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Type:
Solution Manual
Biomaterials The Intersection Resource: of Biology and Materials Science Edition:
2nd Edition
Author(s):
Johnna Temenoff Antonios Mikos
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INSTRUCTOR ’ S MANUAL SOLUTIONS
BIOMATERIALS: THE INTERSECTION OF BIOLOGY ANDMATERIALS SCIENCE 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
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Biomaterials: The Intersection of Biology and Materials Science, 2 nd 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 for expected response of the body to your construct and why this material class may be better 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 1.3 Whether not to thebebiological response a material is acceptable depends thethe specific action wouldor have determined by theto FDA and then that Center would on take lead in the regulatory process, with the other taking a secondary role. 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
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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 so affinity. Protein Y will also be closer to its pI (isoelectric point) at physiological conditions, 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.
Unknown
Concentration 4.9 ng/ml 11.76 ng/ml 31.37 ng/ml
First create the standard curve based on known
concentrations and the area under the curve. The Samples standard curve can be generated in this manner due to A the Beer-Lambert law, which says there a direct B relationship between concentration and is amount C absorption at a given wavelength. Then use the equation of the line generated by the standard curve to calculate the unknown concentrations.
y = 0.1025x R² = 0.996
Series1 Linear (Series1)
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