The Block 6 Renal Physiology Brs W Expl Prt 1 below is the first in the revision quizzes that is designed to ensure you understood all the aspects on the renal system that we covered these last few weeks. Give it a try and keep an eye out for part two of the same.
A high-K± diet
Thiazide diuretic administration
Greater change in intracellular fluid (1CF) volume
Higher positive free-water clearance (CH2O)
Greater change in plasma osmolarity
Higher urine osmolarity
Higher urine flow rate
Normal acid-base status
She is hypoventilating.
The decreased arterial [HCO 3-] is a result of buffering of excess H + by HCO3-.
The decreased blood [K+] is a result of exchangeof intracellular H ± for extracellular K±.
The decreased blood [K+] is a result of increased circulating levels of aldosterone.
The decreased blood [K+] is a result of decreased circulating levels of antidiuretic hormone (ADH).
57 mm Hg
47 mm Hg
37 mm Hg
10 mm Hg
0 mm Hg
Results in reabsorption of less than 50% of the filtered load when the plasma concentration of HCO3- is 24 mEq/L
Acidifies tubular fluid to a pH of 4.4
Is directly linked to excretion of H + as NH4+
Is inhibited by decreases in arterial PcO2
Can proceed normally in the presence of a renal carbonic anhydrase inhibitor
There is net secretion of X.
There is net reabsorption of X.
There is both reabsorption and secretion of X.
The clearance of X could be used to measure the glomerular filtration rate (GFR).
The clearance of X is greater than the clearance of inulin.
Fixed acid production plus fixed acid ingestion
HCO3 - excretion
HCO3- filtered load
Titratable acid excretion
Filtered load of H±
Extracellular fluid (ECF) volume is 1 L
Intracellular fluid (ICF) volume is 1 L
ECF volume is 10 L
ICF volume is 10 L
Interstitial volume is 12.5 L
Clearance of glucose is zero
Excretion rate of glucose equals the filtration rate of glucose
Reabsorption rate of glucose equals the filtration rate of glucose
Excretion rate of glucose increases with increasing plasma glucose concentrations
Renal vein glucose concentration equals the renal artery glucose concentration will occur in a person who
Drinks 2 L of distilled water in 30 minutes
Begins excreting large volumes of urine with an osmolarity of 100 mOsm/L after a severe head injury
Is receiving lithium treatment for depression, and has polyuria that is unresponsive to the administration of antidiuretic hormone (ADH)
Has an oat cell carcinoma of the lung, and excretes urine with an osmolarity of 1000 mOsm/L
1/100 that of A-
1/10 that of A-
Equal to that of A-
10 times that of A-
100 times that of A
Increased filtration fraction
Extracellular fluid (ECF) volume expansion
Decreased peritubular capillary protein concentration
Increased peritubular capillary hydrostatic pressure
Inulin and D20
Inulin and radioactive albumin
Reabsorption is not saturated
Clearance equals inulin clearance
Secretion rate equals PAH excretion rate
Concentration in the renal vein is close to zero
Concentration in the renal vein equals PAH concentration in the renal artery
Higher free-water clearance (CH20)
Lower plasma osmolarity
Lower circulating level of antidiuretic hormone (ADH)
Higher tubular fluid/plasma (TF/P) osmolarity in the proximal tubule
Higher rate of H20 reabsorption in the collecting ducts
A ureteral stone
Dilation of the afferent arteriole
Dilation of the efferent arteriole
Constriction of the efferent arteriole
Here's an interesting quiz for you.