NS exam 5 Biology
C is correct. In the lac operon, the operator is the segment of DNA that binds to the repressor. In the absense of functional repressor/operator binding, the cell will constitutively produce the proteins needed for lactose metabolism. Although this may not be fatal to the cell, it will waste energy if the surroundings lack lactose. A: The promoter is a segment of DNA. It binds to RNA polymerase. B: The operator binds to the repressor. The repressor is coded elsewhere. D: This is false; without functional operator/repressor binding, the cell's lactose-digesting machinery will be "stuck on."
A bacterium has a faulty lac operon in which there is a structural defect in the operator. In this bacterium: A. there is a mutation in a segment of DNA that binds a promoter. B. a missense mutation is found in the gene that codes for the repressor. C. there is a structural problem with a segment of DNA that binds a repressor. D. there will be no proteins available capable of digesting lactose.
A is correct. Given that uracil, not thymine, is incorporated during transcription, the radiolabeled [3H]-thymine would only be incorporated during DNA (and not RNA) synthesis. DNA synthesis only occurs during the S phase of the cell cycle. Therefore, in order for p53 to prevent the uptake of [3H]-thymine, the protein must arrest the cell cycle prior to the S phase at the G1/S regulation checkpoint.
A follow-up experiment assaying for cell cycle arrest with radiolabeled [3H]-thymine indicated that CRC157 cells transfected with pC27-53 did not incorporate [3H]-thymine during DNA synthesis. Based on this information, at what regulation point in the cell cycle does WT p53 arrest cell growth? A. G1/S B. S/G2 C. G2/M D. M/G1
C is correct. Transduction is the process by which DNA is transferred from one bacterium to another by a virus—in the case of the experiment performed in the passage, by the bacteriophage. A: Transformation is the genetic alteration of a cell resulting from the direct uptake and incorporation of exogenous genetic material from its surroundings through the cellular membrane. B: Bacterial conjugation is the transfer of genetic material, ordinarily in the form of a plasmid, between bacterial cells by direct cell-to-cell contact or via a bridge-like connection between two cells. The prototypical conjugative plasmid is the F plasmid, which is transferred from donor to recipient cell via a bridge-like structure known as a pilus. D: Translation is the process of protein synthesis carried out on the ribosomes of cells.
A follow-up experiment revealed that the genetic content of the bacterial cells was altered by the transfer of material from the phage. This process is best described as: A. transformation. B. conjugation. C. transduction. D. translation.
D is correct. The unusual cyclic structure of proline creates a high degree of rigidity in the primary structure of a protein and disrupts the normal formation of alpha or beta secondary structures. Thus proline is much more likely to be found in the unstructured "turn" regions of a protein between the larger secondary structures. A, B, C: Proline is not specifically associated with either alpha helices or beta-pleated sheets.
A particular oncogene product with an unusually high number of proline residues is most likely to use these residues: A. to aid in the formation of alpha helices. B. to be found primarily throughout beta-pleated sheets. C. to aid only in the formation of antiparallel beta-pleated sheets. D. to be found in turn regions of the protein.
B is correct. In DNA sequences, cytosine binds to guanine and adenine binds to thymine. Thus (A+T) + (G+C) = 100%. If A+T = 62%, then G+C = 38%. Therefore A = T = 31%. A, C, D: These answer choices result from miscalculation or confusion regarding base-pairing rules.
A researcher analyzing a genome measures a cytosine composition of 19%. What is the expected adenine composition of this genome? A. 19% B. 31% C. 38% D. 62%
B is correct. The question is asking us what physiological conditions would lead to the same response as what is observed during alcohol consumption. Based on the passage, we know that alcohol decreases the amount of water reabsorbed by the kidney, so it must increase the amount of water excreted in the urine. This is similar to the kidney decreasing water reabsorption, which it does in cases of high blood pressure or low plasma osmolality (our answer here). As a result of this decreased reabsorption, blood volume decreases and blood pressure becomes lower. A, C: If blood pressure is low, then blood volume is likely low as well, as in cases of dehydration. The kidneys would not reduce their reabsorption of water under these circumstances; in fact, they would do the opposite. D: High blood osmolality means that solute concentrations in the blood are high (in other words, the blood is very concentrated). This would provoke the body to attempt to conserve water, not excrete it (as in the case of alcohol consumption).
As described in the passage, the kidney's response to alcohol consumption is most similar to its response to: A. low blood osmolality and low blood pressure. B. low blood osmolality and high blood pressure. C. high blood osmolality and low blood pressure. D. high blood osmolality and high blood pressure.
A is correct. All of the amino acids listed should have a negatively-charged carboxylic acid terminal and a positively-charged amino terminal, so the distinguishing factor here is the side chain. Basic amino acids (of which arginine, or Arg, is the most basic) have nitrogenous side chains with the ability to gain a proton and thus be positively charged. At physiological pH, the side chain of Arg will have a charge of +1, making it the most positive of the given options. B: Methionine has an uncharged side chain, so its net charge will be 0 as opposed to arginine's net charge of +1. C: Watch out here! Asn (asparagine) has a side chain that includes an amide functionality, but amides (unlike amines) are not basic. In fact, amides are neutral, meaning that the side chain of Asn will have a charge of 0 under the described conditions. D: Aspartic acid has a side chain that exists in the COO- form at physiological pH, making it the most negative of the options listed.
Assuming all other conditions are equal, which of the following amino acids is expected to have the most positive charge at physiological pH? A. Arg B. Met C. Asn D. Asp
Blood Glucose Regulation Glucose is the body's main source of fuel under most circumstances, so the regulation of blood glucose levels is critically important in maintaining overall metabolic function. The two main hormones associated with glucose regulation are insulin and glucagon. Insulin is a peptide hormone released by the beta cells of the pancreas in response to high blood glucose levels. Its basic function is to reduce blood glucose levels by promoting the transport of glucose into cells via insulin receptors, which activate membrane-bound glucose transporters. The glucose transported into the cell can be used immediately through glycolysis; alternatively, muscle and liver cells can store the glucose as glycogen, and adipocytes (fat cells) can mobilize fatty acids to store downstream byproducts of glucose metabolism in the form of triglycerides. Insulin upregulates all those processes, as well as protein synthesis. Glucagon is a peptide hormone released by the alpha cells of the pancreas, and its mechanism and function are essentially the opposite of insulin. Glucagon is released in response to low glucose levels and has the effect of increasing blood glucose levels by promoting glycogenolysis and gluconeogenesis in liver cells. Other hormones can affect blood glucose levels as well. Cortisol (the main example of a class of hormones known as glucocorticoids) is released by the adrenal cortex. It is associated with long-term responses to stress and increases blood glucose levels. Epinephrine, which is released by the adrenal medulla and plays a major role in the fight-or-flight response to immediate stress, also raises blood glucose levels. In addition, growth hormone can increase blood glucose levels due to its antagonistic effects on insulin. Nonetheless, on Test Day, if you see a question about blood glucose levels, you should immediately think of the insulin-glucagon pair, unless the passage or question points you specifically in the direction of other hormones that may affect blood glucose levels.
Blood Glucose Regulation
C is correct. Insulin is produced by the beta cells of the pancreas, meaning that RN I is correct. RN III is also accurate; insulin is a peptide hormone, as is characteristic for hormones that have a quick-acting effect (and that end in -in). II: This statement is incorrect because insulin does not directly interact with glucose in the bloodstream; instead, it acts to reduce blood glucose levels by promoting the uptake of glucose into cells. A: RN III is also correct. B: RN II is incorrect, and RN III is correct. D: RN II is incorrect.
Diabetes mellitus is characterized by dysregulation of the hormone insulin. Which of the following statements about insulin is/are correct? I. Insulin is produced by the beta cells of the pancreas. II. Insulin directly acts on glucose in the blood. III. Insulin is a peptide hormone. A. I only B. I and II only C. I and III only D. I, II, and III
C is correct. During spermatogenesis, the main function of Sertoli cells is to nourish the developing sperm cells. These cells are located in the epithelial lining of the seminiferous tubules and are activated by FSH. A: Leydig cells are adjacent to the seminiferous tubules in the testicles. In response to stimulation by LH, they produce testosterone and other androgens. B: Chromaffin cells are catecholamine-secreting neuroendocrine cells of the adrenal medulla. Remember, the catecholamines include epinephrine and norepinephrine. D: Granulosa cells are follicular cells closely associated with the developing female oocyte (egg). They function to convert thecal androgens to estradiol prior to ovulation. After ovulation, they give rise to the corpus luteum and begin producing high levels of progesterone.
During normal development in the seminiferous tubules, what class of Tex11 non-expressing somatic cells is responsible for nourishing sperm cells? A. Leydig cells B. Chromaffin cells C. Sertoli cells D. Granulosa cells
B is correct. Complex II, like complex I, transfers electrons to coenzyme Q. Complex I receives its electrons from NADH, while complex II receives electrons from succinate. A: Complex I transfers electrons from NADH to coenzyme Q. C: Complex III transfers electrons from coenzyme Q to cytochrome c. D: Complex IV transfers electrons from cytochrome c to the final electron acceptor, oxygen.
During oxidative phosphorylation, which membrane complex transfers electrons from succinate to coenzyme Q? A. Complex I B. Complex II C. Complex III D. Complex IV
B is correct. The mesoderm gives rise to the musculo-skeletal system, circulatory system, and kidneys. The spinal cord, is derived from ectoderm. A, C, D: These structures develop from the mesoderm.
Each of the following structures are derived from the mesoderm EXCEPT: A. Kidneys B. Spinal cord C. Triceps D. Circulatory system
C is correct. This is a difficult question that asks us to make inferences based on passage information, so breaking it down helps. The question stem doesn't give us a lot of information, but we should note that ethanol and ethylene glycol are both alcohols, so both can be processed by ADH in the liver. Since ethanol can act as an antidote, it must somehow block the toxic effects of ethylene glycol. If ethanol has a high affinity for ADH, it could act like a competitive inhibitor and prevent the conversion of ethylene glycol to its aldehyde and carboxylic acid. This is exactly what occurs; alcohol's high affinity for ADH keeps ethylene glycol from being converted to glycolic acid, a toxic metabolite. Ethylene glycol itself is likely not very toxic, which we can surmise from the end of paragraph 2, which tells us it is the aldehyde metabolite of ethanol that is toxic. Thus, it is likely that glycol metabolism will have a similar effect.
Ethanol is used as a temporary treatment for poisoning with ethylene glycol (ethane-1,2-diol). The efficacy of alcohol as an antidote is best explained by: A. the high affinity of ethanol for ADH and the toxicity of ethylene glycol. B. the low affinity of ethanol for ADH and the toxicity of ethylene glycol. C. the high affinity of ethanol for ADH and the toxicity of glycolic acid. D. the low affinity of ethanol for ADH and the toxicity of glycolic acid.
Horizontal Gene Transfer in Bacteria Bacteria can engage in horizontal gene transfer. This can be thought of as a way of compensating for the fact that binary fission excludes any genetic variability except for what is introduced by chance. There are three mechanisms of horizontal gene transfer: transformation, transduction, and conjugation. Transformation refers to the ability of some bacteria to absorb genetic material directly from the environment. Early experiments in bacterial genetics revealed that harmless strains of Streptococcus pneumoniae could be made virulent by exposing them to virulent bacteria that had been lysed with heat. Transduction is virus-mediated gene transfer. The lack of nuclei in bacteria makes it relatively easy for viruses that infect bacteria (bacteriophages) to incorporate part of the bacterial genome during their assembly. Bacteriophages can then infect another bacterial cell, taking the genetic material from a previous cell along for the ride, because it can become integrated into the genome of the new cell after infection. Transduction has been widely applied in biotechnology. Conjugation can be thought of as the bacterial equivalent of sexual reproduction, and involves the transfer of a plasmid through a bridge that is created when a sex pilus on one bacterium (often known as F+, which refers to the presence of the fertility factor, or as male) attaches to another bacterium (generally known as F−). During this process, the fertility factor itself is duplicated and transferred, turning the F− bacterium into an F+ cell. Conjugation is a major mechanism contributing to the spread of antibiotic resistance.
Horizontal Gene Transfer in Bacteria
A is correct. The key to answering this question is to recognize that all of the enzymes discussed in the passage are involved in the citric acid cycle, and that therefore high levels of enzyme activity should correspond to high levels of citric acid cycle activity. Citric acid cycle activity, in turn, can be seen as a proxy for rates of oxidative metabolism. Tissues in which intense oxidative metabolism is carried out would require ample blood supply, both to provide oxygen and to remove metabolic waste products. Therefore, the correct answer to this question will be the sample with the highest levels of enzyme activity. Paragraph 2 indicates that the absolute levels of citrate synthase activity were highest in the insect samples and lowest in the crustacean samples. Paragraph 3 provides ratios of the different types of isocitrate dehydrogenase activity, but this will not help on this question because we need to assess absolute levels. Paragraph 3 does reinforce that the absolute levels of enzyme activity are low in crustaceans. Therefore, we can hypothesize that the insect muscle will be the most highly vascularized. B, C: Both the mammalian and fish samples are described as having citrate synthase activity somewhere between the crustacean and insect samples, meaning that they would have an intermediate level of vascularization. D: Paragraphs 2 and 3 indicate that the crustacean samples had very low levels of activity for all of these enzymes, making it likely to be the least vascularized.
If further study of the samples was conducted to assess the extent of vascularization in each muscle type, which sample is likely to be the most vascularized? A. Flight muscle of insects B. Heart muscle of mammals C. Red muscle of fish D. Crustacean muscle
A is correct. Mendel's Law of Independent Assortment states that separate genes which encode separate traits are passed down from parent to offspring independently of each other. Although Mendel induced this law from empirical observations, without a knowledge of molecular genetics, modern research has clarified the molecular basis for this finding. Independent assortment clearly occurs for genes that are physically located on different chromosomes, because it is random whether an organism inherits the maternal or paternal copy of a given chromosome — that is, during meiosis, the orientation of homologous chromosomes on the metaphase plate is random. Furthermore, even for genes on the same chromosome, recombination during meiosis ensures that they are inherited independently in most cases. However, genes that are physically located close to each other on a chromosome show a tendency to be inherited together. Nonetheless, the data presented in paragraph 2 of the passage conform to expected Mendelian distributions, and we are given no data that would suggest linkage, so it is reasonable to expect that there would be no correlation between the genes for these traits. B: The phenotype/genotype distinction is irrelevant to this issue. C: Homologous chromosomes do not necessarily have identical alleles/loci, and even if they did, this would not result in "linkage" in the technical sense of the term, which refers to a greater than 50% chance of alleles being inherited together. D: Homozygosity versus heterozygosity is irrelevant for linkage, which describes to a pattern where alleles have a >50% chance of being inherited together.
Is there a correlation or linkage between wing characteristics and eye color? A. No, because each trait is sorted independently from a genetic perspective. B. No, because phenotypes are distributed differently from genotypes. C. Yes, because the loci are identical when alleles appear on homologous chromosomes. D. Yes if homozygosity is present; no if heterozygosity is present.
Operons
Operons Operons are relatively simple and mechanistic systems that allow a bacterium to respond to changes in its environment by increasing or decreasing the expression of certain genes as appropriate. Operons can be under positive or negative control. In negative control, a repressor prevents transcription by binding to the operator (a sequence upstream of the first protein-coding region), while in positive control, an activator stimulates transcription. The classic examples for the MCAT, the lac operon and the trp operon, both involve negative control, but differ in that the lac operon is inducible and the trp operon is repressible. In a negative inducible operon, the repressor is normally present and the genes are not expressed except under specific conditions. In a negative repressible operon, the genes are usually transcribed, but transcription can be halted by binding of the repressor in appropriate conditions. In the lac operon, when glucose is present but lactose is not, there is no need for the bacterium to express the genes needed for lactose metabolism. Therefore, the repressor is bound to the operator, transcription does not happen, and no lactose metabolism takes place. When lactose is present, its metabolite allolactose releases the repressor from the operator, allowing transcription to happen and some degree of lactose metabolism to take place. However, intense lactose metabolism does not take place unless lactose is present in the relative absence of glucose, meaning that the bacterium really needs to metabolize lactose. In that case, cAMP causes the catabolite activator protein (CAP) to bind to the upstream CAP binding sequence. CAP promotes more intense expression of the genes on the lac operon.
A is correct. The researchers identified candidate genes in part by searching the siblings' genotypes for shared homozygous regions. A relatively rare homozygous allelic variant not associated with a splice site found in all three siblings may be a homozygous gene mutation responsible for DYT2. Since the researchers focused on homozygous regions of DNA, this implies that the researchers believed the inheritance pattern of DYT2 to be recessive. Additionally, because the female siblings' father was neurologically normal, if DYT2 is a recessive condition, it must be autosomal. If the disorder were X-linked recessive, an affected daughter would need to have inherited copies of the mutant X-linked allele from both her mother and her father. In such a situation, her father would show signs of DYT2 because his single X-chromosome would contain the defective allele. B, D: The researchers' search for homozygous tracts common to both siblings indicate that the believed pattern of inheritance was recessive. Dominant conditions require only a heterozygous mutation to be displayed phenotypically. C: Because the female siblings' father was neurologically normal, DYT2 must be autosomal.
Researchers selected candidate genes by identifying homozygous tracts shared by all three siblings. Along with the parental phenotype, what does this suggest regarding the inheritance pattern of DYT2? To identify causative mutations for DYT2, researchers extracted DNA from blood samples obtained from two sisters and a brother affected by DYT2. A. It is recessive because the father was neurologically normal and it is autosomal. B. It is dominant because the mother was neurologically normal and it is autosomal. C. It is recessive because the father was neurologically normal and it is X-linked. D. It is dominant because the mother was neurologically normal and it is X-linked.
D is correct. Facultative anaerobes can produce energy in the presence or absence of O2. In the presence of O2, the bacteria undergo aerobic respiration, which produces approximately 19 times as many ATP molecules per molecule of glucose as does anaerobic respiration. (Note that the fermentation step itself does not yield any ATP; it simply regenerates the NAD+ required to continue glycolysis.) ATP is necessary for many cell processes, including reproduction; thus, conditions that lead to less ATP are expected to lead to a lower growth rate. A: CO2 is not toxic to facultative anaerobes. B: The high-energy product formed under both aerobic and anaerobic conditions is ATP; aerobic conditions simply allow more of it to be produced. C: The Krebs cycle is not entered in the absence of O2.
Several Salmonella species are facultative anaerobes. Assuming that other external conditions are controlled for, would the expected growth rate of a Salmonella colony be slower in the presence or absence of O2? A. In the presence of O2, because aerobic respiration produces CO2, a byproduct that is lethal to facultative anaerobes. B. In the presence of O2, because the final product of aerobic respiration contains more energy than the final product of fermentation. C. In the absence of O2, because the bacteria will need to produce pyruvate decarboxylase, an enzyme required for entrance to the Krebs cycle. D. In the absence of O2, because these conditions result in lower production of ATP, which can fuel binary fission.
B is correct. Eukaryotic cilia and flagella are composed of bundles of microtubules. Note that this differs from prokaryotic flagella, which are formed from the protein flagellin. A: Microfilaments are composed of actin and are found in the cytoplasm of eukaryotic cells, as well as in muscle. C: Intermediate filaments are less dynamic than actin or microtubular filaments. They are not involved in ciliary or flagellar structure. D: Myosin is present in muscle and aids in the process of contraction.
Some eukaryotic cells are covered with small ciliary projections used for absorption, while others contain larger flagella used for propulsion. These cellular structures are composed of: A. microfilaments. B. microtubules. C. intermediate filaments. D. myosin.
A is correct. According to the final paragraph, "the nuclear envelope was intact as expected," but there was "an abnormally small number of tetrads undergoing synapsis." During meiosis, synapsis (and chromosomal crossover) occurs only during prophase I. B, C, D: Synapsis occurs only during prophase I.
The findings of the DNA analysis were confirmed by KpnI digestion of PCR products from both parents. Histological analysis of a testis biopsy revealed meiotic arrest at the pachytene stage. While the nuclear envelope was intact as expected, an abnormally small number of tetrads undergoing synapsis were visible. No post-meiotic germ cells such as spermatids and mature spermatozoa were observed in the seminiferous tubules, consistent with the diagnosis of azoospermia. The histological results of the biopsy of the patient's testicular tissue indicate that meiotic arrest most likely occurred during what stage? A. Prophase I B. Prophase II C. Metaphase I D. Metaphase II
C is correct. Kilodaltons (kDa) are units of size used for proteins. Since CP8 is larger than Zyg_27, it will move more slowly through the polyacrylamide gel that forms the matrix during SDS-PAGE. However, size-exclusion chromatography includes a stationary phase with many small pores, which smaller proteins become trapped in while passing through the column. For this reason, larger proteins (like CP8) travel more quickly during such chromatography procedures, and choice C is correct. A, D: These answer choices incorrectly present how protein size would affect SDS-PAGE and size-exclusion chromatography. Traveling further in SDS-PAGE would be associated with a smaller protein, which would elute more slowly in size-exclusion chromatography. B: This is true of Zyg_27, the smaller protein.
The researchers later used SDS-PAGE and size-exclusion chromatography to separate different mixtures containing both CP8 (a 76-kDa protein) and Zp_127 (a 40-kDa protein). CP8 would be expected to: A. travel farther during SDS-PAGE and elute more quickly during size-exclusion chromatography. B. travel farther during SDS-PAGE and elute more slowly during size-exclusion chromatography. C. travel a smaller distance during SDS-PAGE and elute more quickly during size-exclusion chromatography. D. travel a smaller distance during SDS-PAGE and elute more slowly during size-exclusion chromatography.
C is correct. human endogenous retroviruses, arose through the integration of retroviral material into the genome. It also mentions that HERVs are able to perform reverse transcription, a process that requires the enzyme reverse transcriptase. This enzyme catalyzes the production of DNA from an RNA template. Reverse transcriptase thus must have DNA polymerase activity, since it builds a new DNA strand; it is also RNA-dependent, since it reads an RNA template.
The sequence of a typical HERV is most likely to contain a gene that codes for: says its a retroviruses A. the production of a DNA-dependent RNA polymerase. B. the production of an RNA-dependent RNA polymerase. C. the production of an RNA-dependent DNA polymerase. D. the production of a helicase enzyme.
B is correct. Dithiothreitol (DTT) is a reducing agent often used during SDS-PAGE to further denature proteins by reducing/cleaving disulfide linkages and breaking up the quaternary protein structure (oligomeric subunits). The presence of one band when not exposed to DTT and three bands when exposed to DTT suggests that at least two disulfide linkages are present in the enzyme, and that these linkages hold three separate subunits together. A, C: There is no evidence in the passage or question stem to support either of these statements. D: From the given information, we cannot deduce anything about post-translational modification.
Two separate gel electrophoresis analyses are performed on a sample of purified α enzyme and the following results are obtained: SDS-PAGE: 1 band SDS-PAGE + dithiothreitol: 3 bands Which prediction about the enzyme structure is most strongly supported by these observations? A. Enzyme α contains a high ratio of charged to uncharged residues. B. Enzyme α contains more than one subunit. C. Enzyme α contains a low ratio of charged to uncharged residues. D. Enzyme α is post-translationally methylated.
A is correct. Nearly 95% of the human genome does not code for proteins or RNA. In contrast, the genomes of both prokaryotes and unicellular eukaryotes largely lack introns. In these organisms, most genetic material does code for protein products. B, C, D: These are all characteristics of human cells. The question asked for a trait that is unique to unicellular organisms.
Unlike the cells from which human organs are composed, the cell of a unicellular organism such as algae: A. has a genome where nearly all material codes for protein. B. typically utilizes mitosis for cellular division. C. can perform catabolic reactions to gain energy from macromolecules. D. contains membrane-bound organelles to execute cell functions.
A is correct. As the conjugate base of valproic acid, valproate must be negatively charged. As such, it will associate most strongly with positive species. Of the choices given, only K (lysine) is a basic amino acid and thus will have a positively-charged side chain at moderate pH levels. B, C: Both D (aspartic acid) and E (glutamic acid) are acidic amino acids, with side chains that are negatively charged around physiological pH. These residues would be repelled by, not attracted to, negative species like valproate. D: S (serine) typically has an uncharged, though polar, side chain around physiological pH. A neutral structure will not be as strongly associated to valproate as a positive structure would be.
Valproic acid dissociates in water (negative charge) to form its conjugate ion, valproate. Valproate is expected to associate most strongly with a protein rich in which of the following residues? A. K B. D C. E D. S
B is correct. Not every set of two alleles will have a fully dominant-recessive relationship. In this case, heterozygous mixing of the new allele with the recessive allele will cause the expressed phenotype to be a mix of the dominant trait (full-sized wings) and the recessive trait (no wings), rather than one or the other. "Incomplete dominance" is the correct term for this. A: Purity has no connection to what is observed here - it only has to do with which specific alleles are interacting. C, D: If the new allele received by the offspring was identical to the allele that the offspring received from the other parent (individual I93847, in this case), then the offspring would be either homozygous recessive (and have no wings) or homozygous dominant (and have full-size wings). The offspring would not be heterozygous and would therefore not have shrunken wings.
What best explains the shrunken wings observed in Experiment 3? A. Genetic engineering created a new allele which lacks the purity of a wild-type allele. B. Genetic engineering created a new allele which possesses incomplete dominance. C. Genetic engineering created a new allele which matches the recessive allele in the other parent. D. Genetic engineering created a new allele which matches the dominant allele in the other parent.
D is correct. Double-stranded DNA comes apart (melts) into two single-stranded fragments at elevated temperatures. Because there are three hydrogen bonds holding C and G residues together and only two such bonds between A and T, the more A/T residues in a segment, the lower the melting point. Conversely, the more C/G residues, the higher the melting point. Among the segments listed, choice D has the most A/T residues and thus the lowest melting point. A, B, C: These have more C/G residues, and therefore a higher melting point, than choice D.
Which of the following DNA oligomers, if in a double-stranded alpha helix conformation, would melt at the lowest temperature? A. 5'-CGCGCGTATCGACAAG-3' B. 5'-GGCGCGTATCGACAAG-3' C. 5'-GACGCGTCTCGACGGC-3' D. 5'-GATGCATATCGATAAA-3'
D is correct. In acid dissociation reactions, aqueous H+ is a product. Le Châtelier's principle states that removing one or more products will cause equilibrium to shift toward the product side of the reaction. In other words, if protons are continually removed from the solution, acetic acid will continue to dissociate. A: Increasing the pKa of a species implies that it becomes a weaker acid. Weak acids do not completely dissociate in solution. B: Enzymes allow a reaction to proceed faster, but they do not affect equilibrium concentrations. Also, enzymes by definition catalyze both the forward and reverse reaction of a particular process. C: This would force the equilibrium toward the product side of the reaction, causing more H+ to be present in the solution. However, it would not cause the acetic acid to completely dissociate.
Which of the following factors would be most likely to cause acetic acid to completely dissociate in aqueous solution? A. Higher temperatures, which increase the pKa of the acid B. Enzymes that catalyze the forward reaction C. Continuous addition of acetic acid to the solution D. Continuous removal of protons from the solution
C is correct. Under anaerobic conditions (in the absence of oxygen), pyruvate undergoes fermentation to lactate in the cytoplasm instead of being transported to the mitochondria for conversion to acetyl-CoA. Thus, acetyl-CoA will not be present to enter the Krebs cycle and will not be converted to citrate. A: This is a glycolytic intermediate. Glycolysis does not require oxygen and proceeds similarly under aerobic and anaerobic conditions. B: As a product of anaerobic fermentation, lactate will likely display increased concentrations in the absence of oxygen. D: Pyruvate is formed during glycolysis, which continues regardless of the presence of O2. While pyruvate is later converted to lactate via fermentation, there is no reason to suspect that this would lower its concentrations any more than its conversion to acetyl-CoA under aerobic conditions.
Which of the following molecules would be expected to have the lowest tissue concentrations in active skeletal muscle deprived of O2? A. Glyceraldehyde 3-phosphate B. Lactate C. Citrate D. Pyruvate
D is correct. The inside of the membrane is hydrophobic, so the membrane-spanning domain of a protein will likely consist of more hydrophobic amino acid residues. Only choice D consists of three amino acids all of which have hydrophobic side chains: leucine, isoleucine, and valine all have hydrocarbon R groups as side chains. A, B, C: Each of these choices have charged side-chains (D, E: negative; R: positive) that would not favor the hydrophobic region inside a membrane.
Which of the following segments of amino acids would be most likely to be found in the membrane-spanning domain of the sodium channel in a nerve axon? A. DDR B. EVE C. LAD D. LIV
D is correct. There are many obstacles facing successful gene therapy treatments for cancer, but gene incompatibility is not one of them. Critical genes are often conserved across species, and a wild-type copy of human p53 would certainly be genetically compatible with another human patient. (In fact, many genes are so highly conserved that human homologues can be used interchangeably with their bacterial counterparts.) For this reason, statement I is incorrect. Since statement II is in both remaining choices, we know that it must be correct. Transduction is mediated by viruses, which produce an immunological response when encountered by human cells. Finally, we turn to statement III. In order to arrest the growth of a tumor, nearly all (if not all) of the tumorous cells have must be successfully altered with the gene in question. Simply introducing the new gene to a small number of cells will be ineffective, as the remaining cancerous cells will repopulate the tumor and continue to grow uncontrollably. Thus, statement III is accurate as well.
Which of the following statements describe an obstacle researchers would face in implementing a similar transduction procedure to treat cancerous growth within patients? I. Gene incompatibility between patients with different endogenous copies of p53 II. Patient immunological response to the transduction virus III. Effective virus delivery into the entirety of the tumor A. I only B. II only C. I and III only D. II and III only
C is correct. According to the passage, N75K encodes "... a non-functional substitution of the Ca2+-coordinating residue of the binding sequence within the EF-hand domain 2 of hippocalcin." This suggests that a positively-charged amino acid (K) prevents coordination with Ca2+. As a consequence of this logic, it is most likely that a variant of HPCA with a negatively-charged residue at this location would have an enhanced ability to interact with Ca2+. N75D, in which asparagine is replaced by aspartic acid, is the only such choice. A, B, D: These are not acidic amino acids, so their side chains would not be expected to carry a negative charge.
Which variant of HPCA would be most likely to show an enhanced ability to coordinate with Ca2+? A. N75F B. N75L C. N75D D. N75V
C is correct. Nondisjunction is the failure of chromosomes to separate properly during anaphase I of meiosis or the failure of sister chromatids to separate properly during anaphase II of meiosis. A, B, D: Nondisjunction will not occur in any of these phases.
Women over 35 years of age have an increased risk of nondisjunction due to errors in what phase of meiosis? A. Prophase I B. Metaphase I C. Anaphase II D. Prophase II
crossing over of prophase I
synapsis of meiosis