FL 1 Review Flashcards
Suppose that multiple newly created amino acids interact to build a small protein molecule. The primary structure of that protein is formed when:
The primary structure of a polypeptide refers to the order in which amino acids are connected to one another. That connection consists of peptide bonds, which occur when the lone pair of electrons on one amino-terminus nucleophilically attack the carbonyl carbon of another amino acid’s carboxy-terminus; choice (B) is the correct answer.
In the first step of the reaction proposed in the passage, atomic oxygen is formed from carbon dioxide. Atomic oxygen is extremely reactive, and not found in any significant quantity on Earth’s surface, because:
Atomic oxygen has 8 electrons, 2 inner and 6 valence with an electron configuration of 1s22s22p4. Of the four electrons in the p orbital, 2 are paired in the -1 orbital, while 1 is unpaired in the 0 orbital and another is unpaired in the +1 orbital. These unpaired valence electrons, also known as free radicals, are extremely chemically reactive and explain the lack of [O] on earth’s surface. Choice (A) is correct.
Which of the following will increase the yield of carboxylic acid formed from a nitrile group-containing compound, using a mechanism similar to the Strecker synthesis
Step 6: NH2-CH2-CN + 2H2O → NH3 + NH2-CH2-COOH
In Step 6 we see a nitrile group (-CN) reacting with 2H2O in order to form a carboxylic acid. Even without knowing the mechanism beforehand, we know that H2O has lone pair electrons, and based on the passage, our products have new C-O bonds and no C-N bonds anymore. In fact, H2O acts as a nucleophile and the carbon in -CN acts as an electrophile (as it has its electron density drawn away by the N atom). If we attach another electron-withdrawing group to the carbon, as choice (B) suggests, that carbon will be even more electron deficient, and act as a better electrophile, increasing the yield of carboxylic acid.
Choice (A) is wrong because a Grignard Reagent is a stronger nucleophile than H2O, but remember the goal is to make carboxylic acid, not new C-C bonds, which would be the product of a reaction with Grignard Reagents.
The index of refraction of the vitreous humor is greater than the index of refraction of the aqueous humor, which is greater than the index of refraction of air. How does the speed of light in each of these media compare?
In order to answer this question, recall that the material with the lowest index of refraction will enable light to travel through it at the fastest speed. Therefore, the correct answer choice must list the media in order of increasing indices of refraction, which will correspond to decreasing order of speed of light in the media. Choice (A) is correct.
Which of the following accurately describes the difference between α-D-glucose and β-D-glucose?
α-D-glucose and β-D-glucose are a pair of two anomers, and by definition, anomers differ only in the absolute configuration at their anomeric carbon. The anomeric carbon is the carbon at the center of a hemiacetal group whose configuration will determine whether the -OH group is axial or equatorial in cyclic glucose. This designation matches choice (B).Choice (A) is false; the D in both names refers to the fact that each would rotate polarized light in the same direction (to the right). Choice (C) gets things exactly backwards, as α-glucose has an axial configuration at the anomeric carbon and β-glucose has an equatorial configuration. And choice (D) is wrong because pyranose refers to a 6-membered ring (5 carbons and 1 oxygen) while furanose refers to a 5-membered ring (4 carbons and 1 oxygen). Glucose can take either form (the pyranose form dominates in solution) but the difference is not denoted by α/β.
0.44 moles of AgNO3 and 0.20 moles of MgCl2 are mixed together. What is the mass percent composition of Ag in the product?
The starting number of moles is irrelevant information when calculating the mass percent composition. The products of the reaction are AgCl and Mg(NO3)2. Since the question stem asks about the mass percent composition of Ag, the product of interest is AgCl. It does not matter how many moles of AgCl are formed; there will always be 107.9 g of Ag per mole of product and 35.5 g of Cl per mole of product, since those are the molar masses of Ag and Cl, respectively. The total mass of AgCl is 143.3 g/mol, so the mass percent composition of Ag is 107.9/143.3 or 75.2%. This matches choice (C).
When blood is donated, it is important that no active antibodies are transferred along with the red blood cells. In order to accomplish this, a reducing agent, which breaks the disulfide bonds that maintain the quaternary structure of antibodies, is added to donated blood. This is effective because:
From the question stem, we know that disulfide bonds maintain the quaternary structure of antibodies and that adding a reducing agent will break these disulfide bonds. We need to be aware that quaternary structure exists when multiple polypeptides interact to form a single protein with multiple subunits. We should also know that disulfide bonds are covalent bonds between two thiol groups (R-SH). In the case of proteins, disulfide bonds occur between two cysteine amino acids to form a single unit known as cystine. Lastly, reducing means gaining electrons. Therefore, choice (B) is the right answer.
When an acyl halide reacts with a primary alcohol, which of the following will form?
In this reaction, the carbon of the acyl halide acts as the electrophile; it is having electron density pulled away by both the carbonyl oxygen and the halogen. Alcohol, with its lone pairs of electrons on oxygen, can then attack, pushing up electrons onto the carbonyl oxygen. Those electrons reform a double bond kicking off the halide, which is a good leaving group. Lastly, the oxygen from the alcohol group loses its proton (and positive charge) and we are left with an ester, choice (A).
A conservationist would like to test the acidity of a sample of rainwater by titrating it with 0.05M NaOH. What additional information is needed to calculate the initial number of moles of H2CO3 in the sample?
I. Volume of NaOH used to reach the end point
II. Ka of H2CO3
III. Initial volume of rainwater in the sample
The goal is to find the number of moles of H2CO3, not the concentration. Therefore, it is only necessary to know how many moles of NaOH were required to reach equivalence, which is Roman numeral I. Regarding Roman numeral III, it does not matter what volume of rainwater we start with if we are only concerned with the number of moles, since each mole of OH- added into solution will react with one mole of H+ from the weak acid – the only thing volume would tell us is the concentration of the starting material once we’ve found the initial moles.
Finally, the Ka of the weak acid does not matter, as we are titrating with a strong base. Once OH- is introduced into solution, it will remove the H+ no matter how strong or weak the acid, so Roman numeral II is not relevant. Since statement I is true but statements II and III are not, choice (A) is correct.
The method of carbon dating used to determine age depends upon the assumption that:
Carbon dating uses the fact that 14C is unstable and decays over time, while 12C is stable and does not decay. Let’s say the ratio of 14C to 12C in an organism is known at the moment of death. By measuring the ratio of 14C to 12C in a fossil, the age of the fossil can be determined because the 14C in the dead organism decays over time in a quantifiable way.
This question is best answered through the process of elimination. Choice (A) can be eliminated because all that matters is that the half-life is constant once it is ingested. Choice (B) can be eliminated because all that matters is the ratio of 14C to 12C, which will be constant as long as 12C and 14C are both incorporated into the body in the same way. Choice (C) can be eliminated because this would make it very difficult or impossible to perform carbon dating, since the half-lives of all the atoms would be different, depending on what molecule they were in. This leaves choice (D), which is correct.
Although it is not explicitly stated in the passage, choice (D) is an assumption of carbon dating. It correctly states that the half-life of 14C does not depend upon conditions external to the 14C nucleus. This means that the half-life does not depend on the weather, the amount of 14C present, etc. Because the half-life is a constant with respect to conditions external to the nucleus, it can be used to measure elapsed time accurately. Since measuring time is the goal of carbon dating, choice (D) is a necessary assumption.
2A + B ⇌ 2C + D
Which of the following CANNOT be the mechanism for the rate-determining step of this reaction?
Paragraph 3 states that the slower step of the reaction mechanism follows second-order kinetics. The slowest step of a reaction mechanism is usually the rate-determining step, and a step that involves second-order kinetics must only involve two reactant molecules. Although it is often said that the kinetics of a reaction cannot be derived from its stoichiometry, this is true only of the overall reaction. When a complex reaction has been broken down into a series of elementary reactions, we can then derive the rate law from the slowest of the elementary reactions. Again, In this case, we know that the rate-determining step must consist of an interaction between two molecules. This could be two molecules of A, two of B, or one each of A and B. Choice (D) involves three reactant molecules, and is therefore third-order, and incorrect.
The reaction was found to take place in two steps, the first, which follows second-order kinetics, is slow, and the second is fast. The forward reaction occurs spontaneously even without a catalyst. The reactants and products behave like ideal gases. Changes in equilibrium do not affect the phases of the species involved.
From the information provided in the passage, which of the following can be shown to be true?
Keq > 1
For this question, evaluate each statement one by one, keeping the major points of the passage in mind as a prediction. Paragraph 3 states that the reaction proceeds spontaneously to the right; therefore the equilibrium constant must be greater than one. Thus, choice (A) is the correct answer.
Ozone has a high oxidation potential, and will readily decompose urea: (NH2)2CO + O3→ N2 + CO2 + 2 H2O. In this reaction, the oxidation number of carbon changes from:
It is simpler to figure out the oxidation number of carbon in CO2, so begin there. Oxygen has an oxidation number of -2, and there are two oxygens, for a total of -4. To balance this, the oxidation number of carbon must be +4, and so choice (D) can be eliminated. Next, determine the oxidation number of carbon in urea, by finding the oxidation numbers for all the other molecules. Nitrogen has an oxidation number of -3, and there are two nitrogens, for a total of -6. Hydrogen has an oxidation number of +1, and there are 4 hydrogens, for a total of +4. Finally, oxygen has an oxidation number of -2. Since the molecule is neutral, the oxidation number of carbon must be -6 + 4 - 2 = -4. Since urea is neutral, C will be +4. Match to choice (C).
Citric acid is a weak triprotic acid. Because it has multiple acidic protons, it can:
act as a buffer over a wide pH range over its multiple pKa values.
A buffer resists a change in pH. This relationship can be described using the Henderson-Hasselbalch equation where pH varies least around the pKa of a buffer when there is an equal proportion of weak acid and conjugate base. Because citric acid is triprotic, it has 3 pKa values and 3 different weak acid states (H3A, H2A-, and HA2-). This gives it the ability to act as a buffer across three different pH ranges, which in the case of citric acid overlap one another slightly.
On a globular protein, where are tyrosine and phenylalanine residues most likely to be found?
While tyrosine is slightly more polar than phenylalanine, they both behave similarly and are more likely to be found on the interior of a protein than on the exterior of a protein, because both have hydrophobic (nonpolar) R-groups. Choice (D) is correct. Tyrosine is found on the exterior more often than phenylalanine, but this question does not ask about relative positioning.
Nitrophenols are highly toxic for humans, and are often found contaminating the soil near former factories. Which of the following statements about nitrophenols is supported by the data in the table below?
Solubility per 100 g H2O Melting point (oC)
1-hydroxy-4-nitrobenzene 1.7 114
1-hydroxy-3-nitrobenzene 1.4 97
1-hydroxy-2-nitrobenzene 0.2 44
These compounds have two functional groups that can hydrogen bond: the nitro (NO2) group and the hydroxyl (OH) group. Any two of these functional groups that are in close enough proximity will hydrogen bond, whether they are on different molecules or on the same molecule. How they align depends upon the shape of the molecule, which is different for all three. Furthermore, the melting and boiling points and water solubilities of polar compounds are largely governed by the extent of intermolecular hydrogen bonding.
According to the table, 1-hydroxy-4-nitrobenzene and 1-hydroxy-3-nitrobenzene are more water-soluble and have higher melting points than 1-hydroxy-2-nitrobenzene. This is because 1-hydroxy-2-nitrobenzene tends to form inTRAmolecular hydrogen bonds instead of inTERmolecular bonds; since the NO2 and OH group are on adjacent carbons, they are able to hydrogen bond with each other. That makes the molecules less likely to form intermolecular bonds, so that it takes less energy to make them separate. Thus, they will melt at a relatively low temperature. Likewise, intramolecular hydrogen bonding reduces the molecule’s ability to hydrogen bond with water, making the compound less water-soluble. Therefore, choice (A) is correct.
Which of the following compounds is expected to have the highest boiling point?
Valine has a nitrogen and two oxygen atoms that are capable of hydrogen bonding. It also forms a dipolar ion under appropriate circumstances, and thus is capable of dipole-dipole bonding. As a result, valine is a solid until almost 300°C, at which point it decomposes. Choice (B) is the correct answer choice.
The other compounds would all have lower boiling points than valine. Choice (A), glycine, has a similar structure to valine (and thus similar intermolecular forces), but valine has an extra methyl group and therefore will be heavier and have a higher boiling point. Choices (C) and (D), urea and water, both experience hydrogen bonding, but they have much smaller molecular masses than valine, and have fewer hydrogen-bonding opportunities.