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Biomedical Engineering - Technology for Regenerative Medicine
ese03-ENG
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26.09.2022 Esercitazione 3 (TMR[1]) 1. Regenerative therapy for the mammary gland Consider a therapy for mammary gland regeneration. The therapeutic strategy consists in the isolation of mesenchymal stromal cells from a biopsy of adipose tissue, their expansion in vitro, their differentiation towards adipocytes and their seeding into a solid sphere made of a hydrogel and the orthotopic implant of the cellularized construct. The construct has radius R, and the nutrient concentration outside the construct is constant and equal to C. The oxygen diffusion coefficient in the cellularised construct is equal to D. The minimum concentration to guarantee sufficient oxygenation of cells is c min = C/10. Cells consume oxygen at a constant rate, V. To date, this therapeutic product has been approved for commercialisation and it is in widespread clinical use. The only alternative for mammary reconstruction is implantation of the synthetic breast implants. Answer briefly to the questions in the table: What is the therapeutic product? What are the elements that identify the therapeutic product as a PTC? In what phase of the regulatory process for new PTCs can this PTC be localized? What are the standards that apply in this stage of the PTC process of development? What are the risks associated with this PTC (immunogenic, tumour, teratoma, infection, toxicity)? r R D C 2. Cell expansion in a monolayer A new therapy for osteoarthritis is based on the use of chondrocytes, isolated, expanded in culture and repetitively injected into the patients. Cells are expanded into monolayer on the bottom of petri dishes. The cell monolayer is kept in static culture covered with one layer of culture medium with thickness s. The partial pressure of oxygen present in the atmosphere of the incubator is Po. The flow of oxygen on the cells is known and equal to J c. The diffusion coefficient of the oxygen in the culture medium at 37°C is equal to D, the solubility coefficient of the oxygen in the medium at 37°C is equal to α. The hypothesis of plane symmetry can be assumed for the concentration of oxygen in the medium. This therapy has been successfully tested on the first 10 human patients and the conventional therapy is arthroplasty with an artificial prosthesis. Answer briefly to the questions in the table. What is the therapeutic product? What are the elements that identify the therapeutic product as a PTC? In what phase of the regulatory process for new PTCs can this PTC be localized? What are the standards that apply in this stage of the PTC process of development? What are the risks associated with this PTC (immunogenic, tumour, teratoma, infection, toxicity)? X=s X=0 Classify all the cell sources potentially usable in human patients for this therapy. Cell source (based on immunogenicity) Cell type/s Advantages Criticalities Is it usable respect to the conventional therapy? Why? 1. Calculate how many bioartificial hearts should simultaneously produce a cell factory to satisfy the transplant needs. Hypothesize the decellularization time null (the decellularized hearts are stored), the recellularization time tcell = 10 days/heart and the working days per year twork= 220 days/year. Assume to treat all US patients waiting for transplantation N = 2640. Data: t start = 0 (null decellularization time) t cell = 10 days/heart (recellularization time) t work = 220 days /year (working days) N = 2640 (US patients waiting for transplantation) 2. Cell seeding/dilution Consider to have 10 6 cells diluted in 1 ml of culture medium. Suppose that single adhered cell area is 180 µm 2. We what to cellularized a glass coverslip with an area of 1 cm 2. • Calculate the glass coverslip area in µm 2. • How many cells are needed to seed on the glass coverslip to have a 70% cell confluence? • How many µl must be taken from the mother cell solution? • If the final volume of the cell seeding is 1ml, how many µl of culture medium must be added to the final cell solution? Figura 1 Protocols for cell seeding Data: ������������ 0= 10 6 ������������������������������������������������������������ ������������ ������������ = 1 ������������ 3. Number of focal adhesions Consider a fibroblast in a surface of a microbioreactor in culture medium (µ= 1cP), experiencing a Couette planar flow generated by the upper wall, distant h= 100 µm from the cell and moving with a velocity U M = 400mm/s, as shown in figure. Write the equation for the shear stress acting on the cell. Under the hypothesis that the cell in adhesion can be represented as a flat rectangle (50x100 µm), estimate the minimum number of focal adhesion points that the cell needs to withstand the force generated by the flow (force generated by a single contact adhesion f = 5nN). 4. Organ regeneration To repopulate an adult decellularized monkey heart 10 11 cells are needed. Suppose to use MSC, obtained from the bone marrow and differentiated in vitro. The density of MSC in the aspirate (volume = 0.5ml) is 0.8*10 6 ������������������������������������������������ 3. The efficiency of the cell isolation protocol is low (0.02%). The division time of MSC is 12h. Calculate the expansion time in days. 5. Cell seeding on scaffold Consider different scaffolds with cylindrical shape characterized by the following parameters a) Calculate the number of cells to be seeded on each scaffold to obtain a cell density equal to the density ρ_fin = 10 9 cells ������������������������ b) Assume to isolate cells from a biopsy with a volume V = 8mm 3, ρ _biopsy =10 9 cells ������������������������ . Knowing that the cell division time is t d= 24h, calculate the expansion time needed to obtain enough cells to seed the cells on each scaffold at the previously cited density. SCAFFOLD DIAMETER (mm) Thickness (mm) A 16 0.2 B 6.3 5 C 16 1.2 D 16 2 26.09.2022 Practice Solution 3 (TMR[1]) 1. Regenerative therapy for the mammary gland Consider a therapy for mammary gland regeneration. The therapeutic strategy consists in the isolation of mesenchymal stromal cells from a biopsy of adipose tissue, their expansion in vitro, their differentiation towards adipocytes,their seeding into a solid sphere made of a hydrogel and the orthotopic implant of the cellularized construct. The construct has radius R, and the nutrient concentration outside the construct is constant and equal to C. The oxygen diffusion coefficient in the cellularised construct is equal to D. The minimum concentration to guarantee sufficient oxygenation of cells is c min = C/10. Cells consume oxygen at a constant rate, V. To date, this therapeutic product has been approved for commercialisation and it is in widespread clinical use. The only alternative for mammary reconstruction is implantation of the synthetic breast implants. Answer briefly to the questions in the table: What is the therapeutic product? The cellularized construct consisting in a hydrogel + adipose -derived MSC differentiated in adipocyte What are the elements that identify the therapeutic product as a PTC? The main therapeutic agent is the MSC , obtained after "non-minimal manipulation" consisting in isolation, in vitro expansion seeding in the hydrogel and differentiation In what phase of the regulatory process for new PTCs can this PTC be localized? Commercialization What are the standards that apply in this stage of the PTC process of development? Current Good Manufacturing Practice (cGMP) What are the risks associated with this PTC (immunogenic, tumour, teratoma, infection, toxicity)? Immunological rejection : YES,w e do not know the cell source Tumor formation YES, from the differentiation factors and the hydrogel. Negligible for MSC. Teratoma formation NO, only if we use pluripotent stem cells differentiated in MSC/adipocyte Transmission of infections YES, from the donor material and the cell manipulation Administration of toxic contaminants: YES, from the donor material and the cell manipulation r R D C 2. Cell expansion in a monolayer A new therapy for osteoarthritis is based on the use of chondrocytes, isolated, expanded in culture and repetitively injected into the patients. Cells are expanded into monolayer on the bottom of Petri dishes. The cell monolayer is kept in static culture covered with one layer of culture medium with thickness s. The partial pressure of oxygen present in the atmosphere of the incubator is Po. The flow of oxygen on the cells is known and equal to J c. The diffusion coefficient of the oxygen in the culture medium at 37°C is equal to D, the solubility coefficient of the oxygen in the medium at 37°C is equal to α. The hypothesis of plane symmetry can be assumed for the concentration of oxygen in the medium. This therapy has been successfully tested on the first 10 human patients and the conventional therapy is arthroplasty with an artificial prosthesis. Answer briefly to the questions in the table. What is the therapeutic product? The therapeutic product is the injection of chondrocytes What are the elements that identify the=therapeutic product as a PTC?= The main therapeutic agent are the cells, obtained after "non-minimal manipulation" consisting in cell isolation, expansion in petri dishes where cells can adhere and proliferate In what phase of the regulatory process for new PTCs can this PTC be localized? Human clinical trial of phase I (also known as pilot study) What are the standards that apply in this stage of the PTC process of development? Good Manufacturing Practice (GMP) for the production of the PCT; Good Clinical Practice (GCP) for the clinical trial What are the risks associated with this PTC (immunogenic, tumour, teratoma, infection, toxicity)? Immunological rejection : Y ES, W e do not know the cell source Tumor formation NO, because they are not multipotent cells. Teratoma formation NO, unless we use pluripotent stem cells differentiated in chondrocyte. Transmission of infections ALWAYS, from donor material and any manipulation Administration of toxic contaminants: ALWAYS, from donor material and from any manipulation X=s X=0 Classify all the cell sources potentially usable in human patients for this therapy. Cell source (based on immunogenicity) Cell type/s Advantages Criticalities Is it usable respect to the conventional therapy? Why? Autologous Chondrocytes isolated from the patient and expanded ___________ MSC isolated from the patient and differentiated (in chondrocyte) (ACI/MAC therapy currently in use) Immunological compatibility. NO Risk of tumor ________ Immunological compatibility. Risk of tumor: negligible Limited availability. Limited expandability. Invasive test. Limited availability. Limited expandability. Invasive test. YES Autologous iPS from somatic cells of the patient differentiated (in chondrocytes) Immunological Compatibility. Limitations in redifferentiation protocols Risk of teratoma NO Risk of teratoma Syngeneic Embryonic stem cells isolated from clones of the patient and differentiated (in chondrocytes) Immunological compatibility with the exception of the mitochondrial DNA. Ethical, technical and regulatory limitations for cloned cells. Limitations in differentiation protocols. Risk of teratoma NO Risk of teratoma Allogeneic Chondrocytes isolated from a human donor and expanded Available. Can be Industrialized. NO Risk of tumor Immunogenicity. NO Immunogeni city. Xenogeneic Chondrocytes isolated from a nonhuman donor and expanded Largely available from animal livestock. Can be industrialized. NO Risk of tumor Risk of acute i mmune rejection. Risk of xeno- zoonoses. Limited biological Functionality. NO acute immune rejection. 3. Calculate how many bioartificial hearts should simultaneously produce a cell factory to satisfy the transplant needs. Hypothesize the decellularization time null (the decellularized hearts are stored), the recellularization time t cell = 10 days/heart and the working days per year twork = 220 days/year. Assume to treat all US patients waiting for transplantation N = 2640. Data: t start = 0 (null decellularization time) t cell = 10 days/heart (recellularization time) t work = 220 days /year (working days) N = 2640 (US patients waiting for transplantation) Solution: Number of heart produced in a year (in series): ������������=������������ ������������������������������������������������ ������������������������������������������������������������ =220 ������������������������������������������������ ������������������������������������������������ 10������������������������������������������������ ℎ������������������������������������������������ =22ℎ������������������������������������������������������������ ������������������������������������������������ The number of years required exhausting the waiting list, or the number of hearts to be produced simultaneously are ������������=������������ ������������ =2640 22 = 120 4. Cell seeding/dilution Consider having 10 6 cells diluted in 1 ml of culture medium. Suppose that single adhered cell area is 180 µm 2. We want to cellularize a glass coverslip with an area of 1 cm 2. • Calculate the glass coverslip area in µm 2. • How many cells are needed to seed on the glass coverslip to have a 70% cell confluence? • How many µl must be taken from the mother cell solution? • If the final volume of the cell seeding is 1ml, how many µl of culture medium must be added to the final cell solution? Figure 1 - Protocols for cell seeding Data: ������������ 0= 10 6 ������������ Number of cells that is necessary to seed: ������������ ������������ = ������������������������������������������������ ������������������������ = = 70∗10 6 180 =0.38 10 6 ∼ 39 ∗10 4 ������������ 5. Number of focal adhesions Consider a fibroblast in a surface of a microbioreactor in culture medium (µ= 1cP), experiencing a Couette planar flow generated by the upper wall, distant h= 100 µm from the cell and moving with a velocity U M = 400mm/s, as shown in figure. Write the equation for the shear stress acting on the cell. Under the hypothesis that the cell in adhesion can be represented as a flat rectangle (50x100 µm), estimate the minimum number of focal adhesion points that the cell needs to withstand the force generated by the flow (force generated by a single contact adhesion f = 5nN). Solution The general equation for the wall shear stress is: ������������ = µ ������������������������ ������������ ������������������������ For this case, the velocity gradient can be simplified using the slope of the linear velocity profile, hence the shear stress may be calculated as: ������������ = µ ������������������������ ℎ (where [µ]= cP= 10−3������������������������ ������������ 2 ) The shear force acting on the cell is given by the shear stress times the area that the cell exposes to the flow: ������������ ������������ℎ = ������������� ������������ ℎ ������������ where A is estimated as the area of the cell (rectangle shape): ������������ = 5∗10 −9 ������������ 2. ������������ ������������ℎ = � ������������������������ ℎ ������������ = 10−3������������������������ ������������ 2 4∗10 −1������������ ������������ 1 10 −4������������ 5∗10 −9 ������������ 2 =2 10 −8 ������������ This force is withstood by the focal adhesion points that the cell possesses. Hence, the minimum number of focal adhesions’ points necessary is given by: ������������ ������������ ������������ = ������������ ������������������������������������ = 2∗10−8������������ 5 10 −9������������ =0.4∗10=4 or: ������������ ������������������������������������ =������������ ������������ ������������������������ ℎ������������ =10 − 3 ������������������������ ������������ 2 ∙4 ∗10 −1 ������������ ������������ ∙5∗10 −9 ������������ 2 10 −4 ������������∙5∗10 −9 ������������ =4 6. Organ regeneration To repopulate an adult decellularized monkey heart 10 11 cells are needed. Suppose to use MSC, obtained from the bone marrow and differentiated in vitro. The density of MSC in the aspirate (volume = 0.5ml) is 0.8*10 6 ������������������������������������������������ 3. The efficiency of the cell isolation protocol is low (0.02%). The division time of MSC is 12h. Calculate the expansion time in days. Solution t = t d*d 2 ln ln = ofX X d n = Vol * ������������ The final number of cells is ������������ ������������=10 11 cells The number of cells in the aspirate is N= ������������ * V= 0.8*10 6 ������������������������������������������������������������ ������������������������ 3 * 0.5*10 3mm 3= 0.4 10 9=400 10 6 cells The efficacy of the protocol is 0.02% The number of isolated cell is X i = 0.02%*400*10 6 cells = 2*10 -4*400*10 6 cells = 0.08*10 6 cells ������������������������ ������������������������ = 1011 0.08 10 6= 1.25*10 6 d = ln (1.25 10 6) ln2 = ln (1.25 10 6) 0.69 20.25 ~ 20 The expansion time is ������������ = ������������ * ������������ ������������ = 20*12 = 240ℎ * 1 day 24 ℎ = 10 days 7. Cell seeding on scaffold Consider different scaffolds with cylindrical shape characterized by the following parameters 1) Calculate the number of cells to be seeded on each scaffold to obtain a cell density equal to the density ρ_fin = 10 9 cells ml 2) Assume to isolate cells from a biopsy with a volume V = 8mm 3, ρ _biopsy =10 9 cells ml . Knowing that the cell division time is t d= 24h, calculate the expansion time needed to obtain enough cells to seed the cells on each scaffold at the previous cited density Solution: 1) SCAFFOLD radius (mm) Thickness (mm) Volume (mm 3)= π*(diameter/2) 2 * thickness A = 16/2 0.2 40.2 B = 6.3/2 5 155.86 C = 16/2 1.2 241.27 D = 16/2 2 402.12 Volumes: The cell density required for each scaffold is ρ = ������������������������ ������������ The final number of cells in each scaffold is ������������ ������������=������������∗ ρ ������������ ������������������������ = ������������ ������������ ∗ ρ������������������������������������ =40.2 ������������������������ 3∗10 6������������������������������������������������������������ ������������������������ 3 = 40.2∗10 6 ������������������������������������������������������������ ������������������������������������ = ������������ ������������ ∗ ρ������������������������������������ =155.86 ������������������������ 3∗10 6������������������������������������������������������������ ������������������������ 3= 155.86∗10 6 ������������������������������������������������������������ ������������������������������������ = ������������ ������������ ∗ ρ������������������������������������ =241.27 ������������������������ 3∗10 6������������������������������������������������������������ ������������������������ 3= 241.27∗10 6 ������������������������������������������������������������ ������������������������������������ = ������������ ������������ ∗ ρ������������������������������������ =402.12 ������������������������ 3∗10 6������������������������������������������������������������ ������������������������ 3= 402.12∗10 6 ������������������������������������������������������������ SCAFFOLD DIAMETER (mm) Thickness (mm) A 16 0.2 B 6.3 5 C 16 1.2 D 16 2 ������������������������������������������������ = ρ������������������������������������ ∗ ������������������������������������������������������������������������������������������������������������ 3) Assume to isolate cells from a biopsy with a volume V = 8 mm 3. Knowing that the cell division time is t d = 24h, calculate the expansion time needed to obtain enough cells to seed the cells on each scaffold at the previous cited density New data: ������������ ������������ = 24ℎ ������������ = 8 ������������������������ 3 ρ������������������������������������������������������������������������ = 10 9������������������������������������������������ ������������ ������������������������ = 10 6������������������������������������������������ ������������ ������������������������ 3 Solution: Isolated cells from a single biopsy: ������������ 0= ρ������������������������������������������������������������������������ ∗������������=10 6������������������������������������������������ ������������ ������������������������ 3 ∗8 ������������������������ 3 = 8∗10 6 ������������������������������������������������������������ Doublings: ������������������������= ln ( 40.2∗10 6 8∗10 6 ) ln2 = ln (5.03) 0.69 = 1.61 0.69 =2.3 ������������ ������������=ln ( 155.86∗10 6 8∗10 6 ) ln2 = ln (519.5) 0.69 = 2.97 0.69 = 4.3 ������������ ������������= ln ( 241.27∗10 6 8∗10 6 ) ln2 = ln (30.16) 0.69 = 23.497 0.69 = 4.9 ������������ ������������ =ln ( 402.12∗10 6 8∗10 6 ) ln2 = ln (50.2) 0.69 = 3.92 0.69 =5.6 Cell expansion time: ������������=������������∗������������ ������������ ������������ ������������= 24ℎ∗2.3 = 55.2ℎ = 2.3 ������������������������������������������������ ������������ ������������= 24ℎ∗4.3 = 103.2ℎ = 4.3 ������������������������������������������������ ������������ ������������= 24ℎ∗4.9 = 117.6ℎ= 4.9 ������������������������������������������������ ������������ ������������ = 24ℎ∗5.6 = 134.4ℎ= 5.6 ������������������������������������������������ X0=ρ*V Knowing: d f X X 2 0⋅ = d f X X 2 0 = ( ) 2 ln 2 ln ln d X X d o f = = The doublings are: 2 ln ln =ofX X d And the cell expansion time is: ddt t= . If I want to seed all the scaffolds together: ������������ ������������=������������ ������������������������+������������ ������������������������+������������ ������������������������+������������ ������������������������= 839.25*10 6 ������������ 839.25∗10 6 8∗10 6 ) ln2 = ln (104.9) 0.69 = 4.6 0.69 =6.7 ������������= 24ℎ∗6.7 = 160.8ℎ=6.7