Biomaterials The Intersection of Biology and Materials Science, 2nd Edition Johnna S Temenoff Soluti

Type:

Solution Manual

Biomaterials The Intersection Resource: of Biology and Materials Science Edition:

2nd Edition

Author(s):

Johnna Temenoff Antonios Mikos

INSTRUCTOR’S SOLUTIONS MANUAL

BIOMATERIALS : T HE INTERSECTION OF BIOLOGY AND M ATERIALS S CIENCE SECOND EDITION

J. S. Temenoff Wallace H. Coulter Department of Biomedical Engineering Georgia Tech and Emory University, Atlanta, GA

A. G. Mikos Departments of Bioengineering and Chemical and Biomolecular Engineering Rice University, Houston, TX

Biomaterials: The Intersection of Biology and Materials Science, 2nd ed.

Temenoff and Mikos End of Chapter Problems Solutions Manual

(Note that only the text of the problems and solutions have been included in this solutions manual. For figures associated with particular problems, please see the appropriate chapter in the textbook.)

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Chapter 1 1.1 An artery is a flexible blood vessel that withstands various pressures regulates the flow of blood. A common application of biomaterials is in the fabrication of a vascular stent to prevent re-closure of arteries after angioplasty (removal of plaques that impede blood flow). In many of these products, metals are used and formed into a “chain link” type architecture.

(a) What specific bulk and surface material properties would be required for this application?

A variety of answers are acceptable as long as properly justified, particularly relating to the idea of flexibility (bulk mechanical property) and how the device interacts with blood (surface physical or chemical properties). (b) Explain why this material type and architecture were chosen in terms of the properties you have listed above.

A variety of answers are acceptable as long as properly justified. An example: Metals can be easily formed into complex shapes like that shown here. The chain-link architecture allows for expansibility of the material without fracture, while providing sufficient strength to push against the side of the blood vessel.

1.2 A current area of biomaterials research is in developing a tissue engineered bone replacement to be placed in large bone defects. In general, this approach involves seeding bone cells on a scaffolding material with the idea that this entire construct would be implanted into the site of injury to replace lost bone. (a) Would a natural or synthetic material be best for the scaffold? Justify your answer in terms of the expected response of the body to your construct and why this material class may be better for this application.

Either natural or synthetic materials could be acceptable answers, as long as justified using concepts in 1.4.4. (b) What FDA regulatory pathway would likely be involved in approval of this product?

This would likely be a combination product (device + biologic), so its main mechanism of action would have to be determined by the FDA and then that Center would take the lead in the regulatory process, with the other taking a secondary role.

1.3 Whether or not the biological response to a material is acceptable depends on the specific application (see definition of biocompatibility, Section 1.1). Assume a new material shows activation of the immune system in vivo. Discuss whether or not this result would be acceptable for the following applications and explain your reasoning: .

Variety of answers acceptable if well-justified. Most logical answers are found below. (a) Stem of hip implant Unacceptable since this could lead to long-term “rejection” of an implant that is designed to be permanently integrated into the body. (b) Tissue engineered vascular graft Unacceptable since this could lead to long-term “rejection” of an implant that is designed to be permanently integrated into the body. (c) Carrier for vaccine delivery Acceptable since a heightened immune response is desired overall - this could aid in making vaccine delivery more effective. (d) Coating for implanted electrode Unacceptable since this could lead to long-term “rejection” of an implant that is designed to be permanently integrated into the body. In addition, cellular response could reduce the ability of the electrode to transmit electrical signal to the surrounding tissue. (e) Nanoparticle for cancer therapy May be acceptable since a heightened immune response is one way to jump-start the body’s defenses against cancer cells.

1.4 Fluorine (F) is more electronegative than oxygen (O). (a) If you replace O with F in a compound, which of the three major classes of primary bonds would it affect the most? Explain why you chose this class.

This would affect ionic bonding the most since ionic bonds occur only when atoms have large differences in electronegativity. Adding F would cause stronger bonding in many cases since it is more electronegative and thus there would be a larger attractive force with electropositive elements.

Another acceptable would be that this would affect covalent bonds, since electronegative elements are also found in covalent bonding. Adding F would cause more electron withdrawing from the elements it is bonded to, which would affect the attraction force, and also the polarity of the bond (see below). (b) How would this replacement affect secondary bonding?

Replacement of O with F would cause more electron withdrawing from the elements F is bonded to, which cause the other elements to become more positive and thus the overall molecule to .

become more polar. This will allow for more polar bonds (such as hydrogen bonds) to be formed with neighboring molecules.

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Chapter 2 2.1. Using the diagram below, calculate the Miller indices for the following: a) Plane AGDE (100) b) Plane CDGH (001) c) Plane AGCF (110) d) Point C 0,1,1 e) Point G 1,0,1 Planes (hkl) (no commas) Points h,k,l (no () and with commas) 2.2 Brass is an alloy of copper and zinc. Assume the weight percent composition of brass is 80% Cu and 20% Zn. Would you expect brass to be an interstitial solution or a substitutional solution? Justify your answer based on the Hume-Rothery rules.

Solvent = Copper Atomic radius: 0.1278 nm, Electronegativity: 1.9, Face-centered cubic (FCC).

Solute = Zinc Atomic radius: 0.1332 nm, Electronegativity: 1.6, Hexagonal close-packed (HCP).

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2.

3.

4.

Atomic radii differ by less than 15% (~4%).

Electronegativities are similar (1.9 vs. 1.6).

Valence charges are similar (Cu+/Cu2+ vs. Zn2+).

Solution does not contain a large fraction of solute (20%).

Therefore, this should be a substitutional solution.

2.3 What is the difference in the definition of coordination number between metals and ceramics? Why are these two parameters different for the two classes of materials?

Coordination number in metals refers to the number of nearest neighbor atoms, since this parameter contributes to the overall stability of the structure, whereas coordination number in ceramics refers to the nearest neighbor ions of opposite charge, since this is what contributes to the stability of the crystal in ceramics.

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2.4 Why do point defects occur in groups in ceramics?

The defect cannot affect the overall electroneutrality of the material, so groups of defects must be formed so that both positive and negative charges are affected equally.

2.5 A polymer is made of the following fractional distribution. Calculate M n , M w , and PI for this polymer.

Wi 0.10 0.15 0.35 0.25 0.15 Mi (kg/mol) 100 200 300 400 500 Sum (Σ) Wi Mi (kg/mol) Ni = Wi / Mi xi = Ni / ΣNi xi * Mi wi = Wi / ΣWi wi * Mi

0.1 100 0.001 0.260304 26.03037 0.1 10

0.15 200 0.00075 0.195228 39.04555 0.15 30

0.35 300 0.001167 0.303688 91.10629 0.35 105

0.25 400 0.000625 0.16269 65.07592 0.25 100

0.15 500 0.0003 0.078091 39.04555 0.15 75

1 0.003842 260.3037 320

M n = Σ(xi * Mi) = 260.30 kg/mol M w = Σ(wi * Mi) = 320 kg/mol PI = Mw / Mn = 1.23

2.6. Using the data in Figure 2.35 and assuming first-order diffraction calculate the following: (Note that aluminum has a FCC structure.) (a) the interplanar spacing for peaks corresponding to planes (111), (200), and (420).

First order diffraction: n=1 (111) 2d sin θ = nλ 2d sin 39° = 1.54 Å d = 1.54 Å /(2*sin 39°) d = 1.22 Å (200) 2d sin θ = nλ 2d sin 43° = 1.54 Å d = 1.54 Å /(2*sin 43°) d = 1.13 Å (420) 2d sin θ = nλ 2d sin 116° = 1.54 Å d = 1.54 Å /(2*sin 116°) d = 0.86 Å .

(b) the atomic radius for Al. Using (111) plane: d(111) = a0/ (12 + 12 + 12)0.5 = a0/ (3)0.5 For FCC, a0 = 2r20.5 d(111) = 2r20.5/(3)0.5 ((3)0.5 * d(111))/ (2*20.5) = r ((3)0.5 * 1.22 Å)/ (2*20.5) = r 0.75 Å = r 2.7. Based on the results in Figure 2.47, is 5 seconds of UV light exposure sufficient to polymerize the poly(ethylene glycol) completely?

No. Peaks at 1635 cm-1 and 1410 cm-1 are still visible, indicating that unreacted poly(ethylene glycol) is still present, although these peaks disappear at longer time points.

2.8. In Figure 2.50, higher resonance frequencies are found to the left of the standard peak shown at 0.0 ppm (signal is shifted downfield from the standard). Given this information, provide a possible molecular explanation for the difference in location of the H signal from peak 4 vs the H signal in the solvent.

Possible explanation: The H’s in peak 4 are bound to C, which is bound to another C. However, in the solvent, the H is bound to a C bound to Cl, which is more electronegative than C. Therefore, the more electrons are withdrawn from the C in the solvent peak, which means the H’s are less shielded and a higher resonance frequency (signal shifted downfield) is found for the solvent peak.

2.9. Acetone is preferred by some biomedical device manufacturers as a solvent for their polymers because it is considered by regulatory agencies to have fewer negative effects in the body than other organic solvents. Below is the mass spectrometry data and molecular structure of acetone. Acetone has the molecular formula of C3H6O and a molecular weight of 58. (a) How are particles separated by mass in mass spectrometry?

Charged species are forced through a magnetic field, and lighter species are deflected more than heavier ones. Because of the geometry of the mass analyzer and the strength of the magnet, only ions with a certain mass are allowed to hit the detector, while all others strike the walls of the chamber. By altering the magnetic field in a controlled manner, the detector can record the amount of ions having various masses. (b) What is the large peak at 43? Draw the molecular structure that accounts for this peak. Give a possible explanation why the peak at 43 is larger than the other peaks seen in the mass spectrum. .

The acetone molecule loses a methyl group, leaving either C2H3O+ or C2H3O-:

(In electron-beam ionization, only positive ions are formed.)

The larger peak indicates that C2H3O is detected at the highest rate of all of the fragments. This suggests that the bond between C2H3O and CH3 is the weakest bond and is the easiest to break, resulting in a higher incidence of C2H3O than other fragments. (c) What is the peak at 15? Draw the molecular structure that accounts for this peak.

Theoretically, what is the expected relationship between the intensity of this peak and the one at 43? Why? Do the data above demonstrate this expected relationship? If not, give a possible explanation why the theory and the data do not match.

The acetone molecule loses the C2H3O group, leaving CH3+ or CH3-:

(In electron-beam ionization, only positive ions are formed.)

Theoretically one would expect the C2H3O and CH3 peaks to be the same size, since they are both formed by breaking the same bond.

The data above does not demonstrate this expected relationship.

The smaller peak at 15 suggests that the CH3 ion is not as stable as the C2H3O ion. This would result in the further degradation of CH3 into smaller fragments, such as H’s, which are not visible on the spectrum above.

Also, in some types of mass spectroscopy, both positive and negative ions may form. However, the mass spectrometer can only focus ions of one charge (either positive or negative) at a time.

If C2H3O+ and CH3- are favored over C2H3O- and CH3+, and the mass spectrometer focuses the positive ions, then you would see a larger C2H3O+ peak at 43 than CH3+ peak at 15.

2.10 (a) You perform gel permeation chromatography (GPC) on the polymer found in Problem 4, which contains molecular weights of (a) 100, (b) 200, (c) 300, (d) 400, and (e) 500 kg/mol.

Which of the polymer weights (a-e) would elute through the column first and which would elute last? Explain your choice based on molecular interactions between the sample and the column.

Polymer (e), the heaviest compound, would elute through the column first.

Polymer (a), the lightest compound, would elute through the column last.

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Larger molecules cannot enter the porous beads and remain in the mobile phase, causing them to elute the column quickly. Smaller molecules are retained in the beads (stationary phase) for a longer time, causing them to elute more slowly. (b) You have synthesized two polymers using the same monomer, but with differing finial molecular weights. a) You run the two polymers on the GPC and obtain the following plots: Using the following standards, create the standard curve and calculate the approximate molecular weight of polymer A and polymer B. Molecular Weight 4281 7327 1253 883 6229

Time of Peak 8 3.76 12.4 12.78 5.25

Standard Curve

15

y = ‐0.0014x + 14.021 R² = 1

10

5

0 0

2000

4000

6000

8000

Based on estimates of the time of the peak, the answers will differ. Estimates for A should range from 6-7.25 [or MW of 5729-4657]; For B, from 10 to 10.5 [or MW of 2872 to 2515].

An example: A: 6.25 min; = (6.25-14.021)/-.0014 = 5550.7 B: 10.25 min; (10.25-14.021)/-.0014 = 2693.6 2.11. List all the appropriate techniques from Section 2.5 to determine the following: (a) Chemical composition of an unknown sample (using data to determine the elements/bond types in the unknown sample) IR spectroscopy, nuclear magnetic resonance spectroscopy, mass spectrometry (b) Identification of an unknown sample (comparing with known patterns to determine the unknown sample)

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Xray diffraction, IR spectroscopy, nuclear magnetic resonance spectroscopy, mass spectrometry (c) Size of a unit cell Xray diffraction (d) Molecular weight of a polymeric sample Size exclusion chromatography (e) Concentration of sample UV-Vis spectroscopy, IR spectroscopy

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Chapter 3 3.1. Why does slip occur on planes of high atomic density?

There is less energy required to cause dislocation glide if the bonds to be broken and reformed are physically closer together, so planes with high atomic density are where slip occurs (the result of dislocation glide). Using the caterpillar analogy, it is easier to move sections of the caterpillar’s body if the steps (atoms) are closer together.

3.2 Assume you are exploring using materials as a porogen in for a device made of high density poly(ethylene) in order to improve anchorage in surrounding tissue. In this case, you would plan to insert the porogen during fabrication of the device and then heat the device to remove the porogen. You believe that low density poly(ethylene) may be an appropriate choice, but your boss is not convinced. (a) Using data in Table 3.2, describe the potential concern of use of low density poly(ethylene).

The potential problem is that the melting temperatures are relatively close, so it is possible that the device would soften or melt during heating to remove the porogen. If this is the case, it may change shape and become unusable for the intended application. (b) Of the characterization techniques discussed thus far in the book, which one would be most appropriate to use to address your boss’s concerns? To move forward with low density poly(ethylene) as a viable option, describe the outcome of this test.

DSC should be used and the heating curves, particularly Tm curves, should be examined carefully. If the peaks are separated enough, then there should not be a concern about melting of the device while removing the porogen. However, if they overlap substantially, this is likely not a viable porogen.

3.3 While working for Biomaterials, Inc, you realize your assistant has left an unknown substance contaminating the reference pan in the company’s power-compensated differential scanning calorimeter (DSC). You are extremely upset because you realize that it will be nearly impossible to gather useful data from the run you just made, given that the reference pan was contaminated at the time. Explain why a contaminant in the reference pan will render your newly-collected data useless, based on how a DSC works.

In power-compensated DSC, the sample chamber is heated by one furnace, while the reference chamber is heated by a separate furnace. The chambers are heated at a temperature ramp, while the power required by the furnaces to keep the temperature difference between the chambers at 0 degrees is recorded.

In order to convert the power measurements to heat flow, which is then plotted to determine thermal transitions of your sample, the power from the reference chamber is in essence subtracted from the power required by the sample chamber at each temperature. If the reference is contaminated, this “baseline” subtraction is not accurate, due to the possibility of unknown .

thermal transitions going on within the reference chamber that would alter the power that was required to heat that chamber through that temperature.

3.4 Using the differential scanning calorimetry (DSC) results below for three different polymers, answer the following questions. Note that exothermic reactions are shown as positive on this graph. (a) Identify the approximate thermal transitions for this sample, based on the graph above.

A: Tg = 35°C Tc = 58°C Tm = 83°C B: Tg = 41°C Tc = 58°C Tm = 79°C C: Tg = 29°C Tm = 83°C (b) From the DSC plot, which polymer (A), (B), or (C) is most crystalline? Explain how you drew this conclusion.

Polymer (A) is most crystalline because the area under the melting peak is the largest. This indicates that there are the most intermolecular bonds in this material, and, therefore, more energy is required to melt the material. (c) Describe a possible heating/cooling regimen to improve crystallinity and explain the process from a molecular perspective. For which of the above polymers would this be a good choice, and for which ones would this process not likely work? Explain your rationale.

Crystallinity can be improved by heating the polymer to the crystallization temperature Tc, maintaining this temperature, and then slowly cooling the polymer back to room temperature. At Tc, the polymer chains have sufficient energy to move into a highly ordered crystalline state.

Slow cooling allows the polymer chains sufficient time to become more fully aligned, thus improving crystallinity. Polymer (C) would likely not be a candidate for this process as it possesses no Tc according to the DSC data and thus it is not possible to increase crystallinity through this method. (d) Which polymer (A), (B), or (C) would you expect to have the lowest molecular weight?

Justify your conclusions from a molecular perspective.

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Polymer (B) would have the lowest molecular weight, because it has the lowest melting temperature. In general, melting temperature increases with molecular weight (Figure 3.16).

High molecular weight materials have fewer chain ends, and because chain ends are more free to move than other parts of the macromolecule, more energy is needed to produce the motion required to release the chains from their ordered structure if the chains have a higher molecular weight. (e) Studies later find that one of these polymers is actually more branched that the other two.

Which polymer (A), (B), or (C) is most likely the branched polymer? Explain your reasoning from a molecular point of view.

Polymer (C) is the more branched polymer. Branching prevents alignment of polymer chains, thus decreasing crystallinity.

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Chapter 4 4.1. Prove when a cylindrical isotropic material has constant volume and undergoes a very small deformation, its Poisson’s ratio (ν) is equal to 0.5.

Assume: this material is a cylinder with a radius of r0 (diameter d0= 2r0) and a length of l0. After it undergoes a very small deformation, l = l0 +Δl and r= r0 + Δr Since the volume is constant, Δl * Δr < 0, and V0 = V l0 * 𝜋 *r02 = (l0 +Δl)* 𝜋 *( r0 + Δr)2 l0 * r02 = l0 *r02 +2*l0* r0 *Δr + l0 * Δr2 + Δl*r02 +2*Δl* r0 *Δd + Δl* Δr2 Since it is a very small deformation, l0 * Δr2, 2*Δl* r0 *Δr and Δl* Δr2 can be ignored, -2*l0* r0 *Δr = Δl*r02  ν = -( Δr/ r0)/ (Δl/ l0) = -( 2Δr/ 2r0)/ (Δl/ l0) = -( Δd/ d0)/ (Δl/ l0) = 0.5 4.2. Assume the Young’s modulus of poly(styrene) at 250C is 4.9 × 10 5 Pa, and its Poisson’s ratio is 0.35. Calculate its shear modulus.

G = E/ [2(1+ γ)] = 4.9 × 10 5 Pa / [2 * (1 +0.35)] = 1.8 × 10 5 Pa 4.3. Explain why polycrystalline materials are stronger than the same material made of a single crystal.

When the resolved shear stress exceeds the critical value, slip occurs along that slip plane. For polycrystalline materials, due to the fact that plastic deformation occurs due to deformation of individual grains, this means that the overall force (stress) applied must be sufficient to exceed the critical resolved shear stress along the grain that is the least favorably aligned with the direction of applied force, thus raising the force (stress) needed for deformation of the material overall and causing the material to be stronger.

4.4. The mechanical properties of orthodontic wires are temperature dependent. Force-distance measurements were taken for a wire with a diameter of 1.3 mm. (a) Graph the engineering stress-engineering strain diagrams for both curves. Label the elastic and plastic regions of each plot. Show all work for full credit.

Stress: 𝜎 Area = πr2 = (3.14159)(0.65 mm)2 = 1.3 mm2 Strain: 𝜀

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Length (mm) 100 100.5 101 101.5 102 102.5 103 103.5 104 104.5 105 105.5 106 106.5 107 107.5 108 108.5 109 109.5

Force (N) at 25degC 0 1.5 3 3.7 3.9 4 4.2 4.4 4.5 4.4 4.3 4.2 4.1 4 3.9 3.8 3.7 3.6 3.5 fracture

Force (N) at 37degC 0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 17 17.3 16.8 16.3 fracture

Strain 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085 0.09 0.095

Copyright © 2023 Pearson Education, Inc.

Stress (Mpa) at 25degC 0.00 1.13 2.26 2.78 2.93 3.01 3.16 3.31 3.38 3.31 3.23 3.16 3.08 3.01 2.93 2.86 2.78 2.71 2.63

Stress (Mpa) at 37degC 0.00 1.13 2.26 3.38 4.51 5.64 6.77 7.89 9.02 10.15 11.28 12.41 12.78 13.01 12.63 12.26

14.00 12.00

Plastic

10.00

Elastic

8.00 at 25degC 6.00

at 37degC

4.00 2.00

Elastic

0.00 0

Plastic

0.02

0.04

0.06

0.08

0.1

(b) How does this temperature dependence benefit the orthodontic use of the wire?

At room temperature the wire deforms easily with very little force applied. This allows the orthodontist to form the wire to the appropriate shape for treatment. When the wire is implanted in the mouth and warms to body temperature, the amount of force required to plastic deformation increases significantly—this way the wire will not deform despite the stresses applied to the teeth. Thus the wire is able to maintain the shape desired by the orthodontist and applies force on the teeth to induce them to move. (c) Calculate the modulus of elasticity, yield strength, ultimate tensile strength, and fracture strength of this material at the two different temperatures.

Approximate answers: E(25˚C) = E(37˚C)= = 225.5 MPa σy(25˚C) = 2.3 MPa σy(37˚C) = 12.4 MPa σuts(25˚C) = 3.4 MPa σuts(37˚C) = 13.0 MPa σf(25˚C) = 2.6 MPa σf(37˚C) = 12.3 MPa Note that use of 0.2% strain line should be used to calculate yield strength for each sample. (d) For each of the four parameters that you calculated in c, state whether the parameter changed with the change in temperature or not and briefly (in one sentence each) describe what that indicates about the material properties.

Modulus of elasticity does not change. The material has the same stiffness at either temperature. .

Yield strength changes. At lower temperatures it requires less force to deform the material.

Ultimate tensile strength changes. The material is able to withstand more load at the higher temperatures.

Fracture strength changes. The material fractures at a lower load at lower temperatures. (e) When this material fractures, what will the shape of the fracture surface look like?

Because the material will undergo plastic deformation before it fractures, it will be a ductile fracture rather than a brittle fracture, thus the fracture surface will have a cone-shaped appearance.

4.5. It is known that articular cartilage has viscoelastic properties. In particular, it is known to undergo stress relaxation. Using the definition of stress relaxation and your knowledge of articular cartilage’s role in the knee joint, explain why stress relaxation is actually desirable in articular cartilage.

Cartilage, particularly articular cartilage in the knee, functions to cushion the impact of standing and walking. Cartilage experiences stress relaxation, whereby under constant strain, the stress experienced by the system will decrease over time. This is desirable because a decreased stress when the joint is loaded for a long time (such as during standing) can help preserve the integrity of joints long term by not overloading the underlying bones.

4.6. You are investigating metal alloys for applications in bone replacement. You generate two materials – Alloy A and Alloy B. The following diagram shows the structures of each alloy. (a) Because this material is of interest to the orthopedic field, it is necessary for it to be able to have a high fatigue life. Using the graph below, determine the following parameters: σmax σmin Fatigue life at a stress amplitude of 100 MPa Fatigue strength at 1010 cycles σmax = 150 MPa σmin = 30 MPa Fatigue life of 109 at a stress amplitude of 100 MPa Fatigue strength of 75 MPa at 1010 cycles (b) Would you expect Alloy A or Alloy B to have a shorter fatigue life in the wet state? Why? Alloy A would most likely have the shorter fatigue life because it has more grain boundaries and therefore more dislocations, making the required stress to fracture lower.

Additionally, Alloy A has more reactive surfaces (grain boundaries – see Chapter 3) that are more likely to corrode, leading to reduced fatigue life.

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Chapter 5 5.1. How would the following affect the rate of corrosion of a stainless steel bone fixation plate?

Increase, decrease or not affect? Justify your answer. (a) Presence of proteins on the surface of the plate Increase: proteins can scavenge metal ions, thus altering the balance of the electrochemical reactions on the surface and promote corrosion; also, metals can act as barriers that prevent oxygen from diffusing to surface to form a passive layer, thus increasing corrosion (b) Presence of small scratches on the surface of the implant Increase: creates two different areas on surface (inside scratch and rest of plate), which sets up anodic and cathodic areas; scratches act as stress risers, which increase the energy in atoms near the scratch and make them more susceptible to corrosion (c) Presence of a screw that attaches the plate to the underlying bone Increase: screw and bone plate together make a crevice that encourages corrosion by means discussed in book 5.2 One method to quantify amount of metal corrosion is to assay the amount of metal ions released into the surrounding environment over a certain period of time. Why is this a good method? Justify your answer based on how corrosion works at a molecular level.

Corrosion is the result of redox reactions. At the anode, the metal is oxidized to release electrons and metal ions, while at the cathode, reduction reactions occur by various ions (such as H+) binding to free electrons. Therefore, metal ions are a product of the reaction that comprises corrosion, so monitoring this product is a good way to determine amount of corrosion.

5.3. Dumbbells of the polymer poly-L-lactic acid (PLLA) were stretched a small amount using a mechanical testing device (not to the point of failure) as a way to alter the material properties.

A 1 cm cube of each of these samples (stretched vs. non-stretched) is placed in water. One degrades much more quickly than the other. Which degrades faster and why?

Stretching should cause the polymer to become more crystalline. As the material is drawn, more of the polymer chains become aligned in the direction of the force applied, creating more order within the polymer structure. The drawn/stretched PLLA will degrade more slowly since increased crystallinity will reduce water penetration within the polymer.

5.4. Why does one need a hydrophobic component in a surface-degrading polymer? Explain from a molecular perspective.

A surface degrading polymer is one in which the rate of water penetration into the bulk is less than the rate of hydrolysis of the bonds in the polymer backbone. .

A hydrophobic area is required to decrease water penetration into the polymer bulk because it is thermodynamically unfavorable for water to interact with hydrophobic areas in the polymer.

5.5. You want to achieve controlled, linear drug release from a biomaterial to be used in a wide variety of applications. Provide a rationale for your choices below. (a) Would you choose a synthetic or natural material? (b) Do you think enzymatic or hydrolytic degradation would be preferred. If hydrolytic, would you choose surface or bulk degradation? (c) What is one major chemical or physical property that you would alter to tune the rate of degradation. Explain why at a molecular level this would affect overall degradation.

Synthetic vs. Natural – Either ok, but if pick hydrolytic, that is generally difficult to achieve with natural polymers; similarly, enzymatic degradation is generally easier with naturallybased polymers Enzymatic vs. Hydrolytic degradation – Surface hydrolytic degradation generally best to get 0 order release. Enzyme levels vary from patient to patient, so it is more difficult to predict the rate of release in general One major chemical/physical property altered to tune rate of degradation – Depends on answers above but should be consistent with ideas found in section 5.4.2

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Chapter 6 6.1. You have produced a new material, Metal A. Use the plot below from tensile testing of your material to answer the following questions. (a) You are now developing a second generation biomaterial, Material B. This material will be an alloy of Material A with other elements. In particular, you are thinking of alloying with chromium (it is known that there is a very strong atomic bonding between Cr and Material A). Based on the graph above, draw the expected stress-strain curve for Material B after alloying. Indicate how it is different than Material A and why this would occur from a molecular perspective. What is another advantage of adding Cr to this metal?

Chromium alloying will result in higher elastic modulus in linear region due to the increase bonding between Material A and the Cr, without a large difference in yield strength or brittleness since bond strength primarily affects the elastic portion of the curve (see concepts in Chapter 4).

Cr can also impart corrosion resistance in most areas of the body (neutral pH). (b) Another option for your second generation biomaterial is to use a new thermal treatment on Metal A whereby you quench it so that the temperature is reduced more quickly. On the same graph you drew for (a), draw the expected stress-strain curve for Material A’ after the new thermal treatment. Indicate how it is different than Material A and why this would occur from a molecular perspective.

Faster quenching will result in smaller grains, which will increase the yield strength, but also the brittleness of the material. This is because it will take more force to initiate slip in materials with more grains since one must reach the critical resolved shear stress in the most un-aligned grain, and with more grains, it is more likely that some will be unfavorably aligned. However, since slip is more difficult, this also results in a more brittle material.

Example sketch shown below: Quenched (Material A’)

Cr alloy (Material B)

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6.2 Compare and contrast forming methods in metals, polymers, and ceramic glasses. Why is forming not possible for crystalline ceramics?

Similarities: In forming, one uses plastic deformation to make a shape, often combined with heating to allow easier processing. Drawing processes can be used for all materials.

Differences: For glasses, blowing is a unique form of processing, which is almost a combination of casting and forming techniques. Extrusion is less likely to be used for glasses.

Forging/pressing is not used for polymers.

Crystaline ceramics are too brittle (not enough plastic deformation before fracture) to be formed and instead must be cast into shapes.

6.3 Based on concepts in Chapter 5, explain at a molecular level why radiation sterilization in air resulted in ultra high molecular weight poly(ethylene) degradation. One way that companies have overcome this is to complete radiation sterilization in an inert gas. Why would this be effective? Based on literature, describe one other possible mechanism of overcoming this problem.

Completing this sterilization in air allowed for the formation of free oxygen radicals that attack the polymer backbone (chain scission) using the oxidation reactions shown in Chapter 5. If the radiation is done in another (inert) gas, there will be no radicals formed that would be able to attack the polymer backbone.

One other common solution is to add in free radical scavengers to the UHMWPE prior to sterilization so that any free radicals formed would be immediately quenched and not available to attack polymer chains.

6.4 You are trying to decide between compression and injection molding the acetabular cup of a hip implant from a new polymer. (a) What characteristics of the polymer would you need to know to determine which method is best? Explain your rationale.

One would need to know if the polymer was a thermoset or thermoplastic, as well as the melting point of the polymer compared to the range of heating your injection molding process could achieve (e.g. can you melt the material with the heater that is available?). If the material is a thermoset or cannot be melted, injection molding would not be possible. However, if these criteria are met, then injection molding may be preferred since it is a very fast processing method. (b) What sterilization technique would be best for this implant. Justify your answer.

Steam sterilization is probably not a good choice since that will likely melt the polymer.

Radiation may be a problem due to the development of oxidizing species which would degrade the polymer (see previous question). Ethylene oxide may be the best option, but this may also .

crosslink or otherwise change the molecular structure of the polymer, so additional testing would be required to assure that the sterilization technique chosen did not have deleterious effects on the final device.

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Chapter 7 7.1. Explain why addition of long hydrocarbon chains to a biomaterial surface would be a poor modification to reduce protein attachment.

Protein adsorption is reduced due to steric repulsion of long, hydrophilic chains attached to a biomaterial surface. These chains extend into the aqueous environment around the biomaterial and take up volume, which physically prevents proteins from touching the surface. Since hydrocarbon chains are hydrophobic, there would be no driving force for them to extend into the environment around this biomaterial and thus they would not form this repulsive layer.

7.2. Explain the need for a spacer arm in biological modifications. Why is this not usually needed in other types of (non-biological) surface modifications described in this chapter?

A spacer arm moves the biological molecule away from the surface so that it can freely rotate and obtain an optimal conformation for binding with its partner to initiate the desired biological response (thus maintaining its bioactivity). The success of other modification methods described in this chapter do not depend as much on the specific conformation of the atoms/molecules attached to the surface, so the addition of the spacer arm is not as important.

7.3. Three different surfaces (A, B, and C) were tested for their properties. (a) Reported below are the contact angles and the surface tension for the liquid-vapor surface (γLV) of four different solutions. Determine the critical surface tension (γc) for each material.

Your answer should have two parts: 1) a graph on which you indicate the point that shows critical surface tension for each material and 2) the equation used to calculate exact critical surface tension for each material. Which of these is the most hydrophobic? Why? Solution Water Solution 1 Solution 2 Solution 3 Solution 4

Surface Tension (dyn/cm2) 70 60 55 50 45

Surface A

Angle (Degrees) Surface B

Surface C

95 76 56 39 26

64 43 29 19 15

100 82 63 45 29

γc, critical surface tension, can be determined by plotting γLV against the angle for each material.

First, plot the three materials and the solid-water interface vs. the contact angle in degrees. Use the linear line of best fit to extrapolate to determine the point at which the angle = 0, which corresponds to γc:

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γLV vs. contact angle Surface energy (dynes/cm) 0

10

20

30

40

50

60

70

80

Contact angle (degrees)

0 20

y = 2.0541x ‐ 81.027

40 60 y = 2.9081x ‐ 99.054

80

y = 2.8595x ‐ 101.73

100 120 Surface A

Surface B

Surface C

Linear (Surface A)

Linear (Surface B)

Linear (Surface C)

From this, it is possible to extrapolate each line to the point where the angle is equal to 0 using the following equations: Surface A: angle=2.8595(γLV)-101.73 ->0=2.8595(γc)-101.73 Surface B: angle=2.0541 (γLV)-81.027 ->0=2.0541 (γC)-81.027 Surface C: angle=2.9081(γLV)-99.054 ->0=2.9081(γc)-99.054 Critical surface tension (dyn/cm2)

Surface A 35.57615

Surface B 39.44647

Surface C 34.2748

Surface C is the most hydrophobic because of its low critical surface tension and high contact angle in water. (b) Scientists are interested in coating the surface of one of these materials with an amphiphilic signaling lipid (like that shown below). It is known that the phosphate-containing head group is responsible for signaling activation in target cells. Which material should you recommend for this application and why?

The most hydrophobic material (Surface C) should be recommended so that any adsorbed lipid will be oriented with the signaling head group facing outwards due to interaction of the hydrophobic surface with the hydrophobic tail. This will present the head group to cell surface receptors and induce signaling. Surface C is the most hydrophobic because it has the lowest critical surface tension and the largest contact angle in water (see part a).

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(c) To further investigate the effect of surface chemical composition and behavior in aqueous solutions, you conduct dynamic contact angle analysis with water. Given the results below, determine the contact angle hysteresis.

Surface

Addition/removal of drops 1 2 3 4 1 2 3 4 1 2 3 4

Surface A

Surface B

Surface C

Contact angle (Degrees) Advancing Receding 94.4 82.6 93.9 82.2 93.6 82.4 94.3 82.7 64.1 60.1 64.3 59.8 63.8 59.7 64.0 60.0 99.8 78.2 100.3 78.5 100.1 78.9 99.7 78.4

From this data, contact angle hysteresis can be determined by finding the difference between the two averages for advancing and receding contact angle measurements: Surface

Ave. advancing (degrees) 94.05 64.05 99.975

Surface A Surface B Surface C

Ave. receding (degrees) 82.475 59.9 78.5

Contact angle hysteresis (degrees) 11.575 4.15 21.475

(d) Based on the above dynamic contact angle measurements, would you select a different surface for your answer to part (b)? Explain your reasoning by describing what is occurring molecularly.

Yes, Surface C does not remain the most hydrophobic surface when exposed to an aqueous environment for extended periods of time. Surface C mostly likely undergoes some sort of rearrangement that reduces the hydrophobicity of the surface and leads to contact angle hysteresis. Instead, I would choose Surface A, which remain more hydrophobic in water and undergoes little surface rearrangement. This is apparent by comparing the receding contacts angles for Surfaces A and C (82.475 vs. 78.5). (e) Researchers know that all three surfaces are composed of different self-assembled monolayers (SAMs). The functional group for each SAM is shown below. Note: SAM 3 is a mixture of both SAMs 1 and 2. Given the results above, identify which of these SAMs is which surface and explain your reasoning from a molecular level.

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The most hydrophobic SAM (SAM 1) is Surface A because it displays a high contact angle in water, and remains that way, even after contacting water for a long period (see results of dynamic contact angle experiments) . The most hydrophilic SAM (SAM 2) is Surface B because it displays the lowest contact angle in water. Little surface rearrangement is expected for this surface, given that it is already hydrophilic.

SAM 3 (the mixed surface) is Surface C, which starts hydrophobic, but can undergo surface rearrangement so more of the hydrophobic groups are hidden by the long chain of hydrophilic polymer after contacting water for a longer period (see results of dynamic contact angle).

7.4. (a) In order to create a stable SAM coating, it has been found that you need chlorine in your polymer. Give a possible explanation of why chlorine would be needed – what part of the SAM would the chlorine likely interact with?

Would likely form covalent bond with attachment group to drive assembly of SAM. Mention of attachment group needed for full credit. (b) A colleague would like to use your SAM to modify their polymer, which has no chlorine.

Therefore, they decide to plasma treat their polymer in chlorine gas before SAM modification.

Explain how this treatment could add chlorine to the polymer surface.

A plasma environment is obtained by applying an electric potential across a gas (in this case, the gas would contain Cl). The cathode is the surface that requires treatment and has a negative potential relative to the anode. To travel from the cathode to the anode, electrons must traverse the gas in the chamber and collide with the molecules to form gaseous ions and radicals. These species can interact with the sample and cause a variety of surface reactions. (c) Your colleague decides to perform Electron Spectroscopy for Chemical Analysis (ESCA) analysis on their plasma-treated polymer to confirm surface modification prior to using your SAM. Assuming that the plasma treatment worked, which of the peaks below would represent the modified polymer and which would represent the original polymer? (Assume this is a C1s spectrum). Explain your answer based on the graph and also a detailed explanation of how the ESCA equipment works to generate data.

ESCA works such that a material is hit with an x-ray, which it then absorbs and releases an electron from an innermost atomic orbital. The emitted electrons enter the analyzer chamber and due to the geometry of the analyzer and the difference in voltage between the two walls only electrons with a certain kinetic energy can be collected by the detector, while all other electrons strike the non-detectable areas. We can change which electrons are detected by changing the voltage in the separation chamber. The kinetic energy of the emitted electron is recorded. From the kinetic energy, binding energy can be calculated.

Chlorine, which is very electronegative, should be more electron-withdrawing to electrons in the valence shell, so the positive nuclear charge should be less shielded in the lower shells, which will cause a higher binding energy between the carbon and its inner shell electrons. Therefore the peak representing the modified polymer is on the left. .

7.5 How is it possible to achieve depth profiling with secondary ion mass spectroscopy (SIMS)?

Be sure to discuss the instrumentation of SIMS and how the different parts of the instrumentation work together to produce plots of results.

Primary ions are emitted from the ion gun and strike the surface, which in turn strips off (sputters) the surface layer of atoms. These secondary atoms pass through the mass analyzer (just as in regular mass spectrometry) for separation of the resultant ions based on mass. A detector translates ion impact into an electrical signal that is read by a processor/computer. Dynamic SIMS delivers higher ion dose, so more of the surface is stripped off and analyzed as the experiment goes on. This means that at various time points, the ions being stripped from the surface and passed through the mass analyzer are from deeper layers in the material. Therefore, we can obtain depth profiling from SIMS (mass spectrogram from a specific location at various times, which correspond to various depths.)

7.6. The following images of two different material surfaces were obtained with atomic force microscopy (AFM). Which of the two surfaces below (a or b) would be likely to have more protein attached if placed in the body? Justify your answer. a – the surface is rougher and therefore will act to trap proteins at surface.

7.7. Make a list of advantages and disadvantages of all of the surface patterning techniques described in this chapter.

Direct-write photolithography: Advantages: small spatial resolution for patterns (on the order of nanometers) Disadvantages: low throughput (each sample must be patterned sequentially); high cost of the equipment needed Dip-pen nanolithography: Advantages: small spatial resolution for patterns (on the order of nanometers); technique is very versatile as many different types of biomolecules have been explored as “inks” Disadvantages: low throughput (each sample must be patterned sequentially) Ink-jet printing: Advantages: technique is very versatile as many different types of biomolecules have been explored as “inks;” many samples can be patterned simultaneously; complex patterning of multiple molecules possible; good control of the volume deposited in each spot Disadvantages: patterns smaller than ~10 microns are difficult to achieve Photolithography: Advantages: large areas can be patterned easily; possible to produce complex patterns Disadvantages: patterns smaller than ~10 microns are difficult to achieve

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Microcontact printing: Advantages: technique is very versatile as many different types of biomolecules have been explored as “inks;” quick to produce multiple samples so high throughput Disadvantages: patterns smaller than a few microns are difficult to achieve 7.8. As a biomaterials expert you are asked to consult on surface modification for several companies. For the scenarios below, select the most appropriate type of surface modification and the one type of surface analysis technique that would be the most important to characterize this modification. Provide a rationale for your choices.

A number of answers are possible here. A few are summarized below, with rationales. (a) Company A manufactures metallic implants as replacements for the temporomandibular joint (TMJ, found in the jaw). However, they are concerned about the fatigue properties at the surface given the large loads experienced during chewing. They want to minimize changes to their current production process.

Surface modification: Ion beam implantation Analysis technique: anything that examines the chemical composition of the surface – SIMS, ESCA, FTIR Rationale: Ion beam implantation can occur after the implant has been manufactured and is often used to improve wear and fatigue properties of metals, depending on what ions are implanted. It is important to characterize the chemical composition of the surface to determine to what degree modifications were made, so one of the techniques listed above would be most important. (In reality IR is not often used with metals so one of the other techniques might be more appropriate.) (b) Company B manufactures polymeric membranes to promote guided bone regeneration after tooth loss by encouraging bone cell attachment and migration. The company has discovered a new polymer that can easily be manufactured into a membrane and is cytocompatible and wants to find an appropriate modification. This new polymer can be synthesized by radical polymerization.

Surface modification: a biological modification, either pre-fab or post-fab), or possibly spray coating with a ceramic Analysis technique: anything that examines the chemical composition of the surface – SIMS, ESCA, FTIR Rationale: It would be possible include a biological modification during fabrication since radical polymerization would allow for this – probably the quickest. Alternatively, this could be done post-fabrication, depending on the chemistry employed. This molecule would be specific for attachment of osteoblasts. Regardless of how it was attached to the surface, one needs to confirm that the biomolecule remained on the surface, so chemical characterization is important. .

Spray coating with a ceramic would also be an option, although, in reality, this may affect osteoblast migration/attachment less than a specific biomolecule. However, chemical composition of the surface would also be important in this case to confirm that the coating procedure had worked. (c) Company C manufactures polymeric contact lenses. However, they have found that their current polymer is too hydrophobic, leading to discomfort when the lenses are worn for long periods. The company is therefore looking for a modification that would improve hydrophilicity, but does not want to add any post-processing steps.

Surface modification: Surface-modifying additive Analysis technique: Dynamic contact angle Rationale: If no additional post-processing steps are desired, then the surface modification must be included within the fabrication, so surface-modifying additives are the best choice. In this case, a copolymer with hydrophilic and hydrophobic groups could be used, with the idea that the hydrophilic groups would move to the surface once the material was in contact with the aqueous environment of the eye. Here, hydrophilicity is the most important outcome, so contact angle analysis should be used. In particular, dynamic contact angle measurements would provide information about how quickly the additive moves to the surface of the lens.

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Chapter 8 8.1. Explain how each of the levels of protein structure (primary-secondary-tertiary-quaternary) may affect a protein’s level of adsorption to a hydrophobic biomaterial.

Primary – the number of polar/charged vs. non-polar amino acids in the primary structure would affect the overall affinity for the surface. More non-polar amino acids would increase affinity to the hydrophobic surface.

Secondary – local interactions between amino acids result in secondary structures (helicies, pleated sheets) that may need to be unfolded to expose non-polar amino acids that would have greater affinity to the hydrophobic material. Thus, if there are more and more stable secondary structures, this may result in reduced unfolding and less protein adsorption overall.

Tertiary/Quaternary - interactions between amino acids that are farther apart in the primary sequence result in three dimensional folding of protein subunits (tertiary structure) than can then interact (quaternary structure) to form the final, functional protein. To adsorb to a hydrophobic surface, it is likely that significant unfolding and rearrangement at the tertiary and quaternary levels is needed in order to expose non-polar amino acids that would have greater affinity for the surface.

Thus, greater stability of these structures, which is influenced by the number of S-S bonds in the overall molecule, can prevent unfolding and thus result in less protein adsorption overall.

8.2 You are concerned with protein adsorption to a vascular graft material. Assume that it is known that Protein H, which is rich in histidine, is important in initiating the blood clotting cascade. You have the following choices for your graft material: (1) Poly(ethylene) (contact angle ~90) (2) Poly(ethylene) that has been plasma treated to possess + surface charge (3) Poly(ethylene) that has been plasma treated to possess - surface charge (4) Poly(ethylene) with covalently attached PEG chains on the surface (a) Briefly explain the difference between adsorption and absorption.

Adsorption refers to spontaneous attachment of molecules to a surface, while absorption refers to penetration of molecules into the bulk of another material. (b) Based on the above information and what you know regarding protein adsorption, on which surface (1, 2, 3, or 4) would you be most likely to observe clotting? Explain why you chose the surface you did as well as why you think that there would be less clotting on the other surfaces.

Histidine is a positively charge amino acid; therefore the histidine-rich protein will adsorb most favorably to negatively charged surface 3. Adsorption of the protein initiates the blood clotting cascade, resulting in increased clotting on surface 3. Positively charged surface 2, on the other hand, will repel the histidine-rich protein, and the protein will not adsorb. Some adsorption may occur on hydrophobic surface 1, but will not be as great as adsorption to surface 3 since there are few/no attractive forces between surface 1 and the protein. Steric concerns also play a major role

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in adsorption, so covalently-attached PEG chains on surface 4 would resist protein adsorption and thus blood clotting. (c) Assume that new experiments have revealed that there are actually two types of Protein H, which is comprised of a total of 40 amino acids: Protein H1 has 7 His and 10 Cys residues Protein H2 has 15 His and 5 Cys residues

Adsorbed protein

Protein H1 is dominant and therefore has a higher concentration in the blood. Draw the expected profile of H1 and H2 adsorption to the surface you chose for (b) over time. Explain why your diagram appears at it does based on principles of protein-surface interactions.

H2

H1

Time Protein H1 binds early over H2 because it is present in higher concentration, and diffusion dominates at this point. Over time, protein H2 displaces protein H1, because it has a higher affinity to the surface (higher number of His) and can unfold more (lower number of Cys), allowing for greater surface interactions.

8.3 A neutral, hydrophobic biomaterial surface was incubated with a mixture of Protein X (20%) and Protein Y (80%) OR a mixture of Protein Z (20%) and Protein Y (80%) and the protein compositions at the surface were measured over time. (a) Given the information about the proteins listed below and the graphs of the protein composition at the surface, explain why, in the mixture with Protein Y, Protein X binds to the material and forms a relatively pure monolayer at early timepoints while Protein Z does not.

Protein Protein X Protein Y Protein Z

MW (total) 11.6 kD 67 kD 64.5 kD

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# Cysteines 2 3 10

Isoelectric point (pI) 9.6 6.9 5.8

Protein X is significantly smaller than Protein Y, so it diffuses to the surface and binds more quickly. In contrast, Protein Z and Protein Y are similar in size, so they will both diffuse to the surface at a similar rate, resulting in the initially heterogeneous monolayer. (b) Why does Protein Y displace Protein X from the surface of the material?

Protein Y is larger, so it has more possible points of contact with the material, which can increase its affinity. Protein Y will also be closer to its pI (isoelectric point) at physiological conditions, so it will be uncharged, similar to the surface, which will increase affinity for the surface. (c) Why does Protein Y displace Protein Z from the surface of the material?

Protein Y is marginally larger (more points of contact) and would be closer to its pI at physiological conditions, but moreover Protein Y is likely to have a more flexible protein structure because it has fewer cysteines. Thus, it is more likely to deform at the surface, allowing for contacts that result in the lowest thermodynamic state (i.e. hydrophobic protein residues interacting with hydrophobic portions of the surface, maximizing charge-charge interactions, etc.).

8.4 You have a mixture of two proteins that you are trying to separate via affinity chromatography.

X – contains many serine residues Y – contains many phenylalanine residues (a) You are using a reversed phase system with water as the mobile phase. Which protein would you expect to elute first? Explain based on molecular interactions with the column.

Protein X is more polar than protein Y as it contains more polar amino acids (serine). Therefore, in a reversed-phase system, it will have more affinity for the mobile phase and will elute sooner than protein Y, which will have greater hydrophobic interactions with the non-polar stationary phase. (b) If you switch to using methanol as the mobile phase, will this make it easier or harder to separate the two proteins? Explain based on the table below and molecular interaction with the column.

Solvent List Methanol Water Acetone

Relative Polarity 2 1 0.5

Methanol, according to this table, is even more polar than water, so protein separation should be easier (e.g. there will be a greater difference in the elution times between the proteins in this system compared to the system in part a). This is based on the same molecular interactions as described for part a. .

(c) After optimizing your reversed-phased HPLC system, you run standard samples of protein X. After running the standards, three samples (A, B and C) of protein X of unknown concentration were run through the system as noted below:

Protein [X] (ng/ml) 1.0 7.0 15.0 25.0 50.0

Area Under the Curve (Arbitrary Units) 0.10 0.75 1.60 2.76 5.00

Unknown Samples

Area Under the Curve

A B C

0.5 1.2 3.2

Calculate the concentrations of each of the unknown protein samples. Make all steps explicit in the generation of your answer, including where in the process the Beer Lambert law is used.

First create the standard curve based on known concentrations and the area under the curve. The 4.9 ng/ml standard curve can be generated in this manner due to 11.76 ng/ml the Beer-Lambert law, which says there is a direct 31.37 ng/ml relationship between concentration and amount absorption at a given wavelength. Then use the equation of the line generated by the standard curve to calculate the unknown concentrations.

Unknown Samples A B C

Concentration

y = 0.1025x R² = 0.996

Series1 Linear (Series1)

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8.5 (a) What can you learn about adsorbed protein using an ELISA assay?

An ELISA can give information about the amount of protein present, whether that protein is in a specific conformation (i.e. it will not bind if the protein is denatured), and potentially the orientation of that protein. (b) How do you modify a sandwich ELISA to provide information about a protein adsorbed on a biomaterial surface?

Instead of coating a plate with the primary antibody and binding the protein of interest to the primary antibody, the protein is exposed to the biomaterial of interest and presumably binds (adsorbs) to that surface. The rest of the assay is very similar or the same: incubate with the secondary antibody, wash and add the substrate for the enzyme, quantify the color change. (c) You have two antibodies that are specific for two different regions of the protein that you are studying—one binds to the red region of the protein, the other to the green region. You test the adsorbance of your protein to a biomaterial surface using a modified ELISA. In each of the cases below, would the red antibody bind to the protein, the green antibody bind to the protein, both bind to the protein, or neither bind to the protein? Briefly justify each of your answers.

A: Neither—the protein has denatured, losing the specific shape that the antibodies could recognize.

B: The antibody to the green region would bind, but the antibody to the red region would most likely not be able to bind because the material is blocking the binding site.

C: Both antibodies should have access to their binding sites and the protein does not appear to have deformed, so both should bind.

D: The antibody to the red region should bind, but the antibody to the green region will not have access to its binding site.

8.6. Make a list of advantages and disadvantages of all of the surface patterning techniques described in this chapter.

Affinity chromatography: Advantages: sensitive to small amounts of protein; relatively easy to get quantitative data Disadvantages: cannot be used with proteins adsorbed to surfaces (only in solution) Colorimetric Assay: Advantages: may be used with proteins while on biomaterial surfaces; east to get quantitative data Disadvantages: may require pre-labeling protein with chromophore Fluorescent Assay: Advantages: may be used with proteins while on biomaterial surfaces; easy to get quantitative data Disadvantages: may require pre-labeling protein with fluorophore .

ELISA: Advantages: sensitive to small amounts of protein; may be used with proteins while on biomaterial surfaces; easy to get quantitative data Disadvantages: requires antibody to protein target Western blotting: Advantages: can be used to probe for multiple proteins in the same sample, since they are separated by size; relatively easy to get quantitative data Disadvantages: requires antibody to protein target; may be less sensitive than ELISA; cannot be used with proteins adsorbed to surfaces (only in solution)

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Chapter 9 9.1. Assume that in response to a signal, a cell eventually will secrete protein A. Explain each step that would lead to secretion of protein A, starting in the cell nucleus, and explain in what part of the cell each step occurs. Which single cell analysis technique is appropriate for each step?

1. Transcription in nucleus (formation of mRNA). Microarray/RNA-Seq would be most appropriate (for mRNA generated).

2. Translation on ribosomes in rough ER (mRNA, binds with tRNA, rRNA, and ribosome catalyzes bonds formation to create peptide chain one amino acid at a time).

Microarray/RNA-Seq or single cell proteomics would be most appropriate, depending on if you are interested in the RNAs or the protein product.

3. Post-translational modification in ER lumen and Golgi. Single cell proteomics would be most appropriate.

4. Protein A sent to cell surface in secretory vesicles. Single cell proteomics would be most appropriate.

9.2. (a) Yes-associated protein (YAP) is a transcription factor that is part of a signal transduction pathway that allows cells to respond to changes in their mechanical environment. Below are images of cells on two different surfaces ((i) and (ii)). YAP is shown in green and the nucleus is shown in blue. Based on the actions of transcription factors, which surface has activated this pathway?

Surface (ii) likely activates this pathway because the transcription factor has moved to the nucleus of cells cultured on this surface. Movement to the nucleus is required before transcription factors can bind to regions of DNA and alter their transcription into mRNA, which is the main action of transcription factors. (b) Would confocal microscopy be appropriate for this study? What additional information, if any, would it provide?

Confocal microscopy may be useful, but would not be required for this experiment, since the cells are cultured on a surface, rather than in 3D (such as in a porous scaffold or encapsulated in a hydrogel). Because the cells are cultured only on top of the surface (2D), confocal microscopy would not provide much more information than traditional fluorescent microscopy. (c) Would ChiP-Seq be appropriate for this study? What additional information, if any, would it provide?

ChiP-Seq could be quite helpful in this study. This technique could allow you to learn which portions of the nuclear DNA can be bound by YAP. 9.3. You have developed a new ceramic coating to be used on hip implants and test it in a series of in vitro assays. For each of the parameters listed below, describe 1) how you would measure

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the characteristic and 2) what outcome you would hope for in that particular assay to demonstrate that you have the ideal biomaterial for bone tissue engineering. (a) Cell cytotoxicity Cells could be cultured directly on the material (Direct Contact Assay). After 1-2 days the cells would be stained with a dye that indicates whether the cells are alive or dead (like the MTT assay) and imaged microscopically. An ideal biomaterial would cause minimal cell cytotoxicity, so we would hope to see the majority of the cells surviving. (b) Cell proliferation To study cell proliferation, cells would be cultured on the material for several days and the number of cells on the surface would be counted at several different timepoints. If cell numbers are increasing over time, we know that the cells are dividing. Cell proliferation could be advantageous for integrating the hip implant into surrounding tissues. (c) Cell migration A capillary tube assay could be used to measure the cell migration on the surface of the ceramic.

Cell migration could be beneficial in producing a uniform tissue or more even integration of the hip implant. (d) Cell differentiation Cell differentiation should be measured by looking for cell characteristics known to be present in desired cell types. This can include cell morphology or the expression of proteins known to be markers for the desired cell type. Alternatively, bone tissue can be detected by looking for excretions typical of osteoblasts (i.e. mineral deposition). The ideal hip implant would promote the differentiation of cells into osteoblasts, which would then produce the desired mineralized tissue for integration into the surrounding bone structure.

9.4. You are now investigating a polymer coating that inhibits leukocyte (pro-inflammatory cell) migration with the intent of minimizing inflammation for the stem of your hip implant. You choose to do a capillary tube test to observe leukocyte migration over 4 days. (a) Explain how a capillary tube test is used to measure cell migration.

A cell population is confined in a capillary tube. The opening of the tube is placed against a surface, and cells are allowed to migrate out of the tube over several days, resulting in a fan-like shape. The area of the fan-like shape is measured over time as a quantitative assessment of migration distance.

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(b) The capillary tube test gave the following results over 4 days for leukocytes on coated and uncoated surfaces. Would you recommend using the polymeric coating to inhibit leukocyte migration?

Inhibitory Coating: Area of Cells (mm2) 0 2 3 7 10

Time (d) 0 1 2 3 4

Uncoated Control Surfaces: Area of Cells (mm2) 0 3 7 35 52

Yes, the inhibitory coating appears to inhibit leukocyte migration. The cell area on the inhibitory coating is consistently less than the area on the control surfaces, indicating a reduction of leukocyte migration on coated surfaces. (c) What is the major drawback to this study as a measure of cell migration over 4 days?

Over 4 days, it is difficult to separate the role of proliferation versus migration in the results.

Proliferation can also account for an increase in area, and proliferation is likely to play a significant role over the course of several days. (d) To determine to effect of the coating on other cell types, migration of leukocytes, endothelial cells (to form blood vessels), and osteoblasts (bone-forming cells) were compared on polymercoated surfaces by capillary tube tests. Do these results support the use of this coating for the stem of a hip implant? Justify your answer using the table below. (6 sentences or less) Time (d) 0 1 2 3 4

Leukocytes on Coated Surfaces: Area of Cells (mm2) 0 2 3 7 10

Endothelial Cells on Coated Surfaces: Area of Cells (mm2) 0 2 2 5 7

Osteoblasts on Coated Surfaces: Area of Cells (mm2) 0 1 3 6 9

These results do not support the use of this coating. The inhibitory polymer coating appears to inhibit migration of all cells and not just leukocytes. In the stem of a hip implant, the polymer coating would inhibit migration of endothelial cells and osteoblasts into the stem, thus impeding vascularization and integration of the implant with surrounding bone.

9.5. After the hip implant, you have now decided to develop a different polymer for use as a ligament replacement. Your new material is synthesized by polymerizing monomer X. However, after testing, you found that this polymer has very poor ligament fibroblast cell adhesion. Since you know that fibroblasts like to attach to the extracellular matrix protein collagen, you decide to make a copolymer of X and a monomer Y. Monomer Y has a pendant .

group that will you believe will interact electrostatically with collagen after (co)polymerization and processing. (a) You first tested the contact angle of the copolymers and obtained the results in the following table. Based on these data, what are the relative hydrophophilicity of monomer X and monomer Y? Why? Polymer composition (%Y by weight) 5 10 25 50 75 100

Water contact angle (degree) 120 100 80 60 50 45

Polymer Y is more hydrophilic. When the % Y increases, there are more of these groups existing on the surface of copolymer, which will decrease the water contact angle. This is probably due to the fact that monomer Y is charged since we know it will interact electrostatically with collagen.

(b) You then immerse each of these materials in a collagen solution and rinse gently in water.

You then test the contact angle again obtain the results below. Explain why the water contact angle decreases for all compositions after exposure to collagen. Why is a plateau in contact angle seen?

Polymer composition Water contact angle (%Y) (degree) % Spread fibroblasts* 5 100 10 10 80 30 25 60 50 50 31 90 75 35 90 100 33 90 *This refers to the % of spread cells (e.g. % cells on surface that have a surface area above a threshold value).

The contact angle decreases because of the binding and existence of collagen on the surface of the copolymer. The collagen is more hydrophilic than polymer composition X and Y. Collagen has a high molecular weight. The appearance of plateau may be due to the saturation of bound collagen on the surface. There is not enough space on the surface for binding more collagen molecules, even if there is a greater affinity for the collagen at higher %Y. (c) It was found that 30% fibroblasts spread on the copolymer with 10% monomer Y after treatment with the collagen solution. On the table above, estimate the value of % spread cells on the other surfaces after treatment will the collagen solution. Briefly justify your answer. .

The collagen will facilitate the spreading of fibroblasts. When more collagen molecules are on the surface, a greater percentage of adherent cells will spread. When the amount of collagen molecules have reached saturation on the surface, the percentage of spread cells reaches the maximum and plateaus since there are only so many collagen molecules to bind to. (d) Explain what is happening within the cell to cause cell spreading after the cells are placed on the 100% Y surface.

Cells extend finger-like pseudopodia along the substrate surface. Then integrin receptors in the cell membrane can interact with collagen ligands on the material surface to anchor the cell in place. Spreading then occurs through internal cytoskeletal rearrangement, which results in an overall change of cell shape. (e) Finally, you decide to take the collagen-treated materials and rinse them with a mixed protein solution, rather than water (better representing what would be found in the body), before repeating the experiments above. You obtain the new results below. Fill in your predictions for the rest of the % spread cells column and justify your predictions. What is different versus the case for 5c)?

Polymer composition Water contact angle (%Y) (degree) % Spread fibroblasts* 5 100 5 10 82 5 25 59 5 50 30 5 75 36 5 100 33 5 Because the collagen was only bound through electrostatic interactions, protein exchange occurred with other proteins that are now bound to the surface, but do not interact with fibroblasts to produce spreading. Only a little collagen remains to induce spreading, regardless of copolymer composition. (f) Based on all this data, would you recommend copolymer XY for use as a ligament replacement? If yes, justify your answer. If no, suggest a means to modify this material so it is more suitable for this application.

Should use higher affinity interaction with collagen or covalently tether collagen to surface so no protein exchange occurs.

9.6. Below is a summary of data from a RNA-Seq experiment in which hepatocytes (liver cells) were cultured in 2D (on surfaces), in 3D as cell aggregates (SP-3D), or in 3D on a porous scaffold (ST-3D). In this graph, genes that were upregulated are shown in red, those that were downregulated are shown in green, and those that did not change are shown in grey. Which samples are the most different based on this data? Give a possible explanation based on impact of culture conditions on cell function. .

2D vs SP-3D are the most different (most up and down regulated genes between these two groups).

In SP-3D, there is a great deal of cell-cell contact vs in 2D, there is more cell-material contact, which likely means different types of receptors are being activated, which could lead to changes in downstream gene expression between these two groups.

9.7. This study was conducted with different surface modifications to poly(methyl metacrylate) (PMMA), which is used as a bone cement, among other things. (a) Based on the images below, on which surface epithelial cells prefer to adhere? Cells are stained green.

PMMA-N (b) Subsequently, the cell-laden surfaces were characterized by proteomics. Based on the results below, what is one possible explanation as to why more cells are seen on the surface you chose for (a)?

There are a number of possible answers. However, one explanation could be that a higher percent of the proteins found in the PMMA-N samples were adhesion proteins (12%). Also, there were no apoptosis proteins seen in PMMA-N samples, suggesting less cell death. Overall, more adhesion and less death would result in higher cell numbers on the PMMA-N samples.

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Chapter 10 10.1 List the 4 cardinal signs of inflammation and explain their cellular and molecular causes.

1. Rubor (redness) 2. Tumor (swelling) 3. Calore (tissue heating) 4. Dolore (pain) Rubor (redness) and calore (tissue heating) are results of vasodilation. Tumor (swelling) is a response to increased capillary permeability, clotting, large numbers of granulocytes and monocytes, and cell swelling. Dolore (pain) is a by-product of the release of kinins from the blood clotting cascade.

10.2. Explain the main functions of each of the 4 main types of leukocytes.

1. Granulocytes. These cells can be subdivided into neutrophils, eosinophils and basophils. Their main function is to phagocytose foreign invaders and aid in the inflammatory response.

2. Monocytes. These cells play a central role in the inflammatory response.

3. Lymphocytes/plasma cells. These cells, including T and B cells, function as a part of the acquired immune response. Lymphocytes can be divided into memory cells, which are responsible for a more rapid response if one is exposed to the same pathogen for a second time, and effector cells, which produce antibodies or take other actions to remove foreign invaders.

4. Megakaryocytes. This cell type fragments to produce platelets. These nonnucleated fragments circulate in the blood and aid in blood clotting.

10.3. Explain the role of integrin receptors in transendothelial migration.

Integrins provide the high-affinity interaction between activated neutrophils and endothelial cells: stops rolling  cell arrest on the endothelium, which allows diapedesis through the endothelial cell junctions. (From Chapter 9: Integrins can also promote cell binding to ECM proteins in the tissue, which facilitates migration of the cell from the vascular endothelium (blood vessels) to the wound site.)

10.4.(a) Complete the following table to explain the difference between monocytes and macrophages. Specifically address differences in cell origin/formation, location where these cells appear in the body, time of occurrence in the inflammatory process (hours/days/weeks?), and cell function.

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Monocytes Produced in bone marrow from hematopoietic stem cells/

Macrophages Monocytes enlarge to become macrophages

Location in the body

Bloodstream

Tissue

Time of occurrence in the inflammatory process

Hours after injury

Days (to weeks) after injury

Cell function

Phagocytosis Secretion of chemical mediators Secretion of active species (eg free radicals)

Phagocytosis Secretion of chemical mediators Secretion of active species (eg free radicals) Antigen-presenting cells

Cell origin/formation

(b) What is the main difference in M1-like and M2-like macrophages?

The main difference is in their function. An M1 macrophage’s functions include pathogen clearance and release of reactive oxygen radials, as well as secretion of more pro-inflammatory cytokines. An M2 macrophage’s functions include extracellular matrix deposition, vascular remodeling, and secretion of more anti-inflammatory cytokines.

10.5. List 3 parameters that can be tested as indicators of neutrophil activation and describe why each of these things are important for the overall function of neutrophils in the inflammatory response.

Possible answers include: Adhesion and spreading – Following adhesion, spreading increases neutrophil surface area for better contact to encourage the possibility of phagocytosis.

Viability – Viability is needed for all functions.

Migration – Neutrophils must migrate to site of injury to function locally.

Cell surface markers – Certain receptors/ligands are upregulated during neutrophil activation to improve extravasation.

Cytokine secretion – Particular cytokines are secreted to encourage migration of other cell types and to activate the acquired immune system.

10.6. You are developing a coating for a vascular stent. You decide to test the inflammatory potential of your material by culturing with macrophages. After 48 hours of culture, the culture media was collected to measure the expression levels of various cytokines using ELISA. The following table shows the absorbance data of several Interferon γ standards of known .

concentrations, the media from cells cultured on the new polymeric coating, and the media from cells cultured on the stent without coating (control). (a) Calculate the concentration of Interferon γ in the media from cells on the polymeric coating and control. (Standard curve should be used to calculate exact concentrations.)

Concentration (pg/ml) 5 10 25 50 100 200

Sample standard-1 standard-2 standard-3 standard-4 standard-5 standard-6 polymer coating control

Absorbance 0.322 0.345 0.411 0.542 0.767 1.253

58.0

0.575

78.9

0.675

1.4 y = 0.0048x + 0.2965 R² = 0.9998

1.2

absorbance

1 0.8 0.6 0.4 0.2 0 0

50

100

150

200

250

Concetration (pg/ml)

(b) You also completed ELISAs in the same manner to test the level of other inflammatory cytokines, as shown below. From the point of view of the inflammatory response, would you recommend continuing to develop this material for a vascular stent? Justify your answer based on all of the ELISA data collected. polymer coating control (pg/ml) (pg/ml) IL-12 30 28 TNFα 390 400 IL-1 250 260 Comparing these data, overall, the polymer coating will induce similar inflammation as the stent without coating. As such, the polymer coating will not induce more severe inflammation than stent without coating, which means this polymer coating is reasonable for the application. .

Chapter 11 11.1 Below is an image that depicts fibrous encapsulation of an implanted biomaterial. (a) Next to the numbers below, fill in the names of the cells involved in each stage depicted.

(Each different number represents a different cell type.)

1. Endothelial cell 2. Neutrophil 3. Monocyte 4. Macrophage 5. Foreign body giant cell 6. fibroblast (b) What additional cells would be found in the diagram if the implant caused a chronic inflammatory response?

The main hallmark of chronic inflammation is the presence of lymphocytes, so these cells would need to be present in addition to those listed above.

11.2 What type of resolution occurs with each of the following? (a) degradable drug delivery device (b) in-dwelling vascular shunt for dialysis (c) splinter (d) degradable suture (e) cardiac pacemaker (f) titanium bone plate (a) resorption (b) encapsulation (c) extrusion (d) resorption (e) encapsulation (f) integration 11.3 You are testing a new tissue engineered skin substitute in vivo. What are the most important assays to include to determine if your approach results in repair or regeneration of the skin?

Histological and biochemical assays are the most important to distinguish between repair and regeneration. In both cases, the mechanical properties and other functional properties may be restored. However, only regeneration results in a new tissue with the same biochemical and structural properties as the original tissue.

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11.4 In experiments to evaluate tissue responses to a biomaterial, a control group is often included involving the complete surgical procedure but without implanting the biomaterial. (a) Why is the surgery-only group an important control for these studies?

Surgery by itself will result in tissue injury and cause some inflammation, independent of biomaterial implantation. In this study, inflammation following implantation can be compared to surgery-only controls to determine how much inflammation results from the biomaterial itself.

Additionally, the defect alone will allow assessment of the animal’s native ability to self-heal, without the experimental intervention involving the biomaterial. (b) List the differences in the tissue response at the wound site following implantation of a nondegradable biomaterial as compared to a simple lesion.

Both situations would result in acute inflammation from tissue injury and wound healing response. Unlike a simple lesion, biomaterial implantation often results in a foreign body reaction, the hallmark of which is the presence of foreign body giant cells that attempt to phagocytose the foreign biomaterials. Macrophages and foreign body giant cells may remain surrounding the material for the lifetime of the implant. Granulation tissue maturation around the implant results in fibrous encapsulation, rather than collapse and formation of a scar, which occurs with a lesion.

11.5 You have developed a novel non-degradable biomaterial for controlled drug release. In vitro experiments on a thin slab (20 mm x 10 mm x 2 mm) showed that your material has the release profile shown below (solid line) with linear release over ~30 days. Following implantation in vivo, however, the release rate into the surrounding tissue was severely decreased within the first 4 days (dashed line) and zero release occurred after 12 days. (a) What is the most likely explanation for the difference between the in vitro and in vivo release rates?

It is most likely that fibrous encapsulation of the implant in vivo is limiting diffusion of drug into the surrounding tissue. Fibrous encapsulation is the final stage of healing for nondegradable implants, in which the biomaterial prevents collapse of the capsule surrounding the implant and subsequent scar formation. A fibrous capsule would act as a barrier to diffusion of drug out of the device, thus slowing release. (b) Your company is investigating a degradable polymer as an alternative to the non-degradable material above. In-vitro release studies show the same results as the non-degradable polymer (solid line on graph above). However, when implanted in vivo, it was observed that the polymer quickly (<3 days) formed particulates at the implantation site. Draw the expected drug release from the degradable polymer in vivo and explain why you drew the graph this way.

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Drug Release Rate (ng/hr)

120 100 80 60 40 20 0 0

5

10

15

20

25

30

Time (days) Formation of particulates would increase the thickness of the fibrous capsule that forms. A thicker capsule would cause drug release rate to decrease even more rapidly in vivo.

11.6. Make a list of advantages and disadvantages of all of the in vivo assessment techniques described in this chapter.

Histology/Immunohistochemistry: Advantages: visualization of the tissue reaction at various stages through the identification of specific cell types and/or extracellular matrix molecules Disadvantages: difficult to quantify Transmission Electron Microscopy: Advantages: high magnification; used in conjunction with X-ray analysis, may also be used to determine what types of materials have leached from the sample Disadvantages: difficulty in obtaining the ultrathin sections required; must be imaged under vacuum, so must fix and dehydrate the specimens Scanning Electron Microscopy: Advantages: high magnification; does not require sectioning of the sample Disadvantages: usually must be imaged under vacuum, so must fix and dehydrate the specimens Biochemical Assays: Advantages: quantitative; large number of samples can be examined at one time Disadvantages: requires knowledge of what biomolecules involved in response Mechanical Testing: Advantages: quantitative; may be used to assess tissue integration Disadvantages: consistent sample handling required to allow for comparison of data between studies .

Chapter 12 12.1 Name the four major characteristics of acquired immunity and give a brief explanation of how each is achieved at a cellular/sub-cellular level.

Specificity – each antibody or TCR generated is very specific, and each cell is specific to only one antigen Diversity – rearrangement of genes so each cell has different antibody or TCR Memory – clonal expansion after activation, leading to formation of memory lymphocytes Self/non-self recognition – apoptosis of self-reacting cells during lymphocyte maturation 12.2 Based on what was presented in this chapter, trace the fate of a virus protein that has infected a host cell, as well as the host cell. Including all major cell types and biomolecule types that would respond to this infection.

Virus proteins are reproduced by the cell. As a part of this, the protein is transported to the ER and Golgi, where some of the protein is complexed to MHC I molecules. The resulting complex is trafficked to the cell surface, where the viral peptides are presented. This activates NK cells, or Tc cells, which then expand and mature to CTLs. In either case, perforins are released that lyse the infected cell. Recruited granulocytes then phagocytose cell debris and remaining virus.

12.3 (a) Describe the role of Th cells in humoral and cellular immunity. What cells do Th cells interact with in each case and how do they interact?

Th cells secrete cytokines to activate Tc cells (cellular immunity). They also secrete cytokines to co-stimulate B cells (humoral immunity), leading to their activation. Alternatively, they can bind directly to B cells to provide a co-stimulatory signal. (b) Specifically, explain the role of Th1 vs Th2 cells in humoral and cellular immunity.

Th1 cells usually respond to endogenous antigens and produce cytokines that activate CTLs and induce B cells to produce antibodies that support phagocytosis by granulocytes.

Th2 cells respond to exogenous antigens and secrete cytokines that activate B cells.

12.4 What are the four major actions of antibodies? Which one(s) is likely at play in the response to a bacterial infection? Justify your answer.

Agglutination Precipitation Neutralization Lysis Of those listed above, depending on the type of bacteria, they may be aggulintated so they can’t grow or migrate, or they may be directly lysed. Neutralization is also a reasonable response, depending on the exact actions of the bacteria in the host. .

12.5 Stem cells have been proposed as therapies for many diseases. However, a problem with using stem cells from another person as a therapy is that they can be recognized as foreign and activate the complement system, leading to cell death. a) What pathway is most likely involved in activation of the complement system in this case?

Why? The best answer is: Classical Foreign proteins on the surface of the non-native cells will be recognized by antibodies and start the complement system via the classical pathway.

Also, mammalian cells have enzymes on their surfaces that deactivate C3b, making the alternative pathway much less likely (this is different for synthetic biomaterials- make sure to review how the complement system is activated by biomaterials) b) You have decided to engineer the surface of the stem cells to present an additional biological molecule to help reduce complement activation. What molecule would you put on the cell surface? Justify your answer by relating how this surface molecule interacts with the complement system to cause the desired result.

2 general categories of answers best (could choose either): 1.

Something that blocks antibodies from binding to cells and attaching to C1, which would prevent any of the cascade from starting.

2.

RCA proteins/DAF, or antibodies to a particular complement protein – all of these can stop the cascade at specific points through mechanisms that prevent the cleavage of the next enzyme in the cascade c) Given all the assays so far described in this book, choose one assay to determine if your solution in b) was effective. Describe what outcome you would expect if your solution worked.

2 general categories of answers were best (could choose either): 1. Cytotoxicity assay. Since complement ends in formation of MAC, which lyses cells, if the solution in b) was effective, would see less cell lysis than without treatment. Need details on how a particular cytotoxicity assay works for full credit.

2. ELISA for a product of the complement cascade (such as C3b or C5b). If one stopped the cascade, would expect less cleavage products to be found via ELISA. Need details on how ELISA works for full credit.

12.6 Based on your knowledge of both the innate and acquired immune response, if you wanted to prevent a type IV hypersensitivity reaction, which one cell type would be most important to block and why?

A variety of answers are possible, but the best would likely involve either blocking the step of formation/actions of the TDTH cells as they are key in promoting the hypersensitivity response after the 2nd exposure, or blocking the actions of macrophages, since they cause the tissue irritation/damage seen in type IV responses. .

12.7 You are developing a new material as a part of a synthetic replacement heart valve. (a) Describe the Tier 1 assays would you choose as a first-pass screen for immunotoxicity.

Many Tier I assays are designed to monitor proliferation of lymphocytes in response to an appropriate stimulus that would normally lead to clonal expansion. These are done with B cells and T cells, with the chemical stimuli specific for each. In addition, since NK cells are known to attack tumor cells, a tumor cell line is placed with NK cells in the presence of the biomaterial or extracts, and amount of lysis of the tumor cells is monitored as a measure of the effectiveness of the NK cell response to stimuli. (b) What results would cause you to complete Tier 2 assays and which assay(s) would you choose? Justify your choice.

Variety of answers possible.

T and B cells – results from Tier I would show a lack of proliferation. In this case, Tier 2 would look for cell activation through surface markers, or via ELISA for particular cytokines secreted by the activated cells. If an undesired result from the Tier I assay occurs, then it is important to see if the cells were already stimulated by the biomaterial, or if their activity is suppressed by the biomaterial. Migration assays are also possible, although perhaps would provide less direct results than assays for cell activation.

NK cells – results from Tier I would show lack of ability to lyse tumor cells. Again, surface markers or ELISA as a measure of cell activation would be the preferred Tier 2 assay.

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Chapter 13 13.1. You are developing a polymeric material as a vascular graft. A method explored in the past to reduce blood coagulation is to pre-adsorb the graft with albumin, a non-coagulatory protein found in high concentrations in the blood. (a) Based on your knowledge of protein adsorption, what is the rationale for using albumin to reduce blood coagulation?

If the graft is pre-adsorbed with a protein that is already found in high concentration in the blood, it has already reached equilibrium and is unlikely to desorb. Exchange is possible, but, based on concentrations in the blood, at short times, this would likely be with more albumin. Thus, many of the “spots” on the surface where pro-coagulatory proteins would attach are already taken with protein that will not promote coagulation.

(Note: eventually, all of the albumin may be replaced by pro-coagulatory factors with high affinity (Vroman effect), so, unless the albumin is covalently linked to the surface, this may not be a good strategy in the long term). (b) Below are images taken after exposing platelets to both albumin-coated and non-coated poly(styrene) (PS) surfaces. In addition to images, you want to perform a biochemical assay to determine the activation state of the platelets. What one assay would you choose? What would you expect the results to be on both surfaces? Does it appear that the albumin modification worked? Explain based on characteristics of activated platelets.

Variety of answers possible, although they should be related to ELISAs for factors activated platelets produce (ADP, thrombin, thromboxane A2) or flow cytometry for changes on platelet surface (expression of GPIIb/IIIa).

Neat PS surface should show higher levels in these assays, since spreading is an indicator of platelet activation. Yes, it appears that the albumin coating worked, at least in terms of reducing platelet activation. (c) If you put these two materials under dynamic flow, would you expect any changes the results for either surface? Why or why not?

Variety of answers possible. Possibilities include that flow could shear off the platelets, so could see less on either surface. Similarly, the flow could dilute the concentration of the biomolecule chosen for (b) and make it undetectable. (d) Poly(styrene) is an uncharged, hydrophobic polymer. What is the most likely mechanism by which the coagulation cascade would be started for this material in vivo? Would it be the same mechanism for the albumin-coated material? Justify your answer.

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Since the PS is uncharged, the likely mechanism is the extrinsic pathway, where tissue factor is released by nearby endothelial cells or macrophages. Albumin could add some charge, so intrinsic may play more of a role for the coated material.

13.2 Explain the role of each of the following molecules after injury to a blood vessel. (a) thrombin Cleaves fibrinogen to fibrin, which are main basis of thrombus (b) plasminogen After conversion to plasmin by tissue plasminogen activator, cleaves thrombus as a part of vessel healing (c) α2-macroglobulin Soluble inhibitor of thrombin and plasmin (d) tissue factor Release of tissue factor by macrophages or endothelial cells is the main means of activating extrinisic pathway of blood coagulation (e) Factor XIII Crosslinks fibrin fibers to form mature thrombus 13.3 Why are calcium chelators (agents that bind calcium tightly) often used as anti-coagulants?

Calcium is a key co-factor in a number of steps of each part of the coagulation cascade, so chelating them with an agent that binds tightly will not allow them to bind to their proper substrates, and the coagulation cascade is stopped. Examples include the need for calcium in the last steps of both the intrinsic and extrinsic pathways, along with the need for calcium in formation of thrombin in the common pathway (see Figure 13.1).

13.4 One approach to reduce blood clotting in vascular grafts is to attach heparin-mimetics to the interior surface? Give two potential reasons why this approach might be effective.

Heparin is can bind to anti-thrombin III and increase its effectiveness in binding thrombin and thus acts as a chemical anti-coagulant. In addition, mimicking the heparin sulfate on the endothelium, which makes the graft smoother and thus less likely to adsorb coagulation factors and stimulate coagulation.

13.5. Describe the advantages and disadvantages of in vivo vs. ex vivo testing for a biomaterial intended for use in a blood-contacting application.

In vivo: Advantages: can reproduce exact size, shape, and location of final implant Disadvantages: animal blood chemistry different from humans; hard to control exact blood flow around implant from animal to animal Ex vivo: Advantages: can control and monitor blood flow around implant; easy to take blood for biochemical analysis Disadvantages: cannot reproduce effects of surgical trauma; animal blood chemistry different from humans

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Chapter 14 14.1 What are the major steps in implant infection? Which one would you target in order to reduce device-related infections and why?

4 major steps are: attachment, adhesion, aggregation, dispersion. Usually attachment or adhesion is targeted to reduce implant-related infection. Infections become hard to control after biofilm formation begins.

14.2 You company is developing a new material for a urinary catheter that includes sustained release of antibiotics to reduce infection. Your colleague has performed a zone of inhibition study with several samples (3 formulations, each releasing different antibiotics + 1 control polymer with no antibiotic), but mixed up the labels on the samples. The results of the study are found below. (a) Which sample is likely the control and why?

PS is likely the control as it shows no inhibition. (b) Of the remaining samples, which of the formulations worked the best and why?

SN seemed to be the most effective in that the zone of bacterial inhibition was the largest.

Bacteria

E. coli

Material

Inhibition Zone (mm) Mean ± SD

MR

1.0 ± 2.8

NF

3.1 ± 6.1

SN

9.3 ± 10.0

PS

0.0 ± 0.0

14.3 You are developing a textured titanium-based material to potential reduce orthopaedic implant-related infection. You first characterize the width of the grooves produced on the surface of each sample type using atomic force microscopy (shown as “peak to peak distance” on the graph below). You then perform a static bacterial adhesion assay on each of these samples over 24 hours. Results are found below. (Notes: bacteria are generally 1-2 µm in diameter; fraction of attached bacteria refer to the ratio of amount attached/amount introduced into the system at time = 0.) (a) Explain the effects of groove size on bacterial adhesion. In particular, what may be occurring at groove widths of 0.5 µm and 5 µm to produce the data shown above?

0.5 µm: Grooves are smaller than individual bacterium, so it is difficult to make lasting contact with the surface, leading to lower attachment.

5 µm: Grooves are larger than individual bacterium and they can get caught in the troughs. They may then proliferate, which may explain the ratio of >1 over 24 hours. .

(b) Which size of grooves would you choose to try in vivo? Design an in vivo experiment (including proper controls and outcome measures) to test ability of this surface to reduce implant infection.

One of the samples with 0.5 um would be most promising to test in vivo due to low bacterial adhesion in vitro. In vivo studies could take many forms. The material could be implanted in a rodent in a cage system, or a complete device using this material may be implanted in larger animal models. Bacteria may be introduced via direct placement at the implant site or through injection into the bloodstream. In all cases, a control of the same material without the texture should be included as a control to determine the effect of the texture.

Outcomes: There are many possible choices. Histology or SEM of explants at various time points to examine level of bacterial adhesion. In addition, the device can be explanted, washed, and the bacteria that are extracted from the surface can be counted. The animal’s blood may be tested over the course of the experiment for leukocyte and lymphocyte counts. The amount of bacterial cell wall antibodies found in the blood at different times may be quantified via ELISA techniques, or, a LAL assay could be used if one is interested in gram-negative bacteria.

14.4 (a) What is the main difference between chemical and foreign body carcinogenesis?

The factor that initiates the malignant transformation is the main difference.

Chemical: a chemical leached from a biomaterial is the carcinogen that causes malignant transformation, thought to occur due to mutations in the DNA. Foreign body: a solid material with no chemical carcinogenic activity was observed to cause tumor formation. Exact mechanisms behind this are still under investigation. (b) Would an in vitro assay to detect chromosomal aberrations usually detect chemical, foreignbody or both types of carcinogenesis? Justify based on how this assay works and how that relates to changes that occur in both types of carcinogenesis.

It is likely that this assay would detect changes from chemical carcinogens (if they were large enough to affect the entire chromosome). However, since foreign body carcinogenesis requires implantation in an animal or human before tumors appear, it is unlikely that such an in vitro assay would be a good screening tool for this type of carcinogenesis.

14.5 List 3 proposed mechanisms of how nano-scale materials may cause malignant transformation of a cell.

Many answers possible: 1. An over-expression of reactive oxygen species and pro-inflammatory cytokines by granuloytes interacting with these particles could cause malignant transformation in nearby cells.

2. Small fibers can penetrate the cell membrane and directly cause direct mechanical damage to the nucleus.

3. Small particles they can interact with specific proteins on the cell surface leading to malignant transformation. .

4. Nanoparticles are relatively easily endocytosed compared to larger particles. Once in the cell, a number of interactions can occur with intracellular proteins and organelles that could lead to malignant transformation.

5. They can promote the cell to increase reactive oxygen species, which can then diffuse into the nucleus and damage DNA. 6. If the particles enter the nucleus, they may cause intranuclear protein aggregation that alters DNA replication, or decrease expression of DNA repair genes.

14.6 You are developing a new titanium dioxide (TiO2) sunscreen. It is known that this type of sunscreen provides protection by scattering UV rays, so that less penetrates to the skin, where they can potentially cause mutagenesis. You are developing a new carrier for the TiO2 to further improve scattering properties and you want to determine if this provides additional protection to underlying skin cells. (a) You will be using an Ames test. Explain the principles behind this assay and how you would set up the assay to test your material, including proper control groups.

This technique employs a mutant bacteria line that requires the amino acid histidine for growth.

The bacteria are cultured with the samples in histidine-free media, so only the bacteria able to mutate to a non-histidine-dependent phenotype will be able to survive and proliferate. The number of viable colonies are then quantified to provide an indicator of the mutagenic potential.

In this case, the UV light only should be used as a control, since that is what induces mutagenesis. The experimental samples with only TiO2 and TiO2 + carrier should also be cultured with UV light to see to what degree each sample decreases mutagenesis from the control. (b) You find the following results: Sample Control (UV only) TiO2 + UV TiO2 + carrier + UV

Number viable colonies 3000 1300 500

Based on the results above, did your new carrier work?

Yes. Colonies are only viable in this assay if mutagenesis has occurred, so you would see fewer viable colonies if there was more protection from the mutagenic UV light. In this case, you see 6 times fewer colonies with the carrier than with UV alone, and nearly half as many viable colonies as you did with the TiO2 alone.

14.7 Which part of pathologic calcification is most likely to occur faster in younger patients?

Justify your answer.

In younger patients, metabolism favors deposition of calcium phosphate in bones, so those systemic conditions would promote (increase the rate) of mineral deposition in pathologic

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calcification as well. (It is more likely to affect this part, rather than the rate of crystal growth itself, which depends somewhat more on the local ECM template in the implanted material.)

14.8 You are investigating two hydrolytically-degradable polymers as potential scaffolds for bioresorbable vascular grafts. You implant grafts made from a slow degrading (SD) and fast degrading (FD) polymer in mice. After 8 weeks, you retrieve the samples and complete histology on cross-sections of the blood vessels using von Kossa stain, which stains mineralized matrix black. (Note: arrows show inner and outer layers of the blood vessel in the images below.) (a) Based on the images above, which polymer would you recommend to use? Justify your answer based on the final application for this material.

The slow degrading polymer is worse for pathologic calcification as shown by von Kossa staining (more mineral deposition) and should not be used. Calcification can make the material brittle, which is not desired for a blood vessel that must be able to withstand pulsatile blood flow (needs to be elastic). (b) Matrix Gla protein (MGP) is a known natural inhibitor of calcification, which is thought to act, in part, by binding to the ions and thus preventing deposition of calcium phosphate mineral in tissues. What would you expect if implanted both of these materials in an individual with a genetic mutation that prevented production of MGP? Both would likely show more mineralization, since there is less inhibition of mineral deposition overall in these patients.

14.9 Make a list of the advantages and disadvantages of the assessment methods described in this chapter for assays of pathologic calcification.

Biochemical Assays: Advantages: quantitative Disadvantages: destructive Histology/Immunohistochemistry (some of these from previous chapters where this technique explained): Advantages: easy visualization of the tissue reaction at various stages Disadvantages: can be difficult to quantify; destructive Scanning/Transmission Electron Microscopy (some of these from previous chapters where these techniques explained): Advantages: high magnification Disadvantages: usually must be imaged under vacuum, so must fix and dehydrate the specimens; destructive X-ray Diffraction: Advantages: gives information about structure of the crystals formed Disadvantages: destructive

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X-ray Imaging: Advantages: non-destructive; can be quantitative; may be used to assess tissue integration Disadvantages: difference must exist in the X-ray attenuation coefficient to visualize implant; only 2D image generated Micro-computed tomography: Advantages: non-destructive; 3D images provided; high spatial resolution; can be quantitative; Disadvantages: difference must exist in the X-ray attenuation coefficient to visualize implant; equipment expensive

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