Question #1
Which of the following illustrates the principle of induction in invertebrates?
Answer:
The correct answer is A
How does the ectoderm that will make up the future eye know to develop into an eye? The cells are “told” to develop into an eye because the nearby retina releases chemical signals that direct the development of the cells. This process is called induction. Choice A depicts a similar situation. The notocord releases chemical signals that tell the overlying ectoderm that it is time for it to become the neural tube. None of the other choices depict the process of induction.
Question #2
C6H12O6 + O2 CO2 + H2O
Answer:
The correct answer is D
Respiration begins in the cytoplasm but is completed in the mitochondria. The first step of respiration is glycolysis, which occurs in the cytoplasm. Pyruvate from this reaction is converted to acetyl CoA, which enters the Krebs cycle in the mitochondria. NADH and FADH2 from the Krebs cycle enter the electron transport chain in the cristae, the inner membrane of the mitochondria.
(A) Opposite. The process is begun in the cytoplasm.
(B) Distortion. Glycolysis is not associated with being in the area of the cell membrane.
(C) Distortion. No part of glycolysis occurs in the nucleus.
(E) Distortion. Proteins are created by ribosomes. Glycolysis has nothing to do with ribosomes.
Question #3
Cells that are involved in active transport, such as cells of the intestinal epithelium, utilize large quantities of ATP. In such cells there are
Answer:
The correct answer is C
A cell that is involved in active transport, such as the epithelial cells of the intestine, will require large amounts of ATP. If a cell utilizes large amounts of ATP, it must produce it in many mitochondria.
(A) Distortion. High levels of adenylate cyclase are found in cells that rely heavily on a second messenger system.
(B) Distortion. Many polyribosomes are found in molecules that synthesize high levels of protein.
(D) Distortion. High levels of DNA synthesis occur in cells that are rapidly dividing.
(E) Distortion. High amounts of lysozomes are found in phagocytic cells such as macrophages.
Question #4
In an emergency, an individual with type AB antigen on his red blood cells
Answer:
The correct answer is D
People with the blood type AB are known as the universal recipients. They have both A and B blood group antigens on the surface of their red blood cells. Therefore, their blood will have no antibodies to any of the blood antigens and they can receive blood from type A, B, and O donors.
(A) Distortion. Yes this is true but it is not the only type of blood that the individual can receive.
(B) Distortion. Yes this is true but it is not the only type of blood that the individual can receive.
(C) Distortion. Yes this is true but it is not the only type of blood that the individual can receive.
(E) Distortion. The AB individual is a universal recipient and can receive any type of blood.
Question #5
The least stable free radical is
Answer:
The correct answer is D
A trivalent carbon with an unpaired electron does not possess an octet and can hence be considered electron deficient. The free radical, then, just like carbocations, is stabilized by the presence of groups which can share electrons inductively (alkyl groups) or defuse the electron deficiency through resonance (conjugated systems). The order of free radical stability goes 3°>2°>1°>methyl. Allylic and benzylic radicals are similar to tertiary radicals in terms of stability because of resonance delocalization. The free radicals shown in choice A (benzylic), choice B (tertiary), and choice E (allylic) are all relatively stable, and therefore these three choices are incorrect. The radicals offered in choices C and D are both relatively unstable, but the methyl radical depicted in choice D is more unstable than the primary radical in choice C. Thus choice D, as the most unstable free radical, is the credited choice.
(A) Distortion. Benzylic free radicals are relatively stable due to resonance with the benzene ring.
(B) Distortion. Tertiary free radicals are relatively stable.
(C) Distortion. Primary free radicals are among the least stable free radicals. However, it is not the least stable of all the answer choices.
(E) Distortion. Allylic free radicals are relatively stable due to resonance with the double bond.
Question #6
Which one of the following reactions will be accompanied by an increase in entropy?
Answer:
The correct answer is A
Qualitative thermodynamics questions can be answered by applying the definitions of the various state functions. This question asks about entropy, which can be defined as the energetic equivalent of randomness, or disorder. The question thus translates into “Which of the following will have an increase in disorder as the reaction proceeds from left to right?” Recall that the general order of increasing entropy goes: solid < liquid < ions in solution < gas. Choice A is thus correct since it shows one mole of a highly ordered, crystalline solid, metallic sodium, and one mole of a somewhat well-ordered, hydrogen-bonding liquid, water, transforming into two moles of aqueous ions, Na+ and OH-, and one mole of gas, H2. The relatively well-ordered reactants are replaced by highly disordered products as the reaction proceeds, and a net increase in disorder, or entropy, is the result.
(B) Opposite. Choice B shows the deposition of gaseous iodine. This represents a large decrease in entropy as the random motion of the gas phase molecules is replaced by the highly ordered crystal structure of the solid phase. Therefore, this reaction is accompanied by a decrease in entropy.
(C) Opposite. Choice C shows a reaction between aqueous sulfuric acid and aqueous barium hydroxide; this reaction is simultaneously a precipitation, as the barium ions combine with the sulfate ions to form the ionic solid, barium sulfate, and a neutralization, as the aqueous acid reacts with the aqueous base to form water. Precipitation takes aqueous ions to ionic solid and thus represents a decrease in entropy. Neutralization takes aqueous ions to molecules of liquid and is accompanied by a net decrease in disorder; therefore C is accompanied by a decrease in entropy.
(D) Opposite. Choice D shows the formation of liquid water or, alternatively, the combustion of gaseous hydrogen. By either description, this reaction proceeds with a large decrease in entropy as the mole and a half of mixed gas phase molecules transform into one mole of pure liquid.
(E) Distortion. Choice A is the credited answer choice.
Question #7
The heat of combustion of gaseous ammonia, NH3(g), is 81 kcal/mole. How much heat is released in the reaction of 34 grams of ammonia with excess oxygen?
Answer:
The correct answer is E
This question combines some basic stoichiometry with thermodynamic values for a reaction. We are given the ΔHcomb per mole of ammonia and the mass of ammonia in grams. Applying dimensional analysis, we must convert the mass of ammonia into moles, and then multiply by kcal/mol to find the total number of kcal. Ammonia has a formula weight of 14 + 3(1) = 17 g/mol, therefore:
(34 g)/(17 g/mol) = 2 mol
(2 mol)(81 kcal/mol) = 162 kcal.
Note that it is not necessary to come up with a balanced equation for the combustion reaction to do this problem.
(A) Miscalculation. Random calculation error.
(B) Miscalculation. Random calculation error.
(C) Miscalculation. Random calculation error.
(D) Miscalculation. Failed to multiply ΔHcomb of NH3(g) by the number of moles of NH3(g)
Question #8
Which of the following electron configurations represents the element that would form the most acidic anhydride?
Answer:
The correct answer is E
We can translate the choices into the elements to which the given electron configurations correspond simply by counting up the total number of electrons in each and setting this number of electrons equal to the atomic number of the neutral element:
A. 11 electrons atomic number 11 sodium, Na
B. 12 electrons atomic number 12 magnesium, Mg
C. 13 electrons atomic number 13 aluminum, Al
D. 14 electrons atomic number 14 silicon, Si
E. 15 electrons atomic number 15 phosphorus, P
Recalling that metals form basic anhydrides while nonmetals form acid anhydrides, it follows that the least metallic element will have the most acidic anhydride. Metallic character decreases from left to right across the Periodic Table, and therefore phosphorus will be the least metallic of these elements and will have the most acidic anhydride. Listed below are the balanced hydration reactions for each of the most common anhydrides of the elements in the choices above, and the usual acid/base classification of the hydrate formed:
A. Na2O + H2O 2 NaOH, strong base
B. MgO + H2O Mg(OH)2, base/amphoteric
C. Al2O3 + 3 H2O 2 Al(OH)3, amphoteric
D. SiO2 + 2 H2O Si(OH)4, amphoteric
E. P2O5 + 3 H2O 2 H3PO4, weak acid
(A) Distortion. This anhydride is a very strong base.
(B) Distortion. This anhydride is a base.
(C) Distortion. This anhydride is acidic or basic depending on the situation.
(D) Distortion. This anhydride is acidic or basic depending on the situation.
Question #9
A roulette wheel consists of 38 slots. 36 of these are numbered 1 through 36 and colored red or black so that there are nine red even-numbered slots, nine black odd-numbered slots, etc. These slots occur with equal probability. The two slots marked 0 or 00 are each three times as likely to occur as any one of the other 36. What is the probability that a red even number, a red 23, or a 00 will occur on one roll of the wheel?
Answer:
The correct answer is D
The first step in solving the problem is to determine the likelihood of each slot’s occurrence. Say, then, that the likelihood of each of the 36 regular slots is x. Then the likelihood of each of the two special slots (0 and 00) is 3x. To solve for x, we note that the sum of the probabilities for all 38 slots is 1, so: and . This means that the probability for each special slot is (or , but we won’t reduce the fraction here because we’re going to add it to another in a moment). The probability that (a) a red even number, will occur is , or . The probability that (b) a red 23, will occur is . The probability that (c) 00, will occur is . So the probability that (a) or (b) or (c) will occur is the sum of these probabilities, .
All of the other answers are designed to catch you if you make various simple mistakes. For example, answer choice C is designed to catch you if you decide that the denominator for each probability is going to be 38. The discussion above shows that this is not the case: there are 38 slots, but because they are differently weighted, none of them has a chance of occurring.
Question #10
If 1/5 of those in a high school are freshmen, and 40% of the freshmen are women, and 3/5 of those in the other grades are women, how many men are enrolled in the high school, given that there are 600 students altogether?
Answer:
The correct answer is B
Forty percent is the same as , so the total portion of the student body that are women is exactly two-fifths of the freshmen plus three-fifths of the rest. Since one-fifth of the student body is freshmen, the fraction of all the students that are female is . Fraction multiplication gives us , i.e. . This is the fraction of the whole student body that is female. We will assume that everyone who isn’t female is male, so the portion of the student body that is male is , which is . To turn this fraction into the number of students, multiply it by the total number of students: the number of male students is . Dividing 600 by 25 yields 24, and multiplying 24 by 11 yields 264, our total.
The trickiest of the wrong answers is D, which is the total number of female students. Although all of the figures are given in terms of female students, the question is in fact about the male students. The other answers are reasonable results if you make a couple of simple mistakes.
Question #11
In the first 12 games of the season, the forward on a soccer team scored 20 goals. How many goals must he average in each of the remaining 6 games of the season in order to average 2.0 goals per game overall?
Answer:
The correct answer is D
With 12 games played and 6 remaining, the season is 18 games long. In order to average 2.0 goals per game, the forward must score goals in the season. He’s scored 20 so far, so he needs to score 16 more. With 6 games to go, that’s an average of goals per game. Simplifying this expression we get or goals per game.
Question #12
Susan has $10 with which to buy apples. Apples are priced at 6 for $0.89, or $0.20 each. How much change will Susan receive if she buys as many as she can?
Answer:
The correct answer is E
To get the most out of her money, Susan wants to buy as many apples as she can at the 6 for 89¢ rate, since at less than 15¢ per apple it is a better deal than buying by-the-apple. To see how many sets of 6 she can buy, divide $10 by $0.89, or more to the point, divide 1,000¢ by 89¢ with remainder. The correct answer is 11 with remainder 21¢, that is, , and . So after buying 11 sets of six apples for 89¢ per set, she has 21¢ left over, with which she can buy one more apple at the 20¢-per-apple rate, with 1¢ left in her pocket, which will not afford her any more apples. Her change is therefore $0.01.
Question #13
When a spring is compressed to its minimum length and not permitted to expand
Answer:
The correct answer is A
Compressing a spring requires work, which is then transformed into potential energy. If the spring is not permitted to expand then it is being held motionless, which means it has zero kinetic energy. Zero kinetic energy is also minimum kinetic energy. Because total energy (kinetic plus potential) is conserved, we can say that the point of minimum kinetic energy is also the point of maximum potential energy. Therefore, the correct choice is A.
(B) Opposite. This answer choice says that the potential energy is a minimum. We know this to be false because the spring has been compressed.
(C) Distortion. This answer choice says that the sum of the kinetic and potential energies is zero, which means that the total energy is zero. Again this is false, because the spring has a non-zero potential energy due to its compression.
(D) Distortion. This can be ruled out because of conservation of energy. You can't have both kinetic and potential energies either minimum or maximum at the same time. If kinetic energy is a maximum then potential energy is a minimum and vice versa.
(E) Distortion. This can be ruled out because of conservation of energy. You can't have both kinetic and potential energies either minimum or maximum at the same time. If kinetic energy is a maximum then potential energy is a minimum and vice versa.
Question #14
A 5-N force and a 10-N force act on the same point of an object. The 5-N force acts due west while the 10-N force acts 60° north of east. The resultant force is
Answer:
The correct answer is B
In order to find the resultant of two forces, we have to add the two forces vectorially. The simplest way to do this is to break the forces down into components. We'll use the directional components north, south, east, and west instead of x and y. We're given a 5 N force west and a 10 N force at an orientation of 60° north of east. We can resolve the 10 N force into two forces, one north and the other east. The north component is given by 10sin60 = 10(0.866) = 8.66. The east component is given by 10cos60 = 5. Now we can sum the forces by components. The sum of the east-west components is zero since they each have magnitude 5 and east and west are opposite directions. The only north-south force is from the 10 N force and that's 8.66 N north. Thus, the resultant is just 8.66 N north.
(A) Miscalculation.
(C) Miscalculation.
(D) Miscalculation.
(E) Miscalculation.
Question #15
An object is placed near a biconcave lens as shown below.
The image will form
Answer:
The correct answer is C
The system given in the problem is that of a diverging lens and we know that diverging lenses always produce virtual images. Since virtual images are located on the same side of the lens as the object we can immediately rule out answer choices A and B, both of which have the image on the other side of the lens from the object. To decide between the remaining answer choices we'll need to use the equation 1/o + 1/i = -1/F (Note: F < 0 since we are using a biconcave or diverging lens) .
Solving for 1/i, we have 1/i = -1/F - 1/o = -(1/F + 1/o) . Solving for the common denominator we get, 1/i = -(o/f•o + f/f•o) . If we re-arrange this, it becomes 1/i = -((o + f) /f•o) . We can then invert this to give us i = -(f•o/(o + f) ) .
Notice (f•o/(o + f) ) is less than one. (You can figure this out by substituting simple numbers in for the variables such as F = 1, o = 1.5) . We now have all the components to answer this question. Since i is a negative, we know that the image is on the side of the lens light is coming from, namely the left side. Next, since i turns out to be a fraction, we know it's smaller than F. The only answer remaining that corresponds to both these requirements is answer choice C.
(A) Distortion. Virtual images are located on the same side of the lens as the object so we immediately rule out this answer choice.
(B) Distortion. Virtual images are located on the same side of the lens as the object so we immediately rule out this answer choice.
(D) Miscalculation. Since we know that i turns out to be a fraction, we know it's smaller than F.
(E) Miscalculation. Since we know that i turns out to be a fraction, we know it's smaller than F.
Question #16
Which of the following statements concerning the photoelectric effect is FALSE?
Answer:
The correct answer is A
Let's briefly review the physics of the photoelectric effect. The idea is that electrons will be ejected from a metallic surface when light of sufficient frequency is shone on the surface. From Planck's famous equation E = hf we know that the frequency of light determines its energy. The electrons on the surface require a minimum energy called the work function in order to be liberated from the surface. Thus, the frequency of the light must be large enough so that the light's energy is at least as great as the work function or no electrons will be emitted. By conservation of energy, the kinetic energy of the ejected electrons equals the energy of the light minus the work function.
Answer choice A says that the kinetic energy of the electrons equals the energy of the light waves. From the above discussion we know that the kinetic energy of the electrons equals the energy of the light waves minus the work function. Thus, the kinetic energy of the electrons is always less than the light energy since the work function is always nonzero. Choice A is thus a false statement and is hence the correct choice.
(B) Distortion. This is a true statement.
(C) Distortion. This is a true statement.
(D) Distortion. This is a true statement.
(E) Distortion. This is a true statement.
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