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By the end of this section, you will be able to:
  • Describe the hydrostatic and colloid osmotic forces that favor and oppose filtration
  • Describe glomerular filtration rate (GFR), state the average value of GFR, and explain how clearance rate can be used to measure GFR
  • Predict specific factors that will increase or decrease GFR
  • State the percent of the filtrate that is normally reabsorbed and explain why the process of reabsorption is so important
  • Calculate daily urine production
  • List common symptoms of kidney failure

Having reviewed the anatomy and microanatomy of the urinary system, now is the time to focus on the physiology. You will discover that different parts of the nephron utilize specific processes to produce urine: filtration, reabsorption, and secretion. You will learn how each of these processes works and where they occur along the nephron and collecting ducts. The physiologic goal is to modify the composition of the plasma and, in doing so, produce the waste product urine.

Failure of the renal anatomy and/or physiology can lead suddenly or gradually to renal failure. In this event, a number of symptoms, signs, or laboratory findings point to the diagnosis ( [link] ).

Symptoms of Kidney Failure
Weakness
Lethargy
Shortness of breath
Widespread edema
Anemia
Metabolic acidosis
Metabolic alkalosis
Heart arrhythmias
Uremia (high urea level in the blood)
Loss of appetite
Fatigue
Excessive urination
Oliguria (too little urine output)

Glomerular filtration rate (gfr)

The volume of filtrate formed by both kidneys per minute is termed the glomerular filtration rate (GFR)    . The heart pumps about 5 L blood per min under resting conditions. Approximately 20 percent or one liter enters the kidneys to be filtered. On average, this liter results in the production of about 125 mL/min filtrate produced in men (range of 90 to 140 mL/min) and 105 mL/min filtrate produced in women (range of 80 to 125 mL/min). This amount equates to a volume of about 180 L/day in men and 150 L/day in women. Ninety-nine percent of this filtrate is returned to the circulation by reabsorption so that only about 1–2 liters of urine are produced per day ( [link] ).

Calculating Urine Formation per Day
Flow per minute (mL) Calculation
Renal blood flow 1050 Cardiac output is about 5000 mL/minute, of which 21 percent flows through the kidney.
5000*0.21 = 1050 mL blood/min
Renal plasma flow 578 Renal plasma flow equals the blood flow per minute times the hematocrit. If a person has a hematocrit of 45, then the renal plasma flow is 55 percent.
1050*0.55 = 578 mL plasma/min
Glomerular filtration rate 110 The GFR is the amount of plasma entering Bowman’s capsule per minute. It is the renal plasma flow times the fraction that enters the renal capsule (19 percent).
578*0.19 = 110 mL filtrate/min
Urine 1296 ml/day The filtrate not recovered by the kidney is the urine that will be eliminated. It is the GFR times the fraction of the filtrate that is not reabsorbed (0.8 percent).
110*.008 = 0.9 mL urine /min
Multiply urine/min times 60 minutes times 24 hours to get daily urine production.
0.9*60*24 = 1296 mL/day urine

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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progressive wave
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, 101-321-va - vertebrate form and function ii. OpenStax CNX. Jul 22, 2015 Download for free at https://legacy.cnx.org/content/col11850/1.1
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