Solution Manual for Human Physiology, 16th Edition

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1Fox:Human Physiology, 16thEditionChapter 1 Answers to In-chapter QuestionsTest Your Understanding13.Epithelial membranes can be simple (one cell deep) or stratified (many cells deep).Simple membranes providea barrier that allows selective transport from the externalenvironment to the tissues that underlie the epithelial membrane. Stratifiedmembranes providemore protection but not transport. Stratified squamous keratinizedepithelium of the skin provides the greatest degree of protection.14.Bone, blood, and the dermis are all connective tissues because they have abundantextracellular material. However, they differ on what material comprises the extracellularmaterial and regarding their cells. The extra cellular material of blood is thefluid andplasma. The dermis contains connective tissue fibers. Bone also has connective tissuefibers, but the extracellular matrix is calcified.15.A single negative feedback loop helps to maintainhomeostasis by counteractingchanges in one direction. For example, insulin helps to prevent blood glucose fromgetting too high. A different hormone counteracts too great a fall in blood glucose.When these two negative feedback mechanisms act antagonistically to each other, theyafford a finer degree of control in maintaining homeostasis.16.Insulin is secreted by the pancreatic islets in response to a rise in blood glucose above itsset point. The insulin then acts to lower the blood glucose by promoting the movementof glucose from the blood into tissue cells. As a result, the blood glucose is loweredback to the normal range. The lowering of blood glucose causes the pancreatic islets toreduce their secretion of insulin.17.In phase I clinical trials, the drug is tested on healthy human volunteers. In phase IIclinical trials,the drug is tested on the target population (those for whom the drug isintended). Phase III clinical trials involve a much larger and more diverse testpopulation. Phase IV tests other potential uses of the drug. However, before a drug everreaches clinical trials, its development and early testing require animal research.18.Claude Barnard proposed that the conditions within the body maintain constancydespite challenges to that constancy. This idea led to the concept of homeostasis andthe concept that physiological mechanisms exist to maintain that internal constancy.

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2Test Your Analytical Ability19.Positive feedback mechanisms accentuate changes. Thus, if the body temperature wereto rise, positive feedback mechanisms would make it become even hotter. Conversely, ifbody temperature were to fall, positive feedback mechanisms would make it becomeeven colder. A fall in blood glucose would become an even greater fall in response topositive feedback. Homeostasis could not be maintained and life could not exist underthese conditions.20.Negative feedback mechanismsmay have been acting earlier,but they’re effects don’tbecome evident until 40 minutes after injection, when blood glucose starts to rise. Theinitial blood glucose concentration appears to be restored 120 minutes after theinjection. Quantitative measurements are needed to determinethe effect of insulin andthe effectiveness of antagonistic mechanisms in this instance, and in all cases wherephysiological mechanisms are being evaluated.21.There must be communication between the intracellular fluid andthe extracellular fluidin order for a cellto obtain nutrients and eliminate wastes. There must also becommunication between the interstitial fluid and the blood plasma in order for thesemolecules to be obtained and distributed to the cells and for homeostasis to bemaintained.22.If the blood pressure has fallen so low that the person has collapsed, one would expect,on the basis of homeostasis and negative feedback, that mechanisms would be set inmotion to help raise the blood pressure. Logically, a more rapid and powerful heartbeatwould have that effect and also cause a more rapid pulse.23.Adult stem cells are foundin thered bone marrow, wherethey give rise to the bloodcells. Adult stem cells are also found in skeletal muscles and in the adult brain, as well asin hair follicles and other locations. Unlike embryonic stem cells, adult stem cells cannotform all of the body tissues and a complete organism.Test Your Quantitative Ability24.The set point, as calculated as the average value, is 37.15 degrees.25.The range of values is 1.10 degrees.26.The sensitivity is 0.55 degrees.

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1Fox:Human Physiology, 16thEditionChapter 2Answers to In-chapter QuestionsTest Your Understanding16.Nonpolar covalent bonds are formed when the valence electrons are shared equally.Polar covalent bonds are formed when the valenceelectrons are shared unequally.Ionic bonds are formed by negative and positive ions that are attracted to each other,not by the sharing of electrons.17.Anacidis a molecule that can donate protons (H+) to solution. Abaseis a molecule thatcan directly or indirectly remove protons from the solution.18.Starch in a potato can be hydrolyzed in the digestive system into its glucose subunits,which can be absorbed into the blood. The liver can remove glucose from the blood andthrough dehydration synthesis reaction synthesize glycogen (animal starch). Then theblood glucose fall, hydrolysis of liver glycogen releases free glucose for the blood.19.All fats arelipidsbecause they are nonpolar and not water-soluble. Not all lipids are fats,because fats are triglycerides and there are other categories of lipids, such as steroidsand phospholipids.20.Both fats and oils are triglycerides. Fats are solid at room temperature, whereas oils areliquid. Generally, fats contain more saturated fatty acids then oils. There is evidencethat a diet too high insaturated fatty acids as part ofsaturated fats can contribute toatherosclerosis, and thus cardiovascular disease. This may be because saturated fattyacids help raise the LDL-cholesterol and lower the HDL-cholesterol.21.DNA serves as a template because of complementary base pairing. An adenine mustpair with thymine, and a guanine must pair with a cytosine. One strand of a parent DNAis duplicated in this way, so that the new DNA molecule contains one strand that is“conserved” from the parent DNA and one strand that was newly formed from it.Test Your Analytical Ability22.Theprimary structure refers to the sequence of amino acids in the protein. Thesecondary structure refers toits helical structure, andthe tertiary structure refers to its3D structure. How the polypeptide chain forms its higher order structure is determinedby the particular amino acid functional groups and their positions in the polypeptidechain.

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223.In order to extract the hormone it must be soluble in the fluid used. If the hormone isnot soluble in water, it may be some type of lipid. This idea is supported by theobservation that the benzene (a nonpolar solvent) extract did have a hormonal effect.24.It is chemically correct in the sense that the product is free of cholesterol. It ismisleading in the sense that it could be free of cholesterol yet high in saturated fat andtherefore unhealthy despite the lack of cholesterol.25.Hydrogenation is a method of making butter substitutes that increase fatty acidsaturation and leads to the production of trans fats. Both saturated fats and trans fatsappear to contribute to high blood cholesterol, which is a risk factor in the developmentof atherosclerosis.26.Whenyou cook meat, the heat can cause weak hydrogen bonds and others to break andlead to a change in the consistency of the meat compared to its raw state. However,cooking does not break the much stronger covalent peptide bonds, so it doesn’t lead toa soupof amino acids.Test Your Quantitative Ability27.H2O has a molecular weight of 18. C6H12O6has a molecular weight of 180.28.Fructose must have the same molecular weight as glucose:180.29.Sucrose has a molecular weight of 342.30.Fructose and glucose each have a molecular weight of 180, so the two together wouldbe doubled to 360. However, they come together by dehydration synthesis, so waterwith a molecular weight of 18 is removed. Therefore, 180+180−18 = 342.

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1Fox:Human Physiology, 16thEditionChapter 3Answers to In-chapter QuestionsTest Your Understanding17.The plasma membrane can invaginate to form vesicles inendocytosis andcan performthe opposite function in exocytosis. In certain cells, pseudopods can be extended toallow the cell to move by amoeboid motion. Within the membraneare proteins thatperform a variety of functionsthatdynamically affect cellular activities.18.A nucleosome is a particle composed of histone proteins around which two turns ofDNA are wound. Histone proteins are believed to suppress DNA expression in areaswhere the DNA is highly compacted to form heterochromatin. This function is affectedby acetylation/deacetylation of the histone proteins.19.The genetic code is the sequence of bases in DNA that specify the production of RNA.Messenger RNA in turn codes for proteins, which have many functions, includingenzyme activity. As a result of the production of specific proteins and their actions,thegenetic code thereby influences the structure and function of the body.20.The DNA code determines the structure of its complementary mRNA, and in that waycarries information regarding the structure of proteins. However, in order for thegenetic code as translated into an mRNA code to become realized as a specific sequenceof amino acids, the actions of transfer RNA molecules are needed. In this way, tRNAtranslates the sequence of codons into a sequence of amino acids.21.Proteins that are produced for the cell’s own use are mostly released from the roughendoplasmic reticulum into the cytoplasm.However,some are packaged into vesiclesby the Golgi apparatus and sent to the plasma membrane where they function. All theproteins for secretion are internalized into the ER and then the Golgi apparatus, thensent for exocytosis within vesicles that bud from the Golgi apparatus.22.Thegenomerefers to all of the genes within an organism, whereas theproteomerefersto all of the proteins in an organism. The proteome is coded for by the genome.23.The Golgi complex communicates with the ER, perhaps through the release of vesiclesor perhaps more directly. In this way, proteins produced by the ER are packaged withinthe Golgi complex into vesicles. The vesicles released by the Golgi complex migratetothe plasma membrane,where they fuse with it. This can release secreted products andit can add new material to the plasma membrane.

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224.The two centrioles of a centrosome are required for the production of the microtubulesof the spindle fibers, which attach to the chromosomes during cell division. In anondividing cell, the centrioles of a centrosome organize microtubules to help producenonmotile primary cilia. In the apical membrane of some epithelial cells, thecentrosomes form motile cilia.25.A nondividing cell is in the phase of mitosis called interphase, which is divided intothreephases. In the G1 phase, the DNA is actively directing the affairs of the cell. If a cell isgoing to divide, it replicates its DNA during the S phase. During the G2 phase, the DNAcondenses into shorter, thicker structures,called chromosomes. Cell division occurs asa result of the stages of the mitotic phase. The progression through these steps isregulated in part by proteins called cyclins.26.Oncogenes are those that promote cancer, and tumor suppressor genes help protectagainst cancer. Oncogenes may be genes that normally control cell division andapoptosis, so that the oncogenes promote the division of cancer cells and prevent theirself-destruction. Tumor suppressor genes, such as p53, have the opposite effect.27.Apoptosisis a form of programmed cell death that has characteristic features thatdistinguish it from necrosis. Apoptosis occurs as part of normal, physiological processes,including the normal turnover of epithelial cells and blood cells and for tissueremodeling during embryonic development.28.Epigenetic inheritancerefers to the silencing of specific genes in the gametes thatformed an embryo, so that the expression of those genes is suppressed from onegeneration to another. This is a newly discovered form of inheritance that could explainhow metabolic effects of starvation in the parents could affect the offspring, forexample.Test Your Analytical Ability29.The histone proteins can repress gene expression, in part by affecting how theDNA iswound in the nucleosomes.The proteins influence the degree of compaction of thenucleosomes. The more compact the chromatin, the less access RNA polymerase has tothe sites in the DNA where it must bind to begin transcription. Also, there aretranscription factors, which are proteins that are needed for the expression of specificgenes. Hormones can influence the activity of transcription factors, and thereby affectgenetic expression.30.The tumor suppressor gene p53 can help guard against cancer by either stopping celldivision at a particular checkpoint or by promoting the apoptosis of a mutatedcancerous cell. One possible way that cancer could be treated would be the

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3development of a drug that restored p53 activity in a person with a mutated p53 gene,which occurs in most cancers.31.Destroying all of the lysosomes in all of the cells in a person’s body would have manynegative consequences. Lysosomes are needed for programmed cell death, so all of thephysiological functions dependent on apoptosis would cease. Also, in order for cells toturn over worn out organelles, they need lysosomes to digest them, so they and otherproducts within the cell would accumulate. In addition, the ability of macrophages andother phagocytic cells to function would cease.32.Since actinomycin D did not stop the hormone’s action, genetic transcription (RNAsynthesis) was not required for the action of the hormone. Thus, the action was notdependent of the synthesis of new proteins, such as enzymes. However, the hormone’saction was blockedby puromycin, so that means that the mRNA coding for thatproteinwas already present in the cell. Puromycin blocked the translation of that mRNA so thatthe protein needed for that hormone’s action could not be produced.33.It was once thought that each protein is coded by a different gene, but this cannot betrue if the proteome is larger than the genome. Once way that this could happen is ifthe exons produced by a gene can combine in different ways to form different proteins,or if the intervening sequences (introns) for one protein can function as exons in theproduction of a different protein.34.Short interfering RNA (siRNA) is a short double stranded RNA that can,aftermodification within a RISC particle,bind bycomplementary base pairing to a specificmRNA and cause its destruction.Micro RNA (miRNA) is a short single stranded RNA thatcan enter a RISC particle and bind to part of a number of particular mRNA molecules.This can cause their destruction, but more often just suppresses their expression to adegree.35.Ubiquitin is a 76 amino acid peptide that is added to certain regulatory proteins, ordefective proteins, in order to tag them for destruction. Such tagged proteins can thenbe digested within proteasomes, which are complexes of protease enzymes.Test Your Quantitative Ability36.Since 3 bases comprise one codon for each amino acid, and there are 600 amino acids inthe protein, 1,800 bases are needed (not counting the start and stop sequences andregulatory sequences).37.The gene must have 3 exons in order for it to have 2 introns.

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438.The answer depends on the length of each intron. If we assume, for example, that eachintron is 150 bases, then the combined length of the two is 300 bases. Since the entirecoding sequence is 1,800 bases, this leaves 1,500 for the three exons. If the exons are ofequal length, they must then eachbe500 bases long.

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1Fox:Human Physiology, 16thEditionChapter 4Answers to In-chapter QuestionsTest Your Understanding11.An enzyme is a protein with a specific secondary and tertiary structure that gives it aunique shape and position of active groups to bind in a specific way to its substrates.The tertiary structure is held in place by weak bonds, including hydrogen bonds.Anything that alters this bonding can alter the enzyme’s shape and reduce its action.This 3D structure can be altered and the enzyme function reduced by changes in pH andby increases in temperature.12.An increase in the concentrations of the enzyme’s substrates will drive the reaction tothe right (increase its rate), and a decrease will drive the reaction to the left (decreasingthe reaction rate).Conversely, an increased concentration of products will drive thereaction to the left, decreasing the reaction rate. Consequently, the reaction rate will behigh when the concentration of substrates is high and the concentration of products islow.13.As the reaction products accumulate, they canexert an inhibitory effect on the enzyme,slowing the reaction rate. This is a form of negative feedback, becausea rise in productsinhibitstheir own formation, preventing excessive accumulation of the products.14.According to the first law, energy can be transformed but cannot be created ordestroyed. So,some of the chemical energy released by the metabolism of molecules,such as glucose,can be transformed into the chemical bond energy that bindsphosphate to ADP to form ATP. According to second law, no energy transformation is100% efficient, and so some energy is released as heat. Also,according to second law,energy must be needed to produce a more ordered, lower entropy state of ATP fromthe less ordered, higher entropy state of ADP plus phosphate.15.Shuttling of hydrogen atoms involves transfer of electrons, so the gaining of hydrogen isaccompanied by the gaining of an electron and is a reduction, while the loss of hydrogenis the loss of an electron and an oxidationreaction. NAD and FAD cangain hydrogensand electrons and move from one location (inside the matrix of the mitochondrion) toanother (thecristae) to transfer hydrogens and couple reduction and oxidationreactions.

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216.An inborn error of metabolism often involves single gene mutations that result in singledefective enzymes. Such is the case with albinism and PKU. Both conditions areinherited defects in a single enzyme involved in the metabolism of phenylalanine. PKU iscaused by a defect in the enzyme that converts phenylalanine to tyrosine. Albinism iscaused by a defect in a later enzyme in this pathway, which converts DOPA to melanin.17.Niacin and riboflavin areneededto produce NAD and FAD, respectively. We need NADand FAD to shuttle hydrogens and electrons from the metabolic pathwaysthatbreakdownglucose and other moleculesto the location in mitochondria where most ofthe ATP is produced.We don’t directly get energy from NAD and FAD, but we needthese molecules in order to derive energy (ATP) from food molecules.Test Your Analytical Ability18.Metabolic pathways are frequently branched, with onebranchpromoted by oneenzyme and a different branch promoted by a different enzyme. The enzymes behaveas switches on these branched pathways, determining which direction the path willproceed according to the relative amounts of the two branch point enzymes.19.The LDH enzyme interconverts pyruvic and lactic acids, according to the relativeconcentrations of each. It is normally active within the cell cytoplasmbut can enter theblood when cells die. Once it is inthe blood, it is inactive because conditions in theblood do not favor either reaction.20.First, the plant generally has a pH greater than that of the stomach, so the enzymeswould be unlikely to be very active in the stomach and would likelydenature at thegreat acidity of gastric juice. Since the denatured enzyme is a protein, it would likely bedigested by pepsin in the stomach and other enzymes in the small intestine.21.According to the second law of thermodynamics, energy is lost as heat in every energytransformation, including the energy transformations that occur when animals eatplants or other animals. So,it requires most total biomass of plants to support a smallerbiomass of herbivores, and a greater biomass of herbivores to support a smallerbiomass of carnivores.22.A person who hyperventilates exhales more carbon dioxide than is exhaled in normalbreathing.This lowersthe blood CO2, which lowersthe amount of carbonic acid formedin the blood. The lower than normal formation of carbonic acid lowers the amount ofH+, thereby raising the blood pH.

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3Test Your Quantitative Ability23.The pH optimum of pepsin is close to 2. The pH optimum of salivary amylase is close to7. The pHoptimum of trypsin is close to9.24.Half-maximum activity of salivary amylase appears to occur at a pH of about 6 and a pHof about 8.25.Half-maximum activity of pepsin appears to occur atpH a little over 1, and a pH a littleless than 4.26.Pepsin and salivary amylase appear to have overlapping activities at a pH of about 5.27.Since salivary amylase has a pH optimum near 7, and almost no activity at acidic pHvalues, once the swallowed saliva is mixed with the very acidic gastric juice the activityof salivary amylase should be greatly decreased.

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1Fox:Human Physiology, 16thEditionChapter 5Answers to In-chapter QuestionsTest Your Understanding14.The lactic acid pathway has the advantage that it is shorter than the aerobic pathwayand can quickly generate ATP in the absence of oxygen.However, it only produces a netgain of 2 ATP, which is much less than the ATP produced by aerobic respiration ofglucose and other molecules. It also produces an acid, lactic acid, which can accumulateand lower the pH.15.Pyruvate is formed as the end product of the glycolytic pathway. In this process, NAD isreduced to NADH++ H+. In order to regenerate NAD to continue this process, the twohydrogens are donated to pyruvate to produce lactate. During aerobic respiration, theoxidation of NADH++ H+to NAD is accomplished by the electron transport chain in themitochondria.16.Cyanideinhibits the final step in aerobic respiration, the passage of the electrons tooxygen.When this step is blocked, the components of the electron transport chain stayin the reduced state because they cannot pass the electrons to the next step.As aresult, the electron transport chain cannot accept electrons from NADH. When there isinsufficient NAD+available for the citric acid cycle, it stops. Therefore, the production ofATP must stop, and the cell dies because of lack of sufficient energy.17.Glucose can be converted by glycolysis into 3-phospgoglyeraldehyde, which can beconverted into glycerol. Also, glucose can be converted into acetyl CoA, which can beused to produce fatty acids. The glycerol and fatty acids can then condense to formtriglycerides. End-product inhibition by ATP inhibits the steps of aerobic respiration, sothe intermediate molecules can be funneled into the production of fat.18.The metabolic pathway by which fat is broken down for energy is called beta-oxidation.In this process, each fatty acid chain is successively broken down into two-carbonmolecules that form acetyl CoA. At each step, an ATP and reduced NADH and FADH2 aregenerated, which can donate electrons to the electron transport chain to generate ATPby oxidative phosphorylation. In this way, the breakdown of fatty acids yieldsmanymore ATP than the breakdown of glucose.19.Through the process of oxidative deamination, amino acids can be converted intopyruvate, acetyl CoA, and a few different intermediates of the citric acid cycle. Once inthe aerobic respiratory pathway, these molecules can be metabolized for energy thesame they are when they are derived from glucose. However, the release of the amine

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2groups provides substrates for urea synthesis. So,a starving person who is breakingdown protein for energy will be producing more urea.20.Only the liver can secrete glucose into the blood because only the liver has the enzymeglucose-6-phosphatase to remove the phosphate group to produce free glucose forsecretion. Other organs lack this enzyme, and organic molecules with phosphate groupscannot pass through the plasma membrane.21.The production of lactic acidfrom glucose is termed fermentation by analogy with theway yeast cells convert glucose to ethanol. In both the production of lactic acid and ofethanol, the last electron acceptor is an organic molecule rather than oxygen. In thecase of lactic acid fermentation, the final electron acceptor is pyruvic acid.22.Brown adipose cells have uncoupling protein, located in the inner mitochondrialmembrane, which allows H+to leak back into the matrix and thus not be used foroxidative phosphorylation. As a result, the efficiency of oxidative phosphorylation isreduced (reducing the amount of ATP production) andmetabolic heat productionincreases. This suggests that in white adipose cells, which lack uncoupling protein, thegreater amount of ATP makes metabolism slower by end-product inhibition.23.Three molecules that can be used for gluconeogenesis are lactate, amino acids, andglycerol. Blood lactate levels rise during exercise, and lactate is converted into glucosefollowing exercise (Cori cycle). Amino acids can likewise be converted into glucose whenneeded, as during starvation. Glycerol is a 3-carbon long molecule that can be convertedinto pyruvate when triglycerides are hydrolyzed.Test Your Analytical Ability24.Fat has the highest amount of energy (calories) per weight, so eliminating fat could helpwithreducingher caloric intake andweight loss.However, calories can be obtainedfrom carbohydrates and proteins too, and if the caloric loss from the elimination of fat ismade up for by eating proteins and carbohydrates, weight will not be lost. (In fact, aperson could get hungrier on such a diet and gain weight.) Also, we need to eat somefats and oils to get essential fatty acids and the fat-soluble vitamins.25.If the drug caused the loss of the H+gradient between the intermembrane space andthe matrix, there will be less H+passing through cytochrome oxidase to generate ATP.The cellular ATP would decrease, and with less end-product inhibition, the rate ofaerobic metabolism would increase, generating heat to raise body temperature. Theincreased metabolism of adipose cells would cause weight loss.

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326.First, the NADH generated in the cytoplasm by glycolysis has to be shuttled into themitochondrion to donate electrons to the electron transport chain. There are twopossible shuttles, one in which 1 cytoplasmic NADH becomes1 mitochondrial NADH,and one in which 1 cytoplasmic NADH becomes 1 mitochondrialFADH2. These yielddifferent numbersof ATP. Second,the ATP generated by oxidative phosphorylation inthe mitochondria must be transported into the cytoplasm, and that transport requiresthe use of ATP. These scenarios vary somewhat in different organs, so whenanynumber isgivenforthenumber of ATP produced by oxidative phosphorylation it mustbeconsideredapproximate.27.People who are starving have thin arms and legs because they are breaking down theirmuscle protein for energy. The amino acids travel in the blood to the liver, where theyare used for gluconeogenesis. The liver is the only organ that can secrete glucosederived from stored glycogen or from gluconeogenesis into the blood.These actionsallow the liver to maintain adequate blood glucose during starvation, which is requiredfor the brain’s metabolism and to keep the person alive.28.Chicken proteins never enter your body, because they are hydrolyzed in the GI tract toamino acids. The amino acids from the chicken can then be used for energy or tosynthesize your proteinsaccording to your own genetic code.Similarly, the starch in thebread never enters your body, only the constituent glucose molecules.These can beused within your cells for energy or for the synthesis of your own starch (glycogen).Because chicken proteins and bread starch do not enter your body, it is not strictly truethat you are what you eat.Test Your Quantitative Ability29.A twenty-carbon long fattyacidcan be broken up into 10 acetyl CoA molecules, sinceeach is2 carbons long.30.Since 1 ATP is broken down every time beta-oxidation of a fatty acid releases an acetylCoA, 10 ATP must be broken down in the complete oxidation of a twenty-carbon longfatty acid.31.None. All of the ATP is produced by donation of electrons from NADH and FADH2to theelectron transport chain and oxidative phosphorylation.32.There will be 10 acetyl CoA generated from a twenty-carbon long fatty acid, and 1 NADHand 1 FADH2each time, for a total of 20.33.14 ATP is made per each beta-oxidation, as shown in the figure. So with 10 suchprocesses, the total would be 140 ATP.

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1Fox:Human Physiology, 16thEditionChapter 6Answers to In-chapter QuestionsTest Your Understanding16.In order for osmosis tooccur, two solutions of unequal concentration must beseparated by a semipermeable membrane that is permeable to water but not to one ofthe solutes. When that occurs, the random motion of water molecules will cause the netmovement of water from the side of greater to the side of lower water concentration.17.Simple diffusion occurs in a nonspecific fashion across the plasma membrane.Facilitated diffusion requires protein carriers in the plasma membrane, and displays theproperties of specificity, competition, and saturation that are characteristic of carrier-mediated transport.18.The equilibrium potential for K+is−90 mV, and that for Na+is +60 mV. The restingmembrane potential of neurons is about−70 mV. This value is closer to the equilibriumpotential for K+than that for Na+because the resting membrane is far more permeableto K+than to Na+.19.The Na+/K+pumps transport3 Na+out of the cell for every 2 K+they transport into thecell, thereby adding to the relative negativity of the inside of the resting cell. If gatedNa+channels were to open, Na+would diffuse down its electrochemical gradient intothe cell, making itless negative (and even positive) on the inside. If the gated K+channels opened, K+would diffuse out of the cell and make the inside of the cellmorenegatively charged.20.Glucose and water molecules cannot penetrate the plasma membrane without the aidof specific proteins.Carrier proteins for the facilitated diffusion of glucose can beinserted into the plasma membrane when intracellular vesicles containing these carriersfuse with the plasma membrane. They can be removed from the plasma membranewhen the membrane invaginates to form a vesicle with the carriers in the membrane.Water channels in vesicle membranes can similarly be added or removed to increase ordecrease water permeability.21.The rate of net diffusion across a plasma membrane isinfluenced by the temperature(which is relatively constant), the concentration difference across the membrane, thediffusion distance, and the surface area of membrane through which the diffusingmolecules must pass.Epithelialmembranes specialized for diffusion are simple ratherthan stratified and generally squamous to minimize the diffusion distance. In somelocations,the cells also have microvilli to increase the surface area for transport.

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