Question

What will be the EFP, if BCOP is- 30, CHP is-20 and GHP +70?
A. +20 mm Hg
B. +30 mm Hg
C. +60 mm Hg
D. +120 mm Hg

Answer 1

Hint: The both kidneys of the human contain, the basic and structural unit called nephron, where it has glomerular and tubular part, almost 5 liters of the blood are filtered per time in the kidneys, and both kidney comprise around 1.5 billion of nephrons.

Complete answer:
> The effective filtration pressure is depends upon the arterial blood pressure in the glomerular capillaries GHP=70 mm Hg), colloidal osmotic pressure (BCOP=30 mm Hg) and hydrostatic pressure in the Bowman’s capsule (CHP=20 mm Hg).
Thus the $EFP= GHP-(BCOP+CHP), 70 \;– (20+30) = 20\text{mmHg}.$

> Glomerulus is present as the beginning structure of the nephron, where it contains a network of capillaries called tuft, and this capillary network is supported by mesangial cells, which are nothing but gaps present between the capillary structures. At the start of a glomerulus, there is an afferent arteriole, which is larger and it allows the blood to the filtrate, where capillaries in the glomerulus exit as different arterioles and venules.

> They are smaller in size and create resistance, because of this ultrafiltration happens. At the same time, the endothelium of the capillaries contains the sieve cells, so the molecules which are of that size enter the sieves and enter the Bowman’s capsule.

> The above said the afferent arteriole of the glomerulus is broad in size, so a lot of blood enters daily for the filtration.

Hence, the correct answer is option (A).

Note: The factors, that manage the glomerular filtration rate is major compression, afferent and efferent main conflict, colloidal osmotic pressure of the glomerulus, and the hydrostatic stress of the Bowman’s capsule and sympathetic stimulation.