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Human Physiology 15th Edition By Stuart Ira Fox (Solution Manual)
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Human Physiology, 15e Stuart Ira Fox (Solution Manual) Human Physiology, 15e Stuart Ira Fox (Solution Manual)
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NOTE: For Complete File, Download link at the end of this File)Fox: Human Physiology, 15th 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 keratinized
epithelium 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, they
afford 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 were
to 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 and
the 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. 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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, Fox: Human Physiology, 15th 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. 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 DNA
is 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 its
3D structure. How the polypeptide chain forms its higher order structure is determined
by 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 development
of 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.
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.
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