Reading...
Play button
Play button
Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
9 Cards in this Set
- Front
- Back
|
What is the main function of the renal cortex?
|
It is filtration of large volumes of blood through the glomeruli.
Logically, the renal cortical blood flow is relatively great and and little oxygen is extracted from the blood. |
|
|
Cortical blood flow is about 5ml/g of kidney tissue/min. True/False
|
True.
|
|
|
What is the app. blood floe in the outer medulla?
|
About 2.5mL/g/min. It is app. 0.6mL/g/min in the inner medulla. Metabolic work is being done, especially to reabsorb Na+ in the thick ascending limb of Henle, so larger amounts of O2 are extracted from the blood in the medulla.
|
|
|
Considering the O2 necessity of the medulla, what disadvantage exists for this region of the kidney?
|
This makes the medulla vulnerable to hypoxia if flow is reduced. NO, prostaglandins and many cardiovascular peptides in this region function in a paracrine fashion to maintain the balance between low blood flow and metabolic needs.
|
|
|
What is the theoretical concept behind measuring glomerular filtration rate?
|
It can be measured in intact animals and humans, by measuring the excretion and plasma level of a substance that is freely filtered through the glomeruli and neither secreted nor reabsorbed by the tubules.
|
|
|
Name 1 substance that is employed to measure GFR?
|
A substance for this purpose should be non-toxic and not metabolized by the body. Inulin, a polymer of fructose with a molecular weight of 5200 that is found in dahlia tubers.
|
|
|
How is inulin administered?
|
A loading dose of inulin is given IV, followed by a sustaining infusion to keep the arterial plasma level constant.
After the inulin has equilabrated with body fluids, an accurately timed urine specimen is collected and a plasma sample obtained halfway through the collection. |
|
|
How can inulin clearance be calculated?
|
U(in)= 35mg/ml(concentration in urine)
V= 0.9ml/min(urine flow per unit of time) P(in)= 0.25mg/ml(arterial plasma level) C(in)=U(in)V=35 x 0.9 _______ ________ P(in) 0.25 C(in) = 126mL/min(average approximation). |
|
|
What else can be used for GFR measurment?
|
Clearance of cretinine(C(cr) can also be used, but in primates, and humans, some creatinine is secreted by the tubules and some may be reabsorbed.
|
