Test Questions

A pharmaceutical research team has developed an inhibitor of phospholipase C. Administration of this drug would result in A. Decrease intracellular calcium levels B. Increase intracellular diacylglycerol levels C. Increase intracellular inositol trisphosphate (IP3) levels D. Decrease protein kinase A activity E. Decrease intracellular cAMP levels

(Explanation: Phospholipase C activation causes the formation of IP3 and DAG from membrane phosphatidylinositol bisphosphate. IP3 in turn increases the release of calcium from the endoplasmic reticulum. DAG and calcium activate protein kinase C) A. Decrease intracellular calcium levels - correct answer B. Increase intracellular diacylglycerol levels - DAG levels would decrease following administration of the inhibitor C. Increase intracellular inositol trisphosphate (IP3) levels - IP3 levels would decrease D. Decrease protein kinase A activity - protein kinase A activity is not affected. Protein kinase C activity is decreased E. Decrease intracellular cAMP levels - the cAMP levels are not altered following administration of the inhibitor

A 7-month old infant has been identified to have accumulation of sphingomyelin. Hydrolysis of the accumulated substance would give a. Glucose b. N-acetyl neuraminic acid (NANA) c. Sphingosine d. Glycerol e. Cholesterol

(Sphingomyelin is a sphingophospholipid and its hydrolysis gives sphingosine, fatty acid, phosphate and choline) c. Sphingosine - Correct Answer a. Glucose (is a constituent of glycolipids) b. N-acetyl neuraminic acid (NANA) - Is present in gangliosides c. Sphingosine - Correct Answer d. Glycerol - present in glycerophospholipids and triacylglycerol e. Cholesterol - is not present in sphingomyelin

A number of anti-retroviral and anti-cancer drugs are classified as either nucleoside or nucleotide analogs. Which of the following best describes the difference between a nucleoside analog and a nucleotide analog? A. nucleosides lack a 3' -OH group on the pentose sugar B. nucleotides do not possess heterocyclic base components C. nucleosides lack a 5' phosphate group on the pentose sugar D. nucleotides cannot be incorporated in DNA E. pyrimidines are nucleosides and purines are nucleotides

- This question is testing the difference between a nucleotide (which contains a base, sugar and phosphate) and a nucleoside (which contains a base and sugar) C. nucleosides lack a 5' phosphate group on the pentose sugar - Correct answer A. nucleosides lack a 3' -OH group on the pentose sugar B. nucleotides do not possess heterocyclic base components C. nucleosides lack a 5' phosphate group on the pentose sugar - Correct answer D. nucleotides cannot be incorporated in DNA E. pyrimidines are nucleosides and purines are nucleotides

Individuals of Greek origin often possess a point mutation (AT to AG) in the globin gene that introduces a new 3' splice site into intron 3. This mutation results in -thalassemia. Assuming that a premature stop codon is not generated, which of the following is most likely to occur? A. the beta-globin protein will be shorter B. the beta-globin protein will be longer C. intron 3 and exon 3 will be removed D. exon 3 will be spliced out E. there will be no effect and the beta-thalassemia is caused by something else

- To answer this question, understand that when a new 3' splice site is introduced, the intron will shorter and a few of the bases in the intron would now be translated into a protein that is longer than normal B. the beta-globin protein will be longer - Correct answer A. the beta-globin protein will be shorter B. the beta-globin protein will be longer - Correct answer C. intron 3 and exon 3 will be removed D. exon 3 will be spliced out E. there will be no effect and the beta-thalassemia is caused by something else

A cell with a total osmolarity of 300 mOsm per liter (KCl - ref coeff 1) is placed in the following solution. NaCl 150 mM (reflection coeff. 1) Urea 50 mM (reflection coeff. 0) CaCl2 2 mM (reflection coeff. Ca = 1, Cl = 1) What is the solutions osmolarity, its tonicity and what will happen to the cell volume? A. hyper-osmolar, hyper-tonic, cell will shrink B. hyper-osmolar, iso-tonic, cell will shrink C. hyper-osmolar, hyper-tonic, cell will swell D. hyper-osmolar, hypo-tonic, cell will swell

A

Activation of Beta-adrenoceptors leads to A. activation of adenylyl cyclase B. decrease in cAMP C. activation of guanylyl cyclase D. activation of phospholipase C E. contraction of smooth muscle

A

The expression of many genes is increased when cAMP levels are elevated. How does cAMP affect gene expression? A. It leads to phosphorylation of the transcription factor CREB B. It binds to cAMP response elements in the promoters of genes. C. It increases calcium concentration in the cytoplasm and the nucleus D. It binds to transcription factors and increases their affinity for DNA. E. It binds to a regulatory protein, which then dissociates from the transcription factor and no longer blocks its activity.

A

The following equilibrium potentials are calculated EK = -60 mV ENa = +60 mV Assume the cell membrane can only conduct these two ion species. What would a measured membrane voltage of -30 mV tell you? A. that the membrane was mostly permeable to K but had a significant Na permeability B. that the membrane was exclusively permeable to K C. that the membrane was exclusively permeable to Na D. that the membrane was mostly permeable to Na but had a significant K permeability

A

What is one of the important consequences of repolarization and afterhyperpolarization observed in an action potential? A. it helps to reset the voltage gated sodium channel's gates back to pre stimulated configuration B. it causes K to flow back into the cell and restore K levels C. it allows the K channels to reset their voltage sensitive gates D. it causes the gates of the sodium channel to open so that sodium can flow into the cell

A

What would happen to the Nernst potential for K if intracellular K remains constant but extracellular K doubles. A. the equilibrium potential for potassium would be decreased B. the equilibrium potential for K would not change C. the equilibrium potential for K would be increased D. the polarity of the equilibrium potential would reverse

A

The plasma membrane separates the cell from its neighboring cells and the extracellular or interstitial fluid. The fluidity of the plasma membrane is regulated by a. Free cholesterol which buffers the fluidity changes of fatty acids during temperature changes b. The amount of glucose transporters and other proteins c. Arachidonic acid in phospholipids which will reduce the fluidity d. Palmitic acid which is more fluid than oleic acid e. Free cholesterol which increases the fluidity at the area of the polar head groups

A is correct: the fluidity of the plasma membrane in controlled by free cholesterol, which buffers fluidity changes of fatty acids during temperature changes. Free cholesterol in found in both layers of the plasma membrane b. Incorrect: glucose transporters do not regulate the fluidity of the plasma membrane c. Incorrect: Arachidonic acid in phospholipids will increase the fluidity, not decrease d. Incorrect: Palmitic acid is less fluid than oleic acid e. Incorrect: cholesterol stiffens the plasma membrane at the polar head groups

Substance X is in a chamber (Chamber 1) containing 10L of fluid. The concentration of substance X in this chamber is 100mmol/L. A solid division separates this chamber from another chamber (Chamber 2) containing 5L of fluid. You remove this division leaving only a membrane separating the two chambers. This membrane is permeable to Substance X and nothing else. Which of the following statements are true? A. The concentrations of Substance X in both chambers will eventually equal each other. B. The concentration of Substance X in Chamber 2 will never exceed 33mmol/L. C. The concentration of Substance X in Chamber 2 will never exceed 50mmol/L. D. The concentration of Substance X in Chamber 2 will never exceed 66mmol/L. E. There is no movement of Substance X from Chamber 2 to Chamber 1. Feedback: (Tricky) because there is always some random movements in both directions. F. The net flux of Substance X from Chamber 1 to Chamber 2 is greatest directly after the removal of the solid division. Feedback: flux is proportional to the concentration gradient and the concentration gradient is greatest immediately after removal of the barrier. G. The net flux of Substance X across the membrane is always constant. Feedback: When the barrier is removed the concentration throughout the two chambers is now 66 mmol/L (an additional volume of 5 L has been added to the 10L; if chamber 2 had a volume of 10L then the overall concentration would have been 50 mmol/L)

A, D, F

In the course of investigating the structure and function of a membrane transport protein, a researcher discovered a single base substitution in the gene for the protein. This caused the loss of an important αhelical segment of the protein. The mutation that led to this finding is most likely due to which of the following changes? (Refer to the standard genetic code table) (Please note that a standard table showing the genetic code would be included if this question were on an exam). A. CUC to CCC B. GUU to GCU C. UUG to CUG D. AUC to GUC E. GUG to CUG

A. CUC to CCC - Leucine to Proline - the mutation produces a proline that cannot fit into an alpha-helix - Proline is an alpha helix breaker - Correct answer A. CUC to CCC - Leucine to Proline - the mutation produces a proline that cannot fit into an alpha-helix - Proline is an alpha helix breaker - Correct answer B. GUU to GCU - Valine to Alanine (Branched chain amino acid to alanine - both are hydrophobic amino acids) C. UUG to CUG - Leucine to leucine - this is a silent mutation - no change in structure D. AUC to GUC - Isoleucine to Valine - May have minimal effect on the function as both are branched chain amino acids E. GUG to CUG - Valine to Histidine - Branched chain amino acid is replaced by a positively charged amino acid - this would change the charge on the protein and may affect function

It is widely agreed that cancer is the result of mutations of genes -especially tumor suppressor genes or repair enzyme genes. One type of mutation can occur by the mis-incorporation of incorrect bases during DNA replication. Which of the following is responsible for correcting this type of mis-incorporation errors? A. DNA polymerase B. RNA polymerase C. DNA primase D. Helicase E. Topoisomerase

A. DNA polymerase - Correct answer - it has 3' to 5' exonuclease proof reading activity A. DNA polymerase - Correct answer - it has 3' to 5' exonuclease proof reading activity B. RNA polymerase - Has no proofreading activity C. DNA primase - is involved in synthesis of the RNA primer during replication D. Helicase - is involved in unwinding the DNA double helix E. Topoisomerase - reduces supercoiling in front of the replication fork

Fluoroquinolones are a group of drugs used to treat a broad spectrum of bacterial infections. This group of drugs targets the prokaryote enzyme DNA gyrase. DNA gyrase can be grouped together with which of the following enzymes categories? A. DNA polymerases B. RNA polymerases C. Helicases D. Topoisomerases E. Primases

A. DNA polymerases - Enzymes of replication B. RNA polymerases - Required for transcription C. Helicases - required to separate the two strands of DNA double helix D. Topoisomerases - DNA gyrase (prokaryotic) belongs to the Topoisomerase type II family of topoisomerase enyzmes. E. Primases - Enzymes required for synthesis of the primer in replication

A lab studies a disease gene that leads to loss of function of an enzyme in a metabolic pathway. Southern blot shows that the chromosome structure is unchanged. Northern blot shows that there is no change in the mRNA transcript. Western blot reveals that a smaller than normal protein is formed. Which mutation accounts for this data? A. One base pair insertion in an exon B. Insertion of GCG in an exon C. The codon CCC is mutated to CCG D. Splice site mutation E. 99% reduction in promoter function

A. One base pair insertion in an exon (this is the correct answer) • Will not alter genomic structure • Will not alter mRNA (usually) • Will lead to frameshift, and likely to form an early stop codon - leads to truncated protein

Which of the following drugs will inhibit both DNA replication and transcription in eukaryotes? A. rifamycin B. actinomycin D C. Ciprofloxacin D. Etoposide E. Alpha- amanitin

A. Rifamycin - Is an inhibitor of prokaryotic RNA polymerase and inhibits prokaryotic translation only B. actinomycin D- Correct answer - Intercalates between the base pairs in double helical DNA, distorts the double helix and inhibits DNA replication and transcription in eukaryotes by preventing the formation of single stranded DNA. C. Ciprofloxacin - Prokaryotic DNA gyrase inhibitor D. Etoposide - Eukaryotic Topoisomerase II inhibitor E. Alpha- amanitin - Is a eukaryotic RNA polymerase-III inhibitor - It is an inhibitor of eukaryotic transcription but not a replication inhibitor

In a family with several cases of colon cancer you discover a single nucleotide change where a cytosine has been exchanged for a thymidine. What type of mutation is this? A. Transition B. Transversion C. Frameshift D. Nonsense E. Deletion

Answer A. In this example, a pyrimidine is replaced by a pyrimidine this is a transition mutation. A transversion mutation occurs when a pyrimidine is replaced by a purine, or a purine is replaced by a pyrimidine.

A biopsy was taken from tumor cells from a woman who has advanced ovarian cancer. The cells were analyzed and found to have multiple aberrations, breakages, insertions, fusions and rearrangements. For instance, a portion of chromosome 8 was fused to chromosome 5, and chromosome 1 had a large chunk of chromosome 14 inserted on the q arm. Another aspect of this genome is the appearance of a large amount of extra-chromosomal material. Which type of test would be used to generate data that identifies and displays these chromosomal defects in a single image? A. The cDNA microarray B. G-banded karyotype C. Southern blot D. Spectral karyotype E. ASO probes F. Northern blot

Answer D - Commentary: Spectral karyotype (same as saying SKY FISH, or chromosome painting) is the best answer. This technique provides a snapshot of metaphase stage chromosome structure by "painting" the entire chromosome a single color. This is done by hybridizing FISH probes along the length of each chromosome - each chromosome a different color. This test will reveal structures resulting from different chromosomes fused together. The cDNA microarray is used to measure mRNA expression of many (1,000s,) of genes in a single experiment, the Northern blot is used to measure the expression of a single gene, G-banding karyotype is the standard karyotype, and it may only be used to visualize very large alteration in a chromosome, or to detect aneuploidy. ASO probes are typically used to detect SNPs, or insertions/deletions that are only a few nucleotides. The Southern blot may be used to visualize genomic structure when restriction sites are involved. The Southern is a typically obsolete procedure, since it is much easier to obtain the same data with PCR based methods. Some applications which still require the Southern blot include (1) Methylation Sensitive Restriction Digest followed by Southern to detect methylated cytosines in CpG islands, and (2) detection of genomic regions which are resistant to PCR amplification such as regions which have a very high percentage of GC base pairs

Anne Jeina, arrives in hospital 5 hours after the onset of chest pain. In the ER, blood was drawn for the measurement of various biomarkers. Which of the following biomarker would be most useful at this time? A. Amylase B. Troponin I C. Aspartate transaminase D. Alanine transaminase E. Lactate dehydrogenase-1 F. Alkaline phosphatase G. Creatine kinase-MM

Answer: Troponin I or troponin T could be correct choices Increased serum amylase and lipase levels are indicative of pancreatic disease. Increased alanine aminotransferase and aspartate aminotransferase are indicative of liver disease Increased serum alkaline phosphatase (ALP) is indicative of obstructive liver disease (Gall stone obstructing the common bile duct) Increased Creatine kinase-MM levels in serum indicate skeletal muscle disease/ damage

A reference solution contains a cell impermeant oligosaacharide at 300 mOsm. Which of the following solutions is isotonic with this reference solution? Assume full dissociation of all particles. (sigma = reflection coefficient). A. 600 mM LiCl, sigma = 1 B. 150 mM NaCl, sigma = 1 C. 300 mM KCl, sigma = 1 D. 300 mM glucose, sigma = 0 E. 300 mM urea, sigma = 0

B

David is suffering from hypocalcemia. In response to the low levels of free calcium in his blood his parathyroid gland secretes greater amounts of parathyroid hormone. Elevated levels of parathyroid hormone increase the levels of calcium in his blood. This is an example of: A. positive feedback B. negative feedback C. feed forward control D. set point deviation E. hierarchy of importance of control mechanisms

B

During an action potential when the membrane potential has reached its most positive value, which ion is the membrane most permeable to? A. K+ B. Na+ C. Cl- D. Ca2+ E. Mg2+

B

Ouabain is a Na-K ATPase inhibitor. During an experiment in isloated epithelial cells addition of ouabain inhibits the cellular accumulation of substance X, despite the favorable concentration gradient for diffusion of substance X into the cell. What is the likliest explanation? A. substance X is carried on a facilitated diffusion transporter B. substance X is carried on a Na coupled cotransporter C. substance X is transported by a primary transport pump D. substance X can permeate a transmembrane pore

B

Substance X - is a cation with a valency of 1 and found in greater concentration outside the cell compared to intracellular fluid. Substance X is; A. potassium B. sodium C. chloride D. fixed charge protein E. Calcium

B

The mode of action of NO as a signaling molecule is via A. a rise in cytosolic Ca2+ concentration B. activation of guanylyl cyclase C. activation of PKA D. activation of phospholipase C E. phosphorylation of the transcription factor CREB

B

What mechanism leads to muscle weakness if extracellular K levels are reduced (hypokalemia)? A. the sodium channels undergo depolarizing blockade (inactivation gates locks shut). B. the resting membrane potential is further from threshold and electrical excitability is reduced C. the threshold resets further from the resting membrane potential and reduced electrical excitability D. extracellular K causes inactivation of the voltage gated sodium channel

B

Which of the following signal molecules has to diffuse through the cell membrane to gain access to its receptor? A. Glucagon B. Testosterone C. Epidermal growth factor D. Epinephrine E. Insulin

B

Hepatocytes contain the specific glucose transporter GLUT-2. This transporter is a. Insulin-dependent b. A low-affinity transporter c. Only found in hepatocytes d. Mainly found in RBC e. Only used for uptake of glucose but not for its release from cells

B is correct: GLUT-2 is a low-affinity glucose transporter a. Incorrect: GLUT-4 is insulin-dependent and found in fat cells and muscle b. Correct: GLUT-2 is a low-affinity transporter found in hepatocytes and other cells c. Incorrect: GLUT-2 is also found e.g. in kidney, beta-cells of pancreas and intestinal mucosal cells toward the portal vein d. Incorrect: GLUT-2 is not found in RBC, there we find mainly GLUT-1 e. Incorrect: GLUT-2 facilitates transport in both direction, dependent on concentrations

Phospholipids are components of polar lipids that often can be used to form biomembranes due to their amphipathic character. The group includes cardiolipin which contains totally four esterified fatty acids. What is special about cardiolipin? Cardiolipin is a. Exclusively found in the heart b. Found in high concentration in mitochondria c. A major component of all biomembranes d. Part of the glycocalyx e. A sphingophospholipid

B is correct: cardiolipin is found in high concentration in mitochondria a. Incorrect: cardiolipin is found in mitochondrial membranes of all tissues b. Correct: cardiolipin is necessary in the inner mitochondrial membrane c. Incorrect: cardiolipin is not a major component of all biomembranes d. Incorrect: it is not part of the glycocalyx, it is a phospholipid e. Incorrect: it is a glycerophospholipid, sphingomyelin is a sphingophospholipid

Diet soft drinks have the warning that phenylketonurics should not consume drinks containing aspartame. Aspartame is an artificial dipeptide sweetener that contains phenylalanine. What is correct about this amino acid? A. It contains a hydroxyl group B. It is an aromatic amino acid C. It contains a sulfhydryl group D. It is a dietary nonessential amino acid E. It is as a branched-chain amino acid

B is correct: phenylalanine is an aromatic amino acid a incorrect: a hydroxyl group is found in serine, threonine or tyrosine b. correct, phenylalanine is used to form tyrosine in humans (by hydroxylation) c. incorrect: a sulfhydryl (thiol) group is found in cysteine d. incorrect: phenylalanine is a dietary essential amino acid e. incorrect: branched-chain amino acids are valine, leucine and isoleucine

Allosteric enzymes do not follow Michaelis-Menten kinetics. What is special about the graph plotted with the velocity at the y-axis and the substrate concentration at the xaxis? A. The curve for allosteric enzymes would not lead to a Vmax B. Is sigmoidal due to a cooperative effect of the substrate C. Is shifted to the right when a feed-forward activator is added D. Cannot be used to determine K 0.5 E. Shows only first order kinetics

B is correct: the curve is sigmoidal due to the cooperative effect of the substrate A. incorrect: Vmax is also found for allosteric enzymes B. correct, the substrate is a homotropic effector for the allosteric enzyme C. incorrect: it is shifted to the left when a feed-forward activator is added D. incorrect: K0.5 [S] can be determined from the graph at half Vmax E. incorrect: it shows first order and zero order kinetics

The activity of an enzyme is studied using increasing concentrations of substrate. The data is plotted as reaction velocity in response to increased substrate concentrations. A. The velocity reaches Vmax at the substrate concentration at Km B. represents product formation over time C. is independent of the amount of enzyme D. increases for human enzymes at temperatures above 40° C E. is proportional to substrate concentrations at very high [S]

B is correct: the velocity of a reaction represents product formation over time A. incorrect: Km is the substrate concentration at half Vmax B. correct, velocity describes formation of product in concentration/min, not total amount C. incorrect: the velocity is dependent on the amount of the enzyme. The velocity increases with increasing substrate concentration D. incorrect: human enzymes start to denature above 40° C (high fever) and the velocity decreases E. incorrect: the velocity is proportional to the substrate at low [S] - first order kinetics. At very high [S], the velocity is maximal and is not dependent on [S] - zero order kinetics

Many proteins undergo modification after they have been synthesized, including insulin. Which of the following modifications does the primary insulin translation product undergo in order to produce mature insulin? A. Phosphorylation of serine residues B. Hydrolytic cleavage of specific peptide bonds C. Addition of sugar residues to the protein (glycosylation) D. Chemical modification of some amino acid side chains E. Removal of formyl-methionine groups from the N-terminal end of the protein

B. Hydrolytic cleavage of specific peptide bonds - removal of signal peptide and C peptide - Correct answer A. Phosphorylation of serine residues - This modification occurs in regulatory enzymes by protein kinase A in the presence of glucagon B. Hydrolytic cleavage of specific peptide bonds - removal of signal peptide and C peptide - Correct answer C. Addition of sugar residues to the protein (glycosylation) - Occurs in glycoproteins. Insulin is NOT a glycoprotein D. Chemical modification of some amino acid side chains - like hydroxylation occurs posttranslationally in proline and lysine residues in collagen E. Removal of formyl-methionine groups from the N-terminal end of the protein - this posttranslational modification occurs only in prokaryotes and not in eukaryotes

Glucagon is released by the alpha-cells of the pancreas and signals low blood glucose levels. The glucagon receptor A. Increases guanylate cyclase activity B. Increases intracellular cAMP levels C. activates phospholipase C D. is found in the nucleus of hepatocytes E. Increases intracellular calcium levels

B. Increases intracellular cAMP levels - correct answer - this happens due to activation of adenylate cyclase which increases formation of cAMP A. Increases guanylate cyclase activity - this is the mechanism of action of nitric oxide B. Increases intracellular cAMP levels - correct answer - this happens due to activation of adenylate cyclase which increases formation of cAMP C. activates phospholipase C - this is the signal transduction mechanism for the alpha- adrenergic receptors D. is found in the nucleus of hepatocytes - this is true for steroid-thyroid group of hormones E. Increases intracellular calcium levels - this is the second messenger for the alpha-adrenergic receptors

10. Which of the following drugs will inhibit both DNA replication and transcription in eukaryotes? A. rifamycin B. actinomycin D C. Ciprofloxacin D. Etoposide E. Alpha- amanitin

B. actinomycin D- Correct answer - Intercalates between the base pairs in double helical DNA, distorts the double helix and inhibits DNA replication and transcription in eukaryotes by preventing the formation of single stranded DNA. A. Rifamycin - Is an inhibitor of prokaryotic RNA polymerase and inhibits prokaryotic translation only B. actinomycin D- Correct answer - Intercalates between the base pairs in double helical DNA, distorts the double helix and inhibits DNA replication and transcription in eukaryotes by preventing the formation of single stranded DNA. C. Ciprofloxacin - Prokaryotic DNA gyrase inhibitor D. Etoposide - Eukaryotic Topoisomerase II inhibitor E. Alpha- amanitin - Is a eukaryotic RNA polymerase-III inhibitor - It is an inhibitor of eukaryotic transcription but not a replication inhibitor

A laboratory studying RNA modification treats a mammalian cell culture with an siRNA that inhibits the production of guanylyl transferase enzyme. Which of the following will most likely occur in these cells as a direct result of this experiment? A. DNA replication will no longer occur B. mRNA will no longer be translated into protein C. poly-adenylation of mRNA will be inhibited D. pre-mRNA will not be spliced properly E. transcription will be inhibited

B. mRNA will no longer be translated into protein - This is the correct answer A. DNA replication will no longer occur - This enzyme is not required for replication B. mRNA will no longer be translated into protein - This is the correct answer C. poly-adenylation of mRNA will be inhibited - D. pre-mRNA will no be spliced properly - Splicing is a posttranscriptional modification that requires snRNA and SNRNPs E. transcription will be inhibited - Transcription will not be affected

Nitric oxide is produced in different tissues by nitric oxide synthase from A. lysine B. ornithine C. arginine D. asparagine E. tyrosine

C

Of the following which would increase the membrane permeability to Na+ ? A. increased Na/K pump activity B. increased Na concentration gradient C. increased opening times of Na+ channels D. increased surface area of the membrane E. increased Na+ flux

C

Steroid and thyroid hormones A. Activate protein kinase A B. Activate Ras C. Affect the pattern of gene expression D. Activate guanylyl cyclase. E. Phosphorylate transcription factor CREB

C

The G protein Gs A. Activates guanylyl cyclase B. Is a seven-pass transmembrane receptor protein C. Activates adenylyl cyclase D. Inhibits adenylyl cyclase E. Activates phospholipase C

C

Linkage analysis has implicated a gene involved increased predisposition to cancer. A technique was used that discovered a G to A mutation that was found in a specific family. Which techique revealed this mutation? A. ASO blot B. CGH Microarray C. Sequencing D. Northern blot E. Western

C - commentary - sequencing reveals the DNA code; ASO blot can only be used if the DNA sequence is already known - it cannot be used to discover a sequence variant. CGH microarray reveals copy number variation; Northern blot detects mRNA from a single gene (that is complementary to that probe); Western blot detects protein using an antibody

Membranes contain different transporters. Which of the following is an example of an ABC-transporter? a. SGLT b. GLUT-1 c. CFTR d. GLUT-5 e. Na+/K+ -ATPase

C is correct: CFTR is an example of an ABC-transporter (chloride ion channel) a. Incorrect: SGLT is involved with secondary active transport b. Incorrect: GLUT-1 performs facilitative diffusion c. Correct: CFTR is an epithelial chloride ion channel in the plasma membrane d. Incorrect: GLUT-5 performs facilitated diffusion of mainly fructose e. Incorrect: Na+/K+ -ATPase performs primary active transport

The amino acid residues of serine and threonine contain a hydroxyl group that can be phosphorylated by protein kinase A (which is cAMP-dependent). Tyrosine residues also contain a hydroxyl group but are not substrates for protein kinase A. They can be phosphorylated for example by the tyrosine kinase activities of the insulin receptor. Glutamate, methionine and asparagines do not contain a hydroxyl group and are not phosphorylated. A. Creatine phosphate is formed in the heart by cytochrome C oxidase B. contains two phosphoanhydride bonds C. is grouped as a high-energy compound D. is formed by creatine kinase in cytosol at low ATP levels E. is formed using creatine and inorganic phosphate as substrates

C is correct: Creatine phosphate is grouped as a high-energy compound A. incorrect: it is not formed in the electron transport chain by cyt C oxidase B. incorrect: creatine phosphate contains one energy-rich phosphoamide bond C. Correct, it can be used to form ATP from ADP D. incorrect: at high ATP levels, it can be formed by creatine kinase in cytosol E. incorrect: it is formed from creatine and ATP, inorganic phosphate is not used

A 56-year-old man was seriously injured in an accident. He needs parenteral nutrition, which should include dietary essential fatty acids. Which of the following fatty acids is a dietary essential fatty acid of the omega-6 family? a. Palmitoleic acid b. Oleic acid c. Linoleic acid d. alpha-linolenic acid e. Arachidonic acid

C is correct: Linoleic acid is dietary essential and is of the omega-6 family (18:2, omega-6) a. Incorrect: palmitoleic acid (16:1) can be formed in humans from palmitic acid (16:0) b. Incorrect: oleic acid (18:1) can be formed in humans from stearic acid (18:0) c. Correct: linoleic acid is dietary essential and of the omega-6 family d. Incorrect: alpha-linolenic acid (18:3, omega-3) is also a dietary essential fatty acid but of the omega 3-family. e. Incorrect: arachidonic acid (20:4, omega 6) can be synthesized in humans from linoleic acid

Lab tests in a patient with skeletal muscle dystrophy shows high levels of serum total creatine kinase. What is the true statement regarding this enzyme? A. It catalyses an irreversible reaction B. It is a serum injury marker for bone damage C. It can be used to form ATP D. It has two different subunits and consists of a tetramer E. It is a serum injury marker of MI which peaks after 2 days

C is correct: creatine kinase can be used to form ATP when it uses CrP and ADP A. incorrect: creatine kinase catalyzes a reversible reaction (unusual for a kinase) B. incorrect: it is not a serum injury marker for bone disease, ALP is a marker for bone disease C. correct, at low ATP levels, CrP and ADP are used to form ATP by creatine kinase D. incorrect: CK has two different subunits (M and B) and it consists of a dimer. E. incorrect: it is an early injury marker for myocardial infarction and returns back to baseline after 2 days following an MI

The neurotransmitter GABA can be formed by the decarboxylation of the alpha-carboxyl group of a specific amino acid. Which amino acid is decarboxylated to GABA? A. Glycine B. Tryptophan C. Glutamate D. Aspartate E. Glutamine F. Tyrosine

C is correct: glutamate is decarboxylated to GABA (-aminobutyric acid) a. incorrect: glycine is not decarboxylated b. incorrect: tryptophan is the precursor of serotonin c. correct: Glutamic acid or glutamate is decarboxylated to form GABA d. incorrect: aspartate is not used to form GABA, does not have gamma−carboxyl group e. incorrect: glutamine needs first to be deaminated to form glutamate f. Tyrosine: is a precursor of catecholamines (dopamine, norepinephrine and epinephrine)

Premature babies are often treated with lung surfactant which a. Is a glycosphingolipid b. Is a plasmalogen c. Lines the alveoli of the lungs and prevents collapse d. Contains mainly protein and some phospholipids e. Contains two arachidonic acids in its structure

C is correct: premature babies are often treated with lung surfactant which lines the alveoli of the lungs and prevents collapse a. Incorrect: it is a glycerophospholipid b. Incorrect: it is a specific phosphatidylcholine, not a plasmalogen c. Correct: it is mainly DPPC(dipalmitoylphosphatidylcholine) which is released by Type II alveolar cells d. Incorrect: it is mainly phospholipids and some proteins e. Incorrect: it does not contain arachidonic acid in its structure, but contains two palmitic acids (two saturated fatty acids)

A specific enzyme of intermediary metabolism is regulated by covalent modification of a specific amino acid side chain by protein kinase A in the presence of glucagon. Which amino acid in the enzyme can be phosphorylated by this mechanism? A. Tyrosine B. Glutamate C. Serine D. Methionine E. Asparagine

C is correct: protein kinase A phosphorylates serine residues in specific enzymes.

A 52-year-old man is brought to the emergency room by his wife. He had suffered a mild heart attack two years ago and experienced similar pain during the last night. His ECG does not indicate an acute MI and blood samples are taken. His blood plasma is used to test for both, total serum creatine kinase (CK) and serum CK-MB. What is the reasoning behind this? A. Total serum CK alone is an injury marker for the liver B. Increase of serum CK-MB but not of total CK indicates rhabdomyolysis C. Serum CK-MB larger than about 3 or 5% of total serum CK can indicate an acute MI D. Serum cTnT is also a marker for MI, but accumulates only after 2 days in blood E. Serum levels of CK-MB but not total serum CK are increased after an acute MI

C is correct: serum CK-MB/total CK larger than about 5% can indicate a MI A. incorrect: total CK is not an injury marker for the liver, CK is not found in the liver B. incorrect: rhabdomyolysis (skeletal muscle damage) leads to increase of mainly total CK and little CK-MB C. correct, CK-MB alone can be used, but the % related to total CK is better D. incorrect: serum cTnT and cTnI increase in plasma in several hours following a MI E. incorrect: both, total CK and CK-MB increase in plasma as indication for a MI

A two year old female patient is identified to have Prader Willi syndrome as a result of a 1 Mb deletion of a critical region of chromosome 15. Which approach was most likely used to diagnose the genetic cause of the disorder in the patient? A. ASO blot B. PCR C. Array CGH D. G-band karyotype E. Spectral karyotype

C. In this case, the patient has a deletion of a region of the chromosome. This will be detected by microarray CGH. (Note that this deletion can also be detected by locus specific FISH if the geneticist used a FISH probe for that specific locus (this answer was not offered on this question). Spectral karyotype would not work in this case since the chromosome 15 would just be "painted" a particular color, and the deletion would not be detected.

A bacterium is developed in the laboratory that has a mutation which confers decreased activity in peptide bond formation during ribosomal protein synthesis. The most likely reason for this is a mutation in which of the following? A. The EF-Tu protein B. One of the ribosomal proteins C. One of the ribosomal RNAs D. The Shine-Dalgarno sequence of the mRNA E. The ribosomal protein release factors

C. One of the ribosomal RNAs - Correct answer - catalytic ribosomal RNA participates in the activity of peptidyl transferase A. The EF-Tu protein - is a protein factor required for the elongation step in prokaryotic translation B. One of the ribosomal proteins C. One of the ribosomal RNAs - Correct answer - catalytic ribosomal RNA participates in the activity of peptidyl transferase D. The Shine-Dalgarno sequence of the mRNA - participates in the initiation of prokaryotic translation E. The ribosomal protein release factors - participate in termination of translation

In E. coli, glucose may be converted to chorismate which is a common intermediate in the pathways for histidine, phenylalanine, tyrosine, and tryptophan synthesis. If tryptophan is needed the cell forms an mRNA that allows the production of the five enzymes needed for the trp synthesis pathway to function. Which characteristic of this gene expression is most likely to be correct? A. Five mRNA's are produced for the pathway which allow production of the five enzymes B. One polypeptide is post-translationally processed to produce five enzymes C. Some internal AUG sequences must be recognized as start codons D. Formyl-methionine is incorporated into the polypeptide when the ribosome recognizes the AUG codons on this mRNA E. The AUG start codon on this mRNA is found in an intron

C. Some internal AUG sequences must be recognized as start codons • This is the correct answer - Since there are multiple open reading frames for the ribosome to recognize on a polycistronic mRNA, some of those AUG codons that come after the first AUG must be recognized as 'start' for protein synthesis. • Remember, the first AUG of the protein starts with the formyl-Met, then the internal AUG (for that protein) get the normal Met • The next protein that is produced though, also must start with the formyl-Met

Which of the following eukaryote transcription factors is responsible for activating RNA polymerase II by phosphorylation? A. TFIID B. TFIIB C. TFIIH D. TFIIE E. TATA box binding protein

C. TFIIH - Correct answer. TFIIH phosphorylates RNA polymerase II during the initiation of transcription in eukaryotes

A researcher working for NASA examines a rock sample returned by a space probe sent to the planet Mars. She became very excited after isolating viable microorganisms from the sample, which utilized a genetic code that initially appeared to be the same as that found in living bacterial cells on earth. Five properties of the Martian code are listed below. Which one of the properties of the Martian code shown is different from the terrestrial (earth) genetic code? A. Each codon consisted of three bases B. An amino acid could have more than one codon C. The genetic code was overlapping D. UAA acted as a termination codon E. All proteins started with formyl-methionine

C. The genetic code was overlapping - Correct answer - the genetic code found in all known terrestrial organisms is a non-overlapping code

Activation of which of the following receptors would lead to activation of protein kinase C? A. Insulin receptor B. Glucagon receptor C. alpha1 receptor D. beta1 receptor E. beta2 receptor

C. alpha1 receptor - Correct answer - Activates Phospholipase C which cleaves membrane PIP2 to form IP3 and DAG - which activate Protein kinase C. IP3 also increases intracellular calcium levels A. Insulin receptor - Activates tyrosine kinase and causes autophosphorylation of tyrosine residues on the receptor B. Glucagon receptor - Activates adenylate cyclase which inturn increases cAMP levels resulting in activation of protein kinase A and phosphorylation of key enzymes of metabolism C. alpha1 receptor - Correct answer - Activates Phospholipase C which cleaves membrane PIP2 to form IP3 and DAG - which activate Protein kinase C. IP3 also increases intracellular calcium levels D. beta1 receptor - Activate adenylate cyclase and increase cAMP - activate protein kinase A E. beta2 receptor - Activate adenylate cyclase and increase cAMP

Ciprofloxacin, is a potent antibiotic drug used to treat a broad spectrum of bacterial infections in humans. This drug kills bacteria by inhibiting both DNA replication and transcription of RNA. Which of the following mechanisms is responsible for the antibiotic activity of Ciprofloxacin? A. intercalating with the DNA B. binding and inhibiting DNA polymerase and RNA polymerase C. inhibiting the activity of DNA gyrase D. inducing mutations in genes necessary for bacterial survival E. preventing the formation of phospho-diester bonds during synthesis

C. inhibiting the activity of DNA gyrase - Correct answer A. intercalating with the DNA (Actinomycin D has this mechanism of action) B. binding and inhibiting DNA polymerase and RNA polymerase C. inhibiting the activity of DNA gyrase - Correct answer D. inducing mutations in gene necessary for bacterial survival E. preventing the formation of phospho-diester bonds during synthesis

Sodium butyrate is a drug that disrupts the activity of the enzyme histone deacetylase (HDAC). In general, treatment of a mammalian cell culture system with sodium butyrate would have which of the following effects on the cells? A. increased rates of cell division B. inhibition of protein synthesis C. increased levels of gene transcription D. DNA breakage E. Decreased levels of histone acetylation

Cells treated with sodium butyrate would have histones that are more acetylated than control cells (Sodium butyrate disrupts histone deacetylase activity) - When histones are acetylated the DNA is not wound tightly around the histones - As a result, there is increased rates of gene transcription C. increased levels of gene transcription

Activation of 1 adrenoceptors leads to A. activation of adenylyl cyclase B. decrease in cAMP C. activation of guanylyl cyclase D. activation of phospholipase C E. increase in cyclic GMP

D

After a meal, insulin affects the activity of many metabolic pathways. In which way is the insulin signal transmitted to exert its many effects on target cells? A. Through induction of NO production B. Through release of Ca2+ from intracellular stores C. By direct interaction with G-proteins D. By activation of a tyrosine kinase receptor E. By increasing intracellular cAMP levels

D

Binding of epinephrine to Beta-receptors leads to activation of adenylyl cyclase. Adenylyl cyclase converts ATP to cAMP which, in turn, activates A. guanylyl cyclase. B. smooth-muscle myosin light chain kinase. C. protein kinase C. D. protein kinase A E. phospholipase C

D

Rebecca is running a 10-mile cross-country race on a warm sunny day. Three miles into the race the sympathetic outflow to her cutaneous vessels is decreased by the temperature regulation center of her hypothalamus. This Competition between homeostatic mechanisms can best be described by which term; A. positive feedback B. negative feedback C. feed forward regulation D. hierarchy of response E. mechanism redundancy

D

The G protein Gi A. Activates guanylyl cyclase B. Is a seven-pass transmembrane receptor protein C. Activates adenylyl cyclase D. Inhibits adenylyl cyclase E. Activates Phospholipase C

D

Which of the following is an example of an endocrine signal? A. Nerve growth factor B. Acetylcholine C. Nitric Oxide D. Epinephrine E. Glutamate

D

Consider that the DNA code has been altered in a gene which encodes an enzyme. Which of the following types of gene mutations would be expected to produce the most severe reduction in activity of this enzyme? A. A synonymous mutation in an exon B. A conservative mutation in an exon C. Insertion of a 3 base pair sequence in an exon D. A single base deletion in the first exon E. A missense mutation in an intron F. A transversion mutation in the first base of a codon

D - Commentary: A deletion of 1 base in the first exon would be expected to alter the reading frame of the mRNA. This means that all amino acids that come after that mutation will be something other than what was originally coded for. Furthermore, an early stop codon is likely to be formed by chance. Another event typically happens with these truncating mutations is that any mRNA formed is often degraded while it is being translated in the cytoplasm. This event is called nonsense mediated mRNA decay, and is thought to occur to keep improper mRNAs from being translated

Which of the following hormones is a steroid hormone? A. Insulin B. Glucagon C. Epinephrine D. Cortisol E. Thyroid hormone

D - correct answer - is synthesized from cholesterol (Aldosterone, testosterone, progesterone and estrogen are all synthesized from cholesterol)

Covalent bonds and noncovalent bonds are found in the tertiary structure of proteins. Which of the following bonds is a covalent bond formed by side chains of amino acids? A. ionic bonds between aspartate and lysine B. hydrogen bonds between serine and glutamate C. van der Waals forces found in the alpha-helix D. disulfide bonds between two cysteine residues E. hydrophobic forces between valine and isoleucine

D is correct: disulfide bonds between two cysteine side chains are covalent bonds. In proteins they stabilize the tertiary structure. Ionic bonds, vander Waals forces, hydrogen bonds and hydrophobic forces are broken by denaturation a incorrect: the bonds between aspartate and lysine are noncovalent ionic bonds b incorrect: hydrogen bonds between serine and glutamate are noncovalent bonds c incorrect: van der Waals forces found in the alpha-helix are very weak noncovalent bonds d. correct, disulfide bonds are not broken during conformational change or denaturation e incorrect, hydrophobic forces between valine and isoleucine side chains are noncovalent

Our knowledge about enzymes has increased immensely since the first description of a protein acting as a catalyst. Enzymatic catalysis A. changes the thermodynamics of the reaction B. works without a transition state C. is independent of the pH D. can involve a prosthetic group like FAD E. involves binding of the substrate to its huge active site

D is correct: enzymatic catalysis can involve a prosthetic group like FAD A. incorrect: enzymes do not change the thermodynamics of the reaction B. incorrect: enzymatic catalysis does not work without a transition state C. incorrect: enzymatic catalysis shows a respective pH optimum D. correct, FAD is a bound coenzyme for some oxidoreductases E. incorrect: the active site is small, not large

5-azacytidine is a pyrimidine nucleoside analog that is believed to exert its antineoplastic effects by DNA hypomethylation. What is the effect of this drug? A. It terminates DNA synthesis B. It causes DNA breakage C. It inhibits both DNA and RNA polymerase D. It increases the rate of transcription E. It intercalates with the DNA F. It inhibits ribosomal peptidyl transferase activity

D. It increases the rate of transcription - DNA hypomethylation results in increased rates of transcription. Whereas, methylation of DNA decreases rates of transcription (eg. imprinting) A. It terminates DNA synthesis B. It causes DNA breakage C. It inhibits both DNA and RNA polymerase D. It increases the rate of transcription - DNA hypomethylation results in increased rates of transcription. Whereas, methylation of DNA decreases rates of transcription (eg. imprinting) E. It intercalates with the DNA F. It inhibits ribosomal peptidyl transferase activity

Which of the following drugs is not a nucleoside or nucleotide analog? A. AZT B. Didanosine C. Acyclovir D. Rifampin E. Decitabine

D. Rifampin - Is an inhibitor of prokaryotic RNA polymerase. It is an inhibitor of transcription. It binds to and inhibits RNA polymerase

Fluoroquinolones are a group of drugs used to treat a broad spectrum of bacterial infections. This group of drugs targets the prokaryote enzyme DNA gyrase. DNA gyrase can be grouped together with which of the following enzymes categories? A. DNA polymerases B. RNA polymerases C. Helicases D. Topoisomerases E. Primases

D. Topoisomerases - DNA gyrase (prokaryotic) belongs to the Topoisomerase type II family of topoisomerase enyzmes. A. DNA polymerases - Enzymes of replication B. RNA polymerases - Required for transcription C. Helicases - required to separate the two strands of DNA double helix D. Topoisomerases - DNA gyrase (prokaryotic) belongs to the Topoisomerase type II family of topoisomerase enyzmes. E. Primases - Enzymes required for synthesis of the primer in replication

A 25 year-old female has right upper abdominal pain and signs of acute liver disease. Which of the following serum enzymes would be raised in this patient and confirms these findings? A. Creatine kinase MB B. Creatine kinase MM C. Amylase D. Alanine aminotransferase E. Lactate dehydrogenase -1

D. is correct: alanine aminotransferase in the serum is a marker for liver damage A. incorrect: Creatine kinase MB is increased in patients following myocardial infarction B. incorrect: Creatine kinase MM is increased in patients with skeletal muscle damage/ disease C. incorrect: Amylase is increased in patients with pancreatic disease D. correct, high concentration in liver cytosol, used with other markers E. incorrect: LDH-1 is found in heart and RBC, liver contains LDH-5

A researcher was investigating the activities of various types of G-protein subunits. The G protein Gq: A. Activates guanylyl cyclase B. Causes autophosphorylation of tyrosine residues C. Activates adenylyl cyclase D. Inhibits adenylyl cyclase E. Activates phospholipase C

E

An increase in which one of the following will decrease the water loss from a cell that is placed in a hypertonic solution? A. the number of aquaporins in the cell membrane B. the hydraulic conductivity of the cell membrane C. the temperature D. the surface area of the membrane E. the concentration of impermeable proteins inside the cell

E

If the concentration of a cation inside a cell is 130 mM and outside the cell is 13 mM, what would the equilibrium potential for that ion be? A. +60 mV B. +30 mV C. 0 mV D. -30 mV E. -60 mV

E

Which of the following is a neurotransmitter? A. Nerve growth factor B. Epidermal growth factor C. Insulin D. Glucagon E. Glutamate

E

Which one of the following is an example of a homeostatically controlled variable? A. autonomic nervous system B. thermoreceptors C. heart rate D. hypothalamus E. body temperature

E

Nitroglycerin is used to treat patients with angina pectoris because it releases A. Epinephrine B. Acetylcholine C. IP3 D. Oxygen E. Nitric oxide

E Correct answer. Nitric oxide formed activates guanylyl cyclase - which increases cGMP - which causes smooth muscle relaxation (vasodilation) and increases blood flow

The enzyme lactate dehydrogenase A. contains FAD as prosthetic group B. is a protein dimer with the compostition HM C. is increased in serum in 4 hours after myocardial infarction D. needs vitamin C for its reaction E. cannot form pyruvate at high NADH levels

E is correct: The enzyme lactate dehydrogenase cannot form pyruvate at high NADH levels, as it needs for this direction NAD+ as coenzyme A. incorrect: it does not contain FAD as prosthetic group; it requires NAD+ B. incorrect: it is a tetramer with H and M subunits C. incorrect: it is increased in serum about 2 days following myocardial infarction D. incorrect: needs NAD+ as a coenzyme; not vitamin C E. correct, at high NADH levels lactate cannot be used to form pyruvate. For the formation of pyruvate from lactate NAD+ is needed

The digestion of dietary carbohydrates is very much watched in programs that promote weight loss. One could target the enzymes that degrade complex sugars or one could target the uptake of glucose and galactose into intestinal mucosal cells. This uptake is performed by a. Facilitated transport b. GLUT-4 c. GLUT-2 d. Primary active transport e. Co-transport with sodium ions

E is correct: dietary glucose or galactose are taken up into the intestinal mucosal cells by co-transport with sodium ions a. is incorrect: dietary glucose and galactose are mainly taken up by secondary active transport b. is incorrect: GLUT-4 is mainly found in fat and muscle cells c. is incorrect: GLUT-2 is used for release into the portal circulation d. is incorrect: it is not primary, but secondary active transport e. is correct: either glucose or galactose is transported by SGLT into mucosal cells

A Pharmaceutical company investigates the possible inhibition of a specific enzyme by a new drug. The data are expressed in the Lineweaver-Burk plot. Which of the following is the correct statement regarding such plots? A. It shows a sigmoidal curve that reaches Vmax at high [S] B. It is plotted as initial velocity (Vo) versus 1/[S] C. It results in a reciprocal value of Vmax to be found at the intercept with the x-axis D. It shows a hyperbolic curve that reaches Vmax at high [S] E. It results in a straight line due to plotting 1/Vo to 1/[S]

E is correct: the Lineweaver-Burk plot leads to a graph with a straight line due to plotting 1/V0 to 1/[S] (double reciprocal) a. incorrect: a sigmoidal curve, velocity versus [S] is found for allosteric enzymes b. incorrect: it does not show V0 data versus 1/[S] but instead 1/V0 versus 1/[S] c. incorrect: 1/Vmax is the intercept with the Y-axis and not the X-axis d. incorrect: a hyperbolic curve is found when plotted in the Michaelis-Menten plot e. correct, The X-axis intercept is -1/Km and the Y-axis intercept is 1/Vmax.

A 56-year-old man was admitted to hospital, and was diagnosed as suffering from bilateral bacterial pneumonia. As part of his treatment erythromycin was administered intravenously. The reason that this substance is effective therapeutically is that it interferes with A. Peptide bond formation during translation in bacteria B. Initiation of translation in bacteria C. Initiation of transcription in bacteria D. Bacterial DNA replication E. The translocation step of translation in bacteria

E. The translocation step of translation in bacteria - Correct answer A. Peptide bond formation during translation in bacteria - This is the mechanism of action of chloramphenicol B. Initiation of translation in bacteria - This is mechanism of action of streptomycin C. Initiation of transcription in bacteria - This is mechanism of action of rifamycin D. Bacterial DNA replication - DNA gyrase inhibitors like fluoroquinolones inhibit bacterial DNA replication E. The translocation step of translation in bacteria - Correct answer

Activation of alpha1 adrenoceptors leads to A. activation of tyrosine kinases B. activation of guanylyl cyclase C. inhibition of adenylyl cyclase D. relaxation of smooth muscle E. increase in intracellular Ca2+ levels

E. This occurs due to activation of Gq - activates phospholipase C - forms IP3 and DAG - IP3 increases intracellular calcium levels

A 50-year-old patient has been prescribed a low-dose of aspirin (81 mg per day). What is the mechanism of action of this inhibitor? A. It inhibits acetylcholinesterase B. It reversibly inhibits cyclooxygenase C. It is an inhibitor of HMG CoA reductase D. It is an inhibitor of xanthine oxidase E. It reduces the amount of thromboxane

E. is correct: A low-dose of aspirin is used to reduce the amount of thromboxane A. incorrect: acetylcholinesterase is inhibited by DFP B. incorrect: the inhibition of COX is irreversible, even at low doses C. incorrect: Statins are inhibitors of HMG CoA reductase D. incorrect: Allopurinol is an inhibitor of xanthine oxidase E. correct: mature platelets cannot synthesize new COX

A strand of DNA has the following sequence TAGGCCAATT. What is the sequence of the strand generated by the action of RNA polymerase?

Remember, the standard convention of nucleic acids is to label the strand from 5' TAGGCCAATT 3' direction. 5' TAGGCCAATT 3' - DNA Template strand 3' AUCCGGUUAA 5' - RNA strand synthesized by RNA polymerase (During transcription, if there is an A on the template strand a Uracil is incorporated into the RNA - Remember, RNA does not have Thymine) The sequence of the resulting strand is 5' AAUUGGCCUA 3' (as per the standard naming convention for nucleic acids)

A strand of DNA has the following sequence TAGGCCAATT. What is the sequence of the strand generated by the action of DNA polymerase?

Remember, the standard convention of nucleic acids is to label the strand from 5' TAGGCCAATT 3' direction. 5' TAGGCCAATT 3' - Template strand 3' ATCCGGTTAA 5' - Is the sequence of the strand synthesized by DNA polymerase. The answer is 5'AATTGGCCTA 3' (as per the standard naming convention for nucleic acids)

In some families which have Huntington disease, there is a very closely linked RFLP marker named G8 which is closely linked to the Huntington disease (HD) locus. This marker was very useful in early studies of the Huntington gene, and in molecular diagnosis of patients who were at risk for the disease due to family history. How was this marker useful in the study of Huntington? A. The G8 marker indicated a CAG repeat in intron number 4 of the HD gene B. The presence of the G8 marker indicated that the patient had the trinucleotide repeat C. The G8 marker would always recombine with the HD gene locus D. The G8 marker is physically close the HD gene locus E. The G8 marker provided a direct test for Huntington disease

The answer is D. This is an example of an indirect test. This G8 marker was used in 1982 during the first attempts at positional cloning of the HD gene (Read about the heroic efforts of Nancy Wexler). Now, we would instead characterize a person's HD status by measuring the actual number of CAG repeats in the gene.

An 18 month old male patient is identified to have Prader Willi syndrome as a result of uniparental disomy. In this case, he has lost the paternal contribution of chromosome 15 and instead he has two copies of his maternal chromosome 15. Which approach was used to identify that uniparental disomy caused the disorder in the pateint? A. ASO PCR B. FISH C. SKY FISH D. SNP Chip E. Array CGH

The best answer is D. The point is that the SNP Chip allows haplotype analysis. The idea is to look for a diagnostic technique that would tell us that both chromosomes come from the same parent. The SNP chip will do that for us in a very efficient manner. Recall that about 1 in every 1,000 bp, there is an SNP. So, normally the two chromosome 15's in a normal person will have haplotypes that are derived from each parent; one from the father, and one from the mother. But in our patient his two chromosome 15 copies will appear to both be derived from the mother (as a result of trisomy rescue). Note that there are two possible ways that a ova (or a sperm) may be created that have two copies of chromosome 15: that is non-disjunction in meiosis I or a non-disjunction in meiosis II). If non-disjunction occurs in meiosis I, then the gamete has two different versions of chromosome 15, if the non-disjunction occurred in maternal meiosis II, then the ova will have identical chromosome 15 copies. A. ASO PCR - this would work, but it would not be as efficeint as the SNP chip. Many would have to be tested before you would be convinced that the haplotypes match the mother. But don't worry, we recognize that this is also a correct (sort of) choice and you may be rest assured that you would not find 2 correct answers on the exam B. FISH - this would not work, FISH is going to detect a deletion of DNA - the pateint has two copies of ch 15, the DNA is there, no deletion exists C. SKY FISH - would not work, simply paints each chromosome a different color E. Array CGH - would not work in this case, there is no deletion, our patient instead has inherited two copies of ch 15 from his mom. Uniparental disomy

A group of volunteers have been enrolled for a study of the nutritional importance of essential fatty acids. Alpha-linolenic acid has been eliminated in their diet. Which of these fatty acids cannot be formed in these volunteers? a. Arachidonic acid b. Oleic acid c. Stearic acid d. Eicosapentaenoic acid e. Linoleic acid

This question is testing the omega- 3 vs omega-6 families of unsaturated fatty acids d. Eicosapentaenoic acid (EPA) - Correct answer - belongs to the omega-3 family and can be formed from alpha linolenic acid. Docosahexaenoic acid (DHA) also belongs to the omega-3 family and can be formed from alpha linolenic acid and EPA- Synthesis of both DHA and EPA is reduced when the diet lacks alphalinolenic acid a. Arachidonic acid - is an omega-6 fatty acid. It is not essential and can be synthesized from linoleic acid (omega-6) b. Oleic acid - belongs to the omega-9 family of unsaturated fatty acids and is not essential c. Stearic acid - is an 18 C saturated fatty acid - not essential fatty acid d. Eicosapentaenoic acid (EPA) - Correct answer - belongs to the omega-3 family and can be formed from alpha linolenic acid. Docosahexaenoic acid (DHA) also belongs to the omega-3 family and can be formed from alpha linolenic acid and EPA- Synthesis of both DHA and EPA is reduced when the diet lacks alphalinolenic acid e. Linoleic acid - Is an essential fatty acid belonging to the omega-6 family

A researcher has isolated the glycolipid components of the cell membrane. Which of these macromolecules belongs to this class? a. Ceramide b. Sphingomyelin c. Cerebroside d. Cardiolipin e. Phosphatidyl inositol f. Triacylglycerol

c. Cerebroside (contains sphingosine, fatty and glucose or galactose - it is a glycolipid) - Correct answer a. Ceramide (Ceramide is sphingosine with a fatty acid only) b. Sphingomyelin (contains sphigosine, fatty acid, phosphate and choline - it is a sphingophospholipid) c. Cerebroside (contains sphingosine, fatty and glucose or galactose - it is a glycolipid) - Correct answer d. Cardiolipin (This is a glycerophospholipid present exclusively in inner mitochondrial membranes) e. Phosphatidyl inositol (This is a glycerophospholipid and is a precursor of PIP2 which forms IP3 and Diacylglycerol on hydrolysis during signal transduction) f. Triacylglycerol (is a hydrophobic lipid and is not a constituent of cell membranes. It contains a glycerol and three fatty acids)

Laboratory study of the secondary structure of an enzyme involved in intermediary metabolism contained alpha-helices. What is correct regarding this structure? A. It is stabilized by peptide bonds B. It is stabilized by phosphodiester linkages C. It contains amino acids with their side chains towards the outside of the helix D. It is a structure favored by a high amount of proline residues E. It is retained when the protein is heated to high temperatures

c. It contains amino acids with their side chains towards the outside of the helix a. It is stabilized by peptide bonds - The primary structure is stabilized by peptide bonds. The alpha-helices are stabilized by hydrogen bonds b. It is stabilized by phosphodiester linkages - Phosphodiester linkages are found in nucleic acids (DNA and RNA) c. It contains amino acids with their side chains towards the outside of the helix d. It is a structure favored by a high amount of proline residues - Proline disrupts the formation of alpha helices e. It is retained when the protein is heated to high temperatures - On heating, proteins lose their secondary, tertiary and quaternary structures - only the primary structure is intact

Experimental analysis of the inner leaflet of the plasma membrane revealed the presence of a wide variety of macromolecules in the inner leaflet. Which of these is present at this location? a. Ganglioside b. Phosphatidic acid c. Phosphatidylserine d. Free fatty acids e. Triacylglycerol f. Cholesterol ester

c. Phosphatidylserine - Is present only in the inner leaflet of the membrane - Correct answer a. Ganglioside - is a glycolipid and is present in the outer leaflet of the plasma membrane b. Phosphatidic acid - is not present in the membranes; it is an intermediate in the synthesis of glycerophospholipids c. Phosphatidylserine - Is present only in the inner leaflet of the membrane - Correct answer d. Free fatty acids - are only present in the membrane linked to phospholipids or glycolipids and not in the free state e. Triacylglycerol - not a component of membranes. Storage form of lipid in adipose tissue f. Cholesterol ester - is not present in membranes. Free cholesterol is present in both leaflets of the membrane and regulates membrane fluidity