WGU C785 Biochemistry Unit Exam Questions fully solved & update

Which level of protein structure is disrupted through the hydrolysis of peptide bonds? Quaternary Tertiary Primary Secondary - ANSWER-Primary The primary structure of a protein is the sequence of amino acids held together by peptide bonds. Peptide bonds are formed by dehydration reactions and disrupted by hydrolysis. A mutation in the beta-hemoglobin gene, which results in the replacement of the amino acid glutamate in position 6 with the amino acid valine, leads to the development of sickle cell anemia. The structures of glutamate and valine are shown below. If the beta hemoglobin gene in a patient with sickle-cell anemia were to be edited so that the valine in position 6 was replaced with a different amino acid, which replacement for valine would be expected to have the best clinical outcome, in theory, for the patient? (Assume the valine can potentially be replaced with any amino acid other than glutamate.) - ANSWER-The original amino acid in a healthy patient is glutamate, which is negatively charged. The mutated amino acid is valine, which is non-polar. Valine is causing sickle cell anemia. The best amino acid to replace valine so that the patient is healthy again would be the one most like glutamate, so any negatively charged amino acid. Secondary, tertiary, and quaternary levels of protein structure can all be impacted by exposing a protein to which treatment? Change of a hydrophobic amino acid to a different hydrophobic amino acid Addition of a reducing agent Placement of the protein in a solution with a low pH Increase in the concentration of the protein in solution - ANSWER-Placement of the protein in a solution with a low pH Changes in pH affect hydrogen bonds and ionic bonds. Hydrogen bonds in the backbone of amino acids occur in secondary structure, and both hydrogen bonds and ionic bonds occur in the side chains of amino acids in tertiary structure. An increase in beta-pleated sheet structure in some brain proteins can lead to an increase in amyloid deposit formation, characteristic of some neurodegenerative diseases. What is the primary biochemical process that follows the increase in beta-pleated sheet structure that leads to the development of the amyloid deposits? An increase in glycogen formation in the brain cells Aggregation of the proteins in the brain Secretion of glucagon, leading to excessive ketogenesis An increase in anaerobic metabolism of glucose in the brain - ANSWER-Aggregation of the proteins in the brain This question is describing changes in protein structure. Aggregation occurs when proteins clump together inappropriately, causing plaques like amyloid deposits to accumulate. Which level of protein structure is determined by the sequence of amino acids? Secondary structure Quaternary structure Tertiary structure Primary structure - ANSWER-Primary structure The primary structure of a protein is simply the sequence of amino acids held together by peptide bonds. Which force is most influential in determining the secondary structure of a protein? Hydrophobic effect Disulfide bonding Hydrogen bonding Electrostatic interactions - ANSWER-Hydrogen bonding The secondary structure of a protein is built by hydrogen bonds between the carboxyl groups and amino groups on the backbones of the amino acids. Which amino acid would most likely participate in hydrogen bonds? - ANSWER-Amino Acid structure 4 This is a polar, uncharged amino acid due to the OH group on the side chain. Polar, uncharged amino acids containing oxygen or NH groups make hydrogen bonds. Which portion of the amino acid is inside the box? The box is surrounding the section below the Alpha Carbon - ANSWER-Side Chain The side chain is the variable group of the amino acid, also called the R group. Every amino acid has the same amino group, carboxylic acid group, and an alpha carbon, but the side chain is different. Which pair of amino acids will most likely interact through hydrophobic forces between their side chains? - ANSWER-Both of these amino acids are non-polar and therefore can interact together with a hydrophobic interaction. Please note that the "S" in the amino acid on the right is non-polar, while the "SH" group in answer choice D is polar. The S must have an H to be polar and is otherwise non-polar. Which portion of the amino acid is inside the box? The box is over the Carbon at the Center of the chain - ANSWER-Alpha Carbon The alpha carbon is the central carbon on an amino acid that holds together the other groups of the amino acid. It is always attached to the amino group, the carboxyl group, the side chain, and a single hydrogen. It is part of the backbone of the amino acid and is found in every amino acid. Given the following amino acid structure, what is the strongest intermolecular force it would participate in to stabilize a protein structure? Ionic bond Disulfide bond Hydrogen bond Hydrophobic interaction - ANSWER-Hydrophobic interaction The amino acid pictured only has CH groups in its side chain, and therefore is non-polar. Non-polar amino acids make hydrophobic interactions. Which change would most likely result in a permanent modification of an expressed protein's function? An increase in the pH of a solution in which a protein is dissolved from 6.5 to 8.0, when it is known that the protein has an optimal activity of pH 7.8 A mutation of the gene for a protein that leads to the substitution of a hydrophobic amino acid with a nonpolar amino acid A mutation of the gene for a protein that leads to the substitution of a nonpolar amino acid with a charged amino acid The mutation of a gene for an enzyme involved in protein synthesis following exposure to X-rays, causing the protein not to be synthesized - ANSWER-A mutation of the gene for a protein that leads to the substitution of a nonpolar amino acid with a charged amino acid. The mutation of nonpolar amino acid to a charged amino acid will disrupt the original hydrophobic interaction, permanently changing the function of the protein. Which property of enzymes is illustrated in the final step of the enzymatic cycle? Enzymes are specific. Enzymes increase the reaction rate for a reaction. Enzymes are reusable. Enzymes lower the activation energy for a reaction. - ANSWER-Enzymes are reusable. In the final step of the enzymatic cycle, the product is released and the enzyme is able to bind to a new substrate and begin the cycle again. In the enzyme cycle, which step immediately follows induced fit? Formation of the enzyme-substrate complex Release of the product and enzyme complex Formation of the enzyme-molecule complex Formation of the enzyme-product complex - ANSWER-Formation of the enzyme-product complex The induced fit refers to the conformational change that the enzyme undergoes when it binds to the substrate to form the enzyme-substrate complex. Therefore, the enzymatic cycle step that occurs after the induced fit is the formation of the enzyme-product complex. Which type of inhibition occurs when a particular drug binds to the active site of an enzyme? Competitive Uncompetitive Irreversible Noncompetitive - ANSWER-Competitive Competitive inhibitors compete with the substrate to bind to the active site of the enzyme. Salivary amylase, an enzyme responsible for partial digestion of carbohydrates, has optimum activity at a pH value of 6.8. What is the impact on the activity if the pH is decreased to 4.0? Significantly increase Significantly decrease Slightly decrease Slightly increase - ANSWER-Significantly decrease A drop in pH from 6.8 to 4.0 is a significant change in pH. Recall that the hydrogen bonds and ionic bonds that hold protein structures together can be broken by changes in pH. The disruption in protein structure due to this pH change will also significantly decrease amylase activity. Low levels of glutathione are associated with certain types of ovarian and breast cancers. In the synthesis of glutathione, glutathione accumulates in the cell, binding to an enzyme in the pathway and temporarily preventing the synthesis of glutathione. Which type of inhibition is described by this scenario? Feedback Competitive Allosteric Uncompetitive - ANSWER-Feedback The keywords here are that glutathione accumulates and binds to an enzyme in the pathway to prevent synthesis. Feedback inhibition occurs when a product of a pathway turns into an inhibitor of an enzyme earlier in the pathway. Lipase is an enzyme with an optimum temperature of 98.6°F and an optimum pH of 7.0 in the duodenum in the human body. If a person is experiencing a fever of 99.8°F, what will increase the activity of the lipase enzyme? Decrease in temperature Increase in temperature Decrease of the substrate of the enzyme Increase of pH of duodenum to 8.0 - ANSWER-Decrease in temperature An enzyme will have the highest activity when it is under optimal conditions. In this case, the fever of 99.8 is above the optimal temperature, so lowering the temperature will increase activity. The enzyme glucokinase only binds its substrate glucose and converts glucose into the product glucose-6-phosphate. Which property of enzymes is described by this scenario? Specificity Activation energy Induced fit Reaction rate - ANSWER-Specificity Enzymes have a high degree of specificity. They will bind to one specific class of molecules and usually catalyze only one type of reaction. Low levels of glutathione are associated with neurological, immunological, and cardiovascular impairments. Two enzymatic reactions are involved in the synthesis of glutathione. In the second enzymatic reaction, glutathione synthetase converts glycine to glutathione. What would potentially decrease risks associated with low levels of glutathione? Increasing the amount of glycine in the diet Noncompetitive inhibitor of glutathione synthetase Uncompetitive inhibitor of glutathione synthetase Decreasing the amount of glycine in the diet - ANSWER-Increasing the amount of glycine in the diet One way to increase glutathione levels is to increase the activity of glutathione synthetase. Glutathione synthetase activity can be increased by increasing the amount of substrate, or glycine available. Which class of enzymes impacts protein function by temporarily removing a phosphate? Kinase Phosphatase Polypeptide Lactase - ANSWER-Phosphatase Phosphatases are enzymes that remove phosphate groups from the substrates. How does the activation energy of enzyme-catalyzed reactions compare to those of corresponding uncatalyzed reactions? The activation energy of enzyme-catalyzed reactions are the same as uncatalyzed reactions The activation energy of the enzyme-catalyzed reactions only changes in response to temperature. The activation energy of enzyme-catalyzed reactions are lower than the uncatalyzed reaction The activation energy of enzyme-catalyzed reactions are higher than uncatalyzed reactions - ANSWER-The activation energy of enzyme-catalyzed reactions are lower than the uncatalyzed reaction The activation energy is the amount of energy needed to get the reaction started and over the energy hill to form products. Enzymes reduce the amount of energy needed to start the reaction or lower the amount of energy needed to climb the energy hill. A final product of a four-step metabolic pathway serves as a noncompetitive inhibitor, binding to an enzyme in this pathway and temporarily turning off the pathway. Which enzyme is most likely to be targeted by the inhibitor? The final enzyme in the pathway The first enzyme in the pathway Any enzyme in the middle of the pathway The second enzyme in the pathway - ANSWER-The first enzyme in the pathway The final product's structure is more unlike the substrate for the first enzyme in the pathway than any other, and thus the least likely to fit the active site of enzyme 1. Which of the following could be a potential ramification of a defective nucleotide excision repair (NER) pathway? Thymine dimers would be fixed by DNA polymerase, increasing the probability of developing skin cancer. Thymine dimers would accumulate and increase the probability of developing skin cancer. Thymine dimers, caused by UV light, would be repaired, decreasing the probability of developing skin cancer. Thymine dimers would persist, decreasing the probability of developing skin cancer. - ANSWER-Thymine dimers would accumulate and increase the probability of developing skin cancer. A defective nucleotide excision repair pathway would not be able to repair thymine dimers. The accumulation of thymine dimers would increase the probability of developing skin cancer. A patient has received a large dose of ionizing radiation at his place of employment. Which scenario is accurate? Ligase removes a single damaged base and replaces it with a new nucleotide. Double-stranded DNA breaks are being repaired by nonhomologous end joining, without the use of a homologous template. The homologous recombination pathway is upregulated and is facilitating the removal of mismatched bases. A string of damaged nucleotides is removed and replaced by a new nucleotide sequence in the base excision repair pathway. - ANSWER-Double-stranded DNA breaks are being repaired by nonhomologous end joining, without the use of a homologous template. Due to the large dose of ionizing radiation, the patient has many double-stranded DNA breaks, and undamaged homologous DNA is in short supply. Nonhomologous end-joining will be used to fix these double-strand breaks. Immediately following transcription, mRNA must be processed before it is transported to the cytoplasm to undergo translation. Which statement correctly describes mRNA processing? Alternative combinations of introns can be linked together to produce closely related gene products. Exons are removed and introns are inserted into the mature mRNA sequence. mRNA is folded into beta-pleated sheets and alpha helices to produce a mature mRNA sequence. Introns are spliced out and exons are connected to produce a mature mRNA sequence. - ANSWER-Introns are spliced out and exons are connected to produce a mature mRNA sequence. Introns are removed from the mRNA sequence and the remaining exons are spliced together to create the mature mRNA. A point mutation has altered the amino acid sequence of a neuronal tau protein, causing serine (Ser) at position 202 to be mutated to proline (Pro). Which set of codons below corresponds to this mutation? TCT to CCT AGA to GGA CGA to AGG AGG to CCC - ANSWER-TCT to CCT TCT is a codon for Ser, and CCT is a codon for Pro. TCT to CCT is a point mutation that changes Ser to Pro. Which of the following is required for DNA polymerase to replicate template DNA in PCR? RNA primers tRNA RNA polymerase DNA primers - ANSWER-DNA Primers DNA primers are used in the PCR process. A portion of the HLA gene and its associated amino acid sequence are shown below. If cytosine is deleted from codon 101, what will be the resulting amino acid sequence? Ser Tyr Glu Glu Met Leu Glu Pro Cys Glu His Ala Asp - ANSWER-Glu Met Leu Deletion of cytosine (C) from codon 101 leads to the DNA sequence 5'-GAG ATG CTG AT-3', which corresponds to the amino acid sequence Glu Met Leu. Inheriting mutations in the BRCA-1 or BRCA-2 gene can increase an individual's likelihood of developing breast cancer or ovarian cancer. How can PCR be used to assess an individual's susceptibility to developing these cancers? PCR can be used to produce the mRNA for the BRCA-1 and BRCA-2 genes, which can then be sequenced to look for mutations. PCR primers can be engineered to flank the mutation of interest in the BRCA-1 and BRCA-2 genes, and the PCR product can be sequenced to look for mutations. The PCR process will provide an exponential increase in the amino acid sequence of the gene products. Then the amino acid can be sequenced to look for mutations. PCR primers can be engineered to sequence the BRCA-1 and BRCA-2 genes and determine whether mutations are present. - ANSWER-PCR primers can be engineered to flank the mutation of interest in the BRCA-1 and BRCA-2 genes, and the PCR product can be sequenced to look for mutations. By using primers that flank the mutation of interest, PCR can produce DNA which can be used to determine if the mutation is present. Which template DNA sequence would result in the amino acid proline (Pro)? 5' - TGA - 3' 5' - TGG - 3' 5' - GGT - 3' 5' - CCC - 3' - ANSWER-5' - TGG - 3' The correct answer is 5'-TGG-3'. The corresponding mRNA for 5'-TGG-3' is 3'-ACC-5', which reversed is 5'-CCA-3', a codon for the amino acid Pro. Legumain is an enzyme that is overexpressed in neurons in response to traumatic brain injury (TBI). How does TBI trigger an increase in the expression of the gene that codes for legumain? Nucleosomes are more widely spaced, exposing the legumain gene. DNA polymerase is able to bind to the legumain gene. Nucleosomes become tightly packed together, preventing the binding of transcription factors to the promoter region of the legumain gene. Methyltransferases are up regulated and add methyl groups to the DNA bases in the legumain gene. - ANSWER-Nucleosomes are more widely spaced, exposing the legumain gene. Increased nucleosome spacing leads to increased transcription and translation (i.e. gene expression) of the Legumain gene. Which of the following mRNA sequences can be translated into a peptide sequence containing two amino acids? 5' - GAU CGA UAG UGG - 3' 5' - GAU CGA UGG UAG - 3' 5' - UGG UAC CAG UGA - 3' 5' - GAC CAU GCG GGG - 3' - ANSWER-5' - GAU CGA UAG UGG - 3' The sequence 5' - GAU CGA UAG UGG - 3' contains a stop codon in the 3rd position, therefore, this sequence would produce a peptide sequence with two amino acids. What is the mRNA sequence that would result from this coding DNA sequence: 5' - CAG TTA GAT TCA - 3'? 5' - ACU UAG AUU GAC - 3' 5' - CUG TTU GUT TCU - 3' 5' - CAG UUA GAU UCA - 3' 5' - UGA AUC UAA CUG - 3' - ANSWER-5' - CAG UUA GAU UCA - 3' The coding DNA and mRNA sequences in transcription will be identical, except for any T in the coding DNA sequence having a corresponding to U in the mRNA sequence. The correct answer is 5'-CAG UUA GAU UCA-3'. What is happening to the pH of the lungs due to exhalation? CO2 decreases from exhalation, thus the pH of the lungs increases. CO2 increases from exhalation, thus the pH of the lungs decreases. CO2 decreases from exhalation, thus the pH of the lungs decreases. CO2 increases from exhalation, thus the pH of the lungs increases. - ANSWER-CO2 decreases from exhalation, thus the pH of the lungs increases. The CO2 in the lung decreases as we exhale and the H+ that was on hemoglobin is recombined with the bicarbonate ion which produces CO2 and H2O. The acidic H+ is now part of water, which is neutral, thus increasing the pH. Hemoglobin acts as a buffer to prevent blood from becoming too acidic by binding ___. excess bicarbonate excess oxygen excess H+ excess CO2 - ANSWER-excess H+ H+ causes blood pH to drop and become acidic. Hemoglobin binds to H+ which not only causes O2 to be released to the tissues but also buffers the blood to allow the pH to increase. What property of hemoglobin, not observed in myoglobin, enables hemoglobin to be an effective oxygen delivery molecule? Parabolic curve Bent shape Cooperativity High iron content - ANSWER-Cooperativity Cooperative requires the subunits of hemoglobin to work together so they all release O2 at the same time or are all binding O2 at the same time. Since myoglobin contains only one subunit, there is no other group to "cooperate" with. Therefore, cooperativity is a feature demonstrated by hemoglobin and not myoglobin. Which feature of hemoglobin makes it an effective oxygen transport molecule? It binds oxygen irreversibly. Its affinity for oxygen is regulated by pH. It has a higher affinity for O2 than myoglobin does. It binds O2 readily at low O2 concentrations. - ANSWER-Its affinity for oxygen is regulated by pH. Hemoglobin is a delivery protein. It has an affinity for O2 that is sensitive to a drop in pH in the blood. The drop in pH directly corresponds to a tissue's need for O2 to be delivered. A tissue that has used its O2 will produce a lot of CO2. The CO2 is then converted to bicarbonate ion (HCO3-) and a proton (H+). The increase in H+ lowers the pH causing hemoglobin to release the O2 to the tissue. This way hemoglobin will only deliver O2 when the pH is low, therefore delivering it to tissues in need. This makes for a very efficient delivery protein. Which letter depicts a pulse oximetry measurement of blood with an O2 saturation of 50%? Letter C Letter A Letter B Letter D - ANSWER-Letter D The solid line represents the hemoglobin curve and the O2 saturation at 50% would be the point halfway up the Y-axis (50%) What is happening to the pH of the muscles due to aerobic exercise? CO2 increases from aerobic metabolism, thus the pH of the muscles decreases. CO2 increases from aerobic metabolism, thus the pH of the muscles increases. CO2 decreases from aerobic metabolism, thus the pH of the muscles decreases. CO2 decreases from aerobic metabolism, thus the pH of the muscles increases. - ANSWER-CO2 increases from aerobic metabolism, thus the pH of the muscles decreases. During aerobic exercise, CO2 is released by the tissues and as a result the CO2 increases. This, in turn, results in the CO2 combining with H2O to produce a bicarbonate ion and a proton (H+). This increase in protons, thus lowers the pH. Identify the factors that favor the deoxygenated form of hemoglobin. Excess protons, high carbon dioxide Excess protons, low carbon dioxide Few protons, low carbon dioxide Few protons, high carbon dioxide - ANSWER-Excess protons, high carbon dioxide Favoring deoxygenated hemoglobin means oxygen is being delivered or released from hemoglobin. When the tissues are using O2 to make ATP, CO2 is a byproduct. As CO2 accumulates in the blood it is converted to bicarbonate and H+ (protons). It's the build-up of H+ that lowers the pH and also initiates the release of oxygen from hemoglobin. Thus, high CO2 means excess H+, lower pH, and delivery of O2. During gluconeogenesis, how many ATP molecules are consumed for each glucose molecule produced? 4 2 1 6 - ANSWER-6 It takes 6 ATP molecules to produce glucose through gluconeogenesis. Which molecule signals the liver to release glucose by breaking down glycogen? Insulin Eicosanoids ATP Glucagon - ANSWER-Glucagon Insulin release increases glycogenesis which forms new glucose. Which molecule is regenerated with the production of lactate during anaerobic metabolism? FAD NAD+ ADH FADH2 - ANSWER-NAD+ Fermentation regenerates NAD+ so that it can be used by glycolysis. During aerobic metabolism, acetyl-CoA is produced from which molecule? Glycerol Lactate Pyruvate Oxaloacetate - ANSWER-Pyruvate Acetyl-COA is produced from the breakdown of pyruvate after glycolysis. Why would a marathon runner consume a large meal full of complex carbohydrates the night before her run? To replenish cellular stores of protein To replenish cellular stores of ketone bodies To replenish cellular stores of glycogen To replenish cellular stores of DNA - ANSWER-To replenish cellular stores of glycogen A large amount of carbohydrates can be stored as glycogen and used during exercise. Why must red blood cells rely solely on glycolysis to make ATP? Red blood cells lack mitochondria. Red blood cells absorb too much oxygen. Red blood cells lack cytoplasm. Red blood cells cannot absorb oxygen. - ANSWER-Red blood cells lack mitochondria Red blood cells lack mitochondria and must depend upon anaerobic metabolism to produce ATP. What are advanced glycation end products (AGEs)? The end products of glycogenolysis Glucose molecules linked to lipids or proteins without the need for enzymes The end products of glycolysis Glucose molecules linked to DNA or RNA without the need for enzymes - ANSWER-Glucose molecules linked to DNA or RNA without the need for enzymes The end product of glycogenolysis is glucose. Why are blood glucose levels high in patients with unmanaged Diabetes Mellitus? Cells move too many GLUT4 transporters to their cell membranes. Cells do not move enough GLUT4 transporters to their cell membranes. Ketogenesis is suppressed in the liver. Ketogenesis is stimulated in the liver. - ANSWER-Cells do not move enough GLUT4 transporters to their cell membranes. GLUT4 transporters allow blood glucose to enter cells which then lowers blood glucose levels. What kind of molecule is glycogen? A simple carbohydrate A triglyceride A protein A complex carbohydrate - ANSWER-A complex carbohydrate What happens in the muscle cells when they respond to insulin? Beta-oxidation of fatty acids is increased. Glycogenolysis is increased. GLUT4 transporters move to the surface of the cell. Gluconeogenesis is increased. - ANSWER-GLUT4 transporters move to the surface of the cell. GLUT4 transporters move to the surface of the cell to allow glucose to enter the cell. What is the function of fermentation during anaerobic metabolism? Fermentation consumes the lactic acid produced by glycolysis. Fermentation regenerates the NADH needed for glycolysis. Fermentation consumes the NAD+ produced by glycolysis. Fermentation regenerates the NAD+ needed for glycolysis. - ANSWER-Fermentation regenerates the NAD+ needed for glycolysis. Fermentation produces NAD+. NAD+ is needed as a substrate for glycolysis. How does the proton gradient affect the pH of the mitochondrial matrix and the intermembrane space? The concentration of protons in the matrix is higher than the concentration in the intermembrane space. The concentration of protons in the intermembrane space is higher than the concentration in the matrix. The pH of the intermembrane space is higher than the pH of the matrix. The pH of the intermembrane space is higher than the pH of the cytoplasm. - ANSWER-The concentration of protons in the intermembrane space is higher than the concentration in the matrix. The intermembrane space of mitochondria has a high concentration of hydrogen ions (H+) which makes the pH very low compared to the matrix. Which diet listed below would result in the production of ketone bodies in a healthy individual? A diet rich in fats and very low in carbohydrates A diet rich in carbohydrates and very low in fat A diet rich in protein and very low in fat A diet rich in carbohydrates and very low in protein - ANSWER-A diet rich in fats and very low in carbohydrates Many of our cell- and tissue-types cannot use fatty acids as a fuel source for making ATP because they lack the capacity to carry out beta-oxidation of fatty acids.When our only (or major) source of energy is fatty acids, the liver will do the job of breaking them down into acetyl CoA for those cells that cannot do it for themselves. However, owing to differences between the chemistry of the blood versus the chemistry of the insides of our cells, we cannot ship acetyl CoA through the bloodstream. Therefore, the liver will assemble the excess acetyl CoA it produces (from beta-oxidation of fatty acids) into the ketone bodies, acetoacetate and 3-hydroxybutyrate. You can think of ketone bodies as transport forms of acetyl units that can travel through the bloodstream and be taken up by those cells that cannot break down fatty acids.Those cells will convert the ketone bodies back to acetyl CoA and use that to keep their citric acid cycles going so they can keep making ATP and stay alive. This, after all, is the whole point of this exercise - to enable us to survive when our only energy source is fatty acids. What stimulates beta-oxidation of fatty acids? The insulin signal Low blood lipid levels High blood glucose concentrations The glucagon signal - ANSWER-The glucagon signal Glucagon is the hormone that signals the hungry state. It tells us that our blood sugar level is too low and that we are fasting or starving and need energy. Thus, glucagon signaling switches cells (especially liver cells) to a program of releasing stored energy. It stimulates the breakdown of our storage molecules (glycogen, triglycerides, and fatty acids.) If a person were eating an absolutely fat-free diet, which vitamins would he or she not get enough of? Vitamins A, D, E, and K Vitamins A, C, E, and K Vitamins A, B, C, and D Vitamins A, C, D, and K - ANSWER-Vitamins A, D, E, and K It is important to recall that there are just four fat-soluble vitamins. All the other vitamins that we need in our diet are water-soluble molecules. The fat-soluble vitamins are A, D, E, and K. If we consume no fats in our diet, we will have no source for these vitamins, since they come in along with the fat we consume. What is a function of this molecule? It is used to store energy. It is a hormone. It is used to maintain membrane fluidity. It is used to make nucleic acids. - ANSWER-It is used to maintain membrane fluidity. In the membranes of our cells, cholesterol acts in a manner similar to antifreeze in our cars' radiators. If we add antifreeze to our coolant, the coolant will freeze at a lower temperature and boil at a higher temperature than it otherwise would, all by itself. That is to say, antifreeze EXPANDS the range over which the coolant stays liquid. Cholesterol has the same effect on our cell membranes, expanding the range over which they remain fluid. The presence of cholesterol in our membranes means that they freeze at lower temperatures and melt at higher temperatures than they would without cholesterol. In fact, this function is so crucial that if we did not have any cholesterol, we would die. Luckily, we chordates (~vertebrates) can make all the cholesterol we need so we never need to eat any. Which class of lipid is shown below? Phospholipid Eicosanoid Cholesterol Triglyceride - ANSWER-Triglyceride The correct answer is 'triglyceride.' The key to recognizing a triglyceride is to remember that a triglyceride has three (3) fatty tails attached to a backbone of glycerol. Those three fatty tails are what the "tri" in triglyceride refers to. The "glyceride", obviously, refers to glycerol. In this course, the only molecules we encounter that have three fatty tails are the triglycerides. If a fish raised in cold water were moved to much warmer water, how would it alter its membrane phospholipids? Membrane phospholipids would include the same fatty acids under either condition. Membrane phospholipids would include more shorter-chain, unsaturated fatty acids. It would add more cholesterol to its membranes. Membrane phospholipids would include more longer-chain, saturated fatty acids. - ANSWER-Membrane phospholipids would include more longer-chain, saturated fatty acids. The key to this question is to understand the relationship between the physical/chemical properties of fatty acids and the fluidity of membranes. An important thing to understand about cell membranes is that they are fluid. They are neither solid nor liquid, really, but something in between, almost like a gel. This fluidity allows the things in the membrane, such as membrane proteins, cholesterol, and phospholipids to move around and this movement is essential for staying alive. If we were to suddenly move a goldfish (for example) from freezing water (0 degrees Celsius) to a summertime pond (say, 25 degrees Celsius), that would be a huge problem. This is because the membrane lipids that were appropriately fluid at 0 C will now be TOO liquid at the warmer 25 C. The same is true going the other way too. A membrane that was appropriately fluid at 25 C would suddenly 'freeze-up' and become too solid at 0 C. The bulk of the material that makes up our cell membranes are the fatty acids that are part of the phospholipids from which the membrane is made. We can get a good idea of how temperature will affect membrane fluidity if we remember how temperature affects fatty acid fluidity. A basic relationship you will want to remember is this: The longer a fatty acid molecule is, the higher its melting point. The more double bonds it has, the lower it's melting point. What this means for membrane fluidity is this: When it's warm outside, and we want our membrane not to get too "runny", we want to make our phospholipids with more saturated, longer chain fatty acids. By contrast, when it is colder, we want to use more shorter-chain and unsaturated fatty acids in our phospholipids to keep our membrane from freezing solid. What is the correct chemical formula of the following fatty acid? CH3(CH2)4CH=CH(CH2)4COOH CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7COOH CH3(CH2)3CH=CH(CH2)3CH=CH(CH2)3CH=CH(CH2)7COOH CH3CH2H=HCH2C=CCH2C=C(CH2)7COOH - ANSWER-CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7COOH The given structure and formula match with 18 total carbons and three double bonds at the omega-3, omega-6, and omega-9 position. Which molecule represents the structural formula CH3(CH2)5CH = CH(CH2)7COOH? - ANSWER-The given formula and structure match with a single double bond at the omega-7 position and 16 total carbons. What is the function of the following molecule? It is used to make proteins. It is used to make cell membranes. It is used to make eicosanoid-signaling molecules. It is used to store energy. - ANSWER-It is used to make cell membranes This structure represents a phospholipid. Remember that a phospholipid only has two fatty tails. The third position of the glycerol backbone has a phosphate group attached to it (recognizable by the "P" for the phosphorus atom, surrounded by four 'O's for oxygen atoms). Phospholipids are the major components of our cell membranes. With their water-loving 'head groups (the phosphate group and whatever may be attached to it) and their very hydrophobic fatty tails, phospholipids will self-assemble to form a lipid bilayer, which is the basis for our membranes. The water-loving heads will orient so they can interact with water while the tails will avoid doing so. The net result is a sort of molecular sandwich: the water-loving headgroups are like the bread on either side and the hydrophobic tails face the interior, like the meat of a sandwich. This way, they can interact with each other and avoid interacting with water. Eicosanoids are cell-signaling molecules that are made from arachidonic acid, a 20-carbon polyunsaturated fatty acid. What is the chemical formula for the following fatty acid? CH3(CH2)4C = C(CH2)2C = C(CH2)7COOH CH3(CH2)4CH = CHCH2CH = CH(CH2)7COOH (CH2)3 = (CH2)4 = (CH2)5OO- CH3(CH2)5CH = CHCH2CH = CH(CH2)7COOH - ANSWER-CH3(CH2)4CH = CHCH2CH = CH(CH2)7COOH In order to answer this question, we need to be comfortable with the different ways we draw molecular structures. Because fatty acids are rather simple (boring, almost) molecules that have lots of carbon (C) and hydrogen atoms (H), it can get really tedious to draw them all out in the usual way. To save ourselves the time and trouble of drawing out all those C's and H's, we simply draw the bonds between the carbon atoms. This gives us a "zig-zag" structure. We do not need to draw the hydrogen atoms or the bonds between them since we know that carbon always makes four bonds to other atoms. If a carbon is shown making two bonds, for example, with other carbons (one to the carbon ahead of it in the chain and one to the carbon behind it), then we know it must be making two other bonds to other atoms. If we do not show those bonds or atoms, we agree that means that those atoms are hydrogen. First, we indicate the carbon with the three hydrogens as CH3. Then, the carbon with two hydrogens is written as CH2. Rather than write out CH2CH2CH2CH2, we simply put parentheses around that group and the subscript number outside the parentheses tells us that this occurs 4 times in a row. Next, we come to the carbons in the double bond, each with one hydrogen. We write this as CH and show the double bond as =. The next carbon, of course, is also CH. Then we have another CH2 (but only one, so we don't heed to use parentheses). Then another double bond CH=CH, and 7 more CH2. Finally, we reach the carboxyl group, which we write as COOH. Thus, our answer is CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH In which situation would altering a protein structure lead to a disease state? A mutation in starch can lead to increased blood glucose levels. A mutation in cholesterol can result in heart disease. A mutation in an omega-3 fatty acid can cause a skin rash. A mutation in acetyl-CoA carboxylase can lead to brain damage. - ANSWER-A mutation in acetyl-CoA carboxylase can lead to brain damage. Acetyl-CoA carboxylase is an enzyme, and enzymes are proteins. We know this is an enzyme because the name ends in "-ase". Pancreatic enzymes normally have an optimal pH of about pH 8.0 or above. The pancreatic juices can become acidified in patients with metabolic acidosis. Pancreatic lipase is an enzyme in pancreatic juice that normally begins the process of breaking down triglycerides into fatty acids. Which process would be impaired in patients who have altered pancreatic lipase function? Protein catabolism Catabolism of triglycerides Ketoacidosis Breakdown of starches by salivary enzymes - ANSWER-Catabolism of triglycerides The function of pancreatic enzymes (enzymes are proteins) is described as "breaking down triglycerides". Catabolism is defined as the breakdown of substances to produce energy. Since the pH of the pancreatic juices has become acidic, this will disrupt hydrogen bonds and ionic bonds in the pancreatic enzymes, causing the protein to denature (unfold) and lose their function of breaking down triglycerides. Which coenzymes are needed during beta-oxidation of fatty acids? NAD+, FAD, and coenzyme Q NAD+, FAD, and coenzyme A NADH, FADH2, and coenzyme A NADH, FADH2, and coenzyme Q - ANSWER-NAD+, FAD, and coenzyme A NAD+, FAD, and coenzyme A are substrates needed for beta-oxidation. Using beta-oxidation, how many molecules of acetyl-CoA are available from the fatty acid shown below? 36 18 9 8 - ANSWER-9 Acetyl-CoA contains 2 carbons so the total number of acetyl-CoA produced will be 9 because fatty acid shown contains 18 carbons.