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Flashcards in Week 1 Deck (10)
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1

Blood plasma is collected from both the afferent arteriole and efferent arteriole of a renal cortical glomerulus. Which of the following has the lowest afferent/efferent arteriole concentration ratio?

  1. Albumin
  2. Chloride
  3. Glucose
  4. Potassium
  5. Sodium

Learning objective: Learn critical features of renal anatomy and histology as they relate to function. Understand the normal process of glomerular filtration. SM 192

Explanation: The process of glomerular ultrafiltration creates a tubular fluid that is essentially protein free. Hence, as plasma passes from the afferent arteriole, through the glomerular capillaries to the efferent arteriole, the protein albumin concentration rises as approximately 20% of the fluid is filtered out, leaving the albumin behind, giving an afferent/efferent arteriole concentration ratio less than 1 (A is correct). By contrast, the glomerular capillary membrane is freely permeable to water and other small particles such as glucose, chloride, potassium, and sodium, so their concentrations do not change.

2

A 32-year old woman comes to the office due to fever, dysuria, abdominal pain. She has had several episodes of UTI since her teens. She has no other medical problems. Physical exam reveals mild suprapubic discomfort. A CT scan of the abdomen is obtained and is shown in the image below:

Which of the following prevented the proper ascent of the anomalous organ seen on the CT scan?

  1. Inferior mesenteric artery
  2. Inferior vena cava
  3. Persistent urachus
  4. Superior mesenteric artery

Learning objective covered: Describe how development of the kidney and bladder accounts for clinical conditions such as horseshoe kidney, pelvic and horseshoe kidneys, division of the kidneys/urinary tract, renal agenesis, multiple renal arteries, malrotation of the kidneys, exstrophy of the bladder, bladder fistulas, and urachal cysts, sinuses, and fistulas.  – SM 191

Explanation: The abdominal CT scan shows the kidneys joined at the poles, ie horseshoe kidney. Patients with horseshoe kidney are at increased risk of ureteropelvic junction obstruction and recurrent infection. During kidney development, the metanephros is initially located in the sacral region, then it ascends to vertebral levels T12-L3. When fusion occurs, the central isthmus of horseshoe kidney crosses the midline anterior to the aorta and posterior to the inferior mesenteric artery (IMA). During fetal development, the IMA limits ascent of the horseshoe kidney (A is correct).

Persistent urachus = direct connection between bladder and outside of the body at the umbilicus.

3

Which of the following is a feature of the principal cell in the cortical collecting duct?

  1. Normally generates a lumen-positive transepithelial potential difference
  2. Normally reabsorbs potassium and secretes sodium
  3. Responds to aldosterone with increased water permeability via aquaporins
  4. Utilizes a Na/K ATP-ase in the basolateral cell membrane
  5. Utilizes a Na/K/2Cl cotransport protein in the luminal cell membrane

Learning objective covered: Learn the major functions of the cortical and medullary collecting ducts – SM 195

 

Explanation: The principal cell is the major sodium-reabsorbing and potassium-secreting cell in the collecting duct (B is incorrect). It functions via specific sodium and potassium channels that increase luminal cell membrane permeability to these ions and a sodium/potassium ATPase in the basolateral cell membrane. (D is correct).

The movement of sodium through the luminal membrane creates an excess of negative charges in the tubular lumen (A is incorrect). C is incorrect – this describes action of ADH. Aldosterone regulates sodium and potassium transport. E is incorrect, this transporter is located in the thick ascending loop of Henle.

4

The figure below illustrates the extracellular and intracellular volume-osmolarity status of a patient (dashed lines) and that of a normal subject (solid lines) for comparison. This patient most likely suffers from which of the following conditions?

  1. Chronic vomiting
  2. Iatrogenic fluid overload with 0.9% (isotonic) NaCl
  3. Iatrogenic fluid overload with hypertonic solution
  4. Syndrome of inappropriate antidiuretic hormone secretion (SIADH)

Learning objective covered: Use knowledge of body fluid compartmentalization and osmosis to predict the outcome of fluid losses and gains. SM 192

Explanation: This patient has increased extra and intracellular volumes and a decreased osmolarity. SIADH results in inappropriately low water permeability of the collecting ducts and therefore water retention. As a result, patients with SIADH often present with hypotonic overhydration (D is correct).

Chronic vomiting leads to dehydration (A is incorrect). Fluid overload with isotonic NaCl results in volume expansion without a change in osmolarity (B is incorrect). Fluid overload with hypertonic solution results in volume expansion + increase in osmolarity (C is incorrect).

5

  1. A researcher is studying the effects of various manipulations on kidney blood flow and glomerular filtration. Which of the following is the most likely to decrease renal plasma flow, while increasing the glomerular filtration rate?
    1. Constriction of the afferent arteriole
    2. Dilation of the afferent arteriole
    3. Constriction of the efferent arteriole
    4. Dilation of the efferent arteriole

Learning objective: Understand the normal process of glomerular filtration. Understand the major determinants of GFR. SM 193

Explanation: Renal plasma flow is the volume of plasma delivered to the kidney per unit time. The RPF is provided by the renal blood flow. We want the renal plasma flow to decrease, therefore constriction of either the afferent arteriole or the efferent arteriole will accomplish this. Dilation of either would increase RPF. For GFR to increase alongside the decrease in RBF, we need to constrict the efferent arteriole (C is correct), which will lead to an increase in the glomerular capillary hydrostatic pressure as fluid backs up in the glomerulus.

Constriction of the afferent arteriole causes a decrease in the glomerular hydrostatic pressure and thus a decreased in GFR.

6

  1. A man traveling alone in Northern Africa becomes lost. After several days, he arrives at a village, at which point he has gone more than 24 hours without water. His urine osmolality is 1150 mOsm/L. The majority of the total amount of water filtered by this man’s glomeruli is reabsorbed in which of the following portions of the nephron?
    1. Cortical collecting duct
    2. Distal tubule
    3. Loop of Henle
    4. Medullary collecting duct
    5. Proximal tubule

Learning objective: Learn about the structure and function of the proximal tubule – SM 194

Explanation: The proximal tubules reabsorb > 60% of water filtered by glomeruli, regardless of the patient’s hydration status (E is correct). The water is absorbed iso-osmotically with solutes (Na, glucose etc.). No concentration of urine occurs in this segment.

His increase in urine concentration with water deprivation reflects the actions of ADH, released in response to increased plasma osmolarity. ADH acts to increase the water permeability of the collecting ducts. Up to 20% of the original filtered volume of water can be reabsorbed by the collecting ducts.

7

  1. Researchers working in the field of prematurity and birth defects are investigating the molecular mechanisms underlying embryonic kidney development. They are specifically investigating the signals exchanged between the metanephric diverticulum and metanephric mesoderm that drive their differentiation into the mature kidney. If a toxic insult occurs during development that selectively inhibits the renal structures formed by the metanephric mesoderm, which of the following adult derivates will fail to develop normally?
    1. Collecting ducts
    2. Distal convoluted tubules
    3. Major calyces
    4. Minor calyces
    5. Renal pelvis

Learning objective covered: Describe the significance of the pronephros, mesonephros, and metanephros in development and how the adult kidney develops from the metanephric mesoderm and ureteric bud (metanephric diverticulum).  Identify which parts of the kidney develop from the metanephros and from the diverticulum. SM191

Explanation:

Embryonic kidney development involves the sequential formation of 3 sets of nephric systems termed the pronephros, mesonephros and metanephros. First the pronephros forms, arising from the cephalic portion of the nephrogenic cord, and later completely regresses. Next, the mesonephros forms from the midportion of the nephrogenic cord. The structures of the mesonephros persist in males as the Wolffian ducts, which ultimately form ductus deferens and epididymis. In females, mesonephros regresses and becomes vestigial Gartner’s ducts.

Development of the metanephros = true kidney, begins with formation of the metanephric diverticulum = ureteric bud, which sprouts from the caudal portion of the mesonephric duct. The ureteric bud then penetrates into the intermediate mesoderm to induce formation of the metanephric mesoderm = metanephric blastema.

Ureteric bud gives rise to the collecting system of the kidneys, including collecting tubules and ducts, major and minor calyces, renal pelvis, ureters (Choices A,C,D,E are incorrect).

Metanephric mesoderm gives rise to the glomeruli, Bowman’s space, proximal tubules, loop of Henle and distal convoluted tubules (B is correct).

8

A patient with chronic renal insufficiency has a net functional loss of nephrons. If we assume that production of urea and creatinine is constant and that the patient is in a steady state, a 50% decrease in the normal GFR will cause which of the following to occur?

  1. Decrease plasma urea concentration
  2. Significantly increase plasma Na+
  3. Increase the percent of filtered Na excreted
  4. Not affect plasma creatinine
  5. Significantly decrease plasma K+

Learning objectives covered: Understand how GFR is regulated. Understand the basic concept of clearance. SM 193

Explanation: Both Na+ and K+ excretion are tightly regulated in order to maintain homeostatic plasma concentrations. Thus, as GFR decreases in disease, the percentage of filtered Na+ or K+ excreted increases to maintain a normal amount of Na+ or K+ excretion (C is correct)(B and E are incorrect). Substances like urea (some reabsorption) and creatinine (almost exclusively excreted by glomerular filtration) have no adaptive mechanisms to regulate plasma levels. Thus, a significant decrease in GFR results in significant increases in plasma creatinine and urea (assuming production of both substances remains constant) (choices A and D are incorrect).

9

A 25-year old has completed a long distance bicycle race during a hot, humid day. At the conclusion of the race, he provides a urine sample for testing. Assuming that his fluid intake during the race was 0, in what portion of the nephron is the tubular fluid osmolality the lowest?

  1. Bowman’s capsule
  2. Proximal tubule
  3. Thin descending loop of Henle
  4. Distal convoluted tubule
  5. Collecting duct

Learning objective covered: Understand key mechanisms for transport of sodium, chloride, water, bicarbonate, glucose and organic anions/cations. SM 194

Explanation: The bicycle racer is likely dehydrated from his exercise and ADH would be secreted by the posterior pituitary in response to increased extracellular osmolality. The area of the nephron with the lowest osmolality will be the early distal tubule (choice D is correct), a nephron diluting segment.

The fluid in Bowman’s capsule will be at the same osmolality as the plasma entering the glomerular capillaries (choice A is incorrect).

The fluid at the end of the proximal tubule is isoosmotic with the Bowman’s capsule fluid (B is incorrect) while the fluid in the thin descending loop of Henle is greatly concentrated due to water reabsorption into the hyperosmotic medullary interstitium (C is incorrect). The fluid at the end of the collecting duct can be very concentrated or very dilute, but since ADH greatly increases water reabsorption in this segment, the fluid becomes very concentrated (E is incorrect).

10

 

Nephrologists at a research hospital are investigating physiologic changes that occur in diabetes insipidus. They sample tubular urine at various points along the nephron in experimental animals with physiology similar to humans. In the absence of ADH, tubular fluid from which of the following sample sites will have the highest osmolarity?

  1. Proximal tubule
  2. Descending loop of Henle
  3. Junction between descending and ascending loop of Henle
  4. Ascending loop of Henle
  5. Distal convoluted tubules
  6. Collecting ducts

Learning objective: Learn the major divisions and functions of the loop of Henle. Understand the counter-current multiplier effect of the loop of Henle. SM 195

Explanation: The degree of water reabsorption varies along the different segments of the nephron depending on each segment’s permeability to water, the osmolarity of the renal interstitium and the presence or absence of ADH.

In the proximal tubule, fluid is passively reabsorbed with solutes, thus lumen is isoosmotic with plasma.

The descending loop of Henle is permeable to water, but not solutes. As this segment of the nephron moves into the medullary interstitium, water moves down its concentration gradient from the lumen to the highly osmotic medulla. No reabsorption of electrolytes occurs, thus fluid in lumen becomes hypertonic. In the absence of ADH, tubular fluid is most concentrated at the end of the descending loop of Henle (C is correct).

Thick ascending loop of Henle is impermeable to water. However, electrolytes are reabsorbed by the Na/K/2Cl cotransporter, thus osmolarity of the tubular fluid decreases. The tubular fluid becomes hypotonic by the end of this segment.

Distal convoluted tubule is also impermeable to water, reabsorption of solutes occurs via NaCl transporter, thus tubular fluid increases in hypotonicity.

Water permeability of collecting ducts depends on ADH. In absence of ADH, they are impermeable to water. Without ADH, tubular fluid can become as hypotonic as 50 mOsm/L while solutes continue to be removed.