1. DNA replication is , which al- lows each of the two strands to serve as a for the new strands. b. semiconservative, template 2. Which of the
... [Show More] follow- ing enzymes does NOT assist the DNA poly- merase on the lag- ging strand to overcome its two problems? (Re- call that the DNA poly- merase can only make DNA in the 5'->3' direc- tion, and it must bind a double-stranded nu- cleotide polymer before it can start making its own DNA polymer.) c. Helicase 3. Several components of cigarette smoke, includ- ing benzopyrene, in- sert themselves (inter- calate) into the DNA and lead to sever- al types of mutations such as frameshift mu- tations, including both insertions and deletion. Which of the following repair pathways would DNA replication is semiconservative, meaning that each new duplex has one original (parent) strand and one new strand. Because the two parent strands are separated during replication and the base pairing is predictable, each parent strand can serve as a template for the new strand synthesis. Helicase unwinds the double stranded DNA to allow for replication, but this is not a problem for the DNA polymerase. Nucleotide excision repair is used to repair dele- tions, insertions, and helix-distorting lesions, such as thymine dimers. be used to repair this type of damage? c. Nucleotide Excision Repair 4. Maternal smoking dur- ing pregnancy is haz- ardous yet common in many places. Many studies have associ- ated prenatal smoking to unhealthy physical and psychological out- comes for the baby. Re- searchers know that ma- ternal smoking affects Frame shift mutations are a kind of mutations which result from addition of deletion of a nu- cleotide base resulting in an altered reading frame and ultimately a different protein, than the one the gene originally encoded. Frameshift muta- tions are genetic changes because they alter the DNA sequence, whereas epigenetic changes do not alter the DNA sequence. Epigenetic changes are modifications to genomic structure (not se- quence) that are caused by the external envi- ronment. These environmental factors affect the are epigenetic in nature. overall chromatin structure to allow more or less Which of the following events can be consid- ered epigenetic in na- ture? a. Changes in chromatin structure 5. Blood type is an exam- ple of what type of inher- itance? a. Codominance 6. What is the expected probability that a child will have an autoso- mal dominant disease if their father is heterozy- gous for the allele and "access" to the DNA by gene expression machin- ery to turn the genes "on" or "off". In other words, epigenetics can alter gene expression without changing the underlying DNA sequences. The changes may or may not be heritable, depending on the location and circumstances. The genes that produce the A and B antigen proteins can both be expressed independently, and a heterozygote (someone with both genes) will be produce both A and B proteins - neither will dominate the other. The is an example of codominance. If D is the disease-conferring dominant allele and d is the normal allele, the father has the genotype Dd and the mother's genotype is dd. Each child can only inherit a d allele from their mother, and they have a 50% chance of inheriting the D allele their mother is homozy- gous for the normal al- lele? a. 50% 7. The physical trait of lip protrusion exhibits a characteristic type of inheritance, as shown by the pedigree above. What type of inheritance best describes this in- heritance pattern? a. Incomplete domi- nance 8. The normal sequence of from their father. As a result, the expected proba- bility that their child will inherit the disease is 50%. The correct answer is incomplete dominance. The blending of the large and small lip protrusion into an intermediate, medium lip protrusion, as well as the presence of all three variations in the off- spring, demonstrate a clear example of incom- plete dominance. Silent mutations are those in which the amino a section of the HLA-B27 acid encoded doesn't change as a result of the gene, a genetic mark- er of the inflammato- ry disease Ankylosing spondylitis, is given be- low. Match each muta- tion of the sequence to the type of muta- tion it exhibits. A genet- ic code table is provided for your use in answer- ing the question. 5'- CGG CAG AAU UUA -3' 5'- CAG CAG AAU UUA -3' - Missense mutation 5'- CGG CAG AAA UUU A-3' - Insertion mutation. Missense mutations are those in which the amino acid encoded changes to a different amino acid as a result of the mutation. Nonsense mutations are those in which the amino acid encoded changes to a stop codon as a result of the mutation, yielding a smaller protein. Insertions are the inclusion of extra nucleotides compared to the original sequence. They can re- sult in other mutations, such as nonsense muta- tions. Deletions are the removal of nucleotides com- pared to the original sequence. They can result in other mutations, such as nonsense mutations. 5'- CGG CAG AAC UUA -3' - Silent mutation 5'- CGG CAG AAU UA -3' - Deletion 5'- CGG UAG AAU UUA-3' - Nonsense mu- tation 9. PCR is a powerful tool that can do all of the fol- lowing.... b. detect mutations that lead to disease c. copy small segments of DNA, less than 6kb d. amplify DNA from samples that have just a few cells 10. Which of the following is a required "ingredient" in a PCR reaction? a. DNA nucleotides b. DNA primers c. DNA polymerase 11. Which of the following changes can be detect- ed using PCR? PCR's ability to amplify is powerful, and products can even be generated from samples with just a few cells. PCR is less reliable for amplifying large segments of DNA greater than 6 kb, though some careful changes to the techniques can allow it. The primers used in PCR are made from DNA, rather than RNA. RNA primers are used in DNA replication inside the cell, but the quick degra- dation of RNA makes it less useful for PCR re- actions. Instead, PCR reactions contain primers made of DNA to anneal to the region of DNA that will be amplified and serve as a starting point for DNA polymerase. Differences in DNA sequence can be detect- ed since these changes can alter the ability of primers to anneal to the DNA. They can also be detected by DNA sequencing of the PCR product. a. Differences in DNA se- Epigenetic changes are not detectable via PCR quence b. Insertions c. Deletions 12. A small segment of Kevin's green opsin because they don't affect the DNA sequence. gene is shown be- low. What would be the resulting mRNA se- quence? Kevin's opsin gene at nucleotide positions 936 to 941. 5'-G.C.C.T.A.G-3' (cod- ing strand) 3'-C.G.G.A.T.C-5' (tem- plate strand) c. 5'-GCCUAG-3' 13. The gene for blue opsin protein is locat- ed on chromosome 7, a non-sex chromosome. What kind of inheritance pattern would you ex- pect to see with color blindness due to a mu- tated blue opsin? a. An equal distribution of blue color blindness between males and fe- males. 14. In Kevin's case, a spe- cific kind of mutation in the opsin gene resulted in the premature termi- nation of the translation process. This resulted in a shorter opsin protein This sequence is the same as the coding se- quence, and the "T" bases have been replaced with "U", as is found in RNA. The blue opsin gene is located on a somatic (non-sex) chromosome, and therefore will be in- herited with the same frequency for both sexes. Missense mutations are point mutations that change a single base pair in a codon such that the codon now encodes a different amino acid. In Kevin's case, a nonsense mutation, which is a point mutation that changes a single base pair in a codon to a stop codon resulted in termination than usual. What specif- ic mutation could have caused this? c. Nonsense mutations 15. After consultation with a genetic counselor and an examination of ex- tended family history, the following pedigree was generated. Who did Kevin most likely inherit this mutation from? a. Maternal grandfather 16. The nonsense mutation that causes Kevin's col- or blindness could have resulted from a malfunc- tion of . b. DNA polymerase 17. Primary structure con- sists of the order of in a protein. of the translation signal. This further led to the synthesis of truncated protein. Red-green color blindness occurs more frequent- ly in males than females because these genes are located on the X chromosome, and males inherit only one X chromosome and will express whatev- er alleles are on that chromosome. Females, on the other hand, inherit two X chromosomes, and thus have two alleles for each gene. If one of the alleles is mutated on one X chromosome, there is usually a correct allele on the other that can provide the needed gene. A woman must have two mutated opsin alleles to be color blind, while men need only one. This is known as sex-linked inheritance. Kevin inherited his Y chromosome from his father and his X chromosome from his mother, so his X chromosome with the defective opsin gene most likely came from his maternal grandfather and was passed to him by his mother. Thus, both grandfathers is not correct because his paternal grandfather would have contributed a Y chromosome. Helicase unwinds the DNA during replication, but doesn't add or remove nucleotides. The correct choice is the DNA polymerase. DNA polymerase could have added an erroneous nucleotide, and it's proofreading activity could have missed it. While this type of event is rare, DNA polymerase is known to make a mistake that it doesn't correct about one in one hundred million! Correct Correct! These are held together with bonds that are formed by a reaction. a. Amino acids, peptide, dehydration 18. Several types of side chain interactions sta- bilize the tertiary struc- ture of proteins, includ- ing which of the follow- ing? c. Ion pairs, hydropho- bic interactions, hydro- gen bonds, disulfide bonds 19. Which of the following The stability of a tertiary structure of a protein depends on the various interactions and bonds that occur between the side chains of different amino acids. A secondary structure of protein consists of hy- statements is true about drogen bonds between backbone atoms, includes the secondary structure of proteins? a. It includes alpha he- lices as a common form. c. It includes beta pleat- ed sheets as a common form. d. It involves hydrogen bonding between the backbone atoms. 20. In order to fulfill their function, proteins must fold in proper, three-dimensional con- formations. Which one alpha helices and beta sheets as common forms. Chaperones are helper proteins that ensure prop- er protein folding by stabilizing the polypeptide until the correct structure is fully formed. of the following mole- cules, available in a cell, is likely to help a protein fold properly? b. Chaperone 21. Which of the following is true about a misfolded protein? a. It can be degraded by the cell. b. It can cause protein aggregation. d. It can be the result of denaturation. e. It will lose its normal function. 22. Which of the following amino acids would you expect to find in the in- terior of a protein most often? 23. Which of the following amino acids would you expect to find most of- ten on the exterior sur- face of a protein? 24. Which part(s) of the amino acid below can be involved in peptide bond formation? b. A - Amino group Misfolded proteins can be the result of denatura- tion, are usually degraded by the cell into amino acids, lose their normal function, and may even lead to detrimental protein aggregation, as in the case of Alzheimer's disease. Hydrophobic amino acids want to avoid water, and are most often found in the hydrophobic inte- rior of the protein. While charged and polar amino acids can be found in the interior of the protein, they are also commonly found on the exterior of the protein where they can interact with water. Aspartic acid is a charged amino acid that inter- acts well with water and is often found on the exterior surface of a protein. Peptide bonds are formed between the amino group (A) and carboxylic acid group (B) of amino acids. 25. Which part(s) of the amino acid below con- tributes to tertiary struc- ture stabilization? c. D - R group 26. Drag the name of the protein structure level to the part of the amino acid that primarily con- tributes to that structure level. 27. In the pathway above, which compound could accumulate and trigger feedback inhibition? c. End Product 28. A substrate binds to an enzyme at a specific site, referred to as a(n) . The backbone carboxylic acid is involved in pep- tide bond formation and primary structure. Ter- tiary structure is stabilized primarily by interac- tions between different R groups. Correct! The backbone of the amino acid partic- ipates primarily in primary and secondary struc- ture while the R group stabilizes tertiary and qua- ternary structure. The substrate is the start of the pathway, and therefore cannot inhibit anything before itself. While the amount of substrate could be reduced, and thereby slow product formation, the substrate itself is not part of feedback inhibition. In order to regular the amount of a product from a pathway, cells often use feedback inhibition. When the end product is no longer needed (or needed at a reduced rate), the product begins to accumulate and bind to enzymes in the pathway. This binding can inhibit the action of that enzyme by noncom- petitive inhibition, thus slowing the pathway and reducing the amount of product made. The active site of an enzyme is the site where the substrate specifically binds to the enzyme and the reaction is carried out. b. Active site 29. You are in charge of de- signing a drug that in- hibits the activity of a specific enzyme. An im- portant criteria for the drug selection is to en- sure that the drug is directly in competition with the original sub- strate when binding to the active site of the enzyme. Which of the following kind of in- hibitor would be an ideal choice? a. Competitive inhibitor 30. Which of the following factors can affect the protein folding and ac- tivity of an enzyme? a. pH b. Heat d. Reducing agents 31. One way a cell can avoid overproduction of a mol- ecule is by using a par- ticular type of inhibi- tion in which the mole- cule itself acts as an in- hibitor for an enzyme in its production pathway. This type regulation is A competitive inhibitor competes with the sub- strate for access to enzyme's active site and in- hibits the enzyme. Several factors can denature a protein and affect enzyme activity, including: reducing agents, pH, heat, and salt concentration. The purpose of feedback inhibition in a cell is to prevent overproduction of the product. Feedback inhibition works by inhibiting an enzyme using the product of the very same reaction the enzyme catalyzes or a product from the same enzymatic pathway. known as inhibition. b. feedback 32. In the pathway above, what will happen to the levels of Second Inter- mediate Substrate if En- zyme 1 is inhibited as a result of feedback inhi- bition? c. Decrease 33. What type of enzyme adds a phosphate to an- other molecule? b. Kinase 34. Which of the following are possible effect(s) that phosphorylation/de- phosphorylation can have on the activity of an enzyme? Inhibition of an enzyme earlier in the pathway will prevent the product(s) of that enzyme from forming, thereby preventing the action of subse- quent enzymes and formation of their respective products. For example, inhibition of Enzyme 1 would lead to a decrease in First Intermediate Substrate, which then reduces the levels of Sec- ond Intermediate Substrate because Enzyme 2 has less substrate on which to act. Thus, the levels of Second Intermediate Substrate will decrease as a result of feedback inhibition. Protease enzymes break peptide bonds to digest proteins. Kinases add phosphates to molecules while phosphatases remove them. Kinases are enzymes that use a phosphate donor, usually ATP, to add a phosphate to another molecule. Phosphatases have the opposite role - they remove phosphate groups from molecules. Some enzymes are activated by phosphorylation while others are deactivated by the addition of a phosphate. Adding or removing a phosphate group to an en- zyme affects the activity, and the type of effect is specific to the enzyme. For some enzymes, b. Turn the enzyme "off". c. Turn the enzyme "on". 35. In the enzymatic cy- cle, the enzyme shape is changed when [Sub- strate binds] and re- sumes its original shape when [Product is re- leased] after [Substrate is converted to product]. 36. The rate of an en- zyme reaction can be in- creased by which of the following: b. Decreasing the activa- tion energy of the reac- tion 37. Statins are a class of medications that target the enzyme HMG-CoA reductase in the path- way for cholesterol syn- thesis, as shown below. Compared to a patient that is not on a statin medication, In a patient taking a statin medica- phosphorylation activates the activity of the en- zyme, while other enzymes can be deactivated by phosphorylation. An enzyme is a catalyst, meaning that it can catalyze the same reaction over and over again. Substrate binding can alter the conformation (or shape) of an enzyme, so it must resume its orig- inal shape when product is released, after sub- strate is converted to product, to allow it to per- form the same reaction again. Enzymes do not directly affect the temperature of a cell. The rate (or speed) of a chemical reaction de- pends on the size of the activation energy barrier. Increasing temperature can provide the energy needed to get over the activation barrier, while lowering the temperature will reduce the avail- able energy to overcome the barrier. However, enzymes cannot affect the temperature around them. Instead, enzymes decrease the activation energy of a reaction, thus allowing more reactions to take place at any given temperature. Recall that inhibiting an enzyme generally leads to an increased concentration of the substrate of the inhibited enzyme and a decrease in its product (or products further down the pathway). tion the concentration of cholesterol would [decrease], the con- centration of HMG-CoA would [increase], and the concentration of Acetoacetyl CoA would [stay the same]. 38. Deedra's dad has Alzheimer's Disease (AD) which is caused by a change in structure. d. protein 39. Alzheimer's disease is caused by aggregation of the Amyloid beta pep- tide and tangle forma- tion by the tau protein. What kinds of amino acids are likely to drive the formation of these protein aggregates? c. Hydrophobic 40. Below is a schemat- ic showing the syn- thesis and degradation of acetylcholine, an im- portant neurotransmit- ter. If you were to de- sign a drug to keep the acetylcholine con- AD is a manifestation of neuronal death and dam- age caused by malformed protein structures. Proteins are crucial molecules with various func- tions in the cells. The structure of a protein di- rectly impacts its function. Deedra's father has Alzheimers which is caused by a disturbance in the protein structure of at least two proteins, amy- loid beta and tau. Hydrophobic amino acids strive to avoid water and like to interact with other hydrophobic amino acids. Generally they are found in the interior of a protein. But, when there are too many misformed peptides with exposed hydrophobic surfaces, they tend to attract each other and result in protein aggregation. Acetylcholine is the substrate of the reaction and the desired chemical to preserve. However, since acetylcholine does not control its own produc- tion, it is ideal to target enzymes that can im- pact its concentration. Enzymes that produce ace- tycholine could be targeted for increased pro- duction, but this type of regulation is difficult to achieve with a drug. Instead, drugs usually inhibit centration high, where would you target a drug inhibitor? b. Acetylcholinesterase 41. After learning about specific enzymes. Acetylcholinesterase is the enzyme which di- gests acetylcholine to acetate and choline. Acetyl- cholinesterase inhibition will decrease degrada- tion of the acetylcholine, which will therefore in- crease the concentration of acetylcholine. Sup- plementation with excess acetate or choline may also help increase choline acetyltransferase ac- tivity and perhaps increase acetylcholine concen- trations, but the only productive target for a drug inhibitor in this scheme is acetylcholinesterase. The more we are able to understand the molecu- Alzheimer's disease and lar details of a disease and the enzymes involved one of the drugs used in its treatment, what other drug design strategies could you use to offer a treatment for AD? a. Drugs that block the production of amyloid beta peptide b. Drugs that break up the amyloid plaques d. Drugs that inhibit the aggregation of protein structures 42. Hemoglobin's coopera- tive binding behavior is key to its physiologi- in the disease pathway, the more opportunities one has to target the key molecules. Any of the options here could be a possible treatment in the case of AD. Binding of the first oxygen molecule to a hemo- globin subunit greatly enhances binding of the re- maining oxygen molecules. All subunits communi- cal function. The bind- ing of which of the fol- lowing molecules influ- ences this behavior? b. Oxygen 43. Carbonic anhydrase is an important present in the red blood cells that aids in efficient transportation of carbon dioxide in the form of , from tissues to lungs. b. enzyme, bicarbonate ions 44. Hemoglobin's ability to bind or release oxy- gen depends on the pH of the environment. This behavior is known as the Bohr effect. Consid- ering this, which of the following statements is true? a. Hemoglobin binds to the oxygen at high pH and releases oxygen at low pH. 45. Hemoglobin acts as a buffer and controls the pH of the blood by bind- ing to . cate with each other and work in a unified fashion. This property is referred to as cooperativity. Carbonic anhydrase is an enzyme which cat- alyzes the conversion of carbon dioxide to bi- carbonate ions. However, it does not transport carbon dioxide as is, but rather transports carbon dioxide by converting it to bicarbonate ions. This mechanism enables efficient transport of carbon dioxide by warding off bubble formation in the blood. pH is a measure of H+ ion concentration. High pH indicates a more basic environment in which the H+ ion concentration is low, as seen in the lungs. Under these conditions, hemoglobin binds more oxygen. Low pH indicates an acidic environment in which the H+ ion concentration is high. Such conditions encourage the binding of H+ ions to hemoglobin and stabilizing a form of hemoglobin which decreases the affinity for oxygen, resulting in release of oxygen. H+ ions contribute to the acidity of blood. By bind- ing to these ions, hemoglobin is able to act as a buffer and maintain the pH at an appropriate range. d. H+ ions. 46. In the lungs, the CO2 concentration is [low] and the pH is [high], while in the tissues, the CO2 concentration is [high] and the pH is [low]. 47. In locations where the pH is low, hemoglobin will be in the [T] state, allowing it to [release] oxygen more effectively. 48. Increased levels of 2,3-BPG will [decrease] the affinity for oxygen by binding to and sta- bilizing the [T] state of hemoglobin. The action of 2,3-BPG allows he- moglobin to [release] oxygen more effectively, which is similar to the effect of [low] pH on the action of hemoglobin. 49. In comparison to adult hemoglobin, fetal hemo- globin has a [higher] affinity for oxygen be- cause it [doesn't bind] 2,3-BPG well. 50. Structurally, hemoglo- bin can bind mole- In the lungs, CO2 concentration are low, resulting in decreased carbonic acid and higher pH. This creates a more basic (or alkaline) environment that promotes oxygen binding.In the tissues, CO2 concentration tends to be high, resulting in in- creased carbonic acid which lowers the pH. This creates an acidic environment that promotes oxy- gen release. At low pH, hemoglobin will be in the T, or tense, state to allow it to release oxygen more effectively to oxygen-starved tissues. Increased levels of 2,3-BPG will Incorrect the affinity for oxygen by binding to and stabilizing the Correct state of hemoglobin. The action of 2,3-BPG allows hemoglobin to Incorrect oxygen more effectively, which is similar to the effect of Correct pH on the action of hemoglobin. Fetal hemoglobin has a higher affinity for oxygen than adult hemoglobin because its structure does not bind 2,3-BPG well. As a result, the T state of fetal hemoglobin is not stabilized and it remains more readily in the R state, ready to bind oxygen even at lower concentrations. Hemoglobin has four subunits, all of which contain heme prosthetic groups. Each heme can bind an cules of oxygen, all of which could potentially be replaced by carbon monoxide. b. 4 51. Carbon monoxide out- competes oxygen for attachment to the group of he- moglobin where it is per- manently, covalently at- tached. d. heme 52. As carbon monoxide binds to hemoglobin, the protein subunits change conformation to allow carbon monox- ide to bind faster. This process is called . d. positive cooperativity 53. Carbon monoxide shifts the Oxygen-Hemoglo- bin Dissociation Curve to the left, while carbon dioxide shifts this curve to the right. Based on your knowledge of the Bohr Effect, which of the following statements is true? iron atom, and each iron atom can bind an oxygen molecule resulting in the binding of a total of four oxygen molecules. All four oxygen molecules can be potentially replaced by carbon monoxide. Carbon monoxide has a higher binding affinity than oxygen to the heme group of hemoglobin. Such binding is irreversible. Carbon monoxide does competitively inhibit the overall function of the hemoglobin protein. Note that the question is referring to how the binding of one carbon monoxide molecule influ- ences the conformation of the other subunits and increases the overall binding affinity for the he- moglobin molecule. Such a cooperative behavior exhibited by the subunits is referred to as positive cooperativity. Carbon dioxide, released by the respiring cells, combines with the water molecules of plasma to form carbonic acid. Carbonic acid increases the H+ ion concentration (low or acidic pH). H+ ions stabilize the deoxy-conformation (T state) of he- moglobin and allow oxygen release and delivery to the tissues. On the other hand, carbon monox- ide, which has a very high affinity for hemoglobin than oxygen, binds and maintains hemoglobin in oxy-conformation (R state). Oxy-conformation of HB is amenable to further binding of carbon b. Carbon dioxide de- creases hemoglobin's affinity for oxygen, while carbon monoxide in- creases it 54. Glucose is a monosac- charide. It can be used as the raw material to build which one of the following molecules? d. Glycogen 55. Glycolysis is the conver- sion of glucose to pyru- vate. Is this a catabolic process or an anabolic process? d. Catabolic 56. Which of the following monoxide and shifts the Oxygen-Hemoglobin Dis- sociation Curve to the left. Glycogen is a polysaccharide made up glucose units. Anabolic pathways refer to the synthesis of larger molecules from smaller ones, and they often re- quire energy input from ATP. Catabolic pathways refer to the breakdown of larger molecules into smaller ones and often pro- duce ATP as a result. Glycolysis is a catabol- ic process. During glycolysis, a molecule of glu- cose is broken down into two molecules of pyru- vate and two molecules of ATP are made in the process. Glycolysis final products are two molecules of is an accurate statement pyruvate. about glycolysis? d. The final products of glycolysis are two mole- cules of pyruvate. 57. Glucose is not the only food source that can make acetyl Co-A. 58. In the presence of oxy- gen, the end products Various sources such as amino acids and fatty acids, in addition to glucose, can be catabolized to produce acetyl Co-A. of glycolysis have the potential to enter which one of the following pathways? c. Citric acid cycle 59. In human cells, gly- colysis takes place in the cytoplasm. The prod- ucts of the glycol- ysis will enter the to continue with aerobic respiration. b. mitochondria 60. NADH and FADH2 are two cofactors made in the citric acid cycle. They then donate the In the presence of oxygen, pyruvate forms acetyl Co-A and enters the citric acid cycle, while fer- mentation occurs in the absence of oxygen. The products of glycolysis enter the mitochondria to continue with aerobic respiration. NADH and FADH2 donate electrons to the elec- tron transport chain. they gained in the citric acid cycle to the electron d. electrons 61. Aerobic respiration re- quires oxygen. Oxygen has a role in which of the following pathways: b. Electron transport chain 62. During periods of stren- uous exercise the mus- cle cells can become anaerobic. Without oxy- gen, the electron trans- port chain cannot con- Gluconeogenesis is building of new glucose mol- ecules and is not involved in aerobic respiration. Oxygen has a role as the terminal electron accep- tor in the electron transport chain. In the absence of oxygen, pyruvate is reduced to form lactic acid by fermentation. tinue and the citric acid cycle slows down. In such situations, how do cells make ATP? a. Pyruvate is trans- formed to lactate. 63. The citric acid cycle could be inhibited by high concentrations of [NADH] due to feedback inhibition. 64. Cyanide binds to com- plex IV of the electron transport chain, pre- venting electron trans- port. Which of the fol- lowing would be the most immediate ef- fect(s) of cyanide con- sumption a. Oxygen would not be consumed by the ETC. b. ADP will build up c. Protons will not be pumped to the inter- membrane space 65. In an exercising mus- cle cell under anaer- obic conditions, Incor- rect from glycolysis is turned into --- in or- der to regenerate --- for further rounds of gly- colysis, which produces The citric acid cycle produces large amounts of NADH. If NADH is already abundant in the cell, the enzymes of the citric acid cycle that produce NADH will be inhibited by this product and slow the cycle. Cyanide binds to the last complex of the elec- tron transport chain, complex IV, which usually transfers electrongs to oxygen to create water and keep the flow of electrons going. By blocking complex IV, oxygen will not be consumed, which will prevent electron transport. As a result, NADH cannot be consumed by the ETC and will build up, and protons cannot be pumped into the intermem- brane space of the mitrochondria. As a result, ATP production by ATP synthase will slow. In an exercising muscle cell under anaerobic con- ditions, [pyruvate] from glycolysis is turned into [lactate] in order to regenerate [NAD+] for further rounds of glycolysis, which produces [2] ATP per round for the muscle cell to use. The Cori cycle allows [lactate] to be turned into [glucose] via gluconeogenesis in the liver at the expense of --- ATP per round for the muscle cell to use. The Cori cycle allows --- to be turned into --- via gluconeogenesis in the liver at the expense of --- ATP. This means that, overall, the Cori cy- cle creates a deficit of ---ATP. 66. Which one of the following non-carbohy- drate molecules has the potential to make a glu- cose molecule? a. Amino acids 67. Carbohydrate loading is a common practice among endurance ath- letes. Often, three days before a big endurance event, they are known to eat large amounts of complex carbohy- drates which leads to the storage of glyco- gen in their mus- cles. The stored glyco- gen increases their en- durance by providing a steady supply of glu- cose during the event by the process of . d. glycogenolysis [6] ATP. This means that, overall, the Cori cycle creates a deficit of [4] ATP. Some kinds of amino acids can create sugar mol- ecules by the process of gluconeogenesis. Glycogenolysis is the breakdown of glycogen to release glucose. A further breakdown of glucose results in ATP production which helps with the prolonged physical activity in the athletes. 68. Insulin controls both carbohydrate and fat metabolism. Which of the following state- ments describes the ef- fects of insulin? a. Stimulates fatty acid production and storage c. Stimulates the glu- cose uptake d. Inhibits the glycogen breakdown 69. True or False: The defin- ing characteristic of dia- betes is the lack of suffi- cient insulin production. b. False 70. The complications of di- abetes result from the accumulation of which of the following: b. Advanced Glycation End-products (AGEs) 71. The pancreas can reg- ulate and maintain glu- cose homeostasis by secreting different hor- mones in response to varying blood glucose Insulin is responsible for several controls such as stimulating glucose uptake, inhibiting the glyco- gen breakdown and stimulating fat storage mol- ecule synthesis. Not all cases of diabetes are due to a lack of in- sulin production. Diabetes could result from cells resistance to the insulin. Glycogen is the glucose storage molecule. As such, it does not directly cause any complications in the diabetic patients. Increased glucose levels will result in glycation of proteins, which will affect the function of a protein by making it more stiff and inflexible. Glycation can also lead to additional reactions that crosslink pro- teins together into advanced glycation end-prod- ucts (AGEs), which can impair the function of the proteins and their associated organs. Yes, insulin is secreted by the pancreas in re- sponse to high glucose concentrations. But insulin is already listed for the subject. The other hor- mone secreted by the pancreas in response to glucose levels is the hormone glucagon. levels. The pancreas se- cretes insulin in re- sponse to high blood glucose levels, whereas is se- creted in response to glu- cose levels. b. glucagon, low 72. While treatment with Metformin is beneficial for many individuals with type 2 diabetes, it does increase the risk of lactic acidosis (ac- cumulation of lactate in the blood that results in lower blood pH) in those who take it. What effect of metformin di- rectly contributes to this potential risk? a. Decreased gluconeo- genesis in the liver 73. A risk factor associat- ed with the use of met- formin in the treatment of diabetes is lactic aci- dosis. Acidosis is asso- ciated with [a decrease] in blood pH, due to [an increase] in the concen- tration of hydrogen ions (H+). Given this change in blood pH, hemoglo- Reduction in liver gluconeogenesis can help low- er blood glucose levels in type 2 diabetics. How- ever, it is not without some risk. Recall from the section on the Cori cycle that gluconeogenesis is essential for the conversion of lactate back to glucose after anaerobic fermentation in muscle and red blood cells. Treatment with metformin can, therefore, increase the risk of elevated lactate levels in the blood, a condition known as lactic acidosis because it lowers the blood pH. Acidosis is associated with a decrease in blood pH, due to an increase in the concentration of hydrogen ions (H+). Given this change in blood pH, hemoglobin is more likely to release oxygen. bin is more likely to [re- lease] oxygen. 74. Which one of the follow- ing could result in better outcomes for type 2 dia- betics? c. An increased number of GluT4 transporters in the cell membrane 75. Emma was prescribed metformin which is used to control blood glu- cose levels. One di- rect effect of metformin is inhibition of the ------------------------- path- way in the liver. c. Gluconeogenesis 76. Emma has an A1C level of 7.0% in her blood. A1C is a form of . a. Hemoglobin 77. Myoglobin and hemo- globin each have differ- ent numbers of subunits and this affects their respective oxygen stor- age and delivery capa- bilities. How many sub- units do myoglobin and hemoglobin each have? Gluconeogenesis is creation of more glucose and increasing the rate of gluconeogenesis will raise blood glucose and worsen the symptoms. Type 2 diabetes is often a result of decreased glucose uptake by the target cells. Increasing the translo- cation of GluT4 transporters can help lower blood sugar by allowing more glucose into the target cells. Metformin is a drug of choice for treating Type 2 diabetes. One of the direct actions of the drug in controlling the blood sugar levels is by inhibiting gluconeogenesis in the liver. In the presence of excessive amounts of glucose in the blood, glucose makes a covalent bond to the hemoglobin protein in a reaction known as glycation. A1C is the glycated form of hemoglobin. It's true that myoglobin consists of one subunit. However, hemoglobin consists of four subunits which contribute to its role in oxygen delivery. b. One, Four 78. Addition of Oxygen binding alters the shape of the iron-heme caolmteprsletxh, eansdhathpeereoffotrheeitisroling-hhteambseocroptmiopnlepxro, apn- d th erties as indicated by a color change. c. oxygen 79. Oxygen binding alters the structure of an entire hemoglobin tetramer, so the structures of oxyhe- moglobin and deoxyhe- moglobin are noticeably different. The oxyhemo- globin conformation is specifically referred to as the state, whereas the deoxyhe- moglobin conformation is referred to as the state. b. Relaxed or R and Tense or T 80. Patients with sickle cell anemia have atyp- ical hemoglobin, which will distort the red blood cells into sick- le shape during oxy- gen delivery. The substi- tution of a hydrophilic amino acid with a amino acid in hemoglobin sub- units results in the poly- merization of hemoglo- Oxygen binding causes a marked change in he- moglobin structure. In the presence of oxygen, hemoglobin is in the Relaxed state or R state. In the absence of oxygen, hemoglobin is in the Tense state or T state. Hydrophilic amino acids are water loving amino acids found on the surface of protein with no ten- dency towards polymerization. In case of sickle cell anemia, a hydrophilic amino acid - glutamate is substituted by a non polar-hydrophobic amino acid-valine, leading to polymerization of hemoglo- bin, fibril formation and eventually sickling of the red blood cells. bin, leading to the sick- ling of red blood cells. c. hydrophobic 81. Hemoglobin consists of four protein subunits. Each subunit contains [a heme group] that holds [an iron atom], which can bind to [an oxygen moleule]. 82. When binding hemo- globin or myoglobin, CO binds in place of . a. Oxygen 83. Which one of these mol- ecules is not considered a lipid? a. Vitamin C 84. The acid portion of a fat- ty acid corresponds to which one of the follow- ing groups/molecules? c. COOH 85. Saturated fatty acids have more than unsaturated fatty acids of the same length. Hemoglobin protein consists of four protein sub- units. Each subunit contains a heme group which binds an iron atom. The iron binds to oxygen, which can then be transported by hemoglobin. CO binds in place of oxygen. Vitamin C is a water-soluble vitamin and is not a lipid. CH3 is considered as a methyl group and not an acid group. COOH is a carboxylic acid group. A fatty acid con- sists of a hydrocarbon chain bound to the carbon atom of the COOH. Both saturated and unsaturated fatty acids have the same number of oxygen atoms irrespective of their lengths because the only oxygen atoms are present as a part of the carboxylic acid group. A carbon must always make four bonds. In an unsaturated fatty acid, the presence of double b. Hydrogen atoms 86. How many hydrogens can be accommodated by the carbon (in bold) that is participating in the double bond: -C-C=C-C- a. One 87. True or False: Phospho- lipids that are part of a cell membrane contain only unsaturated fatty acids. b. False 88. Which of the following is the fat storage mole- cule? d. Triglyceride 89. Due to their amphipath- ic nature, phospholipids can form micelles and transport other lipids. The fatty acid chains that make up the phos- pholipids are the ba- sis for their non-po- lar character, and the is/are the bonds between carbon atoms reduces the num- ber of bonds available to bond with hydrogen. Consequently, an unsaturated fatty acid has fewer hydrogen atoms, than a saturated fatty acid of the same length. Recall that carbon can form four bonds. The car- bons that are participating in a double bond will bind to one hydrogen each. Notice that four out of three bond positions are already occupied by the carbon-carbon bonds. Phospholipids that make up the cell membrane contain two fatty acids attached to a glycerol back- bone. An additional phosphate polar group bonds to the molecule and creates a polar end of the molecule. The two long chain fatty acids that are a part of the phospholipid can be of either variety - saturated or unsaturated fatty acids. The specific fat storage molecule is a triglyceride, which consists of a glycerol backbone bound to three fatty acid chains. Phospholipids have a phosphate group attached to the glycerol backbone that is the basis for their polar characteristic. basis of its polar nature. d. phosphate group 90. True or False: Acetyl-CoA can only be made from the beta oxi- dation of fatty acids. a. False 91. Beta oxidation of a C14 fatty acid will make acetyl-CoA units after rounds of beta-oxidation. a. seven, six 92. Mammals and plants have the ability to syn- thesize their fatty acids. Which one of the follow- ing molecules is the pre- cursor to building fatty acids? d. Acetyl-CoA 93. Acetyl-CoA, a central Acetyl-CoA is central to metabolism. Carbo- hydrates, proteins, and fats can all produce Acetyl-CoA, though they use different pathways to make it. Each round of beta-oxidation results in the pro- duction of a two-carbon acetyl-CoA units, so the number of carbons in a fatty acid chain divided by two is how many acetyl-CoA units are made. The number of rounds needed is one less than the number of acetyl-CoA units produced be- cause the last round of beta-oxidation produces two acetyl-CoA units. Thus, a C14 fatty acid will altogether make seven acetyl-CoA units after six rounds of beta oxidation. Acetyl-CoA. Acetyl-CoA, which is the common product of many biomolecules, acts as a precur- sor for the synthesis of fatty acids. It provides the carbon atoms used in the synthesis of fatty acids. Acetyl-CoA combines with oxaloacetate to pro- molecule in metabolism, duce Citrate. Note that citrate is the first product is produced in the mi- tochondrial matrix. How- ever, it is used for the synthesis of fat- ty acids in the cy- tosol. The transport of produced in the citric acid cycle by the same reaction. Cells have the ability to mobilize cit- rate across the mitochondrial membrane. Thus, acetyl-CoA moves out of the mitochondria to the cytosol in the form of citrate. Once in the cytosol, the citrate is converted back to oxaloacetate and the acetyl-CoA from the matrix to the cytosol occurs by the con- version of acetyl-CoA to . a. citrate 94. The inability to tol- erate long-chain fatty acids in the diet could stem from a defect in . c. beta-oxidation of long-chain fatty acids 95. Ketoacidosis is a dan- gerous outcome of un- controlled diabetes. It is a result of the buildup of ketone bodies. What is the primary metabol- ic fuel that results in the build-up of the ketone bodies? c. Fatty Acids acetyl-CoA, and acetyl-CoA is utilized in fatty acid synthesis. All fats, carbohydrates, and proteins make acetyl-CoA, which is the precursor for the fatty acid synthesis. Intolerance towards a certain food stems from the inability to digest a specific bio- molecule in that particular food. Beta-oxidation of long-chain fatty acids. Beta-ox- idation is a pathway that breaks down fatty acids into two carbon acetyl-CoA units that enter the cit- ric acid cycle and electron transport chain, result- ing in ATP production. Beta-Oxidation enzymes can be specific to chain lengths. Thus, a defect in an enzyme that is specific for long chain fatty acid beta-oxidation manifests as an intolerance towards foods containing long chain fatty acids. Glucose uptake by cells is impaired during dia- betes, so glucose is not available for ATP produc- tion. Also, glucose does not contribute to ketone body formation. Acetyl-CoA makes ketone bodies. During dia- betes and under starvation conditions, the pri- mary source for the acetyl-CoA is the fatty acids released from triglycerides. Acetyl-CoA typical- ly enters the citric acid cycle or creates normal amounts of ketone bodies. However, in diabetic conditions, the citric acid cycle intermediates are being drawn out for the gluconeogenesis path- ways. In such a situation, acetyl-CoA has limited oxaloacetate to enter the citric acid cycle. Instead, 96. Below is a diagram of acetyl-CoA creates more ketone bodies, which acidifies the blood. Excessive production of these ketone bodies causes ketoacidosis. The four steps of beta oxidation each involve a the four steps of the beta small molecule other than the fatty acid in the oxidation cycle. Drag the correct small molecule to the step in the path- way to which it con- tributes. 97. Fatty acid synthesis oc- curs in the ---, so its building block mole- process, as shown below: The correct answer is: Fatty acid synthesis oc- curs in the [cytosol], so its building block mole- cule acetyl-CoA must be transported out of the cule acetyl-CoA must be [mitochondria] and into the [cytosol] using the transported out of the --- and into the --- using the [citrate] transport system. The acetyl-CoA is then combined with [carbon dioxide] with the help of --- transport system. The the coenzyme molecule [biotin] in order to form acetyl-CoA is then com- bined with --- with the help of the coenzyme molecule --- in order to form --- in the first com- mitted step of fatty acid synthesis. 98. If Zoey continues on her path of a no-fat diet, she could have a very poor wound healing re- sponse. Which one of the following lipid mole- [malonyl CoA] in the first committed step of fatty acid synthesis. Recall that phospholipids are crucial for cell mem- brane structure and fluidity. Additionally, in re- sponse to certain hormonal signals, certain fat- ty acids in the phospholipids are released from the cell membranes to produce other key lipid molecules such as eicosanoids. Eicosanoids play cules can have an affect several roles in inflammation, fever, immunity and on this response? a. Phospholipid blood coagulation. 99. The nutritionist explains the importance of including diets rich in essential fatty acids. Certain kinds of polyunsaturated fatty acids such as omega-3 and omega-6 fatty acids are essential for health. The body can build most of the saturat- ed fatty acids from the raw materials (acetyl-CoA) The essential fatty acids provided by other metabolites. However, we do are fatty acids. b. Polyunsaturated 100. True or False: Zoey can overcome any and all vi- tamin related deficien- cies by supplementing her zero-fat diet with a daily supplement of mul- tivitamins. b. False not have the enzymes to add double bonds at cer- tain positions (after C9, from the carboxylic acid end) in fatty acids as are found in the omega 3 and omega 6 fatty acids. Consequently, our body needs to obtain them from the diet. Essential fatty acids, as a part of phospholipids, play various key structural and functional roles. Fats are necessary to deliver the fat-soluble vit- amins such as A, D, E and K. Absence of fat in the diet impact the absorption and delivery of the fat soluble vitamins that result in vitamin related deficiencies. [Show Less]