (g/L)1/2 Mv2, the scaling that results after canceling things that are the same on Earth and the Moon is PTM*/PTE* = (gM/gE)1/2 (MM/ME). Ratio = 1 = 1/3 (MM/ME) MM= ME + Extra Mass = 3ME. Thus, Extra Mass = 2ME = 2(70) = 140 kg. 11. The human heart provides about 1-10 watts of power to circulate blood through the body. A 9.6 volt,

2.6 amp-hour rechargeable battery is used for some cordless electric drills. For 100 percent efficient conversion of electrical energy into blood pumping work, estimate how long one of these rechargeable batteries would last when 5 watts of power is consumed. Recall that 1 J/sec = 1 W = 1 A * 1 V.

t * 5 watts = 12.6 volts * 2.6 amp hours; t = 5 hours.

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Chapter 11 1. What pressure is experienced by a diver 10 meters beneath the surface of a lake? Express your answer

in Pascal units (1 N/m2 = 1 Pa). Also express your answer in atmosphere units (1 atm = 105 Pa). How many meters of water are equivalent to the pressure that the atmosphere continually exerts on us?

P = ρgh(Water) + ATM = 103 kg/m3 * 9.8 m/sec2 * 10 m + 105 Pa = 1.98 105 Pa ~ 2 atm. This amounts to ca 2 atm; 10 meters of water is equivalent to one atm as your ear drums will confirm if you do not equalize. 2. For laminar flow in a tube, draw how velocity and shear stress vary with radius. Where are the shear

stress and time spent by a fluid element both the greatest?

r Max τ; greatest time & stress Zero τ

Max τ; greatest time & stress

3. The total flow rate through a 1 cm radius tube of length 1000 cm is 10 liters per minute. What is the

average velocity of the fluid? <v> = Q/Area = 104 cc min-1/(π 1 cm2) = 3.2 103 cm/min = 3.2 m/min = 192 m/h. 4. For laminar flow in a tube of radius R by what factor does the shear stress experienced by a cell

increase if it is located at r = 0.5 R versus r = 0.05R? By what factor does the exposure time to stress increase or decrease when situated closer to the wall versus in the center of the tube?

From Equation 11-5c, τ ~ r. So when located at r = 0.5 R versus r = 0.05, the shear stress is 10-fold higher near the wall. Also recall that v ~ [1 - (r/R)2]. The velocity near the center is 1.33 greater than the value closer to the wall. Thus, when r = 0.05R, the stress is 10-fold greater and time spent is 1.33 greater than when closer to the center (r = 0.5R).

5. Blood packing is a procedure some athletes use to prepare for highly aerobic sports such as marathon

running and cross-country skiing. The procedure entails drawing blood and separating the red cells from plasma months before the event. Just before the event, the red cells are infused back into the athlete in order to increase hematocrit. The theory is that the oxygen-carrying capacity of the athlete’s blood will be increased; hence, endurance and performance will be enhanced. Recently, drugs that increase hematocrit have been used instead of the red cell isolation, storage, and infusion method. Apart from the questionable ethics, from the biofluid mechanics viewpoint, are there potential disadvantages to this strategy?

Yes, the viscosity of blood increases as H increases. Pressure drop increases as viscosity increases. Thus, power consumption for pumping can possibly increase.

6. A two-liter saline bag is elevated 0.5 meter above a patient in bed. The viscosity and density of saline

are 1.2 cP and 1.05 g/cm3, respectively. Saline infusion helps maintain a patient’s fluid and electrolyte balance. The saline flows under the influence of gravity through a sterile, plastic tube of inner radius of 0.20 cm and length, 1 m. To infuse a patient, a 20-gage hypodermic needle can be used of length 2 cm. The needle is inserted into a vein, which allows the saline to drain into the patient.

(a) How long will it take to drain the saline bag when drainage occurs just through the plastic tube?

Assume that the net pressure driving the flow is due to the 0.5 m high column of saline and it remains constant.

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Assume that drainage does not pull a vacuum and the bag’s height is included in the 0.5 m, then the maximum flow is based on 0.5 m of head. The pressure will decrease towards the end so the following time estimate is a minimum value.

Q = ∆PπR4/8µL. ∆P = ρgh = 5.14 103 N/m2. Q = 5.38 10-5 m3/s or 53.8 cm3/s. Drainage time = 2000/53.8 = 37 s.

(b) What is the drain time for infusion when the needle is used to connect the patient to the saline

source, when the same assumption is made? R = 3.3*10-4 m . Q ~ R4. So Q = 53.8 (3.3*10-4 m/2*10-3m)4 = 3.99 10-2 cm3/s. Drain time = 2000/3.99 10-2 cm3/s = 5 104 s = 14 hours. (c) Why may the actual flow rate be less than that calculated in part (b)?

We essentially assumed that the patient’s internal pressure = 1 ATM and the liquid column has 1 ATM imposed upon it. The pressure in blood vessels may exert back pressure. As the bag drains, h, in ρgh, will also decrease. The drop in h will slow the drain rate.

7. A vessel of length L and radius R is replaced by two parallel vessels of equal length, 0.5 L, and equal

radii, 0.2 R. How does the pressure drop of the branched vessels compare to the single vessel alternative?

Although branching can lower pressure drop, too small a radius can offset the advantages as shown below. ∆P = kQL/R4 where k = constant. For the straight vessel, ∆P = kQL/R4. For each branch, ∆P = k*0.5Q*0.5L/(0.2R)4 = 1.56 102 kQL/R4 . The pressure drop across the branches is 1.56 102 −fold greater than the single vessel.

8. Show that if a vessel of length L is replaced by two parallel vessels of length ½L, then RB > 0.707 RU

must be the case in order for the resulting pressure drop of the branched circuit to not exceed that of the single vessel alternative.

∆PU/∆PB = (QU/QB)*(LU/LB)*(RB/RU)4. ∆PU/∆PB > 1 > 2 * 2 *x1/4. x = (0.25)1/4 = 0.707. Thus, as long as RB > 0.707 RU the branched alternative will provide the same pressure drop or less compared to the single vessel.

9. Convince yourself that the Casson equation actually captures the yield stress, variable apparent

viscosity, and then almost Newtonian behavior of blood as strain rate is increased.

(a) Plot τ versus γ for a typical blood sample where H = 0.45, aα = 1.84, and CF = 0.3 g fibrinogen/100 ml. On the same curve, show how a Newtonian fluid such as plasma would behave. (Hint: Expand the Casson equation so that τ appears, as opposed to τ1/2.)

Assuming plasma viscosity is 1.3 cP, τY

1/2 = 0.35 *0.8 = 0.28 dynes/cm2. s = [1.3/0.550.8] ½ = 1.45. τ ½ = 0.28 + 1.45γ½. τ = 0.08 + 0.81 γ½ + 2.1γ. If Newtonian with a viscosity of 1.3 cP, τ = 1.3 γ.

44

0

0.5

1

1.5

2

2.5

3

3.5

0 0.2 0.4 0.6 0.8 1 1.2

Shear Rate

Shea

r Str

ess

BloodPlasma

3

2

1

(b) Show on the curve three key aspects of blood behavior.

1—yield 2—varying viscosity 3—viscosity starting to become apparently constant. 10. Instead of branching into two vessels, assume a vessel of length LU and radius RU is replaced by three

vessels of length LB and radius RB. Derive an equation that predicts how the ratio of pressure drop developed in the three-branch network compares to the single-vessel scenario (∆PB/∆PU).

∆PB/∆PU = (QB/QU)*(LB/LU)*(RU/RB)4 = 0.33*(LB/LU)*(RU/RB)4. If the vessels have equal length, then RB > 0.76 RU.

11. Starting with the velocity distribution for laminar flow (Eq. 11-5b) and τ = -µ dv/dr, show that the

pressure drop depends directly on how large the shear stress is at the wall: ∆P = 2 τ(r = R)L/R. (Note: Here a negative sign links τ and dv/dr because as r increases, v decreases and τ increases.)

v = ∆PR2 [1 - (r/R)2]. 4µL dv/dr = ∆PR2 [-2r/R2].

4µL

When r = R, dv/dr = -∆PR/2µL. Thus, τw =∆PR/2L or ∆P = τw (2L/R). Physically, wall shear stress and pressure drop equate by τw(2πRL) = ∆P(πR2). 12. If someone’s hematocrit was increased from 0.45 to 0.60, estimate how much more power would be

expended by the heart in order to circulate 5 liters of blood every minute. Assume that aα = 1.84 and CF = 0.3 g fibrinogen/100 ml.

Power = Q * ∆P so need to see how ∆P might increase to sustain Q. ∆P increases as apparent viscosity increases. The Casson Equation gives some guidance. τY

½ increases from 0.28 (see above problem) to 0.40. s = [1.3/0.400.8] ½ = 1.64 So to compare situations:

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Η = 0.45: τ ½ = 0.28 + 1.45γ½, τ = 0.08 + 0.81 γ½ + 2.1γ. Η = 0.60: τ ½ = 0.40 + 1.64γ½, τ = 0.16 + 1.3 γ½ + 2.7γ. If γ is order 1, stress ratio is ca 4/3 = 1.3. If γ is order 10, stress ratio is ca 31/23 = 1.3. If γ is order 100, shear stress ratio is 27/21 = 1.3. Unless γ is low where ratio is more like 2, then pretty assumption-insensitive. ∆P is directly proportional to τ, and τ is proportional to viscosity or its apparent value, so estimate that power consumption will increase by 30 percent. 13. A heart-lung machine pumps blood from the patient and oxygenates it; thereafter the blood is returned

to the patient. The tubing that conveys venous blood to the machine needs to be replaced quickly. There are four sterile tubings available in the OR. Which tubing from the choices below would you use to avoid red cell damage and why? Assume the same volumetric flow rate (Q) is pumped regardless of the tubing used.

(a) (b) (c) (d)

Two big ideas in bioflu(how we are designed, nas low as possible. τ ~ ∆PR/L. ∆P ~ L/R4; hence, τ ~ 1/Therefore, the largest ra Now if one wants to gL/average velocity = L/< τ*duration ~ (1/R3)(L/R The largest-radius tube Is average velocity the speaking Stress * duration = τ(r)* Thus, to minimize the pFor the choices, the L/reasonable choices. 14. Which of the choic

(a) Two parallel v(b) Two in-series v(c) Three parallel (d) Three parallel

Two big ideas in bioflu(how we are designed, n∆P ~ QL/R4.

R = 0.5 cm, L = 1 m R = 1 cm, L = 1 m R = 1 cm, L = 1.2 m

R = 1.5 cm, L = 1.5 m

id mechanics are to manage (1) shear stresses (minimize damage) and (2) pressures ot tax organs, clogged arteries, etc). Minimizing damage means keep shear stress

R3. dius (d) is the best choice.

et fancy, damage depends both on stress and duration. Duration will scale to v>. Average velocity = <v> = Q/πR2 ~ 1/R2. The product of stress and duration is

2) = L/R5.

looks even better now.

best metric? Not really. To look at the whole picture more accurately, strictly

[L/v(r)] = [2µL/R] [r*/(1 – r*2)] where r* = r/R.

roduct of shear stress and duration at fixed flow rate (Q), L/R must be minimized. R ratios are: (a) 200 (b) 100 (c) 120 (d) 100. On this basis (b) and (d) are

es below will provide the lowest pressure drop needed to drive flow?

essels: R = 0.2 cm, L = 10 cm. essels: R = 0.2 cm, L = 10 cm.

vessels: R = 0.2 cm, L = 10 cm. vessels: R = 0.15 cm, L = 6 cm

id mechanics are to manage (1) shear stresses (minimize damage) and (2) pressures ot tax organs, clogged arteries, etc). This is an example and application of idea (2).

46

By inspection (b) is worse than (a) because Q(b) = 2 Q(b) and L(b) is 2 L(a). Double whammy on (b). (c) is better than (b) because Q is less in (c). Thus, the only contest is between (c) and (d) ∆P(c)/∆P(d) = (10/6)* (0.15/0.20)4 = 0.55; hence, (c) is best. Here, even though (d) is shorter than (a), its radius is decreased to much which more than offsets the benefit of decreasing L. 15. A Newtonian fluid is flowing down a sloped, flat surface. Imagine rain draining off a sloped roof. The

flow system and spatial dependence of velocity are shown below. v(x) = A [1 – (1 – x )2]

v = velocity A = constant specific to slope angle, etc µ = viscosity x = distance from sloped surface T = fluid layer thickness

T µ

Top of fluid surface--air

x

Fluid layer thickness, T Inclined surface

(a) Based on analogies to other cases, will the shear stress be the greatest when x = 0, 0.25T, 0.5T,

0.75T, or T? Write a sentence that explains your selection. The shear stress (τ) will be greatest when x = 0. Comment: By analogy to flow in a tube/vessel, τ is maximal at stationary boundaries (walls); i.e., at r = R. In this geometry, x = 0 corresponds to a stationary boundary. The velocity will also be zero at x = 0.

(b) Using the given velocity dependence on x, prove mathematically that the value of x you chose in part a (above) corresponds to where the maximal shear stress occurs.

Based on the above analogy, v should be 0 at x = 0. When x/T = 0, v(x) = A/µ[1 – 1] = 0.

As a check, and actually to be more rigorous, τ = µγ = µ dv/dx and τ should be maximal at x = 0.

dv/dx = 0 – 2Aµ-1 [1 – xT-1][-T-1] = 2Aµ-1T-2[1 – xT-1]. x = 0, dv/dx = 2Aµ-1T-2. x = T, dv/dx = 0. Yes, τ is maximal at the wall. 16. Using sketches of how τ depends on γ, contrast two fluids: (1) shear thickening with no yield stress and

(2) shear thinning with a finite yield stress. 17. An artery with a diseased section is shown below; plaque and other deposits (dark blobs) line the inner

wall of the vessel.

Strain rate, γ

2: finite yield and slope of τ vs γ decreases with inc γ. 1: no yield and slope of τ vs γ increases with inc γ.

Shear stress, τ

Blood exit

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An arterial bypass operation is proposed where vessels are surgically removed from another location in the body and then grafted as shown below.

(a) Explain why a patient with clogged arteries exhibits high blood pressure, assuming that blood behaves

like a Newtonian fluid. Cite relevant physical/mathematical relationships when explaining.

Graft Radius

Graft Length

Grafted vessel

Grafted vessel

∆P ~ 1/R4 and effective R decreases. Another interesting answer I got from a student is vessel compliance decreases. So it will stretch (∆R > 0) less after each heartbeat (pressure pulse). Again because R + ∆R is less in a diseased vessel, ∆P increases a lot. (b) If the grafts available to the surgeon vary in their radii and lengths, which type (length-radius

combination) is best to use for the bypass surgery? Explain your reasoning. If they are short (i.e., long enough to span diseased area) and have large radii, they will provide a low resistance shunt around the diseased zone. This argument follows from ∆P ~ L/R4.

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Chapter 12 1. A rough surface has steps that are 20 µm high and long; the steps repeat every 40 µm. The pattern is

shown below. What is the ratio of surface area of the rough-to-perfectly flat surface? Assume the material has a width (W; the dimension into the page) that exceeds the thickness.

20 µm

40 µm

20 µm The surface repeats, so we can take a unit cell approach where a unit is 40 µm in length. For a surface of width, W (into the page), the total surface area in a cell is 40W. For the stepped surface, the surface is 4*20W. The ratio is thus 80/40 = 2. If do not use a cell and work with the whole sketch, find area of smooth surface is 5*40*W. Neglecting the area of the ends, the stepped surface has an area equal to 19*20*W. The ratio is 1.9. An easier way is just to count edge segments since all lateral and vertical step dimensions are equal. For the flat surface there are 10 horizontal runs. For the stepped surface, there are 10 horizontal runs plus 9 vertical rises. So neglecting ends, the ratio is 19/10 = 1.9 ~2. 2. Height profiles measured from four different surfaces are tabulated below. For all the surfaces, the

heights at similar distances along a path of fixed length were measured. The heights are in arbitrary units. How do the values of σs compare? Do any of these profiles illustrate the limitation of the σs measurement, and why other parameters have been defined to quantify surface roughness.

Distance Height Material 1

Height Material 2

Height Material 3

Height Material 4

0 1 0 2 0 1 0 1 0 1 2 1 0 2 1 3 0 1 0 1 4 1 0 2 3 5 0 1 0 0

Material 1: av s = 0.5 and rms roughness = 0.5. Material 2: av s = 0.5 and rms roughness = 0.5. Material 3: av s = 1.0 and rms roughness = 1. Material 4: av s = 1.0 and rms roughness = 1. Materials 1 and 2 are equivalent and could just be phasing in the track measurement. Material 3 has the same pattern as Material 1, but different values of mean height and roughness. Material 4 has the same average height and roughness as Material 3, but shows the presence of a spike and wider range in height. Such a spike represents the possibility of significant asperities. 3. What is the difference between homeostasis and hemostasis? Hemostasis specifically refers to the regulation of blood volume and cellular composition. Homeostasis includes all general regulatory outcomes such as the maintenance of constant body temperature. 4. What does an activated platelet look like and what is one property it acquires after activation? It obtains extensions that foster adhesion. The adhesion can help seal small leaks. Chemical signals are also sent by activated platelets that initiate other sealing steps. 5. Hemophilia is a disease that results when at the genetic level, a mutation has occurred that leads to

factor Xlll being nonfunctional. What is the consequence to blood clotting when factor XIII is absent?

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The fibrin-based physical network cannot be formed so blood loss continues. 6. How can blood coagulation reactions lead to ischaemia? The coagulated blood could shed particles when exposed to shear. These particles could occlude smaller vessels such as those in the brain, thereby starving tissue of oxygen and nutrients. 7. Why could administering too much heparin to a patient recovering from arterial graft surgery prove to

be lethal. The clotting may be suppressed, but bleeding could be excessive from surgical or other wounds. Thus, can compensate for bad biomaterial design only so much with drug intervention. 8. For one cubic centimeter total volume of water, compare the total surface areas when the water

assumes a cubic versus a spherical shape. Which shape with the same total volume has the least surface area?

For a cube, 1 cc = x3; the cube is 1 cm on a side. The total surface area is 6x2 = 6 cm2. For a sphere, 1 cc = 4/3 πr3; r = 0.62 cm. The total surface area is 4πr2 = 4.8 cm2. The sphere has 80% of the area of the cube. 9. Teflon is a hydrophobic material. Would the water-air-Teflon contact angle more likely be 45 or 120

degrees? Teflon is nonwetting, the contact angle should thus be greater than 90 degrees and in this case, 120 degrees is the likely value unless the Teflon has adsorbed contaminants. 10. A solid material prior to treatment has a water-air-material contact angle equal to 40 degrees. A

polymeric material, which will be used to coat the other material, exhibits a water-air-polymer solid contact angle equal to 100 degrees. The coating process involves dissolving the polymer in a solvent. Brushing the solution on the solid material and then allowing for the solvent to evaporate leaves the polymer behind. After the coating process is completed, the contact angle at a water-air-coated solid is measured. Below are different outcomes. For each outcome, provide an explanation for the observation.

(a) Depending on where on the surface the measurement is made, contact angles are either 40 or 100

degrees. The surface coverage is uneven; bare spots exist.

(b) Everywhere the measurement is performed, the contact angle is 110 degrees. Some solvent remains or something has contaminated the surface. 11. What is the difference between an antigen and a hapten? A hapten is a smaller chemical entity and becomes immunogenic only when it combines with a protein. 12. A biomaterial implanted years ago in patients has now been found to be an adjuvant. What may the

outcome for the subset of patients who later underwent total hip replacement surgery? Osteolysis or another immune response may be accelerated due to heightened sensitivity to wear particles. 13. Would a synthetic polymer be a better material for stitching a wound, as opposed to a thread made from

animal-derived protein?

50

A synthetic may less likely resemble a natural carbohydrate or other type of molecule that the immune system recognizes as foreign. 14. An artery is grafted with a synthetic material as shown below. If the synthetic material is not

compliant, sketch how the vessel-graft combination will appear when pressure peaks in the blood circulatory system.

15. A tube is compliant. When the tube is not pressurized, the radius (ro) is 10-2 m. Due to compliance, the

radius (r) will increase linearly with pressure at low to moderate pressures, according to r = ro(m) + 0.005 ∆P(Pa). If water flows through a 0.10 m long segment at a rate 10-5 m3/s, how does the pressure drop needed to sustain flow through the a totally noncompliant (zero stretch) tube compare to that of the compliant tube?

Assuming that the flow is laminar, Equation 11-5a can be used. Because the viscosity and length are the same, they will cancel in the pressure drop ratio. Letting ∆PC and ∆PN equal the pressure drop in compliant and noncompliant tubes, respectively. First find the working relationship and then evaluate the pressure drop that is required for flow in the noncompliant tube. ∆PN/∆PC = (rC/rN)4. ∆PN/∆PC = (10-2 + 0.005 ∆PC)4/10-8. ∆PN = 8µLQ/πR4. ∆PN = [8 (10-3 kg m-1 s-1) 0.1 m (10-5 m3/s)]/ (π 10-8 m4) = 0.25 Pa.

Substituting the pressure value into the ratio equation yields

0.25/∆PC = (10-2 + 0.005 ∆PC)4/10-8.

Rearrangement facilitates trial and error:

2.5 10-9 = (10-2 + 0.005 ∆PC)4∆PC.

Know that ∆PC must be less than ∆PN so try values of ∆PC less than 0.25 Pa until one side equals the other. Alternately could graph the right hand side and see where the curve hits 2.5 10-9.

Find that ∆PC equals about 0.17-0.18 Pa; hence, ∆PN/∆PC = 0.25/0.17 ~ 1.5. The pressure drop is lowered considerably due to compliance.

51

Chapter 13 1. The properties of two drugs are given below. Which one may be safer and best to start treatment with? Drug LD50 (mg drug/kg patient) ED50 (mg drug/kg patient) Mellonzac 0.05 0.02 Burgin 0.1 0.08 Although Burgin has a higher LD50 and may seem safer, its therapeutic index is lower and equals 1.25, versus that of Mellonzac (2.5); hence, start with Mellonzac. 2. What conditions must be satisfied for two different drugs to exhibit similar peak times when they are

orally administered and the model in the text is obeyed? While “true,” the answer is not that the values of kE and kD must be equal. The relative rates also matter, so some scaling may be involved. From Equation (13-7b), find that the time stays the same when kE and kD/kE are fixed. 3. What two conditions must be satisfied for two different drugs to exhibit similar peak concentrations

when they are orally administered and the model in the text is obeyed? From Equation (13-7a), find that the time stays the same when CD0 and kD/kE are fixed. 4. Two different drugs and their absorption and elimination properties are tabulated below. By what

factor will one drug peak faster than the other when oral administration is used? Drug kD (h-1) kE (h-1) Mellonzac 10 1 Burgin 100 10 It will take Mellonzac 10-times longer to peak in concentration. From Equation 13-7b, tMAX = ln (kD/kE)/[kD – kE] the numerators will be the same for each drug, but the denominator will be ten-times larger (100 – 10 = 90) for Burgin than Mellonzac (10 – 1 = 9).

5. The infusion of a drug, which provides a drug at a constant rate, is used instead of oral administration.

The drug is eliminated from the body at a rate that is proportional to concentration.

(a) Construct a picture of a compartment-based, conceptual model from which a mathematical pharmacokinetic model can be developed.

kE CB

Elimination from Body

Constant infusion RI; (Mol or mass/time)

Body

Concentration = CB

(a) Show that the concentration in the body is given by CB = RI [1 – exp(-kE t)] , kE VB

where CB, RI, kE, t, and VB refer to drug concentration in the body, infusion rate (moles/time), elimination rate constant, time, and body volume, respectively.

If the student recalls the enzyme example and/or the first semester of integral calculus, they can solve it.

The balance on the body is

VB dCB/dt = RI – kE VB CB.

52

Separation and integration can now be done:

∫dt = VB [∫dCB/(RI – kE VB CB)], which is subject to when t = 0, CB = 0.

The left hand side is the familiar ∫dx while the right side is the familiar or “lookupable” ∫dx/(a – x).

(c) Sketch how the time profile of drug concentration compares to the case when oral administration is used.

Oral intake results in a concentration peak. In contrast, infusion results in the concentration increasing and then attaining a steady value (as long as infusion is steady and continued).

Infusion Oral

Drug Conc

Time

(d) Blood measurements indicate that the concentration of an infused drug in a patient is half of the

ED50 value. What should be done to bring the concentration up to the ED50 value? According to the model, doubling the existing value of RI will double the ultimate concentration attained.

(e) A drug is infused to attain a level equal to 0.1 of its LD50. Now a chemically modified version of this drug has become available. The new version is no longer degraded by the liver, but it can still be excreted by the renal system. Consequently, the effective value of kE for the new drug is lowered by a factor of ten. The LD50 of the new drug is about the same as the old drug. Based on the pharamcokinetic model for infusion, what could happen to the patient if the new drug is infused at the same rate as the old drug?

We could have a problem. Keeping RI constant while lowering kE by a factor of 10 will raise the ultimate drug concentration by 10-fold. The factor the concentration increases will approach LD50.

6. You can easily solve dy/dx = -2y by using the separation of variables method presented earlier in

Chapter 7. However, the problem represents also a very simple first-order differential equation (FOLDE), which can be solved to yield y(x) by using the general method presented in the Appendix.

(a) Use separation of variables to solve for y as a function of x for the case, y = 1 when x = 0.

dy/y = -2dx. ln(y) = -2x + c. Using y = 1 at x = 0; ln(1) = 0 = -2(0) + c c = 0. y = exp(-2x).

(b) When using a new method, it is helpful to first use it on a simple problem that you know the

answer to. To gain practice at using the general solution method, see if you can get the same answer as in part (a) when you use the general solution method, which entails first obtaining an integrating factor.

53

Writing first in standard form, dy/dx + 2y = 0 indicates that p = 2 and g = 0. µ is thus exp(2x). y = exp(-2x)[∫ exp(2x)* 0 dx + c].

y = exp(-2x)[0 + c] = c exp(-2x).

Using the initial condition, y = 1 when x = 0, requires that c = 1. Therefore,

y = exp(-2x), which is the same answer as obtained in part (a).

54

Chapter 14

1. The percentage of nuclear spins that are aligned with the static magnetic field in a NMR spectrometer is one percent.

(a) What will happen to the percentage aligned and resonant frequency of the nuclei if a spectrometer with a stronger magnet is instead used?

The fraction aligned and resonance frequency will increase as Bo is increased.

(b) For either spectrometer, do you think that the “ON” time required for the second field to tip the magnetization will increase or decrease when the strength of the second field is increased, and what is your reasoning?

The pulse time should decrease because the stronger the second field, the faster the magnetization aligned with the static field will tip.

Note: the pulse time (tP) depends on the strength of the second field (B1). For a “90 degree pulse” tp = ½π γ -1 B1

-1, where γ is the gyromagnetic ratio of the nucleus. Essentially, the stronger B1 is, the less time the rf transmitter needs to be “ON” in order to tip the magnetization aligned with the static field.

2. Based on the questionable presence of lines in the Fourier transformed data, the signal from 100 free induction decays obtained from a cell suspension is weak. What are three possible ways to solve this problem so that the effect of a drug on cell metabolism can be studied?

(i) More FIDS to sum to more signal (note: not strong because the signal to noise ratio is ~ FID½ ). (ii) Use a more concentrated cell suspension. (iii) Use a higher field strength spectrometer to align more spins.

3. Will a highly shielded or low-shielded H nuclei exhibit a lower than average Larmor frequency? The higher the shielding, the less static field that will be seen. Thus, highly shielded nuclei behave as if they are in a lower Bo field strength spectrometer. The Larmor frequency of a highly shielded nucleus will be lower than that of the frequency of a low-shielded nucleus.

4. In a spectroscopy application of magnetic resonance where you seek to watch the chemistry occur

within intact cells, which is better, a strong static magnetic field or one that varies within the sample? A strong field will provide more signal. A spatially varying field will make the analyte abundance relationship to signal strength difficult to establish because a particular species will provide signals at different frequencies depending on the location in the sample. Varying fields are primarily used in imaging applications. A constant Bo field is typically used in spectroscopy applications.

5. A faucet drips at a rate of 1 drop every 2 seconds. What is the range of dripping rate that will be

measured if you count drops over a 5 second observation period? To get a rough answer via thought experiment, assume that the drops hit your hand beneath a faucet every 2 seconds and when a drop hits your hand, a “drip event” is recorded. If you start counting right after the first drop hits, you will count 2 drops over the next five seconds. Instead, when one starts counting just before the first drop hits (or score that drop), 3 hit-events will occur during a 5 second interval. Overall, an expected value is 5/2 = 2.5 (from multiple 5 second observations) so the extremes of 2 and 3 bound the average frequency.

D D D D D D D D

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Overall, there is an inherent uncertainty in the drip rate (i.e. frequency) when the short 5 second window is used.

6. Figure 14.5 (b) is reproduced below. Indicate how the areas other than A, -A, B, and -B cancel.

1

1

product t( )

0.9980470 t0 0.2 0.4 0.6 0.8 1

1

0

1

A

-A -B

B

Product (t)

-D

D

-D

D ½ C ½ C

- C

7. Below is a complex signal. It contains two waves of frequency 3 and 10 Hz. The component signals

have equal amplitudes. If you did not know the component signals and had only the raw data to work with (i.e., below), explain how you would analyze the data to obtain the frequency spectrum (i.e., figure out what the constituent frequencies are). Use words and equation(s) to explain. What would a graph of intensity vs. frequency look like?

2

1.953005

signal t( )

0.9980470 t0 0.2 0.4 0.6 0.8 1

2

0

2Signal (Sum of 1 and 2)

Net Signal(t)

Use a series of test frequencies to probe the composite signal-time data. Can use Equations 14-1 or 14-2. Whenever the integral or summation is not zero, will have found a component frequency. The frequency spectrum will have two lines: one at 3 and 10 Hz of equal height. 8. Below is a spherical object situated in a magnetic field gradient. There is an anomaly in the object,

which does not provide a signal. Plot what the “image” would look like on an intensity versus frequency plot (on right).

Signal Intensity

Low Field High Field

Frequency

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9. Below is an object that is subjected to one-dimensional magnetic resonance imaging. The dark region produces less signal than the light region due to a difference in nuclei concentration. What was the field gradient used to provide the “image” (signal vs frequency plot) shown? Draw your answer on your sketch of the object and be sure to note where the field is high and low.

High Field Strength

Signal Intensity

Frequency

10. Prove the integral below is zero over the interval (0, 1 second) if frequency-1 (v1; Hertz) does not equal

frequency-2 (v2; Hertz). Assume that v1 and v2 are integers. (Hint: use the identity cos(A)cos(B) = ½ [cos(A – B) + cos (A + B)].)

∫cos(v12πt) cos(v22πt) dt = 0. The integration result is given below

sin[(v1 – v2) 2π t]

4π (v1 – v2)

+ sin[(v1 + v2) 2π t]

4π (v1 + v2)

1

0

. Inserting the limits yields

sin[(v1 – v2) 2π ]

4π (v1 – v2)

+ sin[(v1 + v2) 2π ]

4π (v1 + v2)

.

When ν1 does not equal ν2, the left and right terms are zero because sin(2π) = 0 and adding or subtracting whole numbers of v only creates multiples of 2π (i.e., –2, -1, 1, 2… rotations of 360 degrees). Therefore, the integral of the cosine product is zero when ν1 does not equal ν2. When ν1 = ν2, the integral of the cosine product can be looked up; the result is ∫ cos2(2π vt)dt = ½t + [sin (2*2π vt)]/(4*2π v). Evaluating the above for t varying from 0 to 1 yields ½ + [sin (2* v*2π)]/(8π v)= 0.5. For any integer value of ν, the result is the cosine product equals 0.5. Note that changing the amplitudes of the two cosine waves (both were assumed to be one) would alter the numerical value.

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11. An object and a MRI image in one plane are shown below. The object possesses a subvolume (dark region) that emits zero signal. Indicate on the object where the static field (Bo) is low and high (i.e., show the Bo field gradient).

LOW

Intensity

Frequency

HIGH 12. It is desired to obtain a NMR spectrum of a biological tissue sample to determine the abundance of

metabolites. So a spectrum is desired where the signal from each 31P nuclei associated with a given molecule is sought. Despite the best efforts of engineers, the Bo field is not completely uniform. Therefore, when the sample is put into a magnetic field, it is spun as shown below. Explain why the sample is spun and the positive effects on the spectrum.

Sample

Bo direction & local strengths

Rotates about

long axis Rotation axis A major idea in NMR applications is that unlike imaging, where the Bo field is deliberately varied, in spectroscopy a uniform field is required. When Bo is uniform, nuclei will have different Larmor frequencies due to different shieldings. Different Larmor frequencies is advantageous because the P’s in ATP, sugar phosphate, etc. will have different frequencies and thus each can be separately detected and each intensity measured. Spinning the sample evens out the Bo all the nuclei “see.” 13. An inverted “C” chord is played on a piano. Each note is equal in strength. The notes of the inverted

chord are G (low note), C (middle note), and E (high note). (The interval between G-C is greater than C-E.) Sketch (i) what the raw signal looks like (roughly is OK) and (ii) what the Fourier Transformed data look like. Label the axes on the two sketches.

This problem is what FT’s do and why they are useful; FTs convert information into something someone can use.

Decays to zero and one f not evident Intensity Volts

(current)

Note: If one knows a little music/comp, the interval between G-C is greater than C-E by one half step so the line spacing is not equal.

G C EFourier Transformed

Raw Signal

Frequency Time

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