Question 1

Dehydration caused by frequent loose stools is often observed in patients with certain diseases of the gastrointestinal (GI) tract.  Given this, which of the following treatments would be most effective in treating this symptom?

Accepted Answer

11ecba2d_1030_5426_a3bf_09a62f92580a_MD0008_00 The gastrointestinal (GI) tract is composed of the esophagus, stomach, small intestine, and large intestine. Food broken down in the mouth travels down the esophagus and enters the stomach, which produces chyme, a semifluid mixture of partially digested food. Chyme flows into the small intestine, where macromolecules are further broken down and absorbed, along with water, by intestinal epithelial cells. The large intestine absorbs any remaining water and electrolytes. In general, nutrients (ie, solutes) absorbed by intestinal epithelial cells are transported into surrounding interstitial fluid before diffusing into capillaries, increasing the osmotic pressure (solute concentration) in the interstitial fluid and, subsequently, in the blood plasma. Because water generally flows from low to high osmotic pressure via osmosis, increased osmotic pressure within interstitial fluid and blood plasma increases intestinal water absorption. In contrast, some epithelial cells secrete ions into the intestinal lumen, increasing osmotic pressure within the intestine to draw water in and lubricate luminal surfaces. In this scenario, dehydration is caused by frequent loose stools in affected individuals. Loose (watery) stools may be due to decreased water absorption or increased water secretion in the intestinal tract. Chloride (Cl⁻) secretion from intestinal epithelial cells would increase the osmotic pressure of intestinal contents, drawing more water into the intestine and increasing the water content of stool. Therefore, drugs inhibiting luminal secretion of Cl⁻ would likely decrease water secretion into the intestine and make the stools less watery. (Choices A and C) Parasympathetic signals generally enhance peristalsis and motility of intestinal contents through the GI tract. Drugs that enhance parasympathetic activity and, consequently, peristalsis would increase the flow of chyme through the GI tract, decreasing the time available for nutrient and water absorption by intestinal cells and increasing (not decreasing) the excretion of watery stools. (Choice B) A drug that decreases glucose absorption would decrease osmotic pressure of the interstitial fluid and capillary blood plasma surrounding the intestine. Decreased osmotic pressure caused by reduced glucose absorption would likely decrease intestinal water absorption and increase (not decrease) the excretion of water in stool. Educational objective: Most nutrient and water absorption occurs as food moves through the small intestine. The large intestine absorbs any remaining water and electrolytes prior to waste excretion.

Question 2

Passage In the central nervous system, myelin produced by oligodendrocytes (glial cells) functions as an insulating sheath surrounding certain nerve fibers. To identify candidate genes involved in the myelination process, researchers collected oligodendrocytes from zebrafish and purified the messenger RNA (mRNA) transcripts expressed by these glial cells. The mRNA was converted to complementary DNA (cDNA) by reverse transcriptase, and the cDNA was fluorescently labeled and assessed for hybridization on a microarray. The expression of mRNA in other zebrafish cells was similarly measured.Transcripts detected in zebrafish oligodendrocytes at levels greater than 3 times those found in other cells were selected as candidate myelination genes. Some of the detected genes, including plp1a, Sox10, and mbp, were previously known to be specific to oligodendrocytes, validating the procedure. The cDNA of newly identified candidates was amplified by PCR, and the ends were digested with the restriction enzymes EcoRI and XhoI. Genes were then ligated into the multiple cloning site (MCS) of the pSKII vector, shown in Figures 1 and 2, which had also been digested by EcoRI and XhoI. Figure 1 Overview of the pSKII vector (plasmid) Figure 2 The pSKII MCS, showing the restriction sites of several restriction enzymesTable 1 shows the functional parameters of several restriction enzymes. To confirm that the candidate genes originated in oligodendrocytes, complementary RNA (cRNA) strands were synthesized from the cloned plasmids and hybridized to fixed zebrafish sections. One of the genes that localized to oligodendrocytes, known as cldnk, produces a 915 base pair mRNA transcript from two exons. The cldnk gene is involved in tight junction formation and may be required for myelin sheath development around axons. -How would expression of cloned cldnk differ from expression of the endogenous cldnk gene?

Accepted Answer

11ecba2d_102f_b7e5_a3bf_db8777394fc2_MD0008_00 Genes within a eukaryotic chromosome typically consist of exons and introns, both of which are transcribed by RNA polymerase to form pre-mRNA. After transcription, pre-mRNA matures through several processes including 5′ capping by 7-methylguanosine, 3′ polyadenylation, and splicing to remove introns. cDNA is derived from mature mRNA through reverse transcription. Because the mature mRNA has already been spliced (ie, introns were removed), the cDNA will not contain any introns. The cDNA can be cloned into expression vectors such as plasmids, which can be introduced into embryonic cells and integrated into the genome. These cloned genes can then be expressed in the organism. However, because the cloned gene does not contain any introns, its expression will not include splicing. (Choice B) Cloned genes may be expressed at lower levels, the same levels, or higher levels than endogenous genes, depending on several factors, including how the cloning was done, how expression was induced, and the type of cell being observed. (Choices C and D) The cDNA was derived from endogenous mature mRNA, so it will encode the same mature mRNA as the endogenous gene. Because mRNA determines the sequence of the protein to be produced, the cloned gene yields the same protein as the endogenous mRNA from which it was derived. Educational objective: RNA polymerase transcribes both exons and introns to form pre-mRNA. During RNA processing, introns are removed by splicing to yield mature mRNA. cDNA is generated from mature mRNA and does not contain introns, so it is not spliced during expression.

Question 3

Passage In the central nervous system, myelin produced by oligodendrocytes (glial cells) functions as an insulating sheath surrounding certain nerve fibers. To identify candidate genes involved in the myelination process, researchers collected oligodendrocytes from zebrafish and purified the messenger RNA (mRNA) transcripts expressed by these glial cells. The mRNA was converted to complementary DNA (cDNA) by reverse transcriptase, and the cDNA was fluorescently labeled and assessed for hybridization on a microarray. The expression of mRNA in other zebrafish cells was similarly measured.Transcripts detected in zebrafish oligodendrocytes at levels greater than 3 times those found in other cells were selected as candidate myelination genes. Some of the detected genes, including plp1a, Sox10, and mbp, were previously known to be specific to oligodendrocytes, validating the procedure. The cDNA of newly identified candidates was amplified by PCR, and the ends were digested with the restriction enzymes EcoRI and XhoI. Genes were then ligated into the multiple cloning site (MCS) of the pSKII vector, shown in Figures 1 and 2, which had also been digested by EcoRI and XhoI. Figure 1 Overview of the pSKII vector (plasmid) Figure 2 The pSKII MCS, showing the restriction sites of several restriction enzymesTable 1 shows the functional parameters of several restriction enzymes. To confirm that the candidate genes originated in oligodendrocytes, complementary RNA (cRNA) strands were synthesized from the cloned plasmids and hybridized to fixed zebrafish sections. One of the genes that localized to oligodendrocytes, known as cldnk, produces a 915 base pair mRNA transcript from two exons. The cldnk gene is involved in tight junction formation and may be required for myelin sheath development around axons. -What observation could have led researchers to conclude that cldnk is expressed in oligodendrocytes?

Accepted Answer

11ecba2d_102d_46e2_a3bf_7da358058e62_MD0008_00 In genetic studies, hybridization refers to the annealing of two complementary nucleic acid strands. Specifically, mRNA is a single-stranded nucleic acid that is present only in the cytosol of cells that express the corresponding gene. To determine whether a cell expresses a specific gene, tissues may be incubated with a labeled RNA strand that is complementary to the mRNA of interest. This labeled RNA hybridizes with the mRNA in cells that express it, but is washed away from other cells, allowing researchers to visualize which cells express a particular gene. For a cell that expresses a gene of interest, one way to determine the cell type is to identify the presence of another mRNA that is known to be specific to the cell type of interest. The passage gives the names of three genes that are specific to oligodendrocytes: plp1a, Sox10, and mbp. If cldnk mRNA is found in the same cells as the mRNA of any of those genes, then cldnk must be expressed in oligodendrocytes. Therefore, if a probe that is complementary to cldnk mRNA hybridizes to cells that alsoexpress plp1A, then cldnk is expressed in oligodendrocytes. (Choice A) The probe must be complementary to the mRNA of interest for hybridization to occur. If the probe is made of cldnk mRNA, it will be identical, not complementary, to cldnk mRNA in the cell and will not hybridize. (Choice C) The affinity of cldnk cDNA for the microarray will be the same regardless of which cells the cDNA came from. Cldnk cDNA derived from oligodendrocytes binds to a greater extent than the same cDNA derived from other cells because there is more of it present, not because it has higher affinity. (Choice D) Gene expression is mediated by mRNA, which is primarily present in the cytosol. If the probe binds to the nucleus, it would indicate the presence of genomic DNA, which may or may not be transcribed to mRNA. Educational objective: In genetic studies, hybridization is the annealing of two complementary nucleic acids. The presence of a particular mRNA within a specific cell type can be assessed by hybridization to a complementary probe.

Question 4

A physician finds that a patient experiences excessive bleeding after sustaining a cut on the arm.  The excessive bleeding is LEAST likely to be caused by:

Accepted Answer

11ecba2d_1024_bb47_a3bf_8725545c07d2_MD0008_00 A single layer of endothelial cells comprises the inner wall of all blood vessels, forming a barrier that regulates the entry and exit of materials into and out of the bloodstream. In addition, endothelial cells secrete chemical signals that generally prevent blood coagulation (ie, clotting). However, injury to a blood vessel that results in bleeding indicates damage to the endothelium and triggers the following cascade of events: Formation of the platelet plug. Endothelial damage exposes connective tissue (ie, collagen fibers) normally present outside the blood vessel. Circulating cell fragments called platelets (derived from the bone marrow) readily bind these collagen fibers, aggregating to form a platelet plug (clot) that prevents blood flow out of the vessel. In addition, bound platelets and endothelial cells near the site of damage continue to release signals that enhance platelet aggregation. Strengthening of the clot. Clotting factors (mainly synthesized in the liver) are specialized proteins that become activated in response to platelet aggregation and signaling factors outside the vessel. Activated clotting factors induce processes that lead to the formation of the enzyme thrombin. Thrombin induces protein strands (ie, fibrin) to form an adhesive mesh-like structure over the platelet plug, reinforcing the clot. In this scenario, a patient with a cut experiences excessive bleeding, indicating an impaired ability to form clots and prevent blood loss from damaged vessels. This symptom could have several causes, including reduced liver function, which would likely decrease clotting factor production (Choice A), and reduced platelet proliferation (Choice C). In addition, because antibodies mark cellular materials for destruction by immune cells, heightened levels of serum antibodies against clotting factors would promote destruction of these factors, impairing clot formation (Choice D). (Choice B) Macrophages, phagocytic immune cells, engulf and destroy foreign pathogens and damaged cells. Although these cells may destroy pathogens entering the body through the cut, they would not inhibit clot formation and cause the excessive bleeding observed in the patient. Educational objective: Platelets and specialized proteins (ie, clotting factors) work synergistically in response to blood vessel damage, binding to the damaged portion of the vessel and forming a blockage that prevents blood loss.

Question 5

Passage
Within the Actinopterygii (bony fish) lineage, marine teleosts (bony fish with mobile jaws) live in an aquatic environment where there is a higher external concentration of salt and a lower concentration of water relative to the internal concentrations.  Owing to these conditions, salt ions tend to diffuse into the teleost through its skin, whereas water molecules within the organism osmotically traverse the opposite path.Irrespective of lineage, many fish species actively assess their environment and regulate their internal concentration of fluids and electrolytes via a homeostatic process known as osmoregulation.  Teleosts with either glomerular kidneys (eg, eels, sculpin) or aglomerular kidneys that are purely tubular (eg, goosefish, toadfish) have served as experimental subjects for the study of marine teleost osmoregulation.  Analyzing the ionic (ie, NaS1U1P1+S1S1P0, ClS1U1P1−S1S1P0, MgS1U1P12+S1S1P0, SOS1U1B14S1U1B0S1U1P12−S1S1P0) concentration of intestinal fluids, urine, and plasma in each species led to the discovery that the renal tubule of the goosefish has both excretory and reabsorptive functions.Scientists have concluded that marine teleost osmoregulation involves the ingestion of seawater, retrieval of NaCl and water from the intestine, and primary excretion of divalent ions via urine and monovalent ions through the gills (thin barriers between the organism's blood and the aquatic environment).In an experiment to further study osmoregulatory mechanisms, researchers emptied the gut of an eel and sealed the anus to prevent fluid loss.  The eel was transported into seawater that had been treated with the volume marker phenol red.  After 20 hours, the eel's gut contained 2.3 mL of fluid but showed a phenol red concentration equivalent to 12.3 mL of ingested seawater.  Researchers found that the eel lost 2.3 g of weight due to urine output.  They ultimately reported that the eel had lost all the seawater it absorbed through the gut via various mechanisms.
-Which of the following conclusions about the eel can be made based on the experiment in the passage?  (Note: Water density is 1 g/mL.)

Accepted Answer

11ecba2d_1027_a17b_a3bf_612b660615e7_MD0008_00 The passage describes an experiment in which an eel is transferred into seawater that has been treated with phenol red, a volume marker. While the eel was in the treated water, researchers assessed the eel's gut fluid volume, gut concentration of phenol red, urine output, and weight during a 20-hour period. In the given scenario, it is reported that after 20 hours, the eel's gut contains 2.3 mL of fluid, but the gut concentration of phenol red is equivalent to the eel having ingested 12.3 mL of water. Therefore, the eel absorbed 10 mL of water through its gut (12.3 mL − 2.3 mL), and the researchers claim that all the absorbed water was subsequently lost. Of the 10 mL absorbed, the passage states that the eel had a 2.3-g weight loss due to urine output. Because the density of water is 1 g/mL, a 2.3-g weight loss would be equivalent to 2.3 mL being excreted as urine (renal water loss) (Choice D). Consequently, the amount of water lost extrarenally can be calculated using the following equation: 11ecba2d_1027_a17c_a3bf_9d7def281dd3_MD0008_00 Total water loss=renal water loss+extrarenal water loss 11ecba2d_1027_a17d_a3bf_87d39d362157_MD0008_00 10 mL=2.3 mL+extrarenal water loss 11ecba2d_1027_a17e_a3bf_1d28bba8af51_MD0008_00 10 mL-2.3 mL=extrarenal water loss 11ecba2d_1027_a17f_a3bf_fb8eae50f503_MD0008_00 10 mL-2.3 mL=7.7 mL As a result, the eel lost 7.7 mL of water extrarenally. (Choice A) The eel ingested 12.3 mL of water. However, only 2.3 mL of water remained in the gut, which means that only 10 mL (not 12.3 mL) of water was absorbed by the gut. (Choice B) The eel has a gut phenol red concentration equivalent to 12.3 mL of ingested seawater; therefore, the eel drank 12.3 mL (not 10 mL) of seawater. Educational objective: Intestinal fluid ingestion, weight loss as water, and renal and extrarenal mechanisms of fluid excretion must be taken into account when calculating an organism's fluid volume.

Question 6

Passage
Within the Actinopterygii (bony fish) lineage, marine teleosts (bony fish with mobile jaws) live in an aquatic environment where there is a higher external concentration of salt and a lower concentration of water relative to the internal concentrations.  Owing to these conditions, salt ions tend to diffuse into the teleost through its skin, whereas water molecules within the organism osmotically traverse the opposite path.Irrespective of lineage, many fish species actively assess their environment and regulate their internal concentration of fluids and electrolytes via a homeostatic process known as osmoregulation.  Teleosts with either glomerular kidneys (eg, eels, sculpin) or aglomerular kidneys that are purely tubular (eg, goosefish, toadfish) have served as experimental subjects for the study of marine teleost osmoregulation.  Analyzing the ionic (ie, NaS1U1P1+S1S1P0, ClS1U1P1−S1S1P0, MgS1U1P12+S1S1P0, SOS1U1B14S1U1B0S1U1P12−S1S1P0) concentration of intestinal fluids, urine, and plasma in each species led to the discovery that the renal tubule of the goosefish has both excretory and reabsorptive functions.Scientists have concluded that marine teleost osmoregulation involves the ingestion of seawater, retrieval of NaCl and water from the intestine, and primary excretion of divalent ions via urine and monovalent ions through the gills (thin barriers between the organism's blood and the aquatic environment).In an experiment to further study osmoregulatory mechanisms, researchers emptied the gut of an eel and sealed the anus to prevent fluid loss.  The eel was transported into seawater that had been treated with the volume marker phenol red.  After 20 hours, the eel's gut contained 2.3 mL of fluid but showed a phenol red concentration equivalent to 12.3 mL of ingested seawater.  Researchers found that the eel lost 2.3 g of weight due to urine output.  They ultimately reported that the eel had lost all the seawater it absorbed through the gut via various mechanisms.
-Researchers studying marine teleosts claim that the glomerular kidney is 500 times as efficient as the aglomerular kidney in eliminating nontoxic substances.  Similar amounts of nontoxic ferrocyanide are intramuscularly injected into a sculpin and a goosefish of comparable weight.  Given this, which of the following tables showing renal elimination of ferrocyanide after 5 hours would support the researchers' claim?

Accepted Answer

The nephron is the functional unit of the kidney, composed of a glomerulus and tubular segments. The glomerulus, a ball-like network of capillaries, filters blood plasma into the tubular segments of the nephron. The filtered plasma, known as the renal filtrate, enters the tubular segments, where the concentration of nutrients, electrolytes, water, and other substances is regulated via excretory and secretory processes to ultimately produce urine. Based on the passage, some fish species have glomerular kidneys (eg, sculpin) and some have aglomerular kidneys (eg, goosefish). The researchers' claim is that the glomerular kidney is 500 times as efficient as the aglomerular kidney in eliminating nontoxic substances. In the given scenario, a sculpin and a goosefish of comparable weight were injected with similar volumes of nontoxic ferrocyanide. The question asks for a result in the glomerular sculpin and aglomerular eel that would support the researchers' claim regarding kidney efficiency in glomerular and aglomerular species. The only table that supports their hypothesis is Choice D, which shows that 5 hours after injection, the sculpin eliminated 5 percent of the ferrocyanide ion but that the goosefish eliminated only 0.01 percent. The exact level of increased ferrocyanide elimination in the glomerular sculpin compared to the aglomerular goosefish can be calculated as follows: 11ecba2d_1029_4f35_a3bf_ed4382be5567_MD0008_00 % ferrocyanide eliminated by glomerular sculpin% ferrocyanide eliminated by aglomerular goosefish=5%0.01%=51%1100% Dividing one fraction by another fraction is equivalent to multiplying the first fraction by the reciprocal of the second fraction, so the equation can be rewritten as: 11ecba2d_1029_4f36_a3bf_694dc8bd6768_MD0008_00 511100=51×1001=500 Therefore, the results in this table show that the glomerular kidney is 500 times as efficient in eliminating ferrocyanide as the aglomerular kidney. (Choice A) Calculating 0.1/0.005 would show that the glomerular kidney is 20 times as efficient in eliminating ferrocyanide as the aglomerular kidney. (Choice B) Calculating 10/0.05 would show that the glomerular kidney is 200 times as efficient in eliminating ferrocyanide as the aglomerular kidney. (Choice C) Calculating 0.5/0.01 would show that the glomerular kidney is 50 times as efficient in eliminating ferrocyanide as the aglomerular kidney. Educational objective: As the functional unit of the kidney, the nephron consists of a ball-like network of capillaries called the glomerulus. The glomerulus filters blood plasma into the tubular segments of the nephron.

Question 7

Passage In the central nervous system, myelin produced by oligodendrocytes (glial cells) functions as an insulating sheath surrounding certain nerve fibers. To identify candidate genes involved in the myelination process, researchers collected oligodendrocytes from zebrafish and purified the messenger RNA (mRNA) transcripts expressed by these glial cells. The mRNA was converted to complementary DNA (cDNA) by reverse transcriptase, and the cDNA was fluorescently labeled and assessed for hybridization on a microarray. The expression of mRNA in other zebrafish cells was similarly measured.Transcripts detected in zebrafish oligodendrocytes at levels greater than 3 times those found in other cells were selected as candidate myelination genes. Some of the detected genes, including plp1a, Sox10, and mbp, were previously known to be specific to oligodendrocytes, validating the procedure. The cDNA of newly identified candidates was amplified by PCR, and the ends were digested with the restriction enzymes EcoRI and XhoI. Genes were then ligated into the multiple cloning site (MCS) of the pSKII vector, shown in Figures 1 and 2, which had also been digested by EcoRI and XhoI. Figure 1 Overview of the pSKII vector (plasmid) Figure 2 The pSKII MCS, showing the restriction sites of several restriction enzymesTable 1 shows the functional parameters of several restriction enzymes. To confirm that the candidate genes originated in oligodendrocytes, complementary RNA (cRNA) strands were synthesized from the cloned plasmids and hybridized to fixed zebrafish sections. One of the genes that localized to oligodendrocytes, known as cldnk, produces a 915 base pair mRNA transcript from two exons. The cldnk gene is involved in tight junction formation and may be required for myelin sheath development around axons. -Microarrays are chips that contain hundreds of microscopic wells, each of which can detect a distinct nucleic acid. Prior to exposure to cDNA, the wells of the microarray described in the passage most likely contained:

Accepted Answer

11ecba2d_102d_e323_a3bf_c17fca072aa0_MD0008_00 A gene within genomic DNA consists of two strands: a sense strand (coding strand) and an antisense (noncoding) strand. RNA is produced when RNA polymerase uses the antisense strand as a template to make a new nucleic acid. The newly synthesized RNA strand will accordingly have the same sequence as the sense strand of the DNA (with the exception that RNA contains uracil instead of thymine). During the formation of cDNA, the enzyme reverse transcriptase produces a single-stranded DNA molecule using mRNA as a template. The cDNA strand is complementary to the mRNA and therefore is complementary to the sense strand of the genomic DNA (ie, cDNA is an antisense strand). The cDNA in the passage was obtained from zebrafish; therefore, to detect the cDNA, the microarray must contain single-stranded DNA from zebrafish genes. Because cDNA is complementary to the sense strand of genomic DNA, the single-stranded DNA must correspond to (ie, have the same sequence as) the sense strand of zebra fish genes, allowing it to hybridize with cDNA. (Choice A) Double-stranded DNA is already hybridized and would be unable to anneal to the cDNA probe. The cDNA would be washed away from all wells, and no signal would be detected. (Choice B) Proteins perform many different functions within cells, but only a subset (eg, transcription factors) bind to DNA. Accordingly, the cDNA would be unlikely to hybridize with zebrafish proteins, and proteins would not be useful in analyzing gene expression. (Choice C) Antibodies typically bind proteins, not DNA. Antibodies may be used in protein identification techniques such as ELISA but would not hybridize with cDNA. Educational objective: During cDNA synthesis, reverse transcriptase uses mRNA as a template to synthesize a new single-stranded DNA molecule. Because mRNA has the same sequence as the sense strand of genomic DNA (using uracil instead of thymine), cDNA will have the same sequence as the antisense strand. Accordingly, cDNA hybridizes with single-stranded DNA that corresponds to the sense strand of genomic DNA.

Question 8

Passage In the central nervous system, myelin produced by oligodendrocytes (glial cells) functions as an insulating sheath surrounding certain nerve fibers. To identify candidate genes involved in the myelination process, researchers collected oligodendrocytes from zebrafish and purified the messenger RNA (mRNA) transcripts expressed by these glial cells. The mRNA was converted to complementary DNA (cDNA) by reverse transcriptase, and the cDNA was fluorescently labeled and assessed for hybridization on a microarray. The expression of mRNA in other zebrafish cells was similarly measured.Transcripts detected in zebrafish oligodendrocytes at levels greater than 3 times those found in other cells were selected as candidate myelination genes. Some of the detected genes, including plp1a, Sox10, and mbp, were previously known to be specific to oligodendrocytes, validating the procedure. The cDNA of newly identified candidates was amplified by PCR, and the ends were digested with the restriction enzymes EcoRI and XhoI. Genes were then ligated into the multiple cloning site (MCS) of the pSKII vector, shown in Figures 1 and 2, which had also been digested by EcoRI and XhoI. Figure 1 Overview of the pSKII vector (plasmid) Figure 2 The pSKII MCS, showing the restriction sites of several restriction enzymesTable 1 shows the functional parameters of several restriction enzymes. To confirm that the candidate genes originated in oligodendrocytes, complementary RNA (cRNA) strands were synthesized from the cloned plasmids and hybridized to fixed zebrafish sections. One of the genes that localized to oligodendrocytes, known as cldnk, produces a 915 base pair mRNA transcript from two exons. The cldnk gene is involved in tight junction formation and may be required for myelin sheath development around axons. -What is the size of the pSKII plasmid after it is digested by EcoRI and XhoI and the cldnk gene is inserted?

Accepted Answer

11ecba2d_102c_8391_a3bf_57c1be1c5f3f_MD0008_00 Plasmids-small, circular DNA molecules-are commonly present in bacteria but are also found in some eukaryotic cells. In nature, bacteria can exchange plasmids through conjugation. Artificial plasmids are commonly used in laboratories to amplify or express a gene of interest. A gene may be inserted into a plasmid through the following steps: Digest the plasmid with restriction enzymes that only cut the plasmid at the multiple cloning site. If two restriction enzymes are used, any nucleotides between the two cut sites will be removed from the plasmid. Digest the gene of interest with the same enzymes as the plasmid to ensure that the plasmid and the gene will anneal to each other. PCR may be necessary to introduce cut sites into the ends of the gene. Incubate the digested plasmid and the digested gene together, along with DNA ligase to link them. Figure 1 shows that prior to insertion of cldnk, the pSKII plasmid contained 2,958 base pairs. Figure 2 shows that the distance between the EcoRI and XhoI restriction sites is 32 base pairs (17 + 15). These bases will be removed by the digest. The passage states that the cldnk gene produces a processed transcript that is 915 base pairs long, so insertion of cldnk cDNA increases the plasmid size by 915 base pairs. Therefore, the final size of the plasmid after cldnk is inserted can be calculated as 2958 − 32 + 915 = 3841. (Choice B) The plasmid would have 3,856 base pairs if the ClaI site had been used instead of XhoI. (Choice C) The plasmid would have 3,862 base pairs if the KpnI site had been used instead of EcoRI. (Choice D) The plasmid would have 3,873 base pairs if the 32 bases between the EcoRI site and the XhoI site had not been removed. This would happen if only one restriction enzyme (eg, EcoRI) had been used. Educational objective: Plasmids-small, circular DNA molecules that can carry a small number of genes-are often used in laboratories to amplify and express genes of interest. A gene may be inserted into a plasmid by digesting the gene and the plasmid with the same restriction enzymes, followed by ligation.

Question 9

Which of the following mRNA strands contains the longest open reading frame?

Accepted Answer

11ecba2d_1026_41e9_a3bf_b74db9b709ad_MD0008_00 The genetic code consists of sequences of three consecutive nucleotides in DNA or RNA called codons, each of which codes for a specific amino acid or a stop signal during translation. A reading frame is a set of consecutive codons that do not overlap. Any given strand of mRNA has three possible reading frames, depending on which nucleotide is the first in a codon. An open reading frame (ORF) is a portion of a reading frame that can be translated by ribosomes because it contains both a start codon (AUG) and a stop codon (UAA, UAG, or UGA) within the same reading frame. After transcription of DNA to RNA, the resulting mRNA strand is translated in a 5′ to 3′ direction by the ribosome to generate a polypeptide. Translation of the mRNA sequence begins with a start codon, which codes for the amino acid methionine. Translation continues until the ribosome reaches a stop codon in the same reading frame as the start codon. Based on the ORF definition, the mRNA sequence in Choice B has a start codon (AUG) that begins at the third nucleotide and a stop codon (UAG) that begins at the fifteenth nucleotide. There are three additional codons between the start and stop codon of this sequence, making the ORF of this sequence 5 codons long, which is longer than any other ORF shown. (Note that this strand also contains UAA, but this sequence is out of frame, so it is not translated as a stop codon.) (Choice A) This mRNA sequence contains an ORF that is only 3 codons long. (Choice C) The start codon in this sequence is immediately followed by a stop codon, making the ORF only 2 codons long. (Choice D) This mRNA sequence contains an ORF that is only 4 codons long. Educational objective: An open reading frame (ORF) is a set of codons within a strand of mRNA that can be translated by a ribosome. An ORF begins with a start codon (AUG) and ends with a stop codon (UAA, UAG, or UGA). The start and stop codons must be in the same reading frame.

Question 10

Passage
Within the Actinopterygii (bony fish) lineage, marine teleosts (bony fish with mobile jaws) live in an aquatic environment where there is a higher external concentration of salt and a lower concentration of water relative to the internal concentrations.  Owing to these conditions, salt ions tend to diffuse into the teleost through its skin, whereas water molecules within the organism osmotically traverse the opposite path.Irrespective of lineage, many fish species actively assess their environment and regulate their internal concentration of fluids and electrolytes via a homeostatic process known as osmoregulation.  Teleosts with either glomerular kidneys (eg, eels, sculpin) or aglomerular kidneys that are purely tubular (eg, goosefish, toadfish) have served as experimental subjects for the study of marine teleost osmoregulation.  Analyzing the ionic (ie, NaS1U1P1+S1S1P0, ClS1U1P1−S1S1P0, MgS1U1P12+S1S1P0, SOS1U1B14S1U1B0S1U1P12−S1S1P0) concentration of intestinal fluids, urine, and plasma in each species led to the discovery that the renal tubule of the goosefish has both excretory and reabsorptive functions.Scientists have concluded that marine teleost osmoregulation involves the ingestion of seawater, retrieval of NaCl and water from the intestine, and primary excretion of divalent ions via urine and monovalent ions through the gills (thin barriers between the organism's blood and the aquatic environment).In an experiment to further study osmoregulatory mechanisms, researchers emptied the gut of an eel and sealed the anus to prevent fluid loss.  The eel was transported into seawater that had been treated with the volume marker phenol red.  After 20 hours, the eel's gut contained 2.3 mL of fluid but showed a phenol red concentration equivalent to 12.3 mL of ingested seawater.  Researchers found that the eel lost 2.3 g of weight due to urine output.  They ultimately reported that the eel had lost all the seawater it absorbed through the gut via various mechanisms.
-Based on the passage, to confirm that the inorganic composition of marine teleost urine is independent of glomerular function, researchers would most likely have to discover:

Accepted Answer

11ecba2d_1028_3dc0_a3bf_f5c4ae5c0c5d_MD0008_00 According to the passage, marine teleosts such as the eel and the goosefish live in an aquatic environment that, when compared to their internal environment, has more salt and less water. In addition, the passage states that scientists concluded that osmoregulation in marine teleosts involves: Seawater ingestion. Retrieval (absorption) of salt and water from the intestine. Excretion of divalent ions primarily through the urine and monovalent ions through the gills. In the given scenario, researchers are trying to determine whether the inorganic composition of marine teleost urine is independent of glomerular function. To accomplish this, scientists studying teleosts with glomerular kidneys (eg, eels) and aglomerular kidneys (eg, goosefish) would have to assess how kidney structure affects urine production in these two anatomically different fish species. To confirm that the inorganic composition of urine is independent of glomerular function, scientists would have to find similar concentrations of inorganic substances in the urine of both glomerular and aglomerular fish. This finding would signify that inorganic urine composition is the same regardless of whether the fish possess a glomerular kidney. Because divalent ions (eg, Mg²⁺ and SO₄²⁻) are primarily excreted in the urine and monovalent ions (Na⁺ and Cl⁻) are primarily excreted through the gills of marine teleosts, researchers would have to find equally high urinary concentrations of Mg²⁺ and SO₄²⁻ ions in the glomerular eel and the aglomerular goosefish. (Choice A) Finding higher urinary concentrations of Mg²⁺ and SO₄²⁻ in the eel than in the goosefish would mean that the glomerular eel has a greater inorganic urine composition than the aglomerular goosefish. This result would indicate that efficiency of renal Mg²⁺ and SO₄²⁻ excretion is dependent on the better functionality of the glomerular kidney. (Choices B and D) The passage states that monovalent ions (eg, Na⁺ and Cl⁻) are mainly excreted through the gills, not through the kidney. Therefore, measurement of monovalent ions would provide a poor assessment of kidney function. Educational objective: To claim that a particular function is independent of an anatomical structure, researchers must show similar function-related findings in organisms with and without the anatomical structure of interest.

Question 11

A patient is found to have reduced blood flow through pulmonary vessels and excess fluid accumulation in the abdomen and legs.  Which of the following conditions would best explain this patient's symptoms?

Accepted Answer

11ecba2d_1031_1777_a3bf_df97249e0389_MD0008_00 The heart, consisting of four chambers (two atria and two ventricles), continuously contracts to pump blood throughout the circulatory system. The left atrium receives oxygenated blood returning from the lungs and transfers this blood into the left ventricle, a highly muscular chamber that pumps blood throughout the entire body via systemic arteries. Deoxygenated blood returns to the heart via systemic veins and fills the right atrium, which transfers blood into the right ventricle. Right ventricular contraction pumps deoxygenated blood to the lungs via the pulmonary arteries. Specialized valves facilitate unidirectional blood flow through the heart. Atrioventricular (AV) valves allow blood to flow from the atria to the ventricles, and semilunar (SL) valves allow blood to flow from the ventricles into the arteries. In this scenario, a patient exhibits reduced pulmonary blood flow and excess fluid accumulation in the abdomen and legs. These symptoms are best explained by reduced blood flow through the heart valve separating the right atrium and ventricle. This condition would decrease the volume of blood loaded into the right ventricle prior to contraction, reducing blood flow from the right ventricle into the pulmonary arteries. In addition, the reduced volume of blood flowing into the right ventricle and through the pulmonary circuit would cause a backup of blood in the right atrium and systemic vessels. Accordingly, blood volume and hydrostatic pressure in systemic veins would increase, causing excess fluid leakage from systemic vessels into the surrounding interstitial fluid and subsequent fluid accumulation (edema) around tissues. (Choice B) Partial blockage of the valve separating the left ventricle and aorta would likely decrease blood flow through systemic (not pulmonary) vessels. (Choice C) An increased volume of blood pumped from the right ventricle would lead to increased (not decreased) blood flow through pulmonary vessels. (Choice D) Backflow of blood from the left ventricle to the left atrium would cause blood to back up into pulmonary vessels. This would increase the volume of blood and hydrostatic pressure in pulmonary vessels, likely causing excess fluid leakage from vessels of the pulmonary (not systemic) circuit. Educational objective: Deoxygenated blood returning to the heart fills the right atrium, which transfers blood to the right ventricle. The right ventricle pumps blood to the lungs via the pulmonary circuit.

Question 12

A patient found to have an abnormally low number of helper T cells in her blood contracts a bacterial infection.  During the infection, which of the following is most likely NOT affected by the patient's low count of helper T cells?

Accepted Answer

11ecba2d_1025_7e98_a3bf_1385c3e9359e_MD0008_00 The immune response comprises both innate and adaptive immunity. Cells of the innate immune system can rapidly and nonspecifically recognize and destroy foreign antigens. In contrast, the adaptive immune system, subdivided into cell-mediated and humoral immunity, comprises cells that recognize specific antigens and mount more specialized immune responses against invading pathogens. In cell-mediated immunity, T lymphocytes (cells) recognize and mount immune responses against foreign antigens displayed by other cells on specialized surface proteins called major histocompatibility complex (MHC) proteins. MHC proteins on any given cell display fragments of any proteins present within that cell. Accordingly, cells containing foreign pathogens, such as certain immune cells or infected cells, will generally display foreign protein fragments (antigens) on their MHC proteins. Helper T cells are T cells that recognize foreign antigens displayed by MHC proteins of other immune cells, such as: B lymphocytes, cells that bind and engulf a foreign antigen. The antigen is broken down into fragments within the B lymphocyte, and these fragments are transported to be displayed on MHC proteins present on the cell membrane. A helper T cell binds the foreign antigen presented by the B lymphocyte and releases signaling molecules (cytokines) that induce division of the B lymphocyte into many identical cells, some of which differentiate to secrete antibodies against the invading pathogen. macrophages and dendritic cells, which engulf foreign antigens (via phagocytosis) and present these antigen fragments on the MHC proteins found on their cell membrane. Upon binding these antigens, helper T cells release cytokines that enhance other immune responses, including the phagocytotic activity of macrophages and the activation and proliferation of cytotoxic T cells, immune cells that release toxins to induce apoptosis in infected cells. In this scenario, a patient with an abnormally low number of helper T cells contracts an infection. Consequently, a low count of helper T cells would affect the activation of cytotoxic T cells and B lymphocytes (which produce antibodies against the bacterium) (Choices B, C, and D). However, because expression of MHC proteins is solely dependent on a cell's transcriptional and translational machinery, cells would still be able to display bacterial antigens on their MHC proteins regardless of helper T cell count. Educational objective: Helper T cells bind foreign antigens presented by other immune cells and release signaling molecules that enhance immune responses, such as cytotoxic T cell activation and antibody production by B lymphocytes.

Question 13

Several techniques have been developed to treat blood disorders caused by genetic mutations.  Which of the following methods would be most effective in the treatment of a patient homozygous for a deleterious mutation in a gene coding for a subunit of hemoglobin?

Accepted Answer

11ecba2d_1032_9e19_a3bf_3b151679b445_MD0008_00 Hemoglobin is an oxygen-binding protein found within red blood cells (RBCs). Oxygen diffuses into the bloodstream via gas exchange in the lungs, binds hemoglobin, and is transported through the circulatory system to tissues throughout the body. Deleterious mutations affecting hemoglobin genes may negatively impact hemoglobin function, impairing oxygen transport. One method of treating genetic mutations is gene therapy, a technique in which a functional gene is introduced into a patient's cells to replace a mutant gene. The new gene can then be transcribed and translated into functional proteins. A common gene therapy method uses retroviral vectors to insert functional gene copies into the stem cells of patients exhibiting specific genetic mutations. In this scenario, a patient is homozygous for a deleterious mutation in a gene coding for a hemoglobin subunit. Expression of this mutant gene would likely produce a portion of the hemoglobin protein with impaired function. Because RBCs are produced from stem cells in the bone marrow, the most effective treatment for this patient would be gene therapy to introduce and activate a functional hemoglobin gene in bone marrow stem cells. This would likely lead to the production of RBCs that contain hemoglobin with normal oxygen-binding abilities. (Choice A) The mutant gene could not be replaced by a functional copy in mature RBCs because mature RBCs lack nuclei and do not contain DNA. (Choice B) Treatment with a drug that increases RBC production would increase the differentiation of bone marrow stem cells into mature RBCs. However, because this treatment does not introduce a functional copy of the hemoglobin gene, these cells would still express the mutation as well as functionally impaired hemoglobin. (Choice D) A blood transfusion from a patient homozygous for the wild-type hemoglobin gene may introduce RBCs expressing functional hemoglobin into the patient's bloodstream. However, all aging RBCs are eventually destroyed by phagocytic cells (ie, macrophages) in the spleen. Because RBCs with wild-type hemoglobin would eventually be destroyed, the patient would require multiple blood transfusions to maintain a population of healthy RBCs. Educational objective: Gene therapy is a technique in which a functional copy of a gene is introduced (eg, via retroviral vectors) into the cells of a patient with a mutation in that gene.

Question 14

A male African clawed frog (Xenopus laevis) and a female Western clawed frog (Xenopus tropicalis) mate and produce multiple hybrid embryos that die early in development.  Given this, death of the hybrid embryo is most likely caused by:

Accepted Answer

11ecba2d_102a_3997_a3bf_815a6c6111cf_MD0008_00 Hybrid offspring are produced when members of two different species interbreed. The mating of hybrid offspring with members of either parental species may cause genetic leakage, or the transfer of genetic information between species. However, in some cases, two species may be unable to produce hybrid embryos or hybrids may be inviable (unable to survive) or infertile. The inability of two species to produce viable or fertile hybrid offspring is called reproductive isolation, which may be prezygotic (before zygote formation) or postzygotic (after zygote formation). In this scenario, mating of a male African clawed frog (Xenopus laevis) and a female Western clawed frog (Xenopus tropicalis) produces hybrid embryos that die early in development, making this an example of postzygotic reproductive isolation. Mitosis is essential for embryonic growth and production of the organism's somatic cells. Failure of sister chromatids to segregate properly (eg, due to chromosomal nondisjunction) during embryonic mitosis would generate cells containing missing or extra genetic information. Accordingly, the ability of these cells to divide and differentiate into functional somatic cells in the embryo would be impaired. This would inhibit zygote development and likely cause early embryonic death. (Choice B) Sperm and egg incompatibility in the two species would prevent fertilization and the formation of an embryo. Because mating of these frog species produced embryos, the sperm and egg must be compatible. (Choice C) Gametes (eggs, sperm) form via meiosis during gametogenesis. Reduced sperm and egg production in parental species would decrease the probability of fertilization and zygote formation following a mating event. However, the frogs are described as having produced multiple hybrid embryos, and reduced gamete formation in the parental species would not affect the development of any embryos formed after mating. (Choice D) During meiosis I, homologous chromosomes pair up and are segregated into cells that divide to form gametes in meiosis II. This process ensures the production of viable gametes containing correct chromosomal numbers. Errors in this process would impair gametogenesis in the hybrid offspring and potentially lead to infertility. However, impaired gametogenesis in the hybrid offspring would not cause embryonic death. Educational objective: Hybrid offspring are produced from the mating of two different species. However, reproductive isolation (prezygotic or postzygotic) may prevent different species from producing viable or fertile hybrid offspring.

Question 15

A male patient was born with an abnormal opening in the muscular wall separating the two ventricles of the heart.  The opening allows blood to flow from the left to the right ventricle.  If left untreated, this cardiac defect may cause any of the following in the patient EXCEPT:

Accepted Answer

11ecba2d_102a_fce8_a3bf_9573a95c210d_MD0008_00 The four chambers of the heart (two atria and two ventricles) pump blood throughout the circulatory system. Deoxygenated blood returning to the heart through systemic veins fills the right atrium and is pumped into the right ventricle. Contraction of the right ventricle drives blood through the pulmonary artery to the lungs via the pulmonary circuit. Oxygenated blood returning from the lungs via the pulmonary veins enters the left atrium and is pumped into the left ventricle, a highly muscular chamber that pumps oxygenated blood throughout the entire body via systemic arteries. Barriers called septa, composed of both membranous and muscular tissue, separate the two atria and two ventricles, preventing oxygenated and deoxygenated blood from mixing within the heart. In this scenario, the patient was born with an abnormal opening in the muscular wall separating the ventricles of the heart (ie, the ventricular septum) that allows blood to flow from the left to the right ventricle. Because the left ventricle contains oxygenated blood, this opening in the ventricular septum causes oxygenated blood to flow into the right ventricle. If left untreated, this cardiac defect may cause: Increased blood flow into the pulmonary artery. Blood flow from the left to the right ventricle increases the volume of blood in the right ventricle and subsequently increases the amount of blood pumped through the pulmonary artery by right ventricular contraction (Choice A). Increased blood volume in the left atrium. Increased blood flow through the pulmonary circuit from the right ventricle likely increases the volume of blood flowing through the pulmonary veins into the left atrium (Choice B). Low blood oxygen levels in systemic arteries. Because oxygenated blood abnormally flows from the left ventricle into the right ventricle and the pulmonary circuit, the level of oxygenated blood pumped from the left ventricle through systemic arteries decreases (Choice C). (Choice D) The right ventricle pumps blood through the pulmonary artery; therefore, the flow of oxygenated blood into the right ventricle increases, not decreases, the oxygenation of blood pumped into the pulmonary artery. Educational objective: Deoxygenated blood returning to the heart enters the right atrium, is transferred to the right ventricle, and is pumped to the lungs via the pulmonary circuit. The left atrium receives oxygenated blood and transfers this blood to the left ventricle, which pumps oxygenated blood throughout the body.

Question 16

Passage In the central nervous system, myelin produced by oligodendrocytes (glial cells) functions as an insulating sheath surrounding certain nerve fibers. To identify candidate genes involved in the myelination process, researchers collected oligodendrocytes from zebrafish and purified the messenger RNA (mRNA) transcripts expressed by these glial cells. The mRNA was converted to complementary DNA (cDNA) by reverse transcriptase, and the cDNA was fluorescently labeled and assessed for hybridization on a microarray. The expression of mRNA in other zebrafish cells was similarly measured.Transcripts detected in zebrafish oligodendrocytes at levels greater than 3 times those found in other cells were selected as candidate myelination genes. Some of the detected genes, including plp1a, Sox10, and mbp, were previously known to be specific to oligodendrocytes, validating the procedure. The cDNA of newly identified candidates was amplified by PCR, and the ends were digested with the restriction enzymes EcoRI and XhoI. Genes were then ligated into the multiple cloning site (MCS) of the pSKII vector, shown in Figures 1 and 2, which had also been digested by EcoRI and XhoI. Figure 1 Overview of the pSKII vector (plasmid) Figure 2 The pSKII MCS, showing the restriction sites of several restriction enzymesTable 1 shows the functional parameters of several restriction enzymes. To confirm that the candidate genes originated in oligodendrocytes, complementary RNA (cRNA) strands were synthesized from the cloned plasmids and hybridized to fixed zebrafish sections. One of the genes that localized to oligodendrocytes, known as cldnk, produces a 915 base pair mRNA transcript from two exons. The cldnk gene is involved in tight junction formation and may be required for myelin sheath development around axons. -Which of the following diagrams shows the structure of the cloned cldnk gene after digestion with EcoRI and XhoI?

Accepted Answer

11ecba2d_102b_e750_a3bf_6ff69949d945_MD0008_00 Restriction enzymes cleave DNA molecules at specific nucleotide sequences, known as restriction sites. Each restriction enzyme has a unique restriction site that typically consists of a DNA palindrome, in which the sequence of one strand is the reverse of its complement strand. If a restriction enzyme cuts anywhere other than directly in the middle of the palindrome, a few nucleotides will be left on the end of the DNA strand without any complementary bases paired with them. These leftover nucleotides are known as overhangs or sticky ends because they can anneal or "stick to" their complementary bases in another DNA strand. The passage states that the cldnk gene was digested by EcoRI and XhoI. Table 1 shows that EcoRI cuts between the G and the A in GAATTC, leaving an AATT overhang, and XhoI cuts between the C and the T of CTCGAG, leaving a TCGA overhang. Therefore, the digested gene should have an AATT overhang on one end and a TCGA overhang on the other, as shown in Choice B (Note: Because the Xhol overhang is shown as the bottom strand, it is read from right to left [5′ to 3′]). (Choice A) EcoRI breaks the bond between the G and the A in GAATTC, so the G would not be present on the overhang after the digest occurs. Similarly, the terminal C would not be present in the overhang after XhoI digests the CTCGAG sequence. (Choice C) Digestion of one end of the DNA by Xbal (not Xhol or EcoRI) would leave a CTAG overhang (reading from right to left), as shown in this diagram. (Choice D) This diagram shows the wrong overhang for the EcoRI site (GAATT instead of AATT) and the wrong overhang for an Xbal site (TCTAG instead of CTAG). Educational objective: Restriction enzymes cleave DNA molecules at specific sites, known as restriction sites. Each enzyme has a unique restriction site and may cut the DNA such that an overhang, or sticky end, is left on the end of the strand.

Question 17

Scientists studying evolution in yeast found that a certain species underwent a reduction in chromosome number with no loss of coding information.  Given this, which of the following mechanisms best explains how this reduction in chromosome number may have occurred?End-to-end fusion of two chromosomes followed by inactivation of one centromere on the newly formed chromosomeBreakage of a chromosome at the centromere and fusion of the two resulting chromosomal portions to the ends of other chromosomesTransfer of a DNA sequence from one chromosomal arm to a different chromosomal arm

Accepted Answer

11ecba2d_1031_b3b8_a3bf_3d1b9fc72ef5_MD0008_00 Eukaryotic genomes consist of linear chromosomes made up of DNA that can be translated and transcribed into proteins. Each chromosome has a condensed region of repetitive noncoding DNA called the centromere, a structure that facilitates spindle fiber attachment during mitosis and meiosis. In addition, eukaryotic chromosomal ends consist of repetitive noncoding sequences called telomeres that protect coding information from degradation during DNA replication. Alterations to chromosome structure (eg, fusion or breakage) can occur due to DNA mutations, and chromosomal alterations may cause changes in chromosome number within a particular species. In this scenario, scientists studying evolution in yeast, a single-celled eukaryotic organism, discovered a species that underwent a reduction in chromosome number without losing any coding information. Accordingly, this reduction in chromosome number may have occurred via: end-to-end (telomere-to-telomere) fusion of two chromosomes and inactivation of one of the centromeres. This fusion would initially generate a larger chromosome with two centromeres, and inactivation of one of these centromeres would produce a single new chromosome, reducing the chromosome number in the cell by one (Number I). breakage of a chromosome at the centromere and fusion of each chromosomal portion to the ends of other chromosomes. This initial breakage would result in two individual chromosomal portions. Fusion of these portions to other chromosomes would cause the original chromosome to be lost, also reducing the overall chromosome number in the cell by one (Number II). As both chromosomal events involve the transfer of intact genetic material to different chromosomes, neither would lead to a loss of coding information. In addition, because telomeres and centromeres are noncoding sequences, changes to these sequences due to chromosomal fusion or breakage would not likely affect protein-coding information. (Number III) Transposons are small fragments of DNA that can move between different regions of the genome. Movement of a DNA sequence, such as a transposon, from one chromosomal arm to another may alter the nucleotide sequence of each chromosome but would not reduce chromosome number in the yeast cell. Educational objective: Eukaryotic chromosomes have both protein-coding DNA and distinct noncoding regions. Noncoding regions include the centromere (facilitates spindle fiber attachment during cell division) and telomeres (protect chromosomal ends from degradation).

Question 18

Passage In the central nervous system, myelin produced by oligodendrocytes (glial cells) functions as an insulating sheath surrounding certain nerve fibers. To identify candidate genes involved in the myelination process, researchers collected oligodendrocytes from zebrafish and purified the messenger RNA (mRNA) transcripts expressed by these glial cells. The mRNA was converted to complementary DNA (cDNA) by reverse transcriptase, and the cDNA was fluorescently labeled and assessed for hybridization on a microarray. The expression of mRNA in other zebrafish cells was similarly measured.Transcripts detected in zebrafish oligodendrocytes at levels greater than 3 times those found in other cells were selected as candidate myelination genes. Some of the detected genes, including plp1a, Sox10, and mbp, were previously known to be specific to oligodendrocytes, validating the procedure. The cDNA of newly identified candidates was amplified by PCR, and the ends were digested with the restriction enzymes EcoRI and XhoI. Genes were then ligated into the multiple cloning site (MCS) of the pSKII vector, shown in Figures 1 and 2, which had also been digested by EcoRI and XhoI. Figure 1 Overview of the pSKII vector (plasmid) Figure 2 The pSKII MCS, showing the restriction sites of several restriction enzymesTable 1 shows the functional parameters of several restriction enzymes. To confirm that the candidate genes originated in oligodendrocytes, complementary RNA (cRNA) strands were synthesized from the cloned plasmids and hybridized to fixed zebrafish sections. One of the genes that localized to oligodendrocytes, known as cldnk, produces a 915 base pair mRNA transcript from two exons. The cldnk gene is involved in tight junction formation and may be required for myelin sheath development around axons. -Which experiment could confirm that the cldnk gene product is required for myelination?

Accepted Answer

11ecba2d_102e_a674_a3bf_fdf79d07223c_MD0008_00 Knockout organisms have one or more genes that have been made inactive. Genes may be knocked out by introducing mutations that result in a truncated gene product, a frame-shifted gene product, or no gene product, each of which results in genes that produce nonfunctional proteins. The biological function of a gene can be inferred by knocking it out and observing the differences between wild-type organisms and knockout organisms. If cldnk is required for wild-type myelination of oligodendrocytes, then knocking the gene out should result in impaired or abnormal myelination. Therefore, comparing myelination of oligodendrocytes in wild-type zebrafish to myelination in cldnk knockout zebrafish would show whether cldnk is required for proper myelin formation. (Choice A) If cldnk is involved in myelination, it is possible that overexpression of cldnk would cause increased myelination. However, it is also possible that endogenous cldnk already yields maximum activity, in which case excess cldnk would have no effect. Therefore, overexpression of cldnk would not necessarily give any information about its role in myelination. (Choice B) Although cldnk expression in cells other than oligodendrocytes could cause myelin to form in those cells, it is unlikely because other gene products are most likely required for myelination as well. (Choice D) A form of the cldnk gene that inhibits activity of the wild-type gene (ie, a dominant negative gene) could show whether cldnk is required for myelination if it were expressed in oligodendrocytes, but not in cells other than oligodendrocytes. Educational objective: A gene's biological function can be inferred by comparing the differences in organisms with the gene knocked out (inactivated) to wild-type organisms.

Question 19

Passage
Within the Actinopterygii (bony fish) lineage, marine teleosts (bony fish with mobile jaws) live in an aquatic environment where there is a higher external concentration of salt and a lower concentration of water relative to the internal concentrations.  Owing to these conditions, salt ions tend to diffuse into the teleost through its skin, whereas water molecules within the organism osmotically traverse the opposite path.Irrespective of lineage, many fish species actively assess their environment and regulate their internal concentration of fluids and electrolytes via a homeostatic process known as osmoregulation.  Teleosts with either glomerular kidneys (eg, eels, sculpin) or aglomerular kidneys that are purely tubular (eg, goosefish, toadfish) have served as experimental subjects for the study of marine teleost osmoregulation.  Analyzing the ionic (ie, NaS1U1P1+S1S1P0, ClS1U1P1−S1S1P0, MgS1U1P12+S1S1P0, SOS1U1B14S1U1B0S1U1P12−S1S1P0) concentration of intestinal fluids, urine, and plasma in each species led to the discovery that the renal tubule of the goosefish has both excretory and reabsorptive functions.Scientists have concluded that marine teleost osmoregulation involves the ingestion of seawater, retrieval of NaCl and water from the intestine, and primary excretion of divalent ions via urine and monovalent ions through the gills (thin barriers between the organism's blood and the aquatic environment).In an experiment to further study osmoregulatory mechanisms, researchers emptied the gut of an eel and sealed the anus to prevent fluid loss.  The eel was transported into seawater that had been treated with the volume marker phenol red.  After 20 hours, the eel's gut contained 2.3 mL of fluid but showed a phenol red concentration equivalent to 12.3 mL of ingested seawater.  Researchers found that the eel lost 2.3 g of weight due to urine output.  They ultimately reported that the eel had lost all the seawater it absorbed through the gut via various mechanisms.
-Scientists also study freshwater teleosts, which live in an environment where there is more water and less salt outside their bodies than inside.  Osmoregulation in freshwater teleosts would most likely serve to combat their natural tendency to:

Accepted Answer

11ecba2d_1026_de2a_a3bf_23ef15c5a38b_MD0008_00 In the absence of external energy input, molecules in solution will diffuse from areas of high concentration to low concentration. Such flow down a concentration gradient is thermodynamically favorable. Molecules that are separated from an area of low concentration by a membrane will still flow across the membrane down their concentration gradient as long as the membrane is permeable to those molecules. The passage introduces osmoregulation, the homeostatic process by which organisms actively assess their environment and regulate their internal concentration of fluid and electrolytes. In addition, the passage states that marine (saltwater) teleosts inhabit an environment that has a higher concentration of salt and a lower concentration of water molecules than their internal environment. Consequently, it is thermodynamically favorable for salt ions at high external concentration to diffuse through the skin into the low-salt internal environment of the fish. Water will traverse the opposite path via osmosis, moving from the high-concentration environment inside the fish to the lower concentration environment of the surrounding salt water. Osmoregulation uses external energy to reverse these processes and maintain constant salt and water concentrations within the fish. Unlike marine teleosts, freshwater teleosts live in an environment where there is more water and less salt outside their bodies than inside. As a result, salt ions are at relatively high concentration within the teleost and will tend to move to the lower concentration environment of the freshwater. Water molecules will follow the opposite route and move from the higher concentration of the external environment to the lower concentration inside the fish. As a result, the freshwater teleosts must use osmoregulation to combat their natural tendency to lose salt and absorb water. (Choices A and B) The passage states that water and salt move in opposite directions. Because the natural tendency of freshwater teleosts is to lose salt and absorb water, both compounds cannot be absorbed, nor can they both be lost. (Choice C) Saltwater, not freshwater, fish have a natural tendency to absorb salt and lose water. Educational objective: In the absence of external energy input, molecules in solution diffuse from areas of high concentration to low concentration. Osmoregulation provides the energy needed to reverse this process and maintain constant salt and water concentrations within an organism.

Question 20

Bacteria are removed from lymph and blood in the:

Accepted Answer

11ecba2d_1024_1f06_a3bf_331f6ff44eb4_MD0008_00 The lymphatic system is a network of vessels that collect and transport lymph, ultimately draining this fluid into the bloodstream. The lymphatic system is particularly important to the immune response and includes the following lymphoid tissues and organs, which facilitate immune cell proliferation and function: The bone marrow, a spongy tissue lining the inside of bones, is responsible for the production of red blood cells and white blood cells (eg, lymphocytes) from hematopoietic stem cells. The thymus receives immature T lymphocytes migrating from the bone marrow and facilitates their maturation. During this process, any T lymphocytes that recognize self-antigens die, preventing the release of lymphocytes able to mount attacks against the body's own tissues. However, T lymphocytes that do not bind self-antigens mature and enter the blood and lymph, where they bind foreign antigens and mount immune responses. Lymph nodes, organs clustered along lymph vessels, filter lymph as it flows through the lymphatic system. Macrophages housed within lymph nodes identify a variety of pathogens in lymph and destroy them via phagocytosis. In addition, B lymphocytes and T lymphocytes bind specific foreign antigens in lymph, inducing immune responses against invading pathogens. The spleen has extensive capillary networks through which blood is filtered. White blood cells housed within the spleen recognize and bind foreign antigens in the blood, mounting immune responses against these circulating pathogens. In addition, spleen macrophages remove and destroy old and damaged red blood cells. Platelets and other white blood cells are also stored in the spleen and released into the blood as needed. Because lymph nodes filter lymph and the spleen filters blood, bacteria circulating through lymph are removed in the lymph nodes and bacteria circulating through blood are removed in the spleen. (Choices A and C) Although white blood cells are produced in bone marrow, bone marrow cells do not remove bacteria from blood or lymph. (Choice D) Immune cells circulate throughout the body via lymph vessels; however, pathogens are mainly removed from lymph at lymph nodes (not lymph vessels). The thymus is the site of T-cell maturation, but it does not filter blood or lymph. Educational objective: In the lymph nodes, white blood cells remove and mount immune responses against pathogens from the lymph, whereas in the spleen, white blood cells remove pathogens and damaged or old red blood cells from the blood.

Exam 3: Biology

Dehydration caused by frequent loose stools is often observed in patients with certain diseases of the gastrointestinal (GI) tract. Given this, which of the following treatments would be most effective in treating this symptom?

A physician finds that a patient experiences excessive bleeding after sustaining a cut on the arm. The excessive bleeding is LEAST likely to be caused by:

Which of the following mRNA strands contains the longest open reading frame?

A patient is found to have reduced blood flow through pulmonary vessels and excess fluid accumulation in the abdomen and legs. Which of the following conditions would best explain this patient's symptoms?

A patient found to have an abnormally low number of helper T cells in her blood contracts a bacterial infection. During the infection, which of the following is most likely NOT affected by the patient's low count of helper T cells?

Several techniques have been developed to treat blood disorders caused by genetic mutations. Which of the following methods would be most effective in the treatment of a patient homozygous for a deleterious mutation in a gene coding for a subunit of hemoglobin?

A male African clawed frog (Xenopus laevis) and a female Western clawed frog (Xenopus tropicalis) mate and produce multiple hybrid embryos that die early in development. Given this, death of the hybrid embryo is most likely caused by:

A male patient was born with an abnormal opening in the muscular wall separating the two ventricles of the heart. The opening allows blood to flow from the left to the right ventricle. If left untreated, this cardiac defect may cause any of the following in the patient EXCEPT:

Bacteria are removed from lymph and blood in the: