Biomaterials The Intersection of Biology and Materials Science Johnna S. Temenoff Solution Manuals.p

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Type:

Solution Manual

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

1st Edition

Author(s):

Johnna Temenoff Antonios Mikos

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Chapter 1 1.1 One common biomaterial application is the construction of an arterial graft, a device that replaces a section of an artery. An artery is a flexible blood vessel that can withstand varying pressures and regulates the flow of blood. Arteries also provide a smooth interior surface to inhibit blood clotting within the vessel. a. You need to design a vascular graft. List some advantages and disadvantages with each of the three major types of biomaterials. Which would you choose for this application?

Answer: Advantage Ceramics Strong Strong Easy to shape Metals Inexpensive and available Can be flexible, smooth

Disadvantage Rigid Brittle Inflexible

Polymers A polymer would be more suitable for the vascular graft application, since neither metals nor ceramics offer the necessary elasticity. b. What specific material characteristics need to be considered for the arterial graft application?

Answer: Flexibility is very important, and the material should also possess a certain tensile strength. The surface properties of the material like smoothness and hydrophobicity must be also considered in terms of its ability to support adhesion of endothelial cells or not cause damage to platelets. Moreover, the material should be easy to shape. c. Would you use a natural or synthetic material for this application? What are the advantages and disadvantages of each?

Answer: A natural polymer would be more likely to integrate into the surrounding tissue.

However, natural polymers may not possess the necessary mechanical properties for this application. There is also the possibility of evoking an immune response or a pathogen transmission. Synthetic polymers could be manufactured to have the necessary mechanical properties but may not integrate well with native tissue. Either natural or synthetic material is acceptable with proper justification.

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1.2 Various biomaterials can be used for joint replacement applications, such as hip implants (see Fig. 1.4). A hip joint replacement must withstand large forces (standing on one leg results in a load of 2.4 times body weight on the femoral head [19]; jumping and running generate higher forces) normally transferred through the hip joint. It must also allow for proper rotation of the joint. a. Which of the three major types of biomaterials would you use for the femoral stem?

Why? Answer: Metals and ceramics both may be present in the stem of a hip replacement. Metals are typically used in the core of the stem because of their strength and because they are easily molded into complex shapes. Ceramics may be used on the surface of the stem to facilitate integration with the bone. b. Would integration of the femoral stem with the surrounding tissue be an acceptable biological response? Why or why not?

Answer: Integration of the stem with the surrounding bone would be desirable, as it would better fix the implant into the surrounding tissue. Integration at the joint space would be undesirable, as it could impair movement.

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Chapter 2 2.1. You are evaluating a variety of new materials for use in the femoral stem of a hip joint replacement. a. Diagrams of the crystal structures of materials (a) and (b) are shown below. Identify each crystal structure. For each structure calculate the coordination number, derive the relation between r (radius of sphere) and a (length of cube side), and determine the APF of each.

(a)

(b)

Answer: (a) is a simple cubic unit, and (b) is a body centered cubic. (a): Coordination number is 6 (number of nearest neighbor atoms). r/a relation: a=2r (determined by examining edge of cube) Volumeofatomsinunitcell APF = unit cell volume The number of atoms in the unit cell is 1(

4 8x1 8 ), therefore the volume is πr3 3

Unit cell volume is a 3 = 8r 3 4 πr3 π Therefore: APF= 3 = 6 ≅0.52 3 8r (b): Coordination number is 8. r/a relation: Make a triangle that consists of an edge of the cube, a diagonal along the face of the cube, and a diagonal through the cube. The dimension of the edge is a, the dimension of the diagonal through the cube is 4r, and the diagonal through the face (x) needs to be calculated.

To do this, make another triangle on the face of the cube, with 2 sides being a, and the diagonal being equal to what we are looking for (x). Using Pythagorean Theorem, this can be calculated:

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a

4r

a

x

a

x

Diagonal:

Side:

x

a

a

x

4*r

a

a2+a2=x2 x=a 2 Now use Pythagorean Theorem again on the initial triangle.

( )2

a 2 + a 2 = (4r )2 a2 + 2a2= 16r2 3a2= 16r2 4r = a 3 APF = Volumeofatomsinunitcell unit cell volume

⎛4 3⎞ The number of atoms in the unit cell is 2, therefore the volume is: 2 ⎜ πr ⎟ ⎠ ⎝3

Unit cell volume is a3=

Therefore: APF =

3 (4) r

8 3πr3 64r3

3 =

3

π 3 8

≅0.68

3 3

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b. You are given a third material (c), which has not been characterized. What analytical technique would you use to determine the crystal structure and unit cell size?

Answer: X-ray diffraction can determine these material characteristics. c. You determine that material (c) has a hexagonal close-packed structure, which is shown below. Calculate the APF of this material. The c-to-aratio is 1.633, where c is the height and a is the length of each side. (Note: the center layer consists of the equivalent of three r is ). total atoms within the unit cell, and the radius of the atom

(c)

Answer: APF =

Volumeofatomsinunitcell unit cell volume

The number of atoms in the unit cell is 6, therefore the volume is: 6⎛4⎜πr3⎞⎟ ⎝3 ⎠ To calculate the unit cell volume, find the area of the hexagon, and multiply by the height (c). π= 3 3 a2 3 The area of a hexagon is a2 cot 2 2 6 , the volume of the unit cell is 1.63333a3 and since we know c=1.633a 2 and a=2r Therefore: APF 8πr3 ≅0.74 = 1.63312 3 r3

(

)

d. Would you be likely to find interstitial defects in any of these three materials? Why? Answer: It is unlikely to find interstitial defects in structures with large atom size and small spacing between the atoms. The closer the packing, the fewer the defects. Therefore, if interstitial defects were to occur, they would most likely be found in the simple cubic unit structure, which has the lowest APF. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

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2.2 You are examining a copolymer for its potential as a material for a vascular graft. You are trying to determine whether you want a high or low degree of crystallinity in the material. What type of structures for copolymers have a higher probability for crystallization?

Answer: Alternating and block copolymers have a higher affinity for crystallization. The repeatability of the structure induces the formation of order. Random and graft copolymers have a lower affinity for crystallization. The random structure may not allow for crystallization, whereas the graft structure is inclined to hinder secondary interactions between the chains.

Poly(ε-caprolactone) is being considered as a potential material for a vascular graft.

2.3 After a particular batch has been made, someone gives you the following fractional distribution data and asks you to calculateMn, Mw and PI for this polymer.

Wi Mi (kg/mol)

0.10 0.10 0.30 0.40 0.10 25 30 40 70 100

Answer: Use the following equations to generate the numbers in the table below: Ni = Wi / Mi Wi Mi (kg/mol)

Ni xi xiMi wi wiMi

xi = Ni / ∑ Ni

wi = Wi / ∑ Wi

0.10

0.10

0.30

0.40

0.10

25 0.004 0.1856 4 4.6408 8 0.1 2.5

30 0.003 0.1547 0 4.6408 8 0.1 3

40 0.008 0.34807 13.92265 0.3 12

70 0.006 0.26519 18.56354 0.4 28

100 0.001 0.0464 1 4.6408 8 0.1 10

Mn= ∑i xi Mi = 46.41 kg/mol M=∑i wi Mi = 55.5 kg/mol w PI = M/M= 1.20 w n

2.4 You are given four additional polymeric materials (A, B, C, D) to examine for potential as a vascular graft material, and are told to use size-exclusion chromatography to determine the approximate molecular weights of the unknown polymers. You obtain the following information for the first three samples (A, B, C). a. Assuming monodisperse samples, what is the molecular weight of each?

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Standard 1: Molecular weight 50,000 g/mol Standard 2: Molecular weight 40,000 g/mol Standard 3: Molecular weight 35,000 g/mol Standard 4: Molecular weight 20,000 g/mol Standard 5: Molecular weight 10,000 g/mol Unknown A: Peak amount polymer eluted at 7.4 min Unknown B: Peak amount polymer eluted at 8.4 min Unknown C: Peak amount polymer eluted at 9.6 min Answer: See Figure 2.57 in the book.

Amt Polymer (arbitrary units)

14

4

12

1

10

2 3

5

8 6 4 2 0 -2 0

2

4

6

8

10

12

14

Elution time (min)

Std #1: Molecular wt. 50,000 g/mol approx. elution time: 7.0 min. Std #2: Molecular wt. 40,000 g/mol approx. elution time: 7.7 min. Std #3: Molecular wt. 35,000 g/mol approx. elution time: 8.0 min. Std #4: Molecular wt. 20,000 g/mol approx. elution time: 9.0 min. Std #5: Molecular wt. 10,000 g/mol approx. elution time: 9.8 min.

Now create a standard curve:

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Molecular Weight (g/mol) Mol. Weight (g/mo

SEC Standard Curve 100000

y = -14447x + 150908 R2 = 0.9985

10000 0

2

4

6

8

10

12

Time (min)

Use the equation from the standard curve to estimate the molecular weight.

Unknown A: Peak amount polymer eluted at 7.4 min Unknown B: Peak amount polymer eluted at 8.4 min Unknown C: Peak amount polymer eluted at 9.6 min

MW ~ 44,000 g/mol MW ~ 29,500 g/mol MW ~ 12,200 g/mol

b. A problem commonly encountered in SEC is overloading the chromatography column by injection of a highly concentrated sample, which results in broadening of the peak on the chromatogram. You injected a highly concentrated sample of unknown D into the chromatography column. Which of the following curves would result? Pick X or Y and describe why, based on molecular interactions of the sample with the column.

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Answer: Either X or Y is acceptable, as long as a reasonable, molecular-based reason was given.

In reality, Y occurs. This is because, if you add a sample that is highly concentrated, there will be more of both the larger and smaller molecular weight fractions of the sample. The smaller MW fractions will find their way into the pores of the packing material. Therefore, the smaller MW polymers that are in too high of a concentration in the pores act to “trap” each other in the pores and prolong their transition. A larger number of molecules take longer to be eluted than normal; the shoulder is found on the right of the peak.

However, the scenario depicted in X could be explained by the presence of aggregates which would act as high molecular weight (large size) polymers and would be eluted earlier from the column.

2.5

Calculate the Miller indices for the following planes and include the steps necessary to arrive at your conclusion. (Note: x = a axis, y = b axis, z = c axis)

(a) Answer:

(b)

intercepts reciprocals multiplier indices

(x) a 1 1 x2 ( 2

(y) (z) b c 2 ∞ 1/2 0 x2 x2 0) 1

plane B intercepts reciprocals multiplier indices

∞ 0 x2 ( 0

2/3 1/2 3/2 2 x2 x2 3 4)

plane A

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Chapter 3

3.1 Would a perfect metallic crystal be brittle or ductile? Why? In orderAfor Answer: perfect a material metallictocrystal undergo is brittle. plastic deformation, even just a few linear defects must be present. These linear deformations give rise to slip planes, where slip may occur. Materials with many slip planes are ductile, whereas materials with few (or none, in the case of a perfect metallic crystal) are brittle.

3.2 During the normal function of a femoral stem component of a hip joint implant, would you expect slip to occur?

Answer: We do not expect slip to occur in the hip stem because slip is plastic deformation, and if plastic deformation occurs, then the hip stem would lose its shape and function over time.

Materials that undergo plastic deformation under the load-bearing conditions of the implant are generally not chosen for this application.

3.3 Below are three polymers to be examined for use as vascular graft materials: a) Poly(styrene)

b) Poly(acrylonitrile)

c) Poly(tetrafluoroethylene)

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a. Which two do you believe would exhibit the highest and the lowest crystallinity, respectively, and why?

Answer: The highest crystallinity would be with (c), because the lack of any steric hindrance would allow for the chains to get close and form secondary bonds. The lowest would be (a), as the bulky benzene group would prevent chains from coming together and aligning with each other.

b. What analytical technique would you use to determine the crystallinity of each material?

Answer: Differential Scanning Calorimetry (DSC) or X-Ray Diffraction (XRD) could be used to determine thevalues crystallinity the three polymers. c. Assume that the belowofrepresent the Tg and Tm for each of the three materials:

Material T (omC) T og (C)

(a) 38 30

(b) 52 28

(c) 102 45

Based on this information, which of these materials would you select for this application?

Answer: For a vascular graft application, you want the material to be somewhat elastic, and not brittle. Therefore, a material with Tg below physiological temperature and Tm above this temperature would be preferred. This requirement eliminates (c). When examining (a) and (b), it should be noted that the Tm of (a) is very close to body temperature (37°C), and that the body temperature could increase above 38°C, thus causing melting of the polymer.

Hence, (b) is the most suitable for this application.

d. What structural characteristics of these polymers have an effect on the melting and glass transition temperatures?

Answer: The melting transition point will be affected by the amount of secondary bonds between the polymer chains. Crystallinity is largely influenced by the polymer’s structure (bulky side chains and branching would reduce the ability to form crystallites and decrease their Tm). Decreasing the molecular weight also lowers the Tm, since there are more chain ends, which results in less required energy to break any secondary bonds between the chains.

The Tg is affected by the polymer chain’s flexibility, with more flexible chains lowering the Tg. Flexibility is affected by the rotational ability of bonds along the chain’s backbone (C-O bonds rotate more easily than C-C bonds) and the size and reactivity of the side groups (chains with larger side groups will have more difficulty rotating along their backbone).

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e. You are given another polymer to examine for this application. For a preliminary test, you run DSC on a sample, the results of which are shown below (Note: endothermic reactions are shown as positive on this graph). What are the Tm and Tg of the sample?

Would this material be suitable?

Answer: The Tg of the sample is 10-15 °C and the Tm is 95-100 °C. Judging by the guidelines developed in part (c), this material could be used for this application.

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Chapter 4

4.1 You are instructed to choose a class of polymer for use in a vascular graft, with the most significant concern being that the material has the ability to expand and contract in circumference during the pulsatile flow of the blood. a. Which class do you believe would be ideal for this application and why?

Answer: An elastomer would be ideal for this application due to its ability to recover large strains. b.

Design an in vitro experimental setup to examine the susceptibility of the material in a vascular tube form to fatigue along the circumference. You will need to consider a method to simulate the mechanical stresses and physiological conditions the material will in vivo . What are the important parameters to consider in this setup? experience Answer: There are two main setups you could use. The most accurate would be to fully construct a vascular graft and connect it to a recirculating system with a pulsatile pump.

This would create pulsatile flow throughout the graft. The second, more common method would be to create a ring of the material and place it on two posts. The distance between these posts would then expand and contract repetitively, mimicking the expansion/contraction seen in pulsatile flow. In both situations, the experiment would need to be performed at 37°C and probably in a humid or wet environment.

c. You are given another material that could be used for this application, Dacron®, and are instructed to determine the tensile strength of this material. Using a normal dog-bone geometry, you test the polymer at a rate of 1 mm/min and find it has an elastic modulus of 3 GPa. You repeat the test at 5 mm/min. Would you expect the tensile modulus to be on the order of 0.3 GPa or 30 GPa? Why? (Please explain from a molecular perspective) Answer: 30 GPa. With increasing strain rate, there is an increase in modulus (material behaves in a more brittle manner). This is due to the fact that the chains do not have enough time to align if pulled rapidly.

4.2 You want to examine the possibility of using a polymeric material as an artificial load bearing surface for articulating joints. But the deformation of this material under loading is a concern. You have been informed that if the material has a Poisson’s ratio under 0.1, then it can undergo further testing for this application. You are given a cylindrical sample that is 10 mm in diameter and 3 mm in height. During compression testing, you apply a strain of 20% in the axial direction, and the diameter of the cylindrical sample increases by 0.4 mm at its widest point. What is the Poisson’s ratio of the sample? Does it pass or fail the screening test?

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d / d0 εt = 0.4 mm / 10 mm = 0.04 εt = Δ εa = - 0.2 (Negative due to the fact that the sample is in compression, see p. 136-137 in

Answer:

the book)

ν = - tε/ a ε = - (0.04 / - 0.2) = 0.2

Since 0.2 > 0.1, the material cannot undergo further testing.

4.3 As mentioned in problem 1.2 from Chapter 1, if you stand on one leg the load exerted on the hip joint is 2.4 times your body weight. Assuming a simple cylindrical model for the implant, calculate the corresponding stress (in MPa) on a hip implant in a 175 lb. individual with a hip implant with a cross-sectional area of 5.6 cm2. If the hip implant is made of Ti6Al4V (124 GPa elastic modulus), what is the strain for the current loading conditions?

F= Answer: σ A

F = 2.4×175lb = 420lb × 4.448N lb = 1868N and

A=5.6cm2 = 5.6×10−4m 2

1868.16N 2= 3.336MPa Therefore: σ F= = −4 A 5.6×10 m For the strain:

4.4

εσ

=

E

=

0.003336GPa =2.69×10−5or 0.0027% 124GPa

Someone brings you the femoral stem of a hip joint prosthesis that has fractured in its midsection after 15 years in a patient. a. Using your knowledge to date, what is the most likely cause for this failure? Would you expect this fracture to be more brittle or ductile in character?

Answer: Due to the length of time between implantation and failure, this is most likely a fatigue fracture because the fracture occurred after many repeated loading cycles.

Fatigue fractures are brittle. b. After determining that the above material is no longer suitable for a femoral stem, you decide to examine a new material. Tensile testing results are shown below. A cylindrical dog-bone specimen with a diameter of 12.8 mm and a length of 60 mm in the testing region was used. What is the elastic modulus for this material? What is the maximum stress you would want to apply to this material before it would fail for this application?

What is the ultimate tensile strength?

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Force (N) 0 15442 30883 46325 61766 77208 92649 105517 107834 109506 110150 110664 108992 106547

Length (mm) 60.0 0 60.0 6 60.1 2 60.1 8 60.2 4 60.3 0 60.3 6 60.4 2 60.4 8 to stress-strain. To do this, we need to Answer: The force-length data need to be converted 60.5 determine the cross-sectional area A0.

4 ⎞ =π2=π⎛ 0.01282 2 60.6 A0 r 0 2 ⎠ ⎝ Now use the following formulas to convert the data60.6 to stress vs. strain: 6 l − l F 0 σ= A0 and ε=i 60.7 l0 2 Strain 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0.012 0.013

60.7 Stress8(MPa) 0 120 240 360 480 600 720 820 838 851 856 860 847 828

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Plot the data: Stress vs. Strain 1000 900

Stress (MPa

800 700 600 500 400 300 200 100 0 0

0.005

0.01

0.015

St r ain

The elastic modulus is the linear portion of the curve: 720−0 Δσ = 0.006−0 =120GPa E =

Δε

The maximum stress before failure for this application would be the yield stress (you do not want the material to plastically deform). If you draw a line having an offset of 0.002 strain you will find a yield strength around 830 MPa. The ultimate tensile strength is the highest point of stress on the curve: 860 MPa. c. You want to increase the stiffness of this material. Knowing that the material is poured into a cast, removed, and then allowed to cool in a room-temperature, how would you change the processing? Why would this change the modulus of the material?

Answer: The easiest way to increase the stiffness of the material would be to change the processing so that the material is composed of smaller grains. Quenching the material after it has been cast would result in a smaller grain structure, as opposed to the current “air- quenching” step. d. A colleague proposes the idea of using one of the following two shapes instead of the typical dog-bone shape for tensile testing of your modified material. Is either of these a viable option? Why or why not?

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Answer: Neither is a viable option. With the rectangular specimen, the most likely place for fracture will be at the sample-grip interface, due to the added stress. The I-beam shape has sharp corners, which result in localized stress raisers, and will fracture at those locations first.

Therefore, neither sample shape can be used to test properties of the material; instead, results will be an artifact of geometry.

4.5 You are comparing two different processing methods used for forming Ti6Al4V into a femoral stem of a hip joint prosthesis. The following diagram shows the structures that result from the two methods:

Which of the two processing methods results in a material with a higher ultimate tensile strength? Why? Answer: Method A results in a material with a higher ultimate tensile strength. The larger quantity of smaller grains results in more grain boundaries, which inhibits slip in the material and strengthens it. In Method B, there are a lower number of grain boundaries; hence overall slip of the material is more likely to occur.

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Chapter 5 5.1 You are examining a hip implant with the stem made of Ti6Al4V and a femoral head made of CoCr. a. Select two of the following corrosion mechanisms and describe why/how these could be a problem for this application: galvanic corrosion, crevice corrosion, pitting, stress corrosion, fatigue corrosion.

Answer: Galvanic corrosion could be a problem, with the two metals being the anode/cathode and the physiologic environment acting as the salt bridge. Crevice corrosion at the interface between the ball and the neck of the stem could occur. Pitting could occur if there are scratches or small cracks present in the metal (either from manufacturing or implantation). Stress applied to the implant in the form of loading (from walking or other activities) could also result in galvanic corrosion, stress corrosion and fatigue corrosion. b. Of the tests described in Section 5.5, which would you select to examine the potential occurrence of these corrosive events? in vitro Answer: First, perform tests where the materials are placed in an environment with pH and ion content similar to what would be seen in the hip. If no corrosion is observed in vivowould be performed. The implant either macro- or microscopically, then testing would be placed in an animal model and loaded in a manner similar to the final in vitro in vivo environment. In both the and experiments, samples would be taken at set time points to evaluate the effects of time on corrosion. Surface inspection (for color change and cracks), mechanical tests, and chemical assays would be used to determine the extent of in vitro in vivo testing could also be used to estimate corrosion both and.

Electrochemical corrosion rates of the metal.

5.2 You want to compare the degradation susceptibility of two polymeric materials used in vascular grafts.

Poly(ethylene terephthalate):

Poly(tetrafluoroethylene):

a. Which of these materials is more susceptible to hydrolytic cleavage (chain scission)? Why? This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

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Answer: Although both poly(ethylene terephthalate) (PET) and poly(tetrafluoroethylene) are very stable and have been used clinically in the form of a vascular graft, PET would be more susceptible to hydrolysis on the basis of the information presented in Table 5.8 because of the chemical moieties along the backbone (i.e., esters). b. If these materials were manufactured into a tubular shape and used to replace a section of the vascular system, would degradation be more likely to affect the inner surface or outer surface of the graft?

Answer: The answer would depend on the effects of the mechanical forces as well as on those of any degradation products on material degradation. The lumen surface of the graft would experience significant shear due to fluid flow, however there would be no accumulation of any degradation products because of the presence of flow.

5.3 You need to determine which degradable material to use as a tissue engineering scaffold for repair of a critical-sized bone defect. (A critical-size defect is one that would not heal unaided). You are planning to seed the scaffold with cells then implant it into the defect.

While the cells proliferate and generate bone, the scaffold will degrade and create void space into which the new tissue may grow. What degradation method would you prefer for this scaffold material and why? (hydrolytic vs. enzymatic degradation, bulk vs. surface degradation) Answer: Degradation via hydrolysis would be preferred in this application due to the consistency of the aqueous environment between patients, and the lack of necessity for targeting the degradation area (since it is implanted into the defect site, it is already localized, and is accessible by water).

Bulk degradation of the scaffold would also be preferred, since the overall dimensions would be consistent over a period of time, whereas if the material underwent surface degradation the implant could fall out of the defect or loosen before sufficient repair tissue has developed, or any seeded cells could detach from the scaffold.

5.4 You examine the degradation of poly(DL-lactic-co-glycolic acid) foams bothin vitro and in vivo and obtain the following results: In Vitro Weeks MW 0 104300 1 82539 2 65318 3 51690 4 40906 5 32371 6 25617 8 16043 10 10047 12 6292

In Vivo Weeks MW 0 101800 1 74070 2 53894 3 39213 4 28532 6 15105

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What is the molecular weight half-life for each case? Why does this variation occur?

Answer: First, plot the data. The data can be fitted with an exponential curve. Then, determine the half-life according to the curve fit. The half life is defined as the time required for the molecular weight to decrease by half.

In Vitro 120000 100000

MW

80000 - 0.234x

y = 104300e R2 = 1

60000 40000 20000 0 0

2

4

6

8

10

12

14

Weeks

y=104300e−0.234x Initial MW = 104300, therefore, the half of the initial MW is 104300/2 = 52150 Reformat the equation to solve for x, and plug in y=52150:

⎛ y ⎞ ⎟ LN ⎜ ⎝ 104300⎠ = 2.96≈ 3.0weeks x= − 0.234

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In Vivo 120000 100000 80000 MW

- 0.318x

y = 101800e R2 = 1

60000 40000 20000 0 0

1

2

3

4

5

6

7

Weeks

y=101800e−0.318x Initial MW = 101800, therefore, the half of the initial MW is 101800/2 = 50900 Reformat the equation to solve for x, and plug in y=50900: LN⎛ y ⎞ ⎜ ⎝101800⎠=2.18≈2.2weeks x= − 0.318 The reasons for the discrepancy could be due to differences in pH between the two cases (possibly because of an inflammatory response in vivo), mechanical loading or enzymatic degradation in vivo.

5.5

You have developed two hydrogel systems for controlled release of a therapeutic growth factor (GF). In one system the GF is dispersed throughout a synthetic polymer hydrogel material that degrades by hydrolysis. In the second system, the GF is loaded into gelatin microspheres, which are subsequently dispersed throughout the same synthetic hydrogel material used in the first system (in this case the GF is only loaded into the microspheres). You measure the release in vitro of the GF from each system in both (1) phosphate buffered saline (PBS) and (2) PBS containing collagenase (an enzyme that degrades gelatin). The graphs below illustrate the results of the study.

What do the results suggest regarding the GF release mechanism from each system?

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Answer: As the material degrades the growth factor is released. The enzyme (collagenase) speeds the degradation of the hydrogel with the gelatin microspheres; this is indicated by the increase in growth factor release at early time points. This is not seen in the hydrogel without microparticles showing the enzyme does not affect the degradation rate of the bulk hydrogel material. Therefore, the enzyme must speed the degradation of the gelatin microparticles, but not of the hydrogel itself.

5.6 You are comparing two different processing methods used for forming Ti6Al4V into a femoral hip stem. The following diagram shows the structures that result from the two methods.

Which of the two processing methods would result in a material that is more susceptible to corrosion? Why? Answer: Method A results in a material more susceptible to corrosion, due to the higher amount of grain boundaries.

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Chapter 6

6.1

Compare and contrast casting methods for metals, ceramics, and polymers. Also compare the effect of cooling rate (quenching) between metals and polymers.

Answer: For metals, casting consists of pouring hot metal into a mold, and allowing it to cool, after which the mold is removed. For ceramics, a room temperature mixture of ceramic pellets and water/binder is poured into the mold. After drying, the material holds its shape after mold removal, but must be further heated/dried to strengthen. For polymers, one method is injection molding which consists of injecting heated liquid polymer into the mold and allowing it to cool until it can hold its shape. One major advantage of polymer casting is its speed. If using a thermoset, shapes can be created in a few seconds, whereas cooling a large piece of metal in casting can take longer.

As for polymer quenching, if a polymer is quenched in a way that allows for a slow cool, then the crystallinity of the material increases (due to the additional time for the chains to realign), and hence its strength increases. To strengthen a metal, it must be cooled quickly. This results in smaller, more numerous grains, which inhibit slip throughout the material.

6.2

Which fabrication methods lend themselves to the creation of porous metals, ceramics, and polymers? For each of the three material classes, which method would you choose to make a porous structure out of each type of material, and why?

Answer:Techniques to make a porous metal: powder metallurgy (if one chooses not to remove the pores) or rapid manufacturing/selective laser sintering. Porous ceramic: 3D printing. Porous polymer: fiber spinning (e.g. electrospinning), and also certain rapid manufacturing techniques.

The selection of the particular method would depend on the desired application, as porosity and pore size are determined by the processing technique.

6.3

When a hip stem in implanted in the femur it is sometimes held in place with a bone cement, made out of poly(methyl methacrylate). During implantation, the ingredients for the bone cement (consisting mainly of the monomer methyl methacrylate and an initiator) are mixed in the surgical suite and packed into the femur, prior to the insertion of the hip stem of the prosthesis. At this point, the mixture polymerizes in situ. Using this information, and what is provided in the book, is this material a thermoplastic or thermoset? Why?

( Answer: Poly(methyl methacrylate)PMMA) is a thermoplastic. The monomer and the initiator are mixed to initiate the polymerization in situ. The final product consists of linear PMMA chains and lacks any crosslinks.

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6.4

Create a table showing two advantages and two disadvantages of each of the three main sterilization methods (steam, ethylene oxide, and radiation).

Answer: Advantages Quick, easy, inexpensive

Disadvantages Not suitable for some materials due to the high temperature needed for sterilization and potentially detrimental effects of water on the material

Radiation

Rapid, compatible with many materials

Low penetration depth (can only be used on thin materials), reacts with certain materials, very high start up costs

EtO

Works at low Toxic, carcinogenic, time consuming, temperatures, can sterilize requires specialized equipment deep within a material

Steam

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Chapter 7

7.1 You are given a relatively hydrophobic material with PEG chains covalently attached to its surface (with the attachment occurring at one end of each chain). What do you expect to be the effect of PEG attachment on protein adsorption to the material surface?

Answer: Even though the hydrophobic surface may be favorable to protein adsorption, the PEG chains will interfere with this process. Hence, due to steric hindrance, there will be a low degree of protein adsorption (see Figure 7.1).

7.2

You have designed a bone plate that is manufactured via rolling under cold working conditions, and tests show good biocompatibility results for the device. The Manufacturing Department has put in a request to change the processing method to rolling under hot working conditions, in an effort to reduce the total manufacturing cost.

Since the material from which the device is fabricated will not change, do you need to rerun the biocompatibility tests? Why or why not?

Answer: Yes. Surface properties/energies change with different manufacturing techniques, and this can result in variations in protein adsorption on the material surface.

7.3 Someone in your company has developed a material that closely matches the native mechanical properties of bone, and would be an ideal femoral stem of a hip joint prosthesis.

Unfortunately, the material is cytotoxic. Another engineer in the company has developed a coating that can be applied to the material, sealing the cytotoxic material away from the body. Is this a viable option?

Answer: No. While this solution might work in the short term, scratching during implantation or delamination in the long term could cause major cytotoxicity issues.

7.4 A coworker wants to use titanium as the head for a femoral stem of a hip implant. She has asked you to determine a way to improve the wear resistance of this material. What would you do?

Answer: Ion beam implantation could be used to increase the wear resistance of titanium (using nitrogen).

7.5 You are considering two different surface modification techniques (Langmuir-Blodgett films and self-assembled monolayers) to control protein adsorption to the interior surface of an artificial vascular graft material. Which method would be better suited for this particular application? Why? Answer: SAMs usually form covalent bonds with the surface, which are stronger and better suited to the high shear stress environment than LB film deposition.

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7.6 You are considering two biomaterials for a vascular graft application. One has a large quantity of hydroxyl groups exposed on its surface, while the other has fluorine groups exposed. Which material would be more appropriate for this application in terms of protein adsorption?

Answer: The two main factors governing protein adsorption are the surface hydrophobicity and surface charge as well as the corresponding protein properties. Both materials have polar groups on their surface; however, we do not have sufficient information about their surface charge density. Although one would expect that the presence of a large quantity of hydroxyl groups would render that material surface more hydrophilic compared to the one with fluorine groups exposed, one can not draw any conclusion concerning protein adsorption on these materials based solely on their hydrophilicity since there may be differences in surface charge that would also affect protein adsorption. Ideally, if the material is intended to be used as a vascular graft, it should prevent adsorption of certain proteins that may lead to blood clotting. Therefore, the proteins in the surrounding environment must be also examined in terms of their hydrophobicity and charge in order to evaluate their interactions with the materials.

7.7

Given that a water-based liquid has γLV of ~72 dynes/cm, which of the two materials depicted below is more hydrophobic? Explain how the Zisman method is used to determine critical surface tension and how the critical surface tension relates to the surface properties of the material. (Figure adapted with permission from [23]).

Answer: The Zisman method uses the interactions between liquids with different γLV and a

surface to estimate the critical surface tension, or what the surface tension would be when there is complete spreading of the liquid on the surface. This occurs when the energy of the liquid

(γLV) is similar to the surface energy of the material. The hydrophilicity of materials can be

defined with their contact angle with water. Water would spread more on very hydrophilic surfaces. As can be seen from the graph, water would have a smaller contact angle with glass (would spread more on glass since the energy of water is closer to that of glass), whereas its contact angle with Teflon would be larger. Thus, Teflon is more hydrophobic than glass.

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7.8

One potential problem associated with the long-term release of a drug from an implanted material is the initial burst release of significant amounts of the drug. One approach to reduce this burst release is to apply a coating to the material in an attempt to In modulate a studythe performed diffusionby of the Kwok drug. et al. [24] the effectiveness of butyl methacrylate as a coating material was tested. Figure 1 below shows the chemical structure and a ESCA scan of the surface of this coating material. Plasma treatment of its surface was also performed in an attempt to increase its hydrophilicity. Part of the study explored how the power of the plasma-generating equipment affects the surface chemistry (Fig. 2 below). At high power, some of the polymer begins to degrade. Which of the bonds in the polymer is subject to degradation at high power? Explain in terms of the results in Fig. 2. Which power setting would you recommend for surface treating this polymer and why? (Figures adapted with permission from [24].)

Figure 1

Figure 2

Legend: (a) untreated polymer (b-g) plasma-treated polymer: (b) 5W, (c) 10W, (d) 20W, (e) 40W, (f) 50W, (g) 60W This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

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Answer: The oxygenated carbon functionalities (C-O and O-C=O) are more susceptible to degradation under plasma treatment at high powers than the hydrocarbon portion of the molecule (C-Hx). This is shown by the disappearance of the C-O and O-C=O peaks at higher powers in Figure 2. The O-C=O peak is gone after line (d), so anything above 20 W would cause considerable degradation of this portion of the molecule.

7.9

Of the surface analysis methods discussed in this chapter, which require physical contact with the material surface? Which do not?

Answer:

Contact: contact angle, ATR-FTIR, SPM/AFM No contact: light microscopy, ESCA/XPS, SIMS, SEM/TEM

7.10 Self-assembled monolayers (SAMs), a type of surface coating, are composed of three key regions. You have just created a SAM to interact with a glass substrate that you intend to use for protein attachment. Explain the purpose of each region in the molecule and give specific examples of the chemical groups you would expect to find in each region. Would the use of SAMs produce a smooth or rough surface on your substrate, as characterized by atomic force microscopy? Why? Answer: Attachment group forms covalent bond with substrate – silanes are used because of their reactivity with hydroxyl and amine containing materials. Assembly group - van der Waals forces help to stabilize structure - need long hydrocarbon chain (H and C). Functional head group - interacts with environment and, in this case, will allow for attachment of protein, so would either have –NH2 or -COOH for covalent attachment of protein through amine or carboxyl group.

One of the advantages of SAMs is that, due to the forces involved and the fact that they selfassemble in a predictable orientation, they can be used to make molecularly-smooth surfaces.

7.11 The graph below represents the ESCA peak from the 1s carbon of a poly(methyl methacrylate) bone cement. On the graph, draw the 2s carbon peak from the same material and explain why the peak is located in that position.

Answer: Electrons held closer to the nucleus are bound more tightly than electrons farther from the nucleus, thus the 1s electron has a higher binding energy than the 2s electron. Because the binding energy and kinetic energy are related by the equation Eb = hν – Ek, the electron with lower binding energy will reach the detector with higher kinetic energy. Therefore, the peak will be to the right.

Counts

6000

0 1000

1500 Kinetic energy (eV)

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Chapter 8

8.1 Consider a material for a vascular graft application with proteins adsorbed to its surface and a second material with proteins covalently attached to its surface. For which material would you expect the protein layer to be more stable under flow conditions similar to those found in a blood vessel? Why? Answer: Non-covalent bonds which adsorbed proteins form with a surface (hydrophobic or electrostatic interactions) are weaker than covalent bonds; hence there is a higher chance of delamination.

8.2 You are performing an in vitro experiment in which you will expose a material you are considering for a medical device to synovial fluid, which contains the proteins albumin, transferrin, and IgM at concentrations of 5, 0.5, and 0.05 mg/ml, respectively. Each of these components has a particular affinity for your material, with IgM being the highest and albumin being the lowest. a. Describe the kinetics of protein adsorption to your material and how the surface concentration of each protein will change with time.

Answer: Due to the Vroman effect, albumin will initially attach, and eventually be replaced by the IgM, which has a higher affinity for the material. The higher concentration of the albumin results in a greater initial surface concentration via diffusion, but it will eventually be displaced by the proteins with greater surface affinity (first the transferrin, and finally the IgM).

IgM

b. Subsequently in the experiment, more albumin is added to the synovial fluid. What effect will this have on the proteins adsorbed on the surface?

Answer: Addition of albumin, which has less affinity for the material surface, will have minimal effect if IgM is already adsorbed to material surface.

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c. The graph below shows the amount of IgM adsorbed to the material surface as a function of time. What will happen on the material surface with the addition of more IgM?

Answer: As can be seen in the graph, a plateau on the amount of IgM adsorbed has not been yet reached, indicating that the surface is not fully covered with the protein. Hence, more IgM will adsorb to the surface.

8.3 What determiners of protein structure have the largest impact on protein adsorption to a material? Why? Answer: Dehydration, redistribution of charged groups, structural rearrangement of the protein. All these three components play a role in protein adsorption due to their large effect on the Gibbs free energy.

8.4 You are performing an in vitro test to determine protein adsorption to a material. Your testing solution has a pH of 7.2. Partway through your test an assistant accidentally adds some HCl to the solution, resulting in a pH change to 6.7. Will this affect the experiment?

Why or why not? (Explain in terms of the response of proteins to changes in pH.) Answer: Yes, because pH affects the protein structure, which is one of the three major parameters that can affect adsorption. The altered structure may cause the protein to have greater or less affinity for the surface at the new pH. 8.5 You are performing an in vitro test in which you are testing the adsorption of three proteins (X, Y, and Z) to a material. After you have exposed the proteins to the material surface, how do you determine which proteins adsorbed? What assay(s) would be suited to analyze the solution and the material surface?

Answer: For analyzing the solution: use HPLC or a Western blot assay and examine the remaining solution. Then compare the concentration in the solution to the concentration in an unused control.

Whatever is missing should be on the material. For analyzing the material surface: Colorimetric or fluorescent assay.

8.6 a. Describe the similarities and differences between two types of HPLC: size-exclusion chromatography and adsorption chromatography.

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Answer: The description for SEC is in Chapter 2, whereas the description for affinity chromatography is in Chapter 8. In SEC, the molecules that are the largest are not retarded by the stationary phase so they are eluted first. For adsorption chromatography, the size of the molecules is not important, but the affinity for the surface of the solid phase. The surface can be tailored so that either the hydrophobic or hydrophilic molecules are eluted first. b. Assume the graph below was produced using reversed-phase adsorption chromatography. One of the curves represents a polysaccharide (sugar), while the other represents the same polysaccharide covalently linked to poly(styrene) for a drug delivery application. Which curve is associated with which and why? Explain in terms of molecular interactions with the stationary phase of the column.

Answer: X is the polysaccharide because in reversed phase chromatography (non polar stationary phase and polar mobile phase), more hydrophilic molecules are eluted first (more affinity for mobile than stationary phase). c. Now assume that the graph above was produced using gel permeation chromatography (a type of size exclusion chromatography) for the same molecules as described above. Which curve is associated with which and why? Explain in terms of molecular interactions with the stationary phase of the column.

Answer: X is the polysaccharide-poly(styrene) because it is the larger molecule and in SEC, larger MW substances are eluted first (cannot enter pores in packing material like smaller molecules).

8.7 You are interested in surface modification of a material with both a protein and a peptide for a tissue engineering application. However, the affinity of one biomolecule for the material surface is much greater than the other. What surface modification technique(s) could be applied to allow for the presence of both biomolecules on the material surface?

Answer: Surface patterning techniques (microcontact printing or microfluidics-described in Chapter 7) could be used to attach the two biomolecules to the material surface. A pattern could be designed and the biomolecules could be added in two stages, resulting in no interference between the two components.

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Chapter 9

9.1 a. What are the various types of cell membrane receptors and ligands? What type of contacts does each form? (cell-cell or cell-ECM) Answer:

1. Cadherins: cell-cell 2. Selectins: cell-cell 3. Mucins: cell-cell 4. Integrins: both 5. Other cell adhesion molecules (CAMs): cell-cell

b. How do these different receptors and ligands interact with a biomaterial?

Answer: The integrins would be responsible for cell attachment to adsorbed proteins on a biomaterial, and then the other four would moderate cell attachment to this initial layer of cells.

9.2 Some ECM components, such as collagen, can be larger than cells, yet cells are responsible for their formation. How does this occur? Draw a diagram and describe the synthesis of collagen, including all key steps from transcription to formation of fibers in the extracellular matrix. Provide the location at which each of these steps takes place.

Answer: Key steps: Transcription in nucleus Translation on ribosomes in rough ER (mRNA, tRNA, rRNA) Post-translational modification in ER lumen and Golgi Secretion of procollagen to extracellular space Cleavage of procollagen chain to allow for assembly into fibrils, fibers and final crosslinking 9.3 Most tissues in the body are relatively soft compared with bone. What is responsible for the high modulus of bone? How could you apply this knowledge when designing a synthetic scaffold material to replace bone via a tissue engineering approach?

Answer: Mineral crystals between the collagen fibers in the ECM are responsible for the high mechanical modulus of bone. You could use these or similar crystals in a tissue engineering scaffold to help strengthen the scaffold and mimic bone.

9.4

How would you test the cytotoxicity of the following materials in vitro ? (a) metal for a hip implant (b) degradable scaffold for use in tissue engineering

Answer: a) For a metal, you could run a direct contact or agar cytotoxicity assay. b) For a degradable scaffold you could run either one of the previous, but it is also vital to run an elution assay, due to the degradation of the material and the potential cytotoxicity of the degradation components.

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9.5 You have a material that you are testing as a potential tissue engineering scaffold. You have previously attempted to characterize cell attachment to the material using an MTT assay, however the material is interfering with the assay. Name an alternative method to determine the number of cells attached to the material.

Answer: There are three possibilities. 1) Count the cells not attached to the material via the MTT assay, and compare to initial (pre-seeding) cell amount. 2) Radiolabel the cells. 3) Lyse the cells and measure the amount of LDH in the media.

9.6

You decide to use a tissue engineering approach where a scaffold with no cells is implanted, and the surrounding cells will infiltrate the implant and start the regeneration process. To aid in cell infiltration, a chemical agent will be loaded into the scaffold and released upon implantation to attract neighboring cells. How would you determine whether this chemical agent has the desired effect ininan vitro test?

Answer: A Boyden chamber assay could be performed using the cells found around the implantation site and the chemical agent in question.

9.7 Following You are evaluating a synthetic scaffold for repair of cartilage defects. the implantation of the material and the formation of new tissue within the defect, which techniques would you use to determine the protein content of the newly formed tissue?

Answer: For quantitative analysis, an ELISA assay and a Western blot assay (mentioned in Chapter 8) could be used to determine the quantity of specific proteins. Immunostaining could also be used, and would show the spatial distribution of specific proteins (though this technique is semi-quantitative).

9.8 The data below show the effect of the composition of a new block copolymer XY upon the water contact angle and percent cell attachment from a cell adhesion assay in the presence of the protein fibronectin.

Polymer Composition (% X) 5 10 25 50 75 100

Water contact angle (deg) 120 95 80 65 40 15

% Spread Fibroblasts 55 70 90 65 45 20

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a. What is the relative hydrophobicity of the X and Y blocks in the copolymer? Justify your answer based on the data presented above.

100 90 80 70 60

Water contact angle (deg)

140 120 100 80

50 40 30 20 10 0

60 40 20 0 0

20

40

60

80

100

% spread fibroblasts

Answer: X is more hydrophilic. The water contact angle decreases with increasing amount of X. Graphing the data will help visualizing the question:

water contact angle (deg) % spread fibroblasts

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polymer composition (% X)

b. Explain the dependence of cell attachment on copolymer composition. In particular, provide a justification for the amount of cell attachment observed at 5% X, 25% X, and 100% X. Answer: The data suggest that there is a maximum of cellular attachment on the copolymer that contains 25% of the hydrophilic X. When the percentage of X in the copolymer increases above or decreases below 25%, which in turn increases or decreases its hydrophilicity respectively, cell attachment becomes less favorable. We know that for cells to attach on a surface, protein adsorption to that surface is important, and that cells’ receptors recognize certain binding sites on these proteins. When cellular attachment decreases, we can conclude that either protein adsorption on the material surfaces had decreased or that the proteins have been denatured so that their active sites are not recognizable by the cells’ receptors any more (similar idea covered in Figure 8.17).

9.9 The pictures shown below are from an experiment in which osteoblasts (bone forming cells) were cultured on a poly(ethylene glycol)-based hydrogel material with different concentrations of a covalently-linked RGD peptide sequence (designated as Acr-PEGRGD) for 2 or 24 hours. This peptide sequence is known to bind to integrin receptors to promote cell attachment and spreading. (Figure reprinted with permission from [18]).)

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a. Describe two ways in which the RGD peptide could be linked to the surface of the material.

Answer: Preferred methods would include any of those involving covalent attachment of the peptide after forming the hydrogel (post-fabrication reactions) or during the synthesis process (conjugation/synthesis reactions, see Section 7.3 in book). b. The above experiment was performed in the presence of a mixture of blood serum proteins in the cell culture medium. If twice as much of the same protein mixture were added to the culture medium, would cell attachment and spreading be affected for the no- RGD or the 5 mM RGD samples? Explain your reasoning in both cases.

Answer: 2 possible outcomes: 1. No effect on binding to either surface. Due to steric hindrance of PEG at the surface, there will be little protein adsorption from the media, no matter how much protein is present in the Therefore, the media. only reason for cell attachment is receptor interactions with the RGD linked to the hydrogel surfaces and thus the same results will be seen as are shown for the original experiment.

2. A minimal change is expected for the no-RGD sample since this material does not support any cell attachment. For the 5 mM RGD sample, the presence of extra protein in the sample may block some of the receptors interacting with the RGD peptide sequence and thus reduce cell attachment to the material.

NOTE: One could assume negligible protein adsorption on a PEG-modified surface because of steric hindrance (see Fig 7.1) c. Explain why it appears that there is a larger number of cells in the 24-hour image than the 2-hour image for the 5 mM RGD sample.

Answer: Most likely a combination of cell attachment, spreading and proliferation.

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Chapter 10

10.1 Describe the differences between innate and acquired immunity.

Answer: Innate immunity is the source of immunity that one is born with and is relatively nonspecific. The defense provided by innate immunity includes anatomic barriers, physiologic barriers, phagocytic cells and the inflammatory response. Acquired immunity is only triggered after exposure to certain materials (antigens) and is mediated by lymphocytes.

10.2 Of the different components of innate immunity, which is the least likely to have an effect on the response to an implanted biomaterial?

Answer: The anatomic barriers are the least likely to have an effect on the material, since they have been already breached during implantation.

10.3In experiments to evaluate the tissue response to a biomaterial, a control group is often included involving the complete surgical procedure but without implanting the biomaterial.

What is the rationale for the inclusion of this control group, and what is the expected tissue response?

Answer: The control group serves as a baseline for the body’s reaction to the implantation process (since injury occurs during implantation, which will activate the innate immunity).

The researcher can then compare the level of response to the control (such as acute inflammation and macrophage migration) to the levels from the biomaterial test group.

10.4 What are some of implantation?

the phagocytic complications that can occur after material

Answer: There are several complications possible: 1) Macrophages can sequester particles of the material, but then not be able to break them down. When the macrophage dies later, it lyses and releases the particles back into the body’s environment, which will cause more macrophages to attack the particles. This process may be repeated over a long period of time. 2) When the macrophage lyses, other cellular contents such as cytokines are also released into the area. These cytokines can then initiate undesirable reactions, such as the stimulation of fibrous tissue formation. 3) Frustrated phagocytosis (in the case the material is much larger than the cell), where the macrophages and neutrophils release lysosomal enzymes and other products into the environment to attack the large foreign object.

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10.5poly(ethylene You areterephthalate), examining three and poly(propylene)-for materials-poly(tetrafluoroethylene), potential use as a vascular graft.

How would you determine which of the three materials induces the least activation of granulocytes?

Answer: Perform an assay where you culture the materials in the presence of equal in vitro amounts of granulocytes, mark the receptors on the granulocyte surface that indicate activation, and then use a fluorescence-activated cell sorter (FACS) to count the amount of activated cells for each material.

10.6Describe the major steps of neutrophil extravasation and list the important biomolecules involved in each step. Patient X has a rare disease that prevents the expression of selectins on endothelial cells. Patient Y has a related disease that prevents expression of Ig-superfamily CAMs on endothelial cells. What would be expected in each patient in terms of neutrophil extravasation and why?

Answer: The process of neutrophil extravasation can be summarized as follows (Ch. 10.3.1. in the book): 1. Rolling: as the neutrophils are carried along by the blood, they can bind briefly to the vascular endothelium through low-affinity interactions. Selectins are upregulated on inflamed endothelium, facilitating the binding of neutrophils. 2. Activation: as the neutrophils are rolling, they can be activated by chemoattractants, substances that cause the migration of neutrophils toward the damaged tissue. IL-8 and MIP-1b can bind to specific receptors on the neutrophil’s surface and activate intercellular pathways that induce a conformational change in integrin receptors.. The integrins now have a higher affinity for the immunoglobulin superfamily of cell adhesion molecules (CAM) found on the surface of the endothelium. 3. Arrest and adhesion: The strong interaction between integrin receptors on the neutrophil and CAMs presented by endothelial cells stops rolling and causes firm adhesion to the endothelium. 4. Migration: into tissue. The cell accomplishes this through diapedesis, or by squeezing parts of the cell at a time though small spaces between endothelial cells. In both patients the extravasation process will not complete.

Without the selectins, mucins on the neutrophil surface will not be able to bind the endothelial cells, thus preventing “rolling” of the neutrophil along the endothelial surface. Without the expression of Ig-superfamily CAMs, activated neutrophils will not be able to arrest and adhere.

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Chapter 11

11.1 List the stages of wound healing after biomaterial implantation.

Answer: There are three main stages: acute inflammation, chronic inflammation, and granulation tissue formation. Acute inflammation ends when monocytes recruited by the neutrophils mature into macrophages and the number of neutrophils in the tissue decreases.

Fibroblast migration and neovascularization then proceed to create the highly vascularized granulation tissue. Chronic inflammation occurs in the transition between acute inflammation and granulation tissue formation, but can persist in some cases and involve the additional presence of lymphocytes and even granulomas. This type of persistent chronic inflammation is not considered an acceptable outcome after biomaterial implantation.

11.2 List the differences in the tissue response at the wound site following implantation of a nondegradable biomaterial as compared to a simple lesion.

Answer: Foreign body giant cells (FBGCs) will be found around the implant, as the body tries to remove the large foreign object. A fibrous capsule would also potentially form around the foreign material. In a simple lesion, the FBGCs would not be found and the fibrous tissue would eventually collapse to form a scar.

11.3 You have been instructed to design the outer case for an implantable cardioverterdefibrillator, a device that attempts to halt the progression of ventricular fibrillation. What material structure and surface properties can influence the level of the foreign body giant cell response, and what is the effect of each?

Answer: Surface roughness: smoother surface results in a low number of macrophages and no foreign body giant cells (FBGCs), whereas with a rougher surface both macrophages and FBGCs are seen.

Surface area vs. volume ratio: A high ratio will result in more macrophages and FBGCs at the tissue-material interface, whereas a lower ratio will exhibit more fibrous (granulation) tissue formation.

11.4 What type of resolution response occurs with each of the following: a. Splinter b. Degradable tissue engineering scaffold c. Titanium bone screw d. Degradable meniscal arrow (used to repair meniscal tears) e. Breast implant f. Artificial heart valve leaflet g. Catheter h. Surface scar tissue (scab) i. Coral implant for bone repair This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

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Answer:

Extrusion: a, h Resorption: b, d Integration: c, f, i Encapsulation: e, g

11.5 Is tissue regeneration necessary to repair a defect?

Answer: No. As long as the fixed tissue can perform all of its necessary functions, preferably for the life of the patient, then repair is an acceptable solution.

11.6 The company you work for has developed a modified poly(urethane) material for a You havepotential been instructed to examine the body’s inflammatory response to the material and vascular graft application. determine whether an implant can be created such that the inflammatory response would not hinder the implant’s function. What experiment would you conduct to characterize the inflammatory response to the material and to evaluate its suitability for this application?

Answer: The experiment needed would consist of two parts. The first would be a subcutaneous implantation in a small animal model (look at Table 11.3 to determine based on application). If results were favorable, an implantation closer to the implantation site of the intended application with an appropriately processed and shaped material could be performed (in a large animal model, again look at Table 11.3). You would need several time points for both studies. One within 24 hours (acute toxicity), one between 14-28 days (subacute toxicity), one within 90 days (subchronic toxicity), and one after 90 days (chronic toxicity). For the subcutaneous portion, you would only need to test the first two or three time points, due to its preliminary nature and the short lifespan of the animal. At each time point you could perform (depending on application) histology/immunohistochemistry, electron microscopy, biochemical tests (as described in Chapter 9), and mechanical tests. Control groups would also need to be used, to separate the body’s response to the surgical procedure from the response to the material.

11.7 A new polymeric material was being considered for use as part of a drug delivery system.

In order to evaluate the tissue response to the implantation of the material, nonporous compositions in the form of a disc were implanted subcutaneously in an animal model for 4 weeks. Describe the host response to the implantation of the material as a function of time as well as any potential adverse effects.

Answer: Short term response (minutes up to one day): Resident tissue macrophages are the first line of defense. In terms of cell migration into the area, first neutrophils arrive on the scene.

In response to cytokines released by the neutrophils, 5-6 hrs later, monocytes are recruited and mature to tissue macrophages (takes 8 hrs). Neutrophils, phagocytosis, respiratory monocytes burst, and andsecretion macrophages of chemical have similar mediators, actions: but macrophages have a greater killing potential than the other cell types.

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Mid-term response: By 3-5 days, beginning of granulation tissue formation can be seen. This is characterized by migration and proliferation of fibroblasts and vascular endothelial cells to form fibrous This tissuemay andbecapillaries. a part of the foreign body reaction, which is characterized by the presence of foreign body giant cells (fused macrophages) as well as granulation tissue. However, because this implant has a low surface area to volume ratio, fewer FBGCs may be seen and more fibrous tissue may be present (see p. 387). Granulation tissue will mature over weeks to form a fibrous capsule. The size of this capsule depends on the chemical composition of the implant as well as mechanical motion at the implant site.

However, the capsule will probably be thicker over the sharp corners of this implant.

Adverse effects: 1. Because the implant is much larger than the cells, frustrated phagocytosis by neutrophils/macrophages will occur. Over time, this may lead to tissue damage if radicals and enzymes are released continuously.

2. Chronic inflammation leading to the recruitment of immune cells to the site of the implant.

3. Failure or malfunction of the device. Since it is a drug delivery system, the patient may not receive the appropriate drug dosage.

11.8Your company has just developed an electrode to be implanted in the brain to measure electrical activity in terms of electrical current. After implantation in a rat brain in vivo, you record the following results as indicated by the solid line (A) below.

a. Formulate a mechanism to explain these observations.

Answer: A fibrous capsule is formed. This follows the timeline of inflammation; formation of capsule starts around week 1 and matures by weeks 3-4. The fibrous capsule prevents the electrode from measuring the electric activity of the brain. b. After implanting an electrode with its surface modified by covalent attachment of poly(ethylene glycol) in the rat brain, you obtain the data shown by the dashed line above (line B). Why does the surface modification of the electrode alter the results? Would you recommend marketing the surface modified electrode?

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Answer: PEG is a non-conductive polymer, so it will act as an insulator, thereby reducing the electron exchange (current) between the brain tissue and the electrode. Marketing would not be a good idea since the PEG coating prevents the electrode from obtaining accurate (or any) information about electrical activity in the brain.

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Chapter 12

12.1 A degradable scaffold material for bone tissue engineering is implanted into the body.

Describe how the implantation could result in both exogenous and endogenous antigen presentation by the surrounding cells. What type of T cells are activated in each of these scenarios?

Answer: For exogenous, the surrounding cells phagocytose a byproduct of the material itself, and then it is presented with MHC Class II molecules (Fig 12.4b). For endogenous, toxic or carcinogenic properties of the material cause alteration of the surrounding cells, resulting in presentation of the antigen with MHC Class I molecules (Fig. 12.4a). MHC class II are recognized by T helper cells and MHC Class I by T cytotoxic cells.

12.2 If a patient has lost his or her spleen, which type of acquired immunity is most affected?

Answer: Humoral. B cells undergo maturation in the spleen.

12.3a) Describe the four ways that antibodies act in removing pathogens. b) Explain how the complement proteins aid in these actions. c) What immune system mechanism can be used to rapidly produce large amounts of antibodies if needed?

Answer: a) Agglutination, Precipitation, Neutralization, Lysis. See 12.3.2.3 for a summary of each of these. b) Binding of the C1 complement protein to an antibody that is already bound to its antigen results in the activation of the complement classical pathway. This results in the formation of the MAC, so complement mainly helps in lysis actions. c) Memory B cells can be activated upon re-introduction of a particular antigen to produce a clonal population of B effector cells that all produce antibodies specific for that antigen.

12.4 During an in vivo study to determine the efficacy of a new catheter design, you notice that a fibrous capsule has formed around the implant after eight weeks. Monitoring of the animal’s blood via ELISA showed an initial spike in levels of C3 convertase, followed a decrease in levels of this protein complex after week 1. Describe, in detail, a possible molecular/cellular mechanism for the formation of the fibrous capsule observed in this study.

Answer: The spike in C3 convertase in the blood indicates that the complement pathway has been activated. Complement pathway activation is an innate response to the catheter. The ultimate step in complement activation is the formation of the membrane attack complex (MAC) in the target (in this case, the catheter). In the case the target is a cell, the MAC creates pores in the cell, allowing ions to rush into the cell causing cell lysis. However, the MAC cannot form within a biomaterial because the C5b67 cannot properly enter the target. A possible explanation is, since this process is tightly regulated, negative feedback will stop the complement cascade, indicated by the decrease in C3 convertase after one week. Therefore, this will have a lesser direct effect on formation of a fibrous capsule.

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However, another component of the complement pathway is the release of complement fragments into the bloodstream that recruit inflammatory cells to the implant. This ultimately leads to the encapsulation of the implant via the mechanisms described in Chapters 10-11.

12.5What type of hypersensitivity reaction would be most likely observed following implantation of a synthetic, degradable polymer? What type of reaction would be expected with a long-term silicone implant, such as an artificial finger joint?

Answer: Degradable polymer: type III. This is because the fragments of the degrading polymer act as an antigen and form antigen-antibody complexes indicative of a type III hypersensitivity response. Silicone: Type IV. This is a delayed response to the silicone. It is most likely because it is cell-mediated and does not require the use of antibodies. (Type I also acceptable for silicone.)

12.6 What are the main differences between the in vitro techniques to assess the inflammatory response and the immune response to a biomaterial?

Answer: The lymphocyte transformation test (LTT) is used in an immune response study.

Also, the results for cell migration assays will vary between the two responses, with minimal migration of lymphocytes with an immune response and much greater migration of neutrophils/monocytes seen in an inflammatory response.

12.7

What in vivo techniques can be used to determine the immune response to a material?

Answer: Histology/immunohistochemistry, detection of antibodies in the blood, and skin testing are some viable in vivo assays for monitoring immune response.

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Chapter 13

13.1A primary step for blood coagulation resulting from the contact of blood with a biomaterial is protein adsorption to the biomaterial surface. Name some proteins that can initiate coagulation.

Answer: Collagen and von Willebrand factor (vWF) are two proteins that can activate platelets and thus initiate coagulation.

13.2 Describe the main differences between platelets and other mammalian cells.

Answer: Platelets have no nuclei, cannot proliferate, and have a half-life in the body of 8-12 days. Mammalian cells have a nucleus, may proliferate, and can have much longer life spans.

13.3What changes occur within a platelet when it is activated? What differences can be observed in its morphology following activation? Which of the following two micrographs shows activated platelets? (Figures adapted with permission from [6].)

(a)

(b)

Answer: Activated platelets swell and take on an irregular form, extending pseudopodia (as shown in (b)). There is also cytoskeletal contraction, which results in the release of the storage granule contents. Activated platelets can adhere to ECM proteins; they aggregate and secrete bioactive factors, and exhibit coagulatory activity.

13.4 Do you expect blood coagulation to occur following injury of the endothelium of a blood vessel?

Answer: When the endothelium of a blood vessel (its interior wall) is injured, the endothelial ECM molecules are exposed to reactive proteins from the plasma, which is a main mechanism of coagulation initiation (through the intrinsic pathway).

13.5An experimental cardiovascular biomaterial with a positive surface charge was implanted in an animal model. Histological analysis of the harvested implants revealed that coagulation occurred shortly after implantation.

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(a) Which of the two pathways of the coagulation cascade is most likely responsible for this result? (b) How would the coagulation time differ between this and the remaining pathway?

Answer: (a) Extrinsic, (b) Extrinsic: 15 seconds, Intrinsic: 1-6 minutes.

13.6 There are many examples of positive feedback mechanisms within the coagulation cascade to accelerate the coagulation process. What factors prevent coagulation throughout the body?

Answer: There are two main means of coagulation control: physiological factors and chemical factors.

Physiological factors: First, for some parts of the cascade, a surface (phospholipid bi-layer) is needed. Second, the circulating blood can remove and act as a diluting agent for the chemical initiating factors.

Chemical factors: Biochemical inhibitors (such as the heparin/AT III complex) can bind chemical factors that would normally activate the coagulation cascade.

13.7 Would you expect the surface roughness of an implanted cardiovascular biomaterial to affect blood coagulation?

Answer: Yes, the surface roughness is a factor that influences protein adsorption. These proteins in turn could play a role in the coagulation cascade. Additionally, if the material is rough on a larger scale, it could also cause damage to blood cells.

13.8You are considering a new material for use in a vascular graft. Preliminary studies involving subcutaneous implantation of the material in the form of a disc in a rat model are promising. Subsequent in vivo studies with a larger animal model using a vascular graft configuration tested in an ex vivo setup similar to that shown in Fig. 13.4 indicate that the material is not hemocompatible. What is the reason for the different biological response observed in the larger animal model?

Answer: Although subcutaneous implantation can provide valuable information relating to the tissue response to the material, it can not assess the hemocompatibility of a material because the interactions of blood with the material are minimal. Even though there are differences in blood composition between animals and one can expect different hemocompatibility for another animal model, the different biological response in this particular case is mainly because one can not assess hemocompatibility from subcutaneous implantation.

13.9 You are responsible for testing the hemocompatibility of a new poly(ethylene)-based material. You perform an experiment in which you incubate your material with platelets suspended in a mixture of blood plasma proteins containing different amounts of either fibrinogen (Fg) or vWF under either static or flow conditions. Using surface analysis This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

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techniques, you confirm that with increasing amounts of Fg or vWF in the plasma solution, you observe greater adsorption of each protein on the surface of the sample in both static and flow conditions. You then use the LDH assay to determine the number of adherent platelets in each case (results are shown below). (Figure (a) is adapted for educational purposes from [7], with a maintenance of the observed trends. Data appearing in (b) are purely hypothetical and for illustrative purposes only.)

(a)

(b)

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a. How does the LDH assay work? Explain why this assay is a good choice to determine platelet adhesion in this experiment.

Answer: The LDH assay quantifies the amount of an intracellular enzyme (LDH = lactate dehydrogenase) that is present in the media following cell lysis (see section 9.6.1.1). A standard curve is used to relate the amount of LDH to the number of adherent platelets. This assay is appropriate because it makes a quantitative measure and is easily applied to a large number of samples. b. Why are you particularly interested in the proteins Fg and vWF? Explain in terms of the role of these proteins in blood coagulation.

Answer: Both vWF and Fg are potent platelet activators and mediate adhesion of platelets onto a biomaterial, leading to blood coagulation. Fg plays a significant role in blood coagulation because it contains two binding sites which allow for platelet-platelet bridging. c. Your company would like to use this material for a vascular graft application for small diameter blood vessels. You are asked to modify the material so that it releases an agent that deactivates (nullifies the effects of) either adsorbed Fg or adsorbed vWF. Based on the data presented, for this application, which protein is more important to deactivate and why?

Answer: One needs to be concerned only with flow data for a vascular graft application. Under flow conditions, vWF is more effective overall for platelet adhesion than Fg. The use of an agent deactivating vWF would be preferred, since there is only a small concentration range in which Fg will bind platelets under flow, but almost any vWF concentration will promote platelet adhesion. d. Your company has decided to turn its attention towards treating aneurysms (localized, mechanically compromised bulges in blood vessels). One treatment is to encourage clotting around the aneurysm so that blood flow is diverted to nearby vessels instead of stressing the aneurysm. A colleague suggests that a good way to use your material for this application would be to modify its surface to enhance Fg adsorption. Using the data you have already collected, do you agree? Why or why not?

Answer: No. Remember that platelet adhesion is desirable in order to promote coagulation for this application. The data suggest that at higher plasma concentrations of Fg, which result in higher amounts of adsorbed Fg, the packing arrangement of Fg molecules adsorbed on the material surface is affected, thus reducing platelet adhesion under flow conditions at higher plasma concentrations. From the graph, under static conditions, an enhanced Fg adsorption is not expected to have any significant effect on platelet adhesion.

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Chapter 14

14.1 You perform an in vivo study and discover during the first time point after implantation that there are Staphylococcus aureus bacteria at the external surface of the incision site.

Does their presence indicate an infection?

Answer: Not necessarily, since these bacteria are naturally present on the skin.

14.2 Is implant sterilization sufficient to prevent infection following implantation? Why or why not?

Answer: No. There are bacteria naturally already present in the body, and the presence of certain materials can result in a change of these bacteria into virulent organisms.

14.3 You implant a sample material subcutaneously in the back of a rabbit model in an effort to evaluate tissue response. After 24 hours, you notice the presence of an infection in one of the hind legs. Is this infection due to the presence of the implanted material?

Answer: No. It usually takes bacteria a minimum of 2 days to attach to the material surface, aggregate, and disperse from the surface. Since the infection was found at a location other than the implant location at 24 hours, the infection is probably not the result of the material.

14.4 Would the occurrence of tumorigenesis be more or less likely to occur with an implant that is encased in a fibrous capsule (compared with an integrated implant)? Why or why not?

Answer: There is a higher chance of tumorigenesis in the case of an implant encased in a fibrous capsule, as the capsule may help shielding mutated cells from the body’s immune system.

14.5 Preliminary in vitro testing shows that treating a porcine heart valve with a surfactant You decide to design and run an in vivo before implantation reduces calcification. experiment to verify these results. What would you do?

Answer: Implant the treated valves in situ in an animal model, and have a control group consisting of valves not treated with surfactant for comparison. Throughout the study, use noninvasive procedures, such as radiography and microcomputed tomography to monitor the extent of calcification. At the end of the study, harvest the valve and surrounding tissue and assess calcification via histology, colorimetric assays or EDXA analysis.

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14.6The following graph shows the results from a study in which the adhesion of Staphylococcus epidermidis to two different material surfaces was examined. In this study, the “adhesive coefficient” is defined as the percent of the bacteria transported to the biomaterial surface that were retained per unit surface area. The surfaces consisted of either poly(ethylene) that was coated with plasma proteins or poly(ethylene) with adherent platelets. Shear stress was applied to the samples to determine how well the bacteria attached to the experimental surfaces. What do these results indicate? How would you apply this information if you were to design a cardiovascular implant? (Figure adapted with permission from [9].)

Answer: When comparing the results between the two groups, it can be seen that the presence of adherent platelets encourages the attachment of Staphylococcus epidermidis to the poly(ethylene) material. Therefore, one could apply this information by designing a cardiovascular implant that would inhibit platelet adhesion to the material surface in an effort to also minimize bacterial adhesion.

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