NextStep Bio Questions
If the coding strand for a certain gene begins with 5' AGC CTT CGG CTG ACT GGC TGG, which of the following is a primer that researchers could use for reverse transcription PCR amplification? A. 5' AGC CTT CGG CTG ACT GGC TGG 3' B. 5' TCG GAA GCC GAC TGA CCG ACC 3' C. 5' AGC CUU CGG CUG ACU GGC UGG 3' D. 5' CCA GCC AGU CAG CCG AAG GCU 3'
A is correct. PCR uses DNA primers, so we can eliminate C and D because they contain uracil (U) and are therefore RNA. When a question that asks you to figure out a sequence, it's often a good idea to work methodically. B: This is correct primer's complement C, D: This is an RNA primer. Although RNA primers are used in vivo for DNA replication, PCR uses DNA primers
Which of the following is NOT a type of proofreading found in humans? I. Photoreactivation II. Base pair excision III. Homologous recombination A. I only B. I and III only C. II and III only D. I, II, and II
A is correct. Photoreactivation is a prokaryotic process (still used in some eukaryotes) to reverse damage done by UV light, humans use a separate process. Base pair excision repair is seen in both prokaryotes and eukaryotes. Homologous recombination is only seen in eukaryotes, which possess chromosomes. B: Homologous recombination is found to repair DNA damage in eukaryotes and is especially usful in repairing double-stranded breaks C, D: Homologous recombination and base pair excision are found in eukaryotes B, C, D: Only RN I is correct
CK isoforms containing M1 and M2 subunits migrate nearer the anode in gel electrophoresis than do CK isoforms containing the unmodified M subunit. The region near the C-terminus of the unmodified M subunit is enriched in which of the following amino acids? A. Lys, Asn, Arg B. Leu, Lys, Pro C. Lys, Ala, Asp D. Gln, Asp, Glu
A is correct. The passage states that once released from tissue, the CK M subunits are converted into the modified subunits M1 and M2 by successive cleavage of C-terminal amino acid residues. If the modified subunits migrate nearer the positive electrode (which is the anode in gel electrophoresis) than do unmodified subunits, then the modified subunits possess a lower pI amino acids via the removal of higher pI residues. Only choice A- lysine, asparagine, and arginine- contains two higher-PI, basic residues and is the correct answer. (Note that asparagine is not a basic amino acid). All of the other answer choices contain fewer basic residues. B: None of these amino acids are basic C: Of these three amino acids, only Lys is basic D: None of these amino acids are basic; on the contrary, Glu and Asp are acidic
Which of the following could be a treatment to reduce the amount of cholesterol in the body? A. Prevention reabsorption of bile acids once excreted in the body? B. Prevent absorption of free fatty acids by intestinal villi C. Increase availability of pancreatic lipase in the duodenum D. Allow for continued reuse of bile acids once excreted to the duodenum
A is correct. This question is asking us to determine the effects of each of the four options given. If we want to reduce the level of cholesterol in the body, we want to use it to make something else, like bile acids. If bile acids are unavailable to be reused once sent to the duodenum, then cholesterol will have to be continually converted into new bile acids.
How would Schramm and Herick's approaches carry over to anti-viral drugs? A. Neither could work B. Schramm's might work but Hedric's more likely would not C. Hedrick's might work but Schramm's more likely would not D. Either could be effective
A is correct. Viruses are acellular, so they can neither make use of cell signaling (Schramm) nor can they be destroyed by lysing their cell membrane (Hedrick). B, C, D: Be careful! On the MCAT, we have to stay strictly within the bounds of the information given in the passage. Schramm's approach is described a targeting bacterial signaling methods and Hedrick's approach targets the bacterial cell wall. Although we might speculate that the big-picture approaches of interfering with communication and physically targeting the pathogen might carry over to viruses, that goes beyond the scope of what the passage gives us to work with.
Which of the following correctly describes the conversion of AMP to IMP by AMP to IMP by AMP deaminase? A. delta G > 0 and ATP is required B. delta G = 0 and no ATP is required or produced C. delta G < 0 and ATP is required D. delta G < 0 and ATP could be produced
D is correct. A high activation energy barrier usually causes hydrolysis of a high energy bond to be very slow in the absence of an enzyme catalyst. This kinetic stability is essential to the role of ATP and other compounds with these bonds. If ATP would rapidly hydrolyze in the absence of a catalyst, it could not serve its important roles in energy metabolism and phosphate transfer. Phosphate is removed from these molecules only when the reaction is coupled via enzyme catalysis to some other reaction useful to the cell, such as transport of an ion, phosphorylation of glucose, or regulation of an enzyme by phosphorylation of a serine residue. According to the reaction, the Keq > 100. This Keq (>>1), indicates that the free energy change for the reaction is negative. Also, if the reaction is spontaneous, energy (in the form of ATP) is not required to drive it forward. As such, the reaction could be used to generate ATP, a molecule used biologically to store chemical potential energy. A. Since Keq > 1, delta G must be negative (below zero). This relationship comes from the equation delta G naught = -RT ln (Keq). B: For delta G to equal zero, Keq would need to be equal to 1, which is not true here C: ATP is not required for reactions that are already spontaneous (delta G < 0.
Increasing plasma concentration of aldosterone is most likely to be followed by which of the following? A. Increased water reabsorption through increased aquaporin channels in the collecting duct. B. Increased sodium reabsorption in the distal tubule. C. Decreased water reabsorption in the collecting duct D. Decreased plasma calcium concentration
B is correct. Aldosterone is released from the adrenal cortex in response to low blood pressure. Its primary function is to increase sodium reabsorption in the distal tubule and collecting duct. Aldosterone upregulates the sodium-potassium pumps along the lining of the nephron, pumping three sodium ions OUT of the nephron lining (and toward the blood for every two potassium ions it pumps IN (toward the nephron and away from the blood). Since we have a net solute movement out of the nephron, aldosterone also increases the gradient that favors water reabsorption. The human nephron is shown below, with the distal tubule (a primary target of aldosterone) labeled. A: Although aldosterone causes an increase in passive water reabsorption, it does not do so through regulation of aquaporin channels. That's effect of ADH. C: This is the opposite of aldosterone's effect. D: Aldosterone does not have a direct effect on calcium.
Injection of insulin into the bloodstream is LEAST likely to result in which of the following? A. Increased glycogen synthesis B. Decreased lipid synthesis C. Increased esterification of fatty acids D. Decreased gluconeogenesis
B is correct. Insulin is secreted in response to high blood sugar, so an insulin injection would simulate how the body would respond to there being high levels of blood sugar, in which case it would want to stop making more sugar (eliminate choice D), to restore that sugar as glycogen (eliminate A), and to build up fatty acids into fats for storing up energy (eliminate choice C). The thing the body is LEAST likely to do is to stop storing up energy (choice B). In general, think of the function of insulin as causing the body to build up large molecules to store up energy (glycogen, lipids) and to top stop the body from breaking down large molecules to provide energy. A: Again, the high-blood-sugar circumstances that promote insulin release would absolutely coincide with glycogen synthesis, since the excess glucose can be stored in the form form of the glycogen molecule. Choice A is thus not the answer to this LEAST question C: Esterification of fatty acids is analogous to the formation of glycogen from glucose; it packs the fatty acids into molecules for storages rather than using them immediately. This would absolutely occur if glucose were already abundant in the body. D: If blood sugar is already high, there would be little need for gluconeogenesis (production of glucose from non-carbohydrate sources). This choice would also occur and is not the answer to this LEAST question.
All of the following are functions performed by macrophages EXCEPT: A. Antigen presentation B. Opsonization C. Phagocytosis D. Cytokine Release
B is correct. Opsonization is the process by which antibodies bind to and reocgnize antigens on the surface of a pathogen. The antibodies then attract macrophages to phagocytose the invader. A, C, D: These are all functions of macroophages
Which components of cells are physically connected by a gap junction? A. The cytosol of one to the nucleus to the other B. The cytoskeleton of one to the cytoskeleton of the other C. The cytoskeleton one to the plasma membrane of the other D. The plasma membrane to the nucleus of the other
B is correct. The answer is given directly in paragraph 2, which mentions "...two hemichannels (connexons), one embedded in each cell's plasma membrane and anchored to the cytoskeleton of the cells." This means that the cytoskeleton of one cell is connected to the cytoskeleton of another. A schematic of a gap junction is shown below. A, D: Gap junctions don't connect to the nucleus. C: The cytoskeleton of one cell physically connects to the connexons, which then physically link to the cytoskeleton of the next cell. Gap junctions don't bring the cytoskeleton of one cell into contact with the plasma membrane of the neighboring cell.
Which of the following is the most reasonable approximation for the pI of glucagon? A. 2 B. 7 C. 10 D. 12
B is correct. The last part of the passage tells us that we can estimate the pI of a protein using the relative number of acidic and basic side groups. Thus, we must count up these side groups in glucagon. Glucagon has 3 acidic and 4 basic side groups. This tells us that we would expect the pI to be near neutral or slightly basic, making choice B correct. A: This choice is too acidic. There is no reason to think that the acidic side groups of glucagon would have this much sway over the pI. C, D: These choices are much too basic
Based on information in the passage, what type of inhibition best describes the action of NADPH on G6PD? I. Competitive II. Allosteric III. Irreversible A. I only B. II only C. II and III only D. I, II, and III only
B is correct. The passages states that high levels of NADPH inhibit G6PD, and Equation 1 shows that the substrate of G6PD is G6P. Because the structures of G6P and NADPH are very different, it is unlikely that NADPH competes with G6P at the active site; thus it is not competitive inhibition. The passage also never suggests that the inhibition of G6PD is irreversible; rather, it is likely to be dynamic based on the amount of NADPH available. Therefore, NAPDH most probably binds to a site that it is not the active site, which is characteristic of allosteric inhibition. A generic model of allosteric inhibition is included below. A: RN I is incorrect, while RN II is correct C: RN III is incorrect D: RN I and RN III are incorrect
If we start with the molecule below instead of palmitic acid, which step will be interrupted and which enzyme will be used to correct the issue? A. Step 1, phosphatase B. Step 1, isomerase C. Step 2, reductase D. Step 2, kinase
B is correct. The question is asking us to consider the implications of starting with a palmitic acid with a double bond at the 7 carbon. This will mean that after the initial conversion of the acid to a CoA precursor for entrance into the beta oxidation steps shown, we will be left with: Since the presence of an alkene is something that is supposed to be achieved by Step 1, this molecule will interrupt the formation of the double bond in step 1 (eliminate choices C and D). To fix the problem, we need to form a version of the molecule formed by step 1 with its alkene bond, between the alpha and beta carbons instead of where it is now. Since the two molecules have the same atomic ingredients and only differ in the configuration of these atoms, they are isomers. Thus, an isomerase will do the trick nicely. A: A phosphatase would act to remove a phosphate group, and this is not necessary in this scenario C, D: Step 2 is not the step that would be interrupted
If pharmacologists wished to convert morphine into a form available to brain tissue, which of the following changes could be made to its molecular structure to allow for the best chance to use the drug as direct brain treatment? A. Attach a glucose molecule to the nitrogen atom B. Replace the alcoholic protons with acetyl groups C. Lyse the molecule to make it smaller D. Remove all double bonds from the molecule
B is correct. This question asks us to consider the structure of morphine given in the passage. What needs to change for morphine to enter the brain while maintaining its function? We are told in the passage that lipid-soluble molecules are able to pass through, while hydrophilic molecules are not. Morphine is hydrophilic in part because of its hydroxyl groups. So, replacing them with acetyl groups would allow the drug to become more lipophilic; although the C=O bond in the acetylated molecule is polar, it cannot undergo hydrogen bonding, so the acetylated structure is effectively much more lipophilic than the original structure with -COOH groups. Interestingly, diacetylated morphine is heroin. The effects and pharmacology of heroin are somewhat distinct from those of morphine, although they belong to the same general class of opiates, and heroin does indeed more readily cross the BBB. A: This would only make the molecule bigger and do little for its polarity. Additionally, while glucose can pass through the blood-brain barrier, adding glucose to another molecule would not necessarily allow the molecule to utilize glucose transport proteins. C, D: These would destroy the molecule's structure, making it unlikely to maintain its therapeutic function.
What aspects separate single-cross over events from double-crossover events? A. Single-crossover events result in one-way displacement of chromosomal content from one chromosome to another, while double-crossover events always reverse this one-way displacement, resulting in chromosomes identical to the pre-crossover chromosomes B. Single-crossover events occur mitosis when a cell splits into two cells, while double-crossover events can only occur during meiosis when a cell splits into four cells C. Single-crossover events affect only the ends of chromosome arms, while double-crossover events can affect segments in the middle of chromosome arms D. Single-crossover only affect one arm of each chromosome, while double-crossover events affect two arms of each chromosome
C is correct. A double-crossover event is one in which chromosomal arms of homologous chromosomes cross over in two different places along the arm. This results in a section in the middle of each chromosome being exchanged. A simplified schematic of a single- vs. a double-crossover event is shown below. A: A double-crossover event occurs in two different places along the chromosome arm, resulting in a segment exchange, not necessarily a reversal of the original crossover. B: Crossover events only occur during meiosis, not during mitosis. D: Both single- and double-crossover events will only affect one arm of each chromosome
According to the data in Figure 1, what is the probability that a male Robertsonian translocation carrier who mates with a normal female will produce a viable offspring? A. 1/6 B. 1/3 C. 1/2 D. 5/6
C is correct. According to Figure 1, there are 6 possible outcomes for the gamete production in a 14/21 translocation carrier. Going from left to right, we have a normal gamete, a balanced translocation (all genetic info is present; this gamete would give rise to another ROB carrier), trisomy 21, monosomy 14 (because the gamete would only receive a copy of chromosome 14, not chromosome 21, from this parent), and trisomy 14. There are no viable autosomal monosomies, and the only viable autosomal trisomies you should know are trisomy 21, 18, and 13 (though trisomy 8, 9, and 22 can also survive to term). Therefore, of the 6 gametes produced by the man, 3 of them contain genetic information that would produce a viable offspring (two normal, one with Down syndrome). 3/6 = 1/2. A, B, D: These answers are the result of miscalculation.
Which of the following statements is NOT true with regard to the replication of the Srebp1 gene in humans by the eukaryotic DNA polymerases involved? A. The strand that is being copied in the direction of the advancing replication fork is synthesized continuously B. The strand that is being copied in the direction away from the replication fork is synthesized discontinuously C. DNA polymerase responsible for the elongation of the complementary DNA strand read the parental nucleotide sequence in the 5' - > 3' direction only D. Newly synthesized stretches of nucleotide chains are elongated in opposite directions relative to the position of the replication fork
C is correct. DNA polymerases are principally responsible for the elongation of the leading and lagging strands, respectively, during DNA replication. Both polymerases read the parental nucleotide template in the 3' -> 5' direction, adding nucleotides to the growing strand in the 5' -> 3' (antiparallel) direction. This makes choice C an incorrect statement and the correct answer. A: This is accurate; synthesis of the leading strand (which is copied in the direction of the advancing replication fork) by DNA polymerase advances continuously. The lagging strand is synthesized discontinuously via the addition of short RNA primers. B: This is also a true statement and refers to the lagging strand D: The leading and lagging strands are elongated in opposite directions, making this choice accurate
High levels of LDL and OX-LDL increase proportion of cholesterol in cell membranes. If the trisomy 21 data in Figure 2 can be attributed to the effect of cholesterol on these membranes, which of the following statements is most likely true? A. hTERT cells treated with OX-LDL display more rigid membranes than hTERT cells treated with LDL B. Untreated hTERT cells display more rigid membranes than hTERT cells treated with LDL C. At moderate to high temperatures, ethanol increases membrane fluidity D. hTERT cells simultaneously exposed to HDL and LDL display increased membrane fluidity relative to untreated cells.
C is correct. Figure 2 shows that LDL- and OX-LDL have a markedly higher incidence of trisomy 21 than untreated hTERT cells. The question stem also indicates that cells treated with these lipoproteins have more cholesterol in their membranes. At moderate to high temperatures (including normal physiological temperature), cholesterol increases the rigidity of cell membranes by attracting adjacent phospholipid tails. Thus, a more rigid membrane appears to correlate with a higher incidence of triosomy 21. From Figure 3, however, we see that ethanol appears to counteract the trisomy-inducing effects of LDL and OX-LDL. It is therefore reasonable to conclude that ethanol decreases the rigidity (or increases the fluidity) of cell membranes under the conditions in this study. A: Looking at the trisomy 21 data in Figure 2, there appears to be no significant difference between OX-LDL and LDL-treated cells. B: This is the opposite of the conclusion outlined above. LDL increases membrane cholesterol and thus increases membrane rigidity at moderate temperatures. (Note that the opposite is true at very low temperatures; cholesterol fills in the gaps between phospholipid tails, preventing the membrane from becoming too rigid.) D: The study did not include any cells that were exposed to both LDL and HDL at once, so we cannot predict what would happen in that case.
The results in Figure 1 and the information in the passage most strongly support which of the following conclusions? A. Phosphorylation of S156 by protein kinase A promotes the immediate localization of AQP5 to the plasma membrane B. Phosphorylation of S156 promotes the internalization of AQP5 in the short term C. Protein kinase A promotes the internalization of AQP5 in the short term D. 30 minutes of exposure to protein kinase A stimulates the internalization of AQP5, a process that is upregulated when S156 is phosphorylated.
C is correct. From Figure 1, we see that in both cell types, inhibition of protein kinase A results in significantly greater AQP5 expression. This supports the idea that stimulation of PKA activity would decrease membrane AQP5 expression, at least at some point. Paragraph 3 supports this idea with the statement that increased cAMP levels have a biphasic effect, decreasing AQP5 expression in the short term and increasing it in the long term. We should know (by test day) that PKA is stimulated by cAMP, so it makes sense that it would follow this pattern. Additionally, paragraph 3 outlines the difference between translocation to the membrane and internalization from the membrane to the interior of the cell. If cAMP and PKA signaling have the short-term impact of increasing AQP5 internalization, this would bring AQP5 proteins away from the plasma membrane, explaining the results (taken after 30 minutes) shown in Figure 1. A: We have no evidence that protein kinase A is the enzyme that phosphorylates S156. Additionally, it is the inhibition of PKA that appears to drastically increases AQP5 localization to the membrane, as shown in Figure 1. If PKA itself promoted the immediate localization of AQP5 to the membrane, we would expect the opposite of these results. B: Figure 1 shows that the mutant S156E, which simulates the phosphorylation of S156, increases the membrane expression of AQP5. "Internalization" denotes the opposite process- the movement of membrane protein inward, away from the membrane. D: Figure 1 shows that 30 minutes of exposure to a PKA inhibitor promotes membrane expression of AQP5 in both the wild-type and mutant strains. This does not provide sufficient information to clearly predict what would happen after 30 minutes of exposure to PKA itself, and furthermore, if anything, the data in this figure seems to suggest that S156 phosphorylation (here, denoted by the S156E mutant) promotes membrane expression of AQP5, which is the opposite of internalization, as implied by the answer choice.
In cells with elevated low-density lipoprotein levels, ethanol has been shown to act directly on these lipoprotein molecules in a way that decreases their trisomy-inducing effects. On the basis of this information, compared with untreated hTRERT cells, hTERT cells incubated ethanol would most likely: A. Have a higher probability of displaying trisomy 21 B. Have a lower probability of displaying trisomy 21 C. Have a similar probability of displaying trisomy 21 D. Have a higher probability of display trisomy 21 than hTERT cells incubated with HDL and ethanol
C is correct. From Figure 3, we see that the incubation of LDL- and OX-LDL-treated cells with ethanol decreases the incidence of trisomy 21 to levels comparable with the control (hTERT alone). However, do not jump to the conclusion that EtOH must always decrease trisomy 21 incidence. The question stem tells us that ethanol has been observed to act directly on lipoproteins "in cells with elevated low-density lipoprotein levels." We have no reason to believe that hTERT cells not incubated with LDL or OX-LDL would have such elevated levels, so we have no reason to think that ethanol incubation would have any effect on them. Note also that neither the LDL + EtOH nor the OX-LDL + EtOH show trisomy 21 levels significantly below the control levels. It is therefore very possible that ethanol incubation "canceled out" the trisomy-inducing effects of the extra low-density lipoprotein, but would have no effect in cells that did not have high low-density lipoprotein concentrations to begin with. A: Nothing in the results shows that EtOH indicates that EtOH could have a trisomy-inducing effect B: This choice is tempting, since EtOH did not reduce incidence in the LDL- and OX-LDL-treated cells. However, note that EtOH had no significant effect in the cells incubated with HDL alone. This supports the question stem's assertion that EtOH acts selectively in cells with "elevated low-density lipoprotein levels." The hTERT cells incubated with ethanol alone (referenced in the question stem) would have no more reason to display elevated LDL levels than would the hTERT + HDL cells from Figure 3, so we cannot assume that EtOH Would act to decrease the trisomy incidence of cells referenced in the question. D: Figure 3 shows the untreated hTERT cells and the HDL-incubated cells to have similar trisomy 21 levels. It certainly does not show that the untreated hTERT cells have higher trisomy incidence than the HDL-incubated cells, so we have no reason to think that the hTERT + EtOH cells mentioned in this question would have a higher trisomy incidence than HDL + hTER cells.
Scientists who wished to study the metabolic function of cells with balanced translocations while preventing cell replication would be best served by arresting the cells during which phase of the cell cycle? A. Anaphase B. Metaphase C. Interphase D. Prophase
C is correct. If the scientists wanted to prevent cellular replication, they would need to halt cell division (mitosis). Interphase is the stage of the cell life cycle (shown below) that occurs between rounds of division. A, B, D: Prophase, metaphase, and anaphase are all phases of mitosis. If cells were arrested at any of these stages, they would have already begun to replicate.
Which of the following molecules is/are most likely to have selective proteins in the BBB to facilitate its passage into the brain? A. Antibodies B. Starch C. Amino acids D. Urea
C is correct. The correct answer will be a molecule or substance that is essential to brain function. Amino acids are necessary for the production of proteins, which are essential for the function of any cell. A: In general, antibodies are too large to cross the BBB B: While select monosaccharides (e.g. glucose, fructose) can cross the BBB, large polysaccharides like starch are unable to cross D: Urea is a waste product that is unlikely to have specialized proteins for entry into the brain.
Troponin isoenzymes are used as an alternative biomarker in the diagnosis of heart attacks. In which of the following muscle types does the troponin complex function in contraction? I. Skeletal muscle II. Smooth muscle III. Cardiac muscle A. I only B. I and II only C. I and III only D. I, II, and III
C is correct. Troponin is a complex of three proteins (troponin I, troponin C, and troponin T) required for muscle contraction in skeletal muscle and cardiac muscle, but not smooth muscle. I and III are true. Choice C is then correct. A simplified schematic of skeletal and cardiac muscle is shown below. A: RN I is correct, but RN III is also correct B: RN II is incorrect, while RN III is correct D: RN II is incorrect
How many total bile salts can be created? A. 2 B. 4 C. 8 D. 16
C is correct. We are told in the passage that two primary bile acids can be performed, each of which can be dehydroxylated to create four distinct bile acids. To form a bile salt, each of those acids could be combined with either glycine or taurine, resulting in 2x2x2 = 8 possible bile salts
Considering the results of the laboratory testing given in the passage, which of the following diagnoses concerning the patient is most likely correct? A. The patient is unlikely to be suffering a heart attack, because the elevated CK isoenzymes are more closely associated with damage to the brain than with damage to cardiac muscle B. Additional testing is necessary, because the CK isoenzymes that are elevated are not specific for cardiac muscle damage, and may also indicate damage to skeletal muscle C. Additional testing is necessary, because CK isoenzymes that are markers of cardiac damage appeared at normal levels, but may have been elevated prior to the time at which the samples were drawn D: The patient is likely suffering a heart attack, because his CK isoenzyme levels are indicative of recent cardiac muscle damage in a patient with no signs of kidney disease.
D is correct. Compared to reference enzyme activity levels, the patient shows dramatically elevated levels of CK3 (normal: 0-2, patient: 7241) and elevated levels of CK@ (normal: 0.5-6, patient: 219). The patient's CK1 enzyme activity levels are roughly normal (norma: 42-198, patient: 141). According to the passage information, cardiac muscle contains both CK2 and CK3, while skeletal muscle contains only CK3. Thus, the fact that patient has elevated levels of CK3 indicates that he has either cardiac or skeletal muscle damage. Add in the fact that the patient also has elevated CK2 levels then further narrows down the diagnosis to indicate that the patient does have cardiac muscle damage (such elevated CK2 levels would not have come from skeletal muscle damage). A: Neither CK2 nor CK3 contain the B subunit associated with CK1 and the brain B: While muscle injury can account for the presence of elevated CK3 protein levels, it does not account for the presence of elevated CK2 levels C: We have no reason to suspect that we need additional testing, as the elevated CK2 levels indicate cardiac damage
What type of control does siRNA exert on G6PD expression? A. Transcriptional control B. Promotion C. Repression D. Post-transcriptional control
D is correct. Due to its structure, siRNA is only able to bind to other RNA strands, not to DNA or protein. Therefore, it must interfere with gene expression after transcription has already occurred, but before translation. Specifically, it prevents the translation of mRNA corresponding to the target protein. A, B: These controls occur before transcription. In contrast, siRNA prevents mRNA from being translated. C: Repressors are defined as protein molecules that bind with DNA or RNA to prevent eventual translation of a protein. Therefore, siRNA is not technically a repressor.
In what way does the synthesis of phage proteins in the infected cell differ from protein synthesis in a eukaryotic cell? A. Eukaryotic protein synthesis begins prior to completion of transcription B. Eukaryotic protein synthesis occurs in the cytoplasm C. Synthesis of prokaryotic proteins occurs from mRNA molecules coding for a single protein D. Eukaryotic protein synthesis takes place on 80s ribosomes
D is correct. Eukaryotic protein synthesis occurs on 80s ribosomes. These are distinct in their subunit composition from 70s ribosomes, which are found exclusively in prokaryotes. Phage is short for bacteriophage, meaning that a virus produces using the host prokaryotic machinery. A: This is false. Eukaryotic protein synthesis, unlike that which occurs in prokaryotes, does not begin until transcription is completed B: Protein synthesis in both eukaryotes and prokaryotes occurs in the cytoplasm C: mRNA molecules are said to be monocistronic when they contain coding sequences for a single polypeptide. This is the case for most eukaryotic mRNAs. Polycistronic mRNA carries several open rading frames, each of which is translated into a polypeptide. These polypeptides usually have a regulated function, they are often the subunits composing a final complex protein, and their coding sequence are grouped and regulated together in a regulatory region, containing a promoter and an operator. Most of the mRNA found in bacteria are polycistronic.
Heterozygotes for the AAT mutation are significantly less likely to develop emphysema. Which of the following explains this observation? A. An individual must inherit two mutant AAT alleles in order for their genotype to confer protection against the onset of emphysema B. Those with a single mutant allele do not produce sufficient altered neutrophil elatase to cause damage to alveolar walls C. Two defective AAT alleles are required for polymerization and retention of abnormal a1-AT protein D. The presence of a single wild-type AAT allele produces sufficient functional a1-AT to protect respiratory connective tissue
D is correct. The passage states that two defective AAT alleles are required in order to develop emphysema, consistent with the information presented in the question stem. The passage also states that the mutation involves synthesis of defective a1-AT protein that polymerizes in and is retained by hepatic cells, rather than being released (to act in the lungs). This implies that a single wild-type copy of the AAT gene produces sufficient wild-type a1-AT protein to confer protection on the airway from damage by elatase. Choice D is then the best answer A: The passage states that two defective AAT allleles are required in order to develop emphysema. B: The AAT mutation described does not affect the structure of neutrophil elatase produced C: a1-AT protein produced would also polymerize and fail to be secreted from the liver, however, if it at least one wild-type copy gene is present, normal protein is produced
The pedigree below shows the mating between two insects as well selected members of the F1 and F2 generation. The original genotype of P generation for male and female are, respectively: A. PP and PQ B. PP and P'Q C. P'P' and P'Q D. PP' and PQ
D is correct. We see that the female produced both male and female offspring. This means that she passed her Q gene along to her daughter and thus does not have inversion. We can therefore eliminate choices B and C. Choice A would represent a cross between two wild-type insects and as such we do not see the gender disparity in the F2 generation. So choice D only seems fitting.
