Question

The electric capacitance of biological tissues serves as the basis for numerous medical technologies. For example, dielectric spectroscopy glucose reading (DSGR) is a noninvasive diagnostic technique that uses the electric capacitance of blood to estimate blood glucose levels.

A DSGR device records the electric current that results from applying a voltage source to the skin and underlying blood vessels. DSGR can be modeled by the circuit shown in Figure 1, in which V represents the applied potential, C_B represents the capacitance of the blood, and R_E, R_D, and R_B represent the electrical resistances of the epidermis, the dermis, and the blood, respectively.

Figure 1 DSGR circuit model including skin and superficial blood vessel

The capacitance of charged blood may be explained by red blood cell membranes acting as physical barriers that separate electrons introduced into the blood from positively charged ions in the red blood cell cytoplasm. The dielectric constant k of red blood cell membranes varies with glucose concentration (Figure 2) because glucose uptake by red blood cells alters the activity of membrane-bound proteins that regulate the flow of ions into and out of the cell.

Figure 2 Blood dielectric constant vs blood glucose concentration

DSGR readings vary with blood glucose concentration because the time needed to fully charge the blood is related to total blood capacitance. However, interpreting DSGR readings may be complicated by changes in biological variables other than blood capacitance. For example, the resistivity of blood varies in accordance with osmolarity such that changes in diet or hydration status influence the current measured by DSGR devices.

Biological system Circuit model 1.4 DSGR Electrodes Epidermis x1.3 RE Rp Dermis O 1.0 0.9 Blood 2.5 7.5 10 12.5 15 17.5 20 CB Glucose concentration (mM)

Assuming that the size and shape of red blood cells are uniform, by what factor would the quantity of red blood cells need to change to raise the capacitance of a blood sample to 250 % of its original value?

A. 2/5

B. (2/5)²

C. 5/2

D. (5/2)²

Answer 1

Capacitance of blood vessels is given by, C = delta V/delta P Here, C is the capacitance, delta V is the change in volume and delta P is the change in pressure Capacitance and volume are directly proportional therefore, C_2/C_1 = delta V_2/delta V_1 delta V_2/delta V_1 = (250 %)C_1/C_1 delta V_2/delta V_1 = (2.50)C_1/C_1 delta V_2/delta V_1 = 2.5 delta V_2/delta V_1 = 5/2 Therefore, option C is correct