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The Medical College Admission Test (MCAT)

Biological Sciences

C. Thyroid stimulating hormone (TSH) produced by the pituitary gland stimulates the thyroid gland to produce its hormones T₃ (triiodothyronine) and T₄ (tetraiodothyronine or thyroxine). TSH modulates the iodide trapping mechanisms; hypersecretion results in goiter and exophthalmos, whereas hyposecretion leads to diminished thyroid function and lethargy.
B. See explanation for question 143.
C. When thyroxin is administered, one would expect that the iodide uptake would decrease; in graph I it stayed level, whereas in graph III an increase of uptake is exhibited. Both experimental conditions indicate some evidence of thyroid malfunction.
B. A hyperthyroid individual would not be expected to show additional uptake because the gland already is working at an elevated level. When thyroid hormone was administered as in our experiments, normally the uptake should have decreased; it stayed the same however.
A. When thyroxin is administered, TSH production diminishes and RAI uptake should drop. This did not occur in graph I while graph II demonstrates the expected normal. Graph III shows an increase in uptake indicating that the pituitary is not responding to the negative inhibitory feedback that thyroxin elicits; TSH production should decrease at the time of thyroxine administration.
B. Acromegaly is a result of pituitary over-secretion of growth hormone. Lack of iodine will result in goiter development of the thyroid gland. Rickets is due to vitamin D deficiency. A skin rash is not a specific lesion that can be associated with only one specific cause as the others listed.
B. Acidophils produce somatotropic hormone (STH) and luteotropic hormone (LTH) or prolactin. Beta cells produce thyroid stimulating hormone (TSH), adrenocorticotropic hormone (ACTH), and melanocyte stimulating hormone (MSH). Delta cells produce luteinizing hormone (LH) (called interstitial cell stimulating hormone in the male), and follicle stimulating hormone (FSH). Chromophobes are considered resting cells.
D. All statements are correct. The thyroid originates from the foramen cecum region of the tongue. Its structural unit is the follicle, a unit of epithelial cells that surround a colloid space. Colloid is located extracellularly and contains thyroglobulin. T₃ and T₄ are the active thyroid principles and are released into the bloodstream and carried on proteins to the tissues.
B. Acromegaly and (or) giantism is due to over-activity of the alpha cells of the pituitary, which secrete growth hormone. If a person is affected before puberty, he or she will develop into a fairly well-proportioned giant. After maturity, an increase in the size of the hands and feet and massive development of the bones comprising the face are consequences. In the adult, strictly speaking, the term acromegaly must be applied to this condition.
C.

If aortic pressure is unchanged and end diastolic volume increases, the stroke volume or the difference between end diastolic volume and end systolic volume is larger. Ventricular work is equal to:
Work = Pressure • Stroke Volume = Force/Area • Volume = Force • Length
or the area under the curve. The kinetic energy of the ejected blood can be ignored because it represents ≈5% of the total energy and stays constant under most conditions. The area subtended by the pressure-volume loop increases as does the work in this case. Because the contractility of the heart represented by the upper curve remains constant, the end systolic volume also remains constant.
D.

Whereas the pressure to eject blood must increase, the volume of blood the ventricle ejects is less. Therefore, the ventricular work depends on the exact nature of the two curves. Because the ventricle cannot maintain the required pressure at low volumes, the total volume of blood ejected is less and the end systolic volume is increased.
A.

In this case, the ventricle is incapable of generating the same pressure at low volumes. Therefore, the volume of blood ejected is less (i.e., the stroke volume is smaller), and the end systolic volume is correspondingly larger. Note that this question is a play on words. If the stroke volume decreases with a constant end diastolic volume, the volume left in the ventricle after contraction will be more. The ventricular work is reduced because the pressure remains constant while the volume of blood ejected falls so that the product of the two decreases.
C.

The key to this question is to remember which two cases are being compared. The increase in end diastolic volume allows the failing ventricle to increase the stroke volume as compared to the uncompensated case. With a constant pressure, an increase in stroke volume causes an increase in the work (i.e., pressure • stroke volume). Because the aortic pressure is constant and the active pressure curve is unchanged in the two cases, the volume remaining in the ventricle at the point at which the ventricle stops ejecting blood will remain constant.
C.

The key to this question, as in question 155, is that one must remember what conditions are being compared. Increasing the end diastolic volume sufficiently could theoretically allow a failing ventricle to eject a larger stroke volume than is the case in normals. However, the compensation for a failing ventricle usually is not sufficient for this to occur under physiological conditions. Because the stroke volume is indeterminate relative to the normal condition, the ventricular work is also indeterminate given a constant mean aortic pressure. Because by definition a failing ventricle cannot maintain the same pressure at low volumes, the end systolic volume must increase in comparison to that of normal.
B. The definitions of isotonic, hypotonic, and hypertonic need to be recalled. Remember that the structure placed in a medium is surrounded by a semipermeable membrane; the structure is permeable to small particles (e.g., certain inorganic ions and water), but not to large particles (fat and protein molecules). An isotonic environment exerts the same osmotic pull as the medium does on the other side of a semipermeable membrane. It consequently has to possess the same concentration of particles and, therefore, the net gain or loss of water during osmosis is zero. If the red blood cells were placed in a hypotonic solution, hemolysis would occur because a hypotonic solution would exert a lesser osmotic pull than the medium on the other side of a semipermeable membrane (the inside of the RBC); the medium would possess a smaller concentration of particles and consequently would lose water during osmosis. In our case, the red blood cells were hypertonic in relation to the medium and exerted a larger osmotic pull than the medium on the other side of the semipermeable membrane because it possessed a greater concentration of particles and consequently accumulated water during osmosis.
D. In this experiment, urea and glycerol did enter the red blood cell and water consequently followed because it was in greater quantity outside and tried to establish an equilibrium.
D. Cell membranes are described as semipermeable because they allow some materials to pass while they block others. All materials that pass membranes must be either a liquid or dissolved in a liquid. Molecules also must be fairly simple, but molecular size is not the limiting factor because amino acids pass less readily than do many smaller molecules. Osmosis is the passage of liquids through a membrane; usually water through the semipermeable membrane. Diffusion is the passage of molecules from a more concentrated environment to a less concentrated region; equalization is usually the end result.
D. Active transport involves the diffusion of molecules against a gradient; this is an energy consuming phenomenon. The process of moving substances from an area of lower concentration to where they are in a higher concentration is selective, and requires respiration. Osmosis and diffusion are described in the explanation for question 159. Turgor pressure involves the passage of water into a cell at a faster rate than it can leave; the cell becomes plump and filled (turgid). At times the force results in bursting the structures. Plasmolysis is defined as a shrinking of protoplasm due to the loss of water from a cell.
A. Layers of material (probably mucopolysaccharide) secreted by the cell are found on the surface of the cell. The most prominent layer is the basement membrane, or basal lamina. These structures are boundaries and must be traversed by material entering and leaving the cell. Cells must be held together; adjacent cell membranes interdigitate and intercellular cement is utilized. A desmosome is a specialized area of connection between adjacent cellular membranes (macula adherens). A terminal bar is a dense area surrounding the apical cellular surface. It includes the tight junction (zona occludens) and the loose junction (zona adherens). In cardiac muscle, several cardiac muscle cells join end to end at a specialized junctional zone known as an intercalated disc.
B. Some of the proteins embedded in the lipid layers of membranes are shaped to form channels with “gates,” which open only to certain materials or under certain conditions.
A. Osmosis is a process in which solvent passes from an area of lower solute concentration to an area of higher solute concentration. In dialysis solvent and solute both pass through the membrane.
A. The defect cannot be dominant because individual 5 was produced by two normal parents who would have no defective alleles if the defect is dominant (either autosomal or X-linked). The defect is unlikely to be X-linked recessive because unless a new mutation occurred, individual 5 should have received a normal allele for the gene on her X from her father (individual 2) and therefore could not express a recessive X-linked defect. Therefore the most likely possibility (using the assumptions of a single controlling gene and very rare mutations) would be an autosomal gene for which the defective allele is recessive to the normal allele.
C. Unless a new mutation occurred, individual 1 would have to have one normal allele (which produced the normal phenotype) and one defective allele (which was contributed to individual 5 along with a defective allele from individual 2) to produce individual 5's defective phenotype.
B. Individual 4 would have to contain two defective alleles to produce the defective phenotype.
C. Individual 7 would have to contain at least one normal allele to produce her normal phenotype. Her other allele cannot be determined with certainty from the data given. Because her parents are probably both heterozygotes (see above), she has a 1/3 probability of having two normal alleles and a 2/3 probability of having one normal and one abnormal allele.

Therefore she is most likely to contain one normal and one defective allele.
C. Individual 8 would have to contain one normal allele (to produce his normal phenotype) and one abnormal allele (because he contributed one of the two defective alleles in each of offspring 14 and 18).
B. The upper region consists of the fluid portion of the blood, the plasma.
D. The thin “buffy” layer is the white blood cells and platelets.
C. After centrifugation, the packed volume of formed elements was 5.5 ml. As stated previously, most of the formed elements are RBCs.
B. After centrifugation, the volume of packed formed elements of the patient was 5.5 ml. Because hematocrit is defined as the relative volume of formed elements, the hematocrit of this patient is 55% (i.e., 5.5 ml out of 10 ml, or 55% of the total blood volume, was occupied by formed elements).
B. The hematocrit of this patient, 55%, is higher than the normal average value of 45%.
C. Region C consists of RBCs. These cells contain hemoglobin, which binds and carried oxygen.
B. Region B, the “buffy” coat, consists of WBCs and platelets. The WBCs consist of a number of cell types, some of which are responsible for fighting infections.
B. Region B, the “buffy” coat, consists of WBC and platelets. The platelets are responsible for triggering the cascade of events that results in the formation of blood clots.
A. The major component of blood plasma is water.
A. Because the patient's hematocrit is higher than the normal value, the patient has more formed elements, largely RBCs, than a normal individual. Because RBCs are responsible for binding and carrying oxygen, a higher hematocrit might suggest that the individual has a higher physiological demand for oxygen. Such a condition is seen in individuals living at high altitudes where the oxygen concentration of the air is lower (compared to the oxygen concentration of air at sea level).
D. A sarcomere, the area between two Z bands, is the functional unit of muscle; it is the region between two Z lines and consists of an A band and half of two abutting I bands.
C. According to the sliding filament theory (Huxley) the sarcomere response to excitation involves the sliding of thin and thick myofilaments past one another making and breaking chemical bonds with each other as they go. Neither the thick nor thin myofilaments change in length.
A. If we observed this contraction under the light microscope, we would see the narrowing of the H and I bands during contraction while the width of the A band would remain constant.
C. The immediate energy source for contraction is ATP which can be hydrolyzed by actomyosin to give ADP, P₁, and the energy which is in some way associated with cross-bridge motion. The ultimate source of the ATP is the ATP produced by the intermediary metabolisms of carbohydrates and lipids. As mentioned in previous answers the lengths of the myosin filaments would NOT be affected by ATP.
C. Calcium ions released following an action potential in the fiber membrane and T-tubules bind with troponin. Calcium-troponin binding removes the inhibition of actomyosin formation. The sarcoplasmic reticulum concentrates calcium ions (Ca²⁺) within its lumen, but depolarization of the T-tubule membrane induces the nearly terminal cisternae of the sarcoplasmic reticulum to release this Ca²⁺ into the sarcoplasm among the myofilaments. The Ca²⁺ becomes associated with the troponin of the thin myofilament, bringing about contraction.
A. As shown in Figure 1, the percent of neutrophils bound increased from about 20% to about 50% as GMP-140 concentration increased. Tightness of binding, (C) and (D), was not measured, although it could contribute to the increased number of attached cells.
B. Figure 1 indicates that about ¼ of monocytes attached at a GMP-140 concentration of zero.
C. According to Figure 1, coating with GMP-140 caused little change in lymphocyte attachment, increased neutrophil attachment from about 20% up to about 50%, and increased monocyte attachment from 25% to about 75%.
A. Because GMP-140 coating of dishes increased neutrophil attachment (Figure 1), the neutrophils clearly attach to GMP-140. Pretreatment of neutrophils with GMP-140 (Figure 3) decreased attachment of the neutrophils to cultured endothelial cells, indicating that the GMP-140 binding sites on neutrophils are involved in endothelial attachment.
D. Treatment of neutrophils with antibody against CD-18 had no effect on attachment of neutrophils to GMP-140-coated dishes, so CD-18 on neutrophils is apparently not involved.
A. Neutrophils and macrophages (derived from monocytes) are most important in phagocytosis and if these cells could not easily leave vessels in inflammation sites, phagocytosis of bacteria should be impaired. Lymphocytes would not be affected because they do not appear to depend on GMP-140, so natural killer activity (due to large granular lymphocytes), cell-mediated responses (using helper T lymphocytes and cytotoxic T lymphocytes), and humoral responses (using helper T lymphocytes and B lymphocytes) should not be heavily affected. Some reduction in cell-mediated and humoral responses could occur due to reduction in macrophages available to serve as antigen-presenting cells and/or cytokine sources.
B. Enzymes are generally proteins in nature, thus largely polymers of amino acids. Some other material may be present as well. The recent recognition of a small number of nucleic acids (ribozymes) that act as enzymes does not invalidate the question.
C. Enzymes function in reactions by decreasing the energy of activation. The net energy yield (positive or negative) is not affected, and a thermodynamically unfavorable reaction will not proceed simply by addition of an enzyme.
D. Enzymes are considered not to be used up in reaction. Incidental loss or destruction of enzyme does not require replacement however.
A. Covalent modification would not include changes of ions. Phosphorylation/dephosphorylation is often used as a means of regulation of activity of various enzymes. Sulfation/desulfation is seldom if ever employed.
B. The response of most enzymes to being boiled in aqueous solution is substantially decreased activity as a result of enzyme denaturation. (Remember that enzymes are proteins in general.) More moderate increase in temperature increases the activity of an enzyme.
B. The number of possible isomers is four. Note that positions 3 and 4 are chiral and thus may have the hydroxyl group on either side. Thus 2 × 2 = 4.
B. Although there appear at first glance to be three chiral centers, this is not so; the number 3 carbon is not attached to four nonidentical groups. Thus only four isomers appear to be possible, but it can be seen that two of these are identical and thus there are two optically active isomers and one optically inactive isomer.
B. There are two chiral centers.
B. Ribose is found in RNA, contributing to the name, ribonucleic acid. Deoxyribose is found in DNA.
C. As seen in the explanation for question 196, there are only two chiral centers (carbons 2 and 4).
B. DNA is the primary source of information. It is usually considered that DNA directs the synthesis of RNA and that RNA directs the synthesis of protein. DNA also directs its own replication. In some cases (for example the retroviruses), RNA can direct the synthesis of DNA.
C. RNA directs the synthesis of protein (see above) and three nucleotide bases are required to code for each amino acid.
B. Paired strands of DNA are found in the chromosomes of higher animals.
A. None of the cells will have only DNA containing deuterium. Each cell will have two strands of DNA for each chromosome, and one strand of each pair will contain deuterium.
D. As stated above, DNA is required to code for RNA and RNA is required to code for protein. The enzymes that actually carry out the biosynthesis are protein. (It might be argued that DNA is not required, but ultimately DNA is required for RNA synthesis.)
A. This study investigated the response of anti-body produced by a host against the protein of the virus inducing tumor. Statements C and D are contradicted by the evidence. There is no information to either confirm or deny statement B, however, it is clear that Brown Norway rats produced antibody at both time periods and, therefore, statement A is supported by the data presented.
C. From the experimental data, one can conclude that prolactin and LH are both produced by the pituitary, and that the hypothalamus stimulates LH release, but inhibits prolactin release, and that ovarian hormones inhibit LH release, but not prolactin release. Lactogenic hormone or luteotrophic hormone (LTH) or prolactin is secreted by the acidophils (alpha cells) of the pituitary. This hormone: (1) promotes growth of the breast, which has been stimulated already by estrogen and progesterone; (2) promotes and maintains lactation; (3) helps in maintenance of the corpus luteum; and (4) promotes maternal instinct.
C. The experiment was performed to determine the effect of salinity concentrations on the secretory proteins of the urophysis. The data shows that protein 2 was affected more (twice) by the increasing concentration of seawater than protein 1.
D. Careful analysis of the data shows that in the system under investigation, spleen cells and mast cells together are required for antibody production, that spleen cells are probably found in the thymus and that X-irradiation is an important tool in immunological work. Statement C, implying that spleen cells or mast cells are essential to the antibody production against pollen is contradicted by the data because bone marrow and pollen and thymocytes and pollen show a response. Statement D is not supported or contradicted by the experiment, even though mast cells are known to produce histamine and their activity is controlled by certain nasal sprays limiting their effects after exposure to certain substances.
C. Beta cells of the pituitary produce thyroid stimulating hormone, which stimulates the thyroid to produce its hormones T₃ (triiodothyronine) and T₄ (tetraiodothyronine or thyroxin). Iodides consumed in food and water are absorbed and carried to the iodide pool in the extracellular fluid via the circulatory system. Five events are involved in hormone production: (1) trapping iodide; (2) oxidation of iodide to organic iodine; (3) synthesis of hormone; (4) storage as the thyroglobulin moiety; and (5) release of the active principle into the circulation. TSH greatly influences the trapping mechanism; thiocyanates block trapping, whereas thiouracil blocks the oxidation and synthetic steps. The data show that Group 4 received thyroxine, which inhibited TSH production, whereas Group 2 was treated with thicoyanate, which affected the iodine pump. Group 3 received propylthiouracil, an inhibitor of iodine oxidation; in Group 2 iodine is not taken up and no radioactivty can be measured.
B. The five-year total of wheat was 49; for corn it was 56. The ratio is thus 7/8.
C. Scanning of the graphs comparing the number of daylight hours clearly shows that only in greenhouses 2 and 3 are there provided 15 hours of light per day for several weeks, which is necessary for Plant X. If the question would have been that Plant Y will only bear fruit when the period of darkness is 8 hours or less, it could not have born fruit in greenhouse 1. However, if Plant Z will only form flowers and seeds when exposed to days of no more than 15 hours in length, it certainly could not have been native in areas simulated by greenhouses 2 and 3.
A. Blood is composed of cells and plasma. Plasma constitutes about 55% of blood volume, and cellular elements about 45%. The normal red blood cell hematocrit is about 36–45%. Hemoglobin, a complex molecule of iron and protein, is a key element of the RBC; the RBC carries oxygen from the lungs to the tissues and transports carbon dioxide from the tissues to the lungs. In the break-down of hemoglobin, bilirubin is excreted and iron is retained. The purpose of the experiment was to test the effect of different storage conditions on hemoglobin composition as revealed by gel electrophoresis. Although none of the storage conditions were able to maintain hemoglobin in an absoutely unchanged state, storage at −20°C in glycerol allowed the least change from the fresh configuration.
C. The suppressed plasma levels of FSH and LH during the follicular phase of the menstrual cycle are due to the negative inhibitory effects of estrogen being secreted from developing follicles. The small, abrupt rise in plasma estrogen is believed to trigger the ovulatory LH surge (by suspension of negative or positive feedback). During the luteal phase, estrogen and progesterone again establish the negative-feedback suppression and luteal failure occurs on day 26. The higher levels of bound LH in the treated animals indicate that these animals had lower LH levels; the drug seems to inhibit LH release.
B. Examination of the data shows that all three proteins were absorbed at the same rate (2 minutes) by the visceral endoderm; the visceral endoderm appeared to block the transport of protein Y, whereas protein X appeared to be transported through it. The statement that the visceral basement membrane blocked the transport of protein X is not supported by the data because it is found at the next level, namely, the vitelline endothelium. The vitelline endothelium does, however, block protein X from entering the fetal blood.
B. The fiber components of connective tissue add strength and support. They are collagenous, elastic, and reticular in nature; collagen fibers are the most numerous fiber type. The graphs show that labeled proline starts to appear in the collagen in under 2 hours, so B was supported by the evidence. Statements A and C were contradicted by the information given, and D is neither supported nor contradicted.
B. Glycine has no chiral center, but other amino acids in higher animals are ordinarily of the L-series.
C. Carbohydrate monosaccharides found in higher animals are generally of the D-series.
B. Epimers differ in configuration at only one carbon. Enantiomers are nonsuperimposable mirror images. Diastereomers are nonsuperimposable nonmirror images.
C. The number of kcal/g for the classes of food material are: carbohydrates, 4; proteins, 4; and fats, 9. The nucleic acids are not considered as a class of food.

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