NURA 806 Advanced Physiology

Exam 2 Study Guide: Units 5, 6 & 7

Chapter 25:

1. Identify the fluid compartments in the human body and state approximately how much volume each contains.

(p. 286)

Extracellular Fluid: 14 L, (20% of total body weight)

Interstitial Fluid: 11L

Blood Plasma: 3 L

Intracellular Fluid: 28 L (40% of total body weight)

Trasncellular Fluid: 1-2 L(includes fluid in the synovial, peritoneal, intraocular spaces and the CSF)

Note that Total Body Water is about 42 L in a 70kg adult. This is about 60% of body weight.

Note that increase in age and obesity, decreases TBW. This will ultimately affect mechanism of action for anesthesia drugs

1. What ions contribute primarily to osmolarity of extracellular and intracellular fluids?

(p. 288 Table 25-2: would not copy from website)

Refer to this table for any questions on the test. The main ones seem to be Sodium, Potassium, Calcium, Chloride and a whole bunch of others.

Inside the cell: High conc of Potassium , Magnesium, proteins, and Phosphate ions

Extracellular: High conc of Sodium, Chloride, and bicarbonate

1. Differentiate between isotonic, hypotonic and hypertonic solutions.

(p. 291-292)

Isotonic (280 mOsm/L): Has the same concentration of water and constituents as the interior of a cell. If a cell is placed in this solution it will not shrink or swell.

Ex: NaCl & 5% Glucose Soln initially (1st phase- before glucose metabolism)

Hypotonic (200mOsm/L): Solution has a lower concentration of constituents compared to the interior of a cell. If a cell is placed in this type of solution, water will diffuse into the cell until the concentrations inside and outside of the cell are equal, causing the cell to swell.

Ex: 0.45% NaCl, 5% Glucose (after glucose metabolism)

Hypertonic (360mOsm/L): Solution has a higher concentration of constituents compared to the interior of a cell. If a cell is placed in this solution, water will diffuse out of the cell until the water concentrations are equal, causing the cell to shrink.

Ex: 3% NaCl, 21% NaCl

1. You should be able to explain Figure 25-6 on page 293.

Add Isotonic Soln: Addition of an isotonic fluid to the extracellular fluid compartment, there is no change in the osmolarity of the extracellular fluid, only an increase in the volume of the extracellular fluid.

**NO EXCHANGE OF WATER**

Add Hypertonic Soln: Addition of hypertonic solution to the extracellular fluid, the extracellular osmolarity increases and causes osmosis of water out from the cells into the extracellular space. This causes an increase in the extracellular fluid volume, a decrease in the intracellular fluid volume, and an increase in the osmolarity of both spaces.

**WATER MOVES OUT OF CELL**

Add Hypotonic Soln: Addition of a hypotonic solution to the extracellular fluid, the osmolarity of this space decreases. This will cause some of the extracellular water to diffuse into the cells. This will lead to a decrease in the osmolarity of both compartments, and an increases in the cellular fluid compartment. The extracellular fluid volume will be increased only because of the addition of solution to the compartment. The intracellular volume increases to a greater extent.

**WATER MOVES INTO CELL**

1. Review the causes of intracellular edema and of extracellular edema.

(p. 296-297)

Intracellular edema is caused by

1) hyponatremia,

2) depression of the metabolic systems of the tissues, and

3) lack of adequate nutrition to the cells.

If blood flow to tissues is too low to maintain normal cellular function, the ion pumps are depressed. If sodium ions cannot be taken out of the cells, the excess extracellular sodium concentration causes water to diffuse in to the cells.

Inflamed tissues can also cause intracellular edema. This leads to increased membrane permeability to substances that are not normally taken in. This leads to osmosis of water into the tissue cells.

Extracellular edema is caused by abnormal leakage of fluid from the plasma to the interstitial spaces across the capillaries, and failure of the lymphatics to return fluid from the interstium back into the blood (lymphedema). The most common cause is excess capillary filtration.

Lymph edema is caused by blockage of the vessels or loss of the vessels. This can be severe because proteins lost into the interstitial fluid have no other way to be returned to the circulation. Proteins lost to the interstitial space cause a higher osmotic pressure in this space and draw even more water in, making the problem worse.

Page 297 lists a great number of causes of extracellular edema broken down into 4 categories:

1. Increased capillary pressure

2. Decreased plasma proteins

3. Increased capillary permeability

4. Blockage of lymph return

1. What homeostatic “safety” factors are in place to prevent edema?

(p 298-300)

1. Low compliance of the interstitium when interstitial fluid pressure is in the negative pressure range (-3mmHg)

2. The ability of lymph flow to increase 10 to 50 fold, (about 7 mmHg)

3. Washdown of interstitial fluid protein concentration, which reduces interstitial fluid colloid osmotic pressure as capillary filtration increases. (Interstitial fluid colloid pressure about 7mmHg)

Interstitial tissue is physiologically kept at a negative pressure. This creates kind of a suction that holds the tissues together. It takes a lot of fluid to overcome this negative pressure and make the tissues more compliant. With small increases in fluid in the interstitium, the pressure of this space increases markedly, and then forces the collected fluid back into the capillary bed and into the circulation again.

1. Define “effusion.”

Effusion: when edema occurs in the subcutaneous tisssues adjacent to the potential space, edema usually collects in the potential space as well and this fluid is called effusion.

In the abdominal cavity this is called ascites. This can also occur in the pleural cavity, the pericardial cavity, and joint spaces.

Chapter 26: (This is an important chapter)

1. List the homeostatic functions of the renal system.

(p 303-304)

▪ Excretion of metabolic waste products and foreign chemicals

*Urea, creatinine, bilirubin, hydrogen ions, drugs, toxins

(urea= protein metabolism)

(Creatinine = muscle metabolism)

(Bilirubin = hemoglobin metabolism)

(uric acid = nucleic acid metabolism)

▪ Regulation of water and electrolyte balances:

*(Sodium, Water, Potassium, Hydrogen Ions, Calcium, Phosphate, and Magnesium)

▪ Regulation of body fluid osmolality and electrolyte concentrations

▪ Regulation of arterial pressure

*Renin-angiotensin sys, prostaglandins, kallikrein-kinin sys

*Control extracellular Fluid Volume

▪ Regulation of acid-base balance

*Excretes acids (Sulfric and Phosphoric acid only excreted by the kidney)

*Regulate body fluid buffers (bicarb)

▪ Secretion, metabolism, and excretion of hormones

▪ Gluconeogenesis

1. Briefly describe the innervation of the urinary bladder.

(p. 308)

MOSTLY PARASYMPATHETIC (craniosacral). M3 Receptor contracts the detrusor m.

The bladder is innervated by the pelvic nerves, which connect to the spinal cord through the sacral plexus at S2-S3. The pelvic nerves contain both sensory and motor fibers. The sensory fibers detect degree of bladder wall stretch. The motor fibers are parasympathetic fibers, which are connected to the detrusor muscle M3 receptor.

There are skeletal muscle nerves from the pudedal nerve which innervate the external bladder sphincter.

There are sympathetic nerves through the hypogastic nerves connecting to the spinal cord at L2 which inhibit contraction of the bladder wall, and inhibit relaxation of the bladder sphincters.

1. Trace the innervation of the micturation reflex. What higher controls influence this?

Micturition Reflex - You Tube style, your welcome -Dunice

(p. 309-310)

This reflex results from sensory stretch receptors in the bladder wall, especially receptors in the posterior urethra. These signals go to the sacral segments of the spinal cord through the pelvic nerves and the reflexively back again to the bladder through the parasympathetic nerve fibers by way of these same nerves to cause contraction of detrusor muscle.

The impulses follow a positive feedback loop. The stretch receptors sense an increased pressure, causing a contraction, which activates the stretch receptors again, to produce a cycle of small contractions until the pressure is great enough to cause micturition.

Although the micturition reflex is an autonomic spinal cord reflex, it can be inhibited or facilitated by centers in the brain. These centers include the brain stem (mainly in the pons) and several centers of the cerebral cortex which are mainly inhibitory, but can become excitatory. These higher brain centers keep the micturition reflex inhibited until voiding is desired. These centers can keep the external urethral sphincter contracted until voiding is desired.

1. What is the average daily GFR? Why is it so high? (p. 312)

180 L. Total plasma volume is only about 3 L, meaning that the plasma volume is filtered about 60 times a day. This allows the kidneys to rapidly, and precisely control the volume of the plasma fluid and the composition of the body fluids. It also allows for rapid removal of waste products from the body that mainly depend on glomerular filtration for their excretion.

1. Know the layers of the glomerular-capillary membrane (including fenestrae and slit pores).

(p 312-313)

1. the endothelium (innermost layer)

(perforated by thousands of fenestrae- hinder protein passage)

2. basement membrane (middle layer)

(proteoglycan fibrillae having negative charges – hinder protein passage)

3. layer of epithelial cells (podocytes) surrounding the outer surface of the capillary basement membrane (outermost layer)

(Contain slit pores that filtrate passes through with neg. charge – hinder protein passage)

1. Describe the opposing forces that result in net filtration pressure (Figure 26-13 on page 314). How do changes in capillary osmotic and hydrostatic pressures affect GFR?

(p. 314-315)

Forces Favoring Filtration (mm Hg)

Glomerular hydrostatic pressure

60

Bowman's capsule colloid osmotic pressure

0

Forces Opposing Filtration (mm Hg)

Bowman's capsule hydrostatic pressure

18

Glomerular capillary colloid osmotic pressure

32

The net filtration pressure is obtained through the sum of the hydrostatic and colloid osmotic forces that either favor or oppose filtration across the glomerular capillaries. As can be seen in the included diagram, there is a net +10mmHg pressure towards filtration of plasma fluids into the Bowman’s capsule.

There is not much change in the day to day GFR d/t changes in the glomerular pressures listed here. The net glomerular pressures can be decreased through chronic high blood sugars associated with diabetes, which increase the thickness of the basement membrane of the glomerular capillaries and ultimately decrease GFR through decreasing the net filtration pressure through the glomerular capillaries.

The pressure exerted by Bowman’s capsule can become greatly increased by conditions where the urinary tract becomes obstructed. This would markedly decrease GFR.

Increasing glomerular capillary colloid osmotic pressure decreases GFR. An increase of this pressure can be accomplished by decreasing renal blood flow, which effectively increases the fraction of fluid filtered relative the plasma in the filtered fluid.

Increased glomerular capillary hydrostatic pressure increases GFR. This is the main determinant of GFR. Changes in this pressure are related to arterial pressure, afferent arteriole resistance, and efferent arteriole resistance.

1. Describe the effects of constricting the afferent and efferent arterioles on GFR.

(p. 315 under last column)

Afferent Arteriolar

Efferent Arteriolar

Increased Resistance

↓ GFR

↑ GFR, then ↓

Decreased Resistance

↑GFR

↓GFR

Increased resistance of the afferent arteriole reduces glomerular hydrostatic pressure and therefore decreases GFR. The opposite is also true.

Increased resistance of the efferent arteriole increases the resistance to outflow from the glomerular capillaries. This raises the glomerular hydrostatic pressure and increases GFR slightly. It can decrease GFR if the resistance is too great and reduced blood flow through the glomerular capillaries.

1. How are sodium and oxygen related to renal tubular activity?

(p. 316)

Na reabsorption is an active process requiring ATP. Oxygen needed for production of ATP by renal mitochondria. Accounts for 75% of renal oxygen demands.

For sodium to be reabsorbed from the renal tubules, oxygen is required d/t the energy needs of this process. Thus, oxygen consumption of the renal tubules is directly related to the amount of sodium being reabsorbed by the kidneys. This process is also closely related to GFR, which affects the amount of sodium available for reabsorption.

1. What is the effect of norepinephrine, epinephrine and angiotensin II on GFR. Why?

Hormone or Autacoid

Effect on GFR

Norepinephrine (vasoconstrictor)

↓

Epinephrine (vasoconstrictor)

↓

Endothelin (vasoconstrictor)

↓

Angiotensin II (vasoconstrictor, Na retention)

↔ (prevents ↓)

Endothelial-derived nitric oxide (vasodilation)

↑

Prostaglandins (vasodilator)

↑

Norepinephrine and epinephrine, released in SNS responses to decreased blood pressure rarely have an effect on the GFR or constriction of renal arteries, unless the stimulation is very strong related to severe hypotension or brain ischemia.

Both nor-epi and epi cause constriction of both the afferent and efferent arterioles which effectively decrease GFR.

Angiotensin II preferentially constricts this efferent arteriole, rather than the afferent arteriole. Remember that angiotensin II is released in situations where GFR or renal blood flow is already decreased. The constriction of the efferent arteriole in this situation would serve to bring GFR back to a normal level and prevent any further drop in GFR.

1. What is the role of the macula densa in renal autoregulation.

(p 320)

1) Decreases resistance o blood flow in the afferent arterioles ↑GFR

2) Increases renin release from juxtaglomerular cells of the afferent and efferent arterioles, which are the major sites for renin. Leading to Angiotensin II constricting efferent arterioles ↑ GFR

The macula densa is a collection of cells in the distal convoluted tubule in the section that passes near the renal arterioles. These cells sense changes in volume delivery to the distal tubule through mechanisms not well understood. Decreased GFR seems to slow fluid delivery to the loop of Henle causing increased reabsorption of sodium and chloride ions in the ascending loop of Henle, thereby reducing the concentration of sodium chloride at the macula densa cells. These decreases in ion concentrations causes decreased resistance to blood flow in the afferent arterioles, and increases renin release from the juxtaglomerular cells. This leads to increased release of angiotensin II that constricts the efferent arterioles and increases GFR.

Chapter 27:

1. By what pathways are fluids and solutes exchanged between the tubules and blood?

(p. 323-325)

Fluids, like water, are reabsorbed through passive transport through the cellular walls of the renal tubules into the interstitial fluid/space. This fluid is then moved into the capillary beds there d/t the hydrostatic and colloid osmotic forces there. The peritubular capillary beds behave much like any other capillary bed in the body.

Solutes are generally taken across the cell through an active transport system, which can move solutes against a chemical gradient and requires energy. The solutes are moved to the interstitial fluid space just like fluid transport from the renal tubules and are moved into the peritubular capillaries in the same manner as well. This is called primary active transport.

1. What is secondary active transport? Counter transport?

(p.325-326)

Secondary Active Transport: Some molecules, like glucose, are not removed from the renal tubules by primary active transport where they bind to a carrier protein and ATP is used as energy to move the molecule across the cell membrane against a concentration gradient. In secondary active transport, the energy of sodium being transported through an ion pump is used to simultaneously move the glucose molecule into the cell membrane. Thus it is an indirect, energy dependent transport system. The renal tubules are so useful in moving glucose out that the reabsorption rate nearly 100%. The same principle applies to moving amino acids out of the renal tubules.

Counter Transport: This is another type of secondary active transport where one molecule of a substance is exchanged for another molecule of another substance that needs to be excreted. Examples would include moving a sodium ion out of the renal tubules while simultaneously using that energy to move a hydrogen ion into the tubules.

1. What is a “transport maximum?” What solutes do not have a transport maximum? What is the threshold for glucose excretion in the urine?

The transport maximum refers to the maximum rate at which a substance can be reabsorbed through active transport into the circulatory system from the renal tubules. The rate is limited by the saturation of the specific transport proteins involved with reabsorption of that particular substance. This occurs when the tubular load, or amount of substance delivered to the tubules, exceeds the number of transport proteins available for that substance.

Solutes that have a transport maximum include glucose, phosphate, sulfate, amino acids, urate, lactate, plasma protein. Solutes that have a transport maximum for excretion rate include creatine and para-aminohippuric acid.

The threshold for glucose excretion in the urine is about 375 mg/min. (The tubules filter about 125 mg/min and the transport maximum is 375 mg/min, so this only happens with large increases in GFR and/or plasma glucose concentrations that increase the filtered load to above 375mg/min)

1. How is water moved out of the renal tubules? What part of the tubule is impermeable to water?

(p. 328)

Water is moved mostly through the tight junctions of the proximal tubule by osmosis. Most water is reabsorbed through the proximal tubules. As you move down the renal tubule loop, the membrane becomes less and less permeable to water. The ascending loop of Henle is nearly impermeable to water. The ascending loop and the distal tubule can become permeable to water in the presence of ADH.

1. What part of the tubule is sensitive to vasopressin (ADH)?

(p. 328)

The distal tubules, the collecting tubules, and the collecting ducts.

1. How and where do angiotensin II and aldosterone affect the sodium and water balance in the renal tubules?

(p. 332-333 and 338 also chart at top of 338)

These 2 substances act on the late distal tubule and the collecting tubule.

Aldosterone: stimulates the transporters at these sites to reabsorb sodium (followed by water), and initiate the secondary active transport of potassium into the renal tubules.

Angiotensin II: acts in several ways. It stimulates aldosterone secretion. It constricts the efferent arterioles, which causes reduced peritubular capillary hydrostatic pressure, which increases net tubular reasbsorption especially from the proximal tubules. Constriction of the efferent arterioles reduces renal blood flow and increases the concentration of proteins and the colloid osmotic pressure in the peritubular capillaries. This increases the reasbsorptive force at the peritubular capillaries and raises tubular reabsorption of sodium and water. Angiotensin II also directly stimulates the proximal tubules, the loops of Henle, the distal tubules and the collecting tubules in a manner similar to Aldosterone to increase sodium and water reasbsorption.

1. What is the effect of aldosterone on potassium?

(p. 332-333)

Because aldosterone stimulates sodium reasbsorption through a pathway, which exchanges a sodium ion for a potassium ion, the net effect would be a decrease in the total serum potassium level.

1. What does atrial natiruretic peptide do?

(p. 339)

When specific cells in the cardiac atria are stretched due to high fluid volumes, they release this substance. It directly inhibits the reabsorption of sodium and water in the renal tubules, especially the collecting ducts. ANP also inhibits renin secretion, which in turn inhibits angiotensin II and aldosterone secretion, to further inhibit sodium and water reabsorption.

Chapter 28:

1. What is the osmolarity of fluid in the early distal tubule?

(p. 346)

100 mOsm/L

This is regardless of whether ADH is present or absent – fluid in the early distal tubule will always be hypo-osmotic with an osmolarity of about 1/3 that of the plasma

1. What is the maximum osmolarity to which the kidneys can concentrate urine?

(p. 347)

1200-1400 mOsm/L

This is 4 to 5 times the osmolarity of the plasma

1. List the major factors that contribute to the buildup of solute concentration in the renal medulla.

(p. 348)

1. Active transport of sodium ions and co-transport of potassium, chloride, and other ions out of the thick portion of the ascending limb of the loop of Henle into the medullary interstitium.

2. Active transport of ions from the collecting ducts into the medullary interstitium.

3. Facilitated diffusion of urea from the inner medullary collecting ducts into the medullary interstitum.

4. Diffusion of only small amounts of water from the medullary tubules into the medullary interstitium, far less than the reabsorption of solutes into the medullary interstitium.

1. What is the role of the vasa recta in maintaining medullary hyperosmolarity?

There are two special features of the renal medullary blood flow that contribuite to the preservation of the high solute concentrations:

1. Medullary blood flow is very low: sufficient to meet the metabolic requirements of this area of the kidney, but low enough to minimize solute loss from the medullary interstitium.

2. The vasa recta serve as countercurrent exchangers, minimizing washout of solutes from the medullary interstitium.

The vasa recta are the capillaries that provide blood flow through the medullary portion of the kidneys. The capillaries are built in a loop fashion, similar to that of the loop of Henle found in the nephron. This means that as blood passes through the capillaries, it behaves as other capillaries would; water diffuses out into the concentrated medulla, and solutes are taken into the capillaries. As the concentrated solution ascends through the upward part of the loop, water diffuses back in to the capillaries, and solutes diffuse out. This does not contribute to the concentrated state of the medulla, but does slow the removal of these solutes from the medulla.

1. What other solutes beside sodium can the kidneys use to concentrate the urine?

(p. 353 - )

Urea

Hydrogen

1. Where is ADH produced and where is it stored? Describe the body’s feedback system for ADH secretion.

ADH is produced in the hypothalamus and stored in the posterior pituitary gland. When the sodium concentration of the plasma is increased, there are special receptors in the hypothalamus called osmoreceptors that shrink in response to the increased sodium concentrations. This causes nerves to fire down the stalk of the pituitary gland, stimulating the release of ADH. ADH enters the blood stream and is transported to the kidneys where it increases the water permeability of the late distal tubules, cortical collecting tubules, and medullary collecting ducts. The increased water permeability causes increased water reabsoption and excretion of a small volume of concentrated urine.

Chapter 29:

1. Know the factors that move potassium in and out of cells (Table 29-1 on page 362)

After ingestion of a normal meal, ECF K+ levels would rise to lethal levels if the K+ did not move into the cell.

Insulin stimulates K+ uptake into the cell: It is important that there is enough insulin to help move K+ into the cell, for people that are insulin deficient this can cause a problem, injections of insulin can help correct hyperkalemia.

Aldosterone helps with transport of K+ into cells: Aldosterone is stimulated by increased K+ levels, pts with Conn’s disease have an increase of aldosterone secretion thus have a K+ deficiency, also if there is not enough Aldosterone, Addison’s disease, there will be too much circulating K+ as well as too much renal retention of K+

Beta-adrenergic stimulation increases cellular uptake of K+: Increased secretion of catecholamines, mainly epinephrine, can cause movement of K+ into the cell, causing hypokalemia, mainly by activation of the Beta-2 receptors. Conversely, treatment of HTN with Beta-blockers, like Propranolol, causes K+ to move out of the cells and can create a tendency toward hyperkalemia.

Acid-Base abnormalities change K+ distribution: Metabolic acidosis increases EC K+ concentrations, in part by causing the loss of K+ from the cells, whereas metabolic alkalosis decreases EF K+ concentrations. Not totally understood is that H+ ions can shut down the K+/NA+ ATP pump and this decreases cellular uptake of K+ and this can increase ECF K+ levels.

Cell lysis can cause an increase ECF K+ levels: As the cell lyses it leaks K+ levels into the ECF if there is large amt of tissue damage in muscle or RBCs.

Strenuous exercise can cause increased ECF levels from skeletal muscle (another reason to become a couch potato): After prolonged exercise K+ is released from skeletal muscle, usually not an issue unless the exercise is heavy and prolonged and the pt is on B-blockers or has an insulin deficiency. Can lead to arrhythmias or death!

Osmolarity can cause redistribution of K+ to the ECF: Increased Osmolarity causes a flow of water out of the cell, the dehydration of the cell causes an ICF level of K+. This increase promoted secretion of K+ out to the ECF. Decreasing ECF osmolarity has the opposite effect.

1. What controls potassium secretion? What cells secrete potassium? What cells reabsorb potassium?

The primary control is the level of high intake of K+. This high level of K+ excites the secretions of K+ from the distal and collecting tubules. Whatever K+ levels are the distal tubules will absorb or secrete K+.

The primary cells that secrete are the principle cells. They are highly permeable to K+. They make up about 90% of the epithelial cells in the late distal and cortical collecting tubules. Secretion is a 2 part process:

0. K+ is taken up from the interstitum into the cell by the Na+/K+ ATPase pump in the basolateral cell membrane. Moving Na+ out and K+ into the cell. This pump creates a high IC K+ concentration.

0. Then there is passive diffusion of K+ ions from the interior of the cell to the tubular fluid. The high IC concentration of K+ mentioned above creates the driving force of the K+ diffusion to the tubular lumen.

Intercalated Cell can reabsorb K+ during K+ depletion:

0. With severe K+ depletion, there is a cessation of K+ secretion and a net reabsorption of K+ in the late distal and collecting tubules. The process is not completely understood, but a possible hydrogen/Potassium ATPase transport which may exchange of K+ for an H+ ion. This only happens in low ECF K+ levels.

1. State the relationship between acidosis and potassium secretion.

Acidosis increased ECF levels of K+ concentration. The acidosis may shut down the Na+/K+ ATPase pump therefore increasing ECF levels of K+, by decreasing cellular uptake of K+.

1. What hormone regulates plasma calcium concentrations? How?

Parathyroid hormone is the plasma regulator of Ca++ concentrations. Almost all the Ca++ is stored in bones. And only 0.1% is in the ECF and 1.0%is In the ICF.

How?

Parathyroid Hormone (PTH) controls the release and uptake of Ca++ from bone. If the concentration of Ca++ falls below a normal level the parathyroid glands are stimulated by the low concentrations of Ca++. The hormone acts directly on the bones to increase resorption of bone salts(release of salts from the bones) and to release large amts of Ca++ into the ECF. There is only a finite amt of Ca++ that can be released this way so PTH also helps stimulate Vit D to increase the reabsorption of Ca++ in the intestines. PTH also, reacts in the kidneys to increase renal tubular Ca++ reabsorption.

1. What factors alter renal calcium excretion (Table 29-2 on page 369)?

Calcium is both filtered and reabsorbed by the kidney but not secreted! Therefore the rate of renal Ca++ excretions is calculated as: Renal Ca++ excretion = Ca++ filtered-Ca++ reabsorbed. The main factors that alter renal Ca++ excretion are as follows: Inc PTH, dec ECF volume, Dec BP, Inc plasma PO4, metabolic acidosis, and Vit D3.

1. How does PTH affect phosphate excretion?

PTH affects PO4 excretion by two effects:

Promoting bone resorption, thereby dumping large amts of phosphate ions in the ECF from the bone salts

PTH decreases the transport maximum for phosphate by the renal tubules, so a greater proportion of the tubular phosphate is lost in the urine. Whenever plasma PTH is increased, tubular phosphate reabsorption is decreased and more phosphate is excreted.

1. Review the section on pressure natriuresis and diuresis on page 371-72).

The renal-body fluid feedback mechanism operates to prevent continuous accumulation of salt and water in the body during increased salt and water intake. As long as kidney function is normal and pressure diuresis is effective large changes in water and salt intake can be controlled with small changes in the overall hemodynamics of the body.

Chapter 30:

1. What is the relationship of intracellular pH to plasma pH?

Intracellular pH usually is slightly lower than plasma pH because the metabolism of the cells produces acid, especially H2CO3. Depending on the type of cells, the pH of ICF has been estimated to range is 6.0-7.4. Hypoxia of the tissues and poor blood flow to the tissues can cause acid accumulation and decreased intracellular pH. Just as an example, the concentration of H+ that is secreted by parietal cells in the stomach. The H+ concentration is 4 million times greater than the H+ concentration of blood.

1. Be able to describe the bicarbonate buffer system.

This system consists of a water solution that contains 2 ingredients: a weak acid, H2CO3 and a bicarbonate salt, such as NaHCO3(sodium bicarb). H2CO3 is formed by the reaction of CO2 and H2O. The reaction is slow and is reliant on carbonic anhydrase, without it, only small amts of H2CO3 would be formed. The enzyme carbonic anhydrase is abundant in the lung alveoli when CO2 is released, it is also found in the epithelial cells of the renal tubules. These 2 places are where CO2 reacts with H2O to form H2CO3. H2CO3 ionizes easily, b/c it is a weak acid, to form H+ and HCO3-.

The second component of the system, bicarbonate salt, occurs predominantly as NaHCO3 in ECF. NaHCO3 ionizes almost completely to form HCO3- and Na+.

Putting the entire system together, we have (CO2 + H2O -><-H2CO3-><-H+ + HCO3- and Na+). The Na+ that is in this equation is a byproduct from the NaHCO3 and isn’t really involves in the buffering of the acid. Because of the weak dissociation of H2CO3, the H+ concentration is extremely small.

So as H+ ion increase after the release, as in HCl acid, the HCO3 buffers and changes into H2CO3 thus continuing on to disassociate to CO2 and H2O. The increase in CO2 increases respirations to rid the body of the CO2.

1. What are the elements of the phosphate buffer system?

Not as important as the bicarb buffing system, the PO4 buffer system plays a major role in buffering renal tubular fluid and ICF. The main elements of the phosphate buffer system are: H2PO4- and HPO4--. If a strong acid like HCl is added to a mixture of these 2 substances, the H+ ion is accepted by the base HPO4— and converted to H2PO4-. The result of this reaction is that the strong acid, HCL, is replaced by an additional amount of a weak acid, NaH2PO4, and the pH is minimized.

The phosphate buffer is especially important in the tubular fluids of the kidney for 2 reasons: the phosphate usually becomes greatly concentrated in the tubules, thereby increasing the buffering power if the phosphate system; and the tubular fluid usually has a considerably lower pH than the ECF does, bringing the operating range of the buffer closer to pK(6.8) of the system.

The phosphate system is also important in buffering ICF, because the concentration of phosphate in ICF is many times greater than ECF.

1. What other molecules serve as buffers in the blood?

Proteins are among the most prolific buffers in the body. They are mostly found inside the cells. The pH of ICF is slightly lower than the ECF, there is a slight diffusion of H+ and HCO3- though the cell membrane. These ions require several hours to come to equilibrium with ECF, except for rapid equilibrium which happens in RBCs. The diffusion of the elements of the bicarb buffer system cause the pH in ICF to change when there are changes in pH of the ECF. This helps buffer ECF, if ECF pH decreases. It just takes several hours. COOL! Hb, is an important buffer: H+ and Hb = HHb, this equation flow back and forth. Approximately 60-70% of the total chemical buffing of the body fluids is inside the cells, and most of this results from the IC proteins.

1. What is the role of the lungs in maintenance of normal pH?

As seen in question 3 the formation of CO2 in ECF increases respirations, you might have noticed working in the ICU. Just wanted to sound like Paul! Anyway, this elimination of CO2 in mass decreases H+ concentration. As the concentration of CO2 either through formation from the tissues via an increase in metabolic rate or changes in pulmonary ventilation. The lungs will begin to increase Respiration rate to try to decrease CO2 ECF levels. Once that is achieved the RR will decrease to WNL.

1. By what three mechanisms do the kidneys regulate extracellular fluid pH?

Secretion of H+ ions and reabsorption of HCO3- by the renal tubules are closely related. This happens just about in all parts of the tubules except the descending limbs and ascending thin limbs of the Loop of Henle. About 80-90% of the bicarbreabsorption(and H+ secretion) occurs in the proximal tubule, so only a small amount of bicarb flows into the distal tubules and collecting ducts. Lastly, new bicarb is produced along with the reabsorption of the bicarb to offset the acidosis. The kidneys are continuously filtering bicarb and either reabsorbing it for later use if there is acidosis or secreting it if there is an alkalosis.

1. Do your best to understand the hydrogen ion/bicarbonate ion exchange on pp 386-387.

In the tubular lumen sodium bicarb disassociate and the Na+ ion moves into the cell via a membrane protein, as the Na+ is going in, a H+ ion is coming out and associates with the bicarb that the Na+ just left, to form a carbonic acid. These then move a small distance and disassociate to form CO2 and H2O, which then move into the same cell only later. The disassociated CO2 and H2O are rejoined with the help of carbonic anhydrase to reform carbonic acid, then they split apart again, remember this is all happening in the same cell the process began, to become bicarb and H+. The bicarb is then reabsorbed into the renal interstitial fluid and the H+ goes on to do the same process again. This process is the main process to reabsorb bicarb, most of it takes place in the thick segment of the ascending loop of Henle, the epithelial cells of the proximal tubule and the early distal tubule. Since bicarb does diffuse through the cell membrane it must be changes into water and CO2 which both do. Thus the process is to change the shape of the compounds so they can easily diffuse through the cell membrane easily.

1. What is the ammonia buffer system?

The second buffer system in the tubular fluid that is even more important quantitatively than the phosphate buffer system is the ammonia (NH3) and the ammonium ion (NH4+). Ammonium ion is synthesized by Glutamate in the liver from the metabolism of amino acids. Glutamate makes it into the epithelial cells in the same areas as the last process. Once inside the glutamate is transformed into 2 ammonium ions and 2 bicarbs (this process of transformation is not explained in the book. The ammonium ion is exchanged for a Na+ ion and the ammonium ion joins a Cl- ion in the tubular lumen for buffering a secretion. The bicarb and the Na+ are reabsorbed in the interstitial fluid. So with 1 glutamate molecule metabolized, there is 2 ammonium ions and 2 bicarb ions, the 2 NH4+ ion are secreted and the biarb is reabsorbed. This is an important way to regenerate bicarb. As a side bar, an example of the ammonia buffer system’s importance. As ECF H+ concentrations go up, the liver produces more glutamate thus creating more ammonium ions and bicarb to act as buffers. The opposite happens when H+ amounts decrease. Under normal conditions, the ammonia buffer system accounts for about 50% of acid secreted and 50% of new bicarb made. With chronic acidosis, the dominant mechanism by which acid is eliminated is the excretion of NH4+, and is the most important mechanism for generating new bicarb.

Chapter 32:

1. What cells do pluripotentialhemapoietic stem cells give rise to?

(p. 414-415)

These cells are the precursors to all blood cell types. They become erythrocytes, granulocytes (neutrophils, eosinophils, basophils), monocytes, macrocytes, megakaryocytes (platelets), T-lymphocytes, B-lymphocytes.

1. What is the function of erythropoietin and in what organ is it predominately formed?

(p. 416)

In states of low tissue oxygenation, like hypoxia, anemia, poor blood flow, or pulmonary disease, the kidneys primarily are triggered to release a hormone called erythropoietin. This hormone stimulates the production of red blood cells in an attempt to correct the low tissue oxygenation but increasing the total blood oxygenation capacity.

90 percent of erythropoietin is formed in the kidneys and the remaining 10 percent come from the liver.

1. What are the two common chains associated with hemoglobin and how much oxygen can one molecule of hemoglobin bind to?

(p. 417-418)

In each chain of hemoglobin there is a heme chain and a globin chain. The exact formation of each of these chains can be found on p. 417.

Each hemoglobin chain can carry 4 oxygen molecules (or 8 individual oxygen atoms)

1. Iron is transported in the blood loosely bound to another molecule to form what?

(p. 418)

Iron is bound to a substance called apotransferrin immediately after being absorbed from the intestine to form a substance called transferrin. In this form, iron can be released at any tissue site.

1. How is hemoglobin handled in the body once a red blood cell has completed its life span?

(p. 419-420)

Red blood cells last about 120 days in the body, and then when their structural enzymes wear out, they are destroyed by rupturing as they pass through a tight spot in the circulation in the body, usually as they pass through the spleen.

After being destroyed, the hemoglobin is immediately phagocytized by specialized cells. The hemoglobin is metabolized down through a series of steps into a molecule called bilirubin which is released into the plasma, and excreted from the body through the bile (gives poop that nice brown-yellow color).

Chapter 33:

1. Identify the types of white blood cells normally present in blood. Which two are normally present in the greatest numbers?

Neutrophils, eosinophils, basophils, monocytes, lymphocytes, platelets (megakaryoctes are white blood cells found in the bone marrow), and occasionally plasma cells. The 2 that are present in the greatest number are the neutrophils and the lymphocytes.

1. Be able to identify what substances are signals for chemotaxis of white blood cells.

(p. 425)

1. some of the bacterial or viral toxins

2. degenerative products of the inflamed tissues themselves

3. several reaction products of the complement complex activated in inflamed tissues

4. several reaction products caused by plasma clotting in the inflamed area

1. Identify the characteristics that keep normal body structures from being attacked by phagocytes.

1. Most natural structures in the tissue have smooth surfaces, which resist phagocytosis. If the surface is rough for any reason, the likelihood of phagocytosis is increased.

2. Most natural substances of the body have protective protein coats that repel the pahgocytes. Most dead tissues and foreign particles have no protective coats, which makes them subject to phagocytosis.

3. The immune system develops antibodies against infectious agents such as bacteria. The antibodies adhere to the bacteria or foreign material, which makes them especially susceptible to phagocytosis.

1. How do neutrophils and macrophages kill bacteria?

(p. 425-426)

Neutrophils:

attach themselves to the particle to be digested then project pseudopodia in all directions around the particle. The pseudopodia meet on the opposite side of the particle and fuse, creating a complete enclosure for the particle. The enclosure separates itself from the cellular membrane and become a free-floating phagocytic vesicle. A neutrophil can usually do this 3 to 20 times before it becomes inactivated.

Macrophages:

Eat up bacteria in the same manner as neutrophils, but are capable of engulfing up to 100 bacteria, and cells of larger size, even up to the size of a red blood cell if needed. These are considered much more powerful cells in the destruction of bacteria.

1. What are the components of the reticuloendothelialsystem.

(p. 426-428)

This is also called the monocyte-macrophage cell system. Although many macrophages are mobile, many monocytes remain in the tissues they were attracted to and attach themselves there for long periods of time until called upon to provide some sort of local protective function

The system consists of tissue macrophages in the skin and subcutaneous tissues (histocytes), macrophages in the lymph nodes, alveolar macrophages in the lungs, macrophages (kupffer cells) in the liver sinusoids, and macrophages of the spleen and bone marrow.

1. Identify the locations in the body where macrophages reside as part of the antibacterial defenses.

Already answered in previous question. (p.426-428)

(in the skin and subcutaneous tissues (histocytes), macrophages in the lymph nodes, alveolar macrophages in the lungs, macrophages (kupffer cells) in the liver sinusoids, and macrophages of the spleen and bone marrow)

1. Identify the “lines of defense” to inflammation (pages 428-429).

1. Tissue Macrophages: as mentioned previously, the macrophages embedded in the tissues become mobile and begin their phagocytic actions are the site of injury. This happens within minutes of the activation of the inflammatory response. These become the first line of defense for the first hour or so.

2. Neutrophil invasion of the Inflamed Area: inflammatory cytokines cause the neutophils to stick to the walls of the inflamed area (margination), then those same cytokines cause the capillary walls to be leaky so that the neutrophils can enter the tissues, and the cytokines act as chemotaxic agents for the neutrophils.

3. Second macrophage Invasion into the Inflammed Tissue: Macrophages enter the tissue with the neutrophils. These are immature cells and must mature in the infalmmed area over a period of weeks to reach full phagocytic capability.

4. Increased Production of Granulocytes and Monocytes by the Bone Marrow: This takes 3 to 4 days to provide new cells to the tissues, but will continue as long as there is an inflammatory stimulus.

1. What substances are released from basophils?

(p. 430)

The basophils release heparin, histamine, bradykinin, serotonin, lysosomal enzymes, and slow reacting substance of anaphylaxis.

Chapter 34:

1. Differentiate innate immunity from humoral (B-cell) immunity.

Humoral immunity is aka B-cell immunity is by the body developing circulating antibodies, which are globulin molecules in the blood plasma that are capable of attacking an invading agent. This type of immunity doesn’t develop until after an invading organism or toxin is detected in the blood.

Innate immunity is the continual process that keeps our bodies clean of bacteria and other invaders by WBC and cells of the tissue macrophage system. Destruction of swallowed organisms by the acid secretions in the stomach and the digestive enzymes, resistance of the skin to invasion by organisms. Presence in the blood of certain chemical compounds that attach to foreign organisms or toxins and destroy them:

0. Lysosomes, attack bacteria and cause them to dissolute (i.e. explode)

0. Basic polypeptides that react with and inactivate certain types of Gram + bacteria.

0. Using a system of 20 proteins that can be activated in various ways to destroy bacteria, the complement system

0. Natural killer lymphocytes that can recognize and destroy foreign cells, tumor cells and some infected cells.

1. How does the thymus gland ensure that T-lymphocytes will not react to the body’s own tissues or proteins?

The thymus gland controls how T-lymphocytes react to our own body by taking young T-lymphocytes from bone marrow and develop them into having specific reactions again different antigens. This continues until there are thousands of different types of T-lymphocytes with specific reactivities against many thousands of different antigens. The thymus selects which T-lymphocytes will be released by first mixing them with virtually all the specific “self-antigens” from the body’s own tissues. After the T-cells are “familiar” with our own tissue type they can be released for immunity purposes. The other 90% actually get phagocitized b/c they are reactive to our own antigens and the thymus makes sure that they are not released to do more harm.

1. What is the difference in how T-lymphocytes function from B-lymphocytes?

First, instead of the whole cell developing reactivity against the antigen, as in the T-cell, the B-cell actively secrete antibodies that are the reactive agents. These agents are large protein molecules that are capable of combining with and destroying the antigenic substance. Second, the B-cells have even a greater diversity than the T-cells, thus forming many millions of types of B-lymphocyte antibodies with different specific reactivities.

1. What is the function of helper T-cells? What family of proteins do they synthesize?

Most antigens activate both T and B-lymphocytes at the same time. Some T-cells are called “helper cells”. They secrete specific substances called lymphokines that activate the specific B-lymphocytes. This is b/c the B-lymphocytes are unable to form an adequate amount of antibodies.

1. Identify four ways that antibodies work to directly disable invading microorganisms.

Agglutination, in which multiple large particles with antigens on their surfaces, such as RBC or bacteria, are bound together into a clump.

Precipitation, in which the molecular complex of soluble antigen, like tetanus toxin, and antibody becomes so large that it is rendered insoluble and precipitates.

Neutralization, in which the antibodies cover the toxic sites of the antigen agent.

Lysis, in which some potent antibodies are occasionally capable of directly attacking membranes of cellular agents and thereby cause rupture of the agent.

These direct actions of antibodies attacking the antigenic invaders is not strong enough to play a major role in protecting the body against invaders, but work in adjunct to the “complement system”

1. List seven prominent effects of the complement system.

5. Made up of about 20 proteins. The main players are 11 proteins called C1-C9, B and D in figure 34-6 on page 439. All the proteins are normally present but are inactive until the antigen-antibody reaction occurs.

0. Opsonization and phagocytosis: C3b activates phagocytosis by both neutrophils and macrophages, these cells engulf the bacteria. It enhances the number of bacteria that can be destroyed by many hundredfold.

0. Lysis: one of the most important products is the lytic complex, which is a combination of multiple complement factors and designated C5b6789. This has a direct effect of rupturing the cell membranes of bacteria or other invading organisms.

0. Agglutination: The complement products also change the surfaces of the invading organisms, causing them to adhere to one another to form a clump.

0. Neutralization of viruses: The complement enzymes and other products can attack the structures of the viruses and thereby rendering them nonvirulent.

0. Chemotaxis: Fragment C5a initiates chemotaxis of large numbers of neutrophils and macrophages to migrate into tissue area adjacent to the antigenic agent.

0. Activation of mast cells and basophils: Fragments C3a-C5a activate mast cells and basophils to release heparin, histamine and other substances into the local fluids. These substances cause increased blood flow, and increase leakage of fluid and plasma proteins into the tissue and other local tissue reactions that help inactivate of immobilize the antigentic agent, This also happens in inflammation.

0. Inflammatory effects: along with the activation of basophils and mast cells there are other components to the inflammatory process. These(unnamed products) cause another increase to already increased blood flow, cause an larger increase in a capillary leak of proteins and for the interstitial fluid proteins to coagulate in the tissue space, thus preventing movement of the invading organism through the tissue.

1. What is the role of perforins in cytotoxic T-cells?

6. Killer T-cells attack both our cells and bad foreign cells. They do this by binding to the cell they are trying to kill. Once bound the T-cells release a substance called perforins and literally punch round holes in the membrane of the attacked cells. They can kill bind to a cell, deliver their deadly payload and move onto to attack other cells.

1. Identify the immunoglobulin (antibody) most active in allergic reactions.

IgE is the most active immunoglobulin in allergic reactions. This tendency is passed on genetically. These antibodies are also called reagins or sensitizing antibodies, to distinguish them from the more common IgG antibodies. When an allergen enters the body an allergen-reagins reaction takes place. The IgE antibodies have a strong propensity to attach to mast cells and basophils. This causes both cell types to rupture releasing, histamine, heparin, protease, slow-reacting substance of anaphylaxis, eosinphils chemotactic substance, neutrophil chemotacic substance and platelet activating factors.

Chapter 35:

1. Describe what A and B agglutinogens are. Describe what anti-A and anti-B agglutinins are and what stimulates their production in the body. pg. 445-446

0. There are two antigens type-A and type-B that occur on the surfaces of red blood cells in the majority of people. These antigens (often called agglutinogens b/c they often cause blood cell agglutination) are what cause most transfusion reactions. These are inherited agglutinogens and people may have none, one, or both simultaneously.

0. When type-Aagglutinogen is not present in a person’s blood, antibodies called anti-A agglutinins form. This happens because small amounts of type A and B antigens enter the body in food, bacteria, and in other ways, and these substances initiate the development of the anti-A and anti-B agglutinins.

0. Agglutinins are gamma globulins produced by the bone marrow and lymph gland cells. They are mostly IgM and IgG immunoglobulin molecules.

1. What is “agglutination” relative to a blood transfusion?pg. 446

1. When blood is mismatched and anti-A or anti-B plasma agglutinins are mixed with RBC’s that contain A or B agglutinogens, respectively, the red cells agglutinate as a result of the agglutinins’ attaching themselves to the RBC’s. Agglutinins have either 2 binding sites (IgG type) or 10 binding sites (IgM type), and a single agglutinin can attach to 2 or more RBC’s at the same time—causing a process of “agglutination”. The clumps then plug up small vessels throughout the circulatory system. Next, either physical distortion of the cells or attack by phagocytic WBC’s destroys the membranes of the agglutinated cells, releasing hemoglobin into the plasma. This is known as “hemolysis” of RBC’s.

Genotypes

Blood Types

Aggluntinogens

Aggluntinins

OO

O

-

Anti-A and Anti-B

OA or AA

A

A

Anti-B

OB or BB

B

B

Anti-A

AB

AB

A and B

-

1. Which antigen is necessary for a person to be classified “RH Positive”?pg. 447

2. There are six common types of Rh antigens, each of which is called an Rh factor. These types are: C, D, E, c, d, and e. Someone with a C antigen does NOT have a c antigen, but someone missing the C antigen will ALWAYS have the c antigen. This is true for D-d and E-e antigens also. Each person has one of each of the three pairs of antigens.

2. Type D antigen in the most prevalent and more antigenic that other Rh antigens. Anyone who has type D antigen is Rh positive, and anyone who does not have type D antigen is Rh negative.

2. Even if someone is Rh negative, they can still have transfusion reactions from some of the other Rh antigens, they are just usually more mild.

1. State the cause of ErythroblastosisFetalis. pg. 447

3. Erythroblastosisfetalis is also known as hemolytic disease of newborn. It’s a dz of the fetus and newborn child characterized by agglutination and phagocytosis of the fetus’s RBC’s. It is usually caused by a Rh negative mother and a Rh positive father. The baby gets the Rh positive antigen from the father, and the mother develops anti-Rh agglutinins from exposure to the fetus’s Rh antigen. Then, the mother’s agglutinins diffuse through the placenta into the fetus and cause red blood cell aggluntination.

1. Define autograft, isograft, allograft, and xenograft. pg. 449

4. Autograft—transplant of a tissue or whole organ from one part of the same animal to another part

4. Isograft—transplant of a tissue or whole organ from one identical twin to another

4. Allograft—transplant of a tissue or whole organ from one human being to another or from any animal to another animal of the same species

4. Xenograft—transplant of a tissue or whole organ from a lower animal to a human being or from an animal of one species to one of another species

1. Identify three causes of kidney failure after a transfusion reaction. pg. 448

5. Kidney failure can begin within a few minutes to few hours and continue until a person dies of renal failure. Kidney shutdown is the result of three causes:

0. The antigen-antibody reaction of the transfusion reaction releases toxic substances from the hemolyzing blood that results in powerful renal vasoconstriction.

0. The loss of circulating red cells in the recipient, along with the production of toxic substances from the hemolyzed cells from the immune reaction cause circulatory shock. ABP falls and renal perfusion is compromised.

0. The total amount of free hemoglobin released into the circulating blood is greater than the quantity that can bind with haptoglobin (plasma protein that binds small amounts of hemoglobin. The excess leaks through the glomerular membranes into the kidney tubules. If it is only a slight amount, it can be reabsorbed through the tubular epithelium into the blood and will cause no harm. If it is a greater amount, then only a small % is reabsorbed. Water continues to be reabsorbed though, causing the tubular hemoglobin concentration to rise so high that the hemoglobin precipitates and blocks many of the kidneys tubules. Thus—renal vasoconstriction, circulatory shock, and renal tubular blockage together cause the acute renal shutdown.

Chapter 36:

1. What is the function of the glycoprotein coat on the surface of platelets? pg. 451

0. Glycoproteins repulse adherence to normal endothelium, and cause adherence to injured areas of the vessel wall—especially to injured endothelial cells and ever more so to any exposed collagen from deep within the vessel wall.

1. What is the enzymatic function of prothrombin activator? Of thrombin?pg. 453

1. The general mechanisms by which blood clots takes place in three steps:

0. In response to rupture of the vessel of damage to the blood itself, a complex cascade of chemical reactions occurs in the blood involving over a dozen blood coagulation factors. The end result is formation of a complex of activate substances collectively called prothrombin activator.

0. Prothrombin activator catalyzes conversion of prothrombin into thrombin.

1. This occurs in the presence of sufficient amounts of ionic Ca++.

0. Thrombin acts as an enzyme to convert fibrinogen into fibrin fibers that enmesh platelets, blood cells, and plasma to form the clot.

Prothrombin

ProthrombinCa++

activator

Thrombin

FibrinogenFibrinogen monomer

Ca++

Fibrin fibers

Thrombin activated fibrin

stabilizing factor

Cross-linked fibrin fibers

1. What are three stimuli that activate the clotting cascades?pg. 454

2. Trauma to the vascular wall and adjacent tissues

2. Trauma to the blood

2. Contact of the blood with damages endothelial cells or with collagen and other tissue elements outside the blood vessel.

2. In each case—this leads to the formation of prothrombin activator.

1. Know which factors are involved in both clotting pathways (extrinsic and intrinsic).pg. 455-456

Extrinsic: Begins with a traumatized vascular wall or traumatized extravascular tissues that come into contact with the blood. This leads to--

3. Release of tissue factor. Traumatized tissue releases a complex of several factors called tissue factor or tissue thromboplastin—composed mainly of phospholipids from the membranes of the tissue plus a lipoprotein complex that fxns mainly as a proteolytic enzyme.

3. Activation of Factor X—role of Factor VII and tissue factor. Lipoprotein complex of tissue factor further complexes with blood coag Factor VII and, in presence of Ca++ ion, acts on Factor X to form activated Factor X (Xa)

3. Effect of Xa to form prothrombin activator—role of Factor V Activated Factor X combines immediately with tissue phospholipids that are part of tissue factors or with additional phospholipids released from platelets, as well as with Factor V to form the complex prothrombin activator. The proteolytic action of thrombin activates Factor V, which then becomes strong accelerator of prothrombin activation (this is a positive feedback effect of thrombin).

Intrinsic: This is the 2nd mechanism for initiating formation of prothrombin activator—and thus initiating clotting. This begins with trauma to the blood or exposure of the blood to collagen from a traumatized blood vessel. The next step in the process is a series of cascading reactions as follows—

1. Blood trauma first causes (1) activation of Factor XII and (2) release of platelet phospholipids.

1. Activation of factor XI. This reaction requires high-molecular-weight kiniogren and is accelerated by prekallikrein.

1. Activation of Factor IX by activated Factor XI.

1. Activation of Factor X—role of Factor VIII. Factor VIII is what is missing in people with hemophilia.

1. Activation of activated Factor X to form prothrombin activator—role of Factor V. This is the same as the last step in the extrinsic pathway.

1. Identify three factors that prevent clotting in the normal (non-injured or diseased) vascular system.pg. 456

4. The smoothness of the endothelial call surface, which prevents contact activation of the intrinsic clotting system

4. A layer of glycocalyx on the endothelium, which repels clotting factors and platelets, thereby preventing activation of clotting. Glycocalyx is a mucopolysaccharide adsorbed (meaning to undergo or cause to undergo a process in which a substance, usually a gas, accumulates on the surface of a solid forming a thin film, often only one molecule thick) to the surfaces of the endothelial cells.

4. A protein bound with the endothelial membrane, thrombomodulin, which binds thrombin.

2. This slows the clotting process by removing thrombin and also activates plasma protein, protein C, that acts as an anticoagulant by inactivating activated Factors V and VII.

1. What is the function of plasmin? pg. 457

5. Plasminogen is a euglobulin contained in plasma proteins that when activated, becomes plasmin. Plasmin is a proteolytic enzyme that resembles trypsin, the most important digestive enzyme of pancreatic secretion. Plasmin digests fibrin fibes and some other protein coagulants such as fibrinogen, Factor V, Factor VII, and Factor XII. This means that whenever plasmin is formed, it can cause lysis of a clot by destroying many of the clotting factors—sometimes even causing hypocoagulability of the blood.

Chapter 37:

1. Define what pleural pressure, alveolar pressure and trans-pulmonary pressure are.

(p. 466-467)

Pleural Pressure: the pressure of the fluid in the thin space between the lung pleura and the chest wall pleura. This is normally a negative pressure of -5cm at the beginning to inspiration, increasing (negatively) during inspiration to -7.5cm, and returning to baseline with the end of expiration.

Alveolar Pressure: the pressure of the air inside the lung alveoli. When the glottis is open and no air is flowing into or out of the lungs, the pressure in all parts of the respiratory tree, all the way to the alveoli, are equal to atmospheric pressure, which is considered to be zero. To have airflow into the lungs, the alveolar pressure must be slightly negative. During inspiration the alveolar pressure falls to about -1cm H2O. During expiration, the alveolar pressure rises to about +1cm H2O to help expel the air from the lungs.

Trans-pulmonary Pressure: This is the difference between the alveolar pressure and the pleural pressure. It is also a measure of the elastic forces in the lungs that tend to collapse the lungs at each instant of respiration, called the recoil pressure.

1. Define and describe the two forces that determine lung compliance.

(p. 467)

1. Elastic forces of the lung tissue

Determined by elastin and collagen fibers interwoven in the lung tissue. When the lungs are inflated during inspiration, these fibers become stretched and recoil to cause expiration and return the fibers to a resting state.

2. Elastic forces caused by surface tension

When air forms a surface with water, the water molecules are very attracted to one another and the surface of the water is always trying to contract. This is the principle by which raindrops hold together. In the inner surface of the alveoli this same thing is happening. The purpose of this is to try and force air out of the alveoli in the bronchi and out of the lungs. Without the presence of surfactant, which decreases the surface tension and contractile for of the fluid in the alveoli, the alveoli would simply collapse due to their high affinity for each other.

1. Know where surfactant is produced, what it is composed of (not the chemical names) and what it does.

(p. 468)

Surfactant is produced in special surfactant-secreting epithelial cells called type II alveolar cells which constitute about 10 percent of the surface area of the alveoli. It is composed of several phospholipids, proteins, and ions (calcium being the most important ion). It basically reduces surface tension of the fluid on the inner surface of the alveoli to avoid collapse of the alveoli (as explained in the previous question).

1. Identify three fractions that describe the “work of breathing”.

(p. 468 – shaded box)

1. that required to expand the lungs against the lung and chest elastic forces (called compliance work or elastic work)

2. that required to overcome the viscosity of the lung and chest wall structures (called tissue resistance work)

3. that required to overcome airway resistance to movement of air into the lungs (called airway resistance wor

1. Understand the pulmonary volumes and capacities from page 469.

DEFINITION

NORMAL VALUES

PULMONARY VOLUMES

Tidal Volume

Volume of air inspired or expired with each normal breath

500 mL

Inspiratory Reserve Volume

Extra volume of air a person can forcefully inspire beyond the normal tidal volume

3000 mL

Expiratory Reserve Volume

Extra volume of air a person can forcefully expire beyond the normal tidal volume

1100 mL

Residual Volume

The air remaining in the lungs after a full forceful expiration.

1200 mL

PULMONARY CAPACITIES

Inspiratory Capacity

tidal volume + inspiratory reserve volume. Maximum volume a person can inspire

3500 mL

Functional Residual Capacity

Expiratory reserve volume + residual volume. This is the amount of air that remains in the lungs at the end of nomal expiration.

2300 mL

Vital Capacity

Inspiratory reserve volume + tidal volume + expiratory reserve volume. This is the maximum amount of air a person can expel from the lungs after first filling the lungs to their maximum extenet and then expiring to the maximum extent.

4600 mL

Total Lung Capacity

Vital capacity + residual volume. This is the maximum volume to which the lungs can be expanded with the greatest possible effort.

5800 mL

1. Differentiate anatomic from physiologic dead space.

(p. 471-472)

Dead space is the part of the respiratory system that does not participate in gas exchange like the nose, pharynx, and trachea. Anatomic dead space are those areas previously mentioned which do not participate in gas exchange. Physiologic dead space refers to those areas, but also area of the alveoli that do not participate in gas exchange because of disease processes, trauma, or physiologic shunting of blood away from these alveoli. This volume may reach up to 1-2 L.

1. Be able to describe the effect of sympathetic and parasympathetic effects on bronchioles.

(p. 473)

Sympathetic Nervous System: there are relatively few sympathetic nerves in the respiratory system; so direct control of the bronchial tissue is very weak. The bronchial tissue is exposed to circulating catecholamines that cause bronchodilation by stimulating the beta-2 type receptors.

Parasympathetic Nervous System: a few parasympathetic nerves derived from the vagus nerve innervate the lung tissue. When activated, they cause bronchoconstriction.

1. What pulmonary phenomenon is histamine secretion associated with?

(p. 473)

Histamine release in the bronchiole tree causes bronchoconstriction. This is prevalent in asthma in response to airway irritants.

Chapter 38:

1. Describe how the lungs get their blood supply (not pulmonary circulation –how do the lungs themselves get oxygen and nutrients)?

(p. 477)

The respiratory system has a separate circulatory system from the blood supplied to the alveoli for gas exchange. The trachea, bronchial tree, and lung tissue receive arterial blood from the bronchial arteries (branches from the thoracic aorta. The pressure in this system is only slightly lower than the pressure in the aorta.

1. What happens to pulmonary vascular resistance if the O2 concentration in part of the alveoli falls below 70% of normal? What is the significance of this?

(p. 479)

When the concentration of oxygen in the air adjacent to the alveoli falls below 70% of normal the adjacent blood vessels constrict. This is the opposite effect of hypoxia in blood vessels of the systemic circulation. In the pulmonary vasculature, constriction of vessels receiving poor amounts of oxygen redirects blood to vessels receiving adequate amounts of oxygen, thereby making respiration more effective.

1. Differentiate the three ventilation/perfusion zones of the lungs.

(p. 479-480)

Because of the pull of gravity on blood, the top part of the lung receives less blood flow than the bottom of the lung. The capillaries of the alveoli are distended by the pressure of the blood flowing through them, but they are also simultaneously compressed by the alveolar air pressure around them. This leads to 3 different zones (or patterns) or pulmonary blood flow.

Zone 1: No blood flow during any part of the cardiac cycle because the alveolar capillary pressure can never overcome the alveolar air pressure under normal physiologic conditions.

Zone 2: Intermittent blood flow only during the peaks of pulmonary arterial pressure because the systolic pressure is then greater than the alveolar air pressure, but the diastolic pressure is less than the alveolar air pressure.

Zone 3: Continuous blood flow because the alveolar capillary pressure remains greater than alveolar air pressure during the entire cardiac cycle.

The lungs, in an upright position, usually only have zone 2 and 3 type blood flow; zone 2 in the apex of the lungs, and zone 3 in the remainder of the lungs. While a person is lying down, the blood flow to all parts of the lung are primarily zone 3 type flows.

1. How long does blood stay in the pulmonary capillaries at rest? During exercise?

(p. 481)

0.8 sec. normally

0.3 sec during exercise (or periods of increased cardiac output)

1. Which direction is the flow of fluid favored by mean filtration pressure between the pulmonary capillaries and interstitial space? Where happens to the fluid then?

(p. 481-482)

The pressure of the pulmonary capillaries is very low compared to that of the peripheral tissues and are relatively leaky to plasma proteins. The interstitial space of the pulmonary tissues is slightly more negative than the other body tissues. The net pressure is actually almost completely equal, with a 1mm Hg outward force towards the interstitial space, where extra fluid is picked up by the lymphatic system, and a small amount of the fluid evaporates into the alveoli.

The net inward force (towards the capillaries) is 28 mmHg and the outward force (towards the interstitial space) is 29 mm Hg.

1. What is the purpose of the pleural fluid?

(p. 483)

The main purpose of the fluid is to provide a slippery lubricant to reduce friction of the lungs as they continually expand and contract. The fluid is also continually being pumped away from the pleural space to avoid an excess of fluid. This pumping action creates the negative pressure required to keep the lung tissue always slightly expanded to prevent alveolar collapse.

Chapter 39:

1. What determines the solubility coefficient of a gas in solution?

(p. 485-486)

The main determinant of the solubility coefficient of a gas is the degree to which the molecules of the gas are attracted to the molecules of water in the solution. The higher the degree of attraction, the more of the agent that can be dissolved in the solution without building up a high partial pressure.

1. Relate why the concentrations of gases in alveolar air differ from that of atmospheric air.

(p. 487)

There are 4 reasons listed in the book:

1. The alveolar air is only partially replaced by atmospheric air with each breath

-Remember that the functional residual capacity is about 2300 mL, but the tidal volume with each breath is only about 350 mL (accounting for dead space loss), meaning that only one seventh of the total air in the lungs is replaced with each breath.

2. Oxygen is constantly being absorbed into the pulmonary blood from the alveolar air

3. Carbon dioxide is constantly diffusing from the pulmonary blood into the alveoli

4. Dry atmospheric air that enters the respiratory passages is humidified even before it reaches the alveoli

-Water vapor from humidification dilutes all other gases in the inspired air, relatively lowering their concentrations

1. What are four factors that are critical in determining how a gas will diffuse through a biological membrane?

(p. 490-491)

1. The thickness of the membrane

-The respiratory membrane can become thickened by edema of the interstitial space or around the alveoli. This means that gases must diffuse through not only the alveolar membrane but also through the fluids around the alveoli.

2. The surface area of the membrane

-This can be decreased through pathologic processes like emphysema which destroy the alveoli.

3. The diffusion coefficient of the gas in the substance of the membrane

4. The partial pressure difference of the gas between the two sides of the membrane

1. What is the diffusing capacity for oxygen in a resting human? How is it affected by strenuous exercise?

(p. 491)

Resting conditions: 21ml/min/mm (which really equals out to 210 ml of oxygen diffusing through the respiratory membrane each minute. This is equal to the amount used by the body in a resting state).

Strenuos Exercise: 65ml/min/mm (equal to about 715 ml of oxygen diffusing per minute)

This increase is due to several factors:

1. Opening up of many previously dormant pulmonary capillaries or extra dilation of already open capillaries, thereby increasing the surface area of the blood in to which the oxygen can diffuse

2. A better match of V/Q ratio

1. What condition exists when the ventilation/perfusion ratio is equal to zero? To infinity? What are common clinical terms used to describe these conditions?

(p. 492-493)

Zero: this happens when there is no movement of air into or out of the lungs, so the air remaining in the lungs equilibrates in consistency with the partial gas pressures of that of the circulation. This is commonly termed apnea.

Infinity: this happens when there is no capillary blood flow to take in new oxygen from inspired air, or delivery carbon dioxide to be expired, so the air in the lungs equilibrates with the alveolar air concentrations. This is commonly termed asystole.

Chapter 40:

1. Describe how blood becomes saturated with oxygen as it passes through the pulmonary capillary. In what part of the capillary does most of this activity take place?

(p. 496-496)

The alveolar oxygen partial pressure is generally about 104 mm Hg while the venous blood entering the pulmonary capillaries has a oxygen partial pressure of about 40mm Hg. This facilitates rapid diffusion of oxygen from the alveoli into the pulmonary capillary blood. On the arterial end of the capillaries, most of the oxygen is diffused into the blood from the alveoli, and by the time it gets to the venous end of the capillary, the blood oxygen partial pressure matches that of the alveoli and no more oxygen is diffused across the membrane.

1. Why is the O2 concentration of blood leaving the left heart slightly lower than the blood entering the left atrium from the pulmonary veins?

(p. 496)

Most (98%) of the blood entering the left atrium comes from the pulmonary circulation with a oxygen partial pressure of 104 mmHg. The other 2% of blood entering the left atrium comes from the bronchial circulation, which bypasses the gas exchange areas of the lungs to supply oxygen to the lung tissue. This blood comes from the aorta, into the lung tissue then back to the left atrium by way of the pulmonary veins at a oxygen partial pressure of about 40 mm Hg. This lowers the total oxygen partial pressure of this mixture to approximately 95 mm Hg.

1. What is the normal intracellular concentration of oxygen? How much O2 tension is actually needed to prevent cell demise?

(p. 497)

Normal range for intracellular concentration of O2 is 5-40mm Hg. Only 1-3 mm Hg of oxygen is required for chemical processes that use oxygen in the cell.

1. Be able to describe the function of the oxygen-hemoglobin dissociation curve. What causes the curve to shift rightward (the Bohr effect)?

(p. 498-500)

This curve is not as scary as it sounds, or seems to have been in the past.

This curve really just shows that as the partial pressure of oxygen in the blood increases, the amount of oxygen bound to hemoglobin increases as well.

This curve shows us that the oxygen leaving the lungs has the highest concentration of oxygen saturated hemoglobin, because this is where the partial pressure of oxygen is the highest.

The blood returning to the lungs from the tissues has the lowest concentration of oxygen saturated hemoglobin because the oxygen molecules have been delivered to the tissues.

The Bohr Effect: (named for Christian Bohr, father of Neils Bohr; holy smart family Batman. Yeah for Danish people) This effect basically states that in the presence of an acidic blood environment due to increased CO2 levels, increased hydrogen ions, increased blood temperature, or increased BPG (2,3 – biphosphoglycerate), oxygen will be less bound to hemoglobin for any given partial pressure of oxygen in the blood. In short, oxygen is sort of forced away from being bound to hemoglobin so it is more readily delivered to the tissues from the hemoglobin. This is represented on our favorite curve by a shift of the curve to the right and downward.

1. How does hemoglobin serve as a tissue oxygen buffer?

(p. 499)

Being a “tissue oxygen buffer” really just refers to hemoglobin being responsible for stabilizing oxygen pressure in the tissues. To survive, the tissues need about 5 ml of oxygen of every 100 ml of blood delivered to the tissues. Looking back on the oxygen-hemoglobin disassociation curve, to use that amount, the partial pressure of oxygen in the tissues could not be more than 40 mm Hg or the oxygen would not be unbound from the hemoglobin and delivered to the tissues. Hemoglobin sets the upper limit of partial oxygen pressure in the tissues at 40 mm Hg to allow the appropriate amount of oxygen to be delivered to the tissues. (I feel like I really botched this explanation, call with questions: 801-400-8994)

1. What is the significance of the Haldane effect?

(p. 503-504)

Remember the Bohr effect? It stated that when the concentration of CO2 in the blood went up, oxygen was displaced from hemoglobin, increasing the oxygen delivery to the tissues. The opposite of this is also true, and is called the Haldane effect. When oxygen is more bound to hemoglobin, carbon dioxide is more likely to be displaced from the hemoglobin; promoting excretion of carbon dioxide.

This is very important in the pulmonary circulation where the partial pressure of oxygen is very high and is readily bound to hemoglobin (so says the oxy-hemo dis. curve). This highly saturated hemoglobin forces CO2 to be excreted through the alveoli in the pulmonary circulation.

The opposite is true in the tissues as well. Since carbon dioxide levels are high at tissue sites, oxygen is readily released at the tissue sites, and the unbound hemoglobin can be bound to carbon dioxide and transport carbon dioxide to the lungs to be excreted.

Chapter 41: Regulation of Respiration

1. What are the functions of the three primary groups of neurons in the respiratory center of the medulla? P. 505-506

- The respiratory center has several neurons located bilaterally in the medulla oblongata and the pons. The neurons are broken down into three primary sections:

1) Dorsal respiratory group: (dorsal portion of medulla, causes inspiration)

2) Ventral respiratory group: (venterolateral portion of medulla, causes expiration)

3) Pneumotaxic center: (dorsally in the superior portion of the pons, controls rate and depth of breathing)

1) Dorsal respiratory group: This group plays the most important role in control of respiration.

· Most of its neurons are located within the nucleus of the tractussolitarius (NTS) and some in the reticular substance. The NTS is the sensory termination of both the CN IX and X which are responsible for transmitting sensory information from peripheral chemoreceptors, baroreceptors and receptors in the lungs.

· The basic rhythm of respiration is controlled by the dorsal group.

· Even if brainstem transaction occurred above and below the medulla, the dorsal neurons would still emit repetitive bursts of inspiratory neuronal action potentials.

· The nervous signals transmitted from the dorsal group to inspiratory mm’s are not an instantaneous burst of action potentials, rather they begin weakly and increase over about 2s, allowing you to have a steady increase in lung volumes during inspiration, rather than a gasp (this steady increase in inspiratory signal is known as the inspiratory ramp). The dorsal group controls two qualities of the inspiratory ramp: it controls the rate of increase of the ramp signal, and it controls the limiting point at which the ramp suddenly ceases.

2) Ventral respiratory group: (Work only when need overdrive, i.e. exercising) Function in both inspiration and expiration. These neurons are specifically found in the nucleus ambiguus and the nucleus retroambiguus.

· These neurons are inactive during normal inspiration.

· These neurons do not participate in basic rhythmical oscillations that control respiration.

· When your respiratory drive becomes greater than normal, respiratory signals “spill” over to this group from the main dorsal group. As a consequence this area contributes to an extra respiratory drive.

· Electrical stimulation of some of these neurons cause inspiration; and others cause expiration. Therefore, these neurons contribute to both inspiration and expiration.

3) Pneumotaxic center: limits the duration of inspiration and increases the respiratory rate

· Located dorsally in the nucleus parabrachialis of the upper pons

· The primary effect of this center is to limit inspiration, it controls the “switch-off” point of the inspiratory ramp (thus controlling the duration of the filling phase of the lung cycle). Because it “turns off” inspiration, this inadvertently shortens expiration time and the entire time of a respiration, which would increase the rate of breathing

· When the pneumotaxic center signal is strong: inspiration short (0.5s), RR 30-40

· When the pneumotaxic center signal is weak: inspiration long (5s), RR 3-5

1. What is an inspiratory “ramp signal”? p. 505

-See above: dorsal respiratory group

1. Describe what the Hering-Breuer Reflex does. P. 506-507

(Essentially, when the lungs become overinflated, stretch receptors in bronchi send signals to the dorsal group via the vagi to stop inspiration)

-In addition to the signals controlling respirations located in the CNS, there are also sensory nerve signals from the lungs that aid in respiration.

-Within the muscular portions of bronchi and bronchiole walls there are stretch receptors that transmit signals via the vagi into the dorsal respiratory group whenever the lungs are overstretched.

-These signals affect inspiration just like signals from the pneumotaxiccenter, that is when the lungs become overinflated, the stretch receptors activate an appropriate feedback response that turns off the inspiratory ramp and stops further inspiration. This is known as the Hering-Breuer reflex.

1. Where are the chemoreceptors located that sense oxygen tension in the blood? Describe what the blood flow is like to these structures. P. 508-509

-Excess CO2 and H+ in the blood mainly act directly on the respiratory centers, whereas oxygen really doesn’t have much of a direct effect, rather it primarily acts on peripheral chemorecptors in the carotid bodies and aortic bodies that send nervous signals up to the respiratory centers.

-The carotid bodies are located bilaterally in the bifurcations of the common carotid arteries; use CN IX as afferent nerve to dorsal group

-The aortic bodies are located along the arch of the aorta; use CN X to dorsal group

-Each of the chemoreceptor bodies receives its own special blood supply through a tiny artery directly from the adjacent arterial trunk. Blood flow through these bodies is extreme, 20x the weight of the bodies themselves each minute. Therefore, the percentage of oxygen removed from the flowing blood is virtually zero- so the chemoreceptors are exposed at all times to arterial blood, not venous blood. When arterial oxygen levels drop the chemoreceptor is strongly stimulated.

1. Describe the function of the chemosensitive area in the brain stem. Include in your description the relationship of CO2 and H+ relative to the blood-brain barrier. P. 507

-All three of the respiratory centers are not directly affected by changes in blood CO2 and H+ concentrations.

-There is an additional neuronal area of the brain stem, chemosensitive area, which is highly sensitive to blood CO2 and H+ ion concentration.

-Neurons of the chemosensitive area are extremely excited by H+ ions and to a lesser extent by CO2. However, H+ ions do not easily cross the BBB. For this reason, changes in H+ ion concentration in the blood have less effect in stimulating the chemosensitive neurons than do changes in blood CO2.

-Although CO2 has little direct effect stimulating chemosensitive area neurons, it does have a potent indirect effect. CO2 is able to cross into the BBB. CO2 reacts with water in surrounding tissues to form carbonic acid, which then dissociates into a H+ ion and a HCO3 ion, and the H+ ion has a direct potent stimulatory effect on respiration.

-Whenever the blood PCO2 increases so does the PCO2 of the interstitial fluid of the medulla and the CSF. In both of those fluids the CO2 reacts to ultimately produce a H+ ion and excite the chemosensitive area.

-Excitation of the respiratory center by CO2 is high the first few hours, but then it gradually declines over the next couple of days to about 1/5 its effect.

1. How does ventilation increase during strenuous exercise? P. 510-511

-PCO2, pH, and PO2 values do not change much during strenuous exercise because our alveolar ventilation proportionately increases as well.

-The brain, on transmitting motor impulses to the exercising muscles, is believed to transmit at the same time collateral impulses into the brain stem to excite the respiratory center. It is likely that most of the increase in respiration results from neurogenic signals transmitted directly into the brain stem and at the same time neurogenic signals are going out to muscles to contract.

-If these neurogenic brain signals aren’t strong enough, chemoreceptors will play a role