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Test Bank For Human Physiology 16th Edition By Stuart Fox; Krista Rompolski, All 20 Chapters Covered and Verified Solution Manual For Human Physiology 16th Edition By Stuart Fox; Krista Rompolski, All 20 Chapters Covered and Verified 1. Human Physiology 2. Solution Manual 3. Stuart Fox ...

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2024 2024




SOLUTION MANUAL
th
Human Physiology 16 Edition
by Stuart Fox and Krista Rompolski,
All 20 Chapters Covered




Qbank




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Table of Contents
1. The Study of Body Function
2. Chemical Composition of the Body
3. Cell Structure and Genetic Control
4. Enzymes and Energy
5. Cell Respiration and Metabolism
6. Interactions Between Cells and the Extracellular Environment
7. The Nervous System
8. The Central Nervous System
9. The Autonomic Nervous System
10. Sensory Physiology
11. Endocrine Glands
12. Muscle Qbank
13. Blood, Heart and Circulation
14. Cardio Output, Blood Flow, and Blood Pressure
15. The Immune System
16. Respiratory Physiology
17. Physiology of the Kidneys
18. The Digestive System
19. Regulation of Metabolism
20. Reproduction




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th
Fox: Human Physiology, 16 Edition
Chapter 1 Answers to In-chapter Questions

Test Your Understanding

13. Epithelial membranes can be simple (one cell deep) or stratified (many cells deep).
Simple membranes provide a barrier that allows selective transport from the external
environment to the tissues that underlie the epithelial membrane. Stratified membranes
provide more 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 abundant
extracellular material. However, they differ on what material comprises the extracellular
material and regarding their cells. The extra cellular material of blood is the fluid and
plasma. The dermis contains connective tissue fibers. Bone also has connective tissue
fibers, but the extracellular matrix is calcified.

15. A single negative feedback loop helps to maintain homeostasis by counteracting
changes in one direction. For example, insulin helps to prevent blood glucose from
getting 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 its
set point. The insulin then acts to lower the blood glucose by promoting the movement of
glucose from the blood into tissue cells. As a result, the blood glucose is lowered back to
the normal range. The lowering of blood glucose causes the pancreatic islets to reduce
their secretion of insulin.

17. In phase I clinical trials, the drug is tested on healthy human volunteers. In phase II
clinical trials, the drug is tested on the target population (those for whom the drug is
intended). Phase III clinical trials involve a much larger and more diverse test
population. Phase IV tests other potential uses of the drug. However, before a drug ever
reaches clinical trials, its development and early testing require animal research.

18. Claude Barnard proposed that the conditions within the body maintain constancy
despite challenges to that constancy. This idea led to the concept of homeostasis and
the concept that physiological mechanisms exist to maintain that internal constancy.




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Test Your Analytical Ability

19. Positive feedback mechanisms accentuate changes. Thus, if the body temperature wereto
rise, positive feedback mechanisms would make it become even hotter. Conversely, if
body temperature were to fall, positive feedback mechanisms would make it become
even colder. A fall in blood glucose would become an even greater fall in response to
positive feedback. Homeostasis could not be maintained and life could not exist under
these conditions.

20. Negative feedback mechanisms may have been acting earlier, but they’re effects don’t
become evident until 40 minutes after injection, when blood glucose starts to rise. The
initial blood glucose concentration appears to be restored 120 minutes after the injection.
Quantitative measurements are needed to determine the effect of insulin andthe
effectiveness of antagonistic mechanisms in this instance, and in all cases where
physiological mechanisms are being evaluated.

21. There must be communication between the intracellular fluid and the extracellular fluid
in order for a cell to obtain nutrients and eliminate wastes. There must also be
communication between the interstitial fluid and the blood plasma in order for these
molecules to be obtained and distributed to the cells and for homeostasis to be
maintained.

22. Qbank
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 in
motion to help raise the blood pressure. Logically, a more rapid and powerful heartbeat
would have that effect and also cause a more rapid pulse.

23. Adult stem cells are found in the red bone marrow, where they give rise to the blood
cells. Adult stem cells are also found in skeletal muscles and in the adult brain, as well as
in hair follicles and other locations. Unlike embryonic stem cells, adult stem cells cannot
form all of the body tissues and a complete organism.

Test Your Quantitative Ability

24. 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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th
Fox: Human Physiology, 16 Edition
Chapter 2 Answers to In-chapter Questions

Test Your Understanding

16. Nonpolar covalent bonds are formed when the valence electrons are shared equally.
Polar covalent bonds are formed when the valence electrons 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. An acid is a molecule that can donate protons (H ) to solution. A base is a molecule that
can 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 and
through dehydration synthesis reaction synthesize glycogen (animal starch). Then the
blood glucose fall, hydrolysis of liver glycogen releases free glucose for the blood.

19. All fats are lipids because 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 steroids and
phospholipids.

20. Qbank
Both fats and oils are triglycerides. Fats are solid at room temperature, whereas oils are
liquid. Generally, fats contain more saturated fatty acids then oils. There is evidence that
a diet too high in saturated fatty acids as part of saturated fats can contribute to
atherosclerosis, and thus cardiovascular disease. This may be because saturated fatty
acids help raise the LDL-cholesterol and lower the HDL-cholesterol.

21. DNA serves as a template because of complementary base pairing. An adenine must pair
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 Ability

22. The primary structure refers to the sequence of amino acids in the protein. The
secondary structure refers to its helical structure, and the 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 polypeptide chain.




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23. In order to extract the hormone it must be soluble in the fluid used. If the hormone is
not soluble in water, it may be some type of lipid. This idea is supported by the
observation 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 is
misleading in the sense that it could be free of cholesterol yet high in saturated fat and
therefore unhealthy despite the lack of cholesterol.

25. Hydrogenation is a method of making butter substitutes that increase fatty acid saturation
and leads to the production of trans fats. Both saturated fats and trans fats appear to
contribute to high blood cholesterol, which is a risk factor in the developmentof
atherosclerosis.

26. When you cook meat, the heat can cause weak hydrogen bonds and others to break and
lead 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 to a
soup of amino acids.

Test Your Quantitative Ability

27. H2O has a molecular weight of 18. C6H12O6 has a molecular weight of 180.

28. Fructose must have the same molecular weight as glucose: 180.
Qbank
29. Sucrose has a molecular weight of 342.

30. Fructose and glucose each have a molecular weight of 180, so the two together would
be doubled to 360. However, they come together by dehydration synthesis, so water
with a molecular weight of 18 is removed. Therefore, 180 + 180 ! 18 = 342.




Fox: Human Physiology, 16th Edition
Chapter 3 Answers to In-chapter Questions

Test Your Understanding

17. The plasma membrane can invaginate to form vesicles in endocytosis and can perform
the opposite function in exocytosis. In certain cells, pseudopods can be extended to
allow the cell to move by amoeboid motion. Within the membrane are proteins that
perform a variety of functions that dynamically affect cellular activities.



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18. A nucleosome is a particle composed of histone proteins around which two turns of
DNA are wound. Histone proteins are believed to suppress DNA expression in areas
where the DNA is highly compacted to form heterochromatin. This function is affected
by 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, including
enzyme activity. As a result of the production of specific proteins and their actions, the
genetic code thereby influences the structure and function of the body.

20. The DNA code determines the structure of its complementary mRNA, and in that way
carries information regarding the structure of proteins. However, in order for the genetic
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, tRNA translates 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 rough
endoplasmic reticulum into the cytoplasm. However, some are packaged into vesicles
by the Golgi apparatus and sent to the plasma membrane where they function. All the
proteins for secretion are internalized into the ER and then the Golgi apparatus, then
sent for exocytosis within vesicles that bud from the Golgi apparatus.

22. The genome refers to all of the genesQbank
within an organism, whereas the proteome refersto
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 vesicles
or perhaps more directly. In this way, proteins produced by the ER are packaged within
the Golgi complex into vesicles. The vesicles released by the Golgi complex migrate to
the plasma membrane, where they fuse with it. This can release secreted products andit
can add new material to the plasma membrane.

24. 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 a
nondividing cell, the centrioles of a centrosome organize microtubules to help produce
nonmotile primary cilia. In the apical membrane of some epithelial cells, the
centrosomes form motile cilia.

25. A nondividing cell is in the phase of mitosis called interphase, which is divided into three
phases. In the G1 phase, the DNA is actively directing the affairs of the cell. If a cell is
going to divide, it replicates its DNA during the S phase. During the G2 phase, the DNA
condenses into shorter, thicker structures, called chromosomes. Cell division occurs as a
result of the stages of the mitotic phase. The progression through these steps is regulated
in part by proteins called cyclins.



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26. Oncogenes are those that promote cancer, and tumor suppressor genes help protect
against cancer. Oncogenes may be genes that normally control cell division and
apoptosis, so that the oncogenes promote the division of cancer cells and prevent their
self-destruction. Tumor suppressor genes, such as p53, have the opposite effect.

27. Apoptosis is a form of programmed cell death that has characteristic features that
distinguish it from necrosis. Apoptosis occurs as part of normal, physiological processes,
including the normal turnover of epithelial cells and blood cells and for tissue
remodeling during embryonic development.

28. Epigenetic inheritance refers to the silencing of specific genes in the gametes that
formed an embryo, so that the expression of those genes is suppressed from one
generation to another. This is a newly discovered form of inheritance that could explain
how metabolic effects of starvation in the parents could affect the offspring, for
example.

Test Your Analytical Ability

29. The histone proteins can repress gene expression, in part by affecting how the DNA is
wound in the nucleosomes. The proteins influence the degree of compaction of the
nucleosomes. The more compact the chromatin, the less access RNA polymerase has to
the sites in the DNA where it must bind to begin transcription. Also, there are
transcription Qbank
n factors, which are proteins that are needed for the expression of specific
genes. Hormones can influence the activity of transcription factors, and thereby affect
genetic expression.

30. The tumor suppressor gene p53 can help guard against cancer by either stopping cell
division at a particular checkpoint or by promoting the apoptosis of a mutated
cancerous cell. One possible way that cancer could be treated would be the
development 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 many
negative consequences. Lysosomes are needed for programmed cell death, so all of the
physiological functions dependent on apoptosis would cease. Also, in order for cells to
turn over worn out organelles, they need lysosomes to digest them, so they and other
products within the cell would accumulate. In addition, the ability of macrophages and
other phagocytic cells to function would cease.

32. Since actinomycin D did not stop the hormone’s action, genetic transcription (RNA
synthesis) was not required for the action of the hormone. Thus, the action was not
dependent of the synthesis of new proteins, such as enzymes. However, the hormone’s
action was blocked by puromycin, so that means that the mRNA coding for that protein



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was 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 be true
if the proteome is larger than the genome. Once way that this could happen is if the
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 the
production of a different protein.

34. Short interfering RNA (siRNA) is a short double stranded RNA that can, after
modification within a RISC particle, bind by complementary base pairing to a specific
mRNA 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 a degree.

35. Ubiquitin is a 76 amino acid peptide that is added to certain regulatory proteins, or
defective 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 Ability

36. Since 3 bases comprise one codon for each amino acid, and there are 600 amino acids in
Qbank
the protein, 1,800 bases are needed (not counting the start and stop sequences and
regulatory sequences).

37. The gene must have 3 exons in order for it to have 2 introns.

38. The answer depends on the length of each intron. If we assume, for example, that each
intron is 150 bases, then the combined length of the two is 300 bases. Since the entire
coding sequence is 1,800 bases, this leaves 1,500 for the three exons. If the exons are of
equal length, they must then each be 500 bases long.




Fox: Human Physiology, 16th Edition
Chapter 4 Answers to In-chapter Questions

Test Your Understanding




9

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