1Generalized eukaryotic cell test w solutions

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BIOLOGY TOPICAL: Generalized Eukaryotic Cell Test Time: 21 Minutes* Number of Questions: 16 * The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 2 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Rf (261) 105 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) as developed by Generalized Eukaryotic Cell Test Passage I (Questions 1–6) It has been proposed that the nervous tissue of the electric eel, Electrophorus electricus, may be useful in the repair of damaged nerves in humans To isolate the various subcellular structures within the nerve cells of Electrophorus electricus, a researcher uses rate-zonal centrifugation This technique separates the various subcellular structures by size and density After centrifugation, the large dense structures typically form a pellet at the bottom of the test tube, while the lighter, less dense structures remain suspended in the supernatant (the fluid layer) By centrifuging at different speeds and durations, different organelles can be isolated The eel nerve cells are suspended in a 0.25 M sucrose solution, which is isotonic to the cells Most of the cells are broken open by stirring this suspension in a high-speed blender to form a cell homogenate The homogenate is then filtered to remove any unbroken cells The filtered homogenate is centrifuged at increasing speeds four times, with a pellet collected after each step, and the supernatant transferred to each successive tube This procedure is outlined in Figure Centrifuge Pour out: 15,000g x 600g x 10 Filtered Pellet homogenate Supernatant Pellet Pour out: 100,000g x 60 Pellet Supernatant Supernatant Analysis of Pellet showed that it possessed high succinate cytochrome C reductase activity This enzyme oxidizes succinic acid to fumaric acid, both intermediates of the Krebs cycle Therefore, this pellet most likely contains: A B C D mitochondria plasma membrane endoplasmic reticulum cytoplasm To determine the activity of the enzyme lactate dehydrogenase, the compound DPNH is added to each sample of supernatant For every molecule of pyruvate that is reduced to lactate by lactate dehydrogenase, a molecule of DPNH is oxidized: Pyruvate + DPNH → Lactate + DPN In which of the following samples would you expect lactate dehydrogenase activity to be found? Pour out: 300,000g x hr I II III IV Pellet Supernatant A B C D Supernatant Supernatant Supernatant Supernatant I only III and IV only II, III, and IV only I, II, III, and IV Figure GO ON TO THE NEXT PAGE KAPLAN MCAT Which of the following molecules would you expect to find in the pellet containing the plama membrane? O I N N O -O P NH O N NH Which of the following structures would most likely be found in a higher concentration in the plasma membrane of eel nerve cells as compared to the plasma membrane of eel epithelial cells? A B C D Glycoproteins Cholesterol Acetylcholine receptors Phospholipids O OH H H OH H OH O- II O P O- O CH2 O HC O C H2C O C (CH)16 (CH)16 CH3 CH3 O NH2 III N O O -O P O N O O- H H OH H H A B C D H I only II only III only I and III only GO ON TO THE NEXT PAGE as developed by Generalized Eukaryotic Cell Test The enzyme DNase is added to Pellet After this enzymatic digestion, which of the following would most likely be found within the digested pellet? A B C D mRNA Histones Rhodopsin Elongation Factor The researcher believes that Pellet contains ribosomal subunits Which of the following observations would most support this hypothesis? A Pellet was found to contain uncharged tRNA B Pellet was found to contain two molecules with sedimentation coefficients of 40S and 60S C Pellet was found to contain the enzyme peptidyl transferase D Supernatant was found to contain unprocessed proteins Passage II (Questions 7–12) The yeast Schizosaccharomyces pombe divides by fission and is thus an ideal microorganism for the study of cell division in higher eukaryotic cells A scientist believes that she has identified the region of the S pombe chromosome containing the DNA sequence that includes the centromere From this region she has isolated two different fragments of DNA: Fragments I and II The centromere is the region of the chromosome that maintains the attachment of sister chromatids during meiosis I and includes the site of attachment to the microtubules of the meiotic and mitotic spindles During division, the spindle must both assemble and disassemble its tubulin subunits to properly function in the segregation of chromosomes To test which of the two fragments contains the centromere region, the scientist takes a mutant strain of S pombe cells that cannot grow unless adenine is provided in the growth medium, and performs the following set of experiments Experiment Mutant cells are inoculated onto an agar plate containing minimal medium Experiment The normal gene for adenine biosynthesis (adenine+) is inserted into a plasmid, which is then inserted into the mutant cells The cells are then inoculated onto agar plates containing minimal medium (A plasmid is a selfreplicating extrachromosomal DNA fragment.) Experiments 3A and 3B Fragment I and Fragment II are inserted into separate plasmids containing the adenine+ gene The plasmids are then inserted into the mutant cells, which are inoculated onto separate agar plates containing minimal medium If the plasmid contains the DNA fragment with the functional centromere, it will be stably inherited by daughter cells during cell division The experimental results are shown in Table GO ON TO THE NEXT PAGE KAPLAN MCAT Table Expt Mutant S pombe on minimal medium with: 2* 3A* cells alone cells + adenine+ cells + adenine+ + Fragment I 3B* cells + adenine+ + Fragment II *Cells with introduced plasmid Increase in cell # after 1st round of division No Yes Yes Visible colonies after 20 rounds of division No No No Yes Yes Minimal medium contains only salts and sugar The primary reason that the cells used in Experiment cannot grow on minimal medium is that these cells are NOT able to synthesize: A B C D protein essential fatty acids nucleic acids salts Which of the following graphs best represents the four growth curves obtained under the experimental conditions summarized in Table 1? Key: Based on the information in the passage, it can be concluded that: I a drug that inhibits microtubule assembly is added to the growth medium used in Experiment 3B II a drug that inhibits microtubule disassembly is added to the growth medium used in Experiment 3B III a drug that inhibits cytokinesis is added to the growth medium used in Experiment 3B A B C D log cell # 20 # of cell divisions B # of cell divisions 20 D log cell # The results obtained in Experiment 3B can be changed such that they are identical to the results obtained in Experiment I if: C A log cell # Fragment I contains a centromere Fragment II contains a centromere Fragments I and II both contain centromeres the mutant S pombe chromosomes not contain centromeres Expt 3A Expt 3B log cell # A B C D Expt Expt # of cell divisions 20 # of cell divisions 20 I only II only I and II only I, II, and III GO ON TO THE NEXT PAGE as developed by Generalized Eukaryotic Cell Test 1 The cells used in Experiment are most likely able to increase their cell number after one round of cell division because: A the replicated adenine+ plasmid binds to microtubules and segregates to daughter cells during mitosis B the replicated adenine+ plasmid binds to microtubules and segregates to daughter cells during meiosis C the replicated adenine+ plasmid binds to microfilaments and segregates to daughter cells during cell division D the replicated adenine+ plasmid randomly segregates to daughter cells during cell division Questions 12 through 16 are NOT based on a descriptive passage In vitro, the transcription factor SP1 binds nucleic acids with a high affinity Radio-labeled SP1 would most likely be found in all of the following organelle preparations EXCEPT: A B C D nucleoli mitochondria ribosomes Golgi apparati All of the following are involved in directing proteins to their final destination within a cell EXCEPT: A B C D amino acid sequence of the protein vesicles lysosomes endoplasmic reticulum A researcher discovered that in order for the HAP2 protein to elder the nucleus and bind to the CYC1 gene, a molecule of ATP must be consumed Which of the following transport mechanisms must be involved in the movement of HAP2 into the nucleus? A B C D Active transport Facilitated diffusion Passive diffusion through the nuclear pores Endocytosis GO ON TO THE NEXT PAGE KAPLAN MCAT Three funnels, each containing a different concentration of a urea solution, are covered with a semi-permeable membrane that is impermeable to urea The funnels are inverted and placed in a beaker containing a different urea solution What is the concentration of the solution? A B C D 2.0 M urea 1.5 M urea 1.0 M urea 0.0 M urea dH20 1M urea 2M urea dH20 1M urea Urea Solution Urea Solution Initial state Final state 2M urea Which of the following would most likely play a role in cellular adhesion and recognition? A Carbohydrate molecules on the exterior of the plasma membrane B The fluidity of the plasma membrane C Peripheral proteins on the interior of the plasma membrane D The hydrophobic regions of the plasma membrane END OF TEST as developed by Generalized Eukaryotic Cell Test ANSWER KEY: A D B C B 10 KAPLAN B B D C A 11 12 13 14 15 D D C A C 16 A MCAT GENERALIZED EUKARYOTIC CELL TEST EXPLANATIONS Passage I (Questions 1-6) The correct answer is choice A Although the question stem may have scared you a little bit, this question is really quite easy The only thing you needed to get out of the question stem is that Pellet was found to have one of the enzymes of the Krebs cycle So as you can see, this question is simply asking you where the Krebs cycle occurs Well from introductory biology, you should know that the Krebs cycle occurs in the mitochondria, along with electron transport and oxidative phosphorylation Therefore choice A is the correct answer Just to review briefly, in eukaryotic cells, glucose is broken down to pyruvate in the cytoplasm in a series of reactions known as glycolysis The pyruvate then enters the mitochondria, where it is converted into acetyl CoA The acetyl CoA then enters the Krebs cycle All of the NADH and FADH2 molecules generated during these processes are shuttled into the electron transport chain, which is coupled with oxidative phosphorylation to produce ATP Anyway, back to the question As we just learned, glycolysis, not the Krebs cycle, occurs in the cytoplasm, and so choice D is incorrect The other two choices have nothing to with the Krebs cycle at all The plamsa membrane surrounds the cell and acts as a barrier between the cell and the outside environment and regulates the passage of material into and out of the cell So choice B is incorrect Finally, the endoplasmic reticulum, choice C, is involved with protein processing and is the site of ribosomal attachment during translation Thus choice C is also wrong Choice D is the correct answer In order to answer this question you need to understand the basic principle of centrifugation as explained in the passage, as well as the information provided in the question stem From the question stem you know that DPNH is used to monitor the conversion of pyruvate to lactate Although you have probably never heard of DPNH, you should recognize the reaction of pyruvate going to lactate as the fermentation step that occurs in most eukaryotic cells under anaerobic conditions Well, where in the cell does fermentation occur? Glycolysis occurs in the cytoplasm, while the Krebs cycle and electron transport/oxidative phosphorylation occur in the mitochondria So now that we’ve deciphered the question stem, let’s see how centrifugation works From the passage you know that larger objects will form a pellet at the bottom of a tube, while lighter, less dense structures remain suspended in the supernatant You also are told in the passage and shown in Figure 1, that the supernatant from the first tube is transferred to the second, and so on and so on Therefore, cytoplasm will be present in ALL four supernatants and so choice D is the correct answer The correct answer is choice B The question is essentially asking, “the plasma membrane is made up of what molecules?” The plasma membrane is composed of phospholipids, a molecule that has a glyerol backbone attached to two fatty acids and a phosphate group The phospholipids are arranged in a bilayer, which is why the cell membrane is also referred to as the “phospholipid bilayer.” Roman numeral II is a phospholipid, so must appear in the correct answer Answer choice C is the only one that includes Roman numeral II Molecule I is an RNA molecule, and III is a DNA molecule Neither RNA or DNA make up the plasma membrane Another possible molecule could be a protein, as the plasma membrane has receptors and ion channels/pumps studded throughout its length Choice C is the correct answer You’re being asked to deduce what will be found in a higher concentration in the plasma membrane of an eel nerve cell as compared to the membrane of an eel epithelial cell Well, what are the normal components of a generic plamsa membrane? A eukaryotic plasma membrane consists of a phospholipid bilayer So phospholipids are a common element of all plasma membranes and therefore would not be found in a higher concentration in nerve cell membranes Thus choice D is incorrect Cholesterol molecules are embedded in the hydrophobic interior of the bilayer Therefore, cholesterol is also a common component of plasma membranes So choice B is incorrect Glycoproteins, which are proteins that contain carbohydrate components, extend out of the plasma membrane and function in cell adhesion and recognition Therefore, glycoproteins are also common plasma membrane elements, and thus choice A is incorrect So by the process of elimination, choice C is the correct answer if you think about it, this answer makes sense Acetylcholine is one of the principal neurotransmitters responsible for the transmission of a nerve impulse, and its message is transmitted when it binds to acetylcholine receptors found on the plasma membrane of its target cells Since acetylcholine is a neurotransmitter, the plasma membrane of an epithelial cell would not be expected to have a high concentration of acetylcholine receptors But you would most likely find a high concentration of acetylcholine receptors on the plasma membrane of eel nerve cells Therefore, choice C is the correct answer Choice B is the correct answer Although you are not explicitly told what is in Pellet 1, you should have been able to figure it out First of all, you’re told that DNase is added to Pellet Well, you know that the only place DNA is found in 10 as developed by Generalized Eukaryotic Cell Test an animal cell is within the nucleus and the mitochondria And according to the passage, larger, denser organelles will pellet out first Since the nucleus is larger than any mitochondrion, it makes sense that the nucleus would probably be in Pellet 1, while mitochondria would remain in supernatant So now we know that Pellet contains eel nerve cell nuclei What effect will DNase have on the nucleus? DNase is an enzyme that degrades DNA by attacking the phosphodiester bonds that hold it together So adding DNase to the nucleus will destroy the DNA of eukaryotic chromosomes, leaving behind the DNA binding proteins that are an integral part of eukaryotic chromosome structure These DNA binding proteins are known as histones; there are five major types of histones Keep in mind that prokaryotes lack histones Therefore choice B is correct Choice A is wrong because a DNase only degrades DNA, and therefore mRNA would not be an end product of its action RNases degrade RNA Rhodopsin, choice C, is a protein found in the external segments of the rods of the retina Like most proteins, rhodopsin is not located within the nucleus of retinal cells, nor in that of nerve cells Furthermore, DNase activity would not yield rhodopsin Therefore, choice C is incorrect Choice D is also incorrect for the same reason Elongation Factor is a protein that associates with ribosomes during the addition of each amino acid during translation Choice B is the correct answer Since eukaryotic cells were used, the researcher must believe that Pellet contains eukaryotic ribosomal subunits But before we look at the answer choices, let’s briefly review the properties of eukaryotic ribosomes Eukaryotic ribosomes consist of a 60S subunit and a 40S subunit, which come together to form an 80S complex Both of the subunits are comprised of proteins and rRNA Prokaryotic ribosomes, on the other hand, consist of a 30S subunit combined with a 50S subunit, forming a 70S complex Now let’s look at the answer choices Uncharged tRNA, choice A, is typically found in the cytoplasm Uncharged tRNA, which is tRNA without an amino acid, is only transiently associated with the whole ribosome after it has transferred its amino acid to the growing peptide chain Nor would uncharged tRNAs be associated with the individual ribosomal subunits, except for the initiator methionine tRNA, which does bind to the 40S subunit Therefore, choice A definitely does NOT support the researcher’s hypothesis of ribosomal subunits in Pellet Thus, choice A is incorrect The presence of a 40S and a 60S molecule in the pellet would support the presence of the ribosomal subunits Therefore this does support the researcher’s hypothesis And although it DOES support the hypothesis, it in no way PROVES that Pellet contains ribosomal subunits But, so far, this seems like a pretty good answer Let’s check out the remaining two choices The enzyme peptidyl transferase is responsible for the formation of peptide bonds during translation And like the uncharged tRNA molecules, peptidyl transferase is never bound to the ribosome itself, and would only be found in association with the whole ribosome, not its individual subunits Therefore choice C is also incorrect The contents of the supernatant associated with Pellet does not give us any information about Pellet itself, so it neither supports nor contradicts the notion that Pellet contains ribosomal subunits So choice D is also incorrect Therefore, choice B must be the correct answer Passage II (Questions 7-11) The correct answer is choice B To answer this question you must utilize information contained in the passage and the data in Table We see that in Experiment 1, the mutant cells cultured in minimal medium cannot grow And the reason why the mutant strain is unable to grow on minimal medium is that it lacks adenine Remember, the passage specifically states that the mutant strain CANNOT grow in the absence of adenine Although the passage does not explicitly state that the minimal medium used in the experiments lacks adenine, it is the most obvious conclusion that can be drawn given that the mutant strain doesn’t grow in it Minimal medium contains only the bare essentials for a wild-type organism to grow, and since wild-type organisms are able to synthesize adenine on their own, there is no reason for minimal medium to be supplemented with adenine In Experiments 2, 3A, and 3B, the mutant yeast cells that were inserted with plasmids containing the gene that codes for adenine were able to complete at least one round of cell division on minimal medium, according to Table Therefore we can conclude that the mutant yeast cells CAN grow on minimal medium if they contain a plasmid with the adenine+ gene The passage also implies that if a plasmid does not have a centromere, the plasmid will not be stably maintained in the mutant yeast cell colony And if a cell doesn’t have the plasmid, it will die, since it can’t synthesize adenine on its own Only those cells with plasmids containing centromeric DNA and the adenine+ gene will be able to form visible colonies Now that we understand the logic behind the data presented in Table 1, let’s look at the answer choices From the table you see that in Experiment 3A, the yeast with Fragment I produced no visible colonies after 20 rounds of replication This means that Fragment I must NOT contain a centromere, and so choice A and choice C are incorrect In Experiment 3B, however, the yeast with Fragment II DO produce visible colonies after 20 rounds of cell division Therefore Fragment II must contain a centromere, and so choice B is correct Finally, choice D is incorrect because if the mutant yeast’s chromosomes did not contain centromeres, the mutant yeast strain could not live under any circumstances Chromosomes without centromeres would be lost during cell division, and all of the genetic information contained on the chromosomes would likewise be lost And without the genetic information, life is not possible KAPLAN 11 MCAT The correct answer is choice D This question uses the Roman numeral format and requires some information from the passage The passage states that both microtubule assembly and disassembly are necessary for proper chromosome and plasmid segregation This is exemplified by the chromosome movements that occur in metaphase of mitosis During metaphase, the chromosomes, which are attached to microtubules at the centromere, move back and forth across the midline of the cell The microtubules are capable of moving the chromosomes in a back and forth fashion because the microtubules fluctuate between growing longer and then shorter Because the passage tells you that chromosome segregation requires microtubule assembly and disassembly, you can conclude that the experimental conditions described in both Roman numerals I and II would disrupt proper chromosome segregation If chromosome segregation is disrupted, the cells cannot successfully divide and grow, and if these cells were grown on minimal medium, the result would be the same as for Experiment no increase in cell number after the first division, and no visible colonies after 20 divisions The cells in Experiment cannot divide and grow on minimal medium because they have no source of adenine for DNA synthesis Cells grown using the experimental conditions described in Roman numerals I and II would not grow because the microtubules in these cells could not properly perform their roles in chromosome segregation So, because we know that both Roman numerals I and II are correct, we can eliminate choices A and B Let’s look at Roman numeral III Given that cytokinesis is the final stage of cell division, adding a cytokinesis inhibitor to the growth medium would also prevent cell growth on minimal medium, or any kind of media for that matter, and would give the same results as seen for Experiment The correct answer is choice C This question requires that you combine information contained in the passage with outside knowledge The passage states that the mutant yeast cells cannot grow unless adenine is supplied in their growth medium, which implies that this strain has a defect in the adenine biosynthesis pathway The cells used in Experiment did not have adenine+ gene-containing plasmids inserted in them, and as we see in Experiment 1, these cells cannot grow on minimal medium So from the information contained in the passage we can conclude that the mutant cells cannot synthesize adenine and that is why the mutant cells are unable to grow on minimal medium Now that we know this, all we have to is figure out what role adenine plays in cell life Adenine is a nitrogen base contained in DNA and RNA You should also remember that DNA and RNA are nucleic acids Thus, the mutant cells used in Experiment cannot grow because they not have a source of adenine and thus they cannot synthesize the DNA and RNA that is needed for cell division to occur Therefore choice C is the correct answer Let’s look at the wrong answer choices quickly Choice A is incorrect because adenine is not a protein or protein precursor Amino acids are the building blocks of proteins, not nucleotides Choice B is incorrect as well Fatty acids are the building blocks of certain lipids such as fats or triacylglycerides Nothing in the passage suggests that the mutant yeast cells have an inability to synthesize fatty acids Finally, choice D is incorrect because, as is specified in the question stem, minimal medium contains salts Furthermore, living organisms not synthesize salts 10 The correct answer is choice A Let’s start analyzing this question by predicting what the growth curve should be for the cells used in Experiment 3B The cells in Experiment 3B will act like normal, wild-type, yeast cells cultured in growth medium You know this because as evidenced in Table 1, these cells continue to produce visible colonies at their 20th cell division Furthermore, you should recall that the growth curves of microorganisms are sigmoidal, or S-shaped, like the curves shown for Experiment 3B in choices A and C You should have immediately eliminated choices B and D because for continuously exponential growth curves like those depicted in B and D to occur, the yeast would have to continuously grow exponentially, never running out of nutrients, and never having to prepare metabolically for growth Obviously, this is not in the realm of the possible Let’s briefly review why the growth curves for microorganisms are S-shaped When cells are first inoculated onto growth medium and the cell number is very low, the cells must activate the biochemical machinery required for growth and cell division While the cells are gearing up for growth and cell division, the culture is said to be in the lag phase of growth Growth during this period is occurring at a very low rate Once the cells have completed the lag phase, the cells begin to increase their numbers exponentially, which results in the log phase of growth (log is short for logarithmic) This period of exponential growth is a consequence of the fact that each cell in the population divides into two cells At a certain point, the cells will enter the stationary phase of growth During this phase, the growing microorganisms have depleted the growth medium of an essential nutrient, or toxic byproducts of metabolism begin to accumulate, and the growth rate becomes very low, as in the lag phase Anyway, back to the question So, we know that the correct answer is either choice A or C How these two graphs differ from each other? In choice A, the cells in Experiments and 3A reach a higher cell number after 20 cell divisions than the cells in Experiment In choice C, the cells in Experiments 1, 2, and 3A all have a similar small cell number after 20 cell divisions Well, if you look at Table you will realize that choice A is correct The cells in Experiments and 3A will reach a higher cell number than the cells in Experiment 1, because the cells in Experiments and 3A have an adenine+ plasmid Even though the adenine+ plasmid has no centromere, it will most likely be maintained through a small 12 as developed by Generalized Eukaryotic Cell Test number of cell divisions before being lost and will thus allow the cultures in Experiments and 3A to reach an elevated cell number with respect to the culture derived from Experiment As we discussed previously, the mutant cells used in Experiment not contain an adenine+ plasmid and thus they cannot grow at all on minimal medium What I mean by “lost?” Well, according to the passage, plasmids are self-replicating; in other words, they replicate independently of the cell cycle If the plasmid does NOT have a centromere, during mitosis, it will not be able to bind to the mitotic spindle This means that when the cell undergoes cytokinesis, there is no guarantee that either of the daughter cells will inherit the plasmid and its copy or copies In fact, research has shown that such plasmids are inherited only 8-20% of the time Anyway, choice C is incorrect 11 The correct answer is choice D To answer this question you must understand what is occurring in Experiment In Experiment 2, the mutant yeast cells contain an adenine+ plasmid that does not contain a centromere Since the plasmid does not contain a centromere, the replicated plasmid will not be able to bind to the microtubules of the mitotic spindle, and thus the replicated plasmid will randomly segregate to daughter cells By random segregation we mean that since the plasmid CANNOT segregate to daughter cells by normal means, when the cells divide, the plasmid will get distributed to daughter cells by chance, or not at all, and will not be faithfully distributed to ALL daughter cells Due to this random segregation, the plasmid will eventually be lost from almost all progeny cells However, after one round of cell division, most of the yeast cells will still have plasmids, and these cells will be able to complete a second round of cell division, thereby further increasing the cell number of the culture This explains the increase in cell number after one round of division observed in the cells of Experiment [For more detail on how a plasmid gets lost, refer to the explanation to Question 10.] On the molecular level, random segregation of a plasmid means that the replicated plasmid segregates to daughter cells without binding to the mitotic spindle As stated in the passage, a plasmid must contain a centromere in order to bind to microtubules Thus choices A and B are incorrect We can also immediately rule out choice B because the cell division described in the passage is mitotic, not meiotic Meiotic cell division would require mating and sporulation, and the passage gives us no reason to conclude that such processes are occurring Finally choice C is also incorrect Microfilaments are composed of actin molecules and are involved in formation of the contractile ring during cytokinesis, but are not involved directly in the segregation of genetic material via binding to chromosomes or plasmids Discretes (Questions 12-16) 12 The correct answer is choice D From the question stem you know that the transcription factor will bind to nucleic acids This means that SP1 will bind to both DNA and RNA So all you have to to answer this question correctly is figure out which organelle does not contain DNA or RNA Well you should know that the nucleolus is the small dense region of the nucleus where rRNA synthesis occurs Thus, SP1 would bind to material within the nucleolus and so choice A must be incorrect The mitochondria is the organelle, bound by a double membrane, in which the reactions of the Krebs cycle, electron transport, and oxidative phosphorylation take place Mitochondria are the site of the production of the majority of the ATP molecules produced during aerobic respiration In addition to this function, mitochondria contain their own DNA, which means that SP1 will bind to it Therefore choice B is also incorrect Choice C is also incorrect Ribosomes are small organelles composed of protein and rRNA and are the site of translation Therefore SP1 would also bind to ribosomes So choice D is our correct answer The Golgi apparatus is the organelle responsible for the processing, packaging, and distribution of proteins It is not composed of nucleic acids, nor does it process them Since there are no nucleic acids within the Golgi apparatus, SP1 would not bind to it Therefore, choice D is the correct answer 13 The correct answer is choice C In order to answer this question correctly you needed to know the functions of each of the subcellular structures listed as choices Let’s go through each one until we find one that is NOT responsible for directing proteins to their final destinations within cells The Golgi apparatus and endoplasmic reticulum sort proteins according to their destination using signals inherent in the amino acid sequence of proteins For example, the amino acid sequence lys-asp-glu-leu at the N-terminus causes a protein to be transported from the Golgi apparatus to the endoplasmic reticulum So it is obvious that both the endoplasmic reticulum and the amino acid sequence of a protein play a role in protein transport Therefore choices A and D are incorrect Vesicles, choice B, are formed from the fusing of membrane surfaces They are involved in transporting proteins from one location to another by virtue of their ability to move around the cell and through membranes They may fuse with a target membrane and release the protein at its destination So choice B is also incorrect Therefore choice C must be the correct answer Lysosomes are rather small, spherical membrane-enclosed bodies that contain hydrolytic enzymes If a protein were to come in contact with these enzymes it would be degraded So lysosomes are not involved in protein trafficking Therefore, choice C is the correct answer KAPLAN 13 MCAT 14 Choice A is the correct answer The key information in the question stem is that this protein, HAP2, enters the nucleus and a molecule to ATP is consumed in the process All of the other information is just a distraction and I hope that it did not throw you off Furthermore, the specific name of the protein HAP2 is not important in answering the question, and you are not expected to know anything about this molecule So all you have to to answer this question is figure out which type of transport uses ATP Well, passive diffusion is the net movement of dissolved particles down their concentration gradient from a region of higher concentration to a region of lower concentration This process does not require ATP, so choice C is incorrect Facilitated diffusion is the net movement of dissolved particles down their concentration gradient with the help of carrier molecules This process, like passive diffusion, does not require ATP Therefore, choice B is also incorrect Active transport is the net movement of particles AGAINST their concentration gradient with the help of carrier molecules Unlike facilitated diffusion or passive diffusion, active transport requires energy And as you know, ATP stores energy so that cells can perform various functions Thus, choice A is the correct answer Endocytosis, choice D, is a process in which the cell membrane invaginates, forming a vesicle that contains extracellular medium So this process, though ATP-dependent, would not account for a cellular protein moving from the cytoplasm into the nucleus Thus choice D is incorrect 15 The correct answer is choice C To answer this question you need to determine the concentration of the solution in the beaker How can you this? Well if you compare the solution levels in the three funnels in the initial state to the solution levels in the final state, you can determine the concentration of urea in the beaker Since you know from the question stem that the funnels are covered with a semi-permeable membrane that is impermeable to urea, the urea itself will not be able to flow from the solutions in the funnels to the solution in the beaker or vice versa This means that water is the only substance that will move between the funnels and the beaker; in other words, osmosis will occur Osmosis is the movement of water across a semi-permeable membrane from a region of lower solute concentration to a region of higher solute concentration, until the solute concentrations on both sides of the membrane are equal So now that we know that we’re talking about osmosis, let’s look at the diagram The distilled water funnel had a decrease in water For the water to move from the funnel to the solution, the solution must have a higher solute concentration than distilled water Therefore, choice D is incorrect The level of solution in the M urea solution remains unchanged; there was no net movement of water This means that the urea concentration in the beaker must be equal to the urea concentration in the funnel In other words, the solution in the beaker is isotonic to the solution in the funnel Therefore, choice C must be the correct answer The level of solution in the M funnel increases, indicating that water has moved from the beaker into the funnel Thus, the concentration of urea must be greater in the funnel, and so the concentration of urea in the beaker must be less than M Thus, choices A and B are incorrect 16 Choice A is the correct answer This question is about cellular adhesion and recognition This refers to the adhesion of cells to one another and to the recognition of molecules that interact with the cell, such as hormones, antibodies, and viruses So from this you know that we’re dealing with phenomenon of the cell surface, not the cell interior With this in mind you can eliminate choices C and D, since they not refer to elements of the plasma membrane related to its exterior Choice C refers to the side of the plasma membrane that is in contact with neither the exterior cell surface nor the cytoplasm The fluidity of the plasma membrane, choice B, refers to the fact that both lipids and proteins are able to diffuse throughout the entire plasma membrane This fluidity enables the lipid bilayer to act as a solvent for membrane proteins Although some membrane proteins play a role in adhesion and recognition, MOST not, so they would only play a minor role in the actual adhesion and recognition process Thus, choice B is incorrect So by the process of elimination, choice A must be the correct answer Let’s see why In order for a cell to adhere to another cell, or for many molecules to interact with a cell, they must first bind to receptors or ligands on the cell surface Therefore, carbohydrate molecules on the exterior of the plasma membrane are in a perfect position to interact with other cell and molecules, and in fact they Thus, choice A is the correct answer 14 as developed by ... membrane END OF TEST as developed by Generalized Eukaryotic Cell Test ANSWER KEY: A D B C B 10 KAPLAN B B D C A 11 12 13 14 15 D D C A C 16 A MCAT GENERALIZED EUKARYOTIC CELL TEST EXPLANATIONS... log cell # A B C D Expt Expt # of cell divisions 20 # of cell divisions 20 I only II only I and II only I, II, and III GO ON TO THE NEXT PAGE as developed by Generalized Eukaryotic Cell Test. .. the cells in Experiments and 3A reach a higher cell number after 20 cell divisions than the cells in Experiment In choice C, the cells in Experiments 1, 2, and 3A all have a similar small cell
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