MCAT Complete
Practice Passages
Part 2:
MCAT Biology
Drill
MCAT COMPLETE
CHAPTER 7 PRACTICE PASSAGE
Bacterial motility occurs via several mechanisms including bacterial secretions, axial filaments paired with cellular rotation, and most commonly, the use of flagella. Flagellin, a 53-kD
protein functioning as a flagellar subunit, polymerizes to form the helical filament that attaches to a hook and rod. The hook and rod anchor the flagellum by penetrating the outer and in-ner bacterial membranes. The flagellar motor driving rotation consists of proteins called FliG, FliM, and FliN; these form the membrane (M) and supramembrane (S) rings that bridge the periplasmic space and cytosol. MotA and MotB, two transmem-brane proteins, form a ring surrounding the FliG ring within the inner bacterial membrane.
Motor Axis Peptidoglycan Layer Inner Membrane L Ring P Ring S Ring M Ring Cytoplasmic Structure (FliG, FliM, FliN)
Mot A Mot B
Filament Hollow Core
Outer Membrane
Basal Disk (if present)
Inner Membrane Hook
Figure 1 The Flagellar Motor
Interestingly, flagellar rotation is powered by a proton-motive force, not ATP hydrolysis. Protons flow down their concentra-tion gradient from the periplasmic space into the first of two channels found in MotA/MotB, and are transferred to a FliG subunit. This causes the rotation of the FliG ring (as well as the filament and M and S rings), which carries the proton out to the second channel found in the MotA/MotB complex and allows the proton to enter the cytoplasm. This process repeats and results in continuous flagellar rotation and bacterial movement.
The direction of flagellar rotation plays a central role in bacte-rial motility. When rotating in the counterclockwise direction, the intrinsic left-handed sense of the flagella results in forma-tion of a combined flagellar bundle which propels the bacteri-um smoothly forward. When reversed, the flagella separate and spin independently resulting in random bacterial motion known as tumbling. The direction of rotation is dictated by the binding
of the cytoplasmic protein CheY. CheY itself cannot normally bind the flagellar motor; this results in counterclockwise rota-tion. However, when phosphorylated, it binds to the motor and reverses the direction of motion. Bacterial chemotaxis utilizes this mechanism to drive the bacterium up a concentration gradi-ent of an attractant (such as glucose) by increasing the time spent swimming smoothly and decreasing bacterial tumbling.
1. A researcher in a microbiology lab performs a series of experiments isolating transmembrane proteins and discovers a K+/H+ symporter in a multi-flagellated
bacterium. Further experiments suggest that this transmembrane protein allows potassium to flow out of the cell. Given the mechanism for flagellar rotation detailed in the passage, which of the following is true? A) This symporter functions via facilitated diffusion.
B) Cell motility decreases with decreased symporter activity. C) Transport activity increases in low pH solution.
D) Flagellar function increases in hyperkalemic (increased potassium) solution.
2. In an attempt to slow flagellum assembly in order to better study the mechanism, a developing cell culture is treated with a competitive inhibitor for a critical step in flagellum formation. Which of the following is true?
A) Addition of more inhibitor will have no impact on the reaction rate.
B) The inhibitor binds to the same allosteric site as the substrate.
C) The inhibitor has no effect on maximal rate of this critical step.
Drill
| 5 3. Flagellar motility and the electron transport chain both:
A) utilize the same electron carriers.
B) couple the exergonic movement of protons to an endergonic process.
C) generate ATP due to a protomotive force. D) are components of cellular respiration.
4. 2,4-dinitrophenol increases the permeability of protons across membranes. Upon treatment of a bacterium with this compound, how many gross molecules of ATP/ATP equivalents would be produced per glucose molecule? A) 2 ATP
B) 4 ATP C) 6 ATP D) 38 ATP
5. Which of the following is the likely mechanism of a chemoattractant seen in bacterial chemotaxis?
A) Decreased CheY transcription B) Inhibition of CheY phosphorylation C) Increased activity of lateral flagella
D) Decreased activity of CheZ (a CheY phosphatase)
6. A researcher studying bacterial flagella hypothesizes that a series of highly conserved acidic amino acids found on the carboxy-terminal domain of FliG plays a role in proton transport in the above process. Which of the following would likely have the LEAST impact on flagellar function?
A) Base-pair transition resulting in substitution of a hydrophobic amino acid on the carboxy-terminal domain B) Base-pair transversion resulting in substitution of a basic
amino acid on the carboxy-terminal domain
C) Insertion of two base pairs into the 5’ coding region of the FliG gene
D) Missense mutation resulting in the substitution of a buried N-terminus amino acid for a hydrophobic amino acid
MCAT COMPLETE
SOLUTIONS TO CHAPTER 7 PRACTICE PASSAGE
1. B From the passage, flagellar motility relies upon a proton gradient between the cell exterior and interior, and the movement of the flagella is driven by movement of a proton down its concentration gradient into the cytoplasm. Since the K+/H+ symporter moves potassium out of the cell, it must also move protons
out of the cell (symporters move two things in the same direction). Decreased symporter activity would decrease the amount of protons moved out of the cell, thus decreasing the protonmotive force driving movement of the flagellum (choice B is correct). High intracellular potassium concentrations are estab-lished by the Na+/K+ ATPase and this potassium gradient is used to drive the movement of H+ ions against
their gradient. In other words, the symporter allows K+ to flow out of the cell (down its gradient) while
simultaneously moving H+ against its gradient; this is an example of secondary active transport (choice
A is wrong). Since the activity of the transporter is driven by a K+ concentration gradient and serves to
create the H+ gradient, an elevated extracellular proton concentration (that is, low pH solution) would
not increase the activity of this particular transporter (although the increased H+ gradient may facilitate
flagellar rotation; choice C is wrong). An increased extracellular potassium concentration would decrease symporter function because there would be less of a K+ gradient to drive H+ transport. The decreased K+
gradient would result in a failure to establish the proton gradient, and thus decreased flagellar function (choice D is wrong).
2. C Competitive inhibitors bind to the active site of an enzyme (not an allosteric site, choice B is wrong) re-sulting in a decrease in its activity. This results in an apparent decrease in the affinity of the enzyme for its substrate, but does not change the enzyme’s maximal rate of product formation (Vmax) as sufficient quanti-ties of substrate can outcompete the added inhibitor (choice C is correct). Addition of more inhibitor will result in a greater number of active sites being occupied (with inhibitor instead of substrate) which will slow the reaction rate even more (choice A is wrong). Unless the enzyme is saturated, substrate concentra-tion will affect the reacconcentra-tion rate (choice D is wrong).
3. B Both the electron transport chain and flagellar motility, as described in the passage, rely upon a proton-motive force to drive a nonspontaneous (endergonic) process. In the case of the electron transport chain, the exergonic flow of protons down their concentration gradient generates ATP, whereas in flagellar mo-tility the movement of protons results in rotation of the flagellum (choice B is correct). The passage pro-vides no information on electron carriers in flagellar motility (choice A is wrong). Flagellar motility is not a part of cellular respiration and does not generate ATP as the energy from the protomotive force is used to rotate the flagellum rather than form a high energy bond (choices C and D are wrong).
4. C Disruption of the proton gradient causes a failure of the electron transport chain to generate ATP, result-ing in ATP production from glycolysis and the Krebs cycle alone. Note that the Krebs cycle would still be active because the electron transport chain would still be functional; NADH and FADH2 would still be
oxidized, the only thing disrupted is the proton gradient (and thus ATP production). Glycolysis generates a gross of 4 molecules of ATP per mole of glucose (although a net of 2 molecules) and the Krebs cycle produces 2 molecules of GTP per molecules of glucose (choice C is correct).
| 7 5. B Inhibition of CheY phosphorylation would prevent its binding to the flagellar motor, allowing for
in-creased counterclockwise flagellar rotation and inin-creased smooth linear movement (choice B is correct). While decreased CheY transcription would result in more linear movement, the timescale on which this
would occur would be much slower and far less effective (choice B is better than choice A). There is no reason to assume that during linear motion the lateral flagella would be more active than any other fla-gella; all the flagella rotate counterclockwise and form a bundle (choice C is wrong). Decreased activity of a CheY phosphatase would result in more phosphorylated CheY, which would bind to the flagellar motor and increase tumbling (choice D is wrong).
7. D The carboxy-terminal domain is described as highly conserved and possesses acidic amino acids, indicat-ing that these must be important for protein function. Substitution with a hydrophobic amino acid or with a basic amino acid (regardless of the specific mechanism) is likely to have a substantial impact on protein function (choices A and B would affect flagellar function and can be eliminated). The insertion of two base pairs into the 5’ coding region of the gene would result in a frameshift mutation and a significant change in protein structure (choice C would affect flagellar function and can be eliminated). However, buried amino acids are generally hydrophobic, thus substitution for a (presumably different) hydrophobic amino acid is unlikely to have a significant impact on protein structure or function (choice D would have the least impact on flagellar function and is the correct answer choice).
Drill
MCAT COMPLETE
CHAPTER 8 PRACTICE PASSAGE
Adipose tissue is found in two forms: white fat cells, which store energy in the form of triglycerides, and brown fat cells, which oxidize fuels and dissipate energy in the form of heat. When activated, brown fat cells take up fatty acids and glucose from the circulation, as shown in Figure 1. Brown adipose tissue has abundant mitochondria, which, via uncoupling oxidative phosphorylation, release chemical energy in the form of heat. Uncoupling protein 1 (UCP1) causes the inner membrane of the mitochondria to become “leaky”; this dissipates the hydrogen ion gradient by allowing H+ to re-enter the matrix without going through the ATP synthase. UCP1 is regulated by triiodothyronine (T3), which is generated when type 2 5’ deiodinase (D2) removes an iodine from the prohormone thyroxine (T4). A specific adrenergic receptor, known as a β3-adrenergic receptor, stimulates this process. The
β3-adrenergic receptor is a G-protein coupled receptor (GPCR)
and is activated by catecholamines, including epinephrine and norepinephrine. Brown adipose tissue is also rapidly activated by exposure to cold temperatures, and the activity of brown adipose tissue in adults is inversely correlated to age, body mass index (BMI) and fasting plasma glucose levels. While brown fat cells are important for thermoregulation in small mammals and newborn humans, the vast majority of brown fat cells are replaced by white fat cells in adult humans. However, positron emission tomography (PET) with the tracer
18F-fluorodeoxyglucose (18F-FDG), has revealed that there
are commonly multiple depots of brown adipose tissue in the cervical region of adults. This has been seen as a nuisance for
18F-FDG PET scanning, since this technique is used to visualize
the increased uptake of radiolabeled glucose by metabolically active cancers. Metabolically active brown fat has substantial uptake of the radiolabeled glucose, thus leading to false positive results.
Recently, a genetic “master switch” for brown adipose tissue differentiation has been discovered. PRDM16 (PRD1-BF1-RIZ1 homologous domain containing 16) was shown to be selectively expressed in brown fat cells versus white fat cells and stimulates nearly all of the characteristics of brown fat cells. A large majority of genes that are selectively expressed in brown adipocytes are positively regulated by PRDM16. Conversely, PRDM16 expression suppresses the mRNA levels of several genes, such as resistin and serpin3ak, which are
selectively enriched in white fat. It has been further discovered that bone morphogenetic protein 7 (BMP7) activates the formation of brown fat including induction of PRDM16 and PGC-1α [peroxisome proliferator-activated receptor-γ (PPAR-γ) coactivator-1α], as well as increased expression of brown-fat-defining marker UCP1.
Substrates: Free Fatty Acids and Glocose T3 T3 D2 UCP1 T3 T3 CRE TRE Nucleus cAMP T4 D2 T3 Adrenergic neuron β3 Receptor Mitochondria
Figure 1 A Brown Fat Cell. CRE = cAMP Response Element; TRE = Thyroid hormone response element;
D2 = Type 2 5’ Deiodinase; UCP1 = Uncoupling Protein 1.
Adapted from Celi FS. Brown adipose tissue--when it pays to be inefficient. N Engl J Med 2009;360:1553-6 and Seale P, Kajimura S, Yang W, et al. Transcriptional control of brown fat determination by PRDM16. Cell Metab 2007; 6:38-54.
1. A patient with a pheochromocytoma, a tumor of the adrenal gland that leads to hypersecretion of epinephrine, would be expected to have which of the following? A) Decreased brown adipose tissue intracellular levels of T3
B) Decreased brown adipose tissue Krebs cycle activity C) Increased brown adipose tissue levels of D2 mRNA D) Increased fat cell activation, which would lead to
Drill
| 9
2. Which of the following treatments would help decrease the false positive rate of diagnostic 18F-FDG PET scans
for cancers?
I. Warming the room occupied by the study subject
II. Administering beta-blockers (medications that block the β-adrenergic receptors) III. Administering a synthetic version of T4 A) I only
B) I and II C) III only D) I, II, and III
3. Based on the passage, which of the following may be inferred?
A) The regulation of brown fat cells appears to be a potential important therapeutic target for obesity. B) Activation of brown fat cells leads to what may be
considered a “local hypothyroid state”.
C) A patient with diabetes that has frequent hyperglycemia would be likely to have a high state of brown adipose tissue metabolic activity.
D) The PRDM16 gene is selectively present in brown fat cell genomes.
4. Which of the following would be expected in a BMP7-knockout mouse embryo compared to a normal mouse embryo?
A) An increased amount of PRDM16 mRNA
B) A marked paucity of brown fat and almost a complete absence of UCP1 protein
C) An increase in PGC-1α activity
D) A decrease in the expression of the genes resistin and serpin3ak
5. In one experiment, an adenovirus vector was used to increase BMP7 expression in mice. Which of the following would be LEAST likely to occur? A) A reduction in weight gain
B) An increase in brown fat mass C) An increase in energy expenditure
D) An increase in fasting plasma glucose levels
6. Based on Figure 1, thyroid hormone (a peptide hormone) exerts its effect on brown fat cells via a mechanism of action that is most analogous to the mechanism of action of which of the following?
A) Adrenocorticotropic hormone (ACTH) B) Cortisol
C) Antidiuretic hormone (ADH) D) Insulin
MCAT COMPLETE
SOLUTIONS TO CHAPTER 8 PRACTICE PASSAGE
1. C As described in the first paragraph and the figure, increased epinephrine would lead to activation of the β3 receptor, a GPCR, which triggers the production of D2 (choice C is correct). The increased D2 would convert more T4 to T3 (choice A is wrong). This would lead to increased metabolic activity, including, the Krebs cycle (choice B is wrong). This increased metabolic activity, which would ultimately be in the form of uncoupled oxidative respiration, would be expected to lead to weight loss rather than weight gain (choice D is wrong).
2. B The passage states that the increased metabolic activity and uptake of substrates, including glucose, by activated brown fat cells leads to “false positives” on 18F-FDG PET scans for cancers. Therefore,
in order to decrease the uptake of radiolabeled glucose at the brown fat cells, the intervention should decrease brown fat cell metabolic activity. Item I is true: The passage states that exposure to cold tem-peratures rapidly activates brown fat cells; thus warming the room would lead to decreased brown fat activity (choice C can be eliminated). Item II is true: Since activity of the β3 receptor leads to increased brown fat cell metabolic activity, the administration of beta-blockers should lead to decreased activity (choice A can be eliminated). Item III is false: The passage and figure support that increased thyroid hormone (T4) levels lead to increased brown fat cell metabolic activity. Thus the administration of a synthetic T4 would lead to increased metabolic activity and be counterproductive (choice D can be eliminated and choice B is correct).
3. A The regulation of brown fat cells as a potential therapeutic target for obesity is supported in the passage by various statements, including the fact that brown fat cell activity increases metabolism, dissipates energy, and also because it is inversely correlated with body-mass index (BMI). Therefore, choice A is correct. The increased levels of D2 in the brown fat cells leads to increased levels of active T3 hormone; therefore, it is considered to be a “local hyperthyroid state” (choice B is wrong). The passage directly states that the level of brown adipose tissue activity is inversely correlated with the serum glucose levels (choice C is wrong). While the PRDM16 gene is selectively expressed in brown fat cells, the gene is present in the genome of all
cells (choice D is wrong). Remember that all of an individual’s genes are present in their genome, and that specific genes are expressed (i.e., transcribed and translated) in specific tissue types.
4. B As the passage states, BMP7 activates the formation of brown fat, including induction of PRDM16 and PGC-1α. Therefore, a mouse that lacked BMP7 would be expected to not express the PRDM16 gene, which would lead to decreased levels of PRDM16 mRNA (choice A is wrong). Since PRDM16 is the “master switch” for brown adipose tissue differentiation (as stated in the passage), the mouse would lack brown adipose tissue and would not express UCP1 (choice B is correct). Since BMP7 also leads to the expression of PGC-1α, it follows that lack of BMP7 would lead to decreased PGC-1α activity (choice C is wrong). Also, the passage states that PRDM16 inhibits the expression of the genes resistin and serpin3ak. Thus a lack of BMP7, which leads to a lack of PRDM16, would result in an increased expression of resistin
| 11 5. D Since BMP7 induces brown fat cell metabolism, an increase in BMP7 would likely lead to weight loss
due to increased metabolism and uncoupled cellular respiration (choice A is likely and can be eliminated). This would result in increased energy expenditure (choice C is likely and can be eliminated). BMP7 also
induces many genes, including PRDM16, which would lead to brown fat cell differentiation; thus, the total brown fat cell mass would increase (choice B is likely and can be eliminated). However, the passage states that brown fat cell activity is inversely related to fasting plasma glucose; therefore, an increase in brown fat cell activity would lead to a decrease in fasting plasma glucose levels (choice D is not likely and the correct answer choice).
6. B Despite the fact that thyroid hormone is a peptide hormone, Figure 1 supports the fact that this hormone works intracellularly to exerts its effect on the cell at the level of gene expression. This is the mechanism of action primarily used by steroid hormones, such as cortisol (choice B is correct), aldosterone, testosterone, estrogen and progesterone. Peptide hormones, including ACTH (choice A), ADH (choice C) and insulin (choice D), and several others, exert their effect by binding to cell surface receptors and modifying intra-cellular enzyme activity.
Drill
MCAT COMPLETE
CHAPTER 9 PRACTICE PASSAGE
Bacillus anthracis is a gram-positive sporulating rod-shaped
bacterium found extensively in the soil, and it causes the group of diseases collectively known as anthrax. Sporulation allows the bacterium to survive harsh conditions. Depending on the location of the bacteria in the body, an infection with B. anthracis can manifest as anthrax (either cutaneous, pulmonary,
or gastrointestinal) or meningitis. The bacterium possess two virulence plasmids (pXO1 and pXO2), which code for exotoxins and a poly-D-glutamic acid capsule, respectively. When diagnosing a patient with anthrax, the outcome of the disease is highly dependent upon the speed of diagnosis due to the significant difference in mortality observed with the administration of antibiotics.
When identifying B. anthracis as the causative agent,
physicians first attempt to culture and identify the living bacteria. Samples of blood, sputum, feces, or cerebrospinal fluid are cultured on blood agar at 37ºC and if gram-positive chains of rods are observed, further confirmatory tests are performed. These tests include phenotypic typing, gamma phage lysis, real-time PCR assays, direct fluorescent assays, and time-resolved fluorescent assays. Commercially available immunochromatographic tests have recently been developed and are now available.
While cultures of the bacteria are ideal for diagnosis, they are not always possible to obtain and often indirect methods of identification must be used. A recently developed PCR assay targets three loci on the bacterial chromosome and one on each of the virulence plasmids; this allows detection of an infection of as few as 167 total cells. PCR-amplified DNA fragments are identified through gel electrophoresis, in which fragments are separated by size and can be compared to a standard to confirm their respective lengths. Another method detects the antibodies produced by the body against the anthrax toxin protein (PA). This Enzyme-Linked Immunosorbent Assay (ELISA) is commercially available to quantify the level of human anti-PA IgG rapidly; this allows for more directed treatment.
1. A physician collects two blood samples and sends them to the lab for culture and an ELISA. The culture comes back negative but the ELISA is positive for human anti-PA IgG. What is a possible reason for this result? A) The patient is infected with a virus producing similar
symptoms.
B) The patient does not have an active infection.
C) The ELISA results are not conclusive because the test is nonspecific.
D) B. anthracis do not grow in culture.
2. According to the passage, which of the following samples is LEAST likely to result in an anthrax diagnosis?
A) Blood B) Sputum C) Feces
D) Cerebrospinal fluid
3. A new blood sample arrives at a lab and grows a gram-positive bacterium later identified as B. anthracis;
however the ELISA registers a negative result. Is this patient infected with B. anthracis and what most likely
accounted for this result?
A) No; the laboratory sample was contaminated.
B) No; the cultured bacteria were part of the natural flora of the body.
C) Yes; the tested blood sample would register a positive ELISA in approximately seven days.
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| 13
4. In the PCR detection of the bacterium, a laboratory technician obtains a positive result. How many bands should the technician detect on the gel?
A) 2 B) 3 C) 4 D) 5
5. A culture from a sputum sample develops into colonies of round bacterial cells. Which of the following could be the pathogen causing infection in this patient?
A) Streptococcus sp. (streptococcal pharyngitis; “strep
throat”)
B) Mycobacterium tuberculosis (tuberculosis, TB, tubercle
bacillus)
C) Bacillus anthracis (anthrax)
D) Spirillum minus (rat-bite fever)
6. One of the potential complications of inhalation or pulmonary anthrax is the development of hypotension. Which of the following would not cause hypotension in a healthy patient?
A) SA node depression
B) Destruction of the posterior pituitary C) ACTH secreting lung tumor
D) Hemorrhage
MICROBIOLOGY
MCAT COMPLETE
SOLUTIONS TO CHAPTER 9 PRACTICE PASSAGE
1. B Bacteria can be cultured from patient blood only when an active infection is present in the blood. This patient could have already cleared the infection and mounted an immune response, resulting in a positive ELISA but negative culture result (choice B is correct). The ELISA test is very specific (choice C is wrong), and there is no reason to believe that a human would develop the same or similar antibodies against a B.
anthracis infection and a viral infection. Even if the patient was infected with a virus that cause symptoms
similar to a B. anthracis infection, he or she would not have a positive ELISA given the specificity of an-tibodies (choice A is wrong). According to the passage, B. anthracis can be grown in culture (choice D is wrong).
2. D A positive result from a sample of cerebrospinal fluid would result in a diagnosis of meningitis, not an-thrax (choice D is correct). The passage states that an infection with B. anthracis can manifest as cutane-ous, pulmonary, or gastrointestinal anthrax, which indicates that a positive result from samples of blood, sputum, or feces could lead to an anthrax diagnosis (choices A, B, and C can be eliminated).
3. D If the patient has very recently been infected with B. anthracis for the first time, the body may not yet have mounted a detectable immune response (choice D is correct). A significant humoral response in the form of elevated serum antibodies will be detected approximately one week after infection. However, the already-tested blood sample would not register a positive result after seven days, since outside of the body blood lacks the machinery to produce antibodies; a fresh blood sample would be required (choice C is wrong). A positive blood culture could result from contamination, but this is less likely than an active infection (choice A is possible but less likely than choice D). B. anthracis is not part of the body’s normal flora; it is a disease-causing agent (choice B is wrong).
4. D According to the passage, the PCR assay three loci on the bacterial chromosome and one on each of the virulence plasmids (2 total from the plasmids). This means there should be five DNA fragments amplified by PCR, and the technician should have seen a total of five bands on the electrophoresis gel (choice D is correct).
5. A Round bacterial cells are known as cocci. Streptococcus bacteria causing streptococcal pharyngitis, com-monly called strep throat, would develop into a culture of round bacterial cells (choice A is correct).
Myco-bacterium tuberculosis is rod-shaped, as indicated by the “bacillus” in tubercle bacillus (choice B is wrong).
The passage explicitly states that Bacillus anthracis is rod-shaped, and this can also be inferred from the genus Bacillus (choice C is wrong). Spirillum minus is a spiral-shaped bacterium (choice D is wrong).
| 15 6. C Hypotension is a reduction in blood pressure. ACTH stimulates the release of cortisol and aldosterone
from the adrenal cortex, and aldosterone causes an increase in the reabsorption of Na+ from the distal
tubules of the kidneys. This would lead to an increase in blood pressure, or hypertension (choice C does not cause hypotension and is the correct answer choice). SA node depression would result in a decrease in heart rate and a subsequent decrease in blood pressure (choice A is a possible cause of hypotension and can be eliminated). Destruction of the posterior pituitary would lead to a drop in ADH and a subsequent inability to reabsorb water from the collecting ducts of the kidneys. This would result in a decrease in blood volume and thus a drop in blood pressure (choice B is a possible cause of hypotension and can be eliminated). Hemorrhage would also lead to a decrease in blood volume and result in hypotension (choice D can be eliminated).
Drill
MCAT COMPLETE
CHAPTER 10 PRACTICE PASSAGE
Charcot-Marie-Tooth (CMT) disease type 2A, a hereditary primary motor sensory neuropathy, is characterized by normal or mildly reduced nerve conduction velocity and axonal degeneration. Symptoms of the disease generally present in the 20s or 30s and include weakness of the extremities, muscle atrophy, and sensory loss. The disease is caused by a mutation in a kinesin (a cellular transport protein) known as KIF1Bβ. Kinesin is a dimeric motor protein that travels along microtubules in the cell, delivering cargo to the periphery of the cell (anterograde transport) toward the plus-end (growing end) of cellular microtubules. Microtubules are formed from non-covalent interactions between α- and β-tubulin. Movement of the kinesin motor occurs by what is commonly referred to as the “hand-over-hand” mechanism. Beginning with ADP bound to each of the two kinesin heads, neither is bound to the microtubule. The exchange of ADP for ATP on one of the heads causes it to bind to the microtubule. This leads to a conformational change propelling the other kinesin head forward. The unbound head can then exchange ADP for ATP and bind to the microtubule. Simultaneously, the kinesin head first bound to the microtubule hydrolyzes its ATP to ADP causing its release from the microtubule, propulsion
forward, and release of phosphate. This process continues for approximately 100 cycles before both heads are simultaneously bound with ADP allowing for the release of the kinesin. With the mutation of a kinesin, a cell ceases to be able to supply its periphery with macromolecules, organelles, and other intracellular compounds. Other diseases, including primary ciliary dyskinesia, result from failure to transport intracellular structures toward the cell center (retrograde transport) via dyneins, which are structurally and functionally similarly to kinesins, but move in the opposite direction.
1. A scientist develops an in vitro experiment to
observe kinesin movement. Addition of adenosine 5’-γ-thiotriphosphate (a nonhydrolyzable ATP analog) would have what effect on kinesin function?
A) Increased number of cycles per binding occurrence B) Inhibition of the power stroke
C) Decreased free kinesin D) Decreased bound kinesin
2. Skeletal muscle contraction and kinesin function share many similarities. Which of the following is NOT true? A) Both kinesin and myosin interact with components of the
cytoskeleton.
B) A conformational change drives motion in both myosin and kinesin.
C) Myosin and kinesin have similar release mechanisms. D) Both kinesin and myosin possess ATPase activity. 3. A researcher identifies multiple variants of kinesins
with differing functions. Normally the motor domain is found on the N-terminus of the kinesin, but it was discovered that certain kinesins operated in the reverse direction along the microtubule when the motor domain was located on the C-terminus. If a kinesin is observed traveling toward the cell center, which of the following characterize(s) this motion?
I. Kinesin with an N-terminus motor domain and microtubule orientation reversal
II. Kinesin with a C-terminus motor domain and microtubule orientation reversal
III. Kinesin with a C-terminus motor domain A) I only
B) II only C) I and III only D) I, II, and III
4. Symptoms of Charcot-Marie-Tooth include weakness of the extremities, muscle atrophy, and sensory loss. Based on information in the passage, which of the following would be the most likely additional symptom of Charcot-Marie-Tooth disease?
A) Diminished reflexes B) Mental status changes C) Tetanus
Drill
| 17 5. Given the pathogenesis of this disease, which of the
following areas would likely be first affected by Charcot-Marie-Tooth disease?
A) Head B) Thorax
C) Proximal extremities (closer to the midline) D) Distal extremities (farther from the midline) 6. All of the following statements are true regarding
microtubules EXCEPT: A) they form asters during mitosis.
B) they are anchored in the microtubule organizing center (MTOC).
C) they possess quaternary structure. D) they possess a static structure.
MCAT COMPLETE
SOLUTIONS TO CHAPTER 10 PRACTICE PASSAGE
1. C ATP hydrolysis occurs in the kinesin head which is about to be driven forward. Without hydrolysis, the head would remain fixed to the microtubule and would not release (choice C is correct and choice D is wrong), and this would significantly decrease the number of cycles (choice A is wrong). The term “power stroke” is generally used with regard to myosin function, not the driving force propelling the kinesin for-ward (choice B is wrong).
2. C In skeletal muscle, myosin releases actin upon the binding of ATP but kinesin releases the microtubule upon hydrolysis of ATP (choice C is not true and is the correct answer choice). Myosin and kinesin bind actin and microtubules, respectively, and actin and microtubules are both components of the cytoskel-eton (choice A is true and can be eliminated). In both cases, their motion is driven by conformational changes (choice B is true and can be eliminated), and both proteins possess ATPase activity (choice D is true and can be eliminated).
3. C This question is asking about two different components: the kinesin protein and the tracks (microtubules) it is walking along. Kinesins normally travel toward the plus-end of the microtubule, which is normally found at the periphery of the cell. However, when the motor domain is located in the C-terminus, kine-sins travel toward the minus-end of the microtubule, typically found towards the center of the cell (Item III is true and choices A and B can be eliminated). Note that both remaining answer choices include Item I, so Item I must be true: a kinesin moving toward the cell center could have a normal N-terminus motor domain, but could be traveling along microtubules that are in the reverse orientation (that is, their minus-end is found at the center of the cell and their plus-minus-end at the periphery). Item II is false: kinesins with a C-terminus motor domain travel towards the minus-end of the microtubules. However if the microtubule orientation has been reversed, the minus-end would be at the cell periphery, and the kinesin would not be traveling toward the cell center (choice D can be eliminated and choice C is correct).
4. A With loss of sensation and weakness of the extremities, one would expect a loss in deep tendon reflexes (choice A is correct). Tetanus (prolonged sustained contraction) is incredibly unlikely given the loss of strength and atrophy (choice C is not correct). Mental status changes and tooth decay would not be due to the sensorimotor changes seen in this disease (answers B and D are not correct; there is no information in the passage to support these two answer choices).
5. D The longest neurons in the body are found in the distal extremities and would most severely be impacted by a loss of neuronal conduction velocity and axonal degradation (choice D is correct and choice C is wrong). The head and thorax possess relatively short neurons and would be less likely to be impacted (choices A and B are wrong).
6. D Microtubules have a variable structure as a result of their dynamic instability on the growing (plus) end (choice D is a false statement and the correct answer choice). Microtubules form asters in mitosis and are anchored in the microtubule organizing center, or MTOC (choices A and B are true statements and can be eliminated). As microtubules are formed from noncovalently linked alpha and beta tubulin, they pos-sess a quaternary structure (choice C is a true statement and can be eliminated).
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GENETICS AND EVOLUTION
CHAPTER 11 PRACTICE PASSAGE
Whenever the expression of a gene is conditioned by its parental origin, geneticists say that the gene has been imprinted. Genetic imprinting is an inheritance process independent of the classical Mendelian inheritance, as only one parental allele is expressed. For example, paternal imprinting occurs when an allele inherited from the father is not expressed in offspring, while maternal imprinting occurs when an allele inherited from the mother is not expressed. As such, certain mechanisms must be in place to mark and silence imprinted genes.
Previous molecular analyses have repeatedly demonstrated that the mark that conditions the expression of a gene is methylation of one or more CG dinucleotides in the gene’s vicinity. More specifically, methylated DNA is associated with transcriptional repression. For example, in mice, the gene coding for insulin-like growth factor (Igf2) is methylated in the female germ line but not in the male germ line. As such, Igf2 is expressed when it is inherited from the father but not the mother.
During embryogenesis, the methylated and unmethylated states are preserved each time the genome replicates. However, a methylated gene that was inherited from one sex can be unmethylated when it passes through an offspring of the opposite sex. For example, a maternally imprinted gene is unmethylated when male offspring produce gametes, but remethylated when female offspring produce gametes. Thus, imprinting is re-set with each successive generation, and subsequent methylation depends on the sex of the animal. Angelman syndrome is a classic example of an autosomal
disease related to genetic imprinting. The gene implicated in the disease is deleted or mutated on the maternally inherited chromosome, while the paternal gene is silenced secondary to imprinting. As such, the paternal gene cannot compensate for the loss of the gene on the maternal chromosome, because imprinting has turned it off. In contrast, Prader-Willi syndrome is caused by the combination of maternal imprinting and loss of paternally inherited genes in a nearby region of the same chromosome.
The pedigree (Figure 1) illustrates the inheritance pattern of two different diseases—one of which is due to a mutant allele and genetic imprinting, similar to the two diseases discussed above. Assume the following: (1) The imprinted gene in question is located on a somatic chromosome, (2) individual I.a is homozygous recessive for the allele causing Disease 1, (3) individual I.b is homozygous for the normal allele with respect to Disease 1, and (4) individuals not blood-related to Individuals I.a and I.b have a wild type allele rather than either disease-producing allele. Key: = Disease = Disease 2 1 = Disease 2 = Both Diseases I II III IV a b e f c d a b e f g h a b c d a b c Figure 1
1. What is the most likely mode of inheritance for Disease 2? A) Autosomal dominant
B) Autosomal recessive C) X-linked recessive D) Mitochondrial inheritance
2. Is it possible for both men and women to develop Prader-Willi syndrome?
A) Yes, both sexes have an equal probability of disease. B) Yes, both sexes can have the disease, but it is found
predominantly in women because of maternal imprinting. C) No, because it is only maternally imprinted.
D) No, because imprinting is reset with each successive generation.
3. What is the probability that Individual III.d would have Disease I?
A) 0 % B) 25% C) 50%
Drill
MCAT COMPLETE
4. What is the probability Individual IV.c would have both diseases?
A) 0% B) 6% C) 25% D) 57%
5. The gene responsible for Angelman syndrome is UBE3A. In a person with this syndrome, what is the difference between the maternal and paternal chromosome?
A) There is no difference; both homologous genes are methylated.
B) There is no difference; both homologous genes are unmethylated.
C) The maternal gene is methylated and the paternal one deleted or mutated.
D) The paternal gene in methylated and the maternal one deleted or mutated.
6. Which of the following diseases would be consistent with the inheritance pattern of Disease 1?
A) Prader-Willi syndrome B) Color blindness C) Hemophilia D) Angelman syndrome
7. If Individual III.a and III.f had a child, what is the probability they would produce a carrier of Disease 2? A) 0%
B) 25% C) 50% D) 100%
| 21
GENETICS AND EVOLUTION
SOLUTIONS TO CHAPTER 11 PRACTICE PASSAGE
Generation I II III IV Key: = Disease 1 = Disease 2 = Sex Chromosomes = Somatic Chromosome
= Chromosome has mutated/disease allele = Chromosome may/may not have disease allele = allele inactive due to imprinting
X/Y S * +/-I X*Y XX* S*SI XY XY S*SI XXSSI XX S*SI XY XX SSI X*Y SSI XX +/-S*SI XY SSI XX +/-S*SI or S*S*I XY XY XY SSI XX*SSI S*SI SSI SSI XX +/-S*S*I SSI or SS*I S*SI or S*S*I a b e f c d a b e f g h a b c d a b c
Note: Every individual on this pedigree needs to have the S allele that they inherited from their father im-respect to the S allele is determined solely from their mother’s contribution.
1. C
linked recessive diseases (choice C is correct). Autosomal dominant diseases are typically present in every -tions, but show no gender preference (choice B is wrong). Finally, this trait is not showing mitochondrial inheritance because individual III.f has the disease but his mother does not (choice D is false). Note: now that you know Disease 2 is X-linked recessive, Disease 1 MUST be due to some form of genetic imprint-ing, either maternal or paternal. See the pedigree key at the beginning of the solutions.
MCAT COMPLETE
2. A As described in the passage, Prader-Willi is an autosomal syndrome resulting from the inheritance of a maternally-imprinted gene in combination with a deletion/mutation of the homologous paternal gene. Therefore, both genders can theoretically be afflicted with the disease equally (choices C and D are wrong). Maternal imprinting does not confer an increased risk of women being affected, particularly given that the disease is autosomal; all children have the same probability of inheriting an imprinted gene regardless of gender (choice B is wrong and choice A is correct).
3. A The mode of inheritance for Disease 2 is X-linked recessive (see explanation to question #1), therefore Disease 1 must be related to genetic imprinting. The first thing needed is to determine if the trait is an example of paternal imprinting or maternal imprinting. The passage states that Individual I.a is homozy-gous recessive for the allele causing Disease 1, whereas Individual I.b is homozyhomozy-gous for the normal allele with respect to Disease 1 (i.e., homozygous dominant). Since all of the children (II.b, II.c, II.e) display their mother’s (I.a) phenotype in spite of inheriting a dominant allele from their father, this must be an example of paternal imprinting (the father’s genes are silenced). The father of Individual III.d has the dis-ease and the mother is assumed to be homozygous for the normal allele. However, the father’s gene will be silenced due to imprinting, so the child will only express the normal maternal allele. As such, the child has a 0% chance of having Disease 1 (choices B, C, and D are wrong).
4. B Individual III.g (the mother of Individual IV.c) is heterozygous for Disease 1, but has the disease because the chromosome she received from her father (II.f) with the normal allele has been silenced due to im-printing. Therefore, Individual IV.c will have Disease 1 only if III.g passes on the disease carrying allele (a 1/2 probability). Any contributions from III.h (IV.c’s father) can be ignored because they will be silenced by imprinting. With respect to Disease 2, daughters of III.g and III.h will not be affected, since III.h is an unaffected male and will pass a normal X to his daughters; therefore, for IV.c to have Disease 2 it must be male (a 1/2 probability ). Furthermore, III.g must be a carrier of Disease 2 (a 1/2 probability) and must pass it on to IV.c (also a 1/2 probability). Thus, the probability that IV.c has both diseases is determined using the Rule of Multiplication which states: P(A and B) = P(A) × P(B). In this instance, the probability of both diseases = P(Disease 1) × P(Disease 2) = (½) × (½ × ½ × ½) = 1/16 or ∼ 6%.
5. D As stated in the passage, methylation is the mechanism underlying imprinted diseases, whereby methyl-ated genes are repressed. As such, the paternal and maternal chromosomes will differ in terms of meth-ylation patterns (choices A and B are wrong). The passage states that Angelman syndrome is a paternally imprinted disease, thus a segment of the paternal chromosome is silenced (imprinted via methylation) and the maternal gene in this region is deleted or mutated (choice D is correct and choice C is wrong). 6. D Disease 1 is a paternally imprinted disease (see the explanation for Question 4), which is the same
inheri-tance pattern underlying Angelman syndrome (choice D is correct). As stated in the passage, Prader-Willi syndrome is due to maternal imprinting (choice A can be eliminated), and both color blindness and he-mophilia are classic X-linked recessive disorders (choices B and C can be eliminated).
7. C Since Disease 2 is X-linked recessive (see explanation to question #1), and since III.f (the male) has this disease, all of his daughters would receive the disease allele. III.a does not carry the allele; her father (II.b) is unaffected and her mother (II.a) married into the family and does not carry disease alleles. Thus, the probability of producing a carrier of Disease 2 is just the probability of having a female child; one out of two, or 1/2, or 50% (choice C is correct).
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CHAPTER 12 PRACTICE PASSAGE
Anabolic-androgenic steroids were identified and synthesized in the early 1930s, and since that time have been used both therapeutically and a blood doping agent for athletes. As their name suggests, this class of steroids have both anabolic and androgenic effects, and like other steroids, they are chemically derived from cholesterol.
The anabolic effects refer to those that result in increased protein synthesis, bone marrow production, red blood cell production, as well as increased size and number of skeletal muscle cells. The androgenic effects, on the other hand, include those that are involved in the development of masculine characteristics. These include effects such as a thickening of the vocal cords and a deeper voice, increased androgen-sensitive hair growth (such as pubic and chest hair), increased libido, and progression of puberty in children who have not yet reached puberty. The androgenic-anabolic ratio is a measure of which effect is dominant for a certain steroid, and is listed for different steroids in Table 1.
Steroid Ratio Testosterone 1:1 Fluoxymesterone 1:2 Oxymetholone 1:3 Nandrolone decanoate 1:4 Oxyandrolone 1:13
Table 1 Androgenic-anabolic ratio in humans Although notoriously known as performance enhancing drugs,
anabolic-androgenic steroids are also used for medicinal purposes. These can take advantage of the anabolic and androgenic effects of this class of steroids, and the type of hormone used for treatment is chosen depending on the androgenic-anabolic ratio and the desired outcome.
The use of anabolic-androgenic steroids as doping agents in athletes has been a controversial topic for several years. The desired effect of these steroids is due to their anabolic effect, which can give athletes a competitive edge. They are taken for different reasons depending on the sport; endurance athletes benefit from the increased red blood cell count, whereas athletes requiring a lot of strength benefit from the increased lean mass. However, the use of these steroids in athletes has been banned in part due to their adverse effects, which can arise due to the androgenic effects of the steroids. Additionally, anabolic-androgenic steroid use can suppress naturally produced testosterone in men. Non-androgenic side effects of these steroids include increased LDL, hypertension, acne, and liver failure. Although these steroids mimic the effects of naturally produced hormones, they continue to be one of the most detectable blood doping agents.
1. Which of the following is correct about anabolic-androgenic steroids?
A) They are difficult to detect because they are structurally similar to naturally produced hormones.
B) Marathon runners and anemic patients benefit from the same effect of anabolic-androgenic steroids.
C) Medicinal uses only take advantage of the anabolic effects.
D) Androgenic side effects only occur in men.
2. Many of the side effects of anabolic-androgenic steroids result from feedback on the hypothalamic-pituitary axis. Which of the following is true regarding the nature of this interaction?
A) It is a positive feedback cycle that results in an increase in hormone release from the anterior pituitary.
B) It is a negative feedback cycle that results in an increase in hormone release from the anterior pituitary.
C) It is a negative feedback cycle that results in a decrease in hormone release from the anterior pituitary.
D) It is a negative feedback cycle that results in a decrease in hormone release from the posterior pituitary.
3. The pharmacodynamics of steroid hormones and peptide hormones differ due to the different chemical properties of these two classes of hormones. Which of the following is true regarding these differences?
A) Steroid hormones need to be taken intravenously, whereas peptide hormones are readily taken orally. B) Steroid hormones affect posttranslational changes, whereas peptide hormones only affect transcription. C) Steroids begin their action with the production of second
messengers, whereas peptide hormones begin their action when bound to their receptor.
D) Steroids activate their target intracellularly, whereas peptide hormones activate their target extracellularly.
THE NERVOUS AND ENDOCRINE SYSTEMS
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MCAT COMPLETE
4. Which of the following steroids would be best used for a cancer patient whose weight has significantly decreased? A) Oxyandrolone
B) Oxymetholone C) Testosterone D) Fluoxymesterone
5. Which of the following is/are medicinal use(s) of anabolic-androgenic steroids?
I. Induction of puberty
II. Preventing lean muscle mass loss III. Acne treatment
A) I only B) II only C) I and II only D) I, II, and III
6. Anabolic steroids are used by some athletes to increase muscle mass. Which of the following is the best explanation for this effect?
A) They only act to increase the number of muscle fibrils. B) They increase red blood cell count.
C) They increase protein metabolism in muscles and also increase the size and number of muscle cells.
| 25
THE NERVOUS AND ENDOCRINE SYSTEMS
SOLUTIONS TO CHAPTER 12 PRACTICE PASSAGE
1. B Marathon runners would want to use these steroids to enhance endurance via an increased red blood cell count, which is also why anemic patients would take these steroids (choice B is correct). The passage states that these steroids are easily detected (choice A is wrong), and that the type of steroid used medicinally depends on whether the anabolic or androgenic effects are required (choice C is wrong). It is not indicated that the androgenic side effects only occur in men; anabolic-androgenic steroid use can cause typically-male characteristics, such as growth of body hair, to be evident in fetypically-males (choice D is wrong).
2. C Naturally occurring anabolic-androgenic steroids, including testosterone, are products of the hypotha-lamic-pituitary axis. The secretion of gonadotropin releasing hormone from the hypothalamus stimulates the release of LH and FSH from the anterior pituitary, which in turn stimulate testosterone production. A build-up of hormones with similar chemical structures to testosterone will induce negative feedback on the hypothalamic-pituitary axis (choice A is wrong). Negative feedback tends to inhibit the release of hormones at the target organs (choice B is wrong). In the case of anabolic-androgenic steroids, the target organ is the anterior pituitary gland, not the posterior pituitary (choice D is wrong and choice C is correct).
3. D Because they are hydrophobic, steroid hormones can penetrate the cellular membrane to bind to intracel-lular receptors. In contrast, peptide hormones typically bind to receptors on the cell surface and produce second messengers (choice D is correct and choice C is wrong). Also due to their hydrophobicity, steroids can be taken orally because they will readily cross the gastrointestinal membrane. Peptide hormones can be broken down by gastrointestinal peptidases and must be injected (choice A is wrong). Steroid hormones act as gene expression regulators (i.e., they affect transcription) when bound to their receptor, whereas peptides modify existing enzymes (choice B is wrong).
4. A Because the medicinal use would be to increase body weight, a higher anabolic effect versus androgenic ef-fect would be desired. The steroid with the smallest androgenic-anabolic ratio is oxyandrolone (choice A is correct). Oxymetholone and fluoxymesterone have intermediate ratios and should be eliminated (choices B and D are wrong). Testosterone has the highest androgenic-anabolic ratio, which is opposite of the desired effect in this case (choice C is wrong).
5. C Items I and II are true: both induction of puberty and increased lean muscle mass are effects of anabolic-androgenic steroid use, and can be used as medical treatment (choices A and B can be eliminated). Ana-bolic-androgenic steroids can cause acne, making Item III false (choice D can be eliminated; choice C is the correct answer).
6. C The passage states that the anabolic effects of steroids can lead to increased protein synthesis as well as an increase in the size and number of muscle cells (choice C is correct); this confirms that their effects go beyond only increasing the number of muscle fibrils (choice A is wrong). Although endurance athletes benefit from an increased red blood cell count, this does not increase muscle mass (choice B is a true state-ment but does not offer an explanation for increased muscle mass and can be eliminated). The passage states that steroid use decreases testosterone production (choice D is wrong).
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MCAT COMPLETE
CHAPTER 13 PRACTICE PASSAGE
Primary immunodeficiency disorders (PIDs) are categorized by the branch of the specific immune system impacted; thus they can be cell-mediated, humoral, or combined. If a PID is characterized by defects in both B and T cells, it is considered to be a severe combined immunodeficiency disorder (SCID). Patients with this disorder are prone to infections which often spread to the blood with the first serious illness, typically occurring before one year of age.
In order to understand the mechanism of the disease and develop more effective means of treatment, a non-human mammalian model was generated. The SCID mouse lacks the ability to make B or T cells and thus effectively parallels the disorder seen in humans. Because of a recessive mutation on chromosome 16, these mice lack a DNA repair enzyme necessary for the recombination that occurs in the development of cells of the immune system. This animal model has provided a wealth of information about the function of the immune system in general, as well as serving as an important disease model.
Once a mouse model lacking an immune system was established, the possibility of engineering a mouse that contained a human immune system was considered. Such an animal would provide an in vivo model of a human system without utilizing human subjects. This led to the development of the hu-SCID chimera in which human fetal tissue, specifically that of the thymus, was placed in a SCID mouse and then followed by the introduction of human bone marrow stem cells. The result was a mouse lacking its inherent immune system but expressing functional human immune system cells, producing an excellent model in which to study immune system dynamics and disease responses.
Immature Human Immune Cells Immature Human Immune Tissue Mouse Kidneys
Immuno-incompetent SCID Mouse
Figure 1 Representation of cellular components necessary to produce hu-SCID chimera.
1. How could a cell-mediated PID contribute to impaired B cell function?
A) B cells would not receive stimulatory signals from helper T cells and would not be triggered to proliferate.
B) Macrophages would be unable to present antigen to B cells in order to induce appropriate antibody production. C) Abnormally growing cells would fail to display
intracellular peptides and would go unrecognized by B cells.
D) The complement cascade would not be induced so B cells would lack a necessary signal for proliferation. 2. Which of the following takes place in the lymph nodes?
A) Differentiation of B cells
B) Release of circulating antibodies by natural killer cells C) Nonspecific phagocytosis by macrophages
D) Removal of self-reacting B cells
3. In the creation of the hu-SCID chimera, why is it necessary to place human thymus within the mouse?
A) The human thymus ablates any remaining immune system cells within the SCID animal.
B) The human thymus provides a source of lysozyme and thus bolsters innate immune protection.
C) The human thymus acts as the source of self antigens, eliminating auto-reactive cells from those produced by the bone marrow progenitors.
D) The human thymus prepares the mouse’s body to accept the introduction of other tissues and prevents graft rejection.
4. All of the following statements about the SCID or hu-SCID mice are true EXCEPT:
A) by utilizing human bone marrow from a wide variety of sources when creating the hu-SCID chimera, different human autoimmune diseases could be introduced into the SCID mouse.
B) helper T cells of the hu-SCID chimera are not able to be infected by HIV.
C) SCID mice have an extremely low rate of organ transplant rejection.
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| 27 5. Which individual is most likely to present with an immune
system profile like that of a person with SCID? A) A 54-year old man who underwent a thymectomy as an
infant
B) A 3-year old girl undergoing extensive radiation therapy as part of cancer treatment
C) A 27-year old woman positive for the human immunodeficiency virus (HIV)
D) A 41-year old man with Type I diabetes
6. What would be the immune system profile of the offspring of a pair of hu-SCID chimera mice?
A) Offspring would have the genetic profile for SCID and lack any human immune system cells.
B) Offspring would have the genetic profile for SCID and carry some human immune system cells.
C) Offspring might have the genetic profile for SCID and would lack any human immune system cells.
D) Offspring might have the genetic profile for SCID and would carry some human immune system cells.
THE CIRCULATORY, RESPIRATORY, LYMPHATIC,
MCAT COMPLETE
SOLUTIONS TO CHAPTER 13 PRACTICE PASSAGE
1. A The answer needs to link the function of T cells (the cell-mediated arm of the immune system) to that of B cells. B cells receive co-stimulation from helper T cells and the absence of that signal would impair B cell proliferation and function (choice A is the best answer). While macrophages do present antigen, they present to helper T cells, not B cells (choice B is wrong). Abnormally growing cells are targeted by killer T cells, but this does not involve B cells (choice C is wrong). The complement cascade is not required for B cell proliferation (choice D is wrong).
2. C The lymph nodes serve as a site for nonspecific filtration of the lymph by macrophages as well as a site for storage and activation of B and T cells (choice C is correct). B cell differentiation and screening for self-reacting cells takes place in the bone marrow (choices A and D are wrong). Natural killer cells are part of the innate immune system and do not release antibodies (choice B is wrong).
3. C A primary role for the thymus in the immune system is to present self antigens to developing T cells and remove those that react to self antigens. This helps prevent autoimmune responses (choice C is the best answer). The original SCID mouse is better able to avoid graft rejection due to its non-functional im-mune system (choice D is eliminated) and there is no evidence to support the idea that the thymus will kill off native immune system cells within the mouse (choice A is eliminated) or provide the enzyme lyso-zyme (choice B is eliminated).
4. B Since the helper T cells in the hu-SCID chimera come from humans in the form of a bone marrow trans-plant, and since human helper T cells are capable of being infected with HIV, then the helper T cells of the hu-SCID chimera should be able to be infected by HIV (choice B is not true and is the correct answer). One of the nice things about the SCID mouse is that it can be “custom designed” by using dif-ferent human bone marrow donors. If the bone marrow from a normal (non-autoimmune) individual is used, then the hu-SCID chimera would have a normal human immune system, but if the bone marrow from an individual with an autoimmune disease is used, then the hu-SCID chimera could also display that disease (choice A is true and can be eliminated). Since SCID mice lack an functional immune system they are unable to reject transplanted organs (choice C is true and can be eliminated). Finally, the comple-ment system can be activated by antibodies (among other ways of activation), and since the SCID mouse is unable to produce antibodies, it is likely that the SCID mouse would have some degree of compromised complement activation (choice D is true and can be eliminated).
5. B Since the question is looking for a condition most analogous to SCID, the correct answer needs to rep-resent the individual with deficits in both cell-mediated an humoral immunity. Someone who had their thymus removed as a child would be expected to have deficits in cell-mediated immunity to the lack of T-cell education and the potential for more auto-reactive T T-cells, but would not have the full scale failure of SCID (choice A can be eliminated). An HIV patient will experience deficits due to infection of T helper cells. However, there is no additional information to describe the state of the patient’s humoral immunity (choice C can be eliminated). An adult with Type I diabetes has likely been living with the disorder for much of his or her life. Diabetes leads to high blood sugar (and a number of other effects such as kidney failure and poor wound healing), but this is not the same as having a non-functional immune system (choice D can be eliminated). Choice B represents the most profound lack of immune system cells in the most vulnerable patient; radiation will kill the progenitor cells that would normally be creating the im-mune system of a young child.
| 29 6. A This is a two-by-two question, where two decisions must be made to get to the correct answer. Since the
mutation that leads to the SCID condition is caused by a recessive allele, SCID mice must be homozy-gous recessive for the mutation. Thus, if both parents have the SCID phenotype, all offspring will also be SCID (choices C and D can be eliminated). The human cells that are placed in the chimeras do not be-come part of the germ cell line and therefore will not be passed on (choice B can be eliminated and choice A is the best answer).
THE CIRCULATORY, RESPIRATORY, LYMPHATIC,
AND IMMUNE SYSTEMS
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MCAT COMPLETE
CHAPTER 14 PRACTICE PASSAGE
Following mastication, food is shifted by the tongue to the oropharynx where it is swallowed as a bolus and enters the esophagus. Here, the primary peristaltic wave begins by contracting above the bolus and relaxing beneath it in a unidirectional wave, contracting for 8-10 seconds and propelling the food toward the stomach (causing the relaxation of the upper and lower esophageal sphincters). If the entirety of the food does not reach the stomach in this primary wave, secondary peristaltic waves continue where the bolus distends the esophagus and drives the food toward the stomach. The smooth muscle contraction seen during peristalsis in the esophagus occurs via a similar Ca2+-dependent mechanism
as seen in skeletal muscle, with a few notable exceptions. In place of troponin, a Ca2+-dependent protein (calmodulin)
activates myosin light-chain kinase. Phosphorylation leads to activation of the myosin heads, which subsequently allows for cross-bridge formation. To cease contraction, the phosphate is removed by myosin light-chain phosphatase.
Unexplained chest pain seen in patients is often attributed to esophageal spasm. While very limited information is available to characterize a diffuse esophageal spasm (DES), one study has shown that it may be linked to malfunction of endogenous nitric oxide synthesis. This was supported by the improved function of peristalsis and reduction of chest pain in patients treated with glycerin trinitrate (nitroglycerine).
1. Glycerin trinitrate likely forms which of the following in the body to treat diffuse esophageal spasm?
A) Glyceryl trinitrate B) NO
C) NO2 D) NO3−
2. A researcher creates an inducible knockout mouse designed for limiting levels of calmodulin expression in the digestive tract. Which of the following would be a likely effect in the digestive tract due to calmodulin knockout?
A) Diarrhea
B) Increased myosin light-chain phosphorylation C) Dysphagia (painful/difficulty swallowing) D) Increased cross-bridging
3. Researchers have characterized several compounds affecting the activity of esophageal smooth muscle activity. According to the figure, which of the following would lead to smooth muscle contraction in the esophagus? Rostral DMN Caudal Vagus nerve Preganglionic neuron ACh (+) Postganglionic neuron NO/VIP/ATP (-) Smooth muscle (+) ACh (+) ACh/SP
A) Increased acetylcholinesterase release
B) Destruction of the rostral potion of the dorsal motor nucleus (DMN)
C) Addition of a Substance P (SP) agonist D) Increased vasoactive intestinal peptide (VIP)
4. Researchers have shown that peristalsis in the esophagus is preceded by a wave of smooth muscle relaxation (or inhibition of contraction) known as deglutitive inhibition.
What is a possible reason for this wave of inhibition? A) To eat large quantities of food in one sitting
B) To prevent potential esophageal damage associated with overuse
C) To allow for the normal drinking of a glass of water D) To allow for increased food storage capacity in the
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| 31 5. A medical student on her internal medicine clerkship
quickly eats a roll from the cafeteria but, after taking a very large final bite, feels as though she cannot move it down her entire esophagus. She is not having any difficulty breathing but is experiencing sharp pains in the center of her chest every few seconds. The pain increases for several minutes and then suddenly ends. What is the likely source of pain and why is she having difficulty swallowing the roll?
A) Primary peristaltic wave, increased pyloric sphincter tone B) Primary peristaltic wave, increased cardiac sphincter
tone
C) Secondary peristaltic wave, increased pyloric sphincter tone
D) Secondary peristaltic wave, increased cardiac sphincter tone
6. Activation of the sympathetic autonomic nervous system would have what effect on esophageal peristalsis? A) No effect
B) Increased contractile rate C) Decreased contractile rate D) Decreased contractile force
