Module 4 - Physiology: Transport and Exchange

Lecture 22: From Principles of Diffusion, tell how each item below would affect the rate of gas exchange across a membrane -would it: (A) increase diffusion? (B) decrease diffusion? (C) have no change? 1. Increased surface area 2. Steeper concentration gradient 3. Use of ATP pumps

1. A - increased surface area allows for more diffusion because there is more area for it to occur. 2. A - steeper concentration gradient increases diffusion because it allows for more diffusion when going down its concentration gradient. 3. C - diffusion is passive so the use of ATP is not needed.

Lecture 23: Salt-water fish 1. Problem? A. taking in too much salt? B. losing too much salt? Solution? 2. salt: A. conserve salt B. excrete salt 3. water: A. release large quantity of urine B. release small amount of urine

1. A - salt water fish have a problem with taking in too much salt. 2. B - the fish must excrete the salt by pumping it out through active transport in the gills. 3. B - the fish must release small amounts of urine in order to conserve water so it doesn't follow the salt (solute).

Homework 8: Ammonia is likely to be the primary nitrogenous waste in living conditions that include A. lots of fresh water flowing across the gills of a fish B. lots of seawater, such as a bird living in a marine environment C. a terrestrial environment, such as that supporting crickets D. a moist system of burrows, such as those of moles and mole rats

A

Lecture 20: The action of what serves to keep Na+ concentration lower inside this cell (gut epithelium) than outside of it? A. Na+ passive ion channels B. Na+ K+ sodium potassium pump C. Both A and B

B - the Na+/K+ pump helps to take out the Na+ that entered the cell via the co-transport in order to keep the concentration inside lower than the outside.

Homework 8: Compared with the interstitial fluid that bathes active muscle cells, the blood reaching these cells in arteries has a A. higher P-O2 B. higher P-CO2 C. lower pH D. two of these are true of the arriving arterial blood

A - O2 will move from high PO2 to low PO2. Active muscle cells will need more O2 as well.

Lecture 20: What is the evolutionary advantage of a four-chambered heart (mammals) over a three-chambered heart (frog)? A. completes the separation of oxygenated and de-oxygenated blood B. Gets a second push from the heart after being oxygenated C. Both A and B are advantages of this structure over a three-chambered heart.

A - a four-chambered heart has a separation between its oxygenated and de-oxygenated blood. For the three-chambered heart, there is a mix between the oxygenated and de-oxygenated blood.

Lecture 23: Which is most water-soluble? A. Ammonia B. Urea C. Uric acid

A - ammonia

Lecture 23: Which of the three nitrogenous waste molecules is most toxic if it accumulates in cells? A. Ammonia B. Urea C. Uric acid

A - ammonia

Lecture 20: If a vessel is injured (small cut), which one would lose the most blood in the shortest length of time? A. artery B. capillary C. Vein D. 2 of these are the same

A - an artery would lose the most blood because it has the highest velocity (speed) and the highest blood pressure, meaning more blood would come out compared to the veins and capillaries.

Lecture 25: After proton pump action has created an electrical gradient (more positive outside), then in response to that electrical gradient (no more ATP spent): A. Cations like K+ can enter the cell passively B. Anions like NO3-can enter the cell passively C. Water can enter the cell passively. D. Two of these (A, B, C) are true E. All of these (A, B, and C)

A - cations can enter the cell passively because they are going down their electrical gradient. Anions are not able to enter the cell passively, as they would be going up their electrical gradient and would need a co-transporter. And water is not affected by the electrical gradient.

Lecture 18: Which of the following is true, and is a distinction between these two types of hormones? A. Peptide hormones activate signal transduction pathways B. Peptide hormones bind to an intracellular receptor C. Steroid hormones do not require a receptor to activate target cells.

A - peptide hormones activate signal transduction pathways because they bind to receptors that are on the plasma membrane and never enter the cell. Steroid hormones go inside of the cell and bind to receptors that are inside of the cells nucleus and act as a transcription factor. Both require a receptor to activate target cells.

Lecture 25: Plants use ATP to build a proton gradient to power co-transport of sugars into the phloem cell. A. True B. False C. I'm not sure

A - plants do use ATP to build a proton gradient, by pumping H+ ions out of the cell. This proton gradient will then power the co-transport of sugars into the phloem.

Homework 7: Which of the following would be a positive feedback system involving temperature? (A) when the temperature of a system increases, a heater is turned on (B) when the temperature of a system decreases, a heater is turned on (C) when the temperature of a system increases, a cooler is turned on (D) when the temperature of a system decreases, a cooler is turned off (E) none of these represent a positive feedback system.

A - positive feedback amplifies the change that was made. So when the temperature of a system increases, positive feedback would involve the heater turning on to amplify that change and make it hotter.

Homework 8: Why is it advantageous for birds (they fly around a lot) to excrete mainly uric acid? A. uric acid requires less water for excretion and elimination B. uric acid is a digestive waste also, mixed with feces C. uric acid is a simpler, cheaper molecule to synthesize compared to ammonia.

A - put in less energy to secrete waste so that more energy can be used to fly/move.

Homework 8: Which of the following is an adaptive example of active osmoregulation? A. Gill tissues in a salt-water fish use ATP to pump salts out of their blood and release salts into seawater B. Gill tissues in a fresh-water fish use ATP to pump salts out of their blood and release salts into the water C. Eggs of a fresh-water fish are covered with a thick jelly layer to block any exchange of water and solutes. D. Cells in an animal living where salinity changes often (as in an estuary/bay) passively equilibrate salinity with their environment

A - salt-water fish pump salt out of their gills to maintain a balance between the amount of salt and water inside the fish.

Lecture 25: How does the endodermis' Casparian strip prevent the free flow of water into/out of the root's center? A. It blocks flow in the apoplast (cell walls) B. It blocks flow in the symplast(plasmodesmata) C. It blocks flow in both apoplast & symplast

A - the Casparian strip blocks the entry of water that is coming in through the apoplast into the xylem. In order for this water to enter the xylem, it most go through the plasma membrane.

Lecture 25: So, what whole-plant process slows down when stomata are closed?A. upward movement of water in xylem B. downward movement of water in the xylem

A - the upward movement of water from the roots and up the plant within the xylem is slowed down.

Homework 8: Homeostasis typically relies on negative feedback because, on the other hand, positive feedback A. requires a response but not a stimulus B. drives processes to completion rather than to a balance point C. acts within, but not beyond, a normal range D. can decrease but not increase a variable

B

Homework 8: What would be the effect on the volume of urine excreted (eliminated from the body) if there is increased water reabsorption in the excretory tubule? A. urine volume would increase B. urine volume would decrease C. no change in urine volume

B

Homework 8: Which of the following describes the process of filtration in excretory tubules? A. selective transport of water and specific solutes from blood into the tubule B. non-selective transport of water and solutes from blood into the tubule C. selective transport of water and specific solutes from the tubule back into the blood D. non-selective transport of water and solutes from the tubule back into the blood

B

Lecture 24: Put these events in order. Some may occur at same time. Put an "X" if this does not happen in root hairs. A. Water moves into root hair cell by osmosis B. Using ATP, protons are pumped out of root hair cell C. Using ATP, protons are pumped into root hair cell D. Co-transport moves NO3 into root hair against its gradient E. Co-transport moves NO3 out of root hair against its gradient

B - ATP is used to pump H+ ions out of the root hair cell to create an electrical gradient (ion gradient). The inside of the cell is now more negative than the outside. D - H+ binds to co-transporter with the anion NO3 to help pump NO3 into the cell against its electrical gradient (wants to stay will positive but will go to negative). A - water will follow the solutes, NO3, and will enter the cell via osmosis.

Lecture 24: Put these events in order. Some may occur at same time. Put an "X" if this does not happen in root hairs. A. Water moves into root hair cell by osmosis B. Using ATP, protons are pumped out of root hair cell C. Using ATP, protons are pumped into root hair cell D. K+ enters cell through ion channel down its concentration gradient E. K+ enters cell through ion channel down its electrical gradient

B - ATP is used to pump H+ ions out of the root hair cell to create an electrical gradient (ion gradient). The inside of the cell is now more negative than the outside. E - K+ will enter the cell through an ion channel down its electrical gradient, attracted towards the negative charge inside of the cell. The electrical gradient acts as source of energy for the K+ ions to move. A - water will follow the solutes, K+, and will enter the cell via osmosis.

Lecture 18: Which of these is a match between a cell-cell junction and its function? A. Gap junction - block transport in the space between cells B. Gap junction - molecules from one cell move into adjacent cell C. Anchoring junction - block transport in the space between cells D. Anchoring junction - molecules from one cell move into adjacent cell

B - Gap junctions let adjacent cells communication and transfer ions through a tube connecting them. Anchoring junctions/desmosomes are filaments that hold cells together (think of a string), when one is pulled so is the other. Tight junctions prevent interstitial fluids from getting through the gap between two adjacent cells with junctional proteins.

Lecture 20: What is the evolutionary advantage of a 3-chambered heart (amphibians, with two circuits) over a two-chambered heart (fish, one circuit)? A. completes the separation of oxygenated and de-oxygenated blood B. Gets a second push from the heart after being oxygenated C. Both A and B are advantages of this structure.

B - a three-chambered heart allows the blood to be pushed from the heart a second time (second loop) after it has been oxygenated. A two-chambered heart does not give a second push.

Homework 7: Which of the following would we expect to see in transport epithelial tissues with extensive uptake of water? A. membrane proteins that use ATP to pump water against its gradient B. membranes with abundant aquaporins C. membranes with a thinner phospholipid bilayer D. two of these would increase uptake of water

B - abundant aquaporins help the diffusion of water across cell membranes.

Homework 7: Toxins from cholera bacteria cause the intestinal epithelial cells to release Cl- ions and to secrete Na+ ions, K+ ions, and bicarbonate ions out of the cells into the gut cavity. Also, the normal movement of Na+ from the gut into the epithelial cells is blocked. So, what happens in the cholera patient to cause massive diarrhea and dehydration? A. because water follows solutes, more water is taken into intestinal epithelial cells B. because water follows solutes, more water is lost in the feces C. because of high bacterial concentrations in the gut cavity, more water is taken into intestinal epithelial cells D. because of high bacterial concentrations in the gut cavity, more water is lost in the feces.

B - because water follows solutes, more water is taken into intestinal epithelial cells which causes diarrhea and dehydration.

Lecture 18: Which of these is part of the exchange role of the circulatory system in animals? A. exchange between intestine and environment B. exchange between intestine and blood C. exchange between lung and environment D. exchange between excretory system and environment

B - blood is present in the circulatory system and plays an important role in many other systems.

Lecture 21: Most exchange between blood and nearby cells? A. Arteries B. Capillaries C. Veins

B - capillaries have the most exchange because of thinner walls and a lower speed of blood flow which helps to facilitate exchange between the blood and nearby cells.

Lecture 23: Ammonia, Urea, Uric acid. Are these: A. digestive system wastes B. cell metabolism wastes

B - cell metabolism wastes because they come from amino acids.

Homework 7: How does co-transport increase the uptake of specific substances like sugars and amino acids from the gut lumen into the transport epithelium cells? A. Uses ATP to pump substances out of lumen into cells against their concentration gradient B. Uses an ion gradient to pump substances out of lumen into cells against their concentration gradient C. Co-transport symport protein uses ATP to set up ion gradients D. Two of these are how co-transport works in the gut

B - co-transporters do not use ATP and are powered by the ion gradient that is created.

Lecture 19: How does co-transport increase the uptake of specific substances from the gut lumen into the transport epithelium cells? A. Uses ATP to pump substances out of gut lumen against their concentration gradient B. Uses an ion gradient to move substances out of lumen against their concentration gradient C. Co-transport protein uses ATP to set up ion gradients D. Two of these are how co-transport works in the gut

B - for co-transport, the energy source is the ion gradient. ATP is not used.

Lecture 18: Which of the following is an example of the control of blood hormone (H) levels by negative feedback? A. When blood H level rises above a set point, more H is secreted B. When blood H level rises above a set point, secretion of H stops C. When the gland that produces H is activated independently by the brain, more H is secreted D. When the gland that produces H is not activated by the brain, secretion of H stops E. Two of these are examples of negative feedback.

B - negative feedback is when a change in a variable triggers a response that counteracts that change and maintains homeostasis. In this question, blood hormone levels rise above the set point range, which causes secretion of the hormone to stop. This is an example of negative feedback.

Lecture 25: Plants use ATP to pump protons (H+) into the phloem cell. A. True B. False C. I'm not sure

B - plants use ATP to pump protons OUT of the phloem cell.

Lecture 25: In what way does the Casparian strip serve as a selective filter/barrier for substances entering xylem? A. Substance must be able to flow in apoplast (cell walls) B. Substance must be able to cross a cell membrane

B - substances that wish to enter the xylem will have to pass through a cell membrane, as the Casparian strip forces forces all substances to. If it cannot pass through a cell membrane, then it cannot enter the xylem.

Lecture 20: What directly powers the movement of glucose into the gut epithelium cell? A. ATP hydrolysis B. Na+ ion gradient

B - the Na+ ion gradient directly powers the movement of glucose, NOT ATP.

Lecture 20: This co-transporter (Na+-glucose co-transport protein in gut epithelium) moves glucose into the cell against its concentration gradient. Does this protein use ATP directly? A. Yes B. No

B - the protein does not use ATP directly, the movement of glucose is not directly fueled by ATP. Glucose and Na+ are taken into the cell via a co-transport (symport) protein. ATP is then used to power the Na+/K+ pump, which pumps Na+ out and K+ in, that then creates Na+ and K+ ion gradients that power the movement of glucose.

Lecture 19: Where in the digestive tract does the most digestion by hydrolysis occur? A. Stomach B. Small intestine C. Large intestine

B - the small intestine is the major location of hydrolysis, which happens in the gut lumen.

Lecture 25: Plants use ATP to move sugar solutions from sieve tube to sieve tube within the phloem. A. True B. False C. I'm not sure

B - the sugar will be moving from high fluid pressure to low fluid pressure so no ATP is needed.

Homework 7: Many mammals whose diet includes a great deal of leafy plant matter have extra compartments or extensions of their digestive tract that are not present in similar mammals that do not eat leaves. The function of the extra compartments is to A. increase surface area for absorption of nutrients into the blood B. provide an area for their symbiotic bacteria to live C. provide an area to store the leafy matter before digestion D. provide an area to store water required for digestion.

B - this extra compartment is called the cecum and is used to break down cellulose from plants by symbiotic bacteria.

Lecture 25: What happens to transpiration rates when stomata are closed? A. transpiration increases B. transpiration decreases

B - transpiration rates decrease because the pores in which the water evaporates from are closed.

Lecture 20: The pH within the stomach is about 2, and most digestive enzymes in the small intestine have a pH optimum of 7-8. When the stomach's acidic food mass starts to pass into the small intestine a hormone signal is released into the blood. To be effective, that hormone should trigger the release of which of these? (A) acid into the stomach (B) bicarbonate into the small intestine (C) hydrolytic enzymes into the small intestine (D) uptake of glucose into cells (E) aggregation of fats into globules.

B - when the stomach's acidic food mass starts to pass into the small intestine, cells in the small intestine secrete a hormone called secretin into the blood. This will then trigger the pancreas cells to secrete bicarbonate, which will enter the small intestine through a duct, that will neutralize the pH of the acidic food mass.

Homework 8: Of the three main types of nitrogenous wastes, ____ is most soluble in water and ___ is least soluble in water. A. ammonia; urea B. ammonia; uric acid C. urea; ammonia D. urea; uric acid E. uric acid; urea F. uric acid; ammonia

B -ammonia is the most soluble and uric acid is the least.

Homework 7: In a healthy person, after a carbohydrate-rich meal, the production of ________ will increase, causing the uptake of ________ from the blood into target cells. (A) insulin; glucagon (B) insulin; insulin (C) insulin; glucose (D) glucagon; protein (E) glucagon; glucose

C - after eating, blood glucose levels increase. This causes the hormone insulin to be secreted from the pancreas and into the blood. Insulin will then help transport glucose into target cells to then store as glycogen.

Homework 8: Of the three main types of nitrogenous wastes, which is most toxic if accumulates at high levels inside cells? A. urea B. uric acid C. ammonia

C - ammonia will become toxic when in high concentrations within a cell.

Homework 8: Once begun, how does the depolarization signal wave spread through cardiac muscle cells? A. Neural signals from brain to pacemaker B. Spontaneous firing of pacemaker cells C. Gap junctions and desmosomes between cells D. Vibration signals between cells

C - gap junctions allow the ions to pass through to adjacent cells and desmosomes help to link the cells together.

Homework 7: Cardiac muscle cells are connected to each other by combinations of two kinds of cell-cell junctions, desmosomes and gap junctions. These help in all of the following ways EXCEPT for which one? A. desmosomes anchor cells together to strength heart muscle contraction B. desmosomes are protein fibers linking adjacent cardiac cells together C. gap junctions block leakage of interstitial fluids around the cardiac muscle cells D. gap junctions allow ions to flow directly from the inside of one cell to the inside of another cell E. gap junctions function to increase cell-cell communication

C - gap junctions do not block leakage of interstitial fluids around the cardiac muscle cells, that would be tight junctions which aren't present in the cardiac muscle cells.

Lecture 22: Animal respiratory systems are likely to have many adaptations that maximize diffusion because: A. gas exchange readily occurs through the skin & body wall B. membranes have O2 pumps for transporting oxygen. C. gas exchange involves passive transport across membranes D. membranes exchange gases whenever osmosis is occurring

C - gas exchange involves passive transport across membranes, such as from oxygenated blood to the cells.

Homework 7: Respiratory systems are likely to have many adaptations that maximize diffusion because: A. gas exchange readily occurs through the skin and body wall B. membranes have O2 pumps for transporting oxygen. C. gas exchange is limited to passive transport across membranes D. membranes exchange gases whenever osmosis is occurring

C - gas exchange only occurs via passive transport (diffusion) across membrane.

Lecture 24: What do you think is the main source of O2 for aerobic respiration in plant root cells? A. O2 taken into leaves from the air B. O2 produced in photosynthesis C. O2 diffusing in from air pockets in soil

C - in plant root cells, the main source of O2 is from the pockets of air that are in the soil which can then diffuse into the roots.

Lecture 21: Once begun, how does the depolarization wave spread through cardiac muscle cells? A. Neural signals from brain to pacemaker B. Spontaneous firing of pacemaker cells C. Gap junctions and desmosomes between cells D. Vibration signals between cells

C - the gap junctions help to make the cardiac muscle cells synchronized because of the flow of ions between them (tube between cell membrane in which ions can pass through) while the desmosomes help to hold the cells together.

Lecture 25: The action of proton pumps in root hairs (and most everywhere in plants) is to A. Transport H+ ions out of the cell by passive facilitated diffusion B. Transport H+ ions into the cell by passive facilitated diffusion C. Transport H+ ions out of the cell by using ATP. D. Transport H+ ions into the cell by using ATP.

C - the proton pumps are used to transport H+ ions out of the cell against their gradient, which requires ATP. This helps to create an electrical gradient that is more negative inside and more positive outside.

Lecture 21: What initiates the heartbeat? A. Neural signals from brain to pacemaker B. Neural input from spinal column C. Spontaneous firing of pacemaker cells D. Signals from heart ventricle to pacemaker

C - the spontaneous firing of the pacemaker cells initiates the heartbeat. This is because of the unstable resting membrane potential, that is caused by leaky ion channels, which causes the pacemaker cells to fire spontaneously.

Lecture 21: Why do capillaries have the most exchange? A. Stronger walls B. Greater speed C. Thinner walls

C - the thinner walls of the capillaries provide a shorter diffusion distance and therefor can exchange more with nearby cells.

Homework 8: Which of the following characteristics of heart pacemaker cells causes them to fire spontaneously? A. they have gap junctions through which ions can flow into neighboring cells; B. they have receptors for stress hormones such as epinephrine C. they have an unstable membrane potential due to leaky ion channels; D. they receive and summate input from many synaptic connections from neurons E. they have numerous desmosomes serving as anchoring junctions

C - they have an unstable membrane potential due to leaky ion channels which causes them to fire spontaneously.

Lecture 24: What is the name for the major upward pull on water in plants? A. root pressure B. photosynthesis C. transpiration D. translocation

C - transpiration is the major upward pull of water. it is the evaporation of water out of the leaf pores that creates an upward pull (tension) on water in the xylem.

Lecture 22: The process of bringing oxygenated water or air into contact with a gas-exchange surface is called A. respiration B. gas exchange C. ventilation D. diffusion

C - ventilation (breathing) brings more air to the gas exchange surface to be diffused in.

Lecture 20: Na+ and glucose are both solutes. Due to the action of this co-transport (Na+-glucose co-transport protein in gut epithelium), what is expected? A. Water enters the cell by action of an ATP-driven water pump. B. Water moves out of the cell by an ATP-driven water pump C. Water enters the cell because water follows solutes. D. Water moves out of the cell because water follows solutes.

C - water always follows solutes and does not need ATP to do this. Could use aquaporins which are passive diffusion and do not require ATP.

Lecture 19: When insulin is secreted into the blood at high levels, which of these is a response to insulin? A. The secretion of glucagon is halted. B. The further secretion of insulin is halted. C. Target cells activate pathways that use glucose. D. Target cells activate pathways that break down fats

C - when blood glucose levels increase, insulin is secreted into the blood. Insulin will bind to the Glut4 receptor on the plasma membrane of cells that starts a signal transduction pathway that then triggers ALL the responses. More glucose will be taken into cells, make starch made from glucose, speed up glycolysis, and slow down the usage of fat.

Lecture 24: Which happens when stomata are closed? A. CO2 intake decreases; transpiration increases B. CO2 intake increases; transpiration increases C. CO2 intake decreases; transpiration decreases

C - when the stomata are closed, the increase of CO2 is decreased as it cannot come into the plant. This also causes a decrease in transpiration as it cannot function properly, cannot have evaporation out of leaf pores because they are closed.

Homework 8: Cells in kidney tubules have many mitochondria and use a great deal of ATP. Which of the following tubule functions requires a large expenditure of cell energy? A. filtration from blood into the beginning of the excretory tubule B. facilitated diffusion of Na+ ions in reabsorption from tubule to blood C. facilitated diffusion of K+ ions from the tubule back into the blood D. establishing gradients that will power co-transport in reabsorption into blood

D

Homework 7: Which of the following directly increases passive diffusion rates of sugars from the gut cavity into the transport epithelium cells? A. ATP-driven sugar pumps B. increased surface area (microvilli) C. increased specific transport proteins in epithelial membrane D. two of these E. all three of these

D - increased surface area by the microvilli and increased specific transport proteins in the epithelial membrane will both help to increase transport into epithelium cells.

Lecture 19: Which of the following directly increases diffusion rates of sugars and amino acids from the gut cavity into the transport epithelium cells? A. Increased surface area (microvilli) B. Increased food intake C. Increased transport proteins in membrane D. Two of these E. All three of these

D (A & C) - the increased surface area of the microvilli lets more sugars and amino acids into the epithelium cells to then be taken into the blood stream. Increased transport proteins in the membrane allows for passage more materials (sugars and amino acids) to get into the cells and then into the blood stream.

Lecture 22: When hemoglobin in red blood cells is in an area where the PO2 is 40 mm Hg, about what percent of that hemoglobin is bound with O2? A. 0% B. 25% C. 50% D. 75% E. 100%

D - 75%

Lecture 25: In what structural way do fully mature xylem cells differ from phloem cells? A. they have more membrane transport proteins B. they are elongate C. they are tubular D. they are dead

D - fully mature xylem cells are dead while phloem cells are still alive.

Homework 7: The human lung alveolus demonstrates which adaptations to maximize diffusion? A. Short diffusion distance B. Counter current exchange C. Large surface area D. Two of these E. All three of these

D - short diffusion distance due to thin walls and close proximity to capillaries and large surface area.

Lecture 22: A decrease of blood pH from 7.4 to 7.2 causes the hemoglobin binding curve to shift to the right a little. Therefore, under those conditions, hemoglobin will A. release all bound carbon dioxide molecules B. bind more oxygen molecules C. decrease its binding of H+ D. give up more of its oxygen molecules

D - the hemoglobin will give up more of its oxygen and will have less bound to it due to its structural change.

Lecture 19: Which type of cell-cell junctions between cells of the intestinal transport epithelium will aid in highly selective uptake of digested nutrients? A. Desmosomes B. Gap Junctions C. Plasmodesmata D. Tight Junctions

D - the tight junctions that are between the gut epithelial cells prevent interstitial fluids from getting through the space between the adjacent cells and into the blood stream.

Homework 7: The function of tight junctions in epithelial tissue is A. to increase the surface area B. increase the strength and integrity of the tissue layer C. increase the solute uptake into the tissue D. increase the selectivity of transport across the tissue layer E. increase the flow of water across the tissue layer

D - tight junctions help to prevent interstitial fluids from entering into the cell, making it more selective.

Lecture 25: Of the forces contributing to upward movement of water in xylem, which is most important (contributes most force)? A. adhesion B. cohesion C. root pressure D. transpiration

D - transpiration gives the most force to the upward movement of water.

Lecture 22: Blood hemoglobin that has just been in lung capillaries (where PO2=100) is next in a capillary in the liver (PO2=40). When it arrives at the liver capillaries, that hemoglobin will A. take up (bind more) O2 B. give up (unbind) O2

D - when going from a PO2 of 100 to a PO2 40, the O2 saturation of the hemoglobin drops 25% (100% to 75%). This means that the hemoglobin will give up its O2 (unbind essentially) to the liver capillaries.

Lecture 24: Besides light energy, what substances are used up in photosynthesis?A. CO2 B. O2 C. H2O D. CO2 & O2 E. CO2 & H2O

E - CO2 and H2O are used up in photosynthesis.

Homework 7: Secretin is a hormone that controls timing of neutralization of the acidic food mass entering the small intestine from the stomach (through adding bicarbonate). Which best describes the action of the hormone secretin? A. secreted from the hypothalamus to trigger release of bicarbonate from pancreas B. secreted into the small intestine cavity to trigger activation of bicarbonate C. secreted into the small intestine cavity to trigger bicarbonate release from pancreas D. secreted into the blood by the stomach to trigger bicarbonate release from pancreas E. secreted into the blood by small intestine to trigger bicarbonate release from pancreas

E - secretin is secreted into the blood by the small intestine which will then trigger the secretion of bicarbonate from the pancreas. Bicarbonate will then neutralize the acidic food mass coming from the stomach.

Lecture 24: Which of these would cause water to enter a plant cell? A. increased solute concentration inside the cell B. decreased solute concentration inside the cell C. higher fluid pressure inside the cell than outside D. lower fluid pressure inside the cell than outside E. A and C F. B and C G. A and D H. B and D

G - water will follow solutes and will go from areas of high fluid pressure to areas of low fluid pressure.

Lecture 23: Fresh-water fish problem? Solution?

Problem - freshwater fish have a problem with gaining too much water. Solution - keep and gain solutes (salt through active transport in the gills) and release large amounts of urine that are diluted and less concentrated.

Lecture 18: What response in an insulin target cell would help lower blood glucose? a) speed up or slow down glycolysis? b) increase or decrease starch production from glucose? c) speed up or slow down fat metabolism?

a) - speeds up glycolysis due to the increased uptake of glucose by target cells. b) - increased starch production due to the increased uptake of glucose by target cells. c) - slow down fat metabolism because it is the alternative fuel.