BIOLOGY TOPICAL: Microbiology Test Time: 20 Minutes* Number of Questions: 15 * 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 Unq (261) 105 Unp (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 Microbiology Test Passage I (Questions 1–5) Conjugation occurs between bacterial cells of different mating types “Maleness” in bacteria is determined by the presence of a small extra piece of DNA that can replicate independently of the larger chromosome Male bacteria having this sex factor, known as the F factor, are termed F + if the sex factor exists extrachromosomally F+ bacteria can conjugate only with F – bacteria, the “female” counter-parts, which not possess the F factor Genes on the F factor determine the formation of hair-like projections on the surface of the F+ bacterium, called sex pili The pili form cytoplasmic bridges through which genetic material may be transferred The pili also aid the F+ cell in adhering to the F– cell during conjugation During conjugation of an F+ cell with an F– cell, the DNA that is most likely to be transferred to the female is the F factor itself Prior to transfer, the F factor replicates The F– thus becomes an F + by receiving one copy of the F factor, while the original F+ holds on to the other copy If this were the only type of genetic exchange in conjugation, all bacteria would become F + and conjugation would eventually cease However, in F+ bacterial cultures, a few bacteria can be isolated that have the F factor incorporated into their chromosome These bacteria, referred to as Hfr bacteria, may also conjugate with F– cells They not transfer their F factor during conjugation, but they frequently transfer linear portions of their chromosomes The transfer is interrupted by the spontaneous breakage of the DNA molecule at random sites, usually before the F factor crosses to the F– cell This process is unidirectional; no genetic material from the F– cell is transferred to the Hfr cell A strain of the F+ bacteria is resistant to the antibiotic streptomycin If the F+ strain conjugates with a non-resistant F– strain, the F– strain becomes resistant to streptomycin Which of the following best accounts for this observation? A During conjugation, transfer of the F factor induced a spontaneous mutation in the F– cells that conferred streptomycin resistance B During conjugation, the F + cells transfer enzymes that activate the gene for streptomycin resistance in the F– cells C The gene for streptomycin resistance in the F– cells is carried on extrachromosomal F factor DNA D The gene for streptomycin resistance in the F+ cells is carried on extrachromosomal F factor DNA An Hfr strain of E coli was found to donate genes to four F– cells in the following order: F– cells What is the map of these genes on the E coli Hfr chromosome? A C G P T L S A Which of the following will most likely occur when a suspension of Hfr cells are mixed with an excess of F– cells? A Most of the F– cells will become F+ cells B The F– cells will produce sex pili that attach to the Hfr cells C Hfr cells will replicate the F factor independently of their chromosomes D Hfr chromosomal DNA will be transferred to F– cells Gene order TLPA PASG GTLP SAPL A L T P G B S D G S A P L T T P G S L A GO ON TO THE NEXT PAGE KAPLAN MCAT Prior to conjugation, the F factor in F+ bacteria is replicated by the enzyme: A B C D integrase DNA ligase reverse transcriptase DNA polymerase Which of the following conjugations will result in a significant number of new F+ bacteria? I II III IV A B C D F+ and F– F+ and Hfr Hfr and F– F+, F–, and Hfr I only I and IV only I, II, and IV only II, III, and IV only GO ON TO THE NEXT PAGE as developed by Microbiology Test Passage II (Questions 6–10) A strain of penicillin-resistant pneumococci bacteria exists, despite the absence of the bacterial beta-lactamase gene that typically confers penicillin resistance Additionally, half of the cells in this strain are unable to metabolize the disaccharides sucrose and lactose A microbiologist studying this strain discovered that all of the cells in this strain were infected with two different types of bacteriophage, Phage I and Phage II, both of which insert their DNA into the bacterial chromosome To determine if bacteriophage infection could give rise to this new bacterial strain, the microbiologist infected wildtype pneumococci with the two phage Experiment 10µL of Phage I and 10µL of Phage II were added to separate 5mL nutrient broth solutions containing actively growing wild-type pneumococci In addition, a 5mL broth solution containing only wild-type pneumococci was used as a control After 20 minutes of room temperature incubation, the microbiologist diluted 1µ L of the broth solutions in separate 1.999mL aliquots of sterile water She plated these dilutions on three different agar plates containing glucose, sucrose, and lactose, respectively The plates were incubated for 24 hours at 37°C The results are shown in Table Table Phage IPhage IIWild-type infected cells infected cells cells Glucose + + + Sucrose + – + Lactose + – – (+) denotes bacterial growth; (–) denotes no growth Based on the experimental data, which of the following statements is most likely true of Phage I? A Phage I inserted its DNA into the bacterial chromosome, making the bacterial cell wall impervious to the effects of penicillin B Phage I reduced the penicillin on the agar plates, therefore allowing the bacteria to grow C Phage I contained the viral gene that coded for beta-lactamase D Phage I inhibited the growth of the bacteria Which of the following conclusions could be inferred from the data in Table 1? A Phage I inserted its DNA into the region of the bacterial chromosome that codes for the enzymes of glycolysis B Phage I prevented larger molecules such as lactose and sucrose from passing through the bacterial cell wall C Phage II inserted its DNA into the region of the bacterial chromosome that codes for enzymes that digest disaccharides D Phage II utilized all of the sucrose and lactose and starved out the bacteria Plates Which of the following best accounts for the results of Experiment 2? I Both Phage I DNA and Phage II DNA code for beta-lactamase II Both Phage I DNA and Phage II DNA code for enzymes that inhibit tetracycline’s deleterious effects III Both phage disrupted the wild-type bacteria’s ability to resist penicillin IV The wild-type bacteria has no natural resistance to either penicillin or tetracycline Experiment 1µL of the broth solutions from Experiment were again diluted in separate 1.999mL aliquots of sterile water These dilutions were plated on three different agar plates containing tetracycline, penicillin, and no antibiotic, respectively The plates were incubated for 24 hours at 37°C The results are shown in Table Table Phage IPhage IIWild-type infected cells infected cells cells Tetracycline – – – Penicillin + + – No antibiotic + + + (+) denotes bacterial growth; (–) denotes no growth Plates A B C D I only I and IV only III and IV only I, II, and IV only GO ON TO THE NEXT PAGE KAPLAN MCAT Which of the following best describes the appearance of pneumococci, a streptococcal bacteria, when stained and then viewed with a light microscope? A B C D Rod Helical Squamous Spherical In which of the following cycles must Phage I be to produce plaques, which are transparent areas within the bacterial lawn caused by bacterial cell death? A B C D Lytic Lysogenic Mitotic Integration GO ON TO THE NEXT PAGE as developed by Microbiology Test Questions 11 through 15 are NOT based on a descriptive passage 1 Which of the following is true of both a bacteriophage and a retrovirus? A Both can integrate their genetic material into the host cell genome B Both have genes that code for reverse transcriptase C Both can infect human cells D Both are immunosuppressive agents Addition of DNase to a bacterial cell results in hydrolysis of the cell’s DNA, preventing protein synthesis and causing cell death However, certain viruses pre-treated with DNase continue to produce new proteins following infection Which of the following best accounts for this observation? All of the following are true of fungi EXCEPT: A they can exist in either a haploid state or a diploid state B their cells contain a plasma membrane, but no cell wall C they contain masses of hyphae that form the mycelium D they are heterotrophs One form of hepatitis (inflammation of the liver) is caused by a virus The serum of patients with hepatitis may contain the hepatitis B surface antigen, HBsAg, and/or the hepatitis B core antigen, HBcAg Which of the following is most likely true? A HBcAg is probably composed primarily of lipids B HBsAg is probably composed primarily of proteins C Serum concentration of liver enzymes remains the same during acute hepatitis infection D Serum concentration of liver enzymes decreases during acute hepatitis infection A The icosahedral protein coat of these viruses denature DNase B The genomes of these viruses contain multiple reading frames C The genomes of these viruses are comprised of RNA D The genomes of these viruses contain multiple copies of their genes In the treatment of an influenza viral infection, an experimental drug that directly blocks the synthesis of the viral protein coat is discovered This drug most likely: A affects synthesis of host cell proteins, and may therefore cause side effects in the host B has no affect on host cell protein synthesis, and may therefore be an effective treatment C blocks the cell surface receptors to which both the viral protein coat and host hormones bind, and may therefore cause side effects in the host D increases the affinity of antibodies to the viral protein coat antigens, and may therefore be an effective treatment END OF TEST KAPLAN MCAT ANSWER KEY: D D A D B 10 C C B D A 11 12 13 14 15 A C A B B as developed by Microbiology Test MICROBIOLOGY TEST TRANSCRIPT Passage I (Questions 1-5) The correct answer is choice D This is a fairly straightforward question that can be answered with the information given in the passage From the second paragraph we know that Hfr bacteria have the F factor integrated into their chromosome, instead of located in an autonomous circular DNA element in the cytoplasm in other words a plasmid From the passage you should have been able to figure out that when a suspension of Hfr cells is mixed with an excess of F- cells, the Hfr cells attach to the F- cells via pili, conjugation begins, and Hfr chromosomal DNA will be transferred to the F- cells Therefore, choice D is right and choice B is wrong, since it has the things the other way around Since the F factor is part of the bacterium’s main chromosome, the F factor will get replicated along with the rest of the cell’s chromosome prior to conjugation; so choice C is wrong During conjugation, the transfer f Hfr DNA is interrupted by the spontaneous breakage of the DNA molecule at random points Typically, the chromosome is broken before the F factor is transferred to the F- cell Therefore, the conjugation of an Hfr cell with an F- cell does NOT usually result in an F+ cell The F factor usually remains in the Hfr cell Thus, choice A is wrong Again, choice D is the correct answer The correct answer is choice D How can this newfound resistance to streptomycin in a previously nonresistant Fstrain be accounted for? Since resistance to an antibiotic is a genetic thing, the only way that this could have occurred was if the gene for streptomycin resistance was transferred from the F+ cells to the F- cells Well, from the passage you know that during conjugation between F+ cells and F- cells, the extrachromosomal DNA containing the F factor replicates, the F+ cells attach to the F- cells via pili, and the extrachromosomal DNA is transferred to the F- cells The only thing transferred between the two cells is this DNA Thus, the gene for streptomycin resistance must have been carried on the F factor Thus, choice D is the correct answer I hope that you weren’t tricked into picking choice C, which if read too fast, looks kind of like choice D Choice C is wrong because the F- cells were NOT resistant to the antibiotic before conjugation, which implies that they certainly could NOT have carried the gene for streptomycin resistance Choice B can also be ruled out since, as we just said, the F- cells of this strain could not have carried the gene for streptomycin resistance prior to conjugation, which would have had to been the case for choice B to be true And furthermore, we know that only DNA, not enzymes, are transferred from the F + cell during conjugation Choice A is also incorrect because it is not possible to induce a spontaneous mutation A mutation may either be induced by something such as a sex factor or it may be spontaneous, meaning its cause is nothing but pure chance Also, spontaneous mutations are very rare and it would not be possible for an entire strain of F- cells to spontaneously mutate at the same time to confer resistance to the same drug Again, choice D is the correct answer The correct answer is choice A This problem requires you to determine the map of the chromosomal marker genes based on the transfer of genes during conjugation given to you in the question Before you attack this problem, you should visualize what is going on in this question You know from the passage that Hfr cells have the F factor integrated in their chromosome When an F- cell is in close proximity to an Hfr cell, a sex pili reaches out from the Hfr cell to the F- cell, forming a cytoplasm-filled bridge between the two cells Since the F factor is part of the chromosomal DNA it gets replicated along with the chromosomal DNA prior to conjugation, although the replicated DNA is in a linear form It is a segment of this linear DNA that gets transferred to the F- cell The transfer is interrupted by the spontaneous breakage of the DNA molecule at random times The DNA that entered the F- cell will be composed of only those genes that crossed the pili before breakage stopped the transfer Let’s now take a look at the problem There are different combinations of gene order in the four recipient F- cells because the conjugation was disrupted at different times Since the genetic information is transmitted linearly, the order itself is NOT changed Now before we go any further, we can eliminate choices B and D Why? Because as you know from introductory biology, the chromosome in prokaryotic cells is circular Even though the DNA transferred during conjugation is linear, this is the replicated DNA; the original DNA in the donor Hfr cells is circular Therefore, choice B and D must be wrong because they both have the DNA maps of the E coli Hfr chromosome as being linear So now all you have to is decide between the maps in choices A and C To find the map of the circular chromosome, you simply have to fit the fragments from the question stem together by examining the overlapping areas The first segment has the order TLPA The second segment, PASG, overlaps with the P and A from the first segment So the order is now TLPASG The third segment, GTLP, overlaps with the T, L, and P of the order So the chromosome order is now GTLPASG Finally the last segment, SAPL, is already found in our gene order, if you read it backwards, that is Remember that in contrast to a linear chromosome, circular chromosomes don’t have a unique start point and end point Therefore, choice A must be the correct answer because it illustrates the genes in the order TLPASG Or you can read it PASGTL, or SAPLTG; it all depends on your starting point and your direction clockwise or counterclockwise Thus choice C is incorrect and choice A is the correct answer KAPLAN MCAT The correct answer is choice D This question is very straightforward, requiring you to recall the primary enzyme responsible for replicating strands of DNA DNA polymerase DNA polymerase, choice D, is an enzyme that can synthesize a new DNA strain using a template DNA strand Thus, DNA polymerase must be responsible for the replication of the F factor in DNA in F+ cells before conjugation occurs Don’t fall into the trap of thinking, “if it’s too obvious, then it must be incorrect.” There are often questions on the MCAT that are designed to be easy, questions that most people get right So let’s look at the other answer choices Integrase, choice A, is an enzyme that you are not required to be familiar with Integrase is a retroviral enzyme that integrates provirus DNA in host genomes So choice A is incorrect This choice was just thrown in as a distraction Since this is not a common enzyme and it is not discussed in the passage, chances are it’s a wrong choice DNA ligase, choice B, catalyzes the formation of a phosphodiester bond to link two adjacent bases separated by a nick in one strand of double-helical DNA Since DNA ligase does not replicate DNA, choice B is incorrect Reverse transcriptase, choice C, is another retroviral enzyme Reverse transcriptase synthesizes DNA from an RNA template Because DNA replication involves the synthesis of DNA from a DNA template, choice C is also incorrect Again, choice D is the correct answer The correct answer is choice B The key to getting this one right is noting from the passage what is required for a cell to become an F+ , or male, cell Recall that maleness is determined by the presence of a small extra piece of selfreplicating DNA that contains the F factor With this in mind let’s review the listed conjugations In Roman numeral I, F+ bacteria conjugate with F- bacteria This will certainly result in new F+ cells As you know from the passage, the extrachromosomal F factor will be transferred to the F- cell, turning it into a F+ cell, where the F factor DNA will exist indefinitely as an extrachromosomal entity with the ability to replicate autonomously Since Roman numeral I is a correct choice, choice D can be eliminated In Roman numeral II, an F+ cell conjugates with an Hfr cell Since both cell types already contain the F factor, no conjugation will occur Thus no new F+ bacteria will result Thus Roman numeral II is incorrect, and choice C can be eliminated In Roman numeral III, an Hfr cell conjugates with a F- cell As you know from the passage, Hfr cells rarely transfer the F factor to F- cells during conjugation Therefore there will NOT be a significant number of new F+ bacteria and Roman numeral III is incorrect Now if you really had your thinking cap on, you wouldn’t even have examined Roman numeral III, since you previously eliminated choice D, the only choice that contained Roman numeral III In Roman numeral IV, new F+ cells will be formed because the two necessary ingredients F+ cells and F- cells are present Even though Hfr cells cannot be converted themselves or convert F- cells to F+ cells, the F- cells that happen to conjugate with F+ cells will be converted to F+ Therefore Roman numeral IV is also correct So Roman numerals I and IV are correct and choice B is the correct answer Passage II (Questions 6-10) The correct answer is choice C If you look at the answer choices you will see that they all refer to penicillin resistance and Phage I So this means that you need to look at Table for your answer From Table you know that the bacteria infected with Phage I were able to grow on agar plates containing penicillin You also know from the passage and Table that the uninfected wild-type bacterial cells not contain the gene that codes for beta-lactamase, which confers penicillin resistance, since no bacterial growth was observed when the control dilution was incubated in the presence of penicillin So, combining these two pieces of information, you can conclude that Phage I must be responsible for the observed penicillin resistance in Experiment With this in mind, let’s look at the answer choices Right away you should have eliminated choice D, since bacterial growth did occur in both Experiments and when the bacteria were infected with Phage I Choices A, B, and C, however, are all possible ways in which the bacterial cells could have become resistant to penicillin So, how you decide between these three? Well, you’re asked to determine your answer based on the experimental data Thus, choice C is the only one that could be concluded on this basis the penicillin resistance observed in the bacteria infected with Phage I in Experiment must have been conferred by the insertion of Phage I DNA that contained the gene that codes for beta-lactamase Thus choice C is the correct answer The passage does not state the exact mechanism by which this gene functions and therefore you are not expected to know this Perhaps beta-lactamase does function via the mechanisms proposed in either choice A or B by effecting the bacterium cell wall, or by reducing the penicillin on the agar plate However, neither of these can be concluded from the experimental data The only thing you DO know about penicillin resistance is that the gene for beta-lactamase must be present Why? Because the only difference between these bacterial cells and the wild-type cells of the control group is the Phage I infection Therefore, choices A and B are incorrect and choice C is the correct answer The correct answer is choice C From the question stem you know that you will need to interpret the data in Table According to Table 1, both the control bacteria and the bacteria infected with Phage I were able to grow in the presence of all three sugars two of which are disaccharides On the other hand, the bacteria infected with Phage II were only able to grow on glucose This means that Phage II is somehow disrupting the bacteria’s metabolism of sucrose and lactose Since you know 10 as developed by Microbiology Test that Phage I does not have any deleterious effects on bacterial metabolism, as indicated by the ability to grow on all three plates, choices A and B can be eliminated If Phage I DNA was incorporated into the region of the bacterial chromosome that codes for the enzymes of glycolysis, the bacteria would have been unable to grow on glucose, since glycolysis provides the ATP required for growth And if Phage I prevented the entrance of lactose and sucrose into the cells, the bacteria would have been unable to grow on either of these sugars, since the cells would not have had access to metabolic fuel Since neither of these scenarios is the case, we’ve narrowed it down to either choice C or choice D Since viruses are not autonomous life forms, they would be unable to utilize any sugar; viruses not possess the metabolic machinery to metabolize nutrients Thus, choice D is incorrect Insertion of Phage II DNA into the region of the chromosome that codes for the enzymes that digest disaccharides such as sucrose and lactose, WOULD disrupt the metabolism of these nutrients Therefore, you would NOT expect to see any bacterial growth when Phage II-infected bacteria were incubated with either sucrose and lactose, which is exactly what we see happening in Table Thus, choice C is the correct answer The correct answer is choice B From the question stem you know that you need to understand the data in Table to get this question right In Table 2, you see that the control bacteria are unable to grow in the presence of either tetracycline or penicillin Therefore, the wild-type bacteria does NOT have any natural resistance to these two antibiotics and so Roman numeral IV is correct, and choice A can thus be eliminated You can also eliminate Roman numeral III, since we have just established that the wild-type bacteria has no natural resistance to penicillin Therefore, choice C must also be wrong From Table 2, you also know that bacteria infected with either Phage I or Phage II are able to grow in the presence of penicillin, but not in the presence of tetracycline, indicating that both Phage I DNA and Phage II DNA must contain the gene that codes for beta-lactamase, since we know that this is the enzyme that confers penicillin resistance So Roman numeral I must be correct and Roman numeral II must be incorrect Thus, choice D can be eliminated This means that Roman numerals I and IV are correct and therefore choice B is the right answer The correct answer is choice D This is an example of a pure knowledge question tacked onto a passage Bacteria are often categorized on the basis of their shape There are three basic bacterial shapes: spherical, rod, and helical, or spiral Spherical bacteria are known as cocci; rod-shaped bacteria are known as bacilli; and helical bacteria are known as spirochetes or spirilla Thus, a pneumococci bacteria, when stained and viewed under a light microscope, would be spherical in shape Thus, choice D is the correct answer and choices A and B are wrong Helical bacteria, choice B, are the least common of the three groups; rod-shaped bacteria, choice A, include the common E coli Squamous, choice C, refers to the layer of epithelial cells that typically perform protective functions, composing the outer layers of the skin and the lining of the mouth and other mucous membranes So obviously choice C is also incorrect Again, choice D is the correct answer 10 The correct answer is choice A According to the question stem, plaques are transparent areas with a bacterial lawn caused by bacterial cell death, a.k.a lysing The lysing of the bacterial cells is due to eruption of infectious viral particles So basically, all this question is asking you is to determine in which cycle of viral infection does a bacteriophage assemble new viral particles and lyse its host cell Here’s where your knowledge of introductory biology comes into play You should remember that when certain phage infect bacteria, one of two events may occur The viral DNA may enter the cell and set up an infection whereby new viral particles are assembled, and the bacterium lyses; this is know as the lytic cycle Therefore, choice A is correct OR, the viral DNA may become part of the bacterial chromosome, via an integration step Therefore choice D is incorrect Once integrated, the viral DNA replicates with the host chromosome and is passed on to daughter cells This is known as the lysogenic cycle because from time to time, an integrated virus, which is called a prophage, becomes activated and sets up a new lytic cycle As a matter of fact, this passage related to strains of phage that began their infection in a lysogenic stage Remember, the passage specifically tells you that the phage DNA was inserted, or integrated into the bacterial genome Therefore choice B is incorrect As for choice C, mitosis has nothing to with viral infection Mitosis is the nuclear division of eukaryotic cells characterized by chromosome replication and formation of two identical daughter nuclei Thus choice C is incorrect Again, choice A is the correct answer One more thing before we move on to the discretes I hope that you didn’t get too caught up in the very specific details of the experimental protocol It could have really slowed you down if you had gotten too enmeshed in this, especially since none of the questions actually required you to refer back to the protocol Focus on the rationale of the experiments; don’t worry about the minutiae Remember, the passage isn’t going anywhere; you can refer back to it at any time if necessary Discretes (Questions 11-15) 11 The correct answer is choice A In order to get this question right you need to know the general characteristics of two types of viruses: bacteriophage and retroviruses Bacteriophage are viruses that attack only bacteria Bacteriophage typically consist of a head made of a protein coat and a core that contains nucleic acid Bacteriophage also contain a tail made of protein that is specialized for attaching to bacteria Upon infection, bacteriophage can enter one of two cycles In the lytic cycle, the viral nucleic acid enters the bacterial cell and begins using host machinery to produce new virions, which eventually cause cell lysis In the lysogenic cycle, the viral DNA integrates into the bacterial chromosome, replicating with it and being passed on to daughter cells in this inactive form Once an integrated virus, called a prophage, becomes activated, it enters the lytic cycle KAPLAN 11 MCAT Now let’s take a look at retroviruses Retroviruses are RNA-containing viruses that replicate through a DNA intermediate by virtue of a viral-coded RNA-dependent DNA polymerase, known as reverse transcriptase The most notable retrovirus is human immunodeficiency virus (HIV), the causative agent of AIDS Retroviral life-cycle consists of four main events The virus binds to its host, typically an animal cell, and injects RNA and a few viral enzymes The RNA is then converted to DNA by reverse transcriptase, and the DNA then integrates into the host cell’s genome The viral genes are then expressed Finally, the virions are assembled and released from the cell by budding O.K., now that we know the basics of both types of viruses, let’s look at the answer choices Well, we know that both types of viruses can integrate their genetic material into the host cell genome So choice A is the right answer Retroviruses are classified by the presence of reverse transcriptase, which is not present in bacteriophage Thus choice B is incorrect Choice C is also incorrect; as its name implies, bacteriophage only infect bacteria Bacteria, the target of bacteriophage, have no real immune system to suppress, and so choice D must also be wrong Again, choice A is the correct answer 12 The correct answer is choice C From the question stem you know that bacterial cells treated with DNase eventually die due to hydrolysis of their DNA, while certain viruses are unaffected by DNase What’s DNase? It’s an enzyme that degrades DNA by attacking the bonds within a DNA molecule So it makes sense that the bacterial cell would die, because all bacteria store their genetic material in the form of DNA If certain viruses are still able to synthesize proteins following exposure to DNase, then these viruses must be RNA viruses Thus, choice C is the right answer Choice A is wrong because a viral protein coat is not able to denature DNase Denaturation involves the breaking of bonds due to either heat or chemical treatment Choice B is wrong, because while it is true that viral genomes typically contain multiple reading frames so that they can get the most use out of their limited nucleic acid, having multiple reading frames doesn’t prevent DNA hydrolysis Choice D is wrong for the same reason; having multiple copies of a gene is not enough to prevent DNase from hydrolyzing DNA Again, choice C is the correct answer 13 The correct answer is choice A This question tests your basic understanding of viruses Viruses, which are essentially non-living obligate parasites, have several common features: 1) Viruses contain either DNA or RNA, but not both; 2) viruses not have the metabolic machinery for energy production or protein synthesis and are forced to rely on the machinery of their host; 3) viruses are unable to reproduce themselves directly, instead relying on their host’s machinery for replication; and 4) viruses have no membranes to regulate the entry or exit of material So based on this information, you know that if the experimental drug blocks viral protein synthesis, it is likely that it also blocks host protein synthesis, since it is the host’s metabolic machinery that synthesizes viral proteins Thus you would expect this drug to also affect host protein synthesis, which would be considered a side effect of the drug Therefore choice A is the correct answer and choice B is wrong About choice B: just because a drug has an unwanted side effect does not necessarily mean that the drug is not an effective treatment Choice C is incorrect because the blockage of host cell surface receptors would not prevent the synthesis of viral protein, since protein synthesis occurs in the cytoplasm of the host cell Choice D is wrong for the same reason; increasing the affinity of antibodies specific for antigens found on the viral protein coat would not block the synthesis of the viral coat proteins themselves Again, choice A is the correct answer 14 The correct answer is choice B This is a pure knowledge question There is really no reasoning required, so either you knew the answer or you didn’t Examples of fungi include mold, yeast, and mushrooms Fungi are eukaryotic organisms, typically filamentous or, rarely, unicellular The filamentous forms consist basically of continuous hyphae that form mycelium; thus choice C is incorrect since it IS true of fungi and you’re asked to determine which of the choices is NOT true All fungi have chitin-containing cell walls, as well as plasma membranes Therefore choice B must be the correct answer Choices A and D are wrong because they are both true Fungi reproductive cycles often include both sexual and asexual phases, meaning that haploid and diploid states are both possible In addition, fungi are heterotrophs that obtain nutrients through absorption Again, choice B is the correct answer 15 The correct answer is choice B From your knowledge of viral structure you know that the viral coat is the outer surface of a virus and is composed of protein And the viral core, the inside of the virus, contains nucleic acid in the form of either RNA or DNA Therefore, you would expect the core antigen to be composed primarily of nucleic acids and the coat, or surface, antigen to be composed primarily of proteins Remember, an antigen is just something that invokes an immune response in a host organism Thus, choice A is wrong and choice B is the correct answer Let’s just take a look at the other choices quickly From your knowledge of viral replication, you know that the viral coat and the viral core are synthesized separately and then combined to form a virion When a host cell lyses, complete viral particles, as well as any unassembled viral components, namely the coat particle, HBsAg, and the core particle, HBcAg, may be released In addition, the contents of the host cell itself will also be released And, since you’re told that the antigens are found in the serum, it is not unreasonable to expect the enzymes of the host liver cells to be in the serum too In fact, a rise in certain liver enzymes is key to hepatitis diagnosis Therefore, choices C and D can be eliminated Also, since incomplete viral particles are also liberated into the serum, it might be expected that some of these viral products might also appear as antigens, such as viral DNA polymerase for example Again, choice B is the correct answer 12 as developed by ... be an effective treatment END OF TEST KAPLAN MCAT ANSWER KEY: D D A D B 10 C C B D A 11 12 13 14 15 A C A B B as developed by Microbiology Test MICROBIOLOGY TEST TRANSCRIPT Passage I (Questions... (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) as developed by Microbiology Test Passage I (Questions 1–5) Conjugation occurs between bacterial cells of different mating types... only I, II, and IV only II, III, and IV only GO ON TO THE NEXT PAGE as developed by Microbiology Test Passage II (Questions 6–10) A strain of penicillin-resistant pneumococci bacteria exists, despite