Medical Tests MCAT-TEST Online Practice Questions and Exam Preparation

Medical Tests MCAT-TEST Online Questions & Answers

• Question 691:

  Which of following must be present in human's diet to prevent thyroxin deficiency?

  A. Iron
  B. Calcium
  C. Iodine
  D. Phosphorus
  C. Iodine

• Question 692:

  Before birth, the rodent brain is sexually undifferentiated. It is only in the first few days following birth, during a period referred to as the critical period, that the rodent brain differentiates along male or female lines. The hormone testosterone plays a critical role in this development. Specifically, sexual differentiation is determined by the presence of estradiol, an estrogen derivative of testosterone, in certain areas of the brain. Testosterone is converted to estradiol in critical brain cells that contain the enzyme aromatase. To study the effects of testosterone on the neonatal rodent brain, the following experiments were conducted: The above research, combined with additional studies, concluded that testosterone has two "organizational" effects on the male rodent brain: Defeminization Moderate levels of testosterone-derived estradiol during the critical period are sufficient for defeminization of the brain. Defeminization of the rodent brain results in loss of estrogen positive feedback on LH and FSH secretion and the ensuing loss of cyclicity, as well as loss of female sex behavior. Masculinization High levels of estradiol due to high levels of testosterone during the critical period results in masculinization of the brain. Masculinization leads to the induction of male sex behavior including antagonism towards other males and the mounting of females.

  

  Steroid hormones bind to receptors in the nucleus and directly regulate:

  A. the transcription of mRNA.
  B. the translation of protein.
  C. the production of cAMP.
  D. the replication of DNA.
  A. the transcription of mRNA.

  ---

  Explanation/Reference:

  Steroid hormones diffuse into the nucleus and bind to receptors. The steroid/receptor complex binds to DNA and regulates mRNA transcription. In contrast, peptide hormones bind to receptors on the surface of the cell and exert their effects through secondary messenger systems. Choice B is incorrect because protein translation by ribosomes is not regulated by steroid hormones. Choice C is incorrect because cAMP production is generally associated with protein hormones binding to cell membrane receptors. It is part of a single transduction mechanism, acting as a second messenger. Choice D is incorrect because DNA replication is not regulated by steroid hormones.

• Question 693:

  

  The two molecules above can best be described as:

  A. Structural isomers
  B. Diastereomers
  C. Enantiomers
  D. Conformational isomers
  D. Conformational isomers

  ---

  Explanation/Reference:

  The molecule on the left is trans 1, 2-diiodocyclohexane. The molecule on the right is cis 1, 2-diiodocyclohexane. Cis and trans molecules are a type of diastereomer called geometric isomers.

  A: Structural isomers have fundamentally different connectivity between the atoms. So for example if the iodine were on carbons 1 and 3 and then in the other molecule on carbons 1 and 4, that would be structural isomers.

  C: Enantiomers are mirror image reflections of each other, which these molecules are not.

  D: Conformations isomers allow you to freely interconvert between the molecules without breaking any bonds. The classic example is chair and boat cyclohexane. Here, to flip the one iodine from R to S configuration would require breaking a bond in the ring, rotating, and then re-attaching. That requirement for bond-breaking means these are not conformers.

• Question 694:

  Bebop lives! cries the newest generation of jazz players. During the 1980s, musicians like Wynton Marsalis revived public interest in bebop, the speedy, angular music that first bubbled up out of Harlem in the early 1940s, changing the face of jazz. That Marsalis and others thought of themselves as celebrating and preserving a noble tradition is, in one sense, inevitable. After the excesses of experimental or "free" jazz in the 1960s and the electronic jazz-rock "fusion" of the 70s, it is hardly surprising that people should hearken back to a time when jazz was "purer," perhaps even at the apex of its development. But the recent enthusiasm for bebop is also ironic in light of the music's initial public reception.

  In its infancy, during the first two decades of the 20th century, jazz was played by small groups of musicians improvising variations on blues tunes and popular songs. Most of the musicians were unable to read music, and their improvisations were fairly rudimentary. Nevertheless, jazz attained international recognition in the 1920s. Two of the people most responsible for its rise in popularity were Louis Armstrong, the first great jazz soloist, and Fletcher Henderson, leader of the first great jazz band. Armstrong, with his buoyant personality and virtuosic technical skills, greatly expanded the creative range and importance of the soloist in jazz. Henderson, a pianist with extensive training in music theory, foresaw the orchestral possibilities of jazz played by a larger band. He wrote out arrangements of songs for his band members that preserved the spirit of jazz, while at the same time giving soloists a more structured musical background upon which to shape their solo improvisations. In the 1930s, jazz moved further into the mainstream with the advent of the Swing Era. Big bands in the Henderson mold, led by musicians like Benny Goodman, Count Basie and Duke Ellington, achieved unprecedented popularity with jazz-oriented "swing" music that was eminently danceable.

  Against this musical backdrop, bebop arrived on the scene. Like other modernist movements in art and literature, bebop music represented a departure from tradition in both form and content, and was met with initial hostility. Bebop tempos were unusually fast, with the soloist often playing at double time to the backing musicians. The rhythms were tricky and complex, the melodies intricate and frequently dissonant, involving chord changes and notes not previously heard in jazz. Before bebop, jazz players had improvised on popular songs such as those produced by Tin-Pan Alley, but bebop tunes were often originals with which jazz audiences were unfamiliar.

  Played mainly by small combos rather than big bands, bebop was not danceable; it demanded intellectual concentration. Soon, jazz began to lose its hold on the popular audience, which found the new music disconcerting. Compounding public alienation was the fact that bebop seemed to have arrived on the scene in a completely mature state of development, without that early phase of experimentation that typifies so many movements in the course of Western music. This was as much the result of an accident of history as anything else. The early development of bebop occurred during a three-year ban on recording in this country made necessary by the petrol and vinyl shortages of World War II. By the time the ban was lifted, and the first bebop records were made, the new music seemed to have sprung fully-formed like Athena from the forehead of Zeus. And though a small core of enthusiasts would continue to worship bebop pioneers like Charlie Parker and Dizzy Gillespie, many bebop musicians were never able to gain acceptance with any audience and went on to lead lives of obscurity and deprivation.

  According to the passage, which of the following is true about the jazz of the 1920s?

  A. It resembled the jazz played during the first two decades of the century.
  B. It placed greater demands on the improvisatory skills of its soloists.
  C. Its fast tempos foreshadowed those of bebop in the 1940s.
  D. It was primarily dance music.
  B. It placed greater demands on the improvisatory skills of its soloists.

  ---

  Explanation/Reference:

  This seeks a specific, true statement about jazz of the 1920s. The 1920s are covered in paragraph 2 (lines 14?1). The correct answer, choice B, is taken from the fifth sentence of that paragraph, where the author says that Louis Armstrong, one of the two musicians "most responsible for" jazz's rise in popularity, "greatly expanded the creative range and importance of the soloist in jazz." In other words, Armstrong raised the stakes for jazz soloists. They had to improve, to rise above the "fairly rudimentary" improvisations that had been played during the first two decades of the century. So choice B is the correct answer. Regarding choice A, it's clear from paragraph 2 that, in the 1920s, the innovations of people like Armstrong and Henderson really changed jazz, making it more sophisticated and expansive. Choice A says that 1920s' jazz resembled earlier jazz, and this goes against the grain of the author's argument; 1920s jazz was clearly different. Choice (C) is wrong because the author never specifies what kind of tempos were typical of 1920s jazz. The passage never indicates if they were fast or slow, so there's no support for choice (C). And choice D describes jazz music of the Swing Era in the 1930s, not the jazz of the 1920s.

• Question 695:

  The nuclei of certain unstable isotopes will spontaneously decay, producing a more stable nucleus and releasing a particle or quantity of energy. Alpha decay releases a helium nucleus, beta decay emits an electron, while gamma decay is the emission of a high energy photon. Each type of radioactive decay is characterized, in part, by the half-life of the radioactive material--the time required for half of the nuclei in a sample to undergo decay. Examples of such decays are shown in Figure 1.

  

  Figure 1

  A Geiger counter can be used to detect the decay of radioactive materials. A simple Geiger counter consists of a hollow metal cylinder with a wire along its axis. The cylinder is filled with low pressure argon gas and a high voltage difference is

  applied between the wire and the cylinder. When alpha, beta, or gamma radiation passes through the cylinder, it interacts with the gas particles and leads to the formation of ions which cause a discharge between the wire and the cylinder.

  The consequent current may be used to drive a speaker, producing the characteristic clicking sound of the Geiger counter each time a pulse of current occurs. The Geiger counter circuitry is shown in Figure 2.

  

  Figure 2

  The half-life of 32P is 14.28 days. What fraction of an initially pure sample will have decayed after 28.56 days?

  A. 1/4
  B. 1/2
  C. 3/4
  D. 1
  C. 3/4

  ---

  Explanation/Reference:

  

  After two halflives (28.56 days), 3/4 of the sample will have decayed. in the first half life, and

  

  in the second. Only of the sample will remain undecayed. In other words. remains. Therefore is gone.

  

• Question 696:

  The resistance of a resistor is defined as the ratio of the voltage drop across it to the current passing through it. The resistance of a resistor can be measured using the circuit illustrated in Figure 1.

  

  Figure 1

  In the above circuit, a variable voltage source with negligible internal resistance is connected to a resistor. The voltage across the resistor is measured by a voltmeter and the current through the resistor is measured by an ammeter.

  

  Additional resistors may be added to the circuit. The total resistance can be calculated as follows: If and are two resistances of two resistors, then the total resistance is given by = + when the resistors are connected in

  

  series, and by 1/ = 1/ + 1/ when the resistors are connected in parallel.

  

  Circuits similar to the one above are used in the common household appliance known as the toaster. The rate by which energy in the form of heat is dissipated by the resistor equals , where I is the current that passes through the resistor and R is the resistance of the resistor. Energy is dissipated in a resistor because moving electrons collide with atoms in the resistor, causing the atoms to vibrate.

  The variable voltage source in the circuit in Figure 1 is replaced by a battery connected in series with the resistor and ammeter. The battery has a small internal resistance. How will the circuit be affected?

  A. The current measured by the ammeter at a specific voltage will be greater in the circuit with the battery than in the old circuit.
  B. The current measured by the ammeter at a specific voltage will be smaller in the circuit with the battery than in the old circuit.
  C. The resistance of the resistor at a specific voltage will be greater in the circuit with the battery than in the old circuit.
  D. The resistance of the resistor at a specific voltage will be smaller in the circuit with the battery than in the old circuit.
  B. The current measured by the ammeter at a specific voltage will be smaller in the circuit with the battery than in the old circuit.

  ---

  Explanation/Reference:

  The question stem states that the voltage source, which according to the passage has negligible internal resistance, is replaced by a battery with a small, but significant internal resistance. A battery with significant internal resistance can be modeled as a voltage source connected in series with a resistor. Hence, the new circuit is the same as the one in Figure 1, except for the addition of a second resistor (call its resistance Rint) in series with the original resistor (call its resistance R0). To determine how this additional resistance affects the circuit, use Ohm's law V = IR, where V is the voltage, I is the current, and R is the resistance. Solving for I gives I = V/R. Thus, in the original circuit, the voltage source supplies a

  

  

  voltage V and the ammeter measures a current . In the new circuit, the total resistance is given by the addition rule for resistors in series, . Therefore, the current measured by the

  

  ammeter in the new circuit will be , which is clearly smaller than Ioriginal since its denominator is larger. So choice B is correct, and choice A is wrong. Choices C and D are wrong because the resistance of a resistor is a property that does not change with the addition of other resistors to the circuit. The resistance of a resistor remains constant when the voltage across it or the current through it changes.

• Question 697:

  Band theory explains the conductivity of certain solids by stating that the atomic orbitals of the individual atoms in the solid merge to produce a series of atomic orbitals comprising the entire solid. The closely-spaced energy levels of the orbitals form bands. The band corresponding to the outermost occupied subshell of the original atoms is called the valence band. If partially full, as in metals, it serves as a conduction band through which electrons can move freely. If the valence band is full, then electrons must be raised to a higher band for conduction to occur. The greater the band gap between the separate valence and conduction bands, the poorer the material's conductivity. Figure 1 shows the valence and conduction bands of a semiconductor, which is intermediate in conductivity between conductors and insulators.

  

  Figure 1

  When silicon, a semiconductor with tetrahedral covalent bonds, is heated, a few electrons escape into the conduction band. Doping the silicon with a few phosphorus atoms provides unbonded electrons that escape more easily, increasing conductivity. Doping with boron produces holes in the bonding structure, which may be filled by movement of nearby electrons within the lattice. When a semiconductor in an electric circuit has excess electrons on one side and holes on the other, electron flow occurs more easily from the side with excess electrons to the side with holes than in the reverse direction.

  

  Figure 2

  The energy of the band gap for pure silicon is about 1.1 electron volts. If a 1.5-volt electrical potential is connected across a sample of silicon:

  A. the electrons would jump to the conduction band and the silicon would conduct.
  B. the holes in the silicon lattice would move.
  C. the energy of the band gap would be lowered.
  D. the silicon would not conduct.
  A. the electrons would jump to the conduction band and the silicon would conduct.

  ---

  Explanation/Reference:

  Since the band gap is 1.1 eV, an applied voltage in excess of 1.1 V would give the electrons enough energy to cross the band gap and populate the conduction band. (You should know that an electron volt is equal to the energy required to move an electron through a potential difference of one volt.) Choice A is therefore the correct response. Choice B is wrong because pure, undoped silicon doesn't have any holes. The passage states that holes are produced when the silicon is doped with boron choice C is wrong because there is no information in the passage that suggests that the band gap is affected by an external potential. Choice D is wrong because the electrons do have enough energy to overcome the band gap and would move into the conduction band.

• Question 698:

  Historically, two different methods have been used to estimate the fluid pressure in capillary beds.

  Method 1 A glass pipette is inserted into the capillary. The level of blood rising in the pipette is measured and used to calculate the pressure. Alternatively, an inert fluid of density can be placed in the pipette and its height h can be measured. The pressure in the capillary is given by gh, where g is the acceleration due to gravity.

  

  Figure 1

  Method 2 The pressure can be measured indirectly in the following way. A section of gut tissue is removed from a specimen and placed on a beam balance. Blood is circulated through the tissue by a pump. The arterial pressure is then decreased. This leads to a decrease in the capillary hydrostatic pressure in the gut capillaries. The constant osmotic pressure of plasma proteins in the capillary causes absorption of fluid from the gut section which will decrease its weight. To prevent a change in the weight of the gut section, the venous pressure is increased. This tends to increase the capillary pressure, reducing the flow of fluid from the gut tissue into the capillaries. The capillary pressure is thus held constant (and the balance kept level) as the arterial pressure is decreased and the venous pressure increased. The arterial and venous pressures meet at the capillary pressure being measured.

  ( = MRT, where is the osmotic pressure, M the molarity of the solutes, R the universal gas constant, and T the temperature in Kelvin.)

  

  Figure 2

  Assume that a mass (m) of 0.2 kg is placed 25 cm to the right of the fulcrum. A section of gut, initially weighing 0.1 kg is placed 50 cm to the left of the fulcrum. During the experiment, the mass m is seen to descend. In order to maintain the balance level the following action should be taken:

  A. the mass m should be moved away from the fulcrum.
  B. the arterial pressure should be decreased.
  C. the venous pressure should be increased.
  D. an inert fluid with higher density than that of blood should be used in the pipette.
  C. the venous pressure should be increased.

  ---

  Explanation/Reference:

  According to the passage, when fluid flows out of the gut (because the hydrostatic pressure in the capillary bed is lower than the osmotic pressure), the gut will become lighter and will ascend. Thus, the mass m will descend. The question stem indicates that the section of gut has become lighter. Fluid must be returned to the gut section to make it heavier and maintain the balance level. Increasing the venous pressure will increase the hydrostatic pressure in the capillary bed, leading to flow of fluid from the capillary into the gut section. Note that no quantitative analysis (calculation) is necessary to answer this question. Choice A is incorrect because moving the mass m away from the fulcrum will increase the distance between the force applied and the fulcrum, thus increasing the torque produced by the mass m causing it to descend more. The equation for torque is shown below: torque = Frsin where F is the magnitude of the force appliedr is the distance between the force and the fulcrum is the angle between the force and the vector r Choice B is incorrect because decreasing the arterial pressure will decrease the hydrostatic pressure in the capillary bed causing more fluid to be lost from the gut. The gut will become lighter and will ascend, causing the mass m to descend more. Choice D is a distractor choice: the pipette (from Method 1) does not come into play in this set-up, which deals with Method 2.

• Question 699:

  What is the identity of the reducing agent and oxidizing agent, respectively, in the reaction shown below?

  2Cr(OH)3 + 3H2O2 + 4NaOH 2Na2CrO4 + 8H2O

  A. Cr(OH)+ and NaOH
  B. Cr(OH)3 and H2O2
  C. H2O2 and Cr(OH)3
  D. H2O2 and NaOH
  B. Cr(OH)3 and H2O2

  ---

  Explanation/Reference:

  To answer this question, you must identify the atom that is reduced and the atom that is oxidized. In the reaction above, chromium is oxidized from +3 to +6, which makes chromium hydroxide the reducing agent. Alternatively, oxygen in hydrogen peroxide is reduced from -1 to -2, which makes it the oxidizing agent. Thus, the correct answer is B.

• Question 700:

  Ink jet printers produce high resolution output, at a lower cost than laser printers, by generating charged ink droplets which are then deflected onto a sheet of paper by an electric field. Each droplet deflected by the field strikes the paper and forms a tiny dot of ink. While a typical printed letter requires about 100 drops, an ink jet printer is able to produce drops at a rate of 100,000 per second.

  

  The essential elements of the ink jet printer head are shown in Figure 1. The drop generator produces the ink droplets, each with a mass of approximately 1.2 kg and a diameter of approximately 30 m. The drops then enter a

  

  highly precise charging unit which controls the charge q on each droplet to within 2%, with typical charges for drops generated by various ink jet printers ranging from ?.0 C to ?.0 C. The charged droplets are

  

  subsequently passed through the deflecting plates between which a variable electric field is generated. The electronically controlled electric field between the plates is typically varied over a range from 1.0 N/C to 5.4 N/C,

  

  and is used to aim the ink droplet at the paper. .

  

  In which direction would the deflector in Figure 1 deflect the ink drop?

  A. Upwards in the plane of the page
  B. Downwards in the plane of the page
  C. Into the plane of the page
  D. Out of the plane of the page
  E. (0,1) and (2,1)
  F. (0,1) and (0,2)
  G. (1,1) and (3,1)
  H. (2,1) and (1,2)
  A. Upwards in the plane of the page

  ---

  Explanation/Reference:

  An electric field indicates the direction in which a positive charge would be accelerated. The ink droplet is negatively charged and will thus be accelerated in the opposite direction to that of the electric field. Since the electric field in Figure 1 points down, the charged ink droplet will be defelected upwards in the plane of the page. Choice B is incorrect because a positive charge would be deflected downwards in the plane of the page. Choices C and D are incorrect applications of the right hand rule. The right hand rule is used to determine the force on a charge moving through a magnetic field. This problem concerns movement of a charge through an electric field.