FL 1 Review

Question 1

Suppose that multiple newly created amino acids interact to build a small protein molecule. The primary structure of that protein is formed when:

Answer 1

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.

Question 2

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:

Answer 2

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.

Question 3

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

Answer 3

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.

Question 4

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?

Answer 4

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.

Question 5

Which of the following accurately describes the difference between α-D-glucose and β-D-glucose?

Answer 5

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.

Question 6

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?

Answer 6

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).

Question 7

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:

Answer 7

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.

Question 8

When an acyl halide reacts with a primary alcohol, which of the following will form?

Answer 8

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).

Question 9

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

Answer 9

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.

Question 10

The method of carbon dating used to determine age depends upon the assumption that:

Answer 10

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.

Question 11

2A + B ⇌ 2C + D

Which of the following CANNOT be the mechanism for the rate-determining step of this reaction?

Answer 11

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.

Question 12

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?

Answer 12

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.

Question 13

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:

Answer 13

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).

Question 14

Citric acid is a weak triprotic acid. Because it has multiple acidic protons, it can:

Answer 14

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.

Question 15

On a globular protein, where are tyrosine and phenylalanine residues most likely to be found?

Answer 15

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.

Question 16

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

Answer 16

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.

Question 17

Which of the following compounds is expected to have the highest boiling point?

Answer 17

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.

Question 18

An experimenter wants to perform the Gabriel synthesis to create a tyrosine supplement for individuals with PKU. Which of the following is NOT required for the reaction to proceed?

Answer 18

The Gabriel synthesis substitutes an amino group for a halogen on alkyl halides. Phthalimide, choice (A), and the alkyl halide, choice (B), react to form an intermediate which undergoes base-catalyzed, choice (D), hydrolysis. Therefore, ammonia is the only compound out of the four answer choices not required for Gabriel synthesis. Choice (C) is the correct answer.

Question 19

In the absence of a pH meter, which of the following indicator solutions is most appropriate to use for measuring the initial pH of stomach acid by titration with NaOH?

Indicator Color at low pH Transition pH range Color at high pH

Methyl yellow red 2.9-4.0 yellow

Methyl red red 4.2-6.3 yellow

Phenol red yellow 6.6-8.0 red

Indigo carmine blue 11.4-13.0 yellow

Answer 19

The stomach produces HCl, a strong acid. The titrant is NaOH, a strong base. For this titration, the pH of the equivalence point is 7. Locate the region of the table that has 7 within the response range for the indicator. In this case, only one answer choice, phenol red, (choice (C)) contains the appropriate pH.

(If there were not such a good match in choice (C), then the indicator with the narrowest range around 7 would be most appropriate.)

Question 20

Which of the following cellular mechanisms would be directly affected by a vitamin B12 deficiency?

Answer 20

DNA synthesis, and therefore DNA replication, choice (A), would be most affected. B12 deficiency prevents the formation of thymine, a nitrogenous base found in DNA, but not RNA.

Question 21

The following titration curve would be expected from which of the following amino acids?

Answer 21

First, we began with the amino acid in its acidic state at a very low pH. OH–is added and the titration has three plateaus. These plateaus indicate regions where protons are being removed from functional groups on the molecule. The plateaus will occur at a pH equal to the pKa of the functional group. First, the acidic proton of the carboxylic acid group is removed at a low pH, usually around 2. All amino acids have this group. There is also a plateau around a pH of 9. This is the pKa of the –NH3+ group. However, we notice that there is one more flat region at a pH of 6. So, we are looking for an amino acid with an R group with a pKa around 6, indicating that it is weakly acidic. Choice (B), histidine, has an amino-ring structure with a pKa of around 6, and it is therefore the right answer. The aromatic histidine nitrogen (the one shown without a proton in choice (B)) is protonated with a positive charge at acidic (low) pH values. At a pH of around 6, the proton is removed. This is the reaction that gives rise to the plateau around pH 6 in the titration curve

Question 22

Which of the following best expresses the algebraic relationship between molar solubility, x, of Ca3(PO4)2 and its solubility product constant Ksp?

Answer 22

The expression asked for can be derived by setting up the equilibrium expression for the dissolution of calcium phosphate, which will produce 3x moles of calcium ions and 2x moles of phosphate for each x moles of calcium phosphate dissolved. Algebraically, we can rearrange it to solve for Ksp in terms of molar solubility:

Ca3(PO4)2 → 3 Ca2+ + 2 PO43–

Ksp = [Ca2+]3[PO43-]2 = (3x)3(2x)2 = (27x3)(4x2) = 108x5

Question 23

After using polymerase chain reaction (PCR) gene sequencing, an experimenter is able to identify only a small number of distinct fragments. Which of the following sources of error is most likely?

Answer 23

In PCR gene sequencing, dideoxynucleotides terminate replication, while deoxynucleotides allow replication to continue. During normal sequencing, an excess of deoxynucleotides and a very small amount of dideoxynucleotides are used. This allows for an approximately even chance of a dideoxynucleotide appearing anywhere in the gene sequence. Reversing this proportion would lead to very few distinct fragments as the deoxynucleotides are randomly incorporated. Although it can be expected that the total number of fragments is not reduced (compared to an experiment in which the correct proportions of dideoxynucleotides and deoxynucleotides was used), there will only be a few different (distinct) sequences because of the high proportion of chain terminators. Therefore, choice (A) is the correct answer.

Question 24

Which of the following is the correct electronic configuration for radioactive iodine, an isotope used in thyroid imaging?

Answer 24

Isotopes differ in their number of neutrons, not electrons, so we are simply looking for the electronic configuration of iodine. Choices (C) and (D) can be eliminated because they have an incorrect number of electrons (iodine has 53). And since iodine is not a transition element, it is not an exception to the filling order; therefore, the 5s shell will fill with 2 electrons before the 5p shell fills, and choice (B) is correct.
[Kr]4d105s25p5

Question 25

Which of the following is NOT a characteristic associated with peptide bonds?

Answer 25

Peptide bonds are very stable under physiological conditions, which explains the choice of protein as a cellular scaffold. They exhibit low reactivity, and hence choice (B) is the correct answer choice.

Since peptide bonds are resonant, they have partial double bond character, which restricts them to a planar molecular geometry.

Question 26

In some bacteria, lysine is not used as a cell wall component. Which of the following amino acids is most likely to be used instead?

Answer 26

Lysine has an additional amino group that is capable of participating in transpeptidation, which should immediately lead to a prediction of arginine; unfortunately, arginine is not an answer choice. However, the logic is still sound that we are looking for an amino acid which contains an additional amino group. Diaminopimelic acid, choice (D), fits this prediction. We can make these assumptions based on enzyme specificity for a terminal amino group.

Question 27

If the neoprene suit occupied 4 liters on land, approximately how many moles of air would need to be added to the buoyancy compensator to maintain neutral buoyancy at a gauge pressure of 1 atm?

A scuba diver in cool waters must wear an insulated wet or dry suit, which is usually composed of neoprene and contains about 40% air by volume.

Answer 27

The air in the neoprene initially occupied 1.6 L (40% of 4 L). The new pressure is 2 atm (remember, total pressure is atmospheric pressure plus gauge pressure). Using Boyle’s law, we know that the new volume of air in the suit is 0.8 L. Therefore, we require 0.8 L of air from the buoyancy compensator, which would be 0.8/22.4 moles at 1 atm of pressure. Since the pressure is twice as high (2 atm) we require twice as many moles of gas, or 1.6/22.4.

Question 28

Which of the following processes would NOT occur in response to reduced blood insulin levels?

Answer 28

Insulin levels are reduced at times of low blood glucose concentration in order to conserve glucose for the use by the brain. Thus when insulin levels fall, tissues such as muscle and liver decrease glucose uptake and utilization, choice (A), and start using fatty acids instead, choice (D). When insulin levels are low, glucagon levels rise by default and promote conversion of glycogen into glucose, choice (B), to maintain blood glucose levels. The only process that doesn’t occur in response low insulin levels, is choice (C), increased utilization of glucose as fuel.

Question 29

Which of the following is most likely true about the products of hydrolysis of PTH?

Answer 29

Choice (D) is the correct choice because it’s stated that the products of hydrolysis are directly excreted in the urine. In order to be excreted in the urine, a substance must be water soluble. Water insoluble compounds are either stored in the body, converted to water-soluble products, or eliminated in the feces.

Question 30

When plasma Ca2+ exceeds normal physiological levels, which of the following four sets of blood vessels should have the highest levels of the hydrolyzed products of PTH?

Answer 30

If plasma Ca2+ is above normal, the body will respond by excreting Ca2+ and PTH. Excretion occurs at the kidneys, making the afferent arterioles the expect target for PTH’s hydrolyzed products. Haversian canals (in the bones) would be opposite (low levels of PTH action), and hepatic/celiac are both not involved in excretion. Note that while the hepatic portal vein does involve the liver (which breaks down PTH), the hepatic portal vein’s function is to bring blood from the intestinal tract to the liver, not to take metabolites away from the liver after breakdown.

Question 31

What is the net number of ATP molecules synthesized by an obligate anaerobe per molecule of glucose?

Answer 31

This question cannot be answered from the passage so it is a “pseudo-discrete” question. An obligate anaerobe is an organism that must live WITHOUT oxygen in order to survive. Obligate anaerobes produce ATP via fermentation, which includes both glycolysis and the reactions necessary to regenerate the NAD+ necessary for glycolysis to continue. Fermentation leads to a net production of 2 ATP; this ATP is generated during glycolysis. Therefore, an obligate anaerobe will produce 2 ATP per molecule of glucose, which matches with choice (A).

Question 32

Curve C on the graph below represents the normal hemoglobin dissociation curve at 38°C and physiological pH (7.4) . Which curve most likely corresponds to the hemoglobin dissociation curve for a patient suffering from acidosis (low blood pH)?

Answer 32

An increase in the concentration of hydrogen ions decreases hemoglobin’s affinity for oxygen; that is, the binding of hydrogen ions to a molecule of oxyhemoglobin enhances the release of oxygen. According to the question, a patient suffering from acidosis has a decreased blood pH, which means that the concentration of hydrogen ions in the blood is higher than normal. And, as mentioned above, a high concentration of hydrogen ions means that hemoglobin will release oxygen more readily than it does under normal conditions. With reference to the hemoglobin dissociation curve, this means that at a given partial pressure of oxygen in the blood, the percent saturation of hemoglobin with oxygen in a patient with acidosis will be lower than it would be at physiological pH.

In terms of the graph, the hemoglobin dissociation curve corresponding to an acidotic patient would be shifted to the right of the curve for normal pH, and so curve D, choice (D), is correct.

Question 33

Which of the following pieces of evidence would NOT support the hypothesis that mitochondria were once independent bacteria that eventually formed a symbiotic relationship with eukaryotic cells?

Answer 33

Many scientists believe that mitochondria were once independent unicellular entities, possibly prokaryotic in origin, which formed a symbiotic relationship with eukaryotic cells that was mutually beneficial to both parties. This theory is known as the endosymbiotic hypothesis. If mitochondria are believed to have been prokaryotic in origin, then any findings that reveal similarities between mitochondria and bacteria would support the hypothesis. The question asks which choice does NOT support the hypothesis. If choice (B) is true, and mitochondrial ribosomes more closely resemble those found in eukaryotes than those found in prokaryotes, then the hypothesis is not supported.

The other answer choices would all support the hypothesis:

Choice (A). The fact that mitochondrial DNA is circular and not enclosed by a nuclear membrane is a characteristic of bacteria. Bacteria have a single circular chromosome located in a region of the cell known as the nucleoid, which is not bound by a membrane. So (A) SUPPORTS the endosymbiotic hypothesis and is therefore incorrect.

Choice (C). The fact that there are many present-day bacteria that have symbiotic relationships with eukaryotic cells supports the concept that the mitochondrial ancestor could have lived within a eukaryotic cell in a mutualistic relationship because it proves that this idea is feasible and does in fact happen.

Choice (D). The fact that mitochondrial DNA codes for its own ribosomal RNA. provides evidence that at some point, mitochondria existed as free entities, capable of directing their own protein synthesis and cell division, and all the cellular activities associated with independently living organisms.

Question 34

The L protein codes for a nucleic acid polymerase. Based on information in the passage, it most likely synthesizes:

Answer 34

The passage states that the virus does not integrate itself into the host genome, and it is not a retrovirus. Therefore, it does not need to synthesize DNA. Furthermore, it’s stated that the rabies genome is RNA. Therefore, the only plausible answer is choice (D): it must synthesize RNA from RNA.

Question 35

Which of the following is most likely to be true regarding cells in the muscles that spasm in a patient suffering from hydrophobia?

Answer 35

This question is asking for a true statement about the muscle cells that spasm in hydrophobia. The passage states that those cells are in the upper portion of the esophagus, which means that they are striated muscle cells. Striated muscle cells are voluntary, and therefore receive their innervation from the somatic division of the nervous system, thus Choice (D) is the correct answer.

Choice (A). Muscle contraction is dependent on release of calcium, not sodium.

Choice (B). Sarcomeres are a component of striated muscle, and so would be present.

Choice (C). The autonomic division of the nervous system innervates smooth muscle, which makes up the lower two thirds, not the upper portion, of the esophagus.”

Question 36

Recombination frequency

Answer 36

Recombination frequency is used to determine genetic distance, but cannot determine physical distance. If this is unclear, think about two alleles that are very far apart. The passage states that these alleles are likely to crossover multiple times; if a crossover event occurs an even number of times, the recombination frequency will appear to be zero. This thinking shows why choice (B) is correct: the recombination can make the alleles look closer together.

Question 37

During meiosis II of spermatogenesis, nondisjunction of chromosome 4 occurs. The resultant gametes could contain:

Answer 37

Nondisjunction occurs when chromosomes fail to separate. Recall that during meiosis I, the maternal and paternal chromosomes separate; during meiosis II, the sister chromatids from one parent separate. If nondisjunction occurs during meiosis II of spermatogenesis, then you are likely to get two identical chromosomes of either maternal or paternal origin. Therefore, Roman numeral I is correct. However, if crossing over occurred during prophase I, then some of the maternal and paternal DNA was swapped, so two non-identical chromosomes with both maternal and paternal DNA would be found: Roman numeral II is correct. Now, we need to look at Roman numeral III. This is incorrect, since non-identical chromosomes must contain maternal and paternal information: in other words, crossing over between two paternal sister chromatids (if that were to happen) would still leave us with identical chromosomes

Question 38

Which of the following does not derive from the neural crest?

Answer 38

Adrenal medulla
Schwann cells
Melanocyte cells
Erythrocyte cells

Since the neural crest is derived from the ectoderm, knowing all structure types that derive from ectoderm allow you rule out all choices except erythrocytes.

Question 39

Based on the data in Table 1, what is the net charge on the five amino acids listed for S. cerevisiae tubulin at pH 7.4?

Answer 39

To answer this question, it’s necessary to know the charges on the acids listed under S. cerevisiae in Table 1. The basic amino acids—which are positively charged at pH 7.4—are arginine and lysine. The acidic amino acids are aspartic acid and glutamic acid. Since none of these amino acids appear, the net charge on those five amino acids should be zero, which is choice (C).

Question 40

The semi-permeable membrane of the dialysis machine functions in a manner most analogous to which part of the kidney?

Answer 40

The glomerulus, (A), functions like a sieve, allowing the filtration (movement from the circulation into the nephron tubule) of small molecules while blocking the filtration of the plasma proteins. The semi-permeable membrane serves an analogous function in the dialysis machine.

Choice (B) is wrong because the ureter is merely a tube connecting the kidney to the bladder.

Choice (C) is wrong because the descending loop of Henle does not serve a filtration function. The primary action of the descending loop is reabsorption of water.

Choice (D) is wrong because The vasa recta are the capillaries that supply nutrients to the nephron.

Question 41

A patient with renal failure has nephrons which lack the ability to actively secrete or reabsorb any substances. Which of the following actions will the patient’s kidney most likely still be able to perform?

Answer 41

Recall that the fluid in Bowman’s capsule is isotonic to plasma. Without reabsorption or secretion, isotonic urine can still be produced. In other words, this patient’s nephrons filter the blood at the glomerulus, but that initial filtrate then becomes urine without any further modification. This process will still allow excess salt to be removed from the body, because that salt diffuses through the glomerulus, which is still functioning in this patient. Therefore, choice (A) is correct.

Question 42

Which of the following mitochondrial genome characteristics differs most from the characteristics of the nuclear genome?

Answer 42

Almost every base in mitochondrial DNA codes for a product.

Question 43

Which of the following is a common trait of group I and group II introns that makes them different from introns that are not designated into a group number?

Answer 43

Group I and II introns are unique in that they are self-splicing, or do not involve protein enzymes in their splicing reactions. This definition is echoed by choice (D), the correct answer.

Question 44

When action potential induces muscle contraction, which of the following occurs?

Answer 44

An action potential resulting in muscle contraction causes t-tubule depolarization, followed by SR release of Ca2+ in order for Ca2+ to bind to the troponin complex, thereby allowing tropomyosin to expose the myosin-binding sites on actin. This is described by choice (B).