Week 1 Formative Quiz

Question 1

Absence of apoB-100 would result in?

Answer 1

mutation in apoB-100 leads to decreased affinity to LDL-receptor

Question 2

Capillaries with continuous endothelium are best characterized by which of the following?

ability to contract and redirect blood flow

leakiness to leukocytes

prevalence of small endocytic vesicles and caveolae within the endothelium

structure similar to small lymphatic vessels

Answer 2

prevalence of small endocytic vesicles and caveolae within the endothelium

Capillaries have no vascular smooth muscle, so they cannot contract and redirect blood flow. Leukocytes emigrate across postcapillary venules.

Question 3

What bind to desmosomes?

Answer 3

Intermediate filaments

Question 4

Assume that a red blood cell has an intracellular concentration of 290 mOsm of impermeable solutes. The red blood cell is placed in a solution in a beaker. The osmolality of the solution is 580 mOsm. At equilibrium, the volume of the red blood cell is not changed. Which of the following best describes the solution?

All solutes in the solution are impermeable.

hypotonic solution

hypertonic solution

0. 9% NaCl plus 290 mOsm urea
1. 8% NaCl

Answer 4

A - Incorrect: If all solutes in the solution were impermeable with an osmolality of 580 mosM, cell volume would decrease.
B - Incorrect: Cell volume would be increased in a hypotonic solution.
C - Incorrect: Cell volume would be decreased in a hypertonic solution.
D - Correct: This solution has an osmolality of 580 mOsm and the concentration of impermeable solute is 290 mOsm. Urea is a permeable solute.
E - Incorrect: A solution of 1.8% NaCl would cause cell volume to be decreased.

Question 5

How are T-tubules and sarcoplasmic reticulum predominantly arranged?

Answer 5

As a diad

Question 6

A 37-year-old male patient presents to you in your office with complaints of shortness of breath and dizziness. The patient is an avid bodybuilder, and is quite muscular. The patient denies the use of anabolic steroids. All bloodwork is normal, and there are no abnormal valvular sounds. A cardiac ultrasound reveals a marked increase in the size of the heart, particularly in the right ventricle. Which of the following physiological mechanisms may lead to this condition?

• The increased volume of blood passing through the heart triggers the insertion of sarcomeres into the myofibrils. This lengthens each myofibril and allows the heart to pump more blood. This explains the enlarged ventricle.
• There is a large increase in the vasculature surrounding the skeletal muscles. Because of this increased vascular bed there is a decrease in the peripheral resistance to blood flow. Hence his blood pressure will drop leading to the dizziness.
• The increased muscle mass restricts the flow of blood through the muscles by constricting the blood vessels. This increases the diastolic load on the heart. The increase in diastolic load leads the cardiac muscle cells to add myofibrils. The increased myofibrils result in a stiffening of the ventricle due to an increased level of Titan. This leads to stiffening of the ventricle. The heart therefore cannot pump blood efficiently.
• The increased mass of the ventricles lead to an increase in the number and density of gap junctions. These gap junctions result in electrical changes in the heart so that the action potentials will echo throughout the ventricle. This leads to ventricular fibrillation.

Answer 6

The increased muscle mass restricts the flow of blood through the muscles by constricting the blood vessels. This increases the diastolic load on the heart. The increase in diastolic load leads the cardiac muscle cells to add myofibrils. The increased myofibrils result in a stiffening of the ventricle due to an increased level of Titan. This leads to stiffening of the ventricle. The heart therefore cannot pump blood efficiently.

A - Not the best answer. This is true, but it occurs following Aerobic exercise. The enlarged ventricle following aerobic exercise results in a heart that pumps blood more efficiently. The addition of sarcomeres in series does not increase the stiffness of the ventricle. This would not explain the shortness of breath or dizziness.
B - Not the best answer. There is indeed an increase in the vasculature surrounding the muscles. This actually places more resistance on the flow of blood and leads to an increase in blood pressure.
C - This is the best answer.
D - Not the best answer. This would lead to death, not dizziness.

Question 7

Which one of the following congenital heart diseases is characterized by early cyanosis and a right-to-left shunt?

Aortic stenosis, isolated

Patent ductus arteriosus

Tetralogy of Fallot

Atrial septal defect, sinus venosus type

Atrial septal defect, septum primum type

Answer 7

Tetralogy of Fallot

Among the congenital heart defects listed, tetralogy of Fallot is the only one that has a right to left shunt early in life. This shunt causes early cyanosis.

Question 8

Absence of apoB-48 would result in?

Answer 8

would result in the absence of chylomicrons

Question 9

At the peak of a normal action potential the sodium electrochemical energy approaches _________ and the potassium electrochemical energy becomes _________.

zero, greater than 100 mV

the sodium Nernst potential, greater than 100 mV

+60 mV, zero

-100 mV, zero

Answer 9

zero, greater than 100 mV

An ion’s electrochemical energy is the difference between the membrane potential and the ion’s Nernst potential. At the peak of the action potential the membrane potential approaches the sodium Nernst potential (between +40 and +60 mV). Accordingly, Na electrochemical energy approaches zero. Since EK is about -90 mV, the potassium electrochemical energy is greater than 100 mV at the peak of the action potential.

Question 10

Concentration = Mass / Volume
Mannitol is a large molecule that is largely confined to the intravascular space after i.v. injection. Assume that 10 gm of mannitol was injected i.v. to a person and that blood was withdrawn a short time later. The plasma concentration of mannitol was 0.005 gm per ml. If all mannitol remained within the vasculature and none was excreted, which of the following is the best estimate of the plasma volume in this person?

0.5 liter

1 liter

2 liters

5 liters

10 liters

Answer 10

2 liters

The concentration of mannitol in plasma equals 0.005 gm/ml. This concentration equals the mass of mannitol injected (10 gm) divided by the volume of plasma.
Plasma volume = Mass / Plasma concentration = 10 gm / 0.005 gm/ml = 2000 ml = 2 liters

Question 11

The liver plays an important role in the metabolism and elimination of cholesterol from the body. The transport of free cholesterol from peripheral tissues to the liver (reverse cholesterol transport) is dependent on which of the following?

chylomicrons

HDL

IDL

LDL

VLDL

Answer 11

HDL

Reverse Cholesterol Transport: Free cholesterol is removed from tissues by plasma high-density lipoprotein (HDL). It is esterified by LCAT associated with HDL and transported to the liver, where it is eliminated from the body primarily after conversion to bile acids or bile salts.

Question 12

Intravenous infusion of isotonic saline (0.9% NaCl) can be used to correct the sodium deficit associated with dilutional hyponatremia. In this situation, excess water intake decreases plasma sodium below normal levels. Assume that plasma sodium is 110 mEq/liter in a person with dilutional hyponatremia. Which of the following will occur in this person following intravenous infusion of isotonic saline?

decrease in capillary pressure

decrease in intracellular osmolality

decrease in intracellular volume

hypotonic expansion

isotonic contraction

Answer 12

decrease in intracellular volume

Dilutional hyponatremia can occur by excess intake of water, which will initially reduce plasma sodium concentration, decrease ECF osmolality, and increase ECF volume. The decrease in ECF osmolality will cause water to move into cells, increasing ICF volume and decreasing ICF osmolality. I.v. infusion of isotonic saline (290 mOsm) will raise extracellular osmolality towards normal in a person with dilutional hyponatremia. This change will cause water to move out of the ICF.
A - Incorrect: Capillary pressure will be increased due to the increased volume of the ECF.
B - Incorrect: As water moves out of cells, intracellular osmolality will increase towards normal.
C - Correct: ICF volume will decrease.
D - Incorrect: A hypotonic change would increase ICF volume. Infusion of isotonic saline to a person with dilutional hyponatremia will cause ICF volume to decrease.
E - Incorrect: ICF volume is decreased in this situation due to the rise in ECF osmolality following infusion of isotonic saline to a person with dilutional hyponatremia.

Question 13

A 36-year-old male bartender is brought by ambulance to your emergency room because a patron jumped over the bar, grabbed an ice pick, and stabbed him in the chest rather than pay his bar tab at the end of the night. The ice pick entered the chest about 2 cm to the left of the sternum in between the fourth and fifth rib. Upon examining the bartender, you note very little blood is coming from the puncture wound and normal lung sounds from both the right and left lung. However, his heart is beating rapidly at 100 beats per minute, his external jugular veins are bulging, and you have difficulty hearing his heart sounds. You order a PA and lateral chest film because you suspect which of the following?

Hemothorax

Pneumothorax

Cardiac tamponade

Aortic valve stenosis

Deep venous thrombosis

Answer 13

The answer is C. Cardiac tamponade. (Moore and Dalley, pp 140-141.) The ice pick likely penetrated the left ventricle of the heart, causing blood to leak into the pericardial sac. The rapid filling of the pericardial space does not allow the heart to fully expand between contractions leading to increased venous hypertension, thus the filling of the external jugular veins. Since the heart can only pump small quantities of the blood with each beat, it speeds up (tachycardia). The heart sounds and apical heartbeat soften because the blood surrounding the heart absorbs the sounds. A hemothorax (answers a) and pneumothorax (answer b) are unlikely since both right and left lungs sounds are normal and because of the location of the ice pick injury. Aortic valve stenosis (answer d) would not result from a puncture wound. Deep venous thrombosis (answer e) generally occurs in the lower extremity and results in leg pain and is not caused by a puncture wound.

Question 14

In the same image there are specific cellular structures. The cellular structures indicated by the arrows would best be described as:

A Bands

Contraction Bands

I Bands

Intercalated Discs

Z Discs

Answer 14

Contraction Bands

A - A Bands. This is incorrect. The A bands are the Dark bands in the sarcomeres. The A bands are visible in the image, as are the striations. The A bands are composed of the myosin filaments.
B - This is the correct answer. These contraction bands form within cardiac tissue about 12 to 48 hours after injury.
C - I Bands. This is incorrect. The I bands are the light bands in the sarcomeres. They are composed of the Z disk and the actin filaments.
D - Intercalated Discs. This is incorrect. The intercalated discs are the connections between the adjacent cardiac myocytes. There is a distinct possibility that one of the structures may be an intercalated disc (far left arrow), but this is more likely a contraction band forming at the intercalated disc. The other two arrows clearly point to contraction bands.
E - Z Discs. The Z discs are not present in this tissue at this magnification. They are found in the middle of the I bands in the sarcomeres. They are the site where the actin filaments attach.

Question 15

Following the upstroke of the action potential in a nerve, membrane potential moves toward the potassium Nernst potential during repolarization. Which of the following is occurring at this time?

decreasing potassium electrochemical energy

decreasing sodium electrochemical energy

increasing sodium conductance

more negative potassium chemical energy

sodium current is greater than potassium current

Answer 15

decreasing potassium electrochemical energy

Correct answer is A.
A - The potassium electrochemical energy is equal to V – EK. As membrane potential moves more negative during repolarization, V – EK becomes smaller.
B - As V moves more negative during repolarization, the difference between V and ENa becomes greater over time, and so the sodium electrochemical energy increases during repolarization.
C - Sodium conductance increases during the upstroke of the action potential and is responsible for increased sodium current during depolarization.
D - Potassium chemical energy is equal to the potassium Nernst potential. Under normal conditions, an insignificant change occurs in intracellular and extracellular concentrations of sodium and potassium during an action potential, so the Nernst potentials to both ions remain constant. Membrane potential changes during the action potential because of changes in conductance to ions.
E - Sodium current is greater than potassium current during depolarization, and the reverse is true during repolarization.

Question 16

A 16-year-old boy is brought to the ER with dyspnea. His past medical history is significant for an unrepaired atrial septal defect. Physical examination reveals cyanosis, distended jugular veins, hepatosplenomegaly, and a systolic ejection murmur. This patient has most likely developed which of the following complications of congenital heart disease?

aortic aneurysm

myocardial infarction

paradoxical embolism

pneumonia

pulmonary hypertension

Answer 16

pulmonary hypertension

Answer: E Pulmonary hypertension
Patients with uncorrected ASDs may develop narrowing of the pulmonary vasculature, in which case the flow of blood through the defect may be reversed thereby creating a right-to-left shunt. The increase in pulmonary capillary pressure is transmitted to the pulmonary arteries, called pulmonary hypertension, resulting in right-sided heart failure. Complications of ASDs include cyanosis, right ventricular hypertrophy, right heart failure and paradoxical emboli.
This patient’s diagnosis is Eisenmenger syndrome due to atrial septal defect.

Question 17

A mother brings her 2-year-old son to the pediatrician. She states he occasionally appears to be short of breath after running or wrestling with his brother. Also, his preschool teacher notes that she sees him squatting down on the playground. On physical exam you hear a holosystolic murmur. A CXR in your office reveals right ventricular hypertrophy of the heart (giving a vague boot-like appearance). You refer the patient to a pediatric cardiologist. The single greatest factor that determines the clinical outcome associated with the most likely congenital heart defect in this patient is:

The severity of rotation of the overriding aorta

Positioning of the aortic arch

Degree of pulmonary stenosis

Age of patient at diagnosis

Presence of left ventricular hypertrophy

Answer 17

Degree of pulmonary stenosis

Correct answer is C. The clinical outcome of Tetrology of Fallot depends primarily on the severity of the pulmonic stenosis, as this determines the direction of blood flow. If the pulmonary stenosis is mild, the abnormality resembles an isolated VSD and the shunt may be left to right without cyanosis. As the obstruction increased in severity, there is greater resistance to right ventricular outflow. As right sided pressures approach or exceed left sided pressures, right to left shunting develops, producing cyanosis.

Question 18

What will cause myosin to bind?

Answer 18

Myosin will bind whenever the troponin complex does not cover the actin binding site. Activation of myosin by myosin light chain kinase is important in smooth muscle contraction.

Question 19

For a nerve, shifting the sodium channel activation curve toward a more positive potential would be expected to ______________ the rate of depolarization during the upstroke phase of the action potential and __________ the number of resting sodium channels.

decrease, not affect

decrease, increase

increase, decrease

increase, not affect

Answer 19

decrease, not affect

A depolarizing shift in the sodium channel activation curve would not affect the number of resting channels (determined by the position of the inactivation curve in relation to the resting potential) but would reduce the number of sodium channels activated during the upstroke of the action potential. This reduction in active sodium channels would reduce the magnitude of the inward sodium current thereby reducing the rate of depolarization during the upstroke.

Question 20

A burn patient under your care has a plasma osmolarity of 280 mOsm. The osmolarity of the interstitial fluid in the burn area is 279 mOsm compared to 277 mOsm in non-burn areas. Which of the following best completes the following statement? In the Burn Area the osmotic pressure difference across the capillary wall acts to drive fluid __________ across the capillary wall and provides a driving force of __________ mm Hg.

from interstitial to plasma; 20

from plasma to interstitial; 20

from interstitial to plasma; 60

from plasma to interstitial; 60

Answer 20

from interstitial to plasma; 20

Correct answer is A. The fluid osmotic pressure inside the capillary is greater than outside for both the burn area and normal area. Accordingly, the osmotic pressure difference will act to drive fluid inward across the capillary wall in the burn area as well as in the normal area. In the burn area the 1 mOsm concentration difference produces a osmotic pressure difference of about 20 mm Hg.

Question 21

A person with chronic obstructive pulmonary disease had bilateral edema of the lower extremities. Assume that interstitial fluid pressure of the lower extremities is 10 mm Hg, plasma osmolality is 290 mOsm, and interstitial osmolality is 289 mOsm. If the extent of edema is staying constant (i.e., no net fluid movement across the capillary), which of the following is the best estimate of capillary pressure within the lower legs of this person?

10 mm Hg

15 mm Hg

20 mm Hg

30 mm Hg

40 mm Hg

Answer 21

30 mm Hg

Correct answer is D. The difference in solute concentration between plasma and interstitial fluid is 1 mosm/liter. Since each 1 mosm/liter diffence in concentration generates a force equal to 20 mm Hg, the solute difference creates a force for reabsorbtion of water into the capillary of 20 mm Hg. The diffusion of water is from low solute concentration (interstitial fluid) to high solute concentration (plasma within capillaries). For no net movement of water to occur, capillary pressure must be greater than interstitial fluid pressure by 20 mm Hg, so capillary pressure equals 30 mm Hg.

Question 22

A 63-year-old woman is hyponatremic with a plasma sodium concentration of 120 mEq/liter. This person weighs 60 kg and has 17% body fat. Assuming that the desired plasma sodium concentration is 140 mEq/liter, which of the following is the best estimate of this person’s sodium deficit?

685 mEq

700 mEq

715 mEq

730 mEq

745 mEq

Answer 22

Na+ deficit = TBW x (desired [Na+] – present [Na+])

Na+ deficit = (LBM x 0.72) x (desired [Na+] – present [Na+])

Na+ deficit = (60 – 0.17 x 60) x 0.72) x (140 mEq/L – 120 mEq/liter) = 717 mEq

Question 23

A 21 year old woman is hyponatremic with a plasma sodium concentration of 115 mEq/liter. This person weighs 50 kg and has 15% body fat. Assuming that the desired plasma sodium concentration is 140 mEq/liter, which of the following is the best estimate of this person’s sodium deficit?

850 mEq

900 mEq

950 mEq

1000 mEq

1050 mEq

Answer 23

850 mEq

Question 24

An electrode is inserted into a nerve cell to record membrane potential. A drug is added to fluid surrounding the nerve cell, and resting membrane potential changes from -70 mV to -60 mV and remains stable at this value. There is no change in intracellular or extracellular concentrations of sodium or potassium following addition of the drug. Which of the following could account for this effect of the drug?

decrease in fractional conductance to potassium

decrease in sodium conductance of leak channels

increased flux of chloride anion into the nerve cell

increase in the magnitude of repolarizing currents

more negative potassium Nernst potential

Answer 24

decrease in fractional conductance to potassium

Correct answer is A. The drug has caused membrane potential to move to a less negative potential. This could be caused by an increase in magnitude of a depolarizing current or a decrease in the magnitude of a repolarizing current.

A - This is correct because decreased fractional conductance to potassium would cause resting membrane potential to move away from the potassium Nernst potential and towards the sodium Nernst potential.
B - A decrease in sodium conductance of leak channels would cause resting membrane potential to move more negative.
C - An increased flux of an anion into the cell would make resting membrane potential more negative.
D - An increase in the magnitude of repolarizing currents would make resting membrane potential more negative.
E - There was no change in the intracellular or extracellular concentration of potassium, so the potassium Nernst potential did not change.

Question 25

A 25-year-old male has been referred to your clinic by the nurse practitioner at the local CVX pharmacy because of periods of severe chest and arm pain. He also states that he is frequently short of breath and that the symptoms have been getting worse. You have ordered a cardiac stress test and subsequently, a coronary artery perfusion and imaging which reveals significant coronary disease resulting in the placement of 3 stents. Your physical exam of the patient is unremarkable with the exception of the presence of what appear to be xanthomas present on the extensor tendons of both hands. His family history reveals that his father died of a heart attack at the age of 26 and his mother is alive and healthy at age 57. Routine lab work reveals a plasma cholesterol level of close to 700 mg/dL in a fasting blood sample.

Which of the following conditions is the likely underlying cause of this young man’s heart disease?

Familial hyperchylomycronemia

Familial hypercholesterolemia

Familial hypertriglyceridemia

Tangier Disease

A diet high in cholesterol

Answer 25

Correct answer is B, familial hypercholesterolemia or Type II familial hyperlipoproteinemia, is the clear best answer here. In addition to his own high cholesterol levels and cardiovascular disease at such an early age, this young man has a family history on his father’s side of early cardiovascular disease but not on his mother’s side which suggests an autosomal dominant inheritance pattern.
Answer A is not consistent with this patient’s lab work or the properties of Type I hyperlipoproteinemia or lipoprotein lipase deficiency which does not lead to early cardiovascular disease.
Answer C is also not consistent with the patient as there is no history of diabetes, obesity, or alcoholism, etc. noted or increase in VLDL production. While hypercholesterolemia is seen in these patients, cardiovascular disease generally does not manifest itself at such a young age in this condition.
Answer D, Tangier’s Disease or Familial alpha lipoprotein deficiency refers to a lipoprotein deficiency or hypolipoproteinemia and it is characterized by a significant deficiency of HDL which is not noted in this patient. This disease is also characterized by unusual ophthalmic symptoms or appearance and is sometimes referred to as fisheye disease. It is associated with clear and dangerous symptoms in childhood due to lipid soluble vitamin malabsorption and a variety of issues not seen in this patient.
Answer E could be a cause of elevated cholesterol levels but given the family history of this patient, it is clear that an autosomal dominant, condition is the most likely in this case and so this would not be a likely cause.

Question 26

If capillary osmolality is 290 and interstitial fluid is 288, where will fluid flow?

Answer 26

From interstitial fluid to capillary

Question 27

A 34-year-old woman gave birth to her third child; a full term 7 lb 2 oz baby girl. At the second well baby checkup at 3 months the mother reported that the baby girl was more frail than her other children at this age, and seems to have more difficulty breast feeding. The 3-month-old baby girl has a machine like murmur between S1 and S2 heart sounds, best heard in the left suprasternal notch. Additionally, her toes were slightly bluish. Blood pressure taken in both right and left arms and legs were normal for a 3-month-old. A chest plain film showed normal heart size and aortic knob. The physician would order an echocardiogram to mostly likely diagnose which of the following conditions in this 3-month-old girl?

aortic stenosis

coarctation of the aorta

transposition of the great arteries

persistent truncus arteriosus

patent ductus arteriosus

Answer 27

patent ductus arteriosus

Correct answer is E: patent ductus arteriosus which caused both a machine like murmur that is best heard at the left suprasternal notch or in the back between the scapulae, and cyanosis of the toes but not the fingers. Aortic stenosis (answer A) would tend to cause enlargement of the aortic knob (arch due to jetting of blood) and left ventricular hypertrophy (her aortic knob and heart size was normal). Coarctation of the aorta (answer B) would cause hypertension in the upper extremities and hypotension in the legs. A baby girl with transposition of the great arteries (answer C) or persistent truncus arteriosus (answer D) would have symptoms (cyanosis) so severe that she would unlikely live to the 3rd month without medical intervention.

Question 28

A 61-year-old man weighs 80 kg and is determined to have 20% body fat. He receives an i.v. infusion of 3 liters of 5% dextrose. Assume that no fluid losses occur by any route. Which of the following is the best estimate of extracellular fluid volume in this person several hours later when dextrose has been completely metabolized?

15. 4 liters
16. 4 liters
17. 4 liters
18. 7 liters
19. 1 liters

Answer 28

16.4 liters

Question 29

In a nerve cell, the Nernst potential is found to be -90 mV for potassium, +60 mV for sodium, and -60 mV for chloride. The resting potential is -70 mV. Based on this information, you conclude that at resting potential:

chemical energy for chloride is -70 mV.

chloride current is negative.

electrical energy for potassium is +20 mV.

electrochemical energy for sodium is +60 mV.

there is a net repolarizing current.

Answer 29

chloride current is negative.

Question 30

A 78-year old female presents to her primary care physician due to decreased endurance and increased difficulty sleeping at night. She coughs a lot at night if she sleeps flat on her back, but she can sleep when elevated by several pillows. Her lungs sound congested and her external jugular vein is distended. The heart valves sounds immediately adjacent to both sides of her sternum are normal, but there is a mid-systolic click immediately followed by a “whooshing” sound heard at the apex of her heart inferior to her left breast. During cardiac ultrasound the papillary muscles of the left ventricle appeared ruptured. What is the working diagnosis?

aortic valve stenosis

aortic valve regurgitation

mitral valve stenosis

mitral valve regurgitation

tetralogy of Fallot

Answer 30

Answer is D) mitral valve regurgitation. The mid-systolic click immediately followed by a “whooshing” sound at the apex of her heart inferior to her left breast suggests a problem with the mitral valve. The rupture of the papillary muscles of the left ventricle is associated with mitral valve regurgitation rather than C) mitral stenosis. Aortic valve stenosis a. presents as a systolic murmur just to the right of the second intercostal space (location for ausculation of the aortic valve.) If B) aortic valve regurgitation were present, generally there would be a diastolic murmur, which is not present. Aortic valve stenosis is associated with enlargement of the left ventricle (due to increased resistance), which presents as enlargement of the left wall of the heart. In severe cases aortic valve stenosis causes enlargement of the aortic knob due to turbulence created downstream of the stenosis. C) Mitral valve stenosis would be heard as a diastolic murmur at the 4th or 5th intercostal space on the left side of the chest at the midclavicular line. E) Tetralogy of Fallot is unlikely to go undiagnosed in a 78-year old and would present with holosystolic murmur with right ventricular enlargement.

Question 31

A 39-year-old female patient showed marked elevation of plasma triglycerides (~2,500 mg/dl; normal < 150 mg/dl) and chylomicrons. Upon injections with a peptide fragment of which of the following proteins was the hypertriglyceridemia corrected in this patient?

apoA-1

apoC-II

apoE

apoB-100

apoB-48

Answer 31

apoC-II

A. apoA-1 activates lecithin:cholesterol acyltransferase which esterifies cholesterol in HDL
B. apoC-II activates lipoprotein lipase to be able to break down triglycerides
C. apoE LDL receptor ligand; helps to remove lipoproteins from the circulation
D. apoB-100 LDL receptor ligand; major protein of LDL
E. apoB-48 exclusively found in chylomicrons; does not bind to LDL receptor

Question 32

Actin binds to what at the intercalated disc?

Answer 32

Actin binds to adherens at the intercalated discs.

Question 33

During the upstroke of the action potential in a nerve, membrane potential moves toward the sodium Nernst potential. Which of the following is occurring at this time during this upstroke phase?

decrease in potassium chemical energy

decrease in sodium electrochemical energy

more positive sodium chemical energy

net repolarizing current across the membrane

opening of inactivation gates on sodium channels

Answer 33

decrease in sodium electrochemical energy

A. Incorrect: Chemical energy for ions (Nernst potentials) does not change during the action potential. The Na/K ATPase rapidly pumps any sodium ions that enter the cell back out, and pump potassium ions back into the cell. Ion concentrations remain nearly constant during an action potential.
B. Correct: As membrane potential moves closer to ENa during the upstroke of the action potential, the electrochemical potential for sodium (V – ENa) will decrease.
C. Incorrect: The Nernst potential for sodium does not change during the action potential.
D. Incorrect: There is a net depolarizing current during the upstroke of the action potential.
E. Inactivation gates are slowly closing during the upstroke of the action potential.

Question 34

A 100 kg man with 10% body fat receives an i.v. infusion of 3 liters of fluid which is 5% dextrose in 0.9% NaCl. Which of the following is the best estimate of the extracellular fluid volume at equilibrium in this person after the infusion?

15 liters

18 liters

22 liters

25 liters

48 liters

Answer 34

Total body water = Lean body mass x 0.72 = (100 kg – (10% x 100 kg)) x 0.72 = 64.8 liters
ICF volume = (2/3) x TBW = 43.2 liters
ECF volume = (1/3) x TBW = 21.6 liters

The solution infused i.v. is 0.5% dextrose and 0.9% NaCl. At equilibrium, the dextrose will be metabolized so this will be equivalent to infusing 3 liters of 0.9% NaCl. The 3 liters will remain in the ECF, so ECF volume will be 21.6 liters + 3 liters, or 24.6 liters.

Question 35

A 63-year-old man is brought to the emergency department due to weakness in moving his limbs. Earlier in the day, he had eaten a pufferfish which is known to contain tetrodotoxin, a substance which blocks voltage-gated sodium channels. Tests reveal that plasma sodium levels are normal. Which of the following effects in nerves could be caused by tetrodotoxin to account for his symptoms?

decreased rate of rise during depolarization of the action potential

increased chemical energy for sodium

increased conduction velocity of the action potential

increased number of resting sodium channels

more negative threshold

Answer 35

decreased rate of rise during depolarization of the action potential

Since tetrodotoxin blocks voltage-gated sodium channels, sodium current would be reduced during the upstroke of the action potential.
A - Correct: The reduced sodium current will slow the rate of rise of the action potential.
B - Incorrect: Chemical energy for sodium is determined by the ratio of extracellular and intracellular sodium concentrations, and the valence of sodium. These will not be affected by tetrodotoxin.
C - Incorrect: The decreased sodium current will decrease conduction velocity of the action potential.
D - Incorrect: If the number of resting sodium channels was increased, sodium current would be increased. Tetrodotoxin decreases sodium current.
E - Incorrect: By decreasing sodium current, tetrodotoxin would make threshold less negative (more positive), and decrease excitability.

Question 36

Excess water intake decreases plasma sodium below normal levels resulting in dilutional hyponatremia. Isotonic saline (0.9% NaCl) can be used to correct the sodium deficit in this situation. Assume that plasma sodium is 110 mEq/liter in a person with dilutional hyponatremia. Which of the following will occur in this person following intravenous infusion of isotonic saline?

decrease in extracellular osmolality

increase in intracellular osmolality

increase in intracellular volume

hypotonic contraction

isotonic expansion

Answer 36

increase in intracellular osmolality

Although 0.9% NaCl is normally an isotonic solution, it is a hypertonic to this person with dilutional hyponatremia (which decreases osmolality of the extracellular and intracellular fluid). The plasma sodium concentration is 110 mEq/liter, which is about half of the plasma osmolality (which would be about 220 mEq/liter, lower than normal).
A - Infusion of 0.9% saline will raise extracellular osmolality.
B - The increase in extracellular osmolality after the saline infusion will cause water to diffuse out of the cells, causing intracellular osmolality to increase.
C - As water diffuses out of cells, intracellular volume will decrease.
D - The saline infusion expands the volume of the extracellular space.
E - The osmolality of body fluids is still lower than normal.

Question 37

Absence in apoE would result in?

Answer 37

would result in increased levels of VLDL because apoE helps in receptor binding

Question 38

During the 4th though the 7th week of embryonic development the initially single ventricular chamber is being subdivided into right and left ventricles by a muscular ventricular septum. The final growth of the interventricular septum is linked to which other event in embryonic heart development?

growth of the septum primum

formation of the ostium secundum in the septum primum

final septation of the truncus arteriosus

incorporation of the four pulmonary veins into the left atrium

fusion of the superior/anterior and inferior/posterior endocardial cushions

Answer 38

final septation of the truncus arteriosus

Answer is C) final septation of the truncus arteriosus. Thus, as the ventricles are becoming completely separate chambers the outflow of the heart is separating into aortic and pulmonic outflow. This occurs at about the 8th week of development. Growth of the septum primum a. occurs earlier during heart development in the 4th to 6th week of development. The formation of the ostium secundum in the septum primum b. occurs in about the 6th week of development. The incorporation of the four pulmonary veins into the left atrium d. occurs at about the 5th week of development. Fusion of the superior/anterior and inferior/posterior endocardial cushions e. occurs in about the 4th week of development.

Question 39

Leukocyte diapedesis typically occurs where?

across lymphatics exclusively

across post-capillary venules

anywhere in the systemic vasculature

most frequently across fenestrated capillary beds, and occasionally through continuous capillaries

Answer 39

During inflammation, movement of leukocytes from blood into the interstitial space (diapedesis) occurs mainly in post-capillary venules.

Question 40

A pediatrician detects a harsh murmur along the left sternal border in a 3 week old boy. The parents report that the baby gets ‘bluish’ when he cries or drinks from his bottle. Echocardiogram reveals pulmonary stenosis and right ventricular hypertrophy. What is the best diagnosis?

atrial septal defect

coarctation of the aorta

tetrology of Fallot

truncus arteriosus

Answer 40

Answer: C Tetrology of Fallot
Tetrology of Fallot includes 4 anatomic changes: pulmonary stenosis, ventricular septal defect, dextroposition of the aorta, and right ventricular hypertrophy. It is the most common cyanotic congenital heart disease, accounting for 10% of all congenital heart defects. Cyanosis appears shortly after birth or early in infancy due to right-to-left shunting of venous blood from the right ventricle into the dextroposed aorta. Narrowing of the pulmonary artery impedes the entry of blood into the lung, thereby increasing the pressure in the right ventricle.

Question 41

In the eLearning Case, the person with chronic obstructive pulmonary disease had bilateral edema of the lower extremities. Assume that capillary pressure within the lower legs of this person is 30 mm Hg, pressure in interstitial fluid is 0 mm Hg, and plasma osmolality is 290 mOsm. If the extent of edema is staying constant (i.e., no net fluid movement across the capillary), which of the following is the best estimate of osmolality of intracellular fluid?

288 mOsm

288.5 mOsm

290 mOsm

291.5 mOsm

292 mOsm

Answer 41

288.5 mOsm

If the extent of edema is remaining constant, the movement of fluid into the interstitial space from capillaries must be exactly balanced by movement of fluid from the interstitial space into capillaries. Capillary pressure is 30 mm Hg, which drives fluid out of the capillary. Therefore, the force for reabsorbtion of fluid into capillaries must be equal to 30 mm Hg. Each one mOsm difference in between plasma and interstitial fluid develops a force of 20 mm Hg driving water movement. Since capillary pressure is 30 mm Hg, a difference in osmolality of 1.5 mOsm will develop a force of 30 mm Hg. Solute concentration in interstitial fluid must be 1.5 mOsm lower than plasma osmolality, or 288.5 mOsm. Interstitial osmolality will be equal to intracellular osmolality.

Question 42

What would apo-CII absence result in?

Answer 42

would result in elevated chylomicron levels

Question 43

A 42 year old woman weighing 60 kg with 10% body fat has a plasma osmolality of 290 mOsm. She receives an i.v. infusion of 2 liters of 0.45% NaCl. Which of the following best describes the resulting change in body fluids in this person?

hypertonic contraction

hypertonic expansion

hypotonic contraction

hypotonic expansion

isotonic contraction

isotonic expansion

Answer 43

hypotonic expansion

A solution of 0.45% NaCl has an osmolality of 145 mOsm. The intitial effect of infusion of this solution would be to increase the volume of the ECF and to decrease the osmolality of the ECF. The latter effect will cause water to diffuse into cells, reducing osmolality of the ICF and increasing its volume. As water moves out of the ECF, its volume will decrease back towards normal but its volume will still be greater than normal at equilibrium. This infusion will result in a hypotonic expansion resulting in decreased osmolality of the ICF and ECF, as well as an increase in ECF volume.

Question 44

A 47-year old male arrives by ambulance to the emergency department due to left-sided chest pain. He refused to lie down on the ambulance gurney because it made the chest pain worse. He has no other pain. While his heart rate is elevated, his blood pressure is 127 / 82, and all his heart valves sound normal. There does appear to be a diffuse squeaky leather sound, which worsens when the patient is asked to forcefully expel air. His lungs sound normal. What is the working diagnosis?

cardiac tamponade

coarctation of the aorta

congestive heart failure

patent ductus arteriosus

pericarditis

Answer 44

Answer is E) pericarditis. The pain is localized to the left chest and does not radiate to his left arm. The patient’s chest pain is worse when lying flat and when the diaphragm is forced upward during forced expiration both suggest pericarditis. The diffuse squeaky leather sound also suggests pericarditis. All heart valve sounds would be diminished during A) cardiac tampanade (fluid in the pericardial sac). If the patient had B) coarctation of the aorta the upper extremity blood pressure would be elevated, which it is not. If the patient has C) congestive heart failure his lung sounds would not be normal. If the patient had a D) patent ductus arteriosus, then there would be a “machine-like” holosystolic murmur best hear on the upper left sternal border, which is not present.

Question 45

Tests reveal that a 37 year old man is hypernatremic as his plasma sodium concentration is 160 mEq/liter. This person’s total body water is 42 L. If the desired plasma sodium concentration is 140 mEq/liter, which of the following values is closest to the water deficit in this person?

3 liters

4 liters

5 liters

6 liters

7 liters

Answer 45

6 liters

The correct answer is D.

H2O Deficit = Present TBW x { ( Current [Na+] / Normal [Na+]) -1}

H2O Deficit = 42 liters x { (160 mEq/L / 140 mEq/L) -1} = 6 liters

Question 46

You deliver a 6 lbs, 6 oz full term baby girl. Her breathing rate is slightly elevated, and when she cries her lips turn blue. Her toes are also bluish. You immediately send for a pediatric cardiologist who orders an echocardiogram. The sonographer who immediately performs the echo study is convinced that the baby girl has transposition of her great vessels. Which of the following is the most likely additional heart defect that the newborn girl also has?

overriding aorta

ventricular septal defect

ligamentum arteriosum

coarctation of the aorta

aortic aneurysm

Answer 46

ventricular septal defect

The answer is B. Ventricular septal defect. (Moore and Dalley p 151; Sadler p 178.) The additional heart defect the newborn girl also has is a ventricular septal defect. Babies born with transposition of the great vessels normally present with symptoms of cyanosis as the ductus arteriosus closes within the first day of birth. Since in this defect the aorta is sitting on top of the right ventricle and the pulmonary trunk is receiving blood from the left ventricle, blood is being pumped mainly just to the body by the right side and mainly just to the lungs by the left side. The only way oxygenated blood is getting to the body is if the two sides are connected, often by a patent ductus arteriosus. Babies with this defect are immediately but on oxygen, and prostaglandins are also given to help keep the ductus arteriosus open longer than normal. About 25% of the time that transposition of the great vessels is present, there is also a ventricular septal defect, which aids in the intermixing of blood from the two sides of the heart. This condition must be treated surgically. Defects of the aortic arch are normally not present. Ligamentum arteriosum (answer C) would make the problem worse. Coarctation of the aorta (answer D) would make the problem worse. An overriding aorta (answer A) is the second best answer. An aortic aneurysm (answer E) is unlikely to be present in a newborn.

Question 47

A 43 year old woman has been lost in the wilderness for one week. She is brought to the Emergency Department after being found, and is conscious and responsive to questions. Her blood pressure is lower than normal and her plasma osmolality is elevated. Based on this information, which statement is correct about this person?

Interstitial volume in systemic organs is increased.

Intracellular osmolality is decreased.

Intracellular volume is increased.

The best treatment is i.v. infusion of 5% dextrose.

This is an example of an isotonic contraction.

Answer 47

Correct answer is D. This situation is an example of dehydration and this is indicated by the low blood pressure and increased plasma osmolality. Water is lost from the extracellular fluid, which increases it’s osmolality causing water to diffuse out of cells

A - Both interstitial volume and plasma volume would be decreased in this person, resulting in a decreased extracellular fluid volume.
B - The increase in plasma osmolality indicates that extracellular fluid osmolality is increased, which would cause intracellular fluid volume to decrease and increase intracellular osmolality.
C - Water diffuses out of cells when extracellular osmolality is increased, and this would decrease intracellular volume.
D - The person has a water deficit and infusing 5% dextrose is the best treatment to restore water to both the extracellular and intracellular fluid compartments (dextrose will be metabolized).
E - This is an example of a hypertonic contraction.

Question 48

What would an absence of apoA-1 lead to?

Answer 48

absence of functional apoA-1 results in severely reduced levels of HDL

Question 49

A 82-year-old man went to his family practice physician’s office reporting difficulty sleeping without being propped up on three pillows and shortness of breath. He doesn’t report any unusual chest pain. Using a stethoscope the physician listened to his lungs and heart. The lung sounded relatively normal, but when listening to his heart the first heart sound seemed diminished. Specifically at the left mid-clavicular line at the 5th intercostal space the valve sound at that location was soft and abnormal. All other heart valves sounded normal. The patient’s ECG was normal for a 82-year-old. What is the most likely source of the patient’s problem?

cardiac tamponade

pericarditis

occlusion of the right coronary artery

rupture of pectinate muscle fibers

mitral valve prolapse

Answer 49

mitral valve prolapse

Correct answer is E: mitral valve prolapse. The sound of the mitral valve closing is best heard on the left chest wall at the 5th intercostal space at the mid-clavicular line. Of all heart valves the mitral valve is the most frequently damaged. The chordae tendaneae can rupture and cause the valve to flap back into the left atrium during left ventricle systole. Patients with cardiac tamponade (answer A) have shortness of breath (dyspnea), but all of their heart sounds (closure of valves) would be diminished, which was not the case for the 82-year old. Pericarditis (Answer B) often causes a pericardial rub, which was not present in this patient. Occlusion of the right (Answer C) arteries would tend to cause disruption of blood flow, cardiac pain, abnormal ECG, along with shortness of breath, but the patient reports no chest pain, making this unlikely. Rupture of pectinate muscle fibers (answer D) is highly unlikely (they are only present in the atrial chambers) and would have no effect on the mitral valve.

Question 50

Which statement is true for a cardiac muscle cell in the left atria?

The T tubules and sarcoplasmic reticulum are arranged in triads.

The myosin filament cannot bind actin until it is phosphorylated by myosin light chain kinase.

Ca2 -dependent ATPases are present on the sarcoplasmic reticulum.

The actin filaments bind to the desmosomes at the intercalated disk.

The only source of calcium is the release of calcium from the sarcoplasmic reticulum.

Answer 50

Ca2 -dependent ATPases are present on the sarcoplasmic reticulum.

A. The T-tubules and sarcoplasmic reticulum is predominantly arranged as a diad in cardiac muscle. It is a triad in skeletal muscle.
B. Myosin will bind whenever the troponin complex does not cover the actin binding site. Activation of myosin by myosin light chain kinase is important in smooth muscle contraction.
C. These are required for the return of Ca2+ to the sarcoplasmic reticulum following excitation. They are also found on the sarcolemma.
D. Actin binds to the adherens complex at the intercalated disk. Intermediate filaments bind to desmosomes.
E. Calcium also enters through voltage gated calcium channels.

Question 51

During the partitioning of the embryonic atrium from a single chamber into right and left atria, there are two septae that form and three openings. Which is the last opening to form and in which septum does it form?

ostium primum in septum primum

ostium primum in septum secundum

ostium secundum in septum primum

foramen ovale in septum primum

foramen ovale in septum secundum

Answer 51

foramen ovale in septum secundum

Answer is E) foramen ovale in the septum secundum. There is no foramen ovale in the septum primum d.. There also is no ostium primum in the septum secundum b.. The ostium primum in the septum primum a. forms first. The ostium secundum in the septum primum c. is the second opening to form.

Question 52

What are the two sources of calcium release in a cell?

Answer 52

Calcium enters through voltage gated calcium channels and is released by the sarcoplasmic reticulum.

Question 53

Capillary pressure is referring to pressure driving fluid in which direction? What about interstitial fluid pressure?

Answer 53

Driving fluid out of the capillary into the the interstitial space.

Driving fluid from the interstitium to the capillary

Question 54

Assume that a cell has an intracellular potassium concentration of 150 mM and an intracellular sodium concentration of 14 mM. The cell is placed in a solution with a potassium concentration of 15 mM and a sodium concentration of 140 mM. The cell is permeable to only potassium and sodium, and potassium conductance is 20 times higher than sodium conductance. Which of the following is the best estimate of resting membrane potential in this cell?

• 42 mV
• 51 mV
• 54 mV
• 58 mV
• 60 mV

Answer 54

-54 mV