Appendix H
Answers
A-29
APPENDIX H
ANSWERS
(oxygenated) blood to lung tissue.
7.
Te driving force for pulmonary
ventilation is a pressure gradient created by changes in the thoracic
volume.
8.
Te intrapulmonary pressure decreases during inspiration
because of the increase in thoracic cavity volume brought about by the
muscles of inspiration.
9.
Te partial vacuum (negative pressure) inside
the pleural cavity is caused by the opposing forces acting on the visceral
and parietal pleurae. Te visceral pleurae are pulled inward by the lungs’
natural tendency to recoil and the surface tension of the alveolar fluid. Te
parietal pleurae are pulled outward by the elasticity of the chest wall. If air
enters the pleural cavity, the lung on that side will collapse. Tis condi-
tion is called pneumothorax.
10.
Airway resistance is low because (1) the
diameters of most airways are relatively large, (2) for smaller airways there
are many in parallel, making their combined diameter large, and (3) air
has a low viscosity.
11.
A lack of surfactant increases surface tension in
the alveoli and causes them to collapse between breaths. (In other words,
it markedly decreases lung compliance.)
12.
Slow, deep breaths ventilate
the alveoli more effectively because a smaller fraction of the tidal volume
of each breath is spent moving air into and out of the dead space.
13.
In
a sealed container, the air and water would be at equilibrium. Terefore,
the partial pressures of CO
2
and O
2
(P
CO
2
and P
O
2
) will be the same in the
water as in the air: 100 mm Hg each. More CO
2
than O
2
molecules will be
dissolved in the water (even though they are at the same partial pressure)
because CO
2
is much more soluble than O
2
in water.
14.
Te difference
in P
O
2
between inspired air and alveolar air can be explained by (1) the
gas exchange occurring in the lungs (O
2
continuously diffuses out of the
alveoli into the blood), (2) the humidification of inspired air (which adds
water molecules that dilute the O
2
molecules), and (3) the mixing of newly
inspired air with gases already present in the alveoli.
15.
Te arterioles
leading into the O
2
-enriched alveoli would be dilated. Tis response al-
lows matching of blood flow to availability of oxygen.
16.
Both CO
2
and
H
1
increase O
2
unloading by binding to Hb. Tis is called the Bohr ef-
fect.
17.
About 70% of CO
2
is transported as bicarbonate ion (HCO
3
2
) in
plasma. Just over 20% is transported bound to hemoglobin in the RBCs,
and 7–10% is dissolved in plasma.
18.
As blood CO
2
increases, blood pH
decreases. Tis is because CO
2
combines with water to form carbonic
acid. (However, the change in pH in blood for a given increase in CO
2
is minimized by other buffer systems.)
19.
Te ventral respiratory group
of the medulla (VRG) is thought to be the rhythm-generating area.
20.
CO
2
in blood normally provides the most powerful stimulus to
breathe. Central chemoreceptors are most important in this response
(see Figure 22.25).
21.
Te injured soccer player’s P
CO
2
is low. (Recall that
normal P
CO
2
5
40 mm Hg.) Te low P
CO
2
reveals that this is hyperventila-
tion and not hyperpnea (which is not accompanied by changes in blood
CO
2
levels).
22.
Long-term adjustments to altitude include an increase
in erythropoiesis, resulting in a higher hematocrit; an increase in BPG,
which decreases Hb affinity for oxygen; and an increase in minute res-
piratory volume.
23.
Te obstruction in asthma is reversible, and acute
exacerbations are typically followed by symptom-free periods. In contrast,
the obstruction in chronic bronchitis is generally not reversible.
24.
Te
underlying defect in cystic fibrosis is an abnormality in a protein (CF±R
protein) that acts as a membrane channel for chloride ions.
25.
Vital ca-
pacity declines with age because the thoracic wall becomes more rigid and
the lungs lose their elasticity.
Review Questions 1.
b;
2.
a and c;
3.
c;
4.
c;
5.
b;
6.
d;
7.
d;
8.
b;
9.
c, d;
10.
c;
11.
b;
12.
b;
13.
b;
14.
c;
15.
b;
16.
b
Case Study 1.
Spinal cord injury from a fracture at the level of the C
2
ver-
tebra would interrupt the normal transmission of signals from the brain
stem down the phrenic nerve to the diaphragm, and Barbara would be un-
able to breathe due to paralysis of the diaphragm.
2.
Barbara’s head, neck,
and torso should have been immobilized to prevent further damage to the
spinal cord. In addition, she required assistance to breathe, so her airway
was probably intubated to permit ventilation of her lungs.
3.
Cyanosis is a
already been “primed” and has memory cells that are specific for that par-
ticular antigen.
14.
Vaccinations protect by providing the initial encoun-
ter to an antigen—the primary response to that antigen. As a result, when
the pathogen for that illness is encountered again, the pathogen elicits the
much faster, more powerful secondary response, which is generally effec-
tive enough to prevent clinical illness.
15.
IgG antibody is most abundant
in blood. IgM is secreted first in a primary immune response. IgA is most
abundant in secretions.
16.
Antibodies can bring about destruction of
pathogen via “PLAN”—
p
hagocytosis,
l
ysis (via complement),
a
gglutina-
tion, or
n
eutralization.
17.
Class II MHC proteins display exogenous anti-
gens. Class II MHC proteins are recognized by CD4 ± cells (which usually
become helper ± cells). APCs display class II MHC proteins.
18.
Binding
of antigens in the absence of co-stimulators causes the lymphocyte to
develop anergy—a state of permanent unresponsiveness to that antigen.
19.
Helper ± cells are central to both humoral and cellular immunity be-
cause they are required for activation of both cytotoxic ± cells and most
B cells.
20.
Te cytotoxic ± cell releases perforins and granzymes onto the
identified target cell. Perforins form a pore in the target cell membrane,
and granzymes enter through this pore, activating enzymes that trigger
apoptosis (cell suicide).
21.
MHC proteins and blood type antigens (ABO,
etc.) are carefully matched before an organ transplant.
22.
HIV is particu-
larly hard for the immune system to defeat because (1) it destroys helper
± cells, which are key players in adaptive immunity and (2) it has a high
mutation rate and so it rapidly becomes resistant to drugs.
23.
Binding of
an allergen onto specific IgE antibodies attached to mast cells triggers the
mast cells to release histamine.
Review Questions 1.
c;
2.
a;
3.
d;
4.
d, e;
5.
a;
6.
d;
7.
b;
8.
c;
9.
d;
10.
d;
11.
d;
12.
(1)b, g; (2)d, i; (3)a, e; (4)a, e, f, h; (5)e, h; (6)c, f, g
Case Study 1.
Te two kidneys transplanted in this case represent al-
logra²s.
2.
A major histocompatibility complex (MHC) protein is a
type of cell surface protein that the human body uses to recognize
self
and to help coordinate the recognition of
nonself
, or foreign, antigens.
Tese proteins are involved in the display of antigens to ± cells.
3.
Class
I MHC proteins are found on virtually all of the body’s cells, while class
II MHCs are found only on antigen-presenting cells (APCs). Class
I MHCs display antigens for recognition by CD8 ± cells (including
cytotoxic ± cells). Class II MHCs display antigens for recognition by
CD4 ± cells (including helper ± cells).
4.
A foreign MHC protein will
provoke an immune response, so the donor’s and recipient’s MHCs
must match as closely as possible to minimize such an attack. ±issue
typing dramatically reduces the risk of organ rejection due to attack by
the recipient’s immune system.
5.
In this case, the donor and recipi-
ents were not genetically identical. Even with very careful tissue typing
and compatibility testing, there will still be some differences that the
recipients’ immune systems will recognize as foreign. ±o reduce the
risk of organ rejection, the recipients are given drugs to suppress their
immune systems.
Chapter 22
Check Your Understanding 1.
Te structures that air passes by are the
nasal cavity (nares, nasal vestibule, nasal conchae), nasopharynx (with
pharyngeal tonsil), oropharynx (with palatine tonsil), laryngopharynx,
and larynx (with epiglottis, vestibular fold, and vocal fold).
2.
Te epi-
glottis seals the larynx when we swallow.
3.
Te incomplete, C-shaped
cartilage rings of the trachea allow it to expand and contract and yet keep
it from collapsing.
4.
Te many tiny alveoli together have a large surface
area. Tis and the thinness of their respiratory membranes make them
ideal for gas exchange.
5.
Te peanut was most likely in the right main
bronchus because it is wider and more vertical than the le².
6.
Te two
circulations of the lungs are the pulmonary circulation, which delivers
deoxygenated blood to the lungs for oxygenation and returns oxygenated
blood to the heart, and the bronchial circulation, which provides systemic
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