Tissue Engineering - Essay Example

• Biodegradability: Since TE intends to facilitate the body’s cells to eventually replace the implanted tissue-engineered structure over time, scaffolds are not intended to be permanent implants. Thus, Scaff old needs to be biodegradable to allow cells to generate then extra-cellular matrix (Robert, 2013).
• Mechanical property: Relatively strong properties may be needed in cases where the devices may be subjected to weight-loading whereas in another case, relatively less strong properties may be required.

Cells

The key consideration should be the source of the cells. Cells are categorized by their sources. For instance, autologous cells are obtained from the same individual to which they need to be re-implanted; Allogenic cells are got from the body of the donor of the same species while xenogeneic cells are from individuals of another species (Robert, 2013). During cells selection, cells leading to the fewest problems with rejection and pathogen transmission are preferred (mostly autologous). However, autologous cells cannot be obtained if the individual affected is elderly or has suffered severe burns.

Biomaterials

• Biocompatibility: The initial consideration of any biomaterial for TE is biocompatibility; cells must stick, function normally, and move onto the surface and finally through the biomaterial and begin the proliferation process before producing a new matrix. Biomaterial construct must show a negligible immune reaction after implantation to prevent rejection by the body
• Injectability: TE tissues require a precisely designed combination of cells and a specialized scaffold support system, Injectable biomaterials, especially those delivered in aqueous solution are ideal delivery vehicles for both the cells and bioactive factors (Robert, 2013).
• Another consideration for biomaterial should also be resorption rates, concentration, and transparency.

1. When designing a template for repairing a damaged bone, I will ensure that the template is biodegradable, bioresorbable, three-dimensional; which provides mechanical strength to the bone while at the same time providing a guide for regeneration of bone tissue (Robert. 2013). Thus the template will be of biodegradable materials such as poly (L-lactic acid) and has the capacity of being guaranteed porous.

A pore-performing component is mixed with a continuous matrix formed of biodegradable material. The degradation rate of the pore-performing component should be higher than that of the matrix. Differential dissolution/biodegradation provides porosity to the template (Robert, 2013).

Critical considerations for the design of TEMPs for repairing damaged tissue include:

• Biocompatibility
• The Architecture of the template
• Mechanical properties
• Biodegradability

1. The most difficult organ to duplicate via TE is the kidney (Robert, 2013). Kidney has multiple cell types and complex functional anatomy thus making it difficult to reconstruct. This is because obtaining compatible cells and scaffolds is rendered difficult due to the multiple cells and complex anatomy. Thus, most of the cells and scaffolds are rejected by the recipient leading to adverse health problems.

To develop a template for kidneys, I will use immunosuppressive drugs to prevent rejection and health problems thereof.

1.  
2. Can Tissue Engineers use 3D molds to create organs?

Tissue engineers have harnessed the power of stem cells by coming up with 3-D plastic molds known as scaffolds/matrices that resemble organs/body parts. A matrix is squirted by a soup of nutrients causing stem cells to grow into a hunk of tissue that can later be transplanted into a patient.

1. Can optimal cell sources and scaffold-cell interactions be engineered for heart valves?

Outline to finding the answer:

• Get the base valve cells to work with
• Put altered valve cells into a scaffold to incubate the cells
• Put the newly created valve cells to use
• Track the progress.

2. a). The volume of the graft will have a volume of 3 by 2 by 1 which is equal to 6cm3 and 6000mm3. Taking 10000 cells per cubic millimeter, the number of cells will be 6 x 10^9

b). If 112 hours results in generates 0^5 cells, starting from time 0, then starting with n number of cells will take how long to generate k number of cells?

The number of cells per hour will (4.25 x 10^5)/112=3795 per hour.

Thus, if starting number of cells is n and you want to end with k cells, the time taken will be given by (k – n)/3795.

c). the time taken will be given by [ (6 x 10^9) - ( 6 x 10^9 x 0.5)]/3795=790513