FL C/P

Which of the following products would be formed if Compound X were reacted with LiAlH4 without first being converted to a cyclic ketal?

Compound X is discussed in the second paragraph. There, the passage states that lithium aluminum hydride will reduce the ester group to an alcohol and would also react with the ketone if left unprotected. Lithium aluminum hydride will reduce a ketone to an alcohol. Thus, choice (B), which shows both the ketone and ester reduced to an alcohol, is the correct answer.Choice (A). This choice shows the ketone group still intact, so it is wrong.Choice (C). The passage and the diagram of Compound Y both indicate that the ester group is reduced to a CH2OH alcohol group. This choice, which has it converted to a carboxyl group, is wrong.Choice (D). Lithium aluminum hydride is NOT a strong enough reagent to completely remove the oxygen atom and reduce the compound to the alkane form.

A patient's maximum blood pressure (systolic) is 113 mmHg and minimum (diastolic) is 60 mmHg. If his total peripheral resistance is 20.5 mmHg*min/L, what is the difference in his cardiac output between systolic and diastolic pressure? A. 2.59 L/min B. 25.92 L/min C. 108.6 L/min D. 1086.5 L/min.

Correct answer: A Blood pressure = cardiac output * total peripheral resistance. Therefore, the difference in cardiac output can be calculated as: (Psystolic - Pdiastolic) / resistance = 2.59 L/min. Therefore, (A) is the correct answer.

Sources A and B have a frequency of 1000 Hz. If the intensity of Source A is 1000 times greater than that of Source B, what is the difference in relative intensity levels of the two sources? A. 30 dB B. 40 dB C. 1,000 dB D. 10,000 dB.

Correct answer: A Equation 1 equates the relative intensity level, dB, to ten times the log of I over I0. I is the intensity of the sound, and I0 is the reference intensity. The question asks for dBA minus dBB. This equals: With appropriate manipulation of logs:The I0 terms cancel in the quotient, so IA over IB equals 1000, which is given in the question stem. The log of 1000 is 3 since 103 = 1000. Therefore, dβA- dβB= 10 × 3 = 30 decibels. Note that the frequency of the sound does not come into play in this question. Thus, the correct answer is (A).Choice (D). The logarithmic scale is used to reduce the scale in intensity calculations.

A chemist wishing to activate a carboxyl group could effectively do so by: A. reacting it with SOCl2. B. heating the solution. C. adding aqueous base. D. adding excess alcohol and HCl.

Correct answer: A Peptide bonds form by the nucleophilic attack of the amino group of an amino acid on the carboxyl group of another. The reaction of thionyl chloride with a carboxylic acid substitutes a chlorine atom for the carboxyl hydroxyl group, producing an acyl chloride. An acyl chloride is the most reactive of the various kinds of carboxylic acid derivatives, and certainly more so than the corresponding carboxylic acid. This is because the chloride ion is a much better leaving group than the hydroxide anion and is therefore more easily displaced by the nucleophilic attack of the amino group. Thus, choice (A) is the correct response.Choice (B). Heating the solution will have no effect on the structure of the carboxyl group. Increasing the temperature may or may not promote the reaction, depending on the thermodynamics of the reaction, but NOT by activating the carboxyl, which requires some kind of chemical modification.Choice (C). Adding aqueous base to Compound 1 would result in the formation of a carboxylate anion. This would be even less reactive towards an amino nucleophile.Choice (D). This would replace the hydroxyl group by an alkoxide group (RO). This alkoxide group would be a poor leaving group, and therefore the ester would not be reactive enough to form the peptide bond.

Which of the following will affect the frequency of sound waves traveling through air, as measured by a detector? I. Surface area of the detector II. Speed of the source of sound III. Distance from source to detector A. II only B. I and II only C. I and III only D. II and III only

Correct answer: A This question requires application of the Doppler effect: the idea that the speed of both the source and detector of a wave will influence the perceived frequency of that wave. If fine-tuning of that frequency is required, experimenters can change the detected frequency by moving the source at some constant velocity. The surface area of the detector and its distance from the source will only affect the intensity of the wave experienced by that detector. Therefore, (A) is the correct answer.

If the graph of radioactive decay shown below were transferred to logarithmic graph paper, what would the results be? A. A straight horizontal line B. A straight line with a nonzero slope C. A curved line characteristic of linear decay D. A curved line characteristic of exponential decay.

Correct answer: B A logarithmic graph drawn on linear scales translates into a straight line on logarithmic scales. In this case, the line would have a nonzero slope since there is a varying relationship between time and amount of material. Choice (B) is correct.

Suppose an individual with dry skin has come in contact with a high-voltage DC circuit. In an attempt to rescue the individual, another person, with wet skin, makes contact with the first, forming a parallel connection. Consequently, the amount of time required for 10 C of charge to exit the power source will: A. increase. B. decrease. C. remain the same. D. There is not enough information to determine a change in current.

Correct answer: B The first person acts as a resistor with resistance 104Ω. If a second person latches on to the first, he too would be a resistor - one that has been added in parallel. Moreover, the second person has wet skin, so will have a lower resistance. Adding resistors in parallel acts to decrease the overall resistance of a circuit. Therefore, if resistance decreases, total current passing through the circuit will increase. Since current is Δcharge/Δtime, an increase in current implies that less time will be required to pass the same amount of charge. The correct answer is (B).Choice (A), (C) and (D). The decreased resistance decreases the amount of time for charge to pass through a circuit.

The Kyte-Doolittle plot shows the hydrophobicity of the primary structure of a protein from beginning to end where higher values mean more hydrophobic. Given the Kyte-Doolittle plot below, what could regions above the horizontal line represent? A. Regions where transmembrane proteins protrude from the membrane B. Regions where transmembrane proteins transverse the membrane C. Regions where a free floating enzyme interacts with ions D. Regions exposed to the intracellular fluid on a free floating enzyme

Correct answer: B These regions have a high degree of hydrophobicity meaning they likely reside in other regions of high hydrophobicity. The inside of a cell membrane is very hydrophobic as it only contains the hydrophobic tails of phospholipids, so (B) is the correct answer.The rest of the answers constitute areas that should be hydrophilic rather than hydrophobic and thus can't be the correct answers.

Which of the following pH values is likely to favor the form of the alanine molecule shown in the passage? A. 3.0 B. 7.0 C. 9.0 D. It cannot be determined.

Correct answer: C Here, outside knowledge of the relationship between amino acids and pH is essential. As with any nonpolar amino acid, the acidic, neutral, and basic forms of alanine roughly correspond to acidic, neutral, and basic values of pH. At their isoelectric point, amino acids exist predominantly in the zwitterion form: positively charged ammonium groups and negatively charged carboxylate groups. As the pH falls, the carboxylate groups gain hydrogen ions and become neutral, while the ammonium groups retain their positive charges, so the molecules exist predominantly as cations. In contrast, as the pH rises above the isoelectric point, the ammonium groups lose their extra hydrogen to form neutral amino groups, whereas the carboxylate groups remain negatively charged, so that the molecules become anions. The form of alanine shown in the passage is an anion, and as the basic pH value is represented by choice (C), this is the correct answer.Choice (A). At an acidic pH, both the carboxylate and ammonia groups would carry a hydrogen.Choice (B). At neutral pH, alanine would likely exist as a zwitterion.Choice (D). The answer to this question can be determined with outside knowledge of the acid-base characteristics of amino acids.

When light is absorbed by chlorophyll in plants, it is always true that: A. the light energy ionizes the pigment. B. the light provides energy to form new bonds. C. electrons are excited to a higher energy level. D. electrons return to their ground state..

Correct answer: C It is always true that when an atom absorbs energy, its electrons move from their ground state to a higher excited state. Therefore, (C) is the correct answer.As for the incorrect answer choices, (A) can be correct, but isn't always true, since there is not always sufficient energy absorbed to actually remove an electron from an atom. (B) is incorrect because bond breaking requires energy, while bond forming releases energy. Energy would be released from the system if bonds formed; adding energy to the system would not accelerate bond formation but may, in fact, cause the existing bonds to break. Finally, (D) cannot be true, because electrons are in their most stable, lowest energy state when they are in their ground state. Therefore, absorbing energy could not put an electron into its ground state.

In magnetic resonance imaging (MRI), hydrogens in tissues are subjected to a magnetic field. In response, the protons emit radio frequency radiation, which is imaged as a bright area. Which of the following tissues is LEAST likely to show up as a bright region in an MRI scan? A. Kidney B. Eyes C. Femur D. Biceps.

Correct answer: C Since, as stated in the question stem, MRI relies on the imaging of protons, and the brightness is proportional to the number of protons, this question is really asking which tissue is likely to not have very many protons. That is, since most of the protons in biological tissues are found in water, the question is really "which of the following tissues has the lowest water content?" Soft tissues like internal organs (for example, the kidney), eyes, and muscles (like the biceps) will have a higher water content than bones like the femur, so choice (C) is the correct answer.

Over what frequency range is the ear most sensitive? A. 20 to 300 Hz B. 300 to 2,000 Hz C. 2,000 to 4,000 Hz D. Above 4,000 Hz

Correct answer: C The question gives four frequency ranges and asks which one represents the region of highest sensitivity. Therefore, the threshold of hearing curve in Figure 1 will have the answer. The lowest region of the bottom curve is between 2000 and 4000 Hz, so (C) is the correct answer.Choice (A). This range in frequency encompasses a wide range in thresholds of hearing, from 90 dB to about 20 dB. However, it does not include the lowest threshold of hearing, which is less than 0 dB.Choice (B). This range in frequency encompasses a wide range in thresholds of hearing, from about 20 dB to about a little less than 0 dB. However, the threshold of hearing goes even lower between 2,000 and 4,000 Hz.Choice (D). Above 4,000 Hz, the threshold of hearing increases from the minimum it has already achieved.

Which of the following elements would be the best to use for the plates of a defibrillator? A. Fluorine B. Phosphorus C. Lead D. Gold.

Correct answer: D Elements with metallic character are best for conducting electricity and lowering resistance to help shock better. Thus gold, appearing farthest towards the lower left, will be the best option for creating the plates of the defibrillator and (D) is the best choice.

According to the passage, when a current of 70 mA passes through the heart, it will cause: A. less severe damage than a current of 65 mA, since a larger current reduces the risk of death. B. less severe damage than a current of 1.5 A because the effect on the heart would not be as great. C. greater damage than a current of 1.5 A because it will cause the heart to stop, resulting in death. D. the heart to acquire an irregular rhythm, resulting in improper blood flow.

Correct answer: D Recall that a 70 mA current causes more damage to cardiac tissue than a current in excess of 1 A: eliminate (B). Notice, the passage says nothing about a 70 mA current versus a 65 mA current, so eliminate (A). Finally, eliminate (C), since the passage states that current in excess of 1 A can stop the heart, unlike current of 70 mA. The correct answer is (D), which is implied by the passage: a current of 70 mA will cause the heart to contract irregularly, resulting in improper blood flow.

During the dipeptide synthesis described in the passage, is the formation of leucine-leucine dipeptides a concern? A. Yes, because the carbonyl carbon of leucine is also susceptible to electrophilic attack B. Yes, because the phenylacetyl protector of alanine increases the energy of activation C. No, because the amine group of leucine is sterically hindered by its isobutyl side chain D. No, because the carboxyl group of alanine is activated before reacting with leucine

Correct answer: D The condensation reaction between two leucine residues is not a concern because the carboxyl group of alanine is selectively activated before reacting with leucine. (D) is therefore the correct answer.Choice (A). The carbonyl carbon is susceptible to nucleophilic attack.Choice (B). This may be true, but even so, the activation energy is overcome in the formation of alanine-leucine dipeptides, so this does not adequately address the question.Choice (C). While it is true that leucine has an isobutyl side chain, this does not adequately address the question, because nucleophilic attack by leucine is necessary for the formation of both alanine-leucine and leucine-leucine.

Which of the following compounds would best protect the ketone carbonyl group in Compound X?

In Reaction 1, a carbonyl group is reacted with an alcohol to form a ketal. All four choices are alcohols, so the correct answer revolves around " best protect." The passage states in the first paragraph that carbonyls are protected more effectively by conversion to cyclic ketals. As seen in Reaction 1, a non-cyclic ketal is formed when a carbonyl group reacts with the hydroxyl groups of two alcohol molecules. Therefore, it is reasonable to suppose that a cyclic ketal can be formed by the reaction between a carbonyl group and a diol. As the most sterically unhindered 1,2 diol, choice (D) is the best answer.Choice (A). Though this is a 1, 2 diol, it has four isopropyl substituents so it is more sterically hindered, and therefore less reactive.Choice (B). This a trans unsaturated 1,4-diol. The hydroxyl groups are not adjacent, and the mobility of the carbon skeleton is quite limited due to the presence of the trans double bond, therefore it would be impossible to form a cyclic ketal.Choice (C). This choice contains just one hydroxyl group, would form a non-cyclic ketal and can be rejected right away.

The average human's eardrum can be approximated to be half a sphere, 1 cm in diameter. If a sound wave is heard with I = 10-10W/m2, how much energy is transferred to the eardrum every second? A. 1.26 x 10-16 B. 1.57 X 10-14 J C. 1.57 X 10-10 J D. 1.26 X 10-9 J.

Intensity is measured in W/m2 as explained in Paragraph 2. If the eardrum is half a sphere, then its surface area can be measured as (1/2)4πr2. The diameter is 1 cm, so the radius is 0.5 cm, and the surface area is (1/2)(4π) (5 x 10-3 m)2 or approximately 1.5 x 10-4 m2. If intensity is power per area, then power = (intensity)(area) = (1 X 10-10 W)(1.5 x 10-4 m2), or 1.57 x 10-14 W. The question asks about energy, and watts are just joules per second. So in 1 second, 1.57 x 10-14 J are transferred. Thus, (B) is the correct answer.Choice (C) and (D) . These can be eliminated by considering the fact that the intensity is 10-10W/m2 which means for every square meter, 10-10 Joules are transferred every second. The eardrum is significantly smaller than a square meter and so the energy transferred to the eardrum will be correspondingly smaller.