Library Science

Anatomy and Physiology 6th edition

12th June 2007

Anatomy and Physiology 6th edition

posted in Anatomy & Physiology, Library |

Anatomy and Physiology 6th edition was designed to help students develop a solid, basic understanding of anatomy and physiology without an encyclopedic presentation of detail. Great care has been taken to select important concepts and to carefully describe the anatomy of cells, organs, and organ systems. The basic recipe that has been followed for six editions of this text is to combine clear and accurate descriptions of anatomy with precise explanations of how structures function and examples of how they work together to maintain life. To emphasize the concepts of anatomy and physiology, the authors provide explanations of how the systems respond to aging, changes in physical activity, and disease, with a special focus on homeostasis and the regulatory mechanisms that maintain it. Timely and interesting examples demonstrate the application of knowledge in a clinical context.

, , , ,

Chapter 1
1.Student B is correct.Body temperature begins to rise as a result ofexposure to the hot environment.Sweating eliminates heat from the body and lowers body temperature.Body temperature returning to its ideal normal value is an example ofnegative feedback.Student A probably thought that it was positive feedback because sweating continued to increase. Sweating,however,is the response.The variable being regulated by sweating is body temperature.

2.Answer eis correct.Positive-feedback mechanisms result in movement away from homeostasis and are usually harmful.The continually decreasing blood pressure is an example.Negative-feedback mechanisms result in a return to homeostasis.The elevated heart rate is a negative-feedback mechanism that attempts to return blood pressure back to a normal value.In this case,the negative-feedback mechanism was inadequate to restore homeostasis,and medical intervention (a transfusion) was necessary.

3.When a boy is standing on his head,his nose is superior to his mouth.Remember that directional
terms refer to a person in the anatomic position, not to the body’s current position.

4.a.Inferior or caudal
b.Anterior,ventral,or superficial
c.Proximal,superior,or cephalic
d.Medial
5.The esophagus is located in the left-upper quadrant and the epigastric region.The urinary bladder is located in the left-lower and right-lower quadrants and is in the hypogastric region.
6.Answer ais correct.The best way to reach the anterior surface ofthe heart begins with the patient lying on his or her back so that the anterior surfaces ofthe thorax and heart are facing the surgeon.The heart is located in the anterior portion ofthe thoracic cavity within the mediastinum and is surrounded by the pericardial cavity.The pericardial cavity is lined with the pericardial serous membranes,through which a cut can be made to reach the heart.
7.The uterus is located in the pelvic cavity.The pelvic cavity,however,is surrounded by the bones ofthe pelvis and doesn’t increase in size during pregnancy.Instead,as the fetus grows the expanding uterus must move into the abdominal cavity,thereby crowding abdominal organs and dramatically increasing the size of the abdominal cavity.
8.After passing through the left thoracic wall,the first membrane encountered is the parietal pleura. Continuing through the pleural cavity the visceral pleura ofthe left lung and then the left lung are pierced.Leaving the lung the bullet penetrates the visceral pleura,the pleural cavity,and the parietal pleura (remember that the lung is surrounded by
a double-membrane sac).Next the parietal pericardium,the pericardial cavity,the visceral pericardium,and the heart are penetrated. After passing through the abdominal wall,the parietal peritoneum is pierced.In passing through the stomach,the visceral peritoneum, the stomach itself,and the visceral peritoneum on the other side ofthe stomach are penetrated. Because the diaphragm is lined inferiorly by parietal peritoneum and superiorly by parietal pleura,these are the next two membranes pierced.The pole then passes through the pleural space and visceral pleura to enter the lung.

Chapter 2
1.An atom ofiron has 26 protons (the atomic number),30 neutrons (the mass number minus the atomic number),and 26 electrons (because the number ofelectrons is equal to the number ofprotons).Ifan atom ofiron loses three electrons,it has three more protons (positive charges) than electrons (negative charges). Therefore the iron ion has an overall charge of 33,which is represented symbolically as Fe .
2.The formation offree fatty acids and glycerol from a triglyceride is a decomposition reaction because a larger molecule breaks down into smaller molecules.All ofthe decomposition reactions in the body are collectively referred to as catabolism.This reaction can also be classified as a hydrolysis reaction because as part ofthe reaction a water molecule is split into hydrogen, which becomes part ofthe glycerol molecule, and hydroxide,which becomes part ofa fatty acid molecule.
3.The slight amount ofheat functions as activation energy and starts a chemical reaction.The reaction releases a large amount ofheat,thus causing the solution to become hot.
4.Heating (boiling) has destroyed the ability ofthe molecules in one or both ofthe solutions to function in the chemical reaction.This is called denaturation.There are two possibilities as to what is denatured.It could be the reactants themselves or an enzyme that catalyzes the reaction.
5.Muscle contains proteins.To increase muscle mass, proteins must be synthesized from amino acids. The synthesis ofmolecules in living organisms requires the input ofenergy.That energy comes from the potential energy stored in the chemical bonds offood molecules,which is released during the decomposition offood molecules.
6.Remember that pH is a measure ofhydrogen ion concentration.Ifequal amounts ofsolutions A and B are mixed,the resulting hydrogen ion concentration is the average value ofthe two solutions,that is,the pH is (8 2)/2 = 5.A pH
of5 is acidic.This question illustrates an important point:The pH ofa solution can be changed by adding a more acidic or basic solution to it. The sodium bicarbonate dissociates,thereby increasing the amount ofbicarbonate ions in the solution.The bicarbonate ions combine with hydrogen ions to form carbonic acid,which becomes carbon dioxide and water.The decrease in hydrogen ions causes the pH ofthe solution to increase (become more alkaline). As A and B are added to the solution,the enzyme E catalyzes the formation ofC.However,when C binds to the active site ofE,the ability ofE to catalyze the formation ofC is blocked.As more and more C is produced,the rate offormation of C is slowed.Because the reaction ofC with E is reversible,there will always be some E that has a functional (not blocked) active site,and some A will therefore always combine with B. One might try heating the substances because proteins can be denatured and can coagulate (like frying an egg).Another possibility is to try dissolving the substances in water.Most lipids are insoluble in water,while many proteins are either soluble in water or form colloids with water.
Most proteins (i.e.,a typical protein) contain sulfur,which is not found in phospholipids. Typical phospholipids and proteins contain carbon,hydrogen,oxygen,nitrogen,and phosphate.

Chapter 3
1.The cells within the wound swell with water and lyse by the introduction ofa hypotonic solution. This kills potentially metastatic cells that may still be present in the wound.
2.Water moves by osmosis from solution B into solution A.Because solution A is hyperosmotic to solution B,solution A has more solutes and less water than does solution B.Water therefore moves from solution B (with more water) to solution A (with less water).
3.Answer bis correct.Because the solution is isotonic,there is no exchange ofwater.Because the solution contains the same concentration of all substances except that it has no urea,only a net movement ofurea occurs across the membrane.
4.Answer bis correct.At point A on the graph,the extracellular concentration is equal to the intracellular concentration.Ifmovement were by simple diffusion or by facilitated diffusion,at this point the rate ofmovement would be zero. Because it is not zero,it’s reasonable to conclude that the mechanism involved is active transport. 5.A reduced intracellular K concentration reduces across the the concentration difference for K plasma membrane.Thus,the rate at which K diffuses out ofthe cell is reduced,and a smaller charge difference is required across the plasma membrane to oppose the diffusion ofthe K out ofthe cell.The potential difference across the plasma membrane is herefore reduced. Because the drug inhibits mRNA synthesis, protein synthesis is stopped.Ifthe cell releases proteins as they were synthesized,the rate of protein secretion dramatically decreases following the administration ofthe drug.On the other hand,ifthe cell releases stored proteins, the rate ofsecretion at first is normal and then gradually declines. The well-developed rough endoplasmic reticulum is indicative ofprotein synthesis,and a well-developed Golgi apparatus is indicative of secretion.It’s likely that this cell synthesizes and secretes proteins.

Chapter 4
1.The tissue is epithelial tissue,as it is lining a free surface,and the epithelium is stratified because it consists ofmore than one layer.The types of stratified epithelium are stratified squamous, stratified cuboidal,stratified columnar,or
transitional epithelium.The structure ofthe cells in the surface layers enables the determination of a specific tissue type. Flat cells in the surface layer indicate stratified squamous epithelium. Cuboidal cells in the surface layer indicate
stratified cuboidal epithelium,and columnar cells in the surface layer point to stratified columnar epithelium.The surface cells of transitional epithelium are roughly cuboidal with cubelike or columnar cells beneath them. When transitional epithelium is stretched,the surface cells are still roughly cuboidal,but underlying layers can be somewhat flattened.
2.In general,epithelial cells undergo cell division (mitosis) in response to injury,and the newly produced cells replace the damaged cells.If the basement membrane is destroyed,however, nothing is present to provide scaffolding for the
newly formed epithelial cells. Without the basement membrane,there’s not an effective way for the newly formed epithelial cells to repair a structure such as a kidney tubule.Since the basement membranes appear to be mostly present,the person is likely to survive and the kidney will regain most ofits ability to function.

3.Epithelium that functions to resist abrasion is stratified squamous epithelium.The moist stratified squamous pithelium lining the mouth and the keratinized stratified squamous epithelium ofthe skin are examples.The cells at
the surface are flattened,and when scraped away due to abrasion they are replaced by the cells beneath them.In contrast epithelial cells that carry out absorption are either simple cuboidal or simple columnar.Because they are one layer thick,they are more susceptible to damage and are not resistant to abrasion.In addition,these cells are large in volume,which allows them to contain the organelles involved in transport, such as mitochondria to produce ATP in the case ofactive transport.The surface ofthe cells that absorb are likely to contain microvilli,which increases the surface area for absorption.The flat cells that resist abrasion have no microvilli. 4.Glands producing merocrine secretions do so
with no loss ofactual cellular material,whereas glands producing holocrine secretions shed entire cells.The cells rupture and die,and the entire cell becomes part ofthe secretion.You could chemically analyze the secretions for the
types ofmolecules found in cellular organelles. For example,ifphospholipids and proteins normally found in membranes are in the secretion,then the secretion is a holocrine secretion.Ifthe secretion is watery or contains products that are not found in membranes or organelles,it’s a merocrine secretion. 5.The statement is not appropriate.A tissue capable ofcontracting is muscle.Both cardiac muscle and smooth muscle cells are mononucleated,although some cardiac muscle
cells can have two nuclei,and they are both under involuntary control.Cardiac muscle is striated,and smooth muscle is not,however.
6.Histamine is one ofthe mediators of inflammation released in response to tissue damage.Several other mediators of
inflammation,however,are released during inflammation in addition to histamine. Antihistamines might reduce the inflammatory response somewhat,but it’s not likely to have a major effect because ofthe other mediators of
inflammation released at the same time.In certain types ofinflammatory responses,such as allergic responses,histamines are released in large amounts.Under these conditions,antihistamines do reduce the inflammatory response.

Chapter 5
1.Yes,the skin (dermis) can be overstretched due to obesity.
2.The stratum corneum,the outermost layer of the skin,consists ofmany rows offlat,dead epithelial cells.The many rows ofcells,which are continuously shed and replaced,are responsible for the protective function ofthe integument.In infants,there are fewer rows of cells,resulting in skin that’s more easily damaged than that in adults.
3.Melanocytes produce melanin,which protects underlying tissue from ultraviolet radiation. Therefore,we expect melanocytes to be as superficial as possible.Also recall that melanin production varies depending on exposure to the
sun.Response to stimulation is a characteristic of living cells.Thus,melanocytes should be found in the most superficial living layer ofthe epidermis,the stratum basale.
4.When first exposed to the cold temperature just before starting the run,the blood vessels in the skin constrict to conserve heat.This produces a pale skin color.Dilation ofthe skin blood vessels doesn’t occur at this time because the skin has not been exposed to the cold long enough to cause the skin temperature to fall below 15°C. After running for awhile,as a result ofthe excess heat generated by the exercise,the blood vessels in the skin dilate.This results in heat loss and helps to prevent overheating.Increased blood flow through the skin causes it to turn red.After the run,the body still has excess heat to eliminate,so the skin remains red for some time.5.Eyelashes have a short growth stage (30 days) and are therefore short.Fingernails grow continuously but are short because they are cut, broken off,or worn down.
6.Several methods have some degree ofsuccess in treating acne:(a) Kill the bacteria.One effective agent is benzoyl peroxide,found in some acne medications.(b) Prevent blockage ofthe hair follicle.A vitamin A derivative (tretinoin;
Retin-A) has proven effective in keeping the follicular epithelial cells and sebum from building up and closing offthe hair follicle. (c)Unplug the follicle.Some sulfur compounds (Acnederm) speed up peeling ofthe skin and thus unplug the follicle.
7.Probably not,because following removal ofthe nail from the nail fold,it may grow back into the
nail fold and the ingrown toenail would reoccur. One solution is to remove the small part ofthe nail responsible for the ingrown toenail.Prior to this drastic approach,sterile gauze can be placed between the nail and the nail fold to force the nail away from the nail fold.After the nail fold is healed,the gauze can be removed.

Chapter 6
1.Normally bone matrix and bone trabeculae are organized to be strongest along lines ofstress. Random organization ofthe collagen fibers of bone matrix results in weaker bones.In addition, the reduced amount oftrabecular bone makes the bone weaker.Fractures ofthe bone can occur when the weakened bone is subjected to stress.

2.Replacement ofcartilage ofthe epiphyseal plate by bone normally occurs on the diaphyseal side ofthe plate.As growth ceases,the cartilage cells stop dividing and producing new cartilage. Replacement ofcartilage with bone continues from the diaphyseal side,and eventually all of the cartilage ofthe plate is bone.

3.Mechanical stress applied to bone stimulates osteoblast activity,so the patient with a walking cast should heal faster.
4.Osteoporosis is depletion ofbone matrix that results when more bone is destroyed than is formed.Because mechanical stress stimulates bone formation (osteoblast activity),running helps to prevent osteoporosis in the bones being stressed.This includes the bones ofthe lower limbs and the spine.

5.The loss ofbone density results because the bones are not bearing weight in the weightless environment.Therefore osteoblasts are not sufficiently stimulated and bone resorption exceeds bone building.Bone loss can be slowed by stressing the bones using exercises against resistance such as cycling.

6.The kidneys are the site ofproduction ofactive vitamin D (see chapter 5),which is needed for calcium absorption in the small intestine.Kidney failure can result in inadequate vitamin D production,too little uptake ofcalcium,and therefore osteomalacia.
7.Testosterone normally causes a spurt ofgrowth at puberty followed by slower growth and closure ofthe epiphyseal plate.Without testosterone,growth is slower,but proceeds longer,resulting in a taller-than-normal person.

8.Blood vessels in central canals run parallel to the long axis ofthe bone,and perforating canals run at approximately a right angle to the central canals.Thus,perforating canals connect to central canals,which allows blood vessels in the
perforating canals to connect with blood vessels in the central canals.After a fracture,blood flow through the central canals stops back to the point where the blood vessels in the central canals connect to the blood vessels in the perforating canals.The regions ofbone on either side ofthe fracture associated with this lack of blood delivery die.

9.Hyperthyroidism stimulates increased bone breakdown and could cause osteitis fibrosa cystica,a condition in which the bone is eaten away as calcium is released from the bone.The result can be a deformed bone that is likely to fracture.Vitamin D therapy might help because vitamin D promotes an increase in blood calcium levels (see chapter 5) and therefore increased deposition ofcalcium in bone.

Chapter 7
1.An infection in the nasal cavity could spread to adjacent cavities and fossae,including the paranasal sinuses:(1) rontal,(2) maxillary,(3) ethmoidal,and (4) sphenoidal;(5) the orbit (through the nasolacrimal duct);(6) the cranial cavity (through the cribriform plate);and (7) the throat (through the posterior opening ofthe nasal cavity).

2.Falling on the top ofthe head could drive the occipital condyles into the superior articulating processes ofthe atlas,causing a fracture.An uppercut to the jaw would slightly lift the occipital condyles away from the superior articulating processes ofthe atlas and usually doesn’t result in a fractured atlas.Such a blow to the jaw an,however,fracture the temporal bone where it articulates with the mandible.

3.Forceful rotation ofthe vertebral column is most likely to damage the articular processes,especially in the lumbar region,where the articular processes tend to prevent excessive rotation (the superior articular processes face medially and the inferior articular processes face laterally).

4.Weaker back muscles on one side could cause the vertebral column to bend laterally (scoliosis) toward the opposite side.Lordosis can result from pregnancy.As the fetus causes the abdomen to move anteriorly,the thorax and head tend to pull posteriorly,to restore the center ofgravity.This posture increases the lumbar curvature.The same effect can be seen in people who are “pot-bellied.”
5.Ifthe ulna and radius become fused,the radius can no longer rotate relative to the ulna,and,as a result,most ofthe rotation ofthe forearm and hand is lost.

6.Measure from the anterior superior iliac spine (a “stationary”point relative to the limb,which can be easily found as a surface landmark) to the lateral malleolus.The inferior side ofthe foot could also be used,ifthe person is standing on a flat surface.A defect ofthe foot or ankle may occur,however,in which the ankle on one side is elevated.Ifthe length ofthe thigh is the only part to be measured,measure to the lateral epicondyle.

7.The ischial tuberosity is the bony protuberance.
8.Women’s hips are wider than men’s.As the knees are positioned toward the midline the slope of the femur from its proximal end toward its distal end is greater in women,and as a result,women tend to be knock-kneed more often than men.
9.The lateral malleolus extends further distally than does the medial malleolus,thus making it more difficult to turn the foot laterally than medially.The styloid process ofthe radius extends further distally than the styloid process ofthe ulna,thus making it more difficult to cock the wrist toward the thumb (laterally) than toward the little finger (medially).
0.Landing on the heels could fracture the calcaneus.Heavy objects,such as Hefty Stomper, landing on the dorsal surface ofthe foot could fracture the metatarsals or even the tarsals.

Chapter 8
1.Ifthe sternocostal synchondrosis were to ossify, becoming a synostosis,there would no longer be any stretch through the costal cartilage,the thorax could not expand,and,as a result, respiration would be severely hampered. 2.a.Suture,little or no movement.
b.Syndesmosis,some movement.
c.Complex synovial joints:the humeroulnoradial joint is a hinge joint,the radioulnar joint is a pivot joint.All have
considerable movement.
3.a.Flexion and supination
b.Flexion ofthe thigh and extension ofthe leg
c.Abduction ofthe arm
d.Flexion ofthe leg and plantar flexion ofthe
foot
4.The anterior drawer test determines the integrity ofthe anterior cruciate ligament,and the posterior drawer test determines the integrity ofthe posterior cruciate ligament.Unusual movement during the posterior drawer test indicates damage to the posterior cruciate ligament.

Chapter 9
1.Botulism poisoning results from the consumption ofbotulism toxin produced by the bacterium Clostridium botulinum.The toxin binds to presynaptic nerve terminals and prevents the release ofacetylcholine.Thus, action potentials in nerves cannot produce action potentials in skeletal muscles,and the result is paralysis ofskeletal muscles,which explains the difficulty in breathing and swallowing.Other reasonable explanations are that the toxin binds to and blocks the receptors for acetylcholine,that the toxin blocks the entry ofCa into the presynaptic terminal and thus prevents acetylcholine release,or that the toxin specifically prevents entry ofions through Na channels ofskeletal muscle cells.
2.Muscular dystrophy results from gradual atrophy ofskeletal muscle fibers and their replacement with connective tissue.Myasthenia gravis results from the degeneration ofthe receptors for
acetylcholine on the postsynaptic membranes of skeletal muscle cells.Ifan inhibitor of acetylcholinesterase is administered,the result should be an increase in the concentration of acetylcholine in the nerve muscle synapse.Thus,
more acetylcholine is available to bind to acetylcholine receptors.In people suffering from myasthenia gravis,the increased concentration of acetylcholine in the synapse allows acetylcholine to bind a greater percentage ofthe acetylcholine receptors present and causes the muscle contractions to increase in strength.In people who have muscular dystrophy,the muscle contractions don’t increase in strength because muscle atrophy is the cause ofthe weakness.The additional acetylcholine in the neuromuscular synapse has no effect on the weakened muscle fibers.
Placing sarcoplasmic reticulum from skeletal muscle cells into the beaker would remove calcium from the solution because sarcoplasmic from the solution reticulum transports Ca into the sarcoplasmic reticulum.In addition, ATP would have to be added for two reasons:(1) the sarcoplasmic reticulum actively transports calcium and,therefore,requires ATP;and (2) ATP must bind to the heads ofthe myosin molecules before the myosin heads can release from the active sites on the actin molecules.
A lower-than-normal temperature decreases the rate ofall ofthe processes that occur in the lag phase ofmuscular contraction because a lower temperature decreases the rate ofall chemical reactions and the rate ofion diffusion.As a
consequence,the lag phase requires a longer time.
Start with a subthreshold stimulus and increase the stimulus strength by very small increments. Apply the stimulus to the nerve ofmuscle A and muscle B.Ifthe number ofmotor units is the same for both preparations,each time the
stimulus strength is increased the degree of tension produced by the muscles would also increase to the same degree in each muscle.If one muscle has more motor units than the other, the muscle with the greater number ofmotor
units would exhibit a greater number ofseparate increases in tension,and the magnitude ofthe increases in tension would be smaller than those seen in the muscle with fewer motor units.

When a muscle slowly lifts an object,the contraction starts with a small number ofmotor units being stimulated.Each motor unit is stimulated tetanically.As the contraction continues,more and more motor units are recruited to lift the object slowly.To lower the object,the number ofmotor units stimulated tetanically is reduced slowly and the tension
produced by the muscle decreases.In a muscle twitch,a stimulus causes a single action potential in all ofthe muscle fibers responding to the stimulus.The stimulated muscle fibers contract in an all-or-none fashion and then relax.Both
contraction and relaxation occur quickly.
The shape ofan active tension curve for skeletal muscle can be seen in figure 9.20.In contrast,an active tension curve is much flatter for smooth muscle.That is,for each increase in the length of a muscle fiber there is little change in the active tension produced by the smooth muscle fiber.
Smooth muscle has,as one ofits major characteristics,the ability to increase in length without much increase in the tension produced by the smooth muscle cells. 8.Both the 100 m run and weight lifting involve rapid and intense contractions ofskeletal muscles that are completed quickly.These contractions depend on anaerobic metabolism for a significant amount ofthe ATP produced.In contrast,the 10,000 m run involves sustained muscular contractions that are not as rapid,but the slower contractions are repeated many times during the run.Aerobic metabolism produces the majority ofthe ATP for the 10,000 m run.
Anaerobic metabolism is associated with a decrease in creatine phosphate,an increase in creatine,an increase in lactic acid,and a decrease in glycogen,and the enzymes responsible for anaerobic metabolism function more rapidly. Aerobic metabolism is associated with increased enzyme activity in the mitochondria and an increase in carbon dioxide production.Oxygen is used more rapidly during aerobic metabolism.

9.During intense exercise it’s possible to experience physiologic contracture.Being unable to either contract or relax the muscles for a short time while exercising suggests the existence of physiologic contracture.
0.Smooth muscle depends almost entirely on aerobic metabolism to produce the ATP required for muscle ontraction.Ifthe blood supply to smooth muscle fibers is decreased the smooth muscle,therefore,cannot maintain contractions.
1.During the 100 m race Shorty depended on ATP produced by anaerobic metabolism.That produced an oxygen debt at the end ofthe run, which resulted in an elevated rate ofrespiration for a time.During the longer and slower run
most ofthe ATP for muscle contractions was produced by aerobic respiration,and very little oxygen debt developed.Prolonged aerobic respiration is required to “pay back”the oxygen debt.Shorty’s rate ofrespiration was,therefore, prolonged after the 100 m race but not after the longer but slower run.
2.High blood K concentration also results in depolarization ofsmooth muscle plasma membranes.Depolarization ofthe smooth muscle plasma membrane results in increased muscle contractions and increased permeability 2 ofthe plasma membrane to both Na and Ca , which results in further depolarization and an increase in the intracellular concentration of Ca .These changes result in the production of action potentials and muscle contractions.
3.The muscles would contract.ATP would be available to bind to the myosin heads,thus allowing myosin molecules to be released from actin molecules.The cross-bridges would immediately re-form,and complete cross-bridge cycling would result in contraction ofthe muscle fibers.As long as Ca were present at high concentrations in the arcoplasm, contraction of the muscles would occur.Ifthe sarcoplasmic reticulum were intact,ATP would be available to drive the active transport ofCa into the sarcoplasmic reticulum.As the Ca decreased in the sarcoplasm,relaxation would result.Ifthe sarcoplasmic reticulum were not intact,however, into the and could not transport Ca sarcoplasmic reticulum as fast as they leak out,the muscle would remain contracted until it fatigued. 2

14.Hormones can bind to ligand-gated Ca channels,and the channels,in response,open. Ca diffuse into the cell and cause contraction to occur.Only a small amount ofdepolarization 2 results as Ca diffuse into the cell,and since Na channels don’t open,a large change in the resting membrane potential doesn’t occur.

15.In experiment A,the students used anaerobic respiration as they started to run in place,but aerobic respiration also increased to meet most oftheir energy needs.When they stopped,their respiration rate was increased over resting levels because ofrepayment ofthe oxygen debt due to anaerobic respiration.In experiment B almost all ofthe student’s respiration came from anaerobic respiration because the students held their breath while running lace.Consequently,the students had a much larger oxygen debt.The student’s respiratory rate and depth was greater than inexperiment A,or that their respiration rates were elevated for a longer period oftime than in experiment A.

Chapter 10
1. MuscleActionSynergistAntagonist
LongusFlexesRectusMost ofthe
capitisneckcapitisposterior
anteriorneck
Longismuscles
coli
ErectorExtendsInterspinalesMost
spinaevertebralMultifidusanterior
columnSemispinalisabdominal
thoracismuscles
Coraco-AdductsLatissimusDeltoid
brachialisarmdorsiSupra-
Pectoralisspinatus
major
Teres major
Teres minor
Flexes armDeltoidDeltoid
(anterior)(posterior)
PectoralisLatissimus
majordorsi
BicepsTeres major
brachiiTeres minor
Infra-
spinatus
Sub-
scapularis
Triceps
brachii
2.Biceps brachii:Pull-ups with hands supinated Triceps brachii:Push-ups Deltoid:Abduction ofthe arms to shoulder height,with weights in the hands (abduction past shoulder height involves mostly scapular rotation by the trapezius)
Rectus abdominis:Sit-ups to 45 degrees (sit-ups past 45 degrees involve mostly the psoas major) Quadriceps femoris:Extending the legs against a force Gastrocnemius:Plantar flexion ofthe feet against a force,such as toe raises with a weight on the shoulders 3.The brachioradialis originates on the humerus and inserts onto the distal end ofthe radius.The fulcrum ofthis lever system is the elbow joint. With a weight held in the hand,the force, applied between the weight and the fulcrum,is a class III lever system.With the weight on the forearm,the weight is between the force and the fulcrum and is a class II lever system.A greater weight can be lifted ifplaced on the forearm rather than in the hand,but weights placed on the forearm cannot be lifted as far.

4.The muscles that flex the head also oppose extension ofthe neck.In an accident causing hyperextension ofthe neck,these muscles could be stretched and torn.The muscles involved could include the sternocleidomastoid,longus
capitis,rectus capitis anterior,and longus coli. Automobile headrests are designed so that,if adjusted correctly,the back ofthe head hits the headrest during a rear-end accident,thereby preventing hyperextension ofthe neck.

5.The only muscle that elevates the lower eyelid is the orbicularis oculi,which “closes the eye.”With this muscle not functioning,the lower eyelid would droop.The levator anguli oris,which elevates the angle ofthe mouth,was also
apparently affected allowing the corner ofthe mouth to droop.The zygomaticus major may also have been affected,as it inserts onto the corner ofthe mouth (see figure 10.7).

6.The genioglossus muscle protrudes the tongue. Ifit becomes relaxed,or paralyzed,the tongue may fall back and obstruct the airway.This can be prevented or reversed by pulling forward and down on the mandible,thus opening the
mouth.The genioglossus originates on the genu ofthe mandible.As the mandible is pulled down and forward,the genioglossus is pulled forward with the mandible,thus pulling the tongue forward also.

7.The rotator cuffmuscles are the primary muscles holding the head ofthe humerus in the glenoid fossa,especially the supraspinatus.In fact,a torn rotator cuff,which usually involves a tear ofthe supraspinatus muscle,often results in dislocation ofthe shoulder.
8.With the quadriceps femoris paralyzed,the leg could not be extended,and the lower limb could not bear weight unless the knee were passively extended,such as by pushing back on the distal end ofthe thigh with the hand.Walking would be almost impossible,except by taking very small steps and by pushing back on the knee with each
step,or by bracing the knee in an extended position.
9.Speedy has ruptured the calcaneal tendon,and the gastrocnemius and soleus muscles have retracted,thereby causing the abnormal bulging ofthe calfmuscles.Because the major plantar flexors are no longer connected to the calcaneus,
the runner cannot plantar flex the foot,and the foot is abnormally dorsiflexed because the antagonists have been disconnected.

Chapter 11
1.A reduced intracellular concentration ofK causes depolarization ofthe resting membrane potential.Because the intracellular concentration

ofK is reduced,the concentration gradient for potassium from the inside to the outside ofthe plasma membrane is also reduced.Thus the rate at which K diffuse out ofthe cell is reduced, and a smaller charge difference across the plasma
membrane is required to oppose the diffusion of the K out ofthe cell.Therefore,the potential difference across the plasma membrane is reduced,and the cell is depolarized. Because the plasma membrane is much less permeable to Na than to K ,changes in the extracellular concentration ofNa effect the resting membrane potential less than do changes in the extracellular concentration ofK .
Therefore,increases in extracellular Na have a minimal effect on the resting membrane potential.Because the membrane is much more permeable to Na during the action potential, the elevated concentration ofNa in the extracellular fluid results in Na diffusing into the cell at a more rapid rate during the action potential,resulting in a greater degree of depolarization during the depolarization phase ofthe action potential.
Because lithium ions reduce the permeability of plasma membranes to Na ,the Na channels in the plasma membrane tend to remain closed.A normal stimulus causes Na channels to open, allowing Na to diffuse into the cell,thereby resulting in depolarization.The cell is less sensitive to stimuli because the membrane is less permeable to Na .
Smooth muscle cells contract spontaneously in response to spontaneous depolarizations that produce action potentials.One way action potentials can be produced spontaneously is if membrane permeability to Na spontaneously
increases.As a result,a few Na enter the smooth muscle cells and cause a slight depolarization ofthe plasma membrane.The small depolarization can cause voltage-gated Na channels to open,which results in further depolarization,thereby stimulating additional voltage-gated ion channels to open.This positive-feedback cycle can continue until the plasma membrane is depolarized to its threshold level and an action potential is produced.
Action potential conduction along a myelinated nerve fiber is more energy efficient because the action potential is propagated by saltatory conduction,which produces action potentials at the nodes ofRanvier.Compared to an
unmyelinated nerve fiber,only a small portion of the myelinated neuron’s membrane has action potentials.Thus there is less flow ofsodium into the neuron (depolarization) and less flow of potassium out ofthe neuron (repolarization).
Consequently,the sodium–potassium exchange pump has to move fewer ions in order to restore ion concentrations.Because the sodium–potassium exchange pump requires ATP,myelinated axons use less ATP than
unmyelinated axons.
The inhibitory neuromodulator causes the postsynaptic neuron to become less sensitive to excitatory stimuli,probably by causing hyperpolarization ofthe postsynaptic neuron.As a result,the excitatory neurotransmitter released

from the excitatory neuron is less likely to produce postsynaptic action potentials. With aging,there’s a decrease in the amount of myelin surrounding axons,which decreases the speed ofaction potential propagation.At synapses there’s also an increase in the time it takes for action potentials in the presynaptic terminal to cause the production ofaction
potentials in the postsynaptic membrane.It’s believed this results from a reduced release of neurotransmitter by the presynaptic terminal and a reduced number ofreceptors in the postsynaptic membrane. Organophosphates inhibit acetylcholinesterase, thereby causing an increase in acetylholine in the synaptic cleft leading to overproduction of
action potentials,tetany ofmuscles,and possible death resulting from respiratory failure (see chapter 11).Curare is the best antidote because i blocks the effect ofacetylcholine and acts to counteract the organophosphate.Too much
curare,however,could cause flaccid paralysis of the respiratory muscles.Injecting acetylcholine would make the effect ofthe organophosphate worse.Potassium chloride causes depolarization ofmuscle cell membranes,thereby making them
more sensitive to acetylcholine. Ifthe motor neurons supplying skeletal muscle are innervated by both excitatory and inhibitory neurons,then blocking the activity ofthe inhibitory neurons with strychnine results in overstimulation ofthe motor neurons by the excitatory neurons.

Chapter 12
1.Ifthe neuron with its cell body in the cerebrum is an inhibitory neuron and ifit also synapses with the motor neuron ofa reflex arc,then stimulation ofthe cerebral neuron could inhibit the reflex.
2.The phrenic nerve is cut in the thorax,and the surgery is performed while the lung is being removed.
3.The ulnar nerve supplies the medial third ofthe hand,little finger,and medial halfofthe ring finger.The median nerve supplies the lateral two-thirds ofthe palm and thumb,and the surface ofthe index,middle,and lateral halfof
the ring finger.The radial nerve supplies the lateral two-thirds ofthe dorsum ofthe hand.
4.Pulling on the upper limb when it is raised over the head can damage the lower brachial plexus,
in this case,the origin ofthe ulnar nerve.The ulnar nerve innervates muscles that
abduct/adduct the fingers and flex the wrist.
5.The ischiadic nerve has rootlets from L4–S3.
Depending on the rootlet compressed,pain can be felt in different locations.
6.a.Obturator nerve
b.Femoral nerve
c.Ischiadic (tibial) nerve
d.Obturator nerve
e.Obturator nerve,some from femoral nerve

Chapter 13
1.A condition in which a patient looses all sense o feeling in the left side ofthe back,below the upper limb,and xtending in a band around to the chest,also below the upper limb,but where all sensation on the right is ormal,suggests tha the patient’s dorsal roots have been damaged on the left side adjacent to the part ofthe spinal cord supplying that part ofthe body.(The basis ofthis condition is explained more fully in chapter 14.)

2.The skull restricts the growth ofthe brain.The surface area ofthe cerebral cortex increases as more neurons migrate into the cortex and as more synapses are formed.As the cerebral corte increases in area,it becomes folded.This folding allows a greater surface area to be housed in a much smaller volume.
3.IfCSF does not drain properly,the fluid accumulates and exerts pressure on the brain (hydrocephalus).In the eveloping fetus,the ventricles enlarge because ofthe excess fluid pressure.The head also enlarges because the skull bones have not fused.The expansion ofthe head is not sufficient,however,to relieve all the pressure exerted on the developing brain by the expanding ventricles.As a result,the cerebral cortex becomes proportionately thinner as it’s
compressed between the ventricles and skull.In many cases,less gyri form in the cerebral cortex. Brain damage may or may not result,depending on the amount ofexcess CSF,the ventricular pressure generated,and the areas ofthe brain
damaged by the pressure.
4.Enlargement ofthe later al and third ventricles, without enlargement ofthe fourth ventricle, suggests a blockage between the third and fourth ventricles in the cerebral aqueduct.This defect is called aqueductal stenosis and is a common congenital problem.
5.Blood in the CSF taken through a spinal tap indicates the presence ofblood in the subarachnoid space and suggests that the patient has a damaged blood vessel in the subarachnoid space.

Chapter 14
1.The first sensations that occur when a woman picks up an apple and bites into it are visual (special),tactile (general),and proprioceptive (general).The woman holds the apple in her hand and looks at it.The tactile sensations from mechanoreceptors in the hand tell her that the apple is firm and smooth.The proprioceptive sensations originating in the joints ofthe hand tell the woman the size and shape ofthe apple. Visual input also tells her the size ofthe apple,
and that it has a smooth surface,as well as its color.As the woman bites into the apple and begins to hew,proprioceptive sensations from the teeth and jaw provide information as to how widely the jaws must be opened to accommodate the bite and how hard to bite down.Tactile sensations originating in the tongue and cheeks tell the person the location ofthe bite ofapple and its texture as it is moved about in the mouth. Taste sensations (special,chemoreceptor) from the tongue provide information that the apple has characteristics ofbeing both sweet and sour. Olfactory sensations (special) provide more specific information that the “fruity taste”is that ofan apple.
a.The most likely explanation is that the olfactory neurons accommodate and no longer respond to odor stimulus.
b.The fact that one can hear the sound when one tries indicates that the hair cells in the spiral organ have not accommodated and are still able to detect the sound stimulus. Many action potentials arriving in the brain are prevented from causing conscious perception,until we consciously “pay attention”to the stimulus.For example,you may not be paying attention to general conversations in a crowded room or hall, until someone says your name.The sound of
your name leaps out ofthe surrounding babble,and you are suddenly interested in what was being said by the person who spoke your name. The fibers ofthe dorsal-column/medial-lemniscal system carry two-point discrimination and
proprioceptive information.Primary neurons from the right side ofthe body ascend the spinal cord in the dorsal column and synapse with secondary neurons in the medulla oblongata.The secondary neurons cross over in the upper
medulla and ascend through the left side ofthe pons to the thalamus.A patient suffering from a loss oftwo-point discrimination and proprioception on the right side ofthe body as a result ofa lesion in the medial lemniscal system in the pons has a lesion in the left side ofthe pons.
The fibers ofthe lateral spinothalamic tract carry impulses for pain and temperature.A lesion in the area where these fibers decussate results in the bilateral loss ofpain and temperature sensations only at the level ofthe lesion,and
there is no loss ofsensation below the lesion. This occurs because fibers decussating above or below the lesion,as well as tracts that pass lateral to the lesion,are unaffected.
The damaged tracts are the lateral corticospinal tract,controlling motor functions on the right side ofthe body,and the lateral spinothalamic tract for pain and temperature sensations from the left side ofthe body.Damage to these tracts in the right side ofthe spinal cord produces the observed symptoms,because,in the cord,the lateral spinothalamic tract crosses over at the level ofentry,and is,therefore,located on the opposite side ofthe cord from its peripheral
nerve endings,whereas the corticospinal tract lies on the same side ofthe cord as its target muscles. Complete unilateral transection ofthe right side ofthe spinal cord results in loss ofmotor function (lateral corticospinal tract),
proprioception,and two-point discrimination (dorsal-column/medial-lemniscal system) on the same side ofthe body as the lesion,below the level ofthe lesion.Pain and temperature sensations (lateral spinothalamic tract) are lost on the opposite side ofthe body from below the level ofthe lesion.These symptoms describe the Brown-Sequard syndrome.Light touch is not greatly affected on either side because ofthe large number ofcollateral branches in the
anterior spinothalamic tract.
The right cerebral cortex controls the left side ofthe body.The motor cortex has a topographic representation ofthe opposite side ofthe body,with the hand,forearm,arm,and shoulder located approximately in the center of the precentral gyrus.The lesion is therefore in the center ofthe right precentral gyrus ofthe cerebrum.Some grosser control ofthe left-upper limb may still exist because ofthe indirect pathways,but there would be spastic paralysis.
Damage to the cerebellum can result in decreased muscle tone,balance impairment,a tendency to overshoot when reaching for or touching something,and an intention tremor. These symptoms are opposite to those seen with basal ganglia dysfunction.Cerebellar dysfunction exhibits very similar symptoms to those seen in an inebriated person,and the same tests could be applied,such as having the person touch their nose or walk a straight line. Memory storage for the 10 minutes prior to the accident was in short-term memory and was disrupted before it could be transferred to long-term memory.
Additional information:Anytime a person suffers a concussion there’s a possibility that he or she may later develop postconcussion syndrome. Symptoms include muscle tension or migraine headaches,reduced alcohol tolerance,difficulty
learning new things,reduction in creativity and motivation,fatigue,and personality changes,and the syndrome may last a month to a year. Postconcussion syndrome may be the result ofa slowly occurring subdural hematoma,which may
be missed by an early examination.

Chapter 15
1.The first sensations that occur when a person picks up an apple and bites into it are visual.The person holds the apple in his hand and looks at it.Visual input (which stimulates light receptors) tells him the size ofthe apple,and that it has a smooth surface,as well as its color.As the person bites into the apple and begins to chew,taste sensations (chemoreceptors) from the tongue provide information that the apple has characteristics ofbeing both sweet and sour.
Olfactory sensations (hemoreceptors) provide more specific information that the “fruity taste” is that ofan apple.
2.The lens ofthe eye is biconvex and causes light rays to converge.Ifthe lens is removed,then the replacement lens should also cause light rays to converge.A biconvex lens or a lens with a single convex surface would work.Bifocals or trifocals could also be recommended because ofthe loss ofaccommodation.
3.The light reflected by the tapetum lucidum could stimulate photoreceptors and increase the sensitivity ofthe eye to light,which could be an advantage when light levels are low.Because the same light image could stimulate different photoreceptors,however,there is a loss ofvisual acuity and a blurring ofvision.

4.Carrots contain vitamin A (retinoic acid),which can be used to form retinal.Retinal and opsin combine to form rhodopsin,which is found in rods.Rhodopsin is necessary for rods to respond to low levels oflight.Lack ofvitamin A can result in lack ofrhodopsin and night blindness.
5.By looking a few inches to the side,the image of the needle and thread is projected to the periphery ofthe retina,where there is the highest concentration ofrods.The rods function better than cones in low-light intensities.IfJean looks directly at the needle and thread,their image falls on the macula,which has few rods and mostly
cones,which don’t function well in dim light.By looking to the side,however,she is using a part of the retina where the photoreceptor cells are not as densely packed as in the macula,and the image is fuzzy rather than sharp.
6.This phenomenon is called a negative afterimage.While staring at the clock,the darkest portion ofthe image (the black clock) causes dark adaptation in part ofthe retina. That is,part ofthe retina becomes more sensitive to light.At the same time the lightest part ofthe image (the white wall) causes light adaptation in the rest ofthe retina,and that part
ofthe retina becomes less sensitive to light. When the man looks at a black wall,the dark adapted portion ofthe etina,which is more sensitive to light,produces more action potentials than does the light adapted part of the retina.Consequently,he perceives a light clock against a darker background.

7.A lesion ofthe optic chiasma results in visual loss in both the right and left temporal fields,a condition called itemporal hemianopsia,or tunnel vision.Tunnel vision can cause problems for normal functions,such as when driving a car, because the peripheral vision is severely limited. The occurrence ofthis condition can also suggest a much more serious roblem,such as a pituitary tumor,which sits just posterior to the optic chiasma.
8.The most likely area damaged is the spiral organ, where waves result in the production ofaction potentials.The action is much like ocean waves breaking on the shore during a violent storm as compared to those breaking in from a calm
ocean.Specifically,damage likely occurs in the part ofthe spiral organ near the oval window, because it is this part ofthe basilar membrane that vibrates the most in response to high- frequency sounds.
9.Normally,as pressure changes,the auditory tubes open to allow an equalization ofpressure between the middle ear and the external environment.Ifthis doesn’t occur,then the buildup ofpressure in the middle ear can rupture the tympanic membrane,or the pressure can be transmitted to the inner ear and cause sensoneural damage.
0.Normally,airborne sounds cause the tympanic membrane to vibrate,resulting in movement of the middle ear ossicles and the production of waves in the perilymph ofthe scala vestibuli. Vibration ofthe skull bones can also cause
vibration ofthe perilymph in the scala vestibuli.

Chapter 16
1.The sympathetic division ofthe ANS is responsible for dilation ofthe pupil. Preganglionic fibers from the upper horacic
region ofthe spinal cord pass through spinal nerves (T1 and T2),into the white rami communicantes,and into the sympathetic chain ganglia.The preganglionic fibers ascend the sympathetic chain and synapse with postganglionic neurons in the superior cervical sympathetic chain ganglia.The axons ofthe postganglionic neurons leave the ympathetic chain ganglia as small nerves that project to the pupil ofthe eye.
2.Reduced salivary and lacrimal gland secretions could indicate damage to the facial nerves, which innervate the submandibular,sublingual, and lacrimal glands.The glossopharyngeal nerves innervate the parotid glands but not the
lacrimal glands.
3.Cutting the preganglionic fibers in the white rami ofT2–T3 is the best way to eliminate innervation ofthe blood vessels in the skin.Cutting the gray rami at levels T2–T3 is inappropriate because the postganglionic fibers that innervate the hand blood vessels exit from the first thoracic and inferior cervical sympathetic chain ganglia. Cutting the spinal nerves is inappropriate because it eliminates all sensory and motor functions to the area supplied.

4.a.Pelvic nerves
b.Gray rami
c.Vagus nerves
d.Cranial nerves
e.Pelvic nerves
5.The parasympathetic division innervates the heart through the vagus nerves.The postganglionic nerve fibers ofthe vagus nerves release acetylcholine,which reduces heart rate. Methacholine can bind to the same receptors as acetylcholine and reduce heart rate.Side effects result from stimulating other parasympathetic effector organs.For example,stimulating the salivary glands results in increased salivation. Dilation ofthe pupils and sweating are effects
expected from sympathetic stimulation.The muscles ofrespiration are not regulated by the ANS,but they do respond to acetylcholine through somatic neurons.Methacholine would be expected to make contractions ofrespiratory muscles more likely.
6.One would expect mostly parasympathetic effects, because the effects ofacetylcholine are enhanced: blurring ofvision as a result ofcontraction of ciliary muscles,excess tear formation because of overstimulation ofthe lacrimal glands,frequent or involuntary urination because of overstimulation ofthe urinary bladder.Pallor resulting from asoconstriction in the skin is a sympathetic effect that would not be expected because skin blood vessels respond to norepinephrine.Muscle twitching or cramps of skeletal muscles might occur because they normally respond to cetylcholine.
7.Epinephrine causes vasoconstriction and confines the drug to the site ofadministration. This increases the duration ofaction ofthe drug locally and decreases systemic effects. Vasoconstriction also reduces bleeding ifa d
field (an area clear ofblood on its surface) i required. Because normal action potentials are produ the drug doesn’t act at the synapse between preganglionic and postganglionic neurons. Because injected norepinephrine works, sympathetic receptors in the heart are functioning and are not affected by the drug Therefore,the drug must somehow affect th postganglionic neurons.Possibly it inhibits neurotransmitter production or release from postganglionic eurons. Because cutting the white rami ofT1–T4 do affect the action ofthe drug,sympathetic preganglionic neurons in the spinal cord an sympathetic centers in the brain can be rule as a site ofaction.Because cutting the vagus nerves eliminates the effect ofthe drug,the cannot be acting at the synapse between the preganglionic neurons and the postganglion neurons,or between the synapse ofthe postganglionic neuron and the effector orga either division ofthe ANS.The drug must, therefore,excite parasympathetic centers in brainstem,resulting in a decrease in heart raa.Responses in a person who is extremely angry are primarily controlled by the sympathetic division ofthe ANS.These responses include increased heart rate a blood pressure,decreased blood flow to internal organs,increased blood flow to skeletal muscles,decreased contractions the intestinal smooth muscle,flushed sk the face and neck region,and dilation o pupils ofthe eyes. b.For a person who has just finished eatin and is now relaxing,the parasympathet reflexes are more important than sympathetic reflexes.The blood pressur heart rate are at normal resting levels,t blood flow to the internal organs is grea contractions ofsmooth muscle in the intestines are greater,and secretions tha achieve digestion are more active.Ifthe urinary bladder or the colon becomes distended,autonomic reflexes that resul urination or defecation can result.Blood flow to the skeletal muscles is reduced.

Chapter 17
1.Liver and kidney disease increases the concentration ofthis hormone in the blood,and the concentration would remain high for a longer time.The liver modifies the hormone to cause it to be excreted by the kidney more rapidly.In the case ofliver disease,the hormone is not modified and excreted rapidly.Therefore, the concentration becomes higher than normal and the concentration ofthe hormone remains high for longer than normal.A similar result is seen ifthe kidney is diseased and the hormone cannot be excreted rapidly.

2.Secretion ofhormones is usually controlled by a negative-feedback mechanism.Ifa hormone controls the concentration ofa substance in the circulatory system,the hormone is secreted in smaller amounts ifthe substance increases in the
circulatory system.Ifa tumor begins to secrete the substance in large amounts,the presence of the substance has a negative-feedback effect on the secretion ofthe hormone and the concentration ofthe hormone in the circulatory
system is very low. Usually intracellular mediator mechanisms respond quickly,and the effect ofthe hormone is brief. Intracellular receptor mechanisms usually take a long time (several hours) to respond,and their effects last much longer. Ifthe hormone is large and water-soluble,it’s probably functioning through an intracellular mediator mechanism,or if the hormone is lipid-soluble,it’s probably an intracellular receptor mechanism.Ifyou have the ability to monitor the oncentration ofa suspected intracellular mediator and it increases in response to the hormone,or ifyou can inhibit
the synthesis ofan intracellular mediator and it prevents the target cells’response to the hormone, it’s an intracellular mediator mechanism.Ifyou can inhibit the synthesis ofmRNA and this inhibits the action ofthe hormone,or ifyou can
measure an increase in mRNA synthesis in response to the hormone,then the mechanism is an intracellular receptor mechanism.
When the hormone binds to its receptor,the subunit ofthe G protein is released.GTP must bind to the subunit,however,before it can have its normal effect.Ifthe subunit cannot bind GTP,the hormone.therefore,has no effect on the target tissue. Inhibitors ofprostaglandin synthesis reduce prostaglandin synthesis in all tissues,not just in
those tissues in which prostaglandins produce undesirable effects.Symptoms such as inflammation,vomiting,and fever are reduced.
Because prostaglandins also play a role in producing beneficial effects in some tissues, however,these benefits would not occur normally.
Inhibitors ofprostaglandin synthesis may cause labor to be delayed or produce other undesirable responses due to their inhibitory effects on the synthesis ofprostaglandins.
Phosphodiesterase causes the conversion of cAMP to AMP,thus reducing the concentration ofcAMP.A drug that inhibits phosphodiesterase, therefore,increases the amount ofcAMP in cells where cAMP is produced.Therefore,an inhibitor ofphosphodiesterase increases the response ofatissue to a hormone that has cAMP as an intracellular mediator. A short half-life for epinephrine allows epinephrine to produce a short-lived response. The response to a potentially harmful or dangerous situation is terminated shortly after the harmful or dangerous situation passes.If
epinephrine had a long half-life,the heart rate and blood glucose would be elevated for a long time,even ifthe harmful or dangerous situation was very brief.
Because thyroid hormones are important in regulating the basal metabolic rate,a long half-life is an advantage.Thyroid hormones are secreted and have a prolonged effect without large fluctuations in the basal metabolic rate.If thyroid hormones had a short half-life,the basal metabolic rate might fluctuate with changes in the rate ofsecretion ofthyroid hormones.
Certainly the rate ofsecretion ofthyroid hormones would have to be controlled within narrow limits ifit did have a short half-life.
9.Ifliver disease results in a decrease ofplasma proteins to which thyroid hormones bind, higher-than-normal concentrations offree (unbound) thyroid hormones occur in the circulatory system.Because ofthe higher-than-normal concentration ofthyroid hormones that are unbound,the responses to thyroid hormones increase.In addition,the half-life ofthe thyroid hormones is shortened.Thus,as thyroid hormone secretion increases,the concentration ofthyroid hormone also increases.As the thyroid hormone secretion decreases,the concentration ofthyroid hormone also decreases.Thyroid hormones fluctuate in concentration in the circulatory system more than normal.
10.Elevated GnRH levels in the blood as a result of the GnRH-secreting tumor causes down-regulation ofGnRH receptors in the anterior pituitary.This decreases the ability ofGnRH to stimulate the anterior pituitary,and the rate of luteinizing hormone and follicle-stimulating hormone secretion by the anterior pituitary decreases and remains decreased as long as the GnRH levels are chronically elevated.Therefore, the functions ofthe reproductive system controlled by uteinizing hormone and follicle-stimulating hormone decrease.
11.Insulin levels normally change in order to
maintain normal blood sugar levels,despite periodic fluctuations in sugar intake.A constant supply ofinsulin from a skin patch might result in insulin levels that are too low when blood sugar levels are high (after a meal) and might be too high when blood sugar levels are low (between meals).In addition,insulin is a protein hormone that would not readily diffuse through the lipid barrier ofthe skin (see Chapter 5).Estrogen is a lipid soluble steroid hormone.

Chapter 18
1.The hypothalamohypophyseal portal system allows neurohormones that function as releasing and inhibiting ormones,which are secreted by neurons in the hypothalamus,to be carried directly from the hypothalamus to the anterior pituitary gland.Consequently,the releasing and inhibiting hormones are not diluted nor are they destroyed by the enzymes,which are abundant in the kidneys,liver,lungs,and general circulation, before they reach the anterior pituitary.Also the time it takes for releasing and inhibiting hormones to reach the anterior pituitary is less than ifthey
were secreted into the general circulation.

2.A hot environment increases ADH secretion. Because the amount ofwater lost in the form of sweat can be quite large,and because sweat is more dilute than the body fluids,sweating gradually increases the osmolality ofthe body
fluids. The increasing osmolality ofbody fluids stimulates an increase in ADH secretion.Thus,a hot environment can result in increased ADH secretion because ofan increasing osmolality of the body fluids.Increased ADH secretion in a hot environment reduces the amount ofwater lost in the form ofurine.Therefore,water is conserved.
3.Polydipsia and polyuria are consistent with either diabetes mellitus or diabetes insipidus.Diabetes ellitus,however,is consistent with an increased urine osmolality because ofthe large amount of glucose lost in the urine.Diabetes insipidus is consistent with urine with a low specific gravity because little water is reabsorbed by the kidney.
Thus urine has an osmolality close to that ofthe body fluids,and the rapid loss ofdilute urine results in a decrease in blood pressure.Thus polyuria with a low specific gravity is not consistent with diabetes mellitus but is consistent with diabetes insipidus.Administration ofADH reverses the symptoms ofdiabetes insipidus.
Neither polydipsia nor polyuria results from a lack ofglucagon or aldosterone.
4.The symptoms are consistent with acromegaly, which is a consequence ofelevated GH secretion after the epiphyses have closed.Increased GH causes enlarged finger bones,growth ofbony ridges over the eyes,and increased growth ofthe
jaw.The anterior pituitary tumor increases pressure at the base ofthe brain near the optic nerves as it enlarges.The pituitary rests in the sells turcica ofthe sphenoid bone;as it enlarges pressure increases because the pituitary is nearly
surrounded by rigid bone and the brain is located just superior to the pituitary.As the anterior pituitary enlarges because ofa tumor,it pushes superiorly and pressure is applied to the ventral portion ofthe brain.The GH also causes bone deposition on the inner surface ofskull bones, which also increases the pressure inside the skull.
5.Ifhyperthyroidism results from a pituitary abnormality,laboratory tests should show elevated TSH levels in the circulatory system in and T levels. If addition to elevated T3 4 hyperthyroidism results from the production of a nonpituitary thyroid-stimulating substance, laboratory tests should also show elevated T and 3T levels,but TSH levels would be low because of 4 the negative-feedback effects ofT and T on the 3 4 hypothalamus and pituitary gland.
6.The second student is correct.Low levels of vitamin D reduce calcium uptake in the gastrointestinal tract,which results in a decreased blood level ofcalcium ions.As blood calcium levels decrease,the rate ofPTH secretion would increase.Parathyroid hormone increases bone breakdown,which maintains blood calcium levels,even ifvitamin D deficiency exists for a prolonged time. Osteomalacia results because ofthe increased bone reabsorption necessary to maintain normal blood calcium levels.
7.A glucose tolerance test can distinguish between these conditions.The person would consume glucose after a period offasting.Over the next few hours the blood glucose levels in a healthy person increase and then return to fasting levels. The blood glucose levels always remain within the normal range,however.In a person with diabetes,the blood glucose levels increase to above-normal levels and remain elevated for several hours.In a person who secretes large amounts ofinsulin,blood glucose levels would increase,and then they would decrease to below-normal levels within a relatively short time.
8.Because the person is a diabetic and probably is taking insulin,the condition is more likely to be insulin shock than a diabetic coma.To confirm the condition,however,a blood sample should be taken.Ifthe condition is due to a diabetic coma, then the blood glucose levels will be elevated.If the condition is due to insulin shock,the blood glucose levels will be below normal.In the case of insulin shock,glucose can be administered intravenously.In the case ofdiabetic coma, insulin should be administered.An isotonic solution containing insulin can be administered to reduce the osmolality ofthe extracellular fluid.

9.Adrenal diabetes results from elevated and uncontrolled secretion ofglucocorticoid hormones,such as cortisol,from the adrenal gland.Because glucocorticoid hormones increase blood glucose levels,elevated secretion ofthese hormones results in elevated blood glucose levels and symptoms similar to diabetes mellitus. Pituitary diabetes results from elevated secretion ofGH from the anterior pituitary.Elevated GH causes an increase in blood glucose levels and, therefore,produces symptoms similar to diabetes mellitus.Prolonged elevation ofboth glucocorticoids and growth hormone secretion can lead to the development ofdiabetes mellitus ifthe insulin-secreting cells ofthe pancreatic islets degenerate because ofthe prolonged need to secrete insulin in response to the elevated blood glucose levels.
10.Elevated epinephrine from the adrenal medulla promotes elevated blood pressure and increases the work load on the heart,increases the rate of metabolism,and results in increased sweating and nervousness.The risk ofheart attack and
stroke are increased.Elevated cortisol causes hyperglycemia and can lead to diabetes mellitus, a depressed immune system with increased susceptibility to infections,and destruction of proteins leading to tissue wasting.

Endocrine Glands

Chapter 19
1.Because ofthe rapid destruction ofthe red blood cells we would expect erythropoiesis to increase in an attempt to replace the lost red blood cells.
The reticulocyte count would therefore be above normal.Jaundice is a symptom ofhereditary hemolytic anemia because the destroyed red blood cells release hemoglobin,which is converted into bilirubin.Removal ofthe spleen cures the disease because the spleen is the major site ofred blood cell destruction.
2.Blood doping increases the number ofred blood cells in the blood,thereby increasing its oxygen-carrying capacity. The increased number ofred blood cells also makes it more difficult for the blood to flow through the blood vessels, increasing the heart’s workload.
3.Symptoms resulting from decreased red blood cells are associated with a decreased ability ofthe blood to carry oxygen:shortness ofbreath, weakness,fatigue,and pallor.Symptoms resulting from decreased platelets are associated
with a decreased ability to form platelet plugs and clots:small areas ofhemorrhage in the skin (petechiae),bruises,and decreased ability to stop bleeding.Symptoms resulting from decreased white blood cells could include an increased susceptibility to infections.
4.Hypoventilation results in decreased blood oxygen levels,which stimulates erythropoiesis. Therefore,the number ofred blood cells increases and produces secondary polycythemia.

5.Removal ofthe stomach removes intrinsic factor, which is necessary for vitamin B absorption. 12 Therefore,the patient develops pernicious anemia.Lack ofstomach acid can decrease iron absorption in the small intestine and result in iron-deficiency anemia. 6.Vitamin B and folic acid are necessary for 12 blood cell division.Lack ofthese vitamins results in pernicious anemia.Iron is necessary for the production ofhemoglobin.Lack ofiron results in iron-deficiency anemia.Vitamin K is necessary for the production ofmany blood clotting factors.Lack ofvitamin K can greatly increase blood clotting time,resulting in excessive bleeding.

19CardiovascularSystemBlood.pdf

Chapter 20
1.The walls ofthe ventricles are thicker than the walls ofthe atria because the ventricles must produce a greater pressure to pump blood into the arteries.Only a small pressure is required to pump blood from the atria into the ventricles during diastole.The wall ofthe left ventricle is thicker than the wall ofthe right ventricle because the left ventricle produces a much greater pressure to force blood through the aorta than the right ventricle produces to move blood through the pulmonary trunk and pulmonary arteries.
2.During systole,the cardiac muscle in the right and left ventricles contracts,which compresses the coronary arteries. During diastole,the cardiac muscle ofthe ventricles relaxes and blood flow through the coronary arteries increases. The diastolic pressure is sufficient to cause blood to flow through coronary arteries during diastole.
3.A drug that prolongs the plateau ofcardiac muscle cell action potentials prolongs the time each action potential exists and increases the refractory period.Therefore,the drug would slow the heart.A drug that shortens the plateau shortens the length oftime each action potential exists and shortens the refractory period. Therefore,the drug could allow the heart rate to increase further.
4.Endurance-trained athletes have decreased heart rates because their cardiac muscle undergoes hypertrophy in response to exercise.The hypertrophied cardiac muscle causes the stroke volume to increase substantially.The increased stroke volume is sufficient to maintain an adequate cardiac output and blood pressure even though the heart rate is slower.
5.The two heartbeats occurring close together can be heard through the stethoscope,because the heart valves open and close normally during each ofthe heartbeats even ifthey are close together.The second heartbeat,however, produces a greatly reduced stroke volume because there’s not enough time for the ventricles to fill with blood between the first and second contraction ofthe heart.Thus,the preload is reduced.Because the preload is reduced,the second heartbeat has a greatly reduced stroke volume.The reduced stroke volume fails to produce a normal pulse.The
pulse deficit,therefore,results from the reduced stroke volume ofthe second ofthe two beats that are very close together.
6.Aerobic training causes hypertrophy ofthe cardiac muscle in the heart and causes the heart to produce a greater stroke volume as a consequence.The heart rate can decrease while the cardiac output remains the same because
cardiac output is equal to the stroke volume times the heart rate.Ifthe stroke volume increases,the heart rate can decrease and the cardiac output can remain the same.
7.Atrial contractions complete ventricular filling, but atrial contractions are not primarily responsible for ventricular filling.Therefore,ifthe atria are fibrillating,blood can still flow into the ventricles and ventricular contractions can occur.
As long as the ventricles contract rhythmically the heart can pump an adequate amount ofblood even though the atria are fibrillating. Ifthe ventricles undergo fibrillation,however,they cannot fill with blood and cannot function as pumps. Thus the stroke volume will become too low to maintain adequate blood flow to tissues. 8.The results depend on Cee Saw’s response to the conditions ofthe laboratory.First,as Cee Saw’s head is lowered,gravity causes blood pressure in
the carotid sinuses and aortic arch to increase. The increased blood pressure stimulates baroreceptors,which detect the increased blood pressure and send action potentials indicating that blood pressure increased to the cardioregulatory center in the medulla oblongata along sensory nerve fibers.The cardioregulatory center increases parasympathetic stimulation and reduces sympathetic stimulation ofthe heart.Thus the heart rate decreases.Second,if,as her head is lowered,Cee Saw becomes excited, the sympathetic division ofthe ANS becomes
more active.The resulting increase in sympathetic stimulation ofthe heart causes the heart rate to increase.

9.After Cee Saw is tilted so that her head is higher than her feet for a few minutes,the regulatory mechanisms that control blood pressure adjust so that the heart pumps sufficient blood to supply the needs ofher tissues.Ifshe is then tilted so that her head is higher than her feet,gravity would cause blood to flow toward her feet,and the blood pressure in the carotid sinus and aortic arch would decrease.The decrease in blood pressure would be detected by the baroreceptors in these vessels and would activate baroreceptor reflexes.The result is increased sympathetic and decreased parasympathetic stimulation ofthe heart and an increase in the heart rate.The increased heart rate would function to increase the blood pressure to its normal value. 1

10.An ECG measures the electrical activity ofthe heart and does not indicate a slight heart murmur.Heart murmurs are detected by listening to the heart sounds.The boy may have a heart murmur,but the mother does not understand the basis for making such a diagnosis.
11.When both common carotid arteries are clamped,the blood presure within the internal carotid arteries drops dramatically. The decreased blood pressure is detected,and the baroreceptor reflex increases heart rate and stroke volume.The resulting increase in cardiac output causes the increase in blood pressure.
12.Venous return declines markedly in hemorrhagic shock because ofthe loss ofblood volume.With decreased venous return,stroke volume decreases (Starling’s law ofthe heart). The decreased stroke volume results in a decreased cardiac output,which produces a decreased blood pressure.In response to the decreased blood pressure,the baroreceptor reflex causes an increase in heart rate in an attempt to restore normal blood pressure.
However,with inadequate venous return the increased heart rate is not able to restore normal blood pressure.

20CardiovascularSystemTheHeart.pdf

Chapter 21
1.a.Aorta,left coronary artery,circumflex artery, posterior interventricular artery;or aorta, right coronary artery,posterior interventricular artery

b.Aorta,brachiocephalic artery,right common carotid artery,right internal carotid artery; or aorta,left common carotid artery,left internal carotid artery
c.Aorta,brachiocephalic artery,right subclavian artery,right vertebral artery, basilar artery;or aorta,left subclavian artery, left vertebral artery,basilar artery
d.Aorta,left or right common carotid artery, left or right external carotid artery e.Aorta,left subclavian artery,axillary artery, brachial artery,radial or ulnar artery,deep or superficial palmar arch,digital artery (on the right:the brachiocephalic artery would be included)
f.Aorta,common iliac artery,external iliac artery,femoral artery,popliteal artery, anterior tibial artery
g.Aorta,celiac artery,common hepatic artery
h.Aorta,superior mesenteric artery,intestinal branches
i.Aorta,left or right internal iliac artery
2.a.Great cardiac vein,coronary sinus;or anterior cardiac vein
b.Transverse sinus,sigmoid sinus,internal jugular vein,brachiocephalic vein,superior vena cava
c.Retromandibular vein,external jugular vein,subclavian vein,brachiocephalic vein, superior vena cava
d.Deep:vein ofhand,radial or ulnar vein,brachial vein,axillary vein,subclavian vein,brachiocephalic vein,superior vena cava
Superficial:vein ofhand,radial or ulnar vein,cephalic or basilic vein,axillary vein, subclavian vein,brachiocephalic vein, superior vena cava

e.Deep:vein offoot,dorsalis veins offoot, anterior tibial vein,popliteal vein,femoral vein,external iliac vein,common iliac vein, inferior vena cava Superficial:vein offoot,great saphenous vein,external iliac vein,common iliac vein, inferior vena cava;or vein offoot,small saphenous vein,popliteal vein,femoral vein, external iliac vein,common iliac vein, inferior vena cava
f.Gastric vein or gastroepiploic fein,hepatic portal vein,hepatic sinusoids,hepatic vein, inferior vena cava g.Renal vein,inferior vena cava
h.Hemiazygous vein or accessary hemiazygous vein,azygous vein,superior vena cava
3.A superficial vessel would be easiest,such as the right cephalic or basilic vein.The catheter is passed through the cephalic (or brachial) vein and the superior vena cava to the right atrium. Because the pulmonary veins are not readily
accessible,dye would not normally be placed directly into them.Instead,the dye would be placed in the right atrium using the procedure just described.The dye passes from the right atrium into the right ventricle,the pulmonary arteries,the lungs,the pulmonary veins,and into the left atrium.Ifthe catheter has to be placed in the left atrium,it could be inserted through an artery,such as the femoral artery,and passed via the aorta to the left ventricle and then into the left atrium.
4.The viscosity ofthe blood is affected primarily by the hematocrit.As hematocrit increases,the viscosity ofthe blood increases logarithmically,so that even a small increase in hematocrit results in a large increase in viscosity.Greater force is therefore not needed to cause blood to flow through the blood vessels.With the increased blood volume,blood flow through vessels is adequate without an increase in viscosity.
5.The resistance to blood flow is less in the vena cavae for two reasons:first,the diameter ofone vena cava is greater than the diameter ofthe aorta and second,an increased diameter ofa blood vessel reduces resistance to flow (see
Poiseuille’s law).In addition,there are two venae cavae,the superior vena cava and the inferior vena cava,but only one aorta.The blood flow through the aorta and the venae cavae is about equal,but the velocity ofblood flow is much higher in the aorta than it is in the venae cavae.
6.According to Laplace’s law,as the diameter ofa blood vessel increases,the force applied to the vessel wall increases,even ifthe pressure remains constant.The increased connective tissue found in the walls ofthe large blood vessels therefore makes the wall ofthose vessels stronger and more capable ofresisting the force applied to the wall.
7.Veins and lymphatic vessels have one-way valves in them.Massage creates a cycle ofincreasing and decreasing pressure to the veins,which rhythmically compresses them.The compression ofthe veins forces fluid to move out ofthe limb through both veins and lymphatic vessels.The movement offluid through the veins lowers the pressure within the venous end ofthe capillary. Thus the forces that move fluid into the capillaries at their venous ends are greater and they move more interstitial fluid into the capillaries.Compression ofthe lymphatic capillaries also causes more lymphatic fluid to move into the lymphatic vessels.Because there is less fluid in the limb,the edema decreases.
8.The nursing student’s diagnosis was incorrect. Blood pressure measurements are normally made in either the right or left arm,both of which are close to the level ofthe heart.Blood pressure taken in the leg is influenced by pressure created by the pumping action ofthe heart,but the effect ofgravity on the blood,as it flows into the leg,also influences the blood pressure in a substantial way.In this case gravity increases blood pressure from about 120 mm Hg for the systolic pressure to 200 mm Hg.
9.Decreased liver function includes a decrease in the synthesis ofplasma proteins.Consequently the concentration ofplasma proteins decreases, and the colloid osmotic pressure ofthe blood decreases.Therefore,less water moves by
osmosis into the capillaries at the venous ends and the result is edema.
10.Chemoreceptors in the medulla oblongata detect carbon dioxide and the pH ofthe blood.The normal blood levels ofCO and pH stimulate 2 these chemoreceptors,which in turn stimulate the vasomotor center.The vasomotor center keeps blood vessels partially constricted under resting conditions.This basal level ofactivity is called the vasomotor tone.Blowing offCO 2 reduces the blood levels ofcarbon dioxide and increases the pH ofthe body fluids.These changes reduce vasomotor tone and result in vasodilation.Ifa person hyperventilates and blows offCO ,the stimulus to the vasomotor 2 center decreases,which results in a decrease in vasomotor tone.The decrease in vasomotor tone results in a decrease in systemic blood pressure. Ifthe blood pressure decreases enough,the blood flow to the brain decreases and can cause a sensation ofdizziness or can even cause a person to lose consciousness.
11.Epinephrine is secreted from the adrenal medulla in response to stressful stimuli and the epinephrine stimulates responses that are consistent with increased physical activity.
Vasoconstriction ofthe blood vessels in the skin shunts blood away from the skin to skeletal muscles.Vasodilation occurs in blood vessels of exercising skeletal muscles.Blood flow through the exercising skeletal muscles therefore increases.Because epinephrine causes vasodilation ofthe blood vessels ofcardiac muscle,blood flow through the cardiac muscle increases.This response is consistent with the increased work performed by the heart under conditions ofincreased physical activity.
12.The hot Jacuzzi increases Skinny’s skin and body temperature.As a result,the blood vessels ofthe skin dilate.Because the blood vessels dilate, peripheral resistance decreases,causing the blood pressure to decrease. The baroreceptors of the carotid sinus and aortic arch detect the decrease in blood pressure and send action potentials to the cardioregulatory center in the medulla oblongata.As a result,the sympathetic stimulation to the heart increases and the heart rate,in response,increases also.The increased heart rate elevates the blood pressure back to within its normal range ofvalues. Cardiovascular System Peripheral Circulation

Chapter 22
1.Elevation ofthe limb reduces blood pressure in the limb,resulting in less fluid movement from the blood into the tissues (see chapter 21).Thus, the edema is reduced as the lymphatic system removes fluid from the tissues faster that it enters them.Massage moves lymph through the lymphatic vessels in the same fashion as contraction ofskeletal muscle.The application ofpressure periodically to lymphatic vessels forces lymph to flow toward the trunk ofthe body,but valves prevent the flow oflymph in the reverse direction.The removal oflymph from the tissue helps to relieve edema.
2.Normally T cells are processed in the thymus and then migrate to other lymphatic tissues.Without the thymus this processing is prevented.Because there are normally five T cells for every one B cell, the number oflymphocytes is greatly reduced. The loss ofT cells results in an increased susceptibility to infection and an inability to reject grafts because ofthe loss ofcell-mediated immunity.In addition,since helper T cells are involved with activation ofB cells,antibody-mediated immunity is also depressed.

3.That there is no immediate effect indicates there is a reservoir ofT cells in the lymphatic tissue. As the reservoir is depleted through time,the number oflymphocytes decreases and cell-mediated immunity is depressed,the animal is more susceptible to infections,and the ability to reject grafts decreases.The ability to produce antibodies decreases because ofthe loss ofhelper T cells that are normally involved with the activation ofB cells.
4.Injection B results in the greatest amount of antibody production.At first,the antigen causes a primary response. A few weeks later,the slowly released antigen causes a secondary response, resulting in a greatly increased production of
antibodies.Injection A doesn’t cause a secondary response because all ofthe antigen is eliminated by the primary response.
5.Ifthe patient has already been vaccinated,the booster shot stimulates a memory (secondary) response and rapid production ofantibodies against the toxin.Ifthe patient has never been vaccinated,vaccinating now is not effective because there’s not enough time for the patient to develop his or her own primary response.
Therefore,antiserum is given to provide immediate,but temporary,protection. Sometimes both are given:The antiserum provides short-term protection and the tetanus vaccine stimulates the patient’s immune system to provide long-term protection.Ifthe shots are given at the same location in the body,the antiserum (antibodies against the tetanus toxin) could cancel the effects ofthe tetanus vaccine (tetanus toxin altered to be nonharmful).
6.The infant’s antibody-mediated immunity is not functioning properly,whereas his cell-mediated immunity is working properly.This explains the susceptibility to extracellular bacterial infections and the resistance to intracellular viral infections.
It took so long to become apparent because IgG from the mother crossed the placenta and provided the infant with protection.The infant began to get sick after these antibodies degraded.
7.Bone marrow is the source ofthe lymphocytes responsible for adaptive immunity.Ifsuccessful, the transplanted bone marrow starts producing lymphocytes and the baby has a functioning immune response.In this case,there’s a graft
versus host rejection in which the lymphocytes in the transplanted red marrow mount an immune attack against the baby’s tissues,resulting in death.
8.At the first location an antibody-mediated response results in an immediate hypersensitivity reaction,which produces inflammation.Most likely the response resulted from IgE antibodies. At the second location a cell-mediated response results in a delayed hypersensitivity reaction, which produces inflammation.This probably involves the release ofcytokines and the lysis of cells.At the other locations there is neither an antibody-mediated nor a cell-mediated response.
9.The ointment is a good idea for the poison ivy, which causes a delayed hypersensitivity reaction, for example,too much inflammation.For the scrape it’s a bad idea,because a normal amount ofinflammation is beneficial and helps to fight infection in the scrape.
10.Because antibodies and cytokines both produce inflammation,the fact that the metal in the jewelry results in inflammation is not enough information to answer the question.However,the fact that it took most ofthe day (many hours) to develop the reaction indicates a delayed hypersensitivity reaction and therefore cytokines.

22 Lymphatic System and Immunity.pdf

Chapter 23
1.Minute respiratory volume is equal to the respiration rate times the tidal volume.With a respiration rate of12 breaths per minute and a tidal volume of500 mL per breath,normal minute ventilation is 6000 mL/min (12 500).
Rapid (24 breaths per minute),shallow (250 mL per breath) breathing results in the same minute ventilation,that is,6000 mL/min (24 250). Alveolar ventilation rate (V ) is the respiratory A rate (frequency;f) times the difference between
the tidal volume (V ) and dead space (V ). T D Vf(VV ) A T D Normal resting V12(500150) A 4200mL/min In this case ofrapid shallow breathing, V24(250150)2400mL/min A Thus,even though the minute ventilation is the same in both cases,the alveolar ventilation rate is less during rapid,shallow breathing because there’s less effective exchange ofgases between the atmosphere and the dead space.Because there’s less exchange ofgases,the partial pressures ofalveolar gases become closer to the partial pressure ofblood gases.Consequently, the alveolar partial pressure ofO decreases and
2 increases. the alveolar partial pressure ofCO 2 This decreases the concentration gradients for gases,resulting in less gas exchange between alveolar air and blood.

2.We expect vital capacity to be greatest when standing because the abdominal organs move inferiorly,thereby allowing greater depression ofthe diaphragm and a greater inspiratory reserve volume.

3.The hose increases dead space and therefore decreases alveolar ventilation.Ima Diver has to compensate by increasing respiratory rate or tidal volume. Ifthe hose is too long,she won’t be able to ompensate.Furthermore,with a long hose,air is simply moved back and forth in the hose with little exchange ofair between the atmosphere and the lungs taking place. Another consideration is the effect ofwater pressure on the thorax,which decreases compliance and
increases the work ofventilation. In fact,a few feet underwater there’s enough pressure on the thorax to prevent the intake ofair through even a short hose connected to the atmosphere.
4.The increase in atmospheric pressure increases the partial pressure ofoxygen.According to Henry’s law,as the partial pressure ofoxygen increases, the amount ofoxygen dissolved in the body fluids increases.The increase in dissolved oxygen is detrimental to the gangrene bacteria.Because hemoglobin is already saturated with oxygen,the HBO treatment doesn’t increase the ability of hemoglobin to pick up oxygen in the lungs.
5.Compression causes a decrease in thoracic volume and therefore lung volume.
Consequently,pressure in the lungs increases over atmospheric pressure and air moves out of the lungs.Raising the arms expands the thorax and lungs.This results in a lower-than-atmospheric pressure in the lungs,and air moves into the lungs.
6.The victim’s lungs expand because ofthe pressure generated by the rescuer’s muscles of expiration.This fills the lungs with air that has a greater pressure than atmospheric pressure.Air flows out ofthe victim’s lungs as a result ofthis pressure difference and because ofthe recoil of the thorax and lungs.Although the partial pressure ofoxygen ofthe rescuer’s expired air is less than atmospheric,enough oxygen can be provided to sustain the victim.The lower partial
pressure ofoxygen could also activate the chemoreceptor reflex and stimulate the victim to breathe.In addition,the rescuer’s partial pressure ofcarbon dioxide is higher than atmospheric and this could activate the chemosensitive area in the medulla.

7.All else being equal (i.e.,the thickness ofthe respiratory membrane,the diffusion coefficient ofthe gas,and the surface area ofthe respiratory membrane),diffusion is a function ofthe partial pressure difference ofthe gas across the
respiratory membrane.The greater the difference in partial pressure,the greater the rate ofdiffusion.The greatest rate ofoxygen diffusion should therefore occur at the end of inspiration when the partial pressure ofoxygen in the alveoli is at its highest.The greatest rate of carbon dioxide diffusion should occur at the end ofinspiration when the partial pressure of carbon dioxide in the alveoli is at its lowest.

8.Because the partial pressure ofoxygen at high altitudes decreases,a shift to the left is advantageous.Such a shift enables hemoglobin to pick up more oxygen at a lower partial pressure ofoxygen.
9.Cutting the phrenic nerves eliminates contraction ofthe diaphragm.Tidal volume decreases drastically,and death probably results.Cutting the intercostal nerves eliminates raising ofthe ribs and sternum and decreases tidal volume,
unless the diaphragm compensates.Cutting the vagus nerves eliminates the Hering-Breuer reflex and results in a greater-than-normal inspiration. This increases tidal volume.

10.While hyperventilating and making ready to leave your instructor behind,you might make the following arguments:
• Hyperventilation increases the oxygen content ofthe air in the lungs;therefore,you would have more oxygen to use when holding your breath.
• It’s hemoglobin that is saturated.
Hyperventilation increases the amount of oxygen dissolved in the blood plasma.
• Hyperventilation decreases the amount of carbon dioxide in the blood.This makes it possible to hold one’s breath for a longer time because ofa decreased urge to take a breath.
• Hyperventilation activates alveoli not in use
because increasing alveolar oxygen and decreasing alveolar carbon dioxide causes lung arterioles to relax,thereby increasing blood flow through the lungs.

Respiratory System

Chapter 24
1.With the loss ofthe swallowing reflex,the vocal folds no longer occlude the glottis. Consequently,vomit can enter the larynx and block the respiratory tract.
2.Without adequate amounts ofhydrochloric acid, the pH in the stomach is not low enough for the activation ofpepsin. This loss ofpepsin function results in inadequate protein digestion.Ifthe food is well chewed,however,proteolytic enzymes in the small intestine (e.g.,trypsin,chymotrypsin) can still digest the protein.Ifthe stomach secretion ofintrinsic factor decreases, the absorption ofvitamin B is hindered.Inadequate 12 amounts ofvitamin B can result in decreased 12 red blood cell production (pernicious anemia).

3.Even though ulcers are apparently ultimately caused by bacteria,overproduction of hydrochloric acid due to stress is a possible contributing factor.Reducing hydrochloric acid production is recommended.In addition to antibiotic therapy, commonly recommended solutions include relaxation,drugs that reduce stomach acid secretion,and antacids to neutralize the hydrochloric acid. Smaller meals are also advised because distension ofthe stomach stimulates acid production. Proper diet is also important.The patient is also advised to avoid alcohol,caffeine,and large amounts of protein because they stimulate acid production. Ingestion offatty acids is recommended because they inhibit acid production by causing release ofgastric inhibitory polypeptide and cholecystokinin.Stress also stimulates the sympathetic nervous system,which inhibits duodenal gland secretion.As a result,the duodenum has less ofa mucous coating and is more susceptible to gastric acid and enzymes. Relaxing after a meal helps decrease sympathetic activities and increase parasympathetic activities.
4.Lack ofbile due to blockage ofthe common bile duct can result in jaundice (due to an accumulation ofbile pigments in the blood) and clay-colored stools (due to lack ofbile pigments in the feces).Blockage ofthe bile duct causes abdominal pain,nausea,and vomiting. Fat absorption is impaired because ofthe absence of bile salts in the duodenum and a loose, bulky stool would result.Lack offat absorption reduces the absorption offat-soluble vitamins such a vitamin K, resulting in lack ofnormal clotting function.
5.The patient would still be able to defecate. Following a meal the gastrocolic and duodenocolic reflexes could initiate mass movement ofthe feces into the rectum.In the rectum,local reflexes and the defecation reflex (integrated in the sacral level ofthe cord and not requiring connections to high brain centers) would cause defecation.Awareness ofthe need to defecate would be lost (due to loss ofsensory input to the brain) and the ability to voluntarily prevent defecation via the external anal sphincter would also be lost.
6.The accumulation ofmaterials above the site of impaction and the action ofbacteria on the material would result in an increase in osmotic pressure in the area.Water would move by osmosis into the colon above the site of
impaction.Bowel impaction is very dangerous and must be treated quickly.The increased volume and distention ofthe digestive tract above the site ofimpaction causes compression ofthe mucosa.This compression can occlude blood vessels in the mucosa and lead to necrosis. Necrosis ofthe mucosa results in increased permeability ofthe mucosa,thus allowing toxic organisms and substances in the digestive tract to enter the circulation,resulting in septic shock.

24 Digestive System.pdf

Chapter 25
1.In figure 25.2,the Daily Value for saturated fat is listed as less than 20 g for a 2000 kcal/day diet. The % Daily Values appearing on food labels are based on a 2000 kcal/day diet.Therefore,the % Daily Value for saturated fat for one serving of this food is 10% (2/20 .10,or10%).
2.According to the Daily Value guidelines,total fats should be no more than 30% oftotal kilocaloric intake.For someone consuming 3000 kcal/day this is 900 kcal (3000 kcal 0.30).There are 9kcal in a gram of fat.Therefore,the maximum amount (weight) offats the active teenage boy should consume is 100 g (900/9).
3.The % Daily Value is the amount ofthe nutrient in one serving divided by its Daily Value.Therefore the % Daily Value is 10% (10/100 .10,or 10%).
4.The % Daily Value for one serving ofthe food is 10% (see answer to question 3).Since there are four servings in the package,ifthe teenager eats halfofthe food in the package,he consumes two servings.Thus,he eats 20% (10% 2) ofthe
recommended maximum total fat. 5.The protein in meat contains all ofthe essential amino acids and is a complete protein food. Although plants contain proteins,a variety of different plants must be consumed to ensure that all the essential amino acids are included in
adequate amounts.Also,plants contain less protein per unit weight than meat,so a larger quantity ofplants must be consumed to get the same amount ofprotein.
6.Copper is necessary for proper functioning of the electron-transport chain.Inadequate copper in the diet results in reduced ATP production, that is,not enough energy.
7.Fasting can be damaging because proteins are used to produce glucose.The glucose enters the blood and provides an energy source for the brain.This breakdown ofproteins can damage tissues such as muscle and disrupt chemical
reactions regulated by enzyme systems.A single day without food,however,is unlikely to cause permanent harm.
8.Weight is lost when kilocalories used per day exceeds kilocalories ingest per day.About 60% of the kilocalories used per day is due to basal metabolic rate.A person with a high basal metabolic rate loses weight faster than a person
with a low basal metabolic rate,all else being equal.Another factor to consider is the amount ofphysical activity,which accounts for about 30% ofkilocalories used per day.An active person loses more weight than a sedentary person does.
9.Amino acids,derived from ingested proteins,are necessary to build muscles.As Lotta and her friend discovered,excess proteins don’t accelerate this process.Excess proteins can be used as an energy source in oxidative deamination,for the formation ofthe intermediate molecules ofcarbohydrate metabolism,or in gluconeogenesis.Excess proteins are also converted into storage molecules through glycognesis or lipogenesis. Lotta is in positive nitrogen balance because the
amount ofnitrogen she gains from her diet is greater than the amount she loses by excretion. Some ofthe nitrogen in the amino acids she ingests is incorporated into the proteins ofher muscles as they enlarge.
10.No,this approach doesn’t work because he is not losing stored energy from adipose tissue.In the sauna,he gains heat,primarily by convection from the hot air and by radiation from the hot walls.The evaporating sweat is removing heat gained form the sauna.The loss ofwater will make him thirsty,and he will regain the lost weight from fluids he drinks and food he eats.

Nutrition, Metabolism, and Temperature Regulation

Chapter 26
1.The large volume ofhypoosmotic fluid ingested increases blood volume and causes blood osmolality to decrease.The increased blood volume is detected by baroreceptors,and the decreased blood osmolality is detected by osmoreceptors in the hypothalamus.The response to these stimuli is inhibition ofADH secretion. The alcohol in the beer also inhibits ADH secretion.The increased volume inhibits the renin-angiotensin-aldosterone mechanism, which,in turn,inhibits ldosterone secretion.The changes in aldosterone,however,take much longer to influence kidney function than changes
in ADH.As a result ofthese changes a large volume ofdilute urine is produced until the blood osmolality and blood volume return to normal.
2.Once the salt is absorbed,the osmolality ofthe blood increases.The increased osmolality of blood is detected by osmoreceptor neurons in the hypothalamus,thereby stimulating ADH secretion and inhibiting aldosterone secretion.A
small volume ofconcentrate urine is produced as a result,until the excess salt is eliminated and the blood osmolality returns to its normal value.
3.The hypoosmotic sweat loss results in more loss of water than electrolytes.This simultaneously decreases plasma volume and increases blood osmolality,thereby stimulating increased ADH secretion.In addition,the decreased plasma
volume stimulates the renin-angiotensin-aldosterone mechanism,resulting in a decreased glomerular filtration rate and increased aldosterone secretion.The effect ofthe changes is to produce a small amount ofconcentrated urine.
4.The loss ofsweat results in a loss ofwater and electrolytes.Replacing just the water restores blood volume and also decreases blood osmolality.At first,the decreased osmolality inhibits ADH secretion,and dilute urine is produced.As blood volume decreases as a result ofurine production,however,ADH secretion and the renin-angiotensin-aldosterone mechanisms are stimulated.Consequently,urine concentration increases,and only a small amount ofurine is produced.
5.As aldosterone levels decrease,sodium reabsorption in the nephron decreases and, consequently,plasma sodium levels decrease.The sodium is lost in the urine,and water follows the sodium by osmosis.Thus,a large amount of urine that has a high concentration ofsodium is produced.The loss ofwater reduces blood volume,which causes the low blood pressure.As aldosterone levels decrease potassium secretion into the nephron decreases,resulting in an increase in plasma potassium levels.The increased extracellular potassium causes depolarization ofnerve and muscle membranes,
leading to tremors ofskeletal muscles and cardiac arrhythmias including fibrillation.
6.There are several ways to decrease glomerular filtration rate:
a.Decrease hydrostatic pressure in the glomerulus.
1.Decrease systemic arterial blood pressure.
a.Decrease extracellular fluid volume.
b.Decrease peripheral resistance.
c.Decrease cardiac output.
2.Constrict or occlude the afferent arteriole.
3.Relax the efferent arteriole.
b.Increase glomerular capsule pressure.
c.Increase the colloid osmotic pressure ofthe plasma.
d.Decrease the permeability ofthe filtration barrier.
e.Decrease the total area ofthe glomeruli available for filtration.

7.Assume that the ascending limb ofthe loop of Henle and the distal tubules are impermeable to sodium and other ions but actively pump out water.Other characteristics ofthe kidney are assumed to be unchanged.As the urine moves up the ascending limb it becomes hyperosmotic, because sodium remains behind as water is pumped out.Assuming that the collecting ducts are impermeable to sodium,upon reaching the collecting ducts the presence or absence ofADH determines the final concentration ofthe urine.
IfADH is absent,there’s little or no exchange of water as the urine passes down the collecting ducts and a yperosmotic urine will be produced.On the other hand,ifADH is present, water moves from the interstitial fluid into the collecting ducts,thus diluting the urine and producing a hypoosmotic urine.

8.Urea is partially responsible for the high osmolality ofthe interstitial fluid in the medulla ofthe kidney.Since a high osmolality ofthe interstitial fluid must exist for the kidney to produce a concentrated urine,a small amount of urea in the kidney results in the production of dilute urine by the kidney.
9.A low-salt diet tends to reduce the osmolality of the blood.Consequently,ADH secretion is inhibited,producing dilute urine and thus eliminating water.This in turn reduces blood volume and blood pressure.
10.As the loops ofHenle become longer,the mechanisms that increase concentration ofthe interstitial fluid ofthe medulla become more efficient,thus raising the concentration ofthe interstitial fluid.The maximum concentration for urine is determined by the concentration of the interstitial fluid deep in the medulla ofthe kidneys.The higher the concentration of interstitial fluid in the medulla ofthe kidney,the greater the concentration ofthe urine the kidney
is able to produce. Urinary System

Chapter 27
1.When excess glucose is not reabsorbed it osmotically obligates water to remain in the nephron.This results in a large production of urine,called polyuria,with a consequent loss of water,salts,and glucose.The loss ofwater can be compensated for by increasing fluid intake.The intense thirst that stimulates increased fluid intake is called polydipsia. The loss ofsalts can be compensated for by increasing the salt intake.The high glucose levels in the blood would increase the blood osmolality,thus stimulating the secretion ofADH.This increases the permeability ofthe distal convoluted tubule and collecting duct to water.Normally,this would allow reabsorption ofwater from the collecting ducts and thus conserve water.Ifglucose levels in the urine are high enough,however,water loss increases even with high levels ofADH being present.
2.When ADH levels first increase the reabsorption of water increases and urinary output is reduced. This also causes an increase in blood volume and, therefore,an increase in blood pressure.The increased blood pressure increases glomerular filtration rate,which increases urinary output to normal levels.In addition,the increased blood volume inhibits the renin-angiotensin-aldosterone mechanism,inhibits aldosterone secretion,and stimulates natriuretic hormone secretion. These responses also increase urinary output.

3.Elevated ammonia ions in the urine results from an increased secretion ofH .Increased secretion ofH occurs in response to either metabolic or respiratory acidosis.Because an elevated respiratory rate increases blood pH,the most
logical conclusion is that the condition is metabolic acidosis,and the observed increase in respiration rate compensates for the metabolic acidosis by lowering H levels.

4.Diarrhea is one ofthe most common causes of metabolic acidosis,resulting from the loss of bicarbonate ions.Increasing the respiration rate and producing an acidic urine both help to increase the blood pH.
5.Blocking H secretion produces acidosis. Because H are exchanged for Na ,the Na remain in the urine as sodium bicarbonate. This effectively prevents the reabsorption ofHCO 3 and produces an alkaline urine.The blood pH is reduced because H are not being secreted as rapidly by the nephron.The respiration rate increases because ofthe stimulatory effect of decreased blood pH on the respiratory center.

6.Breathing through the glass tube increases the dead air space and decreases the efficiency ofgas exchange.Consequently,blood carbon dioxide levels increase and produce a decrease in blood pH.Compensatory responses include an increased respiration rate and the production of acidic urine.
7.A major effect ofalkalosis is hyperexcitability of the nervous system.Ifthe girl is prone to having convulsions,then inducing alkalosis might result in a seizure.This could be accomplished by having the girl hyperventilate.The resulting lossofcarbon dioxide from the blood causes an increase in blood pH.
8.At high altitudes,we expect stimulation ofthe chemoreceptor reflex and an increase in respiration rate.This could result in hyperventilation,a decrease in blood carbon dioxide,and respiratory alkalosis.The increased secretion ofhydrochloric acid into the stomach could also increase blood pH and contribute to the problem.The kidney produces a more alkaline urine. 27WaterElectrolytesandAcid-BaseBalance.pdf

Chapter 28
1.Removing the testes would eliminate the major source oftestosterone.Blood levels of testosterone would therefore decrease.Because testosterone has a negative-feedback effect on the hypothalamus and pituitary gland,GnRH,FSH,
and LH secretion would increase and the blood levels ofthese hormones would increase.

2.Prior to puberty,the levels ofGnRH are very low because the hypothalamus is very sensitive to the inhibitory effects oftestosterone.Since GnRH levels are low,so are FSH and LH levels.Loss ofthe testes and testosterone production would result in an increase in GnRH,FSH,and LH levels. Because little testosterone is produced the boy would not develop sexually and would have no sex drive. Small amounts ofandrogens would be produced because the adrenal cortex produces some androgens.He would be taller than normal as an adult,with thin bones and weak musculature.His voice would not deepen and the normal masculine distribution ofhair would not develop.
3.Ideally the pill would inhibit spermatogenesis. Using the same approach as in females,inhibition ofFSH and LH secretion should work.It’s known that chronic administration ofGnRH suppresses FSH and LH levels enough to cause infertility, through down-regulation.Lack ofLH can also result in reduced testosterone levels and a loss of sex drive,however.Some evidence indicates that administration oftestosterone in the proper amounts would reduce FSH and LH secretion, thus leading to a reduced sperm cell production. The testosterone,however,maintains normal sex
drive.The technique appears to work for a large percentage ofmales,resulting in a sperm concentration in the semen that’s too low to result in fertilization.The technique is not sufficiently precise,however,to be used as a standard birth-control technique.
4.In a postmenopausal woman the ovaries have stopped producing estrogen and progesterone. Without the negative-feedback effect ofthese hormones the levels ofGnRH,FSH,and LH increase.Removal ofthe nonfunctioning ovaries
in a postmenopausal woman doesn’t change the level ofany ofthese hormones or produce any symptoms not already occurring due to the lack ofovarian function.
5.Answer eis correct.The secretory phase ofthe menstrual cycle occurs after ovulation.It is following ovulation that the corpus luteum forms and produces progesterone.In addition, the progesterone acts on the endometrium ofthe
uterus to cause its maximum development. Progesterone secretion therefore reaches its maximum levels and the endometrium reaches its greatest degree ofdevelopment during the secretory phase ofthe menstrual cycle.
6.The removal ofthe ovaries from a 20-year-old woman eliminates the major site ofestrogen and progesterone production,thereby causing an increase in GnRH,FSH,and LH levels due to lack ofnegative feedback.One expects to see the symptoms ofmenopause such as cessation of menstruation and reduction in the size ofthe uterus,vagina,and breasts.There may also be a temporary reduction in sex drive.
7.It’s clear that estrogen and progesterone administration resulted in a large decrease in the amount ofLH in the plasma the day of ovulation.The differences in plasma LH levels between the groups at other times are very small.
The incidence ofpregnancies suggests that the reduced plasma LH levels may result in no ovulation.
8.The progesterone inhibits GnRH in the hypothalamus.Consequently,the anterior pituitary is not stimulated to produce LH and FSH.Lack ofLH prevents ovulation and lack of FSH prevents development ofthe follicles.LH also is required for maturation offollicles prior to ovulation.Without follicle development, there’s inadequate estrogen production, which causes the hot flash symptoms.

9.GnRH administered either before or after the normal time ofovulation doesn’t result in ovulation,because the anterior pituitary is less sensitive to the effect ofGnRH during those times.Also,follicles in the ovary are not adequately developed.The concentration of GnRH must be controlled carefully because too little results in inadequate FSH and LH being released from the anterior pituitary.Too little FSH and LH fails to cause ovulation.Too much GnRH given at the proper time results in the maturation ofmore than one follicle and the release ofan oocyte from more than one ofthe follicles.Ifthe oocytes are fertilized,multiple pregnancies can result. 28 Reproductive System.pdf

Chapter 29
1.Postovulatory age,the approximate length of time the embryo has been developing,is 14 days less than the time since the last menstrual period (LMP).In this case,the postovulatory age is 30 days (44 14).By this time the neural tube has closed,the somites have formed,the digestive tract is developing,the limb buds have appeared, a tubular beating heart is present,and the lungs are developing.Based on reproductive structures,which are just forming,male and female embryos are indistinguishable at this age.
2.The fever would have occurred on day 21–31 of development,which is during part ofthe time of neural tube closure (days 18–25).Ifthe fever prevented neural tube closure,the child could be born with anencephalus or spina bifida.
3.The limb buds develop in a proximal-to-distal sequence.Ifthe apical ectodermal ridge is damaged during embryonic development when the limb bud is about one-halfgrown,the proximal structures,the arm and forearm, develop normally, but the distal structures,the wrist and hand,do not form normally.Depending on the degree ofdamage,the wrist and hand could be completely absent or underdeveloped.
4.The mesonephric duct system develops,because oftestosterone,to form portions ofthe male reproductive duct system. Without the production ofMullerian-inhibiting hormone, the paramesonephric duct system also develops to form the uterus and uterine tubes.Although ovaries are present,the clitoris may be enlarged because oftestosterone to produce somewhat the appearance ofmale external genitalia.The amount ofmasculinization would depend on the
levels oftestosterone and how long it was administered.High levels oftestosterone over an extended period would completely masculinize the external genitalia.
5.This total Apgar score of5 indicates:appearance (A,0) white or blue;pulse (P,1) low;grimace (G,1) slight;activity (A,1) little movement and poor muscle tone;and respiration (R,2) normal. The white or blue appearance (A,0) is consistent
with a poor circulation indicated by a reduced pulse (P,1).The reduced heart rate,resulting in the low pulse,may indicate a circulatory system problem.The reduced reflexes and motor activity (G,1;A,1) could result from the lack of
oxygen in the muscles resulting from poor circulation.Because the infant has poor circulation despite a normal respiration,clearing the airway (ifobstructed) and administering oxygen are in order.This Apgar score could have
several causes,and additional information is necessary to determine the actual cause. .Suckling the breast stimulates the release of oxytocin from the neurohypophysis (posterior pituitary).Once the oxytocin is in the blood,it travels to both breasts and causes milk letdown.
.Ifboth parents are heterozygous for dimpled cheeks,then the child could receive a recessive gene for no dimples from each parent,resulting in the homozygous recessive condition with no dimples in the cheeks. .It’s not possible at present to determine by phenotype ifa child is homozygous or heterozygous for tongue rolling.Even ifit were possible to determine that the child was heterozygous,that’s not very strong evidence that the recessive allele came from the proposed father. .Hemophilia is a sex-linked trait.Since the father h has hemophilia he must be X Y.Ifthe mother
H H were X X ,all their children would be normal. For halfofthe children to have hemophilia,she H h must be X X .

DevelopmentGrowthAgingandGenetics.pdf

This entry was posted on Tuesday, June 12th, 2007 at 2:03 pm and is filed under Anatomy & Physiology, Library. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Leave a Reply

You must be logged in to post a comment.