Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 541:

  Early experimentation on the single-celled organism Acetabularia led to important discoveries about the role of the nucleus in regulating cell function. Acetabularia is an enormous single cell with three distinct regions: a cap, a root-like rhizoid,

  and a stalk which connects the two. The following experiments were conducted to study the development of the cell:

  Experiment 1

  The stalk of an Acetabularia was cut, fragmenting the cell. The fragment which included the cap died shortly afterwards while the fragment containing the rhizoid regenerated to form a complete Acetabularia.

  Experiment 2

  The nucleus from Acetabularia mediterranea, which has a flat cap, was transplanted into Acetabularia crenulata, which has a tufted cap, following removal of the Acetabularia crenulata nucleus. The Acetabularia crenulata cap eventually

  assumed the flat shape.

  Experiment 3

  The nucleus of Acetabularia mediterranea was removed from the young cell before it first formed a cap. A normal cap formed several weeks later. The cell proved to be inviable and died shortly thereafter.

  Experiment 4

  A young Acetabularia was fractioned into a number of portions before it first formed a cap. Several weeks later, both the portion containing the nucleus and the portion containing the apical tip of the stalk formed caps. The other portions did

  not form caps.

  The differences in cap structure between Acetabularia mediterranea and Acetabularia crenulata are caused by differences in:

  A. genotype.

  B. phenotype.

  C. genus.

  D. phylum.

  Correct Answer: A

  The genotype consists of the genes (DNA) which code for specific proteins. Differences in these proteins lead to a difference in the appearance of an organism. Observable differences in the structure of the organism constitute differences in the phenotype. In this case, the genes coding for different cap structures are part of the cell's genotype. The manifestation of these genes, a flat or a tufted cap, is part of the cell's phenotype. Choice B is incorrect because the difference in phenotype does not cause the difference in cap structure, it is the difference in cap structure. Choice C is incorrect because Acetabularia mediterranea and Acetabularia crenulata do not differ in genus. Their genus is Acetabularia. Their species are mediterranea and crenulata. Choice D is incorrect because Acetabularia mediterranea and Acetabularia crenulata do not differ in their phylum, only in their species. Organisms in the same genus must be in the same phylum (and kingdom, class, etc...).

• Question 542:

  Early experimentation on the single-celled organism Acetabularia led to important discoveries about the role of the nucleus in regulating cell function. Acetabularia is an enormous single cell with three distinct regions: a cap, a root-like rhizoid,

  and a stalk which connects the two. The following experiments were conducted to study the development of the cell:

  Experiment 1

  The stalk of an Acetabularia was cut, fragmenting the cell. The fragment which included the cap died shortly afterwards while the fragment containing the rhizoid regenerated to form a complete Acetabularia.

  Experiment 2

  The nucleus from Acetabularia mediterranea, which has a flat cap, was transplanted into Acetabularia crenulata, which has a tufted cap, following removal of the Acetabularia crenulata nucleus. The Acetabularia crenulata cap eventually

  assumed the flat shape.

  Experiment 3

  The nucleus of Acetabularia mediterranea was removed from the young cell before it first formed a cap. A normal cap formed several weeks later. The cell proved to be inviable and died shortly thereafter.

  Experiment 4

  A young Acetabularia was fractioned into a number of portions before it first formed a cap. Several weeks later, both the portion containing the nucleus and the portion containing the apical tip of the stalk formed caps. The other portions did

  not form caps.

  It can be inferred from the experiments in the passage that development of the cap in Acetabularia is regulated by which of the following mechanisms?

  A. Transcriptional regulation because mRNA in the cytoplasm lies dormant for several weeks before cap formation occurs.

  B. Translational regulation because mRNA is not produced by the nucleus until it is required for cap production.

  C. Translational regulation because mRNA in the cytoplasm lies dormant for several weeks before cap formation occurs.

  D. Neither transcriptional regulation nor translational regulation.

  Correct Answer: C

  Translational regulation occurs when the translation of RNA to protein is a regulated step. In Acetabularia, the mRNA is transcribed and stored in the cap for a number of weeks before being translated. This can be concluded from Experiment 3 (in which the cap is produced a number of weeks after the nucleus is removed) and from Experiment 4 (in which the apical stalk region produces a cap a number of weeks following fragmentation of the cell). Choice A is incorrect because transcriptional regulation occurs when transcription of RNA from DNA is regulated. Choice A indicates incorrectly that transcriptional regulation occurs when mRNA lies dormant. This is translational regulation. Choice B is incorrect because translational regulation occurs when RNA translation to protein is regulated. Choice B indicates incorrectly that translational regulation occurs when mRNA transcription from DNA is regulated. Choice D is incorrect because translational regulation is occurring in Acetabularia.

• Question 543:

  Early experimentation on the single-celled organism Acetabularia led to important discoveries about the role of the nucleus in regulating cell function. Acetabularia is an enormous single cell with three distinct regions: a cap, a root-like rhizoid, and a stalk which connects the two. The following experiments were conducted to study the development of the cell:

  Experiment 1

  The stalk of an Acetabularia was cut, fragmenting the cell. The fragment which included the cap died shortly afterwards while the fragment containing the rhizoid regenerated to form a complete Acetabularia.

  Experiment 2

  The nucleus from Acetabularia mediterranea, which has a flat cap, was transplanted into Acetabularia crenulata, which has a tufted cap, following removal of the Acetabularia crenulata nucleus. The Acetabularia crenulata cap eventually assumed the flat shape.

  Experiment 3

  The nucleus of Acetabularia mediterranea was removed from the young cell before it first formed a cap. A normal cap formed several weeks later. The cell proved to be inviable and died shortly thereafter.

  Experiment 4

  A young Acetabularia was fractioned into a number of portions before it first formed a cap. Several weeks later, both the portion containing the nucleus and the portion containing the apical tip of the stalk formed caps. The other portions did not form caps.

  Acetabularia is a(n):

  A. virus.

  B. prokaryote.

  C. eukaryote.

  D. bacterium.

  Correct Answer: C

  The passage describes Acetabularia as a single cell with a nucleus. Only eukaryotes have nuclei, and other membrane bound organelles.

  Choice A is incorrect because a virus does not have a cell membrane, nucleus, or many of the other components of a cell. A virus is essentially composed of DNA or RNA and a protein coat.

  Choice B is incorrect because prokaryotes lack a nucleus.

  Choice D is incorrect because the term bacteria refers to prokaryotes which, by definition, lack nuclei.

• Question 544:

  Early experimentation on the single-celled organism Acetabularia led to important discoveries about the role of the nucleus in regulating cell function. Acetabularia is an enormous single cell with three distinct regions: a cap, a root-like rhizoid, and a stalk which connects the two. The following experiments were conducted to study the development of the cell:

  Experiment 1

  The stalk of an Acetabularia was cut, fragmenting the cell. The fragment which included the cap died shortly afterwards while the fragment containing the rhizoid regenerated to form a complete Acetabularia.

  Experiment 2

  The nucleus from Acetabularia mediterranea, which has a flat cap, was transplanted into Acetabularia crenulata, which has a tufted cap, following removal of the Acetabularia crenulata nucleus. The Acetabularia crenulata cap eventually assumed the flat shape.

  Experiment 3

  The nucleus of Acetabularia mediterranea was removed from the young cell before it first formed a cap. A normal cap formed several weeks later. The cell proved to be inviable and died shortly thereafter.

  Experiment 4

  A young Acetabularia was fractioned into a number of portions before it first formed a cap. Several weeks later, both the portion containing the nucleus and the portion containing the apical tip of the stalk formed caps. The other portions did not form caps.

  Which of the following conclusions can be logically drawn from the fact that the Acetabularia segment containing the rhizoid regenerated a complete and viable Acetabularia in Experiment 1?

  A. The cell nucleus is located in the rhizoid.

  B. Acetabularia reproduces by budding.

  C. The Acetabularia cap is a vestigial structure.

  D. Cap-coding mRNA is stored in the rhizoid.

  Correct Answer: A

  In order for a cell fragment to survive, it must contain all of the necessary machinery for maintaining life. This requires the production of proteins which are coded for by DNA. In Acetabularia, the region with the rhizoid survives, indicating that the nucleus must be located there. Choice B is incorrect because, while some prokaryotes multiply by budding, eukaryotic cells multiply by mitosis. Also, this information cannot be concluded from the fact that the rhizoid bearing fragment survives. Choice C is incorrect because while there is indication in Experiment 1 that the cap alone is not sufficient for the Acetabularia to survive, there is no indication that the cap is a vestigial (unnecessary) structure. Choice D is incorrect because although the statement may be true, it is not unambiguously proven by the experiment. The experiment demonstrates that cap-coding mRNA is present in the rhizoid region of the fragmented Acetabularia at some point during regeneration. The cap-coding mRNA may have been stored in the rhizoid region when the Acetabularia was cut or it could have been transcribed only after the cell was fragmented.

• Question 545:

  Early experimentation on the single-celled organism Acetabularia led to important discoveries about the role of the nucleus in regulating cell function. Acetabularia is an enormous single cell with three distinct regions: a cap, a root-like rhizoid, and a stalk which connects the two. The following experiments were conducted to study the development of the cell:

  Experiment 1

  The stalk of an Acetabularia was cut, fragmenting the cell. The fragment which included the cap died shortly afterwards while the fragment containing the rhizoid regenerated to form a complete Acetabularia.

  Experiment 2

  The nucleus from Acetabularia mediterranea, which has a flat cap, was transplanted into Acetabularia crenulata, which has a tufted cap, following removal of the Acetabularia crenulata nucleus. The Acetabularia crenulata cap eventually assumed the flat shape.

  Experiment 3

  The nucleus of Acetabularia mediterranea was removed from the young cell before it first formed a cap. A normal cap formed several weeks later. The cell proved to be inviable and died shortly thereafter.

  Experiment 4

  A young Acetabularia was fractioned into a number of portions before it first formed a cap. Several weeks later, both the portion containing the nucleus and the portion containing the apical tip of the stalk formed caps. The other portions did not form caps.

  One explanation for the results of Experiment 4 is that the instructions for forming the cap are stored in the apical tip of the stalk several weeks prior to stalk formation. Which of the following pieces of evidence best supports this explanation?

  A. Isolation of DNA coding for cap-inducing proteins from samples of Acetabularia taken several weeks prior to stalk formation.

  B. Exposure of a young Acetabularia to ribonuclease, which cleaves RNA, blocks cap formation.

  C. Exchange of nuclei between Acetabularia crenulata and Acetabularia mediterranea leads to formation of the cap associated with each nucleus.

  D. mRNA coding for cap-inducing proteins is found to accumulate in the stalk apex.

  Correct Answer: D

  The question stem asks for support of the fact that the information for forming the cap is stored in the apical tip of the stalk. This information is in the form of mRNA which is used as the template for translation of the proteins which coordinate

  cap formation. The other choices may be true but they do not provide evidence for the storage of information in the apical tip.

  Choice A is incorrect because the DNA coding for cap inducing proteins would be present at all times in the Acetabularia genome. This does not provide evidence for the storage of information in the stalk tip. Choice B is incorrect because,

  although exposure of the Acetabularia to ribonuclease and the consequent cleavage of RNA would temporarily block formation of the cap (until more RNA is transcribed), this does not provide evidence for the storage of information in the

  apical stalk segment. Rather, it provides evidence that RNA is the conveyor of information.

  Choice C is incorrect because although the passage describes this exchange of nuclei in Experiment 2, this does not provide evidence for the storage of information in the apical stalk.

• Question 546:

  A cell with a high intracellular K+ concentration, whose plasma membrane is impermeable to K+, is placed in an ATP-rich medium with a low K+ concentration. After several minutes, it is determined that the extracellular concentrations of both K+ and ATP have decreased, while the intracellular K+ concentration has increased. What is the most likely explanation for this phenomenon?

  A. The passively diffused from the medium into the cell.

  B. The entered the cell by way of facilitated transport.

  C. The ATP formed a temporary lipid-soluble complex with the K+, thus enabling the potassium to enter the cell.

  D. The entered the cell by way of active transport.

  Correct Answer: D

  This question requires knowledge of the different mechanisms of transport across a cell membrane and an application of this knowledge to the specific situation described in the question stem. A cell with a high intracellular potassium concentration is placed in a medium with a low potassium concentration and a high ATP concentration. Furthermore, we're told that the cell membrane is impermeable to potassium. If the cell were permeable to potassium, you would expect potassium to flow out of the cell, along its concentration gradient. But, this is clearly not the case. What does happen is that the extracellular concentration of both the potassium and the ATP decreases, while the intracellular concentration of potassium increases. This implies that the potassium moved into the cell, across a membrane that is impermeable to it, and against its concentration gradient. You're asked to choose the most likely explanation for this phenomenon. Choice A says that the potassium passively diffused into the cell, and choice B says that the potassium entered the cell by way of facilitated transport. Since both diffusion and facilitated transport occur ALONG a substance's concentration gradient, not against it, both choices A and B must be incorrect. Furthermore, the potassium could not possibly diffuse across the membrane, because we know that the membrane is impermeable to it. And finally, neither of these choices accounts for the decrease in extracellular ATP concentration. Choice C sounds kind of reasonable, except for two main reasons: First, ATP is not a very lipid-soluble complex with potassium -- to which we're told the membrane is also impermeable -- hereby enabling the potassium to cross the lipid cell membrane and enter the cell. Second, there is no evidence to support the theory that ATP functions as a carrier molecule, shuttling potassium across cell membranes. ATP is the energy currency used by cells. Energy is stored in its high-energy phosphate bonds; and this energy is made available to cells when ATP is hydrolyzed to ADP and AMP. Thus, choice C is also incorrect. Choice D, however, does make sense; it explains both the movement of potassium into the cell against its gradient, as well as the decreased extracellular ATP concentration. Active transport is the movement of a substance against its concentration gradient with the aid of carrier molecules and ENERGY.

• Question 547:

  Destroying the cerebellum of a cat would cause significant impairment of normal:

  A. urine formation.

  B. sense of smell.

  C. coordinated movement.

  D. thermoregulation.

  Correct Answer: C

  To answer this question, you have to be familiar with the functions under the control of the cerebellum. The cerebellum is part of the hindbrain, which is the posterior part of the brain and consists of the pons and the medulla oblongata, in addition to the cerebellum. The cerebellum receives sensory information from the visual and auditory systems, as well as information about the orientation of joints and muscles. In fact, one of the cerebellum's main functions is hand-eye coordination. It also receives information about the motor signals being initiated by the cerebrum. The cerebellum takes all of this information and integrates it to produce balance and unconscious coordinated movement. Damage to the cerebellum could damage any one of these functions. Total destruction of the cerebellum would eliminate all of them. Therefore, choice C is correct because destruction of a cat's cerebellum would seriously impair coordinated movement in the cat. Let's take a look at the other choices. Urine formation, choice A, is the primary function of the kidneys, with a little bit of hormonal regulation to help things out. Choice B, sense of smell, or olfaction, is a function of the cerebrum not the cerebellum. Thermoregulation, choice D, is a function of the hypothalamus, which is a part of the cerebrum.

• Question 548:

  Exocrine secretions of the pancreas:

  A. raise blood glucose levels.

  B. lower blood glucose levels.

  C. regulate metabolic rate.

  D. aid in protein and fat digestion.

  Correct Answer: D

  An exocrine gland is one that excretes its products into tubes or ducts that typically empty onto epithelial tissue, while an endocrine gland is one that releases hormones directly into the bloodstream. The pancreas functions both as an endocrine gland and an exocrine gland. As an endocrine gland it produces and secretes three hormones; insulin, glucagon, and somatostatin. Insulin lowers blood glucose levels by stimulating the uptake of glucose onto tissues, and its subsequent conversion into its storage form. glycogen. So, choice B is wrong. Choice A is incorrect because it's glucagon that raises blood glucose levels by stimulating the conversion of glycogen into glucose. Somatostatin suppresses both insulin and glucagon secretion. Choice C is incorrect because thyroid hormones are involved in the regulation of metabolic rate. As an exocrine gland, the pancreas secretes enzymes that are involved in protein, fat, and carbohydrate digestion; all of its exocrine products are secreted into the small intestine. Pancreatic amylase hydrolyzes starch to maltose; trypsin hydrolyzes peptide bonds and catalyzes that conversion of chymotrypsinogen to chymotrypsin; chymotrypsin and carboxypeptidase also hydrolyze peptide bonds; and finally, lipase hydrolyzes lipids.

• Question 549:

  A certain drug inhibits ribosomal RNA synthesis. Which of the following eukaryotic organelles would be most affected by the administration of this drug?

  

  A. 1

  B. 2

  C. 3

  D. 4

  Correct Answer: A

  This question tests your knowledge of both cell structure and function. You are presented with an unlabeled figure of a cukaryotic cell and asked to determine which of the four numbered structures would be most affected by a drug that inhibits ribosomal RNA synthesis. So basically, you have to know which structure is involved in the synthesis of ribosomal RNA. The nucleolus is the organelle responsible for ribosomal RNA synthesis, and in figure, structure 1 is the nucleolus, and so choice A is correct. Structure 2 is the nucleus; structure 3 is the Golgi apparatus; and structure 4 is a mitochondrion.

• Question 550:

  Compounds containing a hydroxyl group attached to a benzene ring are called phenols. Derivatives of phenols, such as naphthols and phenanthrols, have chemical properties similar to those of phenols, as do most of the many naturally-occurring substituted phenols. Like other alcohols, phenols have higher boiling points than hydrocarbons of similar molecular weight. Like carboxylic acids, phenols are more acidic than their alcohol counterparts. Phenols undergo a number of different reactions; both their hydroxyl groups and their benzene rings are highly reactive. A number of chemical tests can be used to distinguish phenols from alcohols and carboxylic acids.

  

  Thymol, a naturally occurring phenol, is an effective disinfectant that is obtained from thyme oil. Thymol can also be synthesized from m-cresol, as shown in Reaction A below. Thymol can then be converted to menthol, another naturally-occurring organic compound; this conversion is shown in Reaction B.

  Reaction A

  

  Reaction B

  

  Compound ?(C10H14O) dissolves in aqueous sodium hydroxide but is insoluble in aqueous sodium bicarbonate. The proton NMR spectrum of compound X is as follows:

  1.3 (9H) singlet

  4.8 (1H) singlet

  7.1 (4H) Multiplet

  Which of the following is the structure of Compound??

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: B

  All four answer choices are aromatic compounds, so Compound X must be aromatic. The chemical formula for this compound consists of carbons, hydrogens, and oxygens (which means you can eliminate choice C immediately, since that contains bromine atoms). The next piece of information in the question is that Compound X is soluble in aqueous sodium hydroxide, but not in aqueous sodium bicarbonate. This eliminates choice A, which is a substituted benzoic acid which WOULD certainly dissolve in sodium bicarbonate. From our discussion in the previous question, you should remember that the ability to dissolve in aqueous sodium hydroxide, but NOT in aqueous sodium bicarbonate, is characteristic of phenols. This suggests that Compound X is a phenol, which would make choice B correct and choice D, which is a phenyl ether, incorrect. But there's other information in this passage, so you should check through it to see if it supports your preliminary conclusion. Now let's look at the NMR spectrum and see what information we can get from that. The NMR spectrum has three separate peaks. The multiplet of area four has a chemical shift of 7.1, which is very far upfield; this represents the aromatic ring, and is due to the spin-spin coupling of four aromatic hydrogens. The signal at a chemical shift of 1.3 has a peak area of 9 hydrogens; the chemical shift indicates carbon-hydrogen single bonds, and the fact that there are nine hydrogens that are all equivalent means that this represents a tert-butyl group. This would fit EITHER choice B or choice D. And finally, the second singlet peak, whose chemical shift is 4.8 has an area of one. If Compound X were choice D, the single hydrogen peak that you'd get would be from the -CH group of the tert-buyl, next to the ether-group oxygen; this would be much further downfield, probably with a chemical shift of about 2.3 or 2.4, so choice D must be wrong. The fact that the signal from the single hydrogen is shifted much further upfield, to 4.8, indicates that the proton involved is deshielded, which in turn suggests that it's attached to a more elec-tronegative element than carbon. This magnitude of shift is in fact characteristic of a phenolic hydrogen. This agrees with our previous conclusion, that Compound X is choice B, and so B is the correct answer, as we suspected before.