Genetics MCAT Biology Diagnostic Exam 2B

Question 1

A researcher has a population of flies with either white (r) or red eyes (R). A white-eyed female is mated with a red-eyed male. All resultant female flies have red eyes and all the males have white eyes. If two flies from the F1 generation are mated, which of the following would NOT be observed?

A. Exactly half the female F2 flies are pure-breeding.
B. 75% of the F2 offspring will have red eyes, and 25% will have white eyes.
C. Some of the F2 males could have white-eyed daughters, and others cannot.
D. 50% of the F2 female flies are carriers.

Answer 1

B. If the eye-color trait is being inherited differently in males and females, it must be a sex-linked trait. Since both females and males are affected, it must be an X-linked trait. The white-eyed female in the original cross must be XrXr and the male is XRY. All female F1 flies are XRXr (red-eyed) and males are XrY (white-eyed). If these two mate (XRXr × XrY), the F1 offspring will be 25% XRXr (red-eyed females that carry the white-eye allele; choice D would be observed and can be eliminated), 25% XrXr (pure-breeding white-eyed females; choice A would be observed and can be eliminated), 25% XRY (red-eyed males that cannot produce white-eyed daughters) and 25% XrY (white-eyed males that can produce white-eyed daughters; choice C would be observed and can be eliminated). Note that half the F2 generation has red eyes and half has white eyes (choice B would NOT be observed and is the correct answer choice).

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 2

The nuclear envelope makes the interior of the nucleus continuous with what other membrane-bound organelle?

A. Lysosome
B. Golgi complex
C. Endoplasmic reticulum
D. Mitochondria

Answer 2

C. The nuclear envelope subdivides the nucleus and is composed of two lipid bilayers. While the outer side faces the cytoplasm, the inner side faces the interior of the nucleus and is continuous with the endoplasmic reticulum (choice C is correct). The lysosome, Golgi complex, and mitochondria are fully distinct from the nucleus (choices A, B, and D are wrong).

Concepts tested
Cell Biology: Cellular Organelles and Structures

Question 3

typical viral genome:

A. is made of several linear DNA molecules, wrapped around histone proteins.
B. is a circular DNA molecule with one origin of replication.
C. can be made of DNA or RNA, and can be double- or single-stranded.
D. typically contains more thymine and adenine than guanine and cytosine.

Answer 3

C. Choice A describes a typical eukaryotic genome and choice B describes a typical prokaryotic genome (both choices are wrong). Choice C is a true statement, and the correct answer choice. Since viral genomes are so variable, there is no reason to believe choice D is true.

Concepts tested
Microbiology: Viruses and Subviral Particles

Question 4

Which of the following accurately describes the structure of a bacterium?

A. A circular single-stranded DNA genome is located in the cytoplasm.
B. 80S ribosomes can generate multiple peptide chains using one polycistronic mRNA template.
C. Transcription and translation can occur simultaneously, as both occur in the cytoplasm and no post-transcriptional modification is required.
D. Aquaporin proteins in the plasma membrane protect against the osmotic pressure gradient generated by the cell wall.

Answer 4

C. Choice C is a true statement and the correct answer choice. The bacterial genome is a circular, double-stranded DNA molecule (choice A is wrong). While many peptide chains can be made from a polycistronic mRNA, this is done by a 70S ribosome in bacteria. The eukaryotic ribosome is 80S (choice B is wrong). The peptidoglycan cell wall prevents lysis due to osmotic pressure, but doesn’t generate osmotic pressure gradients (choice D is wrong). Many eukaryotic cells have aquaporin water channels in the plasma membrane to facilitate osmosis.

Concepts tested
Microbiology: Bacteria

Question 5

A researcher is tracking two traits in a family. One trait is consistently passed from affected mothers to all offspring. The other trait passes from affected fathers to 0% of sons but 100% of daughters. The respective two traits are mostly likely:

A. autosomal recessive and Y-linked.
B. autosomal dominant and X-linked recessive.
C. mitochondrial and X-linked recessive.
D. mitochondrial and X-linked dominant.

Answer 5

D. All humans receive their mitochondrial genome from their mother’s egg, so mitochondrial traits are passed from mothers to all offspring (choices A and B can be eliminated). If the trait does not pass from fathers to sons, it is likely X-linked. X-linked recessive traits can be masked by a mother’s dominant X chromosome in female offspring. If the trait is showing up in all of their daughters, it is most likely an X-linked dominant trait (choice C can be eliminated and choice D is correct). In this case, when a father passes this X chromosome to a daughter, the allele for this trait is expressed phenotypically.

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 6

All of the following could help prevent cancer EXCEPT:

overexpression of the Ras protooncogene.
activation of initiator caspases upon cellular oxidative damage.
p53 activation in response to accelerated progression through the cell cycle.
A. I only
B. I and II only
C. II and III only
D. I and III only

Answer 6

A. Item I would not help prevent cancer: overexpression of protooncogenes would push the cell inappropriately into cell division (choice C can be eliminated). Item II could help prevent cancer: initiator caspases start the apoptotic pathway when cells are dividing inappropriately (choice B can be eliminated). Item III could help prevent cancer: p53 is a tumor suppressor gene that when active can halt progression through the cell cycle or can trigger apoptosis (choice D can be eliminated and choice A is correct).

Concepts tested
Cell Biology: Cancer

Question 7

The ability to taste phenylthiocarbamide depends on genetics. Humans with the dominant allele taste phenylthiocarbamide as very bitter and make up about 70% of the population. The allele that leads to phenylthiocarbamide being tasteless:

A. is recessive and occurs at a frequency of 0.55.
B. is dominant and occurs at a frequency of 0.70.
C. is recessive and occurs at a frequency of 0.84.
D. is epistatic and occurs at a frequency of 0.30.

Answer 7

A. Since the ability to taste phenylthiocarbamide is dominant we will assign it an allele frequency of p. From the equation for genotype frequency (p2 + 2pq + q2 = 1), individuals who have this trait will make up p2 + 2pq of the population, which the question states is 70%, or 0.70. The recessive allele leads to tasteless phenylthiocarbamide (choices B and D can be eliminated) and occurs at a frequency of q, so individuals who cannot taste phenylthiocarbamide occur at a frequency of q2. Since the question stem tells you p2 + 2pq = 0.70, q2 = 1 – 0.70 = 0.30, and q would be the square root of this number, or approximately 0.55 (choice A is correct and choice C is wrong).

Concepts tested
Genetics: Hardy-Weinberg

Question 8

Staphylococcus aureus is a facultative anaerobic Gram-positive bacterium. This organism has a:

A. thick peptidoglycan cell wall and can use oxygen when it is present.
B. thin peptidoglycan cell wall and dies in the presence of oxygen.
C. thick peptidoglycan cell wall and will never perform aerobic cellular respiration or oxidative phosphorylation.
D. thin peptidoglycan cell wall and can perform fermentation and utilize an electron transport chain.

Answer 8

A. Gram-positive bacteria have a thick peptidoglycan cell wall (choices B and D are wrong). Facultative anaerobes can survive without oxygen, but will use it if it is present (choice A is correct and choice C is wrong). Obligate anaerobes die in the presence of oxygen, and tolerant anaerobes will survive in oxygen but will not use it.

Concepts tested
Microbiology: Bacteria

Question 9

The sodium-potassium ATPase moves three sodium ions out of the cell and two potassium ions into the cell at the cost of one ATP molecule. In the absence of ion leak channels, the ATPase would make the cytoplasm:

A. hypertonic to the extracellular environment.
B. isotonic to the extracellular environment.
C. hypotonic to the extracellular environment.
D. hyperosmotic to the extracellular environment.

Answer 9

C. Since the ATPase moves more ions out of the cell than in, and since in the absence of leak channels these ions have no way to cross the cell membrane, the cytoplasm would ultimately become hypotonic, meaning it has less particles when compared to the extracellular environment (choice C is correct). Hypertonic (or hyperosmotic) would mean that the cytoplasm had more particles than the extracellular environment (choices A and D are wrong), and isotonic would mean they had the same number of particles (choice B is wrong).

Concepts tested
Cell Biology: Osmosis/Diffusion/Colligative Properties

Question 10

Animal viruses differ from prokaryotic viruses in that they:

A. can enter a host cell without specificity for a receptor macromolecule.
B. usually enter the host via membrane-fusion or endocytosis.
C. secrete lysozyme to facilitate entry into the host cell.
D. always bind a receptor on the surface of the host cell, then inject the genomic material into the cytoplasm.

Answer 10

B. All viruses are very particular about the specifics of 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). Receptor binding allows a prokaryotic virus to inject its genome into the host cell (choice D applies to bacterial viruses and can be eliminated). In contrast, animal viruses usually enter the host cell via membrane-fusion (where the envelope of the virus fuses with the host plasma membrane) or endocytosis (choice B is correct). In both cases, receptor binding is still required, and the viral genome is uncoated after entry into the host. Lysozyme is an example of a late gene for prokaryotic viruses, and facilitates viral exit at the end of a lytic cycle. It is not used for viral entry, and in any case, does not apply to animal viruses (choice C is wrong).

Question 11

Autophagy in lysosomes is responsible for:

A. destroying excess secretory products.
B. degrading nonfunctional cellular components.
C. detoxifying cellular waste products.
D. phosphorylating secretory proteins.

Answer 11

B. Autophagy refers to the process by which the cell degrades worn out cellular components (choice B is correct). Excess secretory products are also destroyed in the lysosomes, but by the process of crinophagy (choice A is wrong). Peroxisomes detoxify waste products (choice C is wrong), and post-translational modification of proteins (such as phosphorylation) is the job of the endoplasmic reticulum and the Golgi complex (choice D is wrong).

Concepts tested
Cell Biology: Cellular Organelles and Structures

Question 12

Which of the following is an example of secondary active transport?

A. Na+-glucose cotransporter
B. Ligand-gated ion channel
C. K+ leak channel
D. Na+/K+ ATPase

Answer 12

A. In secondary active transport, energy is used to establish an electrochemical gradient, and that gradient then provides the driving force to move other molecules against their concentration gradient. The use of energy is one step removed from the molecule movement, hence the designation of “secondary.” The Na+-glucose cotransporter moves glucose against its gradient as Na+ moves down its gradient. The Na+ gradient is established by the Na+/K+ ATPase, a primary active transporter (uses ATP directly, choice D is wrong). Ligand-gated ion channels are opened by the binding of a ligand to a receptor; movement of the ion then occurs passively down its gradient (choice B is wrong). K+ leak channels are an example of facilitated diffusion; K+ moves across the membrane, down its gradient, through the channel (choice C is wrong).

Concepts tested
Cell Biology: The Cell Membrane

Question 13

Map distances can be calculated by:

A. sequencing a chromosome.
B. calculating recombination frequencies.
C. determining the length of a chromosome.
D. elucidating the ploidy of a cell.

Answer 13

B. Recombination frequency gives a measure of map distances between genes on the same chromosome (choice B is correct). While sequencing a chromosome can determine how many base pairs apart the genes are, it will not give map unit information directly (choice A is wrong). Determining the length of a chromosome or the ploidy of the cell will give no additional information on map distance (choices C and D are not relevant and can be eliminated).

Concepts tested
Genetics: Linked Genes

Question 14

Warkany syndrome is a perinatal lethal disease caused by nondisjunction. Which of the following is another name for Warkany syndrome?

A. 1p36 deletion syndrome
B. Chromosome 8 trisomy syndrome
C. Alpha-galactosidase A deficiency
D. Congenital hypothyroidism

Answer 14

B. Nondisjunction is a meiotic anaphase separation defect. It leads to too many or too few chromosomes (also known as aneuploidy) in gametes and resultant zygotes. Trisomy is when a cell contains three copies of a given chromosome, instead of two, and is an example outcome of nondisjunction (choice B is correct). There is no reason to believe that aneuploidy would result in the loss of a portion of a chromosome (choice A is incorrect), an enzymatic deficiency (choice C is incorrect) or thyroid hormone deficiency (choice D is incorrect).

Concepts tested
Cell Biology: Mitosis/Meiosis

Question 15

Hemophilia is an X-linked recessive trait in humans. If a carrier female mates with the hemophiliac male, what is the probability they will have a daughter who does NOT have hemophilia?

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

Answer 15

C. If we assign D = normal and d = hemophiliac, the cross in the question stem is XDXd × XdY. If the offspring is female, she will receive the Xd chromosome from her father, and the probability of this is 0.5. To have a normal phenotype, she must receive the XD chromosome from her mother, and the probability of this is 0.5. Overall, the probability of having a normal daughter is (0.5)(0.5) = 0.25, (choice C is correct).

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 16

A geneticist discovers two genes in the zebrafish genome that are 42 cM apart. A homozygous dominant fish is crossed with a homozygous recessive fish, and one of the F1 offspring is testcrossed. Which of the following is true of the F2 fish?

A. 42% have phenotypes like the parental fish.
B. 58% have phenotypes like the parental fish.
C. 21% came from gametes that had undergone crossing over between the two genes.
D. 58% represent recombinant offspring.

Answer 16

B. Let’s assign the two parental fish AABB × aabb. The F1 fish would be AB/ab (in linkage notation) and is mated with a ab/ab fish in a testcross. If the two genes are linked and 42 cM (centimorgans, a measure of distance on the chromosome) apart, 42% of offspring would result from crossing over in the heterozygous parent (generating aB or Ab gametes, choices C and D are wrong) and the remaining 58% would have parental combinations of alleles (AB or ab). In other words, we would expect 29% AB/ab and 29% ab/ab (the parental combinations totaling 58%; choice B is correct and choice A is wrong), and 21% Ab/ab and 21% aB/ab (recombinant combinations totaling 42%).

Concepts tested
Genetics: Linked Genes

Question 17

Animal viruses that reproduce via the productive cycle are sometimes said to be more evolutionarily advanced than those that perform the lytic cycle. Which of the following gives a reason for this?

A. Host genome degradation occurs more efficiently in the productive cycle.
B. Integration of the viral genome into the host genome during the productive cycle ensures extensive viral reproduction.
C. The virus must take over certain host processes in the lytic cycle and this can take time.
D. The host cell can survive the productive cycle because the virus exits the cell by budding.

Answer 17

D. In both the lytic and the productive viral reproduction cycles, the virus takes over parts of the host cell (choice C is wrong). In the lytic cycle, the host genome is commonly degraded to generate free dNTP building blocks. However, the host genome typically stays intact during the productive cycle (choice A is wrong). Viral genome integration into the host genome occurs in the lysogenic cycle, not the lytic or the productive cycles (choice B is wrong). The virus particles are released when the host cell bursts in the lytic cycle, but escape by budding out of the host cell in the productive cycle (choice D is correct).

Concepts tested
Microbiology: Viruses and Subviral Particles

Question 18

If two solutions have the same molarity but Solution X has a greater boiling point elevation that Solution Y, which of the following must be true?

A. Solution X ionizes into more particles than Solution Y.
B. Solution X is an electrolyte and Solution Y is not.
C. Solution Y is hypertonic to Solution X.
D. Water would move across a semipermeable membrane from Solution X to Solution Y.

Answer 18

A. Colligative properties, such as boiling point elevation, freezing point depression, vapor pressure depression, and osmotic pressure only depend on the number of particles in solution. If Solution X has a greater boiling point elevation than Solution Y, it must have more particles than (i.e., be hypertonic to) Solution Y (choice A is correct and choice C is wrong). In osmosis, water moves from a hypotonic solution (e.g., Solution Y) into a hypertonic solution (e.g., Solution X, choice D is wrong). Only knowing boiling point elevation does not tell us anything about the solution’s ability to dissociate, i.e., whether or not it is an electrolyte. Solution X might be 1 M MgCl2 and Solution Y might be 1 M NaCl. Both would be electrolytes, and Solution X would still have more particles (choice B is wrong).

Concepts tested
Cell Biology: Osmosis/Diffusion/Colligative Properties

Question 19

Which of the following is true of phototrophic, mesophilic bacteria?

A. They ingest organic nutrients and can survive cold temperatures.
B. They can be found in hot springs, where they perform photosynthesis.
C. They are often found in deep sea vents, living off hydrogen sulphide.
D. They perform photosynthesis and are common at ambient temperatures.

Answer 19

D. Phototrophs get their energy from the sun and mesophiles live at moderate temperatures such as room or body temperature (choice D is correct). Heterotrophs rely on organic compounds generated by other organisms and psychrophiles live at cold temperatures (choice A is incorrect). Thermophiles live in hot springs (choice B is incorrect). Chemoautotrophs are common in deep sea vents and these organisms build organic molecules from CO2 and use chemicals (commonly hydrogen sulfide) for energy. Since there would be little light in a deep sea vent, this is unlikely to be the home of a phototroph (choice C is incorrect).

Concepts tested
Microbiology: Bacteria

Question 20

Which of the following is the best description of genomic organization at the start of mitosis?

A. Homologous chromosomes connected by a centromere
B. Unlinked sister chromatids
C. Paired homologous chromosomes
D. Sister chromatids connected by a centromere

Answer 20

D. At the beginning of mitosis, the genome has been replicated and consolidated into chromatids which are linked by the centromere (choice D is correct). A centromere does not connect homologous chromosomes (choice A is wrong). While sister chromatids are eventually separated during mitosis, at the start they are linked (choice B is wrong). Chromosomes are not arranged in homologous pairs unless a cell is undergoing meiosis (choice C is wrong).

Concepts tested
Cell Biology: Mitosis/Meiosis

Question 21

The cell cycle can involve all of the following genetic events EXCEPT:

A. replication.
B. transcription.
C. translation.
D. recombination.

Answer 21

D. If recombination is to occur, it happens during prophase I of meiosis, not during the cell cycle, which is loosely composed of mitosis and interphase (choice D is not a cell cycle event and is the correct answer choice). The genome is replicated during S phase (choice A is a cell cycle event and can be eliminated). Transcription and translation (i.e., protein synthesis) happen during G1 and G2 (choices B and C are cell cycle events and can be eliminated).

Concepts tested
Cell Biology: Mitosis/Meiosis

Question 22

Which of the following is true of a uracil auxotroph bacterial strain in the log phase?

A. The cells are rapidly undergoing mitosis and the media must be supplemented with uracil.
B. The cells are slowly cycling through the cell cycle in media lacking uracil.
C. The cells are rapidly undergoing binary fission in uracil-supplemented media.
D. The cells are actively mitotic and are making their own uracil via anabolic pathways.

Answer 22

C. Bacterial cells divide via binary fission and not mitosis (choices A and D are incorrect). If they are uracil auxotrophs, they must be supplied with uracil as they cannot synthesize their own (choice C is correct and choice B is wrong).

Concepts tested
Microbiology: Bacteria

Question 23

What components are required to traffic secretory proteins to the lumen of the endoplasmic reticulum?

A. C-terminal signal sequence, signal recognition particle
B. N-terminal signal sequence, signal recognition particle
C. C-terminal signal sequence, single-strand binding protein
D. N-terminal signal sequence, single-strand binding protein

Answer 23

B. The component constituent to the amino acid chain is the signal sequence on the N-terminus; the trafficking is not done from the C-terminus (choices A and C are wrong). The signal sequence is bound by the signal recognition particle to ensure movement to the ER lumen (choice B is correct). Single-strand binding proteins are involved in DNA replication, not protein trafficking (choice D is wrong).

Concepts tested
Cell Biology: Secretory Pathway

Question 24

The stubble allele is dominant in Drosophila melanogaster, and leads to short, thick bristles. Curly wings are also dominant. Both are autosomal traits. A stubbled curly male is testcrossed and produces four phenotypically distinct groups of offspring. If the same male is crossed with a female of the same genotype, what is the probability of generating a long-bristled straight-winged male or female?

A. 1/2
B. 1/4
C. 1/8
D. 1/16

Answer 24

D. If the male produces offspring with four differing phenotypes in a testcross, he must be heterozygous for both genes (SsCc, where S = stubbled, s = long bristles, C = curly wings, c = straight wings). Crossing that male with a female of the same genotype is SsCc × SsCc. The probability of long bristles (ss) is 0.25. The probability of straight wings (cc) is 0.25. The probability of a long-bristled straight-winged offspring is (1/4)(1/4) = 1/16. Note also that this is a dihybrid cross and offspring are produced in a 9:3:3:1 ratio; 9 dominant/dominant (stubbled/curly), 3 dominant/recessive (stubbled/straight), 3 recessive/dominant (long/curly), and 1 recessive/recessive (long/straight). So only 1 of the 16 has the phenotype in question.

Concepts tested
Genetics: Mendelian Genetics/Probability

Question 25

If a trihybrid (heterozygous for three genes) round worm is crossed to a worm that is AaBBcc, which of the following is true of the F1 offspring? A. Three different phenotypes are possible for the first locus. B. All worms will be homozygous recessive for the second gene. C. 100% of worms will express the recessive phenotype for the third locus. D. There are four possible phenotypes in the offspring

Answer 25

D. The cross in the question stem is AaBbCc × AaBBcc. For the first locus, three genotypes will be possible (25% AA, 50% Aa and 25% aa) but only two phenotypes: 75% dominant A and 25% recessive a (choice A is false and can be eliminated). The worms will be 50% BB and 50% Bb at the second locus (choice B is false and can be eliminated) but will all express the dominant B phenotype. For the third locus, the worms will be 50% Cc and 50% cc, and the phenotypic ratios will be the same (choice C is false and can be eliminated). Therefore, the answer must be D; there are two possible phenotypes at the first locus, one possible phenotype at the second locus, and two possible phenotypes at the third locus. (2)(1)(2) = 4 possible phenotypes. Concepts tested Genetics: Mendelian Genetics/Probability

Question 26

A scientist takes 10 μL of culture containing E. coli cells and puts them in a tube with 4 fmol of a plasmid. The tube is placed on ice for half an hour, then resuspended while in a 42°C water bath for 10 seconds. The bacteria are plated on selective media and grown overnight at 37°C. Which of the following is true? A. The scientist performed transformation and is selecting for plasmid-containing bacteria by using agar plates coated with an antibiotic. B. The scientist performed transformation and is selecting for bacteria that did not take up the plasmid by growing the cells above room temperature. C. The scientist performed transfection and is selecting for bacteria that took up the plasmid using antibiotic selection. D. The scientist performed transduction and any colonies that grow must be the result of genetic exchange.

Answer 26

A. Transformation is the process by which bacterial cells acquire genetic information from the external environment; this is what occurred in this experiment. Transfection is a similar process, but for eukaryotic cells (choice C is wrong). Transduction requires the presence of a lysogenic phage, and there is no indication in the question stem that one was used here (choice D is not supported). Selective media is used for the growth of selected microorganisms. It can contain an antibiotic which will select for any cell that expresses a resistance gene to this antibiotic (choice A is correct). Many strains of bacteria like growing at approximately body temperature and there is no information to support that this temperature will allow for plasmid selection (choice B is wrong). Concepts tested Microbiology: Bacteria

Question 27

Which of the following is the best description of microfilaments? A. Alpha and beta tubulin form dimers, then a sheet of dimers are rolled into a hollow tube. B. Actin monomers form a chain, then two chains are twisted together. C. Actin and myosin fibers overlap in a striated configuration. D. Heterogeneous polypeptides create a filament to provide strong structural support.

Answer 27

B. Microfilaments are composed solely of actin monomers in chains which are then twisted together (choice B is correct). Microtubules are built with both alpha and beta tubulin (choice A is wrong). Intermediate filaments are made with heterogeneous polypeptides (choice D is wrong). An overlapping pattern of actin and myosin fibers refers to the basic unit of skeletal muscle structure (choice C is wrong). Concepts tested Cell Biology: Cellular Organelles and Structures

Question 28

A benefit of the lysogenic viral cycle is that: A. viral proteins are translated with host proteins throughout the life of the host. B. viral hydrolase expression ensures efficient replication of the viral genome after integration into the host genome. C. the host remains alive and the viral genome is replicated with every round of cell division. D. bacterial lysis can release thousands of viral particles at the expense of one host cell.

Answer 28

C. During the lysogenic viral cycle, the viral genome integrates into the host genome but remains mostly silent (choice A is not the best answer). Viral hydrolase breaks down the host genome and this does not occur in the lysogenic cycle (choice B is wrong). With every round of host cell division, the viral genome is replicated (choice C is correct). Bacterial lysis is a characteristic of the lytic cycle, not the lysogenic cycle (choice D is wrong). Concepts tested Microbiology: Viruses and Subviral Particles

Question 29

All of the following are components of cell membranes EXCEPT: A. hydrophilic polar head groups. B. channel proteins. C. steroid hormone receptors. D. hydrophobic nonpolar tails.

Answer 29

C. Since steroid hormones are derived from lipids, they diffuse through cell membranes and their receptors are found inside the cell (choice C is not a component of cell membranes and is the correct answer choice). The receptors for peptide hormones would be found on the cell membrane, but the composition of these hormones is different. The polar head groups and nonpolar tails describe the components of the lipids that are the base component of membranes (choices A and D describe cell membrane components and can be eliminated) and many do have channel proteins established for the selective transport of ions or other small molecules (choice B describes a cell membrane component and can be eliminated). Concepts tested Cell Biology: The Cell Membrane

Question 30

In the Adams family, a father has blood type B+ and his wife has blood type A+. In the Wong family, a mother has blood type B– and her husband has blood type A+. All of the following are true EXCEPT: A. It is possible that all of the Adams family children are universal acceptors and express both the A and B antigens, and the Rh antigen, on erythrocytes. B. It is possible that some of the Wong children are universal donors and express no antigens (A, B, or Rh) on erythrocytes. C. It is impossible for any Adams or Wong family grandparents to have blood type O. D. It is possible that some of the Wong children have blood types different from either parent.

Answer 30

C. If the Adams family parents are pure-breeding (or homozygous), then choice A is true and can be eliminated. All offspring would be IAIBRR and these individuals are called universal acceptors because they can accept all blood types without complications. If the Wong family parents are heterozygous for the ABO blood group (IAi × IBi), and if the Wong husband is heterozygous for the Rh antigen (Rr), then both choices B and D are possible and both can be eliminated (note that the Wong mother must be homozygous recessive for the Rh antigen as she is Rh negative). Some of the children could be blood type O– (iirr) which are universal donors and a blood type different from either parent. In each family, some grandparents could be blood type O (choice C is not true and is the correct answer choice). For example, IBRr (type B+) and iirr (type O–) grandparents could give rise to the Wong family mother. Concepts tested Genetics: Mendelian Genetics/Probability

Question 31

Catalytic receptors operate via which of the following? A. A ligand binds, triggering enzyme activity on the cytoplasmic side of the membrane. B. A ligand binds, triggering the opening of a specific ion channel. C. Binding triggers the release of a second messenger, causing indirect signal transduction and altering the activity of cytoplasmic enzymes. D. Binding triggers a cascade which leads to increasing cytoplasmic calcium levels.

Answer 31

A. Catalytic receptors involve the binding of a ligand in order to impact cytoplasmic enzyme activity (choice A is correct). If binding of the ligand causes an ion channel to open, this is ligand-gating (choice B is wrong). Second messengers are used by G-protein linked receptors (choice C is wrong). A specific example of a G-protein linked receptor, phospholipase C, can lead to increased cytoplasmic calcium (choice D is wrong). Concepts tested Cell Biology: G Proteins/Signal Transduction

Question 32

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 dominant polydactyly in men showed that the frequency of the allele causing polydactyly in this population was 2%? What is the frequency of affected individuals in this population? A. Less than 1% B. 2% C. 3% D. 4%

Answer 32

D. The question states that in this population, polydactyly is caused by a dominant allele. In the Hardy Weinberg equation for allele frequency (p + q = 1), p is the frequency of the dominant allele. If p = 0.02, then q = 0.98. Plugging these numbers into the equation for genotype frequency (p2 + 2pq + q2 = 1), q2 (the frequency of autosomal recessive individuals, and thus UNaffected individuals) is equal to (0.98)2 = 0.96; or in other words, 96% of the population is NOT affected. Therefore, 4% of the population IS affected. Concepts tested Genetics: Hardy-Weinberg

Question 33

Which of the following involves the specific uptake of material into the cell? A. Phagocytosis B. Receptor-mediated endocytosis C. Pinocytosis D. Exocytosis

Answer 33

B. Receptor-mediated endocytosis uses clathrin-coated pits on the cell surface as the basis for endosomes in which to package material that binds to the receptor and bring it into the cell (choice B is correct). Phagocytosis and pinocytosis both involve non-specific uptake (choices A and C are wrong). Exocytosis involves the release of material from the cell using a vesicle (choice D is wrong). Concepts tested Cell Biology: Cellular Organelles and Structures

Question 34

A retrovirus is a: A. (+) RNA virus that undergoes the lysogenic cycle, and so must encode reverse transcriptase in the genome. B. (–) RNA virus that undergoes the lytic cycle, and so must encode reverse transcriptase in the genome. C. (+) RNA virus that undergoes the lysogenic cycle, and so must encode a DNA-dependent RNA polymerase in the genome. D. (–) RNA virus that undergoes the lysogenic cycle, and so must carry a copy of reverse transcriptase.

Answer 34

A. A retrovirus is an example of a (+) RNA virus (choices B and D can be eliminated) that undergoes the lysogenic cycle. To do this, the RNA genome must be converted to DNA by an RNA-dependent DNA polymerase. This enzyme is commonly called reverse transcriptase (choice A is correct and choice C is wrong). Note that the virus isn't absolutely required to carry a copy of this enzyme (although many do), since the (+) RNA genome can be used as a template for translation. Concepts tested Microbiology: Viruses and Subviral Particles

Question 35

Each of the following is a true statement EXCEPT: A. reproductive isolation, both prezygotic and postzygotic, can contribute to speciation. B. homologous structures are present in different species due to a common ancestor. C. humans are parts of domain Eukarya, kingdom Animalia and genus Chordata. D. sexual selection means organisms do not mate randomly, but select their partner based on rituals and physical displays.

Answer 35

C. All the statements are true except choice C. Humans are part of domain Eukarya, kingdom Animalia, phylum Chordata, subphylum Vertebrata, class Mammalia, order Primates, family Hominidae, and our genus and species names are Homo sapien, respectively. Concepts tested Evolution/Speciation: Natural Selection/Speciation

Question 36

Which of the following is a true statement about bacterial conjugation? A. A female strain (F–) can mate with only F+ male strains. B. Hfr bacteria generate a sex pilus to allow conjugation with another Hfr strain. C. While conjugation increases the number of male and Hfr cells, binary fission can increase the number of female cells. D. The F factor is transferred from the female or Hfr cell, through the conjugation bridge.

Answer 36

C. Female (or F–) bacteria can undergo conjugation with both male (F+) and Hfr strains (choice A is wrong). Hfr strains can only mate with female bacteria, not other Hfr strains (choice B is wrong). Conjugation increases the number of F+ cells. However, when bacteria divide by binary fission, plasmid distribution in the two daughter cells is relatively unregulated. This can result in a male parental cell giving both male and female daughter cells (choice C is correct). The conjugation bridge (or sex pilus) is formed by the F+ (or Hfr) cell. During conjugation, nucleic acids (either a plasmid or a chromosome) are transferred to the female cell across this bridge (choice D is wrong). Concepts tested Microbiology: Bacteria

Question 37

Which of the following situations is most conducive to osmosis? A. A solution hypertonic relative to sodium separated by a solvent-permeable membrane from a solution hypertonic relative to potassium, with equivalent ion concentrations in each B. A solution hypotonic relative to sodium separated by a solvent-permeable membrane from a solution hypotonic relative to potassium, with equivalent ion concentrations in each C. A solution isotonic relative to sodium separated by a solvent-permeable membrane from a solution isotonic relative to potassium D. A solution hypertonic relative to sodium separated by a solvent-permeable membrane from a solution hypotonic relative to sodium

Answer 37

D. Osmosis is the passive movement of a solvent down its concentration gradient, but this movement will be impacted by the overall mix of solutes in a given solution. If two solutions are hypertonic relative to two separate ions, but overall have equivalent ion concentration (i.e. are isotonic to each other), osmosis will not occur (choice A is wrong). Similarly, if two solutions are hypotonic relative to two separate ions, but overall have equivalent ion concentrations (are isotonic to each other), osmosis will not occur (choice B is wrong). An isotonic balance will not promote osmosis (choice C is wrong), but a hypertonic solution next to a hypotonic solution (regardless of the specific ion) will (choice D is correct). Concepts tested Cell Biology: Osmosis/Diffusion/Colligative Properties

Question 38

During metaphase I of meiosis: A. spindle microtubules are fully attached to kinetochores on lined-up chromosomes. B. the centrosome is duplicated to facilitate chromosome separations. C. crossing over between homologous chromosomes increases genetic diversity in offspring. D. the actin contractile ring pinches inwards, cleaving the cell into two.

Answer 38

A. The centrosome is duplicated before meiosis, during S-phase (choice B is wrong). Crossing over happens in prophase I, also before metaphase I (choice C is wrong). The spindle (made of microtubules) starts attaching to chromosomes in prometaphase I (in between prophase and metaphase) and is fully attached by metaphase I (choice A is correct; note that the attachment of the spindle fibers and subsequent pulling of the chromosomes is what allows homologous chromosomes to line up across from each other at the metaphase plate). Cytokinesis begins near the end of anaphase and finishes at the end of telophase (choice D is wrong). Concepts tested Cell Biology: Mitosis/Meiosis