MCAT BIO FSQs

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.

A. A solute moves passively down its concentration gradient in a thermodynamically favorable way, increasing entropy.

Mutation of which of the following codons would be most disruptive to the initiation of translation? A. AUG B. UGA C. UUU D. UAA

A. AUG

What steps are necessary to replicate a (-)RNA virus? A. Create a (+)RNA copy of the genome using RNA-dependent RNA polymerase then replicate (-)RNA copies from that template using the same enzyme. B. Create a (+)RNA template using RNA-dependent DNA polymerase then replicate (-)RNA copies from that template using DNA-dependent RNA polymerase. C. Use RNA-dependent DNA polymerase to synthesize a DNA version of the genome and then replicate more RNA copies using the host cell's transcriptional machinery. D. Make more (-)RNA copies directly from the genome using RNA-dependent RNA polymerase.

A. Create a (+)RNA copy of the genome using RNA-dependent RNA polymerase then replicate (-)RNA copies from that template using the same enzyme.

As the level of cellular specialization increases, what cell cycle activity decreases? A. DNA replication B. Cell growth C. Genome packaging D. Genome transcription

A. DNA replication 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.

Hemoglobin is an oxygen-carrying protein found in red blood cells. It is made up of four protein subunits that display cooperative binding. Myoglobin is also an oxygen-carrying protein, however it is found in muscle cells and it is made of only a single protein subunit. How would the saturation curves for hemoglobin and myoglobin compare? A. Hemoglobin would have a sigmoidal curve while myoglobin would have a simple curve. B. Both hemoglobin and myoglobin would have sigmoidal curves, but the curve of myoglobin would be right-shifted compared to the curve of hemoglobin. C. Hemoglobin would have a simple curve while myoglobin would have a sigmoidal curve. D. Both hemoglobin and myoglobin would have simple curves, but myoglobin would display only 1/4 the saturation level of hemoglobin.

A. Hemoglobin would have a sigmoidal curve while myoglobin would have a simple curve.

Which of the following statements best describes pyruvate kinase? A. It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. B. Its quaternary protein structure is most similar to hemoglobin and myoglobin. C. It is involved in gluconeogenesis causing the conversion of pyruvate to glucose during high energy states. D. It has several isoforms due to alternative splicing via the spliceosome and ribosome.

A. It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule.

Which of the following statements about Km is true? A. Km is the substrate concentration required to reach 1/2 Vmax, and is a measure of affinity between an enzyme and its substrate. B. Km is a measure of affinity between an enzyme and its substrate; a high Km indicates a strong affinity. C. Km is the substrate concentration required to reach Vmax, and a low Km indicates a high enzyme-substrate affinity. D. Km is affected by enzyme concentration; if the enzyme concentration is reduced, Km decreases.

A. Km is the substrate concentration required to reach 1/2 Vmax, and is a measure of affinity between an enzyme and its substrate.

Which of the following describes the molecular geometry of a carbon dioxide molecule? A. Linear B. Trigonal planar C. Tetrahedral D. Bent

A. Linear

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

A. Ribosome subunit assembly

In which organ are sickled red blood cells most likely to be hemolyzed? A. Spleen B. Thymus C. Lymph nodes D. Kidneys

A. Spleen

Which of the following differences between RNA and DNA could help explain the differences in secondary structure observed between the two types of nucleic acids? A. The 2' hydroxyl group present on RNA B. The presence of thymine in RNA C. Differing RNA binding proteins present in the nucleus of prokaryotes D. Presence of a single phosphate per nucleotide in the RNA phosphodiester backbone

A. The 2' hydroxyl group present on RNA

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.

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. 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.

The pentose phosphate pathway (PPP) generates NADPH to be used as a reducing agent in biosynthetic reactions such as fatty acid synthesis, and generates ribulose-5-phosphate which can either be converted into ribose-5-phosphate or into glycolytic intermediates. Ribose-5-phosphate can also be generated from glycolytic intermediates without the reduction of NADPH. Which of the following would be favored in a rapidly dividing cell? A. The formation of ribose-5-phosphate B. The reduction of NADP+ to NADPH in the oxidative phase of the PPP C. The conversion of ribulose-5-phosphate into glycolytic intermediates D. The oxidation of glucose-6-phosphate to generate NADH

A. The formation of ribose-5-phosphate

A yeast colony is subject to the mutagen EMS (ethyl methanesulfonate) and subsequently fails to divide. Further analysis reveals excessive supercoiling in the S-phase and failure to progress in the cell cycle. Which of the following was the most likely gene target of EMS? A. Topoisomerase B. Helicase C. DNA ligase D. Single-stranded binding protein

A. Topoisomerase

A barometer is made by erecting an evacuated glass tube over a dish filled with mercury. Atmospheric pressure at sea level causes the mercury to rise inside the tube. If this barometer were moved from sea level to a mountain several kilometers in altitude, the mercury column would most likely: A. fall due to the decreased atmospheric pressure. B. rise due to the decreased atmospheric pressure. C. fall due to the increased atmospheric pressure. D. rise due to the increased atmospheric pressure.

A. fall due to the decreased atmospheric pressure.

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.

A. is made of protein and functions in attachment to the host.

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.

A. lipid and protein content make up equivalent parts of cell membranes.

α-ketoglutarate dehydrogenase (α-KGDH) is an enzyme used in cellular respiration that catalyzes the decarboxylation of α-ketoglutarate (a 5-carbon compound) to succinyl-CoA (a 4-carbon compound). Inherited deficiencies of α-KGDH result in permanent lactic acidosis, among other symptoms, and lead to death at an early age. α-KGDH is mostly likely found in the: A. mitochondrial matrix. B. mitochondrial inner membrane. C. cytosol D. mitochondrial intermembrane space.

A. mitochondrial matrix.

All of the following are examples of a reduction EXCEPT: A. the conversion of glucose to (ultimately) carbon dioxide during cellular respiration. B. the creation of NADH during glycolysis and the Krebs cycle. C. the conversion of oxygen to water along the electron transport chain. D. the creation of ethanol from pyruvate during alcoholic fermentation.

A. the conversion of glucose to (ultimately) carbon dioxide during cellular respiration.

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.

B. (+) RNA viruses must carry an RNA-dependent RNA polymerase.

The polypeptide +NH3-Asp-Lys-Leu-Val-Arg-Glu-Val-Lys-COO- is digested by trypsin. The products will be: A. +NH3-Asp-COO-, +NH3-Lys-Leu-Val-COO-, +NH3-Arg-Glu-Val-Lys-COO- B. +NH3-Asp-Lys-COO-, +NH3-Leu-Val-Arg-COO-, +NH3-Glu-Val-Lys-COO- C. +NH3-Asp-Lys-COO-, +NH3-Leu-Val-Arg-COO-, +NH3-Glu-Val-COO-, Lys D. +NH3-Asp-COO-, +NH3-Lys-Leu-Val-COO-, +NH3-Arg-Glu-Val-Lys

B. +NH3-Asp-Lys-COO-, +NH3-Leu-Val-Arg-COO-, +NH3-Glu-Val-Lys-COO- Trypsin and other serine proteases hydrolyze the peptide bond to the right (toward the COOH-terminus) of basic amino acid residues including lysine and arginine.

β-Oxidation is a means of creating acetyl-CoA from fatty acids. This acetyl-CoA can then enter the Krebs cycle. Each turn of the β-oxidation cycle produces one acetyl-CoA molecule and a fatty acid two carbons shorter than it was at the beginning of the cycle. Additionally, one NADH and one FADH2 are generated per turn. Lauric acid is an 12-carbon saturated fatty acid. Including those produced in the Krebs cycle, how many total NADH and FADH2 molecules would be generated from the complete β-oxidation of lauric acid and subsequent entry of the acetyl-CoA into the Krebs cycle? A. 24 NADH and 12 FADH2 B. 23 NADH and 11 FADH2 C. 12 NADH and 12 FADH2 D. 11 NADH and 11 FADH2

B. 23 NADH and 11 FADH2 (Each turn of the β-oxidation cycle produces one acetyl-CoA and a fatty acid two carbons shorter than before. Lauric acid, with 12 carbons, would ultimately produce 6 acetyl-CoA. If lauric acid enters the β-oxidation cycle, then after one turn of the cycle we would have one acetyl-CoA and a 10C fatty acid. After two turns of the cycle we would have two acetyl-CoA and an 8C fatty acid. This would continue in this manner until after the 5th turn of the cycle we would produce our 5th acetyl-CoA molecule, and all that would be left over would be another two-carbon acetyl-CoA molecule (the 6th acetyl-CoA). This last 2-carbon molecule does not need to go through the β-oxidation cycle again, so only 5 turns of cycle are necessary. Thus, 5 NADH and 5 FADH2 would be generated during β-oxidation. When the 6 acetyl-CoA molecules go through the Krebs cycle, and additional 18 NADH would be generate (3 per turn of the Krebs cycle) and and additional 6 FADH2 would be generated (1 per turn), for a total of 23 NADH and 11 FADH2.)

If the absolute pressure of a gas is increased from 3 atm to 4 atm at constant volume, then the absolute temperature of the gas will increase by: A. 25%. B. 33%. C. 67%. D. 75%.

B. 33%. Assuming the validity of the Ideal-Gas law, PV = nRT, an increase in pressure by a factor of 4/3 at constant volume will result in an increase in temperature by the same factor (since P is proportional to T if V and n are constant). Multiplying the temperature by 4/3 = 133% implies an increase by 33%.

Some amino acids can be converted to pyruvate via several biochemical pathways. Pyruvate can then enter the cellular respiration pathways, either by decarboxylation to acetyl-CoA or by carboxylation to oxaloacetate. For a single pyruvate molecule, first converted to acetyl-CoA, then traveling through the Krebs cycle, how many NADH molecules are produced? A. 3 B. 4 C. 6 D. 8

B. 4 (The decarboxylation of pyruvate to acetyl-CoA nets 1 NADH, and as that acetyl-CoA travels through the Krebs cycle, an additional 3 NADH are generated, resulting in a total of 4 NADH per pyruvate (choice B is correct and choices A, C, and D are wrong).)

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.

B. After the lag phase, the culture will start growing rapidly. 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).

A converging glass lens forms a real image of a red object at a distance equal to twice its focal length. If a blue object is placed adjacent to the red object, will its image form closer to the lens or farther from the lens than the image of the red object? A. Farther, because blue light is refracted more due to its shorter wavelength B. Closer, because blue light is refracted more due to its shorter wavelength C. Farther, because blue light is refracted less due to its longer wavelength D. Closer, because blue light is refracted less due to its longer wavelength

B. Closer, because blue light is refracted more due to its shorter wavelength

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.

B. Dry ear wax is an autosomal recessive trait and hairy ears is Y-linked.

Which one of the following compounds would be most soluble in ethanol? A. Methane, CH4 B. Formic acid, HCOOH C. Stearic acid, CH3(CH2)16COOH D. 1-Chlorooctodecane, CH2Cl(CH2)16CH3

B. Formic acid, HCOOH

Which of the following statements is true? A. When ADP levels are high, pyruvate kinase is inhibited. B. Fructose-2,6-bisphosphate levels are reduced when glucose levels are low; this helps to drive gluconeogenesis forward. C. The phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate is an important regulatory step in gluconeogenesis. D. High levels of citrate inhibit fructose-1,6-bisphosphatase to prevent gluconeogenesis when Krebs intermediates are plentiful.

B. Fructose-2,6-bisphosphate levels are reduced when glucose levels are low; this helps to drive gluconeogenesis forward.

It has been suggested that one reason tumor cells have novel metabolism is to generate the additional biomolecules required to increase biomass, which is needed to support high rate of proliferation. Which is the best explanation of how this could occur? A. Fructose-6-phosphate is shuttled from glycolysis to the pentose phosphate pathway to support both nucleotide and fatty acid biosynthesis. B. Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides. C. High glycolysis rates can support the pentose phosphate pathway, which generates ribose-2-phosphate for nucleotide catabolism. D. Glutamate is converted into glutamine to provide the cell with an amino acid precursor, thus powering translation.

B. Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides.

Which of the following best explains why the isoelectric point of HbS is higher than that of HbA? A. The side chain of glutamic acid is less acidic than that of valine. B. Glutamic acid is isoelectric at a lower pH than is valine. C. Valine is isoelectric at a lower pH than is glutamic acid. D. Glutamic acid has a net charge of 0 at its isoelectric point.

B. Glutamic acid is isoelectric at a lower pH than is valine. The isoelectric point is the solution pH at which an amphoteric molecule (a molecule that can act either as an acid or a base) has a net electric charge of zero. Choice A is eliminated since it is a false statement; glutamic acid is more acidic, not less acidic than valine. C is also a false statement: Glutamate would require more acidic pH to neutralize its charge. D is a true statement but is not the correct choice. The fact that glutamate has a net charge of zero at its isoelectric point would be true of any molecule since this is the definition of an isoelectric point. The best answer is B. The negative charge of glutamic acid would require a more acidic pH to neutralize, making the isoelectric point of HbA (with glutamic acid) lower than that of HbS (with valine).

Which of the following are enzymes directly activated by G-protein linked receptors? I. Adenylate cyclase II. Phospholipase C III. Tyrosine kinase A. I only B. I and II only C. II and III only D. I, II, and III

B. I and II only

Which of the following viruses should be able to reproduce successfully if they carry RNA-dependent RNA polymerase into their host cell? I.(+) RNA virus II.(-) RNA virus III.ds DNA virus A. I only B. I, II, and III C. II only D. I and II only

B. I, II, and III

Protein synthesis is primarily regulated at the level of transcription. Which of the following could lead to an increase in protein synthesis in a typical eukaryotic cell? I. Substitution of a strong promoter for a weak promoter when more of the protein is required II.Inhibition of repressor binding to a gene regulatory region III. Inhibition of an enhancer in a gene regulatory region A. I only B. II only C. I and II only D. II and III only

B. II only

Which of the following could lead to cancer? I.An overexpression of initiator caspases, leading to apoptosis II.A mutation in the tumor suppressor gene p53 that prevents the expression of the p53 protein III. 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

B. II only 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).

A graduate student attaches a fluorescent tag to β-galactosidase and performs site-directed mutagenesis to generate a mutation in the operator of the lac operon. The mutation completely prevents repressor binding. Which of the following is the most likely observation following mutagenesis of these cells? A. Increased fluorescence compared to control due to repressor inhibition, only in the presence of lactose B. Increased fluorescence compared to control due to failed repressor binding, in the presence or absence of lactose C. Decreased fluorescence compared to control, only in the presence of lactose D. Decreased fluorescence compared to control, in the presence or absence of lactose

B. Increased fluorescence compared to control due to failed repressor binding, in the presence or absence of lactose (If the mutation prevents the repressor from binding to the operator, this would result in constitutive expression of the genes in the lac operon, including the fluorescent β-galactosidase (choices C and D are wrong). This would occur whether or not lactose was present (choice B is correct and choice A is wrong))

A graduate student in a yeast lab that studies double-strand break (DSB) repair has a mutant strain that is unable to complete repair via nonhomologous end joining. Which of the following is true about this strain? A. It can only repair DSB with minimal specificity. B. It is able to form a joint molecule when repairing DSBs. C. It will accumulate many chromosomal aberrations over time. D. It can specifically repair DSB in a DNA polymerase-independent manner.

B. It is able to form a joint molecule when repairing DSBs.

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.

B. Linked alleles tend to be inherited together.

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.

B. Natural selection acts on genetic diversity in populations and causes evolution.

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.

B. The new E. coli strain must have acquired random genomic mutations such as deamination.

A strain of bacterium was infected with a phage which caused all of the bacteria it infected to become resistant to the antibiotic streptomycin. Which of the following is the most likely explanation? A. The streptomycin resistance is conferred by a viral gene that allows the virus to replicate in the presence of antibiotic. B. The phage is lysogenic and carries a gene from other bacteria. C. The phage is lysogenic and alters the composition of the cell wall. D. When the phage inserts into the genome, it mutates the resistance gene.

B. The phage is lysogenic and carries a gene from other bacteria.

Proteins are polymers of amino acids that play diverse roles in the body. Which of the following statements about proteins is true? A. Proteins cannot contain more than 1000 amino acids. B. The side chains of individual amino acids can have a significant effect on protein structure. C. Protein synthesis always occurs in the 5' to 3' direction. D. Protein functions include structure, transport, enzymatic catalysis, and emulsification of fats.

B. The side chains of individual amino acids can have a significant effect on protein structure.

Which of the following is LEAST characteristic of prions? A. They can be associated with gene mutations. B. They are not contagious. C. They are usually found in the nervous system. D. They are resistant to degradation by heat or chemicals.

B. They are not contagious.

Which of the following statements about lipids is NOT true? A. Phospholipids are amphipathic and are used in the formation of lipid bilayers. B. Triglycerides are formed by joining three fatty acids to a glycerol molecule, and the three fatty acids are always the same. C. Saturated fats are solids are room temperature because their hydrocarbon chains can pack together tightly. D. Cholesterol is a lipid formed from several hydrocarbon rings; it is used in the derivation of steroids such as estrogen, cortisol, and vitamin D.

B. Triglycerides are formed by joining three fatty acids to a glycerol molecule, and the three fatty acids are always the same.

Which of the following describes a mechanism by which a viroid can cause disease? A. The capsid of the viroid binds to the exterior of the cell and the DNA is injected into the cell. The viroid enters the lysogenic cycle, and is replicated along with the host genome. B. Viroids can serve as siRNAs that silence expression of genes that are necessary for cell function. C. Viroid RNA codes for abnormally folded proteins that cause host proteins to misfold as well. D. The capsid of the viroid binds to the exterior of the cell and the DNA is injected into the cell. The viroid enters the lytic cycle, and the cell bursts, releasing new viroids.

B. Viroids can serve as siRNAs that silence expression of genes that are necessary for cell function.

A highly proliferating cell would most likely: A. overexpress hexokinase and fructose-1,6-bisphosphatase. B. express high levels of the lactate transporter and the glutamine transporter. C. power cell growth by running the electron transport chain and oxidative phosphorylation. D. upregulate pyruvate dehydrogenase kinase and downregulate phosphofructokinase.

B. express high levels of the lactate transporter and the glutamine transporter.

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.

B. facilitated diffusion.

During spermatogenesis, spermatids: A. are frozen in meiosis II until after fertilization. B. have already undergone meiotic recombination. C. have four copies of the genome per cell. D. have no nucleus.

B. have already undergone meiotic recombination. Spermatids are the sperm precursors that have completed meiosis but have not yet fully matured. A mature sperm has fully completed meiosis, and this is true of spermatids as well; it is the ova that are frozen in meiosis II until after fertilization (eliminate choice A). Recombination occurs in both oogenesis and spermatogenesis during meiotic prophase I, which the spermatid already completed (choice B is correct). The spermatid has passed through two reductive divisions in meiosis to end up with only one copy of the genome (eliminate choice C). The cell does have a nucleus. In fact, the sperm has virtually no cytoplasm, but the genome is still packaged into a nucleus (eliminate choice D).

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.

B. involve a virus using host cellular machinery to replicate the viral genome and capsid.

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.

B. the viral genome is silent, but replicated along with the host genome. 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).

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.

B. there are two chromatids per chromosome and the cell is haploid.

f a silent mutation occurs at position 6 in the β chain of hemoglobin, what will be the overall charge of the resulting tryptic peptide in a solution buffered to pH 5? A. -2 B. -1 C. 0 D. +1

C. 0 By determining the charge states on each individual amino acid in the tryptic peptide at pH 5, the overall charge on the peptide can be calculated. Recall that when the pH of a solution exceeds the pKa of a functional group, that group is deprotonated. This applies only to the acidic residues in the peptide. So, from left-to-right: the N-terminal Val bears a +1 charge on its α-amino group; His bears a +1 charge on its side chain; Leu, Thr, and Pro are uncharged; each Glu bears a -1 charge on its side chain; and the C-terminal Lys is neutral overall (+1 on its side chain and -1 on its α-carboxyl group). The sum total of these charges [(+1) + (+1) + (-1) + (-1) + (0)] is 0, or answer choice C.

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

C. 0.1 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.

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

C. 0.125 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).

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

C. 0.25

A biochemist uses a sensitive assay to quantify the amount of energy required to translate an unknown protein. Given that he found 404 ATP equivalents of energy to be consumed in the process, how long was the peptide of interest? A. 99 amino acids B. 100 amino acids C. 101 amino acids D. 404 amino acids

C. 101 amino acids (While initiation requires only 1 GTP, tRNA loading requires 2 high energy phosphate bond equivalents per amino acid/tRNA pair, and elongation for each subsequent amino acid requires 2 GTP. Termination requires 1 GTP. This can be summarized in the equation 4n, where n is the number of amino acids in a peptide chain, and the equation represents the number of high energy bonds required to make the peptide. In this question, you are given the total quantity of energy consumed and need to determine the length of the peptide. If 4n = 404, then n = 101 (choice C is correct).)

Which of the following is the anticodon sequence on the tRNA for the start codon? A. 5'-AUG-3' B. 5'-UAC-3' C. 5'-CAU-3' D. 5'-GUA-3'

C. 5'-CAU-3' The start codon on the mRNA would be 5'-AUG-3', therefore, the anticodon on the tRNA would be the complementary sequence: 3'-UAC-5' (or alternatively written, 5'-CAU-3').

Prions do not follow the principles of the Central Dogma and are referred to as self-replicating proteins. Which of the following best describes their means of replication? A. A disease-causing version of the protein recruits loaded and activated tRNAs to synthesize new prions. B. A regular version of the protein automatically becomes disease-causing when produced in neurons. C. A disease-causing version of the protein acts as the folding template to cause a regular version of the protein to become misshapen. D. A regular version of the protein undergoes a nonsense mutation to induce a change to the disease-causing version.

C. A disease-causing version of the protein acts as the folding template to cause a regular version of the protein to become misshapen.

Cytokinesis occurs during which two phases of mitosis? A. Prophase and metaphase B. Metaphase and anaphase C. Anaphase and telophase D. Telophase and prophase

C. Anaphase and telophase

HIV is a +RNA retrovirus whose genome can be directly translated to form a variety of viral proteins even before insertion into the host's genome. Which of the following enzymes would be needed AFTER insertion into the host genome to transcribe the mRNA for viral proteins? A. RNA dependent DNA polymerase B. DNA dependent DNA polymerase C. DNA dependent RNA polymerase D. RNA dependent RNA polymerase

C. DNA dependent RNA polymerase The enzyme needed to transcribe the mRNA for viral proteins AFTER insertion into the host genome is DNA dependent RNA polymerase. The first part of the enzyme name ("___ dependent") describes what type of nucleic acid the enzyme uses as a template, and the second part of the enzyme name ("___ polymerase") describes the type of nucleic acid the enzyme synthesizes. Since this enzyme is synthesizing mRNA, it must be an RNA polymerase. The question asks for the enzyme needed AFTER the viral genome is inserted into the host genome. In order to be inserted into the host genome, the viral genome must have been converted to its DNA version by reverse transcriptase (an RNA dependent DNA polymerase). If the viral genome is now in its DNA version and is inserted in the host genome, the enzyme transcribing it must be DNA dependent (use DNA as a template).

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

C. Directly coupling the movement of a molecule against its concentration gradient to ATP hydrolysis

Which of the following statements is NOT true about competitive inhibition? A. The Km of the uninhibited reaction is lower than the Km of the inhibited reaction. B. Competitive inhibitors bind at the active site of an enzyme. C. Even if you greatly increase substrate concentration, the Vmax of the inhibited reaction will never reach the same Vmax of the uninhibited reaction. D. Competitive inhibitors can resemble the transition state of a reaction.

C. Even if you greatly increase substrate concentration, the Vmax of the inhibited reaction will never reach the same Vmax of the uninhibited reaction.

Perceiving the color of a pH indicator requires what type of visual processing? A. Depolarization of cone cells to trigger hyperpolarization of bipolar neurons. B. Depolarization of cone cells to trigger depolarization of bipolar neurons. C. Hyperpolarization of cone cells to trigger depolarization of bipolar neurons. D. Hyperpolarization of cone cells to trigger hyperpolarization of bipolar neurons.

C. Hyperpolarization of cone cells to trigger depolarization of bipolar neurons.

Polysaccharides can be used for many different functions. Which of the following is/are polysaccharides that are used primarily for glucose storage? I. Starch II.Glycogen III.Cellulose A. I only B. II only C. I and II only D. I, II, and III

C. I and II only

Which of the following could explain the Warburg effect? I. Metabolic differences allow cancer cells to adapt to hypoxic (oxygen-deficient) conditions inside solid tumors. II. Cells with low proliferation rates often have a high ratio of glycolysis to mitochondrial respiration. III. Some oncogenic changes shut down the mitochondria because these organelles are involved in apoptosis, which would result in cell death. A. I only B. III only C. I and III only D. I, II, and III

C. I and III only Item I is true: tumors can grow quite quickly and often the inside of tumors don't have sufficient blood supply. This can lead to hypoxic conditions inside the tumor. Since glycolysis and fermentation don't require oxygen, they could facilitate cell growth in anaerobic conditions (choice B can be eliminated). Item II is false: the passage is about highly proliferative cells that use glycolysis and fermentation to power growth so a statement about slow growth does not explain the Warburg effect discussed in the passage (choice D can be eliminated). Item III is true: if a tumor cell is relying on glycolysis and fermentation instead of mitochondrial cell respiration, it will have less use for the mitochondria in general. Since the mitochondria can initiate apoptosis, less reliance on this organelle could lead to apoptosis resistance, conferring a survival advantage to the cancerous cell (choice A can be eliminated and choice C is correct).

Coronaviruses are enveloped, +RNA viruses that typically cause respiratory infections. A coronavirus was implicated in the SARS epidemic of 2002. Which of the following statements are true about Coronavirus? I.They can be cultured in any type of cell as long as an RNA dependent RNA polymerase is included II.They can be cultured in animal cells. III.Their genome likely has a poly-A tail. A. II only B. I and III only C. II and III only D. I only

C. II and III only

Which of the following does NOT occur during starvation? I.β-oxidation in the mitochondrial matrix provides acetyl-CoA to feed into the Krebs cycle. II. Ketone bodies are converted into acetyl-CoA and the acetyl-CoA is converted into glucose. III.Fatty acid synthesis in the cytoplasm produces NADH to drive electron transport and oxidative phosphorylation. A. I only B. II only C. II and III only D. I, II, and III

C. II and III only

Eukaryotes running aerobic respiration net 30 ATP per glucose, while prokaryotes net 32 ATP. Why? A. Prokaryotic glycolysis does not require the input of 2 ATP for the phosphorylation of glucose and fructose-6-P. B. Prokaryotes generate more pyruvate from glucose than do eukaryotes. C. In eukaryotes, the electrons from glycolytic NADH must be shuttled from the cytosol into the mitochondrion, and bypass the first proton pump. D. Eukaryotes have a more efficient ATP synthase.

C. In eukaryotes, the electrons from glycolytic NADH must be shuttled from the cytosol into the mitochondrion, and bypass the first proton pump.

Which one of the following statements concerning the SN2 reaction mechanism is true? A. It proceeds best with tertiary substrates and is a two-step mechanism. B. It proceeds with retention of stereochemistry. C. It proceeds best with primary substrates and is a one-step reaction. D. It proceeds through a carbocation intermediate.

C. It proceeds best with primary substrates and is a one-step reaction.

A bacteriologist initiated an E. coli culture from one E. coli cell and hypothesized that some of the progeny in the culture would be genetically different from the original parent cell. Is this hypothesis true? A. Yes; bacteria are capable of undergoing genetic recombination through a variety of mechanisms. B. Yes; bacteria reproduce sexually, and the progeny of any one cell are genetically distinct from the parent cell. C. No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell. D. No; bacteria can reproduce only by meiosis, which ensures preservation of the genome.

C. No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell.

An experiment with a previously unidentified pathogen is done using guinea pigs and various means of inoculation are attempted. Within a few days, the guinea pigs begin to demonstrate a loss of coordination along with other neurological symptoms. Could the organism be a prion? A. Yes; loss of coordination is specific to prion infections. B. No; prions do not infect guinea pigs. C. No; the onset of symptoms was too rapid. D. Yes; direct inoculation is required to contract a prion infection.

C. No; the onset of symptoms was too rapid. An experiment with a previously unidentified pathogen is done using guinea pigs and various means of inoculation are attempted. Within a few days, the guinea pigs begin to demonstrate a loss of coordination along with other neurological symptoms. No, the organism could not be a prion because the onset of symptoms was too rapid. Prion incubation times are long; in humans, it can take years or even decades. The onset as described in the question is too rapid, even in the small animal model of the guinea pig. Prions can infect all sorts of animals so guinea pigs are not specifically immune and loss of coordination can be a sign of all sorts of disorders and infections. There an many means by which prion infection can be acquired, including mutation.

Following the binding of a loaded tRNA to its codon during translation, which of the following steps occurs next? A. Translocation of the ribosome along the mRNA transcript B. Dissociation of the tRNA present at the P site C. Peptide bond formation D. A release factor binds

C. Peptide bond formation

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.

C. The boiling point elevation of Solution B is twice the boiling point elevation of Solution A.

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.

C. The female bacterium could end up female (F-), male (F+), or Hfr. 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.

Which of the following is specific to an animal virus and NOT to a bacteriophage? A. Destruction of the host cell during the productive cycle B. Infection of the host cell occurs via penetration of genome through host cell membrane C. Uncoating of a genome from the capsid coat within the host cell cytoplasm D. Lysis of the cell membrane after the lytic cycle

C. Uncoating of a genome from the capsid coat within the host cell cytoplasm

A FISH probe should be made of: A. an antibody that binds to a DNA epitope, covalently linked to a fluorescent segment of DNA. B. a segment of double-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorescent protein. C. a segment of single-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorochrome. D. a single-stranded piece of RNA.

C. a segment of single-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorochrome.

Phosphofructokinase (PFK) catalyzes the phosphorylation of fructose-6-P in the third step of glycolysis. If ATP levels in the cell are high, ATP can bind to PFK (at a site other than the active site) to inhibit the reaction. This is most accurately described as: A. competitive inhibition. B. negative feedback. C. allosteric regulation. D. positive feedback.

C. allosteric regulation.

Fluoroquinolones are a new class of antibiotics that are extremely effective against a wide range of bacteria. It has been determined that fluoroquinolones enter cells along with water via a passive transport mechanism. After exposing both bacterial and human cells to high concentrations of fluoroquinolones, a researcher discovered that, in the bacterial cells only, the DNA was nicked and supercoiling had been disrupted. Of the following, the most likely explanation for this is that fluoroquinolones: A. are unable to diffuse into human cells. B. interfere with DNA polymerase, which is not present in human cells. C. interfere with DNA gyrase, which is not present in human cells. D. interfere with DNA helicase, which is not present in human cells.

C. interfere with DNA gyrase, which is not present in human cells.

Tryptophan, an essential amino acid found in banana, turkey, and milk proteins, can induce sleep in some people. Warm milk causes greater sleepiness than cold milk because heating the milk: A. reduces the solubility of tryptophan in the milk. B. causes hydrolysis of lactose, releasing tryptophan. C. releases free tryptophan from proteins, causing more rapid intestinal absorption. D. increases the rate of absorption of tryptophan by the stomach.

C. releases free tryptophan from proteins, causing more rapid intestinal absorption.

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.

C. requires organic molecules as a carbon and energy source and cannot synthesize leucine.

A virologist is working to classify a new pathogen, and believes that it is a viroid. If the new pathogen is a viroid, an analysis of the pathogen should show that: A. the pathogen has a complex capsid. B. the ratio of adenine to guanine in the genome is 1:1. C. the genome contains no thymine. D. the genome codes for numerous proteins.

C. the genome contains no thymine. Viroids are RNA, and thus, there would be no thymine in its genome. Viroids do not have capsids and do not usually code for proteins. Viroids are single stranded; the ratio of A to G nucleotides is not 1:1.

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

D. 0.51 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.

Which of the following conditions is LEAST likely to lead to a spontaneous reaction? A. A reaction with a large negative ΔH, run at a low temperature, that has a large increase in entropy. B. A reaction with a small positive ΔH, run at a low temperature, that has a large increase in entropy. C. A reaction with a small positive ΔH, run at a high temperature, that has a large increase in entropy. D. A reaction with a large positive ΔH, run at a low temperature, that has a large decrease in entropy.

D. A reaction with a large positive ΔH, run at a low temperature, that has a large decrease in entropy.

Which of the following forms of eukaryotic post-transcriptional modification is most critical for ribosome association with an mRNA transcript? A. Substitution of uracil for thymine B. Intron splicing C. Addition of the 3' poly-A tail D. Addition of the 5' cap

D. Addition of the 5' cap

Which of the following is NOT a means by which prion infection can typically occur? A. Inheritance B. Spontaneous mutation C. Consumption D. Aerosol contact

D. Aerosol contact

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.

D. Animal viruses can enter their host via endocytosis, but this requires a specific receptor on the host surface.

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

D. Cis stack, medial stack, trans stack, anchor to cell membrane

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.

D. Cyanobacteria have photosynthetic machinery in the chloroplast and cytoplasm.

A biologist designs a fluorescent form of DNA helicase which emits visible light at the edge of spreading replication forks. She then images an unknown cell and observes several dozen fluorescent puncta. Which of the following best characterizes the cell and why? A. Prokaryotic: multiple locations of active replication are visible on a single chromosome. B. Prokaryotic: multiple locations of active replication are visible on multiple chromosomes. C. Eukaryotic: multiple locations of active replication are visible on a single chromosome. D. Eukaryotic: multiple locations of active replication are visible on multiple chromosomes.

D. Eukaryotic: multiple locations of active replication are visible on multiple chromosomes.

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

D. Exocytosis

A researcher has discovered a novel frameshift mutation in pyruvate carboxylase and hypothesizes that this could affect tumor metabolism. Which of the following would be the best experiment to perform to either confirm or disprove her hypothesis? A. Isolate genomic DNA from a cancer cell line and sequence the gene for pyruvate carboxylase, to confirm which amino acid is altered in the mutant form. B. Generate an animal model that expresses the mutant form of pyruvate carboxylase, and determine how the proton gradient is used by ATP synthase isolated from these animals. C. Perform western blot analysis on lysates from a brain tumor to confirm the mutation causes increased enzymatic activity. D. Generate a cell line that expresses the mutant form of pyruvate carboxylase and measure lactate secretion, and glucose and glutamine uptake compared to control cells that express the normal form of pyruvate carboxylase.

D. Generate a cell line that expresses the mutant form of pyruvate carboxylase and measure lactate secretion, and glucose and glutamine uptake compared to control cells that express the normal form of pyruvate carboxylase.

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.

D. Half the flies have red eyes and half have white eyes, regardless of sex. 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).

Which of the following intermolecular attractions will exhibit the greatest strength? A. London dispersion forces B. Induced dipole interactions C. Instantaneous dipole interactions D. Hydrogen bonds

D. Hydrogen bonds The various intermolecular forces, in order of decreasing strength, are the following: Hydrogen bonding > Dipole-Dipole interactions > Dipole- Induced dipole interactions > Induced dipole-Induced dipole interactions (London forces).

Enzymes are proteins that catalyze specific chemical reactions within a cell. Which of the following statements about enzymes is/are NOT true? I.A common means of regulating enzyme activity is through phosphorylation of the enzyme. II. Enzymes increase the energy of activation for a reaction, thereby making it go faster. III. Enzymes shift the equilibrium of a reaction towards products. A. I only B. II only C. I and II only D. II and III only

D. II and III only

Which of the following is an example of reciprocal regulation of glycogen metabolism? A. High levels of citrate stimulate phosphofructokinase and inhibit fructose-1,6-bisphosphatase. B. High levels of AMP inhibit glycolysis and stimulate gluconeogenesis. C. Glucagon inhibits glycogen phosphorylase and stimulates glycogen synthase. D. Insulin stimulates glycogen synthase and inhibits glycogen phosphorylase

D. Insulin stimulates glycogen synthase and inhibits glycogen phosphorylase

Viroids are circular pieces of single-stranded RNA, approximately 200-400 bases in length. What induces the folding of these subviral particles? A. Use of gyrase B. Formation of β-helices C. Double-bond linkages D. Many regions of self-complementarity

D. Many regions of self-complementarity

Symptoms of a prion disease can include problems with motor coordination and even with sleep. This indicates that the pathological protein is destroying what type of cell? A. Hair cells of the inner ear B. B cells capable of producing antibodies against it C. Skeletal muscle cells D. Neurons

D. Neurons

If a patient with cystic fibrosis receives a double-lung transplant from a non-cystic fibrosis donor, would the new lungs be expected to develop cystic fibrosis? A. Yes, once you have cystic fibrosis, it develops in every organ of the body. B. Yes, since the primary defect is with respiratory secretions. C. No, because the infectious causes of the disease will be removed when the old lungs are taken out. D. No, since cystic fibrosis is due to a gene defect, the cells of the new lungs will have the normal CFTR gene.

D. No, since cystic fibrosis is due to a gene defect, the cells of the new lungs will have the normal CFTR gene. Cystic fibrosis is a genetic disease based on the abnormal protein CFTR. In a set of lungs from a person without cystic fibrosis, the lung cells presumably have normal-functioning CFTRs. Therefore, the new lungs should not be subject to the development of cystic fibrosis (eliminate choices A and B). Cystic fibrosis is a multi-organ disease since the CFTR is used in secretions from several glands, but this does not cause normal CFTRs to become abnormal. The primary defect is the protein CFTR, not the pulmonary secretions or an infectious cause (choice C is incorrect and choice D is correct).

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.

D. Nondisjunction in anaphase II led to aneuploidy in some but not all gametes.

An error was made during DNA replication that ultimately resulted in the synthesis of an mRNA with guanine substituted for cytosine in the codon UCA. This would represent which of the following types of mutation? A. Insertion B. Silent mutation C. Missense mutation D. Nonsense mutation

D. Nonsense mutation

A mixture of aspartate and phenylalanine is separated into its component molecules by thin layer chromatography on a silica plate eluted with benzene. Which of the following best explains why the separation occurs? A. Aspartate will move farther with the mobile phase, because it has a polar side chain. B. Aspartate will move farther with the mobile phase, because it has a nonpolar side chain. C. Phenylalanine will move farther with the mobile phase, because it has a polar side chain. D. Phenylalanine will move farther with the mobile phase, because it has a nonpolar side chain.

D. Phenylalanine will move farther with the mobile phase, because it has a nonpolar side chain. The side chain on aspartate is -CH2COO-, which is very polar, while the side chain on phenylalanine is -CH2Ph, which is nonpolar. Since "like dissolves like," and the mobile phase is nonpolar, the molecule which is less polar (phenylalanine, in this case) will move farther with the mobile phase.

Which of the following statements regarding RNA molecules is NOT true? A. RNAs can act as enzymes to catalyze reactions. B. Some RNAs have more than four different types of bases. C. Some RNAs are synthesized in the nucleolus. D. RNAs are insusceptible to alkaline hydrolysis.

D. RNAs are insusceptible to alkaline hydrolysis.

Which of the following transport mechanisms allow for infection of animal cells by viruses? A. Secondary active transport B. Simple diffusion C. Ligand-gated channels D. Receptor-mediated endocytosis

D. Receptor-mediated endocytosis

Mitochondria share what characteristic with prokaryotic cells? A. Presence of membrane-bound organelles B. Maternal inheritance pattern C. Lack of ribosomes D. Single chromosome

D. Single chromosome

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.

D. The black lab and golden lab parents will have black, brown, and golden puppies, with golden coat color being the most common. 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).

If a fully saturated solution of AgI, with precipitate present, were treated with NaCl instead of NaI, which of the following observations is likely? A. As NaCl is added, all precipitates are dissolved into the aqueous solution. B. The decrease in [AgI] is even more drastic than with the addition of NaI in Figure 1. C. There is no change in the amount of undissolved AgI. D. The concentration of [I-] increases.

D. The concentration of [I-] increases. Unlike NaI, NaCl does not have a common ion with AgI and will therefore NOT cause a decrease in the solubility for AgI with increasing concentration (eliminate choice B). The following will act as a competing reaction when [Cl-] concentrations become sufficiently large: Ag+ (aq) + Cl- (aq) → AgCl (s) With this in mind, there will be no situations wherein the solution is free of precipitate (eliminate choice A). As the dissolved [NaCl] concentration increases, AgCl will be precipitated from solution, which will enable additional AgI to dissolve (eliminate choice C). The increased dissolution of AgI will cause the increase in [I-], even as [Ag+] levels remain low.

Which one of the following statements is true regarding PCl3? A. The phosphorus has sp2 hybridization, and the molecule has a trigonal pyramid shape. B. The phosphorus has sp3 hybridization, and the molecule has a tetrahedral shape. C. The phosphorus has sp2 hybridization, and the molecule has a trigonal planar shape. D. The phosphorus has sp3 hybridization, and the molecule has a trigonal pyramid shape.

D. The phosphorus has sp3 hybridization, and the molecule has a trigonal pyramid shape.

What cellular process can viroids exploit in order to replicate their genomes? A. Use of ribosomes B. Use of primase C. Use of DNA polymerase D. Use of transcriptional machinery

D. Use of transcriptional machinery

Some vitamins are essential to humans because they act as precursors of: A. auxins. B. glucose. C. enzymes. D. coenzymes.

D. coenzymes.

The diaphragm plays an important role in respiration. During inspiration, the diaphragm: A. relaxes, causing alveolar pressure to drop below atmospheric pressure. B. contracts, causing alveolar pressure to rise above atmospheric pressure. C. relaxes, causing alveolar pressure to rise above atmospheric pressure. D. contracts, causing alveolar pressure to drop below atmospheric pressure.

D. contracts, causing alveolar pressure to drop below atmospheric pressure.

All viruses must: A. have an equal ratio of thymine:adenine. B. have an envelope. C. code for DNA polymerase. D. have a capsid.

D. have a capsid.

As an ideal fluid flowing in a narrow pipe passes from a region of cross-sectional diameter d to a region of cross-sectional diameter d / 2, the flow speed of the fluid will: A. decrease by a factor of 4. B. decrease by a factor of 2. C. increase by a factor of 2. D. increase by a factor of 4.

D. increase by a factor of 4.

High levels of ATP would: A. stimulate phosphofructokinase and inhibit pyruvate kinase, thus stimulating glycolysis. B. stimulate phosphofructokinase and stimulate pyruvate kinase, thus stimulating glycolysis. C. inhibit phosphofructokinase and stimulate pyruvate kinase, thus inhibiting glycolysis. D. inhibit phosphofructokinase and inhibit pyruvate kinase, thus inhibiting glycolysis.

D. inhibit phosphofructokinase and inhibit pyruvate kinase, thus inhibiting glycolysis.

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

D. microtubule 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).

Transposons are mobile genetic elements that can "jump" around the genome, causing chromosomal aberrations. All of the following are true of transposons EXCEPT: A. a chromosomal inversion can be caused by two transposons in the opposite orientations. B. chromosomal deletions and translocations can be caused by two transposons in the same orientation. C. if the protein coding region of a gene is disrupted by a single transposon, then proteins levels will likely decrease. D. the insertion of a single transposon into the promoter or regulatory region of a protein coding gene will always decrease protein levels.

D. the insertion of a single transposon into the promoter or regulatory region of a protein coding gene will always decrease protein levels.

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.

D. thin peptidoglycan cell wall and is rod-shaped.

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.

D. two parents with photic sneeze reflex that have normal and affected offspring in a 1:3 ratio, respectively.

The activation free energy (GTS) is composed of the enthalpy (HTS) and entropy (STS) of activation by the same relation as free energy, enthalpy and entropy are related for energy minima (GF, HF and SF). Which of the following is likely true regarding the transition from the folded state (F) to the transition state (TS)? A. ΔH (F → TS) < 0 B. ΔG (F → TS) < 0 C. ΔS (F → TS) < 0 D. ΔH (F → TS) > 0

D. ΔH (F → TS) > 0 Choice B can be immediately eliminated, as the free energy of activation will always be higher than the free energy of either minimum it connects. To eliminate choice C it is helpful to remember that proteins are highly ordered macromolecules. In fact, just the term "unfolding" implies that there is an orderly arrangement being undone, and hence it is very unlikely that such unfolding would require a negative change of entropy. This, on its own, means that choice D is correct as ΔG is positive from the folded state to the transition state and ΔS is also positive (ΔG = ΔH - TΔS). Still, reasoned further, the bonding (enthalpic) components that hold a folded protein together must overcome the entropic penalty associated with this ordering. Therefore, ΔH (TS → F) must be very negative, meaning the opposite process is very positive, and choice D is correct.

The bond between the two sugar molecules below is best characterized as a/an: glucose - fructose A. ionic bond. B. β-glycosidic linkage. C. phosphodiester bond. D. α-glycosidic linkage.

D. α-glycosidic linkage.