USMLE : United States Medical Licensing Examination Step 1

USMLE Dumps
USMLE Braindumps
USMLE Real Questions
USMLE Practice Test
USMLE Actual Questions


USMLE
USMLE
United States Medical Licensing Examination Step 1
https://killexams.com/pass4sure/exam-detail/USMLE


QUESTION 834
A 68-year-old postmenopausal female with a history of osteoprosis and essential hypertension is placed on the thiazide diuretic chlorothiazide, which
has a beneficial action toward both conditions. Which letter in below figure depicts the tubular location of epithelial cells containing a / cotransport
protein inhibited by thiazide diuretics?

www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications


A. A
B. B
C. C
D. D
E. E
Answer: D
Section: Physiology
Explanation
Explanation:
The epithelial cells of the early portion of the distal tubule contain a / cotransporter that is
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
inhibited by thiazide diuretics which promote a diuresis and a natriuresis, and which secondarily, promote increased renal reabsorption of filtered
calcium. Choices A and B denote the proximal and thin loop of Henle segments, respectively; which do not have the properties of the early distal tubular
segment. Likewise, choices C and E denote the thick ascending loop of Henle and the collecting duct, which also do not have the same properties as
the early distal tubular segment.
QUESTION 835
Worldwide, one of the most common parasitic infections is schistosomiasis. Assume a complete blood cell count is performed in a patient with this
condition. Which of the following blood cells would most likely be present in elevated amounts?
A. eosinophils
B. erythrocytes
C. monocytes
D. neutrophils
E. platelets
Answer: A
Section: Physiology
Explanation
Explanation:
Eosinophils normally constitute about 2% of all blood leukocytes. These cells are phagocytic and exhibit chemotaxis. Their primary role is in fighting
parasitic infection. In schistosomiasis the eosinophil attaches to the immature parasite and, by releasing various cytotoxic compounds, is able to kill the
parasite. Erythrocytes (choice B) are the most numerous blood cells. They are not leukocytes, but function in blood gas transport. Erythrocyte number
would not be increased in schistosomiasis. Monocytes (choice C) are leukocytes found in the blood. They are immature cells which migrate to the
tissues, where they mature into macrophages. These cells are very active in phagocytosis and play a prominent role in the inflammatory response. They
are not, however, particularly important in parasitic infections. Neutrophils (choice D) are another type of leukocyte that, though essential to combat
bacterial infections, are not particularly sensitive to parasitic challenge. Platelets (choice E) are not leukocytes. The platelet is a cell fragment that
functions in hemostasis.
QUESTION 836
After total thyroidectomy as a result of a car accident, a 47-year-old woman is positive for Trousseau sign. It is manifested as carpopedal spasm
occurring after an inflated blood pressure cuff is held above systolic blood pressure for a few minutes. Which of the following would best describe her
postoperative condition when compared to preoperative?
A. decreased deep tendon reflexes
B. higher serum calcitonin levels
C. lower serum calcium levels
D. lower serum phosphate levels
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
E. shorter Q-T interval of her ECG
Answer: C
Section: Physiology
Explanation
Explanation:
A positive Trousseau sign indicates increased neuromuscular irritability, which is caused by low serum calcium levels. Apositive Trousseau sign or a
positive Chvostek sign (spasms of facial muscles) can often diagnose hypocalcemia before other gross manifestations develop. Serum calcium is
regulated by parathyroid hormone produced by the parathyroid glands. During the woman's accident, the glands, or their vasculature, were most likely
damaged enough so that they could not be saved during the total thyroidectomy. If total thyroidectomy occurs as a consequence of thyroid cancer,
damage to all four parathyroid glands would be considered an operative mishap. A consequence of hypocalcemia is hyperreflexia and not hyporeflexia
(choice A). Calcitonin is secreted by parafollicular or "C" cells in the thyroid gland. Since no other cells of the body make this hormone, serum calcitonin
levels do not rise (choice B), but instead go to zero postoperative. Low serum phosphate levels (choice D) typically signify secondary
hyperparathyroidism or low dietary phosphate intake. In the current case, high or unchanged serum phosphate would be expected. The Q-T interval of
the ECG represents the time for both ventricular depolarization and repolarization to occur. It therefore roughly estimates the duration of an average
ventricular action potential. Hypocalcemia leads to a long Q-T interval, not a short Q-T as in choice E. In severe cases, this can lead to arrhythmias,
hypotension, and heart failure.
QUESTION 837
A 60-year-old woman is admitted to the hospital with a fever and severe diarrhea for the last 24 hours. Cultures of blood, cerebrospinal fluid, urine, and
stool are all negative for pathogens. The profile of gut hormones reveals elevated levels of VIP. An analogue of which of the following would most likely
lower her VIP levels?
A. erythromycin
B. histamine
C. motilin
D. somatostatin
E. trypsin
Answer: D
Section: Physiology
Explanation
Explanation:
VIP is a neurotransmitter in the brain and in the parasympathetic nerves of the digestive tract. It also acts as a hormone. VIP has a secretin-like effect
on the pancreas. It increases the volume of water and bicarbonate output and affects GI blood flow and motility. All this contributes to severe secretory
diarrhea in the case of VIP overproduction. Somatostatin is the best choice because it has a broad range of inhibitory effects, inhibiting GI secretions,
slowing GI motility, and reducing splanchnic blood flow. An analogue of somatostatin, called octreotide, was found useful for treatment of syndromes
related with VIP overproduction. Octreotide has a longer half-life than physiologic somatostatin. Erythromycin (choice A) is known to have antibiotic
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
features. This is not the best choice, since no pathogens were found. Histamine (choice B) is a physiologic stimulant of gastric hydrochloric acid
secretion, potentiating the effects of acetylcholine and gastrin. Motilin (choice C) stimulates gastric motility, which is contraindicated for diarrhea. Trypsin
(choice E) is a protease released by pancreatic cells, breaking peptide bonds as well as converting many other proteolytic enzymes to their active form.
All of the negative choices will not affect, and in fact may even worsen the VIPrelated symptoms.
QUESTION 838
Malignant hyperthermia is a relatively rare genetic disorder, found in approximately one in 10,000 individuals. Affected individuals can exhibit a life-
threatening response to inhalation anesthetics, which entails elevated core body temperature, skeletal muscle rigidity, and elevated blood carbon
dioxide levels. Based on these observed symptoms, which of the following is most likely present in malignant hyperthermia?
A. decreased release of acetylcholine at the neuromuscular junction
B. increased activation of myosin light chain kinase
C. increased activation of skeletal muscle potassium channels
D. increased activation of the calcium release channels of the sarcoplasmic reticulum
E. inhibition of skeletal muscle sodium channels
Answer: D
Section: Physiology
Explanation
Explanation:
Malignant hyperthermia is caused by a mutation to the calcium release channel of the skeletal muscle sarcoplasmic reticulum (the socalled ryanodine
receptor). Increased calcium release into the myoplasm causes strong, sustained skeletal muscle contraction. This causes excessive ATP hydrolysis,
increased metabolic activity to replenish ATP levels, heat generation, and carbon dioxide production. Decreased acetylcholine release would decrease
muscle contraction (choice A). Myosin light chain kinase is involved in contractile activation in smooth muscle (choice B). Activation of K channels would
hyperpolarize muscle fibers and tend to decrease susceptibility to contractile activation (choice C). Inhibition of sodium channels would diminish the
ability of skeletal muscle to generate action potentials and thus decrease excitability (choice E).
QUESTION 839
Below figure shows the conversions of cholesterol into the hormones C and D within follicular cells of the ovary (large arrows) and the regulation of
these processes (small arrows) by pituitary hormones (A, B, E), when binding to their receptors on the cell surfaces (dark squares). Which of the letters
in the figure best represents the hormone the concentration of which in serum changes in the following way?
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications


A. A
B. B
C. C
D. D
E. E
Answer: D
Section: Physiology
Explanation
Explanation:
It represents the estrogen estradiol, which is produced from androgen by granulosa cells of the ovaries and released into the capillaries. A woman's
estradiol serum concentration roughly reflects the activities of her ovaries. Estradiol levels rise during the follicular phase of the menstrual cycle (days 0
13). They reach peak levels shortly before ovulation on day 1314, and drop back during the luteal phase (days 1428). They reach menstrual/follicular
levels at the end of the luteal phase unless there is a pregnancy. After menopause, the ovaries atrophy and estradiol levels become very low. During
reproductive cycles, androgens are the main substrate for estrogen synthesis in the granulosa cells. The androgen-estrogen conversion is under the
influence of FSH (choice E). FSH concentrations are high after menopause compared to their premenopausal concentrations. The source of androgens
is theca cells. They produce it from cholesterol under the influence of LH (choice B). LH concentrations, like FSH, are high postmenopausal. Both, FSH
and LH levels are used as a diagnostic tool to determine menopause. Granulosa cells also produce progesterone (choice C) from cholesterol.
Progesterone serum levels are low during follicular phase of the menstrual cycle and high during the luteal phase of the menstrual cycle. The transition
of cholesterol to progesterone is under the influence of LH (choice A).
QUESTION 840
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
During a marathon attempt a runner collapses and is admitted with severe acute dehydration. Which of the following is most likely to occur in this
patient?
A. decreased baroreceptor firing rate
B. decreased plasma osmolarity
C. high renal water excretion
D. low plasma ADH levels
E. low water permeability of collecting duct tubular cells
Answer: A
Section: Physiology
Explanation
Explanation:
Acute dehydration results in decreased plasma volume, cardiac output, and arterial pressure, which leads to a lower firing rate of the baroreceptors.
Plasma osmolarity increases (choice B) because more water than salt is lost in sweat. The increase in plasma osmolarity leads to increased ADH
secretion (choice D). High plasma ADH levels cause increased water permeability of collecting duct cells (choice E). Therefore, more water is
reabsorbed by the kidneys and renal water excretion is low (choice C).
QUESTION 841
When glucose is chronically elevated in poorly controlled diabetes mellitus, nonenzymatic glycosylation of various proteins occurs. The change of which
of the following substances is most commonly monitored as indicator for the efficiency of blood glucose control?
A. hemoglobin A1c (HbA1c)
B. lipoprotein A.
C. modified albumin
D. myoinositol
E. sorbitol
Answer: A
Section: Physiology
Explanation
Explanation:
The hallmark of poorly controlled diabetes mellitus is elevated blood glucose, which causes unphysiological glycosylation of proteins. An important long-
term measure of blood glucose control in patients with diabetes mellitus is to monitor the modification of hemoglobin A to form glycated hemoglobins.
HbA1c is the major subfraction, and determination of HbA1c is usually achieved by ion- exchange or gel electrophoresis. The level of glycated
hemoglobins in the blood is directly related to the average blood glucose levels over the life span of the hemoglobin in the circulation. Since the half- life
of red blood cells is about 120 days, a single determination of glycated hemoglobin reflects the average blood glucose level during the preceding 812
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
weeks. Lipoprotein A. (choice B) is a lipoprotein particle implicated in atherosclerosis and thrombosis. Albumin (choice C) is the most abundant plasma
protein, but is not significantly affected by glycosylation. Myoinositol (choice D) is a signaling molecule the decrease in response of which to elevated
sorbitol has been suggested as a complication of diabetes. Sorbitol (choice E) is another sugar derivative, unrelated to HbA1c that is believed important
in causing other diabetic complications such as cataracts and peripheral neuropathy.
QUESTION 842
Apatient on intensive care is ventilated with a frequency of 12 per minute and a tidal volume of 0.6 L. His arterial pH increases to >7.5. What is the most
reasonable action to correct this respiratory alkalosis?
A. decrease dead space
B. decrease tidal volume
C. increase minute ventilation
D. increase oxygen fraction
E. use positive end-expiratory pressure (PEEP)
Answer: B
Section: Physiology
Explanation
Explanation:
Respiratory alkalosis is due to hyperventilation, which lowers . Decreasing tidal volume will
reduce alveolar ventilation and correct the respiratory alkalosis. Assuming a dead space of 150 mL, alveolar ventilation in this patient is 450 mL × 12/
min = 5400 mL/min. If the tidal volume were decreased from 600 to 300 mL and the frequency increased from 12 to 24 per minute, then the alveolar
ventilation would decrease to 150 × 24/min = 3600 mL/min even though the minute ventilation (12 × 600 mL/min = 24 × 300 mL/min) remains
unchanged. The fraction of (choice D) in the respiratory
air does not affect respiratory volumes or frequencies in a mechanically ventilated patient. Increasing minute ventilation (choice C) or decreasing dead
space (choice A) would increase alveolar ventilation and worsen respiratory alkalosis. PEEP (choice E) is positive pressure applied during the expiratory
phase to prevent the collapse of alveoli and to increase FRC of the lungs. It is used primarily to improve arterial oxygenation in severely hypoxic
patients.
QUESTION 843
Below figure illustrates the extracellular and intracellular volumeosmolarity status of a patient (broken lines) and that of a normal subject (solid lines) for
comparison. This patient most likely suffers from which of the following conditions?
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications

A. adrenal insufficiency
B. chronic vomiting
C. iatrogenic fluid overload with 0.9% NaCl
D. iatrogenic fluid overload with hypertonic solution
E. syndrome of inappropriate hypersecretion of antidiuretic hormone (SIADH)
Answer: E
Section: Physiology
Explanation
Explanation:
This patient has increased extra- and intracellular volumes and a decreased osmolarity. SIADH results in inappropriately low water permeability of the
renal collecting duct tubular cells and inappropriate water retention. As a result, patients with SIADH often present with hypotonic overhydration. Adrenal
insufficiency (lack of aldosterone) (choice A) and chronic vomiting (choice B) lead to dehydration. Fluid overload with isotonic NaCl (choice C) results in
volume expansion without change in osmolarity. Fluid overload with hypertonic solution (choice D) results in volume expansion with increased
osmolarity.
QUESTION 844
Gamma-aminobutyric acid (GABA) is an amino acid that functions as a neurotransmitter in the central nervous system. GABA typically causes
increased chloride conductance and functions as an inhibitory transmitter. Assume that the equilibrium potential for chloride (ECl-) in a particular cell is -
80 mV and that application of GABA inhibits the cell without any change in resting membrane potential. What is the resting membrane potential of the
cell?
A. +80 mV
B. 0 mV
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
C. -70 mV
D. -80 mV
E. -90 mV
Answer: D
Section: Physiology
Explanation
Explanation:
This problem addresses two issues: (1) the mechanism of action of inhibitory neurotransmitters and (2) the relationship of equilibrium potential and
membrane potential. First, an inhibitory neurotransmitter acts by increasing conductance of an ion the equilibrium potential of which is either equal to or
more negative than the cell resting membrane potential. If the equilibrium potential of the ion is more negative than the resting membrane potential,
increasing the conductance of that ion will hyperpolarize the membrane--will generate an inhibitory postsynaptic potential (IPSP). Second, if the
equilibrium potential of the ion is equal to resting membrane potential, increasing the conductance of that ion will "clamp" the membrane potential more
tightly at its resting level--this will make the cell less excitable. In this problem, the membrane potential must equal -80 mV D.. If, and only if, the resting
membrane potential equals the equilibrium potential for an ion, will an increase in the conductance to that ion not cause a change in membrane
potential.
QUESTION 845
The striped area in below figure indicates a red light on the receptive fields of a red on-center, green off-center surround retina ganglion cell. How will
the light spot be perceived?

A. brown (red + green)
B. green
C. no color
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
D. red
E. reddish green
Answer: D
Section: Physiology
Explanation
Explanation:
A light spot exclusively hitting the center would be perceived as deep red. Some of the light hits the inhibitory surround area. In this area, light is
processed coming from cones that are best excited by green light (middle wavelength) and inhibited by red light (long wavelength). In the below figure
the on-response dominates the off-response so that the perceived light is different from deep red, but not green (choice B). At the retina ganglion level,
the trichromate signals from the cones are transferred into the neuronal stage of color processing, the opponent color stage. The cone photoreceptors
are linked together to form opposing color pairs, in this case red versus green. Activation of one member of the pair inhibits the activity in the other so
that a color pair cannot be seen at the same location. We cannot experience reddish green (choice E). Physiological color image formation differs from
the physical process of coloring a piece of paper with a red and green pencil (choice A), one on top of the other, called subtractive color mixture. It starts
with the presence of all colors of light and then dyes subtract some of the reflected light so that adding more dyes results in a darker image such as
brown. For physiological color mixing, the presence of all colors is perceived as white, the absence as black (choice C).
QUESTION 846
Creatinine clearance is often used to evaluate glomerular function. Which of the curves in below figure best represents the relationship between plasma
creatinine concentration and creatinine clearance in a normal healthy person?

A. A
B. B
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
C. C
D. D
E. E
Answer: C
Section: Physiology
Explanation
Explanation:
Creatinine clearance is independent of plasma creatinine concentration; otherwise, creatinine would not be a useful measure of GFR. Clearance is
defined as the amount of plasma that delivered the excreted substance, and for a substance that is neither actively secreted nor reabsorbed by the
kidneys; its clearance equals the amount of plasma filtered through the glomerular membrane. All creatinine contained in that amount of plasma is
excreted by the kidney, no matter what the concentration of creatinine in that plasma volume was. Because in the normal person a small amount of
creatinine is secreted by the renal tubules, clearance at low plasma concentrations is slightly higher than at elevated plasma concentration (slight initial
upward bend of curve). Curve Adescribes the relationship between creatinine plasma concentration and renal excretion of creatinine. Note that
excretion and clearance are not synonymous. Curves B and D describe the clearance of a substance that is secreted and filtered, or filtered and
reabsorbed, respectively. At large plasma concentration, the active transporters become saturated and the clearance of these substances approaches
the creatinine clearance. Curve E depicts an improbable event with relatively increased clearance at both low and high concentrations of a substance,
although there is an independent linear clearance at intermediate substance concentrations.
QUESTION 847
The patient is a 43-year-old male. He is anemic, with a hemoglobin level of 12.2 g/dL (normal is 15.5 g/dL). The erythrocytes are microcytic (MCV = 70
fL, with normal MCV = 80100 fL). Which of the following would most likely be present in this patient?
A. acute bleeding
B. folate deficiency
C. iron deficiency
D. vitamin deficiency
E. vitamin K deficiency
Answer: C
Section: Physiology
Explanation
Explanation:
Microcytic anemia can often be associated with defective hemoglobin synthesis. In the case of iron deficiency, heme synthesis is impaired due to the
lack of iron. Acute bleeding (choice A) is normally associated with erythrocytes of normal size, since the anemia is due to simple loss of blood, with
plasma expansion to maintain total volume. Both vitamin and folate (choices B and D) are
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
required for DNA synthesis. Due to the extremely active cell division required to sustain the erythrocyte pool, impaired DNA synthesis first normally
manifests as a lack of erythrocyte production. Anemia due to vitamin B12 or folate deficiency is macrocytic. Vitamin K (choice E) is required for
synthesis of several clotting factors. Lack of vitamin K often presents as a problem with coagulation.
QUESTION 848
Which of the following statements concerning total body energy storage is correct?
A. Most of the body's energy store is held as carbohydrate.
B. Most of the body's energy store is held as lipid.
C. Most of the body's energy store is held as plasma glucose.
D. Most of the body's energy store is held as protein.
E. Total body's energy storage approximately equals resting metabolic rate.
Answer: B
Section: Physiology
Explanation
Explanation:
Lipid is the most concentrated form of energy storage, holding 9.4 kcal/g. For a typical 70 kg human over 130,000 kcal is typically stored as fat. Storage
of energy as available protein (choice D) is about 20,000 kcal, while storage as carbohydrate (choice A) is about 3,000 kcal. Though blood glucose
(choice C) is critically important, in particular as a source of energy for the brain, blood glucose does not provide a significant fraction of total body
energy storage. While total body energy stores typically exceed 150,000 kcal, resting metabolic rate (depending on age and body mass) is about 2100
kcal/day (choice E).
QUESTION 849
During a brain surgery, during which the patient remained conscious, a part of the DC/ML (dorsal column, medial lemniscal) system was exposed and
stimulated. Which of the following sensations might the patient most likely experience?
A. a feeling as if his fingers are warming up
B. a feeling as if something were touching his index finger
C. a tingling sensation
D. mild pain but he is unable to explain exactly where it hurts
Answer: B
Section: Physiology
Explanation
www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications
Explanation:
The dorsal column-medial lemniscus (DC/ML) system is the sensory pathway that transmits conscious proprioceptive information with fine gradations
such as a fine touch on the index finger to the cerebral cortex. The brain interprets any signals coming along the "labeled lines," the nerves that carry
messages from the sensory organ to the cortex, as being information about a particular sense. Hence, activation of certain neurons in the somatic
sensory system can cause a sensation as if it originated from mechanical pressure on the finger (choice E). Thermal sensations (choice A), tingling
(choice C), and crude undetermined pain (choice D) are sensory modalities that are processed in the spinothalamic or anterolateral system.



www.vceplus.com - Download A+ VCE (latest) free Open VCE Exams - VCE to PDF Converter - VCE Exam Simulator - VCE Online - IT Certifications

User: Svetlana***** To ensure my success in the USMLE exam, I sought assistance from Killexams.com. I chose them for their excellent evaluation of the exam concepts and regulations, consumer-friendly and resourceful material. Their practice tests removed all the problems associated with the exam topics and contributed generously to my education, enabling me to be successful in the exam.

User: Oliver***** I had a tough time studying for the usmle exam, as I did not have enough time for coaching. To find a way out, I took help from the practice tests and the professional Certification guide. The practice tests were top-notch, and they handled all the topics smoothly and pleasantly. I was able to get through most of them with little effort and responded to all the queries in only eighty-one minutes and received a 97 mark. I felt virtually glad and would like to thank Killexams.com for their valuable steering.

User: Maximiliano***** I am delighted to inform you that I have successfully passed my usmle certification exam with a solid average score, thanks to Killexams exam preparation material. Killexams is an excellent source for anyone preparing for an IT certification exam. It not only helps you pass but also ensures that you learn and become a successful professional. In fact, everyone in my IT company knows about Killexams and has used or heard of their materials.

User: Liev***** I am thrilled to announce that I passed my usmle exam last week and even passed another exam this month. Many people agree that practice tests are a great way to study, whether Its for the exam or for improving your knowledge. During my exams, I encountered many questions, and I was pleased to know that I knew all the answers.

User: Tassa***** The usmle exam training from killexams.com is outstanding. I bought it because I could not find any books or PDFs to prepare for the exam. This practice exam gives you the right questions, just like the ones you will see on the actual test. No irrelevant information or inappropriate questions. I highly recommend killexams.com to anyone planning to take the usmle exam.

---