B

(7 pts) A human leukocyte with an intracellular osmolarity of 300 mosm/L is just suspended in 0.6 M (mol/L) urea. If the membrane permeability of urea is I x 10* cm/sec and the surface area of the cell is i x 10* cm: (A) Use the Fick's Law of Diffusion to determine magnitude and direction of the urea flux. (5 pts) (B) Briefly describe what will happen to the cell. (Molar mass of urea is 60.06 g/mol) (2 pts)

(A) [Urea), - 0.6 mol/L-60.06 x 0.6 g/1000cm'-3.6 x 1o'g/ em [Urea),-0 g/em P-Ix 10* cm/sec A-Ix 10° cm (2 pts) J= PA(C. 10* - C) 6giving the right equation J= (1 x cm/sec) i x 10 cm' 3.6 x 10g - 2 pts) cm -0) J- 3.6 x 10" g/sec (2pts) (Alternative answer: 6 x 10 molisec) Direction into the cell (Ipt)

(B) Briefly describe what will happen to the cell. (Molar mass of urea is 60.06 g/mol) (2 pts)

(B) Initially, the solution is hypertonic with respect to the interior of the cells and cell will shrink. As urea diffeses into the cell, the interior of the cells will first be isotonic then be hypertonic with respect to the bathing solution and water will flow in, causing the cell first stop shrinking then to swell until it bursts.

7. True or False: (A) The greater the magnitude of the receptor potential generated by a stimulus, the greater the amplitude of the action potentials the receptor potential induces. (B) The process by which sensory receptors change various forms of energy into electrical energy is called translation. (False)

(False) (False)

5. Cells with an intracellular osmolarity of 300mosm/L. are suspended in the following solutions. Describe the sequence of events that take place. (2 pts each)

(a) 0.3M urea (membrane permeable) No volume change initially, but as urea diffuses in and reaches equilibrium, the intracellular Osmolarity will increase, causing water to flow in. Osmotic equilibrium will never be established and cell will burst. (b) 0.3M glucose (membrane impermeable) No volume change and cell will remain stable. (c) 0.6M urea Initially, the solution is hypertonic with respect to the interior of the cells and cell will shrink. As urea diffuses into the cell and equilibrate, the interior of the cells will be hypertonic with respect to the bathing solution and water will flow in, causing the cell to swell until it bursts. (d) 0.6M glucose Because of the solution is hypertonic with respect to the interior of the cells, cells will shrink to half their initial sizes. When intracellular osmolarity reaches 600 mOsm, extracellular solution becomes isotonic, and cells will finally stop shrinking.

10. Contrast the difference between graded potentials and action potential.

* Graded potentiah (GP have veriehle amplitnule: actien potentials (AP are all-none ence hreshald depolaricenie seched AP ampliude is independent of the initiating vent • G can be summed: AP cannat he sammed threshald: AP have a shreshald thar is mally abar 15 l'depatericed Gh relastive heve to ne she resting patential. GP here ne refractery peried: AP hane beh an absalte ende reledive refractery peried • Gh are he condwn depalarisation pessively A ere and is ampliyied decrementelly: tcestnt ralur conducted at each ectively peint and aleng the nithout decremem membrene. duration of G varies • for The e given ceill ope ander cestant condiens with he ininiening conditionN he daration af AP is censtant • en A GP cen evershoet hea he kyperpolarication phase in becomes nearans cel depelaricanien era pesitive, inside kyperpolarioenion with an repecet AP te eanide) alvays a and an depolerisation, ufer wi • Gh are spentenesly: Ibeoshald initieted AP by ere enviranmental iniieted y snimali membrane (receporsi, depolarizetion neuremansmiers ie, G hat fynapnes, depolarie er te The mechanism for eliciting GP dependi on ligend-geted channets er other chemicel or physical changen: ahe mechanim for As depends on volage-guted channe

3. List five characteristics of enzymes.

1. An enzyme undergoes no net chemical change as a result of catalyzing a reaction. 2 The binding of substrate to an ezyme's active site has all the characteristics-chemical specificity, affinity, competition, and saturation-of a ligand binding to a protein. 3. An enzyme increases the rate of a chemical reaction but does not cause a reaction to occur that would not occur in its absence. 4. Some enzymes increase both the forward and reverse rates of a chemical reaction, and thus do not change the chemical equilibrium that is finally reached. They only increase the rate at which equilibrium is achieved. 5. An enzyme lowers the activation energy of a reaction but does not alter the net amount of energy that is added to or released by the reactants in the course of a reaction.

2. Describe how the electrical signal is transduced from the presynaptic neuron to the postsynaptic neuron through neurotransmitter release at the synapses.

Action potential-> Voltage gated Ca2+ channels open-> Ca+ in --> neurotransmitter vesicles fuses with presynaptic membrane --> neurotransmitters release to synaptic cleft --> neurotransmitters bind to receptors on the postsynaptic membrane --> ligand gated ion channels open --> postsynaptic potential

1. Suppose you graduated with honors and you are interviewed for your dream job at a company that makes artificial organs. The interviewer asks you what the kidney is made up of. What would you say?

Any attempt to describe how the kidney is made up of fnctional units called Nephron, which contains four different tissuelcell types (Figure I-I), would be acceptable. We will learn about kidney toward the end of the semester.

4. Give one example about how homeostasis is maintained in the body other than the ones described in class (temperature and blood glucose level). Show the feedback loops.

Any reasonable examples showing feedback loops will be fine. We will discuss various cases of homeostasis in greater details.

3. Contrast the postsynaptic mechanisms of excitatory and inhibitory synapses.

At an excitatory synapse, the postsynaptie response to the released neurotransmitter is a depolarization (an excitatory postsynaptic potential-EPSP) because binding of neurotransmitter to its receptor causes ligand-sensitive channels for small. positively charged ions to open. Even though the permeahility of the postsymaptic membrane is increased to both Na" and K", both the electrical and concentration gradients favor Na" influex, while the electrical gradient opposes the concentration gradient for K. Therefore, there is net influx of Na At inhibitory synapses. the binding of neurotransmitter to its receptor results in a hyperpolarizing graded potential (an inhibitory postsynaptic potential-PSP) or a stahilization of the membrane potential at its existing value. The activated receptor at such synapses causes ligand-sensitive channels for Cl- and'or K+ to open. (In the cases where only Cl- channels open, a hyperpolarizing potential is not recorded but the membrane is more stabilized at its resting level than normal, because the increased membrane permeability to Cl- makes it more difficult for other ion types to change the membrane potential.) Increasing the permeability to K+ causes the postsynaptic memibrane potential to become closer to the K+ equilibrium potential.

5. Secondary active transport mechanisms A. Hydrolyze ATP B, Couple movement of one material down its electrochemical gradient with another solute moving up its electrochemical gradient C. Cannot concentrate solutes D. Are facilitated diffusion mechanisms E. Are not saturable

B, Couple movement of one material down its electrochemical gradient with another solute moving up its electrochemical gradient

1. Homeostasis A. Refers to rigid constancy of the internal B. Refers to dynamic constancy of the internal environment C. Is maintained primarily by positive feedback D. Both refers to rigid constancy of the internal environment and is maintained primarily by positive are correct E. Both refers dynamic constancy of the internal environment and is maintained primarily by positive feedback are correct

B. Refers to dynamic constancy of the internal environment

3. Most of the water in the human body is found in A. The interstitial fluid compartment B. The intracellular fluid compartment C. The plasma compartment D. The total extracellular fluid compartment

B. The intracellular fluid compartment

List the four characteristics of a protein-binding site, and give brief description for each.

Chemical specificity- protein binding sites bind to specific ligands affinity- Meaure of strength of attraction between recpetor and ligand saturation- limit at which reversible binding site Is fully occupied by a ligand competition- compete for sites

2. The interview asks further you list 10 organ systems of the body and give one-sentence descriptions of their functions.

Circulatory: Transports blood throughout the body. Digestive: Digests and absorbs nutrients and water; eliminates wastes. Endoczine: Regulates and coordinates many activities in the body. Immune: Defends the body against pathogens, returns interstitial fluid to the blood, and forms white blood cells. letogumentary: Protects the body against injury, dehydration, and pathogens; helps regulate body lemperature. Lymphatic: Collects extracellular fluid for return to circulation. Participates in immune defenses. Musculoskeletal: Supports and protects the body and allows for body movement. Produces blood cells. Nervous: Regulates and coordinates many activities of the body: detects and responds to changes in the internal and external environments; allows for states of consciousness, learning, memory. emotions, etc. Reproductive: Produces germ cells (eggs in females, sperm in males). In females, provides a mutritive environment for the developing embryo and fetus, and nutrition for the infant after birth. Respiratory: Exchanges oxygen from the air with carbon dioxide to and from the cells of the body by means of the blood and helps to regulate hydrogen ion concentration in the body fluids. Urinary: Regulates the plasma concentration of minerals and water and excretes organic wastes.

1. Allosteric proteins A. Contain more than one kind of binding site B. Undergo a change of shape when a ligand binds to the regulatory binding site C. Are always activated when a modulator molecule binds to the regulatory site D. Both contain more than one kind of binding site and undergo a change of shape when a ligand binds to the regulatory binding site are correct E All of the choices are correct

D. Both contain more than one kind of binding site and undergo a change of shape when a ligand binds to the regulatory binding site are correct

2. In the concept of homeostasis, the "internal environment" refers to the A. Cytoplasm of the cells in the body B. Nucleoplasm, the fluid in the cell's nucleus C. Mitochondrial matrix D. Extracellular fluid E. Intracellular fluid

D. Extracellular fluid

6. The stoichiom etry of the Na,K-ATPase is A. 3 Na+ ions are pumped into the cell with 2 K+ ions being pumped out B. 3 Na+ ions move in and 1 Ca2+ ion moves out C.I Na+ ion moves out with K+ ion that moves in D. 3 Na+ ions are pumped out every of the cell with 2 K+ ions being pumped in E3 Na+ ions are pumped out of the cell with each K+ ion being pumped in

D. Na+ ions are pumped out of the cell with 2 K+ ions being pumped in pumped

5. The reason solutions for injection or infusion into people normally contain either 150 mM NaCl or 300 mM glucose is that these A. Solutes are necessary for metabolism B. Solutions are hypertonic to the blood and cells C. Solutions are hypoosmotie to the blood and cells D. Solutions are isotonic to the blood and cells E. Solutions contain penetrating solutes

D. Solutions are isotonic to the blood and cells

solutions NaCl or mM glucose A. Solutes are necessary metabolism B. Solutions are hypertonic to the blood and cells C. Solutions are hypoosmotic to the blood and cells D. Solutions are isotonic to the blood and cells E. Solutions contain penetrating solutes

D. Solutions are isotonic to the blood and cells

11. (8 pts ) Reproduce an action potential as membrane potential vs. time and label its various parts in terms of being depolarized, repolarizing, and after-hyperpolarized. Briefly describe how the action potential is generated.

Depolarization by a graded potential opens voltage-gated Na" channels. The higher extracellular Na" concentration and the negative membrane potential favor the influx of Na", which carries a positive charge into the cell. This depolarizes the cell further which opens more Na" channels which allows more Na" to enter. This positive feedback mechanism accounts for the "spike" or upstroke of the action potential. Na" channel inactivation prevents further influx of Na". As the slower, voltage-gated K" channels open, the efflex of K" begins. K" efflux is favored by the positive membrane potential and the higher intracellular K" concentration. K" effhex repolarizes the cell.

11. When you roll down a long hill, and then try to stand immediately upon reaching the bottom, you discover that you are dizzy. Why?

During the roll down the hill, the fluid in the horizontal canals is accelerated until eventually it rotates with the same angular velocity as the body as a whole. When you stop at the bottom, the fluid in the canal continues to rotate, gradually decelerating to the new zero rotation. The continuous rotation of the endolymplh stimulates the hair cells, so that you think you are rotating when you are in fact not rotating. This discrepancy of signals is what makes you dizzy you prepare your body for rotation when there is none.

4. Shivering in response to a cold draft is an example of A. A homeostatic mechanism B. Negative feedback C. A physiological reflex D. A homeostatie mechanism and negative feedback E. All of the choices are correct

E. All of the choices are correct

3. Which of the following is NOT a characteristic of biological ion channels? A. They display specificity B. They often display gating C. They are passive D. They are integral proteins E. They are carriers

E. They are carriers

10. List the sequence of events that occurs between the entry of a sound wave into the external auditory canal and the firing of action potentials in the cochlear nerve.

Entry of a sound wave into the external auditory canal causes the tympanic membrane to vibrate at the same frequency as the sound wave. This vibration causes the chain of bones in the air-filled middle car - the malleus, incus, and stapes - to vibrate. The stapes connects to a membrane (the oval window) that separates the air-filled middle car from the fluid-filled inner ear, the cochlea. Movement of the bone chain transmits and amplifies the vibration of the tympanic membrane to the oval window, which sets up pressure twaves in the fluid of the scala vestibuli. These oves cause vibrations in the cochlear duct, setting up pressure waves in the fluid there, which in turn cause vibrations in a specific part of the basilar membrane. (Ench part of the basilar membrane vibrates maximally in response to one particular sound frequency.) As this membrane vibrates, the sound receptor cells - the hair cells in the organ of Corti - move in relation to the overhanging tectorial membrane. Movement of the hair cells' stereocilia against the tectorial membrane bends them, opening ion channels in the hair cell's plasna membrane, which generates a depolarizing receptor potential. Hair cell depolarization leads to the release of the neurotransmitter glutamate, which activates protein receptors on the peripheral ends of the afferent neurons. These neurons, which coalesce to form the cochlear branch of the vestibulocochlear nerve (cranial nerve VIII), will fire action potentials when the graded potential in their peripheral membrane reaches threshold.

3. Contrast feedforward and negative feedback.

Feedforward regulation initiates an adaptive response in the body in anticipation of a change in the environment. Feedforward regulation happens before homeastasis has been disrupted and helps to minimize fluctwations and speed up the response. Negative feedback occurs only after there has been a change in the environment. Negative feedback mechanisms come into play to restoe the variable toward its original set point.

7. Contrast feedforward and negative feedback. (4 pts)

Feedforward regulation initiates an adaptive response in the body in anticipation of a change in the environment. Feedforward regulation happens before homeostasis has been disrupted and helps to minimize fluctuations and speed up the response. Negative feedback occurs only after there has been a change in the environment. Negative feedback mechanisms come into play to restoe the variable toward its original set point.

6. Describe how resting membrane potential is established and maintained.

First, the action of the Na+/K+-ATPase pump sets up the concentration gradients for Na+ and K+ (Figure 6-13a). Then there is a greater flux of K+ out of the cell than Na+ into the cell (Figure 6-13b). This is because in a resting membrane there are a greater number of open K+ channels than there are Na+ channels. Because there is greater net efflux than influx of positive ions during this step, a significant negative membrane potential develops, with the value approaching that of the K+ equilibrium potential. In the steady-state, the flux of ions across the membrane reaches a dynamic balance (Figure 6- 13c). Because the membrane potential is not equal to the equilibrium potential for either ion, there is a small but steady leak of Na+ into the cell and K+ out of the cell. The concentration gradients do not dissipate over time, however, because lon movement by the Na+/K+-ATPase pump exactly balances the rate at which the ions leak through open channels.

9. Briefly describe how resting membrane potential is established and maintained.

First, the oction of the Nor/K+-ATPase pump sets up the concentration gradients for No+ and K+ (Figure 6-13a). Then there is a greater flux of K+ out of the cell than No+ into the cell (Figure 6-13b). This is becouse in a resting membrane there are a greater number of open K+ channels than there are No+ channels. Because there is greater net efflux than influx of positive ions during this step, a significant negative membrane potential develops, with the value approaching that of the K+ equilibrium potential. In the steady-state, the flux of ions across the membrane reaches dynamic balance 6-13c). Because the membrane potential is not equal to the equilibrium potential for either ion, there is a small but steady leak of No+ into the cell and K+ out of the cell. The concentration gradients do not dissipate over time, however, becouse ion movement by the No/Ke-ATPase pump exactly balances the rate at which the through open channels.

2. How does regulation of protein activity by covalent modulation differ from that by allosteric modulation?

In covalent modulation a chemical group, usually a phosphate, is bound to a side chain on the protein, a process called phosphorylation (when phosphate is involved). This binding, which requires the mediation of an enzyme called a protein kinase, adds a negative charge to the protein and causes a change of shape in its ligand-binding site, This change of shape remains until the phosphate is removed by another enzyme called a phosphoprotein phosphatase.

1. Describe the propagation of an action potential. Contrast this event in myelinated and unmyelinated axons.

In unmyelinated axons, action potentials are propagated by local current flow from the site of one action potential to an adjacent part of the aron membrane. Positive charge flows from the depolarized area in both directions along the intracellular membrane (and negative charges do the same along the outside of the membrane). In the "forward" direction, this current flow depolarizes the adjacent membrane to threshold, and an action potential is generated. In the "backward" direction, the membrane is still in the absolute refractory period, so it cannot be stimulated to generate another action potential. In myelinated axons the same process occurs, except that current cannot flow across the membrane where there is myelin because myelin has a very kigh resistance to flow. Instead, the current flows along the membrane from one interruption in the myelin (a node of Ranvier) to the next. At the nodes, the current flow causes depolarization to threshold and an action potential is generated. This is called saltatory conduction, and it is considerably faster than action potential propagation in unmyelinated aNons

6. How does the nervous system code information about stimulus intensity?

Increasing the intensity of a stimulus increases the magnitude of the receptor potential, and this leads to an increased frequency of action potentials. Thus, the frequency of action potentials from a single receptor is one signal. In addition, receptors on adjacent branches of the same afferent neuron will begin to respond as stimulus intensity increases, and the action potentials generated by them will add to the train of action potentials in the sensory unit. Stronger stimuli also uswally affect a larger area and thus "call in " more sensory units responsive to the same stimulus, a process called recruitment

9. Application of the Fick's Law: In the kidney, urea can diffuse between blood and proximal tubular fluid. Suppose the urea concentration of blood is 10 mg/100 mL. The urea concentration of proximal tubular fluid is 20 mg/100 mL. If the permeability to urea is I x 10 cm/sec and the surface area is 100 cm, what are the magnitude and direction of the urea flux. (Note: I ml lem')

JUreapad

dissociate ken and keer usually vary according the instantaneous concentrations of A, B and AB, as well as the nature of the interaction. However, when the reaction reaches equilibrium, the amount of new complexes formed at any given time equals the amount of complexes that dissociate into A and B: konf AB)-ka[A][B] (1) Define the equilibrium dissociation constant, Ka, shown in the lecture slides. (2) If concentrations are measured in molar units (M; mol/L), time in seconds (s), work out the units for kon, ko and Ka. (3) You are a bioengineer at a pharmaceutical company. You are developing a drug to treat cancer, and the drug must interact with a tyrosine kinase 1000 times stronger than the kinase's natural substrate, (i.e., affinity of the drug should be 1000 times higher than the substrate). You conducted experiments and found that at equilibrium, the concentrations for the tyrosine kinase, its natural substrate, and their complex were 10*M, 10ʻM and 10M, respectively. And then, you performed another experiment and substitute the substrate with your drug; and at equilibrium the concentrations for the tyrosine kinase, drug, and the kinase-drug complex were 5 x 10 M, 4x 10°M, and 10*M. Will you be satisfied with the performance of the drug? If not, considering what you have learned about affinity in physiology class, how would you change the structure of the drug to make improvements?

Kd of tyrosine kinase and its natural substrate: 10'M 10ʻM/ 10'M - 10M Kd of tyrosine kinase and the drug: 5 x 10'M 4x 10'M/10 M-2x 10'M 10ʻM/2x 10'M - 500, therefore, it hasn't reached your goal of 1000 times stronger of affinity. To achieve higher affinity, modify the structure of the drug in a way negatively that its shape matches better with that of you the may binding site. If the binding contains charged or positively charged amino acids, consider modify the drug so it carries an opposite charge. If the binding site is hydrophobic, consider increase the hydrophobicity of the drug. All these will achieve stronger attraction between drug and its target molecule, and hence a stronger affinity.

12. Describe the similarities between pain and the other somatic sensations. Describe the differences.

Like the receptors for the other somatic senses, pain receptors (nociceptors) are specific - to noxious, damaging, or potentially damaging stimuli. Like information about the other somatic senses, information about pain is carried by both specific and nonspecific pathways, with the specific pathways going primarily to the somatosensory cortex on the opposite side of the brain from the side of the body in which the receptor is located. Pain differs from the other somatic senses in several ways. After transduction of the first noxious stimuli into actionn potentials in the afferent neuron, a series of changes occur in conponents of the pain pathway components - respond including the ion channels in the nociceptors themselves - that alter the way these to subsequent stimuli, a process called sensitization. These changes can result in an increased sensitivity to painful stimuli, called hyperalgesia, or to decreased sensitivity, called analgesia. Decreased sensitivity is a result of activation of descending pathnways that inhibit synaptic transmission between the afferent nociceptor neurons and the secondary ascending neurons and from interneurons along the pain pathway. Moreover, more than any other sensation, pain can be altered by past experiences, suggestion, emotions, and the simultaneous activation of other sensory modalities. Thus, the level of pain experienced is not solely a plhysical property of the stimulus.

5. What property of molecules allows them to change their 3 dimensional shape?

Molecules are not rigid structures. Atoms can rotate around their covalent bonds to form different shapes.

4. Explain how synapses allow neurons to act as integrators; include the concepts of facilitation (i.e., axo- axonic synapse), and temporal and spatial summation.

Neurons act as integrators because they receive information in the form of synaptic activity from more (often many, many more) than one presynaptic cell, a property known as convergence. The amount of depolarization caused by the discharge of neurotransmitter from a single excite excitatory synapse is only about 0.5 mV, much less than the -15 mV depolarization needed to reach threshold. Thus, activation of a postsynaptic neuron to the point that it will generate action potentials requires summation of many excitatory synaptic events. This occurs in two ways: The same synapse continues to be repeatedly activated before the previous EPSPS or IPSPS have died away-temporal summation; and more than one synapse may be activated simultaneously or within a short time of the first synapse-spatial summation. The postsynaptic cell integrates IPSP: as well as EPSPS. In order for the postsynaptic cell to become activated to threshold, excitatory synaptic activity must be considerably greater than inhibitory synaptic activity. Facilitation refers to one presynaptic cell's (cell A) positive influence on another presynaptic cell (cell B) through an axo-axonic synapse. If the neurotransmitter from cell A causes cell B to release more neurotransmitter to postsynaptic cell C when an action potential reaches cell B's axon terminal, then cell A has increased cell B's synaptic effectiveness by presynaptic facilitation. (Another presynaptic cell with an axo-axonic synapse with cell B could decrease the amount of neurotransmitter released by cell B. This is presynaptic inhibition.)

8. Describe the general mechanism of lateral inhibition and explain its importance in sensory processing.

Often, a stimulus will activate multiple receptors and thus many afferent neurons. The neurons closest to the center of the stimulus have a higher initial firing frequency than neurons on the edge of the stimulus. The central neurons can further decrease the firing of the lateral neurons by activating inhibitory interneurons. While lateral afferent neurons can also inhibit the central neurons, it has mnch less of an effect due to their lower initial firing frequency. The overall effect is that lateral inhibition helps us accurately locate a stimulus by removing the input from the peripheral regions,

8. Briefly describe the different types of cellular membrane transport processes. (4 pts)

Simple Diffusion: down concentration, polar vs nonpolar, ion channels Mediated transport: specificity, affinity and saturation • Facilitated diffusion: no energy • Active transport: energy. primary vs. secondary Cytosis • Endocytosis: Pinocytosis, phagocytosis and receptor-mediated phagocytosis . Exocytosis

7. What are the major substrates entering the Krebs cycle, and what are the products formed? Why oxygen is needed in the Krebs cycle.

Substrates: acetyl coenzyme A, amino acid intermediates Products: 2 CO: per molecule of acetyl coenzyme A, some intermediates used to synthesize amino acids and other organic molecules The Kreb's cycle depends upon the oxidized (hydrogen free) forms of NAD+ and FAD accept electrons from the carbohydrate intermediates. The process of oxidizing the coenzymes so that they can be used by the Kreb's cycle again depends upon oxidative phosphorylation in which molecular axygen acts as the terminal electron acceptor.

13. In what ways are the sensory systems for taste and olfaction similar? In what ways are they different?

Taste and olfaction are part of the special sensory afferent division of peripheral nervous system. The receptors of both senses respond to chemicals by producing a receptor potential. In addition, olfaction contributes to the sense of taste of foods. The two systems differ in mmany ways. Taste modalities fall into 5 categories, but there are 1000 or more modalities of odor, or odorants. Upon depolarization by binding to a food chemical, the receptor cells for taste secrete neurotransmitter at synapse with its afferent neuron. The receptors for olfaction are extensions of the afferent (olfactory) nerve. The neural pathnways for taste are simtilar to those for the other special senses in that they project to the thalanus and then to primary sensory area of cerebral cortex (a region of the parietal lobe that is close to the "mouth" region of the somatosensory cortex). The neural pathuays for olfaction are different from those of all of the other special senses: The axons of the olfactory receptor cells synapse with structures called olfactory bulbs that lie on the undersurface of the frontal lobes. Information from the olfactory bulbs is passed to the olfactory cortex and parts of the limbic system. The olfactory system is the only sensory system that does not synapse in the thalamus prior to reaching the cortex.

6. Describe the types of interactions that determine the conformation of a polypeptide chain.

The conformation of a polypeptide is its three-dimensional shape. It is determined by (1) hydrogen bonds between portions of the chain or with surrounding water molecules; (2) ionic bonds between polar and ionized regions along the chain; (3) attraction between nonpolar (rydrophobic) regions; (4) covalent bonds, called disulfide bonds, between the side chains of the amino acid cysteine (not all polypeptides hae disulfide bonds); and (5) van der Waas forces, which are very weak and transient electrical interactions betueen the orbiting electrons in the outer shells of two atoms that are in close proximity to each other. Hydrogen bonds between the hydrogen linked to the nitrogen in one peptide bond and the exygen in another occur at regular interoals along the chain and coil it into a helical shupe (alpha helix). Hydrogen bormds between peptide bonds running parallel to each other can force a straight structure called a beta pleated sheet. Some proteins, called multimeric proteins, consist of more than one polypeptide chain.

9. What will happen when the photoreceptors (rod/cone cells) are exposed to light? What changes take place in neurotransmitter release from the rods or cones?

The photopigments on photoreceptor membranes are made up of a protein component (opsin) and a chromophore (retinal). The rods and ench of the three cone types have different opsins, which make each of the four receptor types sensitive to different ranges of light wwelengths. When light falls upon the chromophore, the light energy is absorbed by the chromophore, causing the chromophore to change shape. This subsequently triggers the closure of cGMP-gated ion channels and membrane hyperpolarization. In the dark, rod and cone membranes are relatively depolarized and the cells secrete neurotransmitter (glutanmate) at their symapse with bipolar cells. When the photoreceptors are exposed to light, they become hyperpolarized and decrease their secretion of glutamate.

5. The International Olympic Committee has approached you for help in making sprint races fair. In these races, sprinters are started by three commands: "On your marks", "Get set" and "Blam !" A blank pistol report starts the runners. Winners of these races are often determined by a scant hundredth of a second, or 10 msec, over the course of 100 m. The slightest advantage by a sprinter can mean victory. Ir a sprinter anticipates the gun, just slightly, he can win the race. In an unfair race, when a runner starts before the gun, the race is stopped by a second blank pistol report. The field is allowed one false start, and the next false start results in the disqualification of the runner who first begins. The IIOC wants an automatic determination of when a false start occurs, and their question to you is: what is the minimum reaction time for a sprinter to come off the blocks after hearing the pistol report? How would you determine this reaction time? Knowing what you do about the pathways involved, synaptic delays and propagation velocities, estimate the minimum reaction time.

The reaction time requires sound transduction in the ear, conduction to the cochlear nuclei, symaptic delay in the cochlear nuclei, conduction to the inferior colliculus and synaptic delay there, conduction to the medial geniculate nucleus and synaptic delay, and conduction to the auditory cortex and synaptic delay there. The distance is not long: perhaps 0.2 m. At a conduction velocity of 50 m s the total conduction time is perhaps 4 ms. The synaptic delays are all about 0.7 ms, so the total in audition is perhaps another 2.8 ms. Some central processing is necessary, but we have no idea how many synapses are involved in transforming the auditory signal into a decision to start running. Let's estimate it as another 10 ms. The pathway from the motor cortex to the muscles requires conduction down the upper motor neuron to the lower motor neuron, synaptic delay to the lower motor neuron, conduction to the muscle and synaptic delay to the muscle. The total conduction to the muscle can be estimate by the conduction velocity and the distance. The distance can be about 1.5 m; with a conduction velocity of 50 ms this is a delay of 30 ms; the synaptic delays add another 1.4 ms.

6. A graded potential in a membrane results in an electric current along an adjacent area of membrane and this current diminishes with distance from the site of the initial potential change. True False

True

6. What are the end products of glycolysis under aerobic and anaerobic conditions? How many molecules of ATP can form from the breakdown of one molecule of glucose under each condition? Why is it more efficient to store energy as fat than as glycogen?

Under aerobic conditions (per molecule of glucose): 2 pyruvate + 2 ATP + 2 NADH + 2H +2 H0 Under anaerobic conditions (per molecule of glucose): 2 lactate + 2 ATP + 2 H0 Total ATP: under aerobic conditions: 34 to 38 ATP under anaerobic conditions: 2 ATP The amount of ATP formed by the catabolism of I g of fat is about 2.5 times greater than the amount formed from I g of carbohydrate.

8. What are the end products of glycolysis under aerobic and anaerobic conditions? How many molecules of ATP can form from the breakdown of one molecule of glucose under each condition? (5 pts)

Under aerohic conditions (per molecule of glucose): pyruvate anaerohic + 2 ATP + 2 NADH + 2H" + 2 H0 Under conditions (per molecule of glucose): 2 lactate + 2 ATP + 2 H:0 Total ATP: under aerobic conditions: 34 to 38 ATP under anaerobic conditions: 2 ATP

4. Facilitated diffusion differs from simple passive permeability in all of the following EXCEPT A. Facilitated diffusion is saturable, passive permeability is not B. Facilitated diffusion uses metabolic energy, passive permeability snot C. .Facilitated Facilitated dittusion is specific, passive permeability is not diffusion displays competitive inhibition, passive permeability does not E. Facilitated diffusion uses carrier, passive permeability does not

permeability B. Facilitated diffusion uses metabolic energy, passive permeability snot

10. List the sequence of events that occurs between the entry of a sound wave into the external auditory canal and the firing of action potentials in the cochlear nerve.

vibrate at air-filled middle membrane (the oval ear, the cochlea. tympanic membrane vestibuli. These fluid there, which of the basilar membrane this membrane vibrates, relation to the overhanging tectorial tectorial membrane bends them, opening ion channels in the hair cell's plasna membrane, generates a depolarizing receptor potential. Hair cell depolarization leads to the neurotransmitter glutamate, which activates protein receptors on peripheral ends of the afferent neurons. These neurons, which coalesce to form the cochlear branch of the vestibulocochlear nerve (cranial nerve VIII), will fire action potentials when the graded potential in their peripheral membrane reaches threshold.