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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.
Passage I (Questions 1–5)
2 . A strain of the F+ bacteria is resistant to the
antibiotic streptomycin. If the F+ strain conjugates Conjugation occurs between bacterial cells of with a non-resistant F– strain, the F– strain becomes different mating types. “Maleness” in bacteria is resistant to streptomycin. Which of the following 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 A . During conjugation, transfer of the F factor
factor, are termed F+ if the sex factor exists induced a spontaneous mutation in the F– cells extrachromosomally. F+ bacteria can conjugate only with that conferred streptomycin resistance.
F– bacteria, the “female” counter-parts, which do not B . During conjugation, the F+ cells transfer
possess the F factor. Genes on the F factor determine the enzymes that activate the gene for streptomycin formation of hair-like projections on the surface of the F+ bacterium, called sex pili. The pili form cytoplasmic C . The gene for streptomycin resistance in the F–
bridges through which genetic material may be transferred.
cells is carried on extrachromosomal F factor The pili also aid the F+ cell in adhering to the F– cell during conjugation. During conjugation of an F+ cell D . The gene for streptomycin resistance in the F+
with an F– cell, the DNA that is most likely to be cells is carried on extrachromosomal F factor transferred to the female is the F factor itself. Prior to transfer, the F factor replicates. The F– thus becomes anF+ by receiving one copy of the F factor, while theoriginal F+ holds on to the other copy.
3 . An Hfr strain of E. coli was found to donate genes to
If this were the only type of genetic exchange in conjugation, all bacteria would become F+ andconjugation 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 do not transfer their F factor during conjugation, but they frequently transfer linear portions oftheir chromosomes. The transfer is interrupted by the What is the map of these genes on the E. coli Hfr 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.
1 . Which of the following will most likely occur when
a suspension of Hfr cells are mixed with an excess of A . Most of the F– cells will become F+ cells.
B . The F– cells will produce sex pili that attach to
C . Hfr cells will replicate the F factor independently
D . Hfr chromosomal DNA will be transferred to F–
4 . Prior to conjugation, the F factor in F+ bacteria is
A . integrase.
B . DNA ligase.
C . reverse transcriptase.
D . DNA polymerase.
5 . Which of the following conjugations will result in a
A . I only
B . I and IV only
C . I, II, and IV only
D . II, III, and IV only
Passage II (Questions 6–10)
6 . Based on the experimental data, which of the
following statements is most likely true of Phage I? A strain of penicillin-resistant pneumococci bacteria exists, despite the absence of the bacterial beta-lactamase A . Phage I inserted its DNA into the bacterial
gene that typically confers penicillin resistance.
chromosome, making the bacterial cell wall Additionally, half of the cells in this strain are unable to impervious to the effects of penicillin.
metabolize the disaccharides sucrose and lactose. A B . Phage I reduced the penicillin on the agar plates,
microbiologist studying this strain discovered that all of therefore allowing the bacteria to grow.
the cells in this strain were infected with two different C . Phage I contained the viral gene that coded for
types of bacteriophage, Phage I and Phage II, both of which insert their DNA into the bacterial chromosome.
D . Phage I inhibited the growth of the bacteria.
To determine if bacteriophage infection could give rise tothis new bacterial strain, the microbiologist infected wild-type pneumococci with the two phage.
7 . Which of the following conclusions could be inferred
10µL of Phage I and 10µL of Phage II were added to separate 5mL nutrient broth solutions containing activelygrowing wild-type pneumococci. In addition, a 5mL broth A . Phage I inserted its DNA into the region of the
solution containing only wild-type pneumococci was used bacterial chromosome that codes for the enzymes as a control. After 20 minutes of room temperature incubation, the microbiologist diluted 1µL of the broth B . Phage I prevented larger molecules such as
solutions in separate 1.999mL aliquots of sterile water.
lactose and sucrose from passing through the She plated these dilutions on three different agar plates containing glucose, sucrose, and lactose, respectively. The C . Phage II inserted its DNA into the region of the
plates were incubated for 24 hours at 37°C. The results are bacterial chromosome that codes for enzymes that D . Phage II utilized all of the sucrose and lactose and
8 . Which of the following best accounts for the results
(+) denotes bacterial growth; (–) denotes no growth.
I. Both Phage I DNA and Phage II DNA code 1µL of the broth solutions from Experiment 1 were II. Both Phage I DNA and Phage II DNA code again diluted in separate 1.999mL aliquots of sterile water.
for enzymes that inhibit tetracycline’s These dilutions were plated on three different agar plates containing tetracycline, penicillin, and no antibiotic, III. Both phage disrupted the wild-type bacteria’s respectively. The plates were incubated for 24 hours at 37°C. The results are shown in Table 2.
IV. The wild-type bacteria has no natural resistance to either penicillin or tetracycline.
B . I and IV only
C . III and IV only
D . I, II, and IV only
(+) denotes bacterial growth; (–) denotes no growth.
9 . Which of the following best describes the appearance
of pneumococci, a streptococcal bacteria, whenstained and then viewed with a light microscope? A . Rod
B . Helical
C . Squamous
D . Spherical
1 0 . In which of the following cycles must Phage I be to
produce plaques, which are transparent areas withinthe bacterial lawn caused by bacterial cell death? A . Lytic
B . Lysogenic
C . Mitotic
D . Integration
1 4 . All of the following are true of fungi EXCEPT:
NOT based on a descriptive
A . they can exist in either a haploid state or a
B . their cells contain a plasma membrane, but no
1 1 . Which of the following is true of both a
C . they contain masses of hyphae that form the
A . Both can integrate their genetic material into the
D . they are heterotrophs.
B . Both have genes that code for reverse
C . Both can infect human cells.
1 5 . One form of hepatitis (inflammation of the liver) is
D . Both are immunosuppressive agents.
caused by a virus. The serum of patients withhepatitis 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.
1 2 . Addition of DNase to a bacterial cell results in
B . HBsAg is probably composed primarily of
hydrolysis of the cell’s DNA, preventing protein synthesis and causing cell death. However, certain C . Serum concentration of liver enzymes remains
viruses pre-treated with DNase continue to produce the same during acute hepatitis infection.
new proteins following infection. Which of the D . Serum concentration of liver enzymes decreases
following best accounts for this observation? A . The icosahedral protein coat of these viruses
B . The genomes of these viruses contain multiple
C . The genomes of these viruses are comprised of
D . The genomes of these viruses contain multiple
1 3 . In the treatment of an influenza viral infection, an
experimental drug that directly blocks the synthesis ofthe viral protein coat is discovered. This drug mostlikely: 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 aneffective treatment.
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 theirchromosome, 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 ofthe bacterium’s main chromosome, the F factor will get replicated along with the rest of the cell’s chromosome prior toconjugation; 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 remainsin 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 F- strain 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 betweenthe two cells is this DNA. Thus, the gene for streptomycin resistance must have been carried on the F factor. Thus, choice Dis 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 theycertainly 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 havehad 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. Amutation 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 tospontaneously 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 shouldvisualize 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 replicatedalong 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 beforebreakage 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 itselfis NOT changed. Now before we go any further, we can eliminate choices B and D. Why? Because as you know fromintroductory biology, the chromosome in prokaryotic cells is circular. Even though the DNA transferred during conjugation islinear, this is the replicated DNA; the original DNA in the donor Hfr cells is circular. Therefore, choice B and D must bewrong because they both have the DNA maps of the E. coli Hfr chromosome as being linear. So now all you have to do isdecide between the maps in choices A and C. To find the map of the circular chromosome, you simply have to fit thefragments from the question stem together by examining the overlapping areas. The first segment has the order TLPA. Thesecond 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 linearchromosome, circular chromosomes don’t have a unique start point and end point. Therefore, choice A must be the correctanswer because it illustrates the genes in the order TLPASG. Or you can read it PASGTL, or SAPLTG; it all depends onyour starting point and your direction--clockwise or counterclockwise. Thus choice C is incorrect and choice A is the correctanswer.
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 anew 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 beincorrect.” There are often questions on the MCAT that are designed to be easy, questions that most people get right. So let’slook at the other answer choices. Integrase, choice A, is an enzyme that you are not required to be familiar with. Integrase is aretroviral enzyme that integrates provirus DNA in host genomes. So choice A is incorrect. This choice was just thrown in asa distraction. Since this is not a common enzyme and it is not discussed in the passage, chances are it’s a wrong choice. DNAligase, choice B, catalyzes the formation of a phosphodiester bond to link two adjacent bases separated by a nick in one strandof double-helical DNA. Since DNA ligase does not replicate DNA, choice B is incorrect. Reverse transcriptase, choice C, isanother retroviral enzyme. Reverse transcriptase synthesizes DNA from an RNA template. Because DNA replication involvesthe 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 self- replicating 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 existindefinitely 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 haveexamined 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 correctand choice B is the correct answer.
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 2 for your answer. From Table 2 you know that thebacteria infected with Phage I were able to grow on agar plates containing penicillin. You also know from the passage andTable 2 that the uninfected wild-type bacterial cells do not contain the gene that codes for beta-lactamase, which conferspenicillin resistance, since no bacterial growth was observed when the control dilution was incubated in the presence ofpenicillin. So, combining these two pieces of information, you can conclude that Phage I must be responsible for theobserved penicillin resistance in Experiment 2. 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 1 and 2 when the bacteria were infected with Phage I. Choices A, B, and C, however, are all possible ways in which the bacterial cells couldhave become resistant to penicillin. So, how do you decide between these three? Well, you’re asked to determine your answerbased on the experimental data. Thus, choice C is the only one that could be concluded on this basis--the penicillin resistanceobserved in the bacteria infected with Phage I in Experiment 2 must have been conferred by the insertion of Phage I DNA thatcontained the gene that codes for beta-lactamase. Thus choice C is the correct answer. The passage does not state the exactmechanism by which this gene functions and therefore you are not expected to know this. Perhaps beta-lactamase doesfunction via the mechanisms proposed in either choice A or B--by effecting the bacterium cell wall, or by reducing thepenicillin on the agar plate. However, neither of these can be concluded from the experimental data. The only thing you DOknow about penicillin resistance is that the gene for beta-lactamase must be present. Why? Because the only differencebetween these bacterial cells and the wild-type cells of the control group is the Phage I infection. Therefore, choices A and Bare 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 1.
According to Table 1, both the control bacteria and the bacteria infected with Phage I were able to grow in the presence of allthree sugars--two of which are disaccharides. On the other hand, the bacteria infected with Phage II were only able to grow onglucose. This means that Phage II is somehow disrupting the bacteria’s metabolism of sucrose and lactose. Since you know that Phage I does not have any deleterious effects on bacterial metabolism, as indicated by the ability to grow on all threeplates, choices A and B can be eliminated. If Phage I DNA was incorporated into the region of the bacterial chromosome thatcodes for the enzymes of glycolysis, the bacteria would have been unable to grow on glucose, since glycolysis provides theATP required for growth. And if Phage I prevented the entrance of lactose and sucrose into the cells, the bacteria would havebeen unable to grow on either of these sugars, since the cells would not have had access to metabolic fuel. Since neither ofthese scenarios is the case, we’ve narrowed it down to either choice C or choice D. Since viruses are not autonomous lifeforms, they would be unable to utilize any sugar; viruses do 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 thatdigest disaccharides such as sucrose and lactose, WOULD disrupt the metabolism of these nutrients. Therefore, you wouldNOT expect to see any bacterial growth when Phage II-infected bacteria were incubated with either sucrose and lactose, whichis exactly what we see happening in Table 1. 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 2 to get this question right. In Table 2, you see that the control bacteria are unable to grow in the presence of either tetracycline orpenicillin. Therefore, the wild-type bacteria does NOT have any natural resistance to these two antibiotics and so Romannumeral IV is correct, and choice A can thus be eliminated. You can also eliminate Roman numeral III, since we have justestablished that the wild-type bacteria has no natural resistance to penicillin. Therefore, choice C must also be wrong. FromTable 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 codesfor beta-lactamase, since we know that this is the enzyme that confers penicillin resistance. So Roman numeral I must becorrect and Roman numeral II must be incorrect. Thus, choice D can be eliminated. This means that Roman numerals I andIV 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 spirochetesor 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 thethree groups; rod-shaped bacteria, choice A, include the common E. coli. Squamous, choice C, refers to the layer of epithelialcells that typically perform protective functions, composing the outer layers of the skin and the lining of the mouth and othermucous membranes. So obviously choice C is also incorrect. Again, choice D is the correct answer.
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. Sobasically, all this question is asking you is to determine in which cycle of viral infection does a bacteriophage assemble newviral particles and lyse its host cell. Here’s where your knowledge of introductory biology comes into play. You shouldremember that when certain phage infect bacteria, one of two events may occur. The viral DNA may enter the cell and set upan 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. Thereforechoice 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, becomesactivated and sets up a new lytic cycle. As a matter of fact, this passage related to strains of phage that began their infection ina lysogenic stage. Remember, the passage specifically tells you that the phage DNA was inserted, or integrated into thebacterial genome. Therefore choice B is incorrect. As for choice C, mitosis has nothing to do with viral infection. Mitosis isthe nuclear division of eukaryotic cells characterized by chromosome replication and formation of two identical daughternuclei. 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 theexperimental protocol. It could have really slowed you down if you had gotten too enmeshed in this, especially since none ofthe questions actually required you to refer back to the protocol. Focus on the rationale of the experiments; don’t worry aboutthe minutiae. Remember, the passage isn’t going anywhere; you can refer back to it at any time if necessary.
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 typicallyconsist of a head made of a protein coat and a core that contains nucleic acid. Bacteriophage also contain a tail made of proteinthat is specialized for attaching to bacteria. Upon infection, bacteriophage can enter one of two cycles. In the lytic cycle, theviral nucleic acid enters the bacterial cell and begins using host machinery to produce new virions, which eventually cause celllysis. In the lysogenic cycle, the viral DNA integrates into the bacterial chromosome, replicating with it and being passed onto daughter cells in this inactive form. Once an integrated virus, called a prophage, becomes activated, it enters the lytic cycle.
Now let’s take a look at retroviruses. Retroviruses are RNA-containing viruses that replicate through a DNA intermediate byvirtue of a viral-coded RNA-dependent DNA polymerase, known as reverse transcriptase. The most notable retrovirus ishuman immunodeficiency virus (HIV), the causative agent of AIDS. Retroviral life-cycle consists of four main events. Thevirus binds to its host, typically an animal cell, and injects RNA and a few viral enzymes. The RNA is then converted toDNA 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 isincorrect. Choice C is also incorrect; as its name implies, bacteriophage only infect bacteria. Bacteria, the target ofbacteriophage, have no real immune system to suppress, and so choice D must also be wrong. Again, choice A is the correctanswer.
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 thatdegrades DNA by attacking the bonds within a DNA molecule. So it makes sense that the bacterial cell would die, because allbacteria store their genetic material in the form of DNA. If certain viruses are still able to synthesize proteins followingexposure to DNase, then these viruses must be RNA viruses. Thus, choice C is the right answer. Choice A is wrong becausea viral protein coat is not able to denature DNase. Denaturation involves the breaking of bonds due to either heat or chemicaltreatment. Choice B is wrong, because while it is true that viral genomes typically contain multiple reading frames so thatthey 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 hydrolyzingDNA. Again, choice C is the correct answer.
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 notboth; 2) viruses do not have the metabolic machinery for energy production or protein synthesis and are forced to rely on themachinery of their host; 3) viruses are unable to reproduce themselves directly, instead relying on their host’s machinery forreplication; and 4) viruses have no membranes to regulate the entry or exit of material. So based on this information, youknow that if the experimental drug blocks viral protein synthesis, it is likely that it also blocks host protein synthesis, sinceit is the host’s metabolic machinery that synthesizes viral proteins. Thus you would expect this drug to also affect hostprotein synthesis, which would be considered a side effect of the drug. Therefore choice A is the correct answer and choice B iswrong. About choice B: just because a drug has an unwanted side effect does not necessarily mean that the drug is not aneffective treatment. Choice C is incorrect because the blockage of host cell surface receptors would not prevent the synthesisof 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 theviral coat proteins themselves. Again, choice A is the correct answer.
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 formmycelium; 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 asexualphases, meaning that haploid and diploid states are both possible. In addition, fungi are heterotrophs that obtain nutrientsthrough absorption. Again, choice B is the correct answer.
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 ofeither RNA or DNA. Therefore, you would expect the core antigen to be composed primarily of nucleic acids and the coat, orsurface, antigen to be composed primarily of proteins. Remember, an antigen is just something that invokes an immuneresponse in a host organism. Thus, choice A is wrong and choice B is the correct answer. Let’s just take a look at the otherchoices quickly. From your knowledge of viral replication, you know that the viral coat and the viral core are synthesizedseparately and then combined to form a virion. When a host cell lyses, complete viral particles, as well as any unassembledviral components, namely the coat particle, HBsAg, and the core particle, HBcAg, may be released. In addition, the contentsof the host cell itself will also be released. And, since you’re told that the antigens are found in the serum, it is notunreasonable to expect the enzymes of the host liver cells to be in the serum too. In fact, a rise in certain liver enzymes iskey to hepatitis diagnosis. Therefore, choices C and D can be eliminated. Also, since incomplete viral particles are alsoliberated into the serum, it might be expected that some of these viral products might also appear as antigens, such as viralDNA polymerase for example. Again, choice B is the correct answer.

Source: http://c.t.gt/kaplan%20mcat/Biology%20Microbiology.pdf

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