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Biomedical Engineering - Technology for Regenerative Medicine

ese04-ENG

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1 WHOLE LUNG Tissue Engineering Petersen et al. [“Tissue-Engineered Lungs for in Vivo Implantation”, 2010] decellularized murine lungs and then recellularized them with epithelial cells and micro-vascular lung endothelial cells isolated from neonatal rat lungs. Suppose to design a specific bioreactor to mimic certain features of the human adult lung environment, including vascular perfusion (red arrows), which delivers oxygenated and glucose- rich medium to the lungs, and cyclic inflation-deflation of the alveoli with a liquid medium, mimicking ventilation (green arrows). Bioreactor sketch Exemplification of lung structure at alveolar level 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)? 2 Classify the cell sources potentially usable in human patients for this therapy. Cell source Cell type/s Advantages Criticalities Can be used? NB: conventional therapy: drugs administration 3 EXERCISE A Knowing that the vascular medium flow rate Q m is equal to 5 ������������ ������������������������������������, the arterial concentration c 1 of glucose is 4.86 mM [ P. Pezzati, M. Tronchin, G. Messeri. Biochimica clinica, 2004, vol. 28, n. 5-6 ] and the venous concentration c 2 is 5% lower, assess the total consumption of glucose. Determine also the total consumption of oxygen knowing that the difference of oxygen concentration between arterial and venous blood is 2 mM. Consider that the alveolo-capillary oxygen transfer ( ������������������������) is 310 ������������������������ ������������������������������������ and the molar volume of oxygen is equal to 17.36∗10 −3 ������������3 ������������������������������������ . EXERCISE B The lung is composed of over 40 types of cells including cells of the epithelium, endothelium, interstitial connective tissue, blood vessels, hematopoietic and lymphoid tissue, and mesothelial cells composing the pleura. A typical human lung biopsy contains approximately 45% type II epithelial, 2.5% type I epithelial cells and 15% endothelial cells. From histological analysis, the total cellular density (Nv) after 8 days of culture is 404∗10 6������������������������������������������������ ������������ ������������������������ 3 and we assume a human pulmonary volume of 3.5 l. Compute the number of cells for each cell type and the cellular consumptions of glucose and oxygen. (remember that the cellular consumptions is expressed in [ ������������������������������������������������ ℎ∗10 6������������������������������������������������������������ ]). EXERCISE C The lungs were cultured for 8 days; we know that the cell division for type II and type I epithelial cells occurs in 12 hours, while it occurs in 36 hours for endothelial cells. If the venous concentration c 2 of glucose decreases more than 10% with respect to c 1 (see data given at point A), hypoglycemia occurs. Determine the maximum initial number of cells that avoids hypoglycemia after 8 days of culture. 1 WHOLE LUNG Tissue Engineering Petersen et al. [“Tissue -Engineered Lungs for in Vivo Implantation”, 2010] decellularized murine lungs and then recellularized them with epithelial cells and micro -vascular lung endothelial cells isolated from neonatal rat lung s. Suppose to design a specific bioreactor to mimic certain features of the human adult lung environment, including vascular perfusion (red arrows), which delivers oxygenated and glucose - rich medium to the lungs, and cyclic inflation -def lation of the alveoli with a liquid medium, mimicking ventilation (green arrows) . Bioreactor sketch Exemplification of lung structure at alveolar level Answer briefly to the questions in the table: What is the therapeutic product? The decellularized murine lungs + neonatal rat lung epithelial cells and micro -vascular lung endothelial cells 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 , manipulation , expansion and seeding in the biological matrix In what phase of the regulatory process for new PTCs can this PTC be localized? In vitro What are the standards that apply in this stage of the PTC process of development? Good Laboratory Practice (GLP) What are the risks associated with this PTC (immunogenic, tumour, teratoma, infection, toxicity)? Immunological rejection : YES , b ecause cell sources is not autologous , and biological matrix composing the scaffold. Tumour formation : YES, endothelium and epithelium are rich with multipotent stem cells and for the biological matrix composing the scaffold (possible presence of multipote nt cells) . Teratoma formation: NO, because we do not us e re-differentiated pluripotent cells Transmission of infections YES, due to donor material (cell and scaffold) or from any manipulation Administration of toxic contaminants : YES, due to any manipulation. 2 Classify the cell sources potentially usable in human patients for this therapy. Cell source Cell type/s Advantages Criticalities Can be used? NB: conventional therapy: drugs administration Autologous Lung epithelial and endothelial cells , isolated from the patient (and expanded ) Immunological compatibility Limited availability and expandability. Invasive procedure. Risk of tumour from multipotent cells in the endothelium NO Autologous iPS from somatic cells of the patient, re - differentiated (in lung epithelial and endothelial cells ) Immunological Compatibility. Limitations in re -differentiation protocols Risk of teratoma NO Syngeneic Embryonic stem cells isolated from clones of the patient and differentiated (in lung epithelial and endothelial cells ) Immunological compatibility except for mitochondrial DNA. Ethical, technical, and regulatory limitations for cloned cells. Limitations in differentiation protocols. Risk of teratoma . NO Allogeneic Lung epithelial and endothelial cells isolated from a human donor (and expanded ) Fairly available, considering only the use of cells that present the correspondence of human leukocyte antigens (HLA) Can be Industrialized. Risk of tumour from multipotent cells in the endothelium Risk of immune reactio n. NO Xenogeneic Lung epithelial and endothelial cells isolated from a non -human donor (and expanded ) Largely available from animal livestock. Can be industrialized. Risk of acute immune rejection. Risk of xeno -zoonoses. Limited biological functionality. Risk of tumor from multipotent cells NO 3 EXERCISE A Knowing that the vascular medium flow rate Q m is equal to 5 � �������� , the arterial concentration c1 of glucose is 4.86 mM [P. Pezzati, M. Tronchin, G. Messeri. Biochimica clinica, 2004, vol. 28, n. 5 -6] and the venous concentration c2 is 5% lowe r, assess the total consumption of glucose . Determine also the total consumption of oxygen knowing tha t the difference of oxygen concentration between arterial and venous blood is 2 mM. Consider that the alveolo -capillary oxygen transfer (������������) is 310 �� �������� and the molar volume of oxygen is equal to 17 .36 ∗10 −3 �3 ��� . DATA: Qm=5 � �������� c1g= 4.86 mM [1M=1 ��� � ] c2g= c1g-5% c1g ������������ =310 ml/min =310 �� 3 �������� . (alveolo -capillary oxygen transfer) c1-O2-c2-O2=∆�������2= 2mM O2 molar volume= 17 .36 ∗10 −3 �3 ��� . Compart imental model: Vol ∙∂c ∂t= Q1∙c1 − Q2∙c2+ p− v SOLUTION: The equation of the compartmental model is: ������� ������� �������� = ��������1�− ��������2�+ ��− �� Since • There is no glucose accumulation ������� ������� = 0 • There is no production of glucose within the tissue ��=0. The equation simplifies into: ������� ∗∆�= �� �2�= �1�− 0.05 �1�= �1�∗0.95 = 4.86 ���� � ∗0.95 = 4.62 ���� � ∆�= �1�− �2�= (4.86 − 4.62 ) ���� � = 0.24 ���� � ��= ������� ∗∆�= 5 � �������� ∗0.24 ���� � = 1.2 ���� �������� 4 To calculate the total oxygen consumption, we may use again a compartmental model, with the following adaptation s: ������� ������� �������� = ��������1−������2− ��������2−������2+ ��2− ��2 Since there is no accumulation : ������� ������� = 0. 0 = ��������1−������2− ��������2−������2+ �������2− �������2= �������∆�������2+ �������2− �������2 �������2= ������� ∗∆�������2+ �������2 We may model the oxygen transfer from the alveoli to the capillaries as a “distributed oxygen production ”. Oxygen production must be converted from volumetric to molar units: �������2= ������������ ���������� ������ = 310 �� 3 �������� ∗10 −6� 3 �� 3 17 .36 ∗10 −3 � 3 ��� = 17 .86 ���� �������� Applying the compartmental model, the total oxygen consumption can then be computed: �������2= ������� ∗∆�������2+ �������2 �������2= 5 � �������� ∗2∗10 −3��� � + 17 .86 ∗10 −3��� �������� = 27 .86 ���� �������� 5 EXERCISE B The lung is composed of over 40 types of cells including cells of the epithelium, endothelium, interstitial connective tissue, blood vessels, hematopoietic and lymphoid tissue, and mesothelial cells composing the pleura . A typical human lung biopsy contain s approximately 45% type II epithelial , 2.5% type I epithelial cells and 15 % endothelial cells. From histological analysis, the total cellular density (Nv ) after 8 days of culture is 404 ∗10 6����� �� 3 and we assume a human pulmonary volume of 3.5 l. Compute the number of cells for each cell type and the cellular consumptions of glucose and oxygen . (remember that the cellular consumptions is expressed in [ ��������� ℎ∗106����� ]). DATA 45% type II epithelial cells 2.5% type I epithelial cells 15% endothelial cells ttot = 8 days Nv = 404 ∗10 6����� �� 3 total cell density after 8 days Volhp = 3.5l human pulmonary volume [1l=1dm 3=1000 cm 3] N.B. Data to be used from the exercise A : �������� = 1.2 ���� �������� �������2= 27 .86 ���� �������� SOLUTION : Compute the total cell number: �������,���������� = ������� �������� ℎ�= 404 ∗10 6����� �� 3 ∗3.5∗10 3 �� 3= 1.414 ∗10 12 ����� And then the fractions for each cell type: Xf_epII = 0.45*X f,total = 1.414 ∗10 12 ∗0.45 = 6.36 ∗10 11 ����� Xf_epI = 0.025*X f,total = 1.414 ∗10 12 ∗0.025 = 3.54 ∗10 10 ����� Xf_endo = 0.15*X f,total = 1.414 ∗10 12 ∗0.15 = 2.12 ∗10 11 ����� Compartimental model: Vol ∙∂c ∂t= Q1∙c1 − Q2∙c2+ p− v Global consumption  [µ��� ℎ , µ��� � ] Volumetric consumption V [µ��� �� ∗ℎ, µ��� ��3∗ℎ] Cellular consumption V cell [ µ��� 106 ���� ∗ℎ] V= Vol Vcell= X ( where X is the number of cells) 6 Assume that all cell types have the same individual consumption, then you can write: �������,���� = 15% �������,���� + 2.5% �������,�� ������+ 45% �������,�� ������������+ 37 .5% �������,��ℎ�� ����� Where the last term takes into account all other cells (those that mainly make up the parenchyma). The oxygen consumption in ��������� ℎ∗10 6����� for each cell type is : �������������������� �������������� �������������� = �� �������������������������������� = 0.051 ��������� 106 ����� ∗ℎ ������������2������������� ���������������������������� = �������2 �������,���������� = 1.182 ��������� 10 6 ����� ∗ℎ 7 EXERCISE C The lungs were cultured for 8 days; we know that the cell division for type II and type I epithelial cells occurs in 12 hours, while it occurs in 36 hours for endothelial cells. If the venous concentration c 2 of glucose decreases more than 10% with respect to c 1 (see data given at point A) , hypoglycemia occurs. Determine the maximum initial number of cells that avoids hypoglycemia after 8 days of culture . DATA ttot = 8 days td,ep = 12h td,endo = 36 h c2,g_critical = c 1,g - 10% c1,g c1,g = 4.86 mM Qm=5 � �������� Nv = 404 ∗10 6 ����� �� 3 cell density after 8days Volpulmonary = 3.5l ��= 1.2 ���� �������� Xf,epII = 1.414 ∗10 12 ∗0.45 = 6.36 ∗10 11 ����� Xf,epI = 1.414 ∗10 12 ∗0.025 = 3.54 ∗10 10 ����� Xf,end o = 1.414 ∗10 12 ∗0.15 = 2.12 ∗10 11 ����� SOLUTION : The logical steps to solve this exercise are: ������������= ������� 2� � = 1 �� ∗���� Xf=N f*Vol �∶ �������= ��,�� ∶�������,�� W here ��,�� is calculated using the compartment model equation as in previous exercises. Expansion time t cr = d cr*t d Number of passages ��� = ��(�������,������� �������,�������) ��2 Number of cells X f=N f*Vol 8 Indeed, c onsidering ������� ∗∆�= �� And b y using : �2,�� = �1− 0.1�1= 0.9�1= 0.9∗4.86 mmol l = 4.37 mmol l ∆��� = �1− �2,�� = (4.86 − 4.37 )���� � = 0.49 mmol l The consumption critical value is ��,�� = ������� ∗∆��� = 5 � �������� ∗0.49 ���� � = 2.45 mmol min Considering a proportionality of glucose consumption with cell density , we may compute the critical cell density based on the current, noncritical, condition �∶ �������= ��,�� ∶�������,�� �������,�� = ��,�� ∗������� � = 2.45 ���� �������� ∗404 ∗10 6����� �� 3 1.2 mmol min = 8.15 ∗10 8����� �� 3 = 815 ∗10 6����� �� 3 Therefore, we may derive the number of cells at which the critical condition occurs: �������,�� = �������,�� ∗������� �������������������� = 8.15 ∗10 8����� �� ∗3500 �� = 2.85 ∗10 12 ����� And, separating epithelial cells from endothelial cells: �������,�������� +������������,�� = (0.025 + 0.45 )�������,�� = (0.025 + 0.45 )∗2.85 ∗10 12 ����� = 1.36 ∗10 12 ����� �������,���� ,�� = 0.15 �������,�� = 0.15 ∗2.85 ∗10 12 ����� = 4.28 ∗10 11 ����� The relation between X i and X f is: �������,�� = 2� ������������ The number of cell divisions, d, is equal to � = 1 �� ∗���� where t d is the time of cell division and t tot the time of expansion. ��������������� = ��������������������� = 1 12ℎ ∗8 �������������� = 16 ������������������������ ����� = 1 36ℎ ∗8 �������������� = 5.33 ������������������������ Thus, the initial maximum amounts of cells to not reach the hypoglycemic condition in 8 days is : 9 ������������,�������� ,�� = �������,�������� +������������,�� 2��� = 1.36 ∗10 12 ����� 216 = 20 .8∗10 6����� ������������,���� ,�� = �������,���� ,�� 2����� = 4.28 ∗10 11 ����� 25.33 = 10 .6∗10 9�����