Microbiology MCAT Biology Diagnostic Exam 2A

Question 1

During prophase II of meiosis:

A. there are two chromatids per chromosome and the cell is diploid.
B. there are two chromatids per chromosome and the cell is haploid.
C. there is one chromatid per chromosome and the cell is diploid.
D. there is one chromatid per chromosome and the cell is haploid.

Answer 1

B. Cells undergoing meiosis are haploid after the first cytokinesis event (choices A and C are wrong). In prophase II, the sister chromatids have not separated yet (choice B is correct and choice D is wrong).

Concepts tested
Cell Biology: Mitosis/Meiosis

Question 2

A scientist has had a stationary-phase culture of E. coli at 4°C for several weeks. If she takes a small amount of this culture and puts it into 100 mL of new media (shaking at 37°C overnight), which of the following will be observed?

A. The culture will immediately start growing exponentially and will be in the stationary phase by morning.
B. After the lag phase, the culture will start growing rapidly.
C. No bacterial growth will be observed because most of the original culture was dead.
D. The culture will be mostly transparent in the morning, as the cells will be in stationary phase.

Answer 2

B. Bacteria are very hardy cells. While the original culture was in stationary phase, all it takes is more food to get the cells growing again (choice C is wrong). Initially, the new culture will undergo a lag phase as the bacteria replicate their genomes and get ready to divide (choice A is wrong), but then the cells will start undergoing binary fission and the culture density will increase exponentially. This is termed the log phase (choice B is correct). If the culture were to reach stationary phase again after undergoing logarithmic growth, it is more likely to be turbid (not clear) due to the high concentration of bacteria. Furthermore, depending on media and other conditions, we cannot be certain of the culture reaching the stationary phase again overnight (choice D is wrong).

Concepts tested
Microbiology: Bacteria

Question 3

The viral capsid:

A. is made of protein and functions in attachment to the host.
B. is coded for and translated by host machinery.
C. is a complex mixture of amino acids and rRNA.
D. can form many shapes and is composed of peptidoglycan.

Answer 3

A. Only choice A is a true statement. While the host translates viral proteins, the viral genome must code for them (choice B is wrong). There is no rRNA component to the protein capsid (choice C is wrong). While the viral capsid can take many shapes, it is not made out of peptidoglycan; this is what forms the bacterial cell wall (choice D is wrong).

Concepts tested
Microbiology: Viruses and Subviral Particles

Question 4

Which of the following provides the best description of diffusion?

A. A solute moves passively down its concentration gradient in a thermodynamically favorable way, increasing entropy.
B. A solute moves actively down its concentration gradient in a thermodynamically favorable way, increasing entropy.
C. A solute moves passively down its concentration gradient in a thermodynamically favorable way, decreasing entropy.
D. A solute moves actively down its concentration gradient in a thermodynamically favorable way, decreasing entropy.

Answer 4

A. Diffusion is the passive movement of a solute down its concentration gradient (choices B and D are wrong). In order for the movement to be passive, it must be thermodynamically favorable and will increase the entropy or measurement of disorder in the system (choice C is wrong and choice A is correct).

Concepts tested
Cell Biology: Osmosis/Diffusion/Colligative Properties

Question 5

Polydactyly (extra fingers or toes) is a congenital physical anomaly that can be caused by recessive or dominant alleles. It is found more frequently in blacks than in whites, and more frequently in men than in women. A study on autosomal recessive polydactyly in black women showed the incidence of this condition to be 10 in 1000 births. What is the frequency of the allele causing polydactyly in this population?

A. 0.001
B. 0.01
C. 0.1
D. 1.0

Answer 5

C. If 10 in 1000 births have polydactyly, then the condition is found at a frequency of 1% in this population. In the Hardy-Weinberg equation for genotype frequency (p2 + 2pq + q2 = 1), q2 is the frequency of the autosomal recessive genotype. Since the question states that polydactyly (at least in this study) is an autosomal recessive disorder, q2 = .01, so q (the frequency of the recessive allele) equals 0.1.

Concepts tested
Genetics: Hardy-Weinberg

Question 6

Cytokinesis occurs during which two phases of mitosis?

A. Prophase and metaphase
B. Metaphase and anaphase
C. Anaphase and telophase
D. Telophase and prophase

Answer 6

C. Cytokinesis refers to physical process of cell division once the replicated genome has been appropriately divided (choices A, B, and D are wrong). This division starts during anaphase and is completed during telophase at which time the nuclear membranes will re-form (choice C is correct).

Concepts tested
Cell Biology: Mitosis/Meiosis

Question 7

Which of the following is a true statement?

A. The best way to increase evolutionary fitness is to have many offspring, regardless of whether they survive to adulthood.
B. Natural selection acts on genetic diversity in populations and causes evolution.
C. Changes in gene expression are the basis of evolution in populations.
D. Divergent selection drives a population closer to the average trait.

Answer 7

B. Only choice B is a true statement. Fitness is defined as how well an organism passes its alleles to future generations. If few offspring survive to reproductive age, this does not correspond to increased fitness (choice A is false). Changes in allele frequency—not gene expression patterns—are the basis for evolution in both species and populations (choice C is false). Divergent selection removes members near the average and favors traits at the extremes. When a population is driven closer to the average trait, stabilizing selection has occurred (choice D is false).

Concepts tested
Evolution/Speciation: Natural Selection/Speciation

Question 8

A researcher isolates Gram-negative bacilli bacteria. This organism has a:

A. thick peptidoglycan cell wall and is spiral-shaped.
B. thin chitin cell wall and is round-shaped.
C. thick cellulose cell wall and is rod-shaped.
D. thin peptidoglycan cell wall and is rod-shaped.

Answer 8

D. Gram-negative bacteria have a thin cell wall (choices A and C are wrong) and an outer membrane which leads to a light pink staining. The bacterial cell wall is made of peptidoglycan; note that the plant cell wall is made of cellulose and the fungal cell wall is made of chitin (choice B is wrong). Bacilli means rod-shaped, cocci means round, and spirochetes or spirilla mean spiral-shaped (choice D is correct).

Concepts tested
Microbiology: Bacteria

Question 9

A researcher dissects testes from a mutant mouse and isolates individual gametes. Flow cytometry analysis shows that there are three populations of cells. Population I has 20 chromosomes, Population II has 19 chromosomes, and Population III has 21 chromosomes. Which of the following is most likely?

A. The meiotic cells are unable to complete meiosis I.
B. Mitosis of the spermatogonia is occurring very slowly.
C. Nondisjunction in meiosis I resulted in an abnormal karyotype in gametes.
D. Nondisjunction in anaphase II led to aneuploidy in some but not all gametes.

Answer 9

D. Nondisjunction during meiosis I leads to two cells with too many chromosomes (n + 1) and two cells with too few (n – 1). Thus, nondisjunction in meiosis I would only produce two populations of cells (choice C is not correct). Nondisjunction during meiosis II leads to two cells with the correct number of chromosomes (n), one cell with an extra chromosome (n + 1) and one cell that is missing a chromosome (n – 1). This could account for the three populations of cells isolated by the researcher (choice D is correct, note that aneuploidy is just an abnormal number of chromosomes). There is no information to support the fact that meiosis is not being completed (choice A is wrong), or that there is limited mitosis of spermatogonia (this would lead to fewer gametes, but not a change in chromosome number; choice B is wrong).

Concepts tested
Cell Biology: Mitosis/Meiosis

Question 10

The viral genome integrates into the host genome during the lysogenic cycle. After this:

A. the host genome is not expressed, due to virus-encoded repressor proteins.
B. the viral genome is silent, but replicated along with the host genome.
C. the virus genome excises and activates once the host cell is dead or dying.
D. excision of the viral genome is very precise and occurs only when the host cell is under stress.

Answer 10

B. In the lysogenic cycle, the viral genome inserts into the host genome. The host genome continues to be transcribed and translated, but the viral genome is silent due to viral-encoded repressor proteins (choice A is wrong). Both the host and viral genomes are copied during DNA replication (choice B is correct). The viral genome is activated and excised when the host cell is stressed, and will enter the lytic or productive cycle. If the virus waits until the host cell is dead, it will certainly not be able to reproduce (choice C is wrong). Finally, excision of the viral genome is relatively imprecise. Some of the host genome can also be excised and packaged into viral particles. This new bit of DNA can be transferred to the next host on subsequent infection (transduction, choice D is wrong).

Concepts tested
Microbiology: Viruses and Subviral Particles

Question 11

Which of the following describes a trafficking pathway through the Golgi complex?

A. Trans stack, medial stack, cis stack, anchor to cell membrane
B. Pick up via endocytosis, cis stack, medial stack, trans stack
C. Medial stack, cis stack, trans stack, release via exocytosis
D. Cis stack, medial stack, trans stack, anchor to cell membrane

Answer 11

D. The portion of the Golgi complex closest to the endoplasmic reticulum is its cis stack so this is the first step in trafficking (choice D is correct). From there, proteins travel to the medial stack and finally the trans stack, the Golgi component farthest from the endoplasmic reticulum. At this point, the protein can be released via exocytosis or anchored to the cell membrane, depending on its function (choices A, B, and C are wrong).

Concepts tested
Cell Biology: Cellular Organelles and Structures

Question 12

As the level of cellular specialization increases, what cell cycle activity decreases?

A. DNA replication
B. Cell growth
C. Genome packaging
D. Genome transcription

Answer 12

A. Cellular specialization typically leads to a decrease in cell division, i.e., highly differentiated cells do not divide. There is no point to replicating DNA (the S phase of the cell cycle) if division is not going to occur (choice A is correct). Though not dividing, specialized cells can still grow (choice B is wrong), genomes may be packaged or unpackaged as needed in order to access required genes (choice C is wrong), and transcription (and translation) of needed cellular products will also continue (choice D is wrong). Choices B, C, and D all occur during the G1 phase of the cell cycle.

Concepts tested
Cell Biology: Mitosis/Meiosis

Question 13

Eye color is a sex-linked trait in the fruit fly Drosophila melanogaster. A pure-breeding red-eyed female is mated with a pure-breeding white-eyed male. All offspring have red eyes. If an F1 female was backcrossed, which of the following would be observed in the F2?

A. All female flies are red-eyed and all male flies are white-eyed.
B. Half the female flies have white eyes and the other half have red eyes; all male flies have white eyes.
C. All female flies are white-eyed and the males are 50% red-eyed and 50% white eyed.
D. Half the flies have red eyes and half have white eyes, regardless of sex.

Answer 13

D. If all the F1 flies have red-eyes, you know the red eye trait is dominant. Let’s assign R = red and r = white. The parental (P) cross is female XRXR × male XrY. The F1 flies would be XRXr (red-eyed females that carry the white allele but don’t express it) and XRY (red-eyed males). The next cross in the question is between a XRXr female and the XrY male from the parental generation (remember that a backcross is when an individual is crossed to a previous generation). The resultant F2 flies would be 25% XRXr (red-eyed females), 25% XrXr (white-eyed females), 25% XRY (red-eyed males) and 25% XrY (white-eyed males). You can see that regardless of sex, half the offspring will have white eyes and half will have red eyes (choice D is correct).

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 14

Mitochondria share what characteristic with prokaryotic cells?

A. Presence of membrane-bound organelles
B. Maternal inheritance pattern
C. Lack of ribosomes
D. Single chromosome

Answer 14

D. Both mitochondria and prokaryotes have a single, distinct chromosome (choice D is correct). Mitochondria are a type of membrane-bound organelle, but do not contain such structures (choice A is wrong). Maternal inheritance refers to the inheritance of the mitochondria and its genome exclusively from the mother (as are all organelles); such a pattern does not exist in prokaryotes (choice B is wrong). Both mitochondria and prokaryotes have ribosomes (choice C is wrong).

Concepts tested
Cell Biology: Cellular Organelles and Structures

Question 15

The lytic and productive viral cycles are similar in that they both:

A. destroy the host cell to allow viral particle release.
B. involve a virus using host cellular machinery to replicate the viral genome and capsid.
C. require the viral genome to be integrated into the host genome.
D. involve expression of hydrolase to generate dNTP building blocks, and lysozyme to allow viral release.

Answer 15

B. While there are many similarities between the lytic and productive viral cycles, there are many differences also. In both cycles, the virus takes over certain parts of the host cell, to allow production of viral proteins and to replicate the viral genome (choice B is correct). At the end of a lytic cycle the host cell lyses and dies to allow release of viral particles, however in the productive cycle, viruses bud out of the host cell; this allows the host to survive, and generates viruses with an outer envelope (choice A is only true of the lytic cycle and can be eliminated). The viral genome integrates into the host genome in the lysogenic cycle (choice C is not true of either the lytic or productive cycles and can be eliminated). Hydrolase is a lytic cycle early gene, and lysozyme is a lytic cycle late gene (choice D is only true of the lytic cycle and can be eliminated). Hydrolase degrades the host genome to generate dNTP building blocks. Lysozyme degrades the bacterial cell wall, allowing bacterial lysis.

Concepts tested
Microbiology: Viruses and Subviral Particles

Question 16

A family pedigree is being constructed for the Laing family. The hairy ear phenotype is consistently passed from fathers to sons. Having dry ear wax has skipped a few generations, but affects Laing males and Laing females at the same frequency. Which of the following is most likely?

A. Dry ear wax is an X-linked recessive trait and hairy ears is a mitochondrial trait.
B. Dry ear wax is an autosomal recessive trait and hairy ears is Y-linked.
C. Dry ear wax is an X-linked dominant trait and hairy ears is Y-linked.
D. Dry ear wax is autosomal dominant and hairy ears is X-linked recessive.

Answer 16

B. If dry ear wax skips generations, it must be a recessive trait. Classically dominant traits cannot be hidden and then show up in future generations (choices C and D can be eliminated). Furthermore, equal distribution among sexes likely rules out a sex-linked trait; dry ear wax must be an autosomal recessive trait (choice A can be eliminated and choice B is correct). Note that traits that pass from fathers to sons (such as the hairy ear phenotype in this question) are coded by the Y chromosome.

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 17

Which of the following could lead to cancer?

An overexpression of initiator caspases, leading to apoptosis
A mutation in the tumor suppressor gene p53 that prevents the expression of the p53 protein
A mutation in the protooncogene Ras that prevents the expression of the Ras protein
A. I only
B. II only
C. I and II only
D. II and III only

Answer 17

B. Item I is false: apoptosis is programmed cell death, triggered when cell damage is critical. Apoptosis is used to destroy potential cancer cells before they become a problem (choices A and C can be eliminated). Both remaining answer choices include Item II so it must be true: p53 is a tumor suppressor gene, halting cell growth and division when the cell is damaged, or triggering apoptosis if the damage is too severe. Preventing the expression of p53 would prevent these effects and could lead to cancer. Item III is false: protooncogenes are genes that when damaged or mutated, become oncogenes, triggering inappropriate cell division and cancer. If Ras protein expression is prevented, then cancer is less likely (choice D can be eliminated and choice B is correct).

Concepts tested
Cell Biology: Cancer

Question 18

In lab mice, agouti coat color is dominant over chinchilla fur, and black eyes are dominant over pink eyes. If a pure-breeding male agouti black-eyed mouse is crossed with a pure-breeding female chinchilla pink-eyed mouse, then one of the male pups is backcrossed, what is the probability of having a black-eyed chinchilla female F2?

A. 0.5
B. 0.875
C. 0.125
D. 0.75

Answer 18

C. Let’s assign alleles as A = agouti, a = chinchilla, B = black, and b = pink. The original cross is AABB × aabb. All F1s are AaBb and a male from this group is then backcrossed (must be to the female parent, so AaBb × aabb). The probability of getting chinchilla fur (aa) is 0.5. The probability of getting black eyes (Bb) is 0.5. The probability of having a female mouse is 0.5. Therefore, overall the answer is (0.5)(0.5)(0.5) = 0.125 (choice C is correct).

Concepts tested
Genetics: Mendelian

Question 19

A virologist dips his pipette tip into a plaque on a bacterial plate, then into an actively growing culture of E. coli cells. A new strain of E. coli results. Which of the following best explains what occurred?

A. The new E. coli strain is the result of transduction with a lysogenic phage.
B. The new E. coli strain must have acquired random genomic mutations such as deamination.
C. Antibiotics or toxins in the plaque exerted selective pressure on the bacterial culture.
D. The plaque contained bacterial cells that underwent conjugation with the second culture.

Answer 19

A. A plaque is a clear area on a plate otherwise covered in bacterial cells. It can be caused by addition of a toxin, antibiotic, or virus; each of these can kill bacteria and would generate a clear area. Since whatever was in the plaque did not outright kill the actively growing culture, it is not likely a toxin or antibiotic (choice C is wrong). Most likely, the plaque was a region of dead bacteria due to viral infection, and contains active virus. Infection of new cultures with active virus can sometimes result in new strains, if the virus transfers some DNA from its previous host; this process is known as transduction (choice A is the most likely scenario). Since a plaque is typically an area of dead bacteria, choice D is wrong; dead bacteria cannot undergo conjugation. Random genomic mutations are possible but less likely than transduction (choice B is wrong).

Concepts tested
Microbiology: Bacteria

Question 20

Alpha tubulin and beta tubulin form dimers which are linked to form a protein sheet. This sheet is rolled into a tube which is the basis for a(n):

A. intermediate tubule.
B. microfilament.
C. intermediate filament.
D. microtubule.

Answer 20

D. The alpha and beta tubulin proteins are used to compose microtubules (choice D is correct). The rolled sheet is then anchored in the microtubule organizing center; the free end can be lengthened or shortened as needed. Microfilaments are composed of actin (choice B is wrong), intermediate filaments are a mix of heterogeneous polypeptides (choice C is wrong), and intermediate tubules are not an actual cytoskeletal component (choice A is wrong).

Concepts tested
Cell Biology: Cellular Organelles and Structures

Question 21

A researcher hypothesizes that the photic sneeze reflex (a condition where exposure to the sun causes uncontrollable sneezing) is due to a dominant autosomal allele. To test this hypothesis, he finds a family that has this trait in some individuals and not others, and generates a pedigree for analysis. To confirm his hypothesis, the researcher should look for:

A. a pattern where two normal individuals have a child with photic sneeze reflex.
B. two parents with photic sneeze reflex, that have only affected children.
C. a testcross between two normal individuals with affected parents.
D. two parents with photic sneeze reflex that have normal and affected offspring in a 1:3 ratio, respectively.

Answer 21

D. Choice A would support the opposite hypothesis, that photic sneeze reflex is recessive (recessive conditions typically skip generations; choice A is wrong). Choice B could be true whether the trait was dominant or recessive, assuming both parents were homozygous (choice B is wrong). A testcross between two individual with the same genotype wouldn’t be helpful, and besides, humans are resistant to participating in testcrosses (choice C is wrong)! Two parents with photic sneeze reflex would have three photic sneeze reflex children for every one normal child, if both parents were heterozygotes and the photic sneeze reflex allele was dominant (choice D is correct).

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 22

A membrane is impermeable to a solute, but the solute moves passively across the membrane, down its concentration gradient, using a specific integral membrane protein. This describes:

A. simple diffusion.
B. facilitated diffusion.
C. primary active transport.
D. secondary active transport.

Answer 22

B. The use of an integral membrane protein makes this facilitated diffusion (choice B is correct) since simple diffusion does not require a transport protein (choice A is wrong). Isotonic and hypotonic do not describe types of diffusion, but rather refer to comparative solute concentrations between two locations (choices C and D are wrong).

Concepts tested
Cell Biology: The Cell Membrane

Question 23

Which of the following is a hallmark of linkage?

A. Linked genes function in similar cell processes and pathways.
B. Linked alleles tend to be inherited together.
C. Linked loci are found in similar locations on homologous chromosomes, but can be far apart on sister chromatids.
D. Linked genes must be less than 50 map units apart on different chromosomes.

Answer 23

B. Linkage occurs when two genes are on the same chromosome (choice D is wrong) and less than 50 map units apart. This leads to decreased recombination, and linked alleles being inherited together (choice B is correct). For example, if genes J and N are linked, and a woman is genotype (Jn/jN), she will make more Jn and jN gametes (parental) than jn and JN gametes (recombinants). Linked genes do not have to function in similar pathways (choice A is wrong). A given gene (and any others it may be linked to) is always found in the same location on chromosomes and chromatids (choice C is wrong).

Concepts tested
Genetics: Linked Genes

Question 24

Which of the following is true regarding viral entry into a host cell?

A. Viral attachment is relatively random, and this helps viruses evolve quickly.
B. Attachment can also be called eclipse, and penetration can also be called adsorption.
C. Prokaryotic viruses enter their host via receptor-mediated membrane fusion.
D. Animal viruses can enter their host via endocytosis, but this requires a specific receptor on the host surface.

Answer 24

D. All viruses are very specific regarding their host cell. This specificity comes from the requirement of a virus to bind a host cell receptor to allow viral entry (choice A is wrong). Attachment can also be called adsorption, and penetration can also be called eclipse (choice B is wrong). Prokaryotic viruses lack an envelope and therefore cannot undergo membrane fusion with their host. Instead, they rely on injection of viral contents (choice C is wrong). Animal viruses can enter their host cell via membrane-fusion (where the envelope of the virus fuses with the host plasma membrane) or endocytosis (choice D is correct). In both cases, receptor binding is still required, and the viral genome is uncoated after entry into the host. A few animal viruses insert their genome without being taken up entirely, but this is rare.

Concepts tested
Microbiology: Viruses and Subviral Particles

Question 25

Which of the following are enzymes directly activated by G-protein linked receptors?

Adenylate cyclase
Phospholipase C
Tyrosine kinase
  	A.  I only
  	B.  I and II only 
  	C.  II and III only
  	D.  I, II, and III

Answer 25

B. Item I is true: adenylate cyclase can be directly activated by stimulatory (Gs) G-protein linked receptors (choice C can be eliminated). Item II is true: phospholipase C is another enzyme that can be directly activated by G-proteins (choice A can be eliminated). Item III is false: some tyrosine kinases are cytoplasmic and can be activated as part of the rise in cAMP due to activation of adenylate cyclase by a G-protein. However, this is an indirect activation by a G-protein, and the question specifically asks which are directly activated (choice D can be eliminated and choice B is correct).

Concepts tested
Cell Biology: G Proteins/Signal Transduction

Question 26

If a Hfr cell mates with an F– bacterium, which of the following is true?

A. Both cells will become Hfr strains.
B. The Hfr bacterium becomes female (F–), while the female strain becomes Hfr.
C. The female bacterium could end up female (F–), male (F+), or Hfr.
D. The mating is impossible since only F+ bacteria (male) can perform conjugation with F– (female) cells.

Answer 26

C. The Hfr strain has the F factor in the genome, male (F+) bacteria contain the F factor as a plasmid, while the female bacteria (F–) has no F factor. Both male (F+) and Hfr strains can mate with female strains (choice D is wrong). The Hfr cell keeps a copy of its genome (choice B is wrong) and passes a copy to the female cell, through the conjugation bridge (or sex pilus). The female cell can receive a portion of the Hfr chromosome, or a copy of the whole thing (if the cells stay connected for a longer time). The F factor is the final gene transferred during conjugation; if the F factor doesn’t make it over to the female cell, she will stay female (but may acquire new genetic traits). If the F factor is transferred, it can end up in a plasmid (the cell would then be male, F+) or the genome (the cell would then be Hfr). While choice A is possible, choice C is more likely.

Concepts tested
Microbiology: Bacteria

Question 27

Black coat color is dominant to brown coat color in Labrador retrievers. A second gene (E or e) controls expression of the fur pigment gene, where the dominant allele allows pigment expression and the recessive allele prevents pigment expression. Labradors lacking black or brown pigment are referred to as “golden”. If a dihybrid (heterozygous) male is bred to a homozygous recessive female, which of the following would be expected?

A. The black lab and brown lab parents will have all black lab puppies.
B. The black lab and golden lab parents will have black, brown, and golden puppies in approximately equal proportions.
C. The black lab and brown lab parents will have half black and half golden puppies.
D. The black lab and golden lab parents will have black, brown, and golden puppies, with golden coat color being the most common.

Answer 27

D. Let’s assign B = black and b = brown. For the second gene, E allows pigment expression and e does not (this leads to the golden labs). The cross in the question stem is BbEe (a black lab) × bbee (a golden lab), so choices A and C can be eliminated. The genotypic ratio of the puppies would be 25% BbEe (black lab), 25% Bbee (golden lab), 25% bbEe (brown lab) and 25% bbee (golden lab). Overall, 25% of the puppies would have black fur, 25% would have brown fur, and 50% would have golden fur (due to pigment expression being turned off).

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 28

Which of the following has a vesicle fuse with the plasma membrane in order to release material from the cell?

A. Phagocytosis
B. Clathrin-coated pits
C. Pinocytosis
D. Exocytosis

Answer 28

D. Exocytosis is the process by which the cell uses vesicles to move material out of the cell (choice D is correct). Phagocytosis and pinocytosis are large-scale and small-scale uptake, respectively (choices A and C are wrong), while clathrin-coated pits are involved in receptor-mediated endocytosis (choice B is wrong).

Concepts tested
Cell Biology: Secretory Pathway

Question 29

Which of the following is a FALSE statement about (+) RNA viruses and (–) RNA viruses?

A. Both types of viral genomes must encode an RNA-dependent polymerase.
B. (+) RNA viruses must carry an RNA-dependent RNA polymerase.
C. Viral proteins can be directly translated from a (+) RNA genome but not a (–) RNA genome.
D. Injection of a (+) RNA genome into a host cell would result in infective activity.

Answer 29

B. (+) RNA viral genomes can serve as templates for transcription, but (–) RNA genomes are complementary to a transcript template (choice C is true and can be eliminated). As a consequence, both must encode an RNA-dependent RNA polymerase to replicate the genome (choice A is true of (+) and (–) RNA viruses and can be eliminated), but a (–) RNA virus must also carry a copy of this protein, to generate functional mRNA from the genome (choice B is false and the correct answer choice). Since a (+) RNA genome can serve as a template for protein production (i.e. mRNA), if a (+) RNA viral genome was injected into a host cell, it would be immediately infective (choice D is true and can be eliminated).

Concepts tested
Microbiology: Viruses and Subviral Particles

Question 30

A geneticist testcrosses a dihybrid (heterozygous for two genes) mouse and notices the double dominant and double recessive phenotypes are present at 8% and 10% frequency, respectively. Which of the following is the best conclusion from these data?

A. The dominant allele of one gene is linked to the recessive allele of the other, and the two genes are approximately 20 map units apart.
B. The dominant alleles of the two genes are linked and the two genes are 80 map units apart.
C. The recessive alleles of the two genes are approximately 20 map units apart and are not linked.
D. The two genes are linked and 80 map units apart, but it cannot be determined which alleles are being inherited together.

Answer 30

A. Let’s assign the dihybrid mouse a genotype of AaBb. The testcross would be AaBb × aabb and the expected ratios of offspring would be 25% AB/ab, 25% Ab/ab, 25% aB/ab and 25% ab/ab (in linkage notation). The question stem says AB/ab is present at only 8% and ab/ab is present at only 10%. These two combinations must represent the recombinant offspring given their low percentages, Ab/ab and aB/ab must be the parental combination of alleles, and the genotype of the dihybrid parent must have been Ab/aB. In other words, the two genes are linked and the dominant allele of one gene is linked to the recessive allele of the other. To determine how far apart the genes are, we can calculate Rf, where Rf = (# recombinants / total) x 100% = [(8% + 10%) / 100%] x 100% = 18% recombination frequency = 18 cM = 18 map units. Overall, the best answer is choice A.

Concepts tested
Genetics: Linked Genes

Question 31

Which of the following is an INACCURATE statement about bacterial cell structure?

A. The peptidoglycan cell wall protects against osmotic pressure gradients.
B. The 30S/50S ribosome performs translation in the cytoplasm.
C. ATP synthesis occurs across the plasma membrane.
D. Cyanobacteria have photosynthetic machinery in the chloroplast and cytoplasm.

Answer 31

D. All these statements are true and accurate except choice D. Cyanobacteria perform photosynthesis, but since bacteria contain no membrane-bound organelles (including chloroplasts), the photosynthetic machinery is located in the cell membrane and the cytoplasm (similar to the cell respiration machinery). Note that it is thought that chloroplasts in plants and algae evolved from an endosymbiotic relationship with cyanobacteria.

Concepts tested
Microbiology: Bacteria

Question 32

Red-green colorblindness is an X-linked recessive trait in humans. If a carrier female mates with a normal male, what is the probability they will have a colorblind son?

A. 0.75
B. 0.5
C. 0.25
D. 0.125

Answer 32

C. If we assign D = normal and d = colorblind, the cross in the question stem is XDXd × XDY. The probability of receiving the colorblind allele from the mother is 0.5. The probability of receiving the Y chromosome from the father is 0.5. Therefore, the probability they will have a colorblind son is (0.5)(0.5) = 0.25 (choice C is correct).

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 33

The nucleolus is a defined sub-region of the nucleus responsible for what cellular function?

A. Ribosome subunit assembly
B. Protein synthesis
C. ATP production
D. Lipid breakdown

Answer 33

A. The nucleolus assembles ribosome subunits from folded rRNA and various protein components (choice A is correct). Protein synthesis happens on the ribosomes themselves (choice B is wrong). ATP production is the job of the mitochondria (choice C is wrong) while peroxisomes break down lipids not needed by the cell (choice D is wrong).

Concepts tested
Cell Biology: Cellular Organelles and Structures

Question 34

All of the following statements about the cell membrane are true EXCEPT:

A. lipid and protein content make up equivalent parts of cell membranes.
B. as lipid saturation increases, lateral movement of proteins decreases.
C. lipids and proteins move two-dimensionally, but do not tend to invert.
D. lipids in the membrane can often be found grouped in structures called rafts.

Answer 34

A. Cell membranes are composed of a lipid bilayer where the outer facing head groups are polar and the inner facing, intercalating tails are nonpolar. This creates the hydrophobic barrier that is the hallmark of membranes in aqueous environments. Cell membranes are interspersed with proteins, but the number of proteins varies based on the cell type so it is not always equal to the lipid content (choice A is false and is the correct answer choice). Increased lipid saturation makes the membrane more rigid, thus decreasing the ability of proteins within it to move (choice B is true and can be eliminated). To invert, lipids or proteins would have to drag their hydrophilic regions (the parts interacting with the polar edges of the membrane) through the hydrophobic center of the lipid bilayer, and this is not thermodynamically favorable (choice C is true and can be eliminated). Lipid rafts are a common component of membranes and a property of the mosaic model (choice D is true and can be eliminated).

Concepts tested
Cell Biology: The Cell Membrane

Question 35

Which of the following best describes primary active transport?

A. Directly coupling the movement of a molecule down its concentration gradient to ATP hydrolysis
B. Indirectly coupling the movement of a molecule down its concentration gradient to ATP hydrolysis
C. Directly coupling the movement of a molecule against its concentration gradient to ATP hydrolysis
D. Indirectly coupling the movement of a molecule against its concentration gradient to ATP hydrolysis

Answer 35

C. Active transport is used to move molecules against their concentration gradients (choices A and B are wrong), hence the need for energy in order to make such movement possible. In primary active transport, that movement is directly linked to the hydrolysis of ATP (choice C is correct and choice D is wrong).

Concepts tested
Cell Biology: Osmosis/Diffusion/Colligative Properties

Question 36

Alcohol flush reaction is due to an accumulation of acetaldehyde after alcohol consumption and results in a red face and neck. It is due to a dominant missense polymorphism that encodes the enzyme acetaldehyde dehydrogenase (ALDH2); this allele, ALDH2*2, occurs at a frequency of 0.3 in the human population. What is the proportion of individuals in this population that have a red face after drinking alcohol?

A. 0.09
B. 0.21
C. 0.49
D. 0.51

Answer 36

D. The question stem says ALDH2*2 is dominant so we will assign it A; the normal allele is recessive and we will assign it a. In the equation for allele frequency (p + q = 1), p = A = 0.3, so q = a = 1 – 0.3 = 0.7). Since this trait is dominant, both homozygous dominants (AA) and heterozygotes (Aa) will express the phenotype. From the equation for genotype frequency (p2 + 2pq + q2 = 1), the proportion of AA in the population is p2 and the proportion of heterozygotes is 2pq. Therefore the answer is (0.3)2 + 2(0.3)(0.7) = 0.09 + 0.42 = 0.51, or 51% (choice D is correct). Alternatively 1 – q2 can be used in place of the above.

Concepts tested
Genetics: Hardy-Weinberg

Question 37

Solution A is 1 M glucose. Solution B is 1 M NaCl. Which of the following is true?

A. Solutions A and B have the same osmotic pressure.
B. If the two solutions were placed on opposite sides of a semipermeable membrane, the volume of Solution A would increase.
C. The boiling point elevation of Solution B is twice the boiling point elevation of Solution A.
D. The freezing point depression of Solution A is twice the freezing point depression of Solution B.

Answer 37

C. Although both solutions have the same molarity, they do not have the same number of particles. Because NaCl ionizes in solution into two particles, a sodium ion and a chloride ion, a 1 M NaCl solution actually contains 2 moles of particles. Osmotic pressure is directly related to the number of particles in a solution, so if Solution B has twice the number of particles than Solution A, they do not have the same osmotic pressure (choice A is wrong). Because Solution B has more particles, it has less room for water, and Solution A has the higher “water concentration.” Thus, if they were placed on opposite sides of a semipermeable membrane, water would flow by osmosis from Solution A (fewer particles and more water) to Solution B (more particles and less water). The volume of Solution B would increase (choice B is wrong). Boiling point elevation and freezing point depression are also related to the particle concentration and the van’t Hoff factor. Solution B, with twice the number of particles as Solution A, would have twice the boiling point elevation (choice C is correct) and twice the freezing point depression (choice D is wrong).

Concepts tested
Cell Biology: Osmosis/Diffusion/Colligative Properties

Question 38

A chemoheterotrophic bacteria that is a leucine auxotroph:

A. makes its own food from organic molecules and can survive without leucine supplements.
B. builds organic molecules from CO2 and must be given leucine supplements to survive.
C. requires organic molecules as a carbon and energy source and cannot synthesize leucine.
D. performs photosynthesis to obtain energy and can build organic molecules from glucose.

Answer 38

C. Chemotrophs get their energy from chemicals, not light (choice D is wrong). Heterotrophs rely on organic molecules made by other organisms; they cannot make them from CO2 (choice B is wrong). A leucine auxotroph cannot synthesize the amino acid leucine; to survive, leucine must be provided for the organism in its medium (choice A is wrong and choice C is correct).

Concepts tested
Microbiology: Bacteria