Test Bank for Human Physiology An Integrated Approach 7th Edition By Dee Unglaub Silverthorn

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Human Physiology: An Integrated Approach, 7e, (Silverthorn)

Chapter 1   Introduction to Physiology

 

1) Physiology is the study of

1. A) the structure of the body.
2. B) the tissues and organs of the body at the microscopic level.
3. C) growth and reproduction.
4. D) the normal function of living organisms.
5. E) the facial features as an indication of personality.

Answer:  D

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.1

Bloom’s Taxonomy:  Knowledge

 

2) The literal meaning of the term physiology is knowledge of

1. A) organs.
2. B) nature.
3. C) science.
4. D) chemistry.
5. E) math.

Answer:  B

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.1

Bloom’s Taxonomy:  Knowledge

 

3) Because anatomy and physiology have different definitions, they are usually considered separately in studies of the body.

1. A) True
2. B) False

Answer:  B

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.1

Bloom’s Taxonomy:  Knowledge

 

 

4) The following is a list of several levels of organization that make up the human body.

1. tissue
2. cell
3. organ
4. molecule
5. organism
6. organ system

 

The correct order from the smallest to the largest is

5. A) 2, 4, 1, 3, 6, 5.
6. B) 4, 2, 1, 6, 3, 5.
7. C) 4, 2, 1, 3, 6, 5.
8. D) 4, 2, 3, 1, 6, 5.
9. E) 6, 4, 5, 2, 3, 1.

Answer:  C

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.2

Bloom’s Taxonomy:  Knowledge

5) “Glucose is transported from blood into cells because cells require glucose to meet their energy needs.” This type of explanation is

1. A) mechanistic.
2. B) theological.
3. C) teleological.
4. D) metalogical.
5. E) scatological.

Answer:  C

Section:  Function and Mechanism

Learning Outcome:  1.4

Bloom’s Taxonomy:  Comprehension

 

6) “Glucose is transported from blood into cells by transporters in response to insulin.” This type of explanation is

1. A) mechanistic.
2. B) theological.
3. C) teleological.
4. D) metalogical.
5. E) scatological.

Answer:  A

Section:  Function and Mechanism

Learning Outcome:  1.4

Bloom’s Taxonomy:  Comprehension

 

 

7) Which of the following is a buffer zone between the outside world and most of the cells of the body?

1. A) cell membrane
2. B) red blood cells
3. C) intracellular fluid
4. D) extracellular fluid
5. E) All of the answers are correct.

Answer:  D

Section:  Homeostasis

Learning Outcome:  1.7

Bloom’s Taxonomy:  Comprehension

 

8) Which of the following is one of Cannon’s “internal secretions”?

1. A) hormones
2. B) nutrients
3. C) water
4. D) inorganic ions
5. E) None of the answers are correct.

Answer:  A

Section:  Homeostasis

Learning Outcome:  1.6

Bloom’s Taxonomy:  Knowledge

9) The study of body function in a disease state is

1. A) necrology.
2. B) physiology.
3. C) microbiology.
4. D) pathophysiology.
5. E) histology.

Answer:  D

Section:  Homeostasis

Learning Outcome:  1.6

Bloom’s Taxonomy:  Knowledge

 

10) Homeostasis is the ability of the body to

1. A) prevent the external environment from changing.
2. B) prevent the internal environment from changing.
3. C) quickly restore changed conditions to normal.
4. D) ignore external stimuli to remain in a state of rest.
5. E) prevent excessive blood loss.

Answer:  C

Section:  Homeostasis

Learning Outcome:  1.6

Bloom’s Taxonomy:  Comprehension

 

 

11) Oxytocin is a hormone that is released in response to cervical dilation. It in turn causes more uterine contractions that will further dilate the cervix. Which type of feedback loop does oxytocin trigger?

1. A) negative feedback
2. B) positive feedback
3. C) local control
4. D) nociceptive feedback

Answer:  B

Section:  Control Systems and Homeostasis

Learning Outcome:  1.16

Bloom’s Taxonomy:  Comprehension

 

12) How genetics influences the body’s response to drugs is called

1. A) pharmacokinetics.
2. B) pharmacogenetics.
3. C) pharmacogenomics.
4. D) pharmacodynamics.
5. E) pharmageddon.

Answer:  C

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.1

Bloom’s Taxonomy:  Knowledge

13) A physician basing clinical decisions on primary research published in biomedical literature is doing ________ medicine.

1. A) evidence-based
2. B) traditional
3. C) alternative
4. D) whimsical
5. E) holistic

Answer:  A

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Comprehension

 

14) A study in which a participant acts as an experimental subject in part of the experiment and a control in another part of the experiment is called a ________ study.

1. A) double-blind
2. B) crossover
3. C) meta-analysis
4. D) retrospective

Answer:  B

Section:  The Science of Physiology

Learning Outcome:  1.19

Bloom’s Taxonomy:  Knowledge

 

 

15) The Internet database for molecular, cellular, and physiological information is called the ________ Project.

1. A) Human Genome
2. B) Physiognomy
3. C) Physiosome
4. D) Physiome
5. E) Manhattan

Answer:  D

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.1

Bloom’s Taxonomy:  Knowledge

 

16) A placebo is

1. A) any drug being tested in a clinical trial.
2. B) any drug in a class of drugs commonly used as pain relievers.
3. C) a drug or treatment that is expected to have no pharmacological effect.
4. D) a nutritive and respiratory organ in fetal development.
5. E) a hole in a cavity wall through which an organ protrudes.

Answer:  C

Section:  The Science of Physiology

Learning Outcome:  1.20

Bloom’s Taxonomy:  Comprehension

17) A technique used to resolve contradictory results in scientific studies is

1. A) meta-analysis.
2. B) retrospective analysis.
3. C) prospective analysis.
4. D) cross-sectional analysis.
5. E) longitudinal analysis.

Answer:  A

Section:  The Science of Physiology

Learning Outcome:  1.19

Bloom’s Taxonomy:  Knowledge

 

18) A scientifically logical guess is a

1. A) model.
2. B) theory.
3. C) hypothesis.
4. D) law.
5. E) variable.

Answer:  C

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Knowledge

 

 

19) If a scientific model is supported or verified repeatedly by multiple investigators, it may become a

1. A) model.
2. B) theory.
3. C) hypothesis.
4. D) law.
5. E) variable.

Answer:  B

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Knowledge

 

20) Place these terms in the typical sequence in which they appear in the process of scientific inquiry: experimental data, theory, model, observation, hypothesis, replication.

1. A) experimental data, theory, model, observation, hypothesis, replication
2. B) replication, hypothesis, experimental data, theory, model, observation
3. C) theory, observation, experimental data, hypothesis, replication, model
4. D) observation, replication, model, experimental data, hypothesis, theory
5. E) observation, hypothesis, experimental data, replication, model, theory

Answer:  E

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Comprehension

21) You are interested in learning more about Parkinson’s disease, a neurological disorder that primarily affects motor function. Which is the best source to begin your investigation?

1. A) Ask.com
2. B) MedlinePlusPubMed
3. C) public library
4. D) physiology textbook
5. E) a physician

Answer:  B

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Knowledge

 

22) Which of the following systems does NOT exchange material with the internal and external environments?

1. A) respiratory system
2. B) circulatory system
3. C) digestive system
4. D) urinary system
5. E) All of the above.

Answer:  B

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.3

Bloom’s Taxonomy:  Knowledge

 

23) The human environment is terrestrial, dry, and highly variable. However, our bodies expend enormous amounts of energy maintaining a constant internal environment. Studying why our bodies do this is what kind of scientific endeavor?

1. A) mechanistic
2. B) translational
3. C) teleological
4. D) anatomical
5. E) meterological

Answer:  C

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.4

Bloom’s Taxonomy:  Comprehension

 

24) Individuals with Type I diabetes do not make enough insulin. Which of the following would be a mechanistic explanation of how insulin is used by the body?

1. A) Cells need insulin because glucose will not cross the cell membrane.
2. B) Insulin is a hormone involved in glucose transport.
3. C) Insulin binds to its receptor which triggers the movement of glucose transporters to the cell membrane.
4. D) Since all cells need glucose, insulin is required.
5. E) Without insulin most cells in the body would be unable to produce enough ATP.

Answer:  C

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.4

Bloom’s Taxonomy:  Comprehension

25) Excretion is a function of the body. Which of the following would be considered excretion?

1. A) Movement of sodium from the intestines to the bloodstream.
2. B) Movement of glucose from the kidney to the blood stream.
3. C) Movement of potassium from kidney cells into one’s urine.
4. D) Movement of salt from sweat glands to the surface of the skin.
5. E) Movement of oxygen from the lungs to the blood stream.

Answer:  D

Section:  Homeostasis

Learning Outcome:  1.10

Bloom’s Taxonomy:  Application

 

26) What is a nocebo effect?

Answer:  It is the phenomenon whereby a patient who has been informed of the side effects of a drug he is taking is more likely to experience some of the side effects than an otherwise similar patient receiving the same drug who has not been so informed.

Section:  The Science of Physiology

Learning Outcome:  1.20

Bloom’s Taxonomy:  Knowledge

 

 

27) List the key concepts or themes in physiology.

Answer:  See Table 1.1 in the chapter.

Section:  Themes in Physiology

Learning Outcome:  1.5

Bloom’s Taxonomy:  Knowledge

 

28) Adaptive significance is an important concept in physiology because it describes the

1. A) importance of a highly variable external environment.
2. B) physiological functions that promote an organism’s survival.
3. C) ability of an organism to monitor and restore its internal state to normal conditions when necessary.
4. D) similarities between ancient and modern marine organisms.
5. E) parameters necessary to maintain a constant internal environment.

Answer:  B

Section:  Function and Mechanism

Learning Outcome:  1.4

Bloom’s Taxonomy:  Comprehension

 

29) You conduct an experiment on twenty 18-year-old male subjects to see how various intensities of exercise influence heart rate. Which of the following is/are considered an independent variable?

1. A) age of subjects
2. B) sex of subjects
3. C) intensity of exercise
4. D) heart rate
5. E) More than one of the answers is correct.

Answer:  C

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

30) You conduct an experiment on twenty 18-year-old male subjects to see how various intensities of exercise influence heart rate. Which of the following is/are considered a dependent variable?

1. A) age of subjects
2. B) sex of subjects
3. C) intensity of exercise
4. D) heart rate
5. E) More than one of the answers is correct.

Answer:  D

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

 

31) Why are physiology and anatomy frequently studied together?

Answer:  This is discussed in the “Physiology Is an Integrative Science” section of the chapter.

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.1

Bloom’s Taxonomy:  Application

 

32) You want to display data on the finish times of the 10 fastest race horses in a single race at the Kentucky Derby.

Which type of graph would be best to display this information?

1. A) bar graph
2. B) line graph
3. C) scatter plot

Answer:  A

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

33) You want to display data on the finish times of the 10 fastest race horses in a single race at the Kentucky Derby.

What would the labels be for the graph axes?

Answer:  The x-axis is horse name or number; the y-axis is finish time in minutes.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

34) A horse runs 10 races, each a mile long, during a 6-month period, and you are interested in determining if the horse’s race time changes with experience. You set up a graph to display the race finish times of this horse.

Which type of graph would be best to display the race finish times of this horse?

1. A) bar graph
2. B) line graph
3. C) scatter plot

Answer:  B

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

35) A horse runs 10 races, each a mile long, during a 6-month period, and you are interested in determining if the horse’s race time changes with experience. You set up a graph to display the race finish times of this horse.

What would the labels be for the graph axes?

Answer:  The x-axis is race number or date; the y-axis is finish time in minutes.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

 

36) There are 10 cloned horses, born on the same day, with identical chromosomes. They are each subjected to the same physical training regimen, but given daily injections of different concentrations of a particular vitamin. They all run the same race. You set up a graph to explore a relationship between race finish time and vitamin dose.

Which type of graph is best to explore a relationship between race finish time and vitamin dose?

1. A) bar graph
2. B) line graph
3. C) scatter plot

Answer:  C

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

37) There are 10 cloned horses, born on the same day, with identical chromosomes. They are each subjected to the same physical training regimen, but given daily injections of different concentrations of a particular vitamin. They all run the same race. You set up a graph to explore a relationship between race finish time and vitamin dose.

What are the labels for the graph axes?

Answer:  The x-axis is vitamin dose; the y-axis is finish time in minutes.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

38) What is the difference between a peer-reviewed article and a review article?

Answer:  A peer-reviewed article describes original research by one author (or group of authors working together) that has gone through a screening process in which a panel of qualified scientists evaluate the work. A review article is a summary (usually a collection of published research that was previously peer-reviewed, usually from more than one independent lab) that discusses a particular topic in the field.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

39) What is the major problem with the deconstructionist view of biology?

Answer:  Return to the topic of function and process. The deconstructionist view of biology predicted that once we uncovered the sequence of the human genome, the inner workings of the human body would be revealed. In reality, it is possible to know HOW a gene codes for a particular protein without knowing WHY that protein exists. Our knowledge of the human genome is only a piece of the puzzle.

Section:  Physiology Is an Integrative Science

Learning Outcome:  1.1

Bloom’s Taxonomy:  Comprehension

 

40) Sarah has just flown around the world in the last 48 hours. She is having trouble sleeping, a condition known as insomnia. How do you think Sarah’s long flights and her insomnia are related to biological rhythms?

Answer:  Our sleep-wake cycle is a biological rhythm that lets our body know when it is time to rest. Most likely Sarah has ignored the signals like sleepiness, changes in body temperature, and mood that her body is sending. By ignoring these rhythms she has disrupted the cycle and the body is struggling to maintain homeostasis.

Section:  Control Systems and Homeostasis

Learning Outcome:  1.17

Bloom’s Taxonomy:  Application

 

41) Why is it necessary to label the axes of a graph?

Answer:  A graph with no axis labels is meaningless—without knowing what trend is being illustrated, there is no communication of scientific information.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Comprehension

 

42) Why is it necessary to space grid marks on a graph proportionally to the quantity measured (example: each square represents one centimeter)?

Answer:  If this is not done, a trend would be obscured or even misrepresented.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Comprehension

 

43) Explain why the prefix homeo- is used in the term homeostasis. Why do some physiologists prefer the term homeodynamics over homeostasis?

Answer:  The prefix homeo-, meaning like or similar, is used to indicate that the body’s internal environment is maintained within a range of acceptable values rather than a fixed state. Some physiologists argue that the term homeodynamics better reflects the small but constant changes that continuously take place in the internal environment, as opposed to homeostasis, which erroneously implies lack of change.

Section:  Homeostasis

Learning Outcome:  1.6

Bloom’s Taxonomy:  Comprehension

 

44) Explain why animals are used in research. Are there any limitations to the application of animal data to human physiology? Could these limitations be addressed using cell or tissue culture, or computer simulations?

Answer:  (Note to instructor: This may be a good question to ask early in the semester, then again toward the end, after the organ systems have been covered.) There is a brief discussion of using humans or animals in research in the chapter. This question is intended to stimulate students to think about how science is done, how data are generated, and how the process is challenged by social issues. Generally, there are limitations to the usefulness of computer simulations and cell/tissue culture systems for the same reason that nonhuman animal data are not 100% applicable to human physiology. How human organ systems perform may be different in very subtle ways from corresponding systems in other species. Cells in culture are in an artificial environment, and while much has been learned from such systems, it has also been noted that the behavior of cells in culture is not identical to cells in a living body. Furthermore, cells cultured from established lines can change over time, becoming less like the original cells from which they were derived, and presumably less like normal cells. Computer simulations are valuable, but are only as good as the data entered, and given that we don’t know everything there is to know about physiology, we can’t write a perfect computer program. All three approaches are useful, but for different reasons, and therefore one research system does not completely substitute for another, nor is it appropriate to abandon one entirely.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

45) You conduct an experiment on twenty 18-year-old male subjects to see how various levels of exercise influence heart rate. Explain why only 18-year-old males were used as subjects.

Answer:  An important part of scientific inquiry is to remove sources of variation from among subjects. By choosing subjects of one gender in a particular age group, it is easier to determine that the dependent variable (heart rate, in this case) depends only on the independent variable, level of exercise. This also allows a study to have fewer participants, assuming that subjects were randomly assigned to a level of exercise. If subjects were of random ages and genders, data would have to be collected from many more individuals.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

46) Use the following terms to develop a concept map:

brain, sensory neuron, an eye, foot, soccer ball, motor neuron

Answer:  Eye sees soccer ball.

Sensory neuron sends visual information.

Brain receives information and formulates a plan.

Motor neuron carries action information.

Foot muscle contracts and the ball is kicked.

Section:  Control Systems and Homeostasis

Learning Outcome:  1.12

Bloom’s Taxonomy:  Application

 

47) Provide an example of a control system. Be sure to include the three main components: an input signal, a controller, and an output signal.

Answer:  Variable. One example is blood glucose concentration. The input signal is a blood glucose concentration outside of the normal range, the controller is the pancreas, and the output signal is release of either insulin or glucagon.

Section:  Control Systems and Homeostasis

Learning Outcome:  1.12

Bloom’s Taxonomy:  Knowledge/Application

 

48) Write a teleological explanation for why heart rate increases during exercise. Now write a mechanistic explanation for the same phenomenon.

Answer:  Teleological: Heart rate increases because the increased activity of skeletal and cardiac muscles requires increased delivery of blood contents such as oxygen and glucose. Mechanistic: Heart rate increases in response to signals from the brain (pacemaker cells of the heart are stimulated by the nervous system).

Section:  Function and Mechanism

Learning Outcome:  1.4

Bloom’s Taxonomy:  Application

 

49) What is a hypothesis? What are the steps involved in following the scientific method? How does one distinguish the dependent variable from the independent variable in an experiment? How are each of these represented on a graph?

Answer:  This is discussed in “The Science of Physiology” section of the chapter and in Figure 1.15.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Comprehension

 

50) You are designing a study to assess the effects of a new treatment for hypertension. What ethical considerations would you employ when monitoring your progress?

Answer:  Major considerations should involve assessing the efficacy of the treatment such that the control group patients are not deprived as well as ensuring that the experimental treatment is not less effective than the standard treatments.

Section:  The Science of Physiology

Learning Outcome:  1.19

Bloom’s Taxonomy:  Analysis

 

 

51) You are designing a study to assess the effects of a new drug treatment for hypertension. In your study of this drug’s efficacy in treating hypertension, your subjects are white males, ages 40 to 60 years. Is your study applicable to all people? Explain.

Answer:  Possibly, but not necessarily. There are gender differences in appropriate therapies because of physiological effects of higher testosterone in males compared to females, for example. Drugs are often not tested in children, and children also have a different hormonal environment than adults (again, sex hormones are a good example, because their levels are low until just before the onset of puberty). There are also racial differences in effectiveness of therapies, and while it is a contentious issue as to whether these represent genetic or socioeconomic influences, they should be considered.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

52) High cholesterol levels have been shown to be a contributing factor to heart disease and death due to cardiovascular disease for many decades. In the 1970s, scientists used this information to develop a hypothesis that giving a medicine to reduce blood cholesterol levels could reduce the chances of developing cardiovascular disease or dying from cardiovascular disease. They tested a group of people living in a town called Framingham, Massachusetts. This study became known as the Framingham Study, and it is very well known because it did not support the hypothesis that giving cholesterol-lowering medications would reduce the risk of developing or dying from cardiovascular disease. Does this mean that high cholesterol is not a risk factor for heart disease? What does this demonstrate about the scientific process, especially as it pertains to human studies? You can find a copy of the study online and read it, if necessary.

Answer:  This demonstrates the difficulty in doing human research because, even though elevated cholesterol levels are a risk factor for cardiovascular disease, reducing cholesterol levels without addressing the reason those levels were high in the first place may not have the expected effect on reducing heart disease. Human testing on hypotheses is important because humans don’t always respond to treatments like other animals do, they may actually respond quite differently and each person may respond differently from the rest. It is why we need to test each hypothesis in circumstances as similar to the actual real group that would be treated.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

Use the table and graph below to answer the following questions.

 

Table 1.1

 

Figure 1.1

 

53)  List all of the errors in Figure 1.1.

Answer:

1. The units of concentration are labeled as M when they should be mg.
2. The x-axis is in decreasing order of concentration.
3. The graph needs a legend.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

54) What is the reason for using a line graph to express the results of this study?

Answer:  Line graphs are commonly used when the independent variable (x-axis) is a continuous phenomenon. In this study the concentration of epinephrine is a continuous function. The line allows for interpolation (i.e., estimating values between the measured values).

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

55) Use Table 1.1 to graph the data appropriately. What can you CONCLUDE based on the new figure?

Answer:  Graphs should address the errors in Figure 1.1.

This small sample suggests that an increase in epinephrine concentration increases the average heart rate of Sprague-Dawley rats.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Application

 

Use the table and graph below to answer the following questions.

 

Table 1.2

 

Figure 1.2

 

56) Summarize the data shown in Figure 1.2.

Answer:  The systolic pressure of both genders increases with age. Under age 40, the systolic pressure of males is higher than that of females. After age 40, the systolic pressure of females is higher than that of males. The greatest rate of increase is from ages 50 to 70 in both genders. Blood pressure declines after age 70.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

57) Referring to Table 1.2, what general trend in systolic blood pressures is seen as both men and women increase in age?

Answer:  The systolic pressure of both genders increases with advancing age.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

58) Referring to Figure 1.2, at approximately what age do men begin to show higher systolic blood pressures than women? At what age does this trend reverse?

Answer:  From age 10 to 40, male pressures are higher; after age 40, female pressures are higher.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

59) You are doing an experiment to determine if caffeine consumption affects reaction time.

1. Which is the dependent variable?
2. Which is the independent variable?
3. Briefly describe some ways you might manipulate the independent variable.
4. Name three stimuli you could use, and how you might measure reaction time for each.
5. Write an appropriate hypothesis for this study.
6. You compute the following average values from your experiment. What would be a logical conclusion for these data?

Average caffeine consumer’s reaction time: 400 ms

Average noncaffeine consumer’s reaction time: 650 ms

1. Sketch a simple graph to convey these results to your classmates. What kind of graph did you choose? Why? Which variable did you plot on the x-axis? Which one did you plot on the y-axis? Why?
2. Do the results of this experiment support the hypothesis you chose?

Answer:

1. Reaction time
2. Caffeine consumption
3. Vary the amounts of caffeine consumed; vary the source, for example, use coffee, pills, cola drinks, and/or chocolate; vary both the amounts and sources.
4. Answers will vary. Example: a computer-based timer could measure the time elapsed between the subject’s detecting the appearance of an object on the computer monitor and depressing a key on the keyboard. Auditory or touch stimuli could be used, too.
5. Depending on the answer to C, could choose: “Consumption of caffeine decreases reaction time” or similar statement.
6. Consumption of caffeine improves reaction time by 250 ms, on average.
7. Bar graph; allows comparison of the average of two groups. The x-axis: group, caffeine or none. The y-axis: reaction time in milliseconds.
8. Yes, in case of hypothesis written in D.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

 

Following is a table of data collected from one section of an 8 a.m. physiology lab. There were 20 students present, 10 men and 10 women. Information collected from the students included their height, weight, age, gender, and resting pulse rate. In addition, the students were surveyed to see if they smoked cigarettes, considered themselves “regular exercisers,” if they had consumed caffeine the morning of the lab, and if they had eaten breakfast that day. A “y” or “n” (yes or no) was recorded to indicate their answers. Each student did “jumping jacks” for 5 minutes and recorded the time required to regain their resting heart rate, which is listed on the table as “recovery time.” Finally, each student participated in an exercise designed to measure their reaction time (in milliseconds) in catching an object dropped by a lab partner according to specified criteria. Use this table to answer the following questions. Ignore statistical problems caused by small sample size, and so on.

 

Table 1.3

 

Figure 1.3      

 

 

For these questions, the data were separated and analyzed by gender.

 

60) Refer to Table 1.3 and Figure 1.3 (bar graph).

1. Write a hypothesis regarding gender and weight.
2. What is the dependent variable? What is the independent variable?
3. Based on the data in the graph above, what is your conclusion?
4. Why is a bar graph a good choice for presentation of these data? Would another type of chart be as effective?

Answer:

1. Males weigh more than females.
2. Weight depends on gender; thus weight is dependent, gender is independent.
3. Males weigh more than females.
4. Bar graph allows comparison of the average of two groups. No.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

61) Refer to Table 1.3.

1. Write a hypothesis regarding gender and recovery time.
2. What is the dependent variable? What is the independent variable?
3. Create a graph using the averages from the data table. Based on these data, what do you conclude?

Answer:

1. A prediction such as “Males recover from exercise more quickly than females” would be appropriate.
2. The independent variable is gender; the dependent variable is recovery time.
3. A bar graph such as the one below is appropriate. In this study, men recovered from exercise more quickly than women.

 

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

 

62) Refer to Table 1.3.

1. Write a hypothesis regarding the effects of breakfast consumption on reaction time.
2. What is the dependent variable? What is the independent variable?

Answer:

1. A prediction such as “Eating breakfast prior to testing improves reaction time of subjects (compared to subjects who did not eat breakfast)” is appropriate.
2. The independent variable is breakfast consumption; the dependent variable is reaction time.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

63) Refer to Table 1.3.

1. Disregarding gender, write a hypothesis that expresses the relationship between weight and height.
2. What is the dependent variable? What is the independent variable?
3. From the data in Table 1.3, construct a graph that examines this relationship.

Answer:

1. A prediction such as “As height increases, weight increases” would be appropriate.
2. The dependent variable would be weight, the independent variable is height.

C.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

 

64) Table 1.3 shows data on various factors that may or may not be related to resting pulse rate, time to recovery to resting pulse rate after a few minutes of exercise, and reaction time measured by how quickly a student could press a keyboard key after seeing a computer-generated prompt. For each question below, write a testable hypothesis, identify the dependent and independent variables, sketch an appropriate graph of the results, and draw a conclusion from the data presented in the table. Discuss your results.

1. Does caffeine consumption have an effect on resting pulse rate?
2. Does age play a role in resting pulse rate? Does weight?
3. Is there a relationship between eating breakfast and recovery time?
4. Is there a relationship between reaction time and height?
5. Do women who smoke show differences in their resting pulse rates compared to female nonsmokers or to male smokers and nonsmokers?
6. Does regular exercise have an effect on resting pulse rate?

 

Answer:  Answers will vary, but examples follow (conclusions written here are based on cursory examination of graphed data—no statistical tests of significance were performed).

1. Hypothesis: Caffeine consumption increases heart rate.

Independent variable: caffeine consumption.

Dependent variable: resting pulse rate.

Conclusion: Mean pulse rates between caffeine-drinking (68 bpm) and control subjects 73 bpm) are similar (large variation between individuals); hypothesis rejected.

1. Hypothesis: Pulse rate is lower in older people and is higher in heavier people.

Independent variables: age and weight.

Dependent variables: resting pulse rate.

Conclusion: Pulse rate was similar in all groups; hypothesis rejected.

1. Hypothesis: People who ate breakfast have a faster reaction time.

Independent variable: breakfast consumption.

Dependent variable: pulse rate.

Conclusion: People who ate breakfast had a faster reaction time (168.7 msec vs. 180.5 msec); hypothesis supported.

1. Hypothesis: There is no relationship between height and reaction time.

Independent variable: height.

Dependent variable: reaction time.

Conclusion: Reaction time did not vary with height; hypothesis supported.

1. Hypothesis: Smokers of both genders have a higher resting pulse rate than nonsmokers of either gender, and males and females are affected equally.

Independent variables: smoking and gender.

Dependent variable: pulse rate.

Conclusion: There was no difference in pulse rate in any of the groups (70.4 bpm in nonsmokers vs. 70.3 bpm in smokers); hypothesis rejected.

1. Hypothesis: People who exercise regularly have a lower resting pulse rate.

Independent variable: exercise.

Dependent variable: pulse rate.

Conclusion: Regular exercise had no effect on resting pulse rate (68.9 bpm in nonexercisers vs. 71.8 bpm in exercisers); hypothesis rejected.

 

Discussion may cover issues such as the effect of small sample size, use of adults of limited age range, lack of control over treatments (Were the subjects honest about age, eating breakfast, consuming caffeine, smoking, and exercising? Were the quantitative data of height and weight determined in the lab using the same equipment and same data collector?), the value of statistical analysis, and so on. It is likely that students will be surprised by some of the results and could make erroneous conclusions. For example, pulse rate may vary with age, but without including children and senior citizens in the sample population, this trend would be missed.

Section:  The Science of Physiology

Learning Outcome:  1.18

Bloom’s Taxonomy:  Analysis

 

 

65) The law of mass balance states:

1. A) if a substance is to remain constant any gain must be offset by an equal loss.
2. B) that homeostasis can be maintained when the load of a substance is continuously lost.
3. C) if one is to survive they must have a certain amount of mass.
4. D) that all matter is neither created or destroyed.
5. E) that all substances in the body have equal mass.

Answer:  A

Section:  The Science of Physiology

Learning Outcome:  1.8

Bloom’s Taxonomy:  Knowledge

66) Mass balance involves determining the total amount of a substance in the body. We can determine the rate of production (i.e. Mass Flow) of this substance by which of the following formulas?

1. A) intake + production – excretion – metabolism.
2. B) (amount of substance / min) × (concentration of the substance)
3. C) volume of flow / (amount of substance / min)
4. D) (concentration of a substance) / volume flow
5. E) (concentration of a substance) × (volume/min)

Answer:  E

Section:  The Science of Physiology

Learning Outcome:  1.9

Bloom’s Taxonomy:  Knowledge

 

67) ________ are kept within normal range by physiological control mechanisms which are used if the variable strays too far from its ________.

1. A) Setpoints, regulated variable
2. B) Independent variables, steady state
3. C) Regulated variables, setpoint
4. D) Dependent variables, lowest value
5. E) Steady state values, integrating center

Answer:  C

Section:  The Science of Physiology

Learning Outcome:  1.13

Bloom’s Taxonomy:  Knowledge

 

68) The vasodilation of blood vessels surrounding muscles due to the production of carbon dioxide during exercise is an example of which of the following?

1. A) neural control
2. B) long-distance control
3. C) reflex control
4. D) local control
5. E) hormonal control

Answer:  D

Section:  The Science of Physiology

Learning Outcome:  1.14

Bloom’s Taxonomy:  Comprehension

 

69) Which of the following are used to keep our systems at or near their setpoints?

1. A) positive feedback loops
2. B) response loops
3. C) feedback loops
4. D) open control loops
5. E) feedforward control loop

Answer:  C

Section:  The Science of Physiology

Learning Outcome:  1.15

Bloom’s Taxonomy:  Knowledge

 

Human Physiology: An Integrated Approach, 7e, (Silverthorn)

Chapter 5   Membrane Dynamics

 

1) Which body fluid compartment contains high levels of K+, large anions, and proteins?

1. A) plasma only
2. B) interstitial fluid only
3. C) intracellular fluid only
4. D) both plasma and intracellular fluid
5. E) both plasma and interstitial fluid

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Knowledge

 

2) Which body fluid compartment contains higher levels of Na+, Cl-, and HCO3-?

1. A) plasma only
2. B) interstitial fluid only
3. C) intracellular fluid only
4. D) both plasma and intracellular fluid
5. E) both plasma and interstitial fluid

Answer:  E

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Knowledge

 

3) All of the following are types of mediated transport EXCEPT one. Identify the exception.

1. A) facilitated diffusion
2. B) primary active transport
3. C) simple diffusion
4. D) secondary active transport

Answer:  C

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Comprehension

 

4) Bulk flow is fluid flow as a result of a(n) ________ gradient.

1. A) concentration
2. B) electrical
3. C) pressure
4. D) Two of the answers are correct.
5. E) None of the answers are correct.

Answer:  C

Section:  Transport Processes

Learning Outcome:  5.5

Bloom’s Taxonomy:  Comprehension

 

5) Water is a polar molecule, yet it easily moves through the nonpolar portions of cell membranes. Which transport process is responsible?

1. A) facilitated diffusion
2. B) simple diffusion
3. C) uniport
4. D) symport
5. E) antiport

Answer:  B

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Comprehension

 

6) Permeability is a property of

1. A) membranes.
2. B) ions.
3. C) solutes.
4. D) solvents.
5. E) proteins.

Answer:  A

Section:  Transport Processes

Learning Outcome:  5.5

Bloom’s Taxonomy:  Knowledge

 

7) The term cellular (metabolic) energy indicates any biological process requiring

1. A) energy in any form.
2. B) ATP.
3. C) thermal energy.
4. D) chemical energy.
5. E) thermal energy and chemical energy.

Answer:  B

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Knowledge

 

8) What are the two extracellular fluid compartments in the body?

1. A) intracellular and plasma
2. B) plasma and interstitial
3. C) interstitial and intracellular
4. D) plasma and the fluid portion of the blood
5. E) None of the answers are correct.

Answer:  B

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Knowledge

 

9) Saturation occurs when

1. A) molecules are moved by the use of vesicles.
2. B) the energy required to move molecules results from a high-energy bond.
3. C) a group of carrier proteins is operating at its maximum rate.
4. D) a preference of a carrier protein for a substance is demonstrated based on the differing affinities of the carrier for the substrates.
5. E) a carrier molecule has the ability to transport only one molecule or a group of closely related molecules.

Answer:  C

Section:  Protein-Mediated Transport

Learning Outcome:  5.9

Bloom’s Taxonomy:  Knowledge

 

10) The means by which a cell transports large molecules out of the cell is called

1. A) phagocytosis.
2. B) endocytosis.
3. C) exocytosis.
4. D) diffusion.
5. E) active transport.

Answer:  C

Section:  Vesicular Transport

Learning Outcome:  5.10

Bloom’s Taxonomy:  Knowledge

 

11) Which of the following is a way for solutes in an aqueous solution to move from an area of high solute concentration to an area of low solute concentration?

1. A) only facilitated diffusion
2. B) only osmosis
3. C) only active transport
4. D) both facilitated diffusion and osmosis
5. E) None of the answers are correct.

Answer:  A

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Comprehension

 

12)  In an epithelium, the apical membrane is also known as the ________ membrane.

1. A) basolateral
2. B) mucosal
3. C) serosal
4. D) basement
5. E) nictitating

Answer:  B

Section:  Epithelial Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Knowledge

13) Hyposmotic solutions

1. A) have higher concentrations of solutes than hyperosmotic solutions.
2. B) have lower concentrations of solutes than other hyposmotic solutions.
3. C) have the same concentration of solutes as hyperosmotic solutions.
4. D) have lower concentrations of solutes than hyperosmotic solutions.
5. E) None of the answers are correct.

Answer:  D

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Comprehension

 

14) Which of the following statements about the Na+/K+ pump is FALSE?

1. A) It transports Na+out of the cell and K+into the cell.
2. B) It is present in neurons.
3. C) Its activity requires the expenditure of metabolic (cellular) energy.
4. D) It transports Na+and K+in a 1:1 ratio.

Answer:  D

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

15) Which of the following statements about the resting membrane potential is TRUE?

1. A) It is normally equal to zero volts.
2. B) The inside of the membrane is positively charged compared to the outside.
3. C) It results, in part, from the concentration gradients for Na+and K+.
4. D) It is due in part to the presence of extracellular proteins.

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

 

16) Voltage-gated (voltage-dependent) channels and antiport carriers are both types of

1. A) structural proteins.
2. B) enzymes.
3. C) transporters.
4. D) receptors.

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Comprehension

 

 

17) The resting membrane potential in a typical nerve cell is approximately

1. A) +70 mV.
2. B) -70 mV.
3. C) +35 mV.
4. D) -35 mV.
5. E) 0 mV.

Answer:  B

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Knowledge

18) Compared to the outside surface, the inside of a resting cell membrane is

1. A) positively charged.
2. B) negatively charged.
3. C) electrically neutral.
4. D) continuously reversing its electrical charge.
5. E) positively charged whenever the sodium-potassium pump is active.

Answer:  B

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Knowledge

 

19) Caveolae and clathrin-coated pits are both used in

1. A) endocytosis.
2. B) exocytosis.
3. C) phagocytosis.
4. D) All of the answers are correct.
5. E) None of the answers are correct.

Answer:  A

Section:  Vesicular Transport

Learning Outcome:  5.10

Bloom’s Taxonomy:  Comprehension

 

20) As the charge on the membrane of a typical neuron approaches 0 from -70 mV, the cell is

1. A) only repolarizing.
2. B) only hyperpolarizing.
3. C) only depolarizing.
4. D) only becoming more difficult to stimulate.
5. E) hyperpolarizing and becoming more difficult to stimulate.

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Comprehension

 

 

21) The ion that plays a key role in initiating electrical signals in neurons is

1. A) K+.
2. B) Na+.
3. C) Cl-.
4. D) Ca2+.

Answer:  B

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Knowledge

 

22) Which of the following is NOT involved in creating the resting potential of a neuron?

1. A) diffusion of potassium ions out of the cell
2. B) diffusion of sodium ions into the cell
3. C) resting membrane permeability for sodium ions greater than potassium ions
4. D) resting membrane permeability for potassium ions greater than sodium ions

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

23) Passive transport refers to a process that requires

1. A) no energy at all.
2. B) no cellular energy.
3. C) no pressure gradient.
4. D) no concentration gradient.
5. E) no electrical gradient.

Answer:  B

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Comprehension

 

24) Which of the following is NOT true of diffusion in the human body?

1. A) Diffusion occurs faster at higher temperatures.
2. B) Smaller molecules take longer to diffuse than larger ones.
3. C) Net movement of molecules occurs until the osmolarity is equal.
4. D) Diffusion is rapid over short distances and slower over longer distances.

Answer:  B

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

 

 

25) Gated channels for sodium ions may include

1. A) mechanical gates, which respond to pressure.
2. B) chemical gates, which respond to ligands.
3. C) voltage gates, which respond to electrical signals.
4. D) All of the answers are correct.
5. E) None of the answers are correct.

Answer:  D

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

 

26) When a neuron changes its ion permeability from the resting state,

1. A) a variety of gated ion channels may open or close.
2. B) Na+channels may open, allowing Na+to enter the cell.
3. C) K+channels must open, allowing K+to enter the cell.
4. D) only a variety of gated ion channels may open or close and Na+channels may open, allowing Na+to enter the cell.
5. E) a variety of gated ion channels may open or close, Na+channels may open, allowing Na+to enter the cell, and K+ channels must open, allowing K+ to enter the cell.

Answer:  D

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Comprehension

 

Match the membrane protein with its function.

 

1. transfer signals from the extracellular environment to the cytoplasm of the cell
2. form cell-to-cell connections
3. bind to molecules to facilitate entry to or exit from the cell
4. ligands bind to these proteins and are changed by the protein

 

27) structural proteins

Answer:  B

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

28) enzymes

Answer:  D

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

29) receptors

Answer:  A

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

30) transporters

Answer:  C

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

 

Match the transport process to its description.

 

1. active transport
2. passive transport

 

31) the movement of molecules from an area of high concentration to an area of low concentration

Answer:  C

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Comprehension

 

32) the movement of molecules via proteins embedded in the cell membrane; requires ATP

Answer:  A

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Comprehension

 

33) the movement of molecules against the concentration gradient

Answer:  A

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Knowledge

34) tends to create an equilibrium state

Answer:  B

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Comprehension

 

 

Match the transport process to its description.

 

1. simple diffusion
2. facilitated diffusion
3. both
4. neither

 

35) the movement of molecules from an area of high concentration to an area of low concentration

Answer:  C

Section:  Diffusion, Protein-Mediated Transport

Learning Outcome:  5.7, 5.8

Bloom’s Taxonomy:  Knowledge

 

36) the use of ATP to move molecules

Answer:  D

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

37) a form of mediated transport

Answer:  B

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

38) conform(s) to the properties of specificity, competition, and saturation

Answer:  B

Section:  Protein-Mediated Transport

Learning Outcome:  5.9

Bloom’s Taxonomy:  Comprehension

 

 

Match the type of transport with its description.

 

1. secretion
2. paracellular transport
3. transcellular transport
4. absorption

 

39) between adjacent cells

Answer:  B

Section:  Epithelial Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Comprehension

40) from an organ’s lumen to the extracellular fluid

Answer:  D

Section:  Epithelial Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Comprehension

 

41) in one side of a cell and out the other

Answer:  C

Section:  Epithelial Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Comprehension

 

42) movement from the extracellular fluid into the lumen of an organ

Answer:  A

Section:  Epithelial Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Comprehension

 

 

Match the terms to changes presented, assuming a resting membrane potential of -70 mV. Answers may be used once, more than once, or not at all.

 

1. electrical polarization
2. hyperpolarization
3. depolarization
4. repolarization
5. more than one of the answers

 

43) to -50 mV from resting potential

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Knowledge

 

44) to -70 mV from -50 mV

Answer:  D

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Knowledge

 

45) to -90 mV from resting potential

Answer:  B

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Knowledge

 

46) to +30 mV from resting potential

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.11

Bloom’s Taxonomy:  Knowledge

47) to -70 mV from -90 mV

Answer:  D

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Knowledge

 

48) any value other than 0 mV, regardless of relationship to resting potential

Answer:  A

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Knowledge

 

 

Match the potential or potential change with the causative circumstances. Assume ion movements are net movements. Answers may be used more than once or not at all.

 

1. resting membrane potential
2. hyperpolarization
3. depolarization
4. repolarization
5. more than one of the answers

 

49) Na+ enters the cell

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Comprehension

 

50) K+ leaves the cell

Answer:  E

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Comprehension

 

51) Cl- enters the cell

Answer:  B

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Comprehension

 

52) membrane potential is 0 mV

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Knowledge

 

53) This type of membrane protein extends all the way through the cell membrane into both the extra- and intracellular fluids: ________.

Answer:  membrane-spanning

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

54) The membrane proteins that catalyze reactions that take place on the external or internal surface of the cell are ________.

Answer:  enzymes

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

55) The membrane proteins that bind to a ligand and act in the body’s chemical signaling system are ________.

Answer:  receptors

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

56) The membrane proteins that provide for support of the cell membrane and allow for cells to connect to each other are ________.

Answer:  structural proteins

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

57) The membrane proteins that change shape and bind with specific molecules to transport them across the cell membrane are ________.

Answer:  carrier proteins

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

58) Carrier proteins operating at their maximum rate are said to be ________.

Answer:  saturated

Section:  Protein-Mediated Transport

Learning Outcome:  5.9

Bloom’s Taxonomy:  Knowledge

 

59) The three types of gated channels are ________, ________, and ________.

Answer:  chemically gated channels, voltage-gated channels, mechanically gated channels

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

60) Membrane protein pores that can be opened and closed are called ________ channels.

Answer:  gated (regulated)

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

61) The Na+-K+-ATPase pumps (this number of) ________ Na+ ions ________ (into/out of) the cell and (this number of) ________ K+ ions ________ (into/out of) the cell.

Answer:  3 Na+, out of, 2 K+, into

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Knowledge

 

62) A pump that helps maintain an electrical gradient, such as the Na+-K+-ATPase is a(n) ________ pump.

Answer:  electrogenic

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Comprehension

 

63) Channel proteins that allow water to pass are called ________.

Answer:  aquaporins

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

64) Membrane protein pores that are essentially always open are called ________ or ________ channels.

Answer:  open, leak (either order)

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

65) Which of the following is a unique characteristic of glucose as a solute in biological systems?

1. A) It enters the interstitial fluid before going into the cell.
2. B) It is freely penetrating and can pass in and out of the cells at any time.
3. C) 100% of it is absorbed into the cell from the extracellular fluid.
4. D) It is converted into dextrose inside the cell.

Answer:  C

Section:  Osmosis and Tonicity

Learning Outcome:  5.4

Bloom’s Taxonomy:  Comprehension

 

66) The two extracellular compartments in the body are ________ and ________.

Answer:  plasma, interstitial fluid

Section:  Osmosis and Tonicity

Learning Outcome:  5.2

Bloom’s Taxonomy:  Knowledge

 

67) The walls of the ________ separate the two extracellular fluid compartments.

Answer:  circulatory system

Section:  Osmosis and Tonicity

Learning Outcome:  5.2

Bloom’s Taxonomy:  Knowledge

 

 

68) The ability of a carrier molecule to transport only one specific molecule or a group of closely related molecules is called ________.

Answer:  specificity

Section:  Protein-Mediated Transport

Learning Outcome:  5.9

Bloom’s Taxonomy:  Comprehension

69) At rest, nerve cells have an unequal distribution of ions on either side of the cell membrane, producing the ________.

Answer:  resting membrane potential

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

 

70) At rest, nerve cells have a voltage of ________ mV.

Answer:  -70 (Note to instructor: All nerve cells are different, so you may wish to accept a range of similar values instead of insisting on precisely this value.)

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Knowledge

 

71) A cell membrane that is selectively permeable

1. A) randomly chooses which substances will pass through.
2. B) can change which substances pass through by changing its lipid and protein content.
3. C) is impermeable to all substances but water.
4. D) will only allow substances in or out if their concentration in the cell is above or below a certain point.

Answer:  B

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

72) Fick’s law of diffusion states that the rate of diffusion across a membrane is

1. A) proportional to surface area and membrane thickness, but inversely proportional to concentration gradient.
2. B) proportional to concentration gradient, surface area, and membrane permeability.
3. C) proportional to membrane permeability, but inversely proportional to concentration gradient and surface area.
4. D) proportional to membrane thickness and surface area.

Answer:  B

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

 

 

73) Which of the following would increase the rate of diffusion across a cell membrane?

1. A) a decrease in the surface area of the membrane
2. B) a decrease in the concentration gradient
3. C) a decrease in membrane permeability
4. D) a decrease in membrane thickness

Answer:  D

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

74) Pinocytosis and potocytosis are types of

1. A) endocytosis.
2. B) exocytosis.
3. C) phagocytosis.
4. D) endocytosis and exocytosis.
5. E) exocytosis and phagocytosis.

Answer:  A

Section:  Vesicular Transport

Learning Outcome:  5.10

Bloom’s Taxonomy:  Comprehension

 

75) Sodium ions are more concentrated in the extracellular fluid than in the intracellular fluid. This is an example of

1. A) electrical disequilibrium.
2. B) osmotic equilibrium.
3. C) chemical disequilibrium.
4. D) failed homeostasis.

Answer:  C

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

 

76) The inside of a resting cell is slightly negative relative to the outside. This is an example of

1. A) electrical disequilibrium.
2. B) osmotic equilibrium.
3. C) chemical disequilibrium.
4. D) failed homeostasis.

Answer:  A

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

 

 

77) Which property of diffusion best helps explain the necessity of the circulatory system in multicellular organisms?

1. A) Molecules move from an area of higher concentration to an area of lower concentration.
2. B) Diffusion can take place in an open system or across a partition that separates two systems.
3. C) Diffusion is rapid over short distances but much slower over long distances.
4. D) Diffusion rate is inversely related to molecule size.

Answer:  C

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

78) Facilitated diffusion and active transport differ in that

1. A) facilitated diffusion uses cell membrane proteins to move substances, whereas active transport does not.
2. B) facilitated diffusion uses a substrate to bind to a protein carrier, whereas active transport does not.
3. C) ATP is necessary for active transport, but not for facilitated diffusion.
4. D) facilitated diffusion moves water across the cell membrane, whereas active transport does not.
5. E) potential energy is required for active transport but not for facilitated diffusion.

Answer:  C

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

79) Water will always move from ________ areas to ________ areas, if there are no impermeable barriers.

1. A) hyperosmotic, hyposmotic
2. B) hyposmotic, hyperosmotic
3. C) isosmotic, hyposmotic
4. D) hyperosmotic, isosmotic

Answer:  B

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Comprehension

 

80) Substances that readily dissolve in water and do not readily dissolve in lipids are

1. A) hydrophobic and lipophobic.
2. B) hydrophilic and lipophobic.
3. C) hydrophobic and lipophilic.
4. D) hydrophilic and lipophilic.

Answer:  B

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

 

 

81) Which of the following molecules can move across the phospholipid bilayer by simple diffusion?

1. A) lipids
2. B) steroids
3. C) water
4. D) lipids and water
5. E) All of the answers are correct.

Answer:  E

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

82) A cell that is permeable to Solute X is placed into solution containing a higher concentration of X. Diffusion occurs until equilibrium is attained. At this time,

1. A) there is no further movement of Solute X across the membrane.
2. B) there is no further change in concentration of Solute X.
3. C) Both of the statements are correct.
4. D) Neither of the statements is correct.

Answer:  B

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Application

 

83) Cations will ________ each other.

1. A) attract
2. B) repel
3. C) not react with

Answer:  B

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

 

84) Anions will ________ each other.

1. A) attract
2. B) repel
3. C) not react with

Answer:  B

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

 

 

85) The cell membrane acts as a good

1. A) electrical insulator.
2. B) electrical conductor.
3. C) electrical gradient.
4. D) source of ions.

Answer:  A

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

 

86) Describe the distribution of water in the body’s fluid compartments.

Answer:  67% is intracellular.

33% is extracellular; of that 75% is in the interstitial fluid and 25% is in the plasma.

Section:  Osmosis and Tonicity

Learning Outcome:  5.2

Bloom’s Taxonomy:  Comprehension

87) Explain why transporting epithelial cells are said to be polarized. What does it mean when a nerve cell is said to be polarized?

Answer:  Transport epithelia have cells that are polarized with respect to distribution of transport proteins in the apical and basolateral membranes, i.e., different types of transporters in the two membranes. Nerve cells are electrically polarized, meaning that the inside of the membrane is charged relative to the outside.

Section:  Epithelial Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Comprehension

 

88) Explain the term resting membrane potential difference.

Answer:  See “The Resting Membrane Potential” section of the chapter.

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Comprehension

 

89) Evaluate the validity of this statement: “The extracellular and intracellular fluid compartments have a stable solute composition that is in equilibrium.”

Answer:  This statement is only partially true. The two compartments have stable solute compositions, but they are not in equilibrium. We use the term dynamic disequilibrium to describe this relationship.

Section:  Osmosis and Tonicity

Learning Outcome:  5.2

Bloom’s Taxonomy:  Comprehension

 

 

90) Explain how the body can be in a state of osmotic equilibrium and chemical disequilibrium.

Answer:  Osmotic equilibrium occurs because water moves freely between most cells and the extracellular fluid. Water will continue to move across membranes into more highly concentrated compartments until the concentrations (solute/volume) are equal, hence osmotic equilibrium. Osmotic equilibrium does not take into account what particles are present in each compartment, just the total number. The key is that water moves freely, but the solutes do not. Na+ and Cl- are more highly concentrated in the ECF while K+ and many anions are more highly concentrated inside the cell. Each ion is in chemical disequilibrium because it is not evenly distributed between the two compartments. Although the compartments are chemically different (chemical disequilibrium), when all solutes in one compartment are compared to all the solutes in another compartment they have the same total concentrations of solutes (are in osmotic equilibrium).

Section:  Osmosis and Tonicity

Learning Outcome:  5.1

Bloom’s Taxonomy:  Application

 

91) Explain why the composition of the phospholipid bilayer determines how readily water passes through it.

Answer:  The phospholipid bilayer is a fluid mosaic and, depending on the function of the cell, contains various ratios of phospholipids, cholesterol, and proteins. Water molecules slip between the spaces between the fatty acid tails. Membranes with higher levels of cholesterol are less permeable to water because cholesterol fills these spaces.

Section:  Transport Processes

Learning Outcome:  5.5

Bloom’s Taxonomy:  Application

92) How do most polar molecules move through a cell membrane? Explain why water, a polar molecule, is able to cross the nonpolar portion of a cell membrane.

Answer:  Most polar molecules must be assisted by a protein, because the molecule will not interact with the nonpolar phospholipid tails. Examples are facilitated diffusion and active transport. Because water is very small and electrically neutral, it is able to diffuse between the phospholipid tails.

Section:  Protein-Mediated Transport

Learning Outcome:  5.7

Bloom’s Taxonomy:  Application

 

93) Water can cross a cell membrane by a variety of means. List at least three. Do water molecules cross a membrane through the same molecules as other solutes? Explain. Are all cells equally permeable to water? Explain.

Answer:  Water can cross through the phospholipid molecules, through special water channels called aquaporins, and through open or leak channels (pores) that also transport ions. Water can move through pores as a solvation shell around ions or independent of ions, because the watery interstitial fluid is continuous with the watery cytosol when pores are open. Some cells are more permeable to water, especially those with less cholesterol in the membranes and those with a high density of pores.

Section:  Osmosis and Tonicity

Learning Outcome:  5.1

Bloom’s Taxonomy:  Application

 

94) Explain the differences between channel proteins and carrier proteins and why cells need both.

Answer:  Channel proteins allow more rapid transport, but are not as selective. Carrier proteins are slower because of the conformation change. They are also more selective and can move larger molecules than channel proteins. Carrier proteins never allow free exchange across the membrane because they never create a continuous passage between the inside and outside of the cell.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

 

95) Briefly explain the difference between open channels and gated channels.

Answer:  This is discussed in the “Protein-Mediated Transport” section of the chapter.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

 

96) Explain the difference between Fick’s law of diffusion and the flux of a molecule.

Answer:  Fick’s law of diffusion determines the rate of diffusion. The flux of a molecule is the rate of diffusion per unit surface area of membrane.

Fick’s: Rate of diffusion = concentration gradient × membrane permeability × surface area

Flux = concentration gradient × membrane permeability

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Application

97) Name two ways the selectivity of a channel is determined.

Answer:

1. the diameter of the central pore
2. the electrical charge of the amino acids that line the channel

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

 

98) Compare and contrast facilitated diffusion and active transport.

Answer:  Both involve binding of substrate to a carrier, but facilitated is passive, moving solutes down their concentration gradients, whereas active requires ATP and can move solutes against their concentration gradients.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

 

 

99) Distinguish between these statements, explaining what is correct or incorrect about each, and what requires clarification. Relate your answers to the energy hill concept from the previous chapter.

1. Passive transport requires energy.
2. Active transport requires energy.
3. Vesicular transport requires energy.

Answer:  All forms of transport require energy, because something is being moved.

1. Passive transport uses the thermal energy present in the living cell to move molecules in the energetically favorable downhill direction (with concentration gradients).
2. Active transport uses the energy transferred by the ATP molecule, to move molecules in the energetically unfavorable uphill direction (against concentration gradients).
3. Vesicular transport uses the energy of the ATP molecule also, to move large molecules or large quantities of molecules.

Section:  Transport Processes, Protein-Mediated Transport, Vesicular Transport

Learning Outcome:  5.6, 5.8, 5.10

Bloom’s Taxonomy:  Application

 

100) Distinguish between the following terms: cotransport; antiport; symport.

Answer:  Cotransport is the moving of more than one kind of molecule at one time. Antiport is cotransport of two or more solutes in opposite directions across the membrane. Symport is cotransport of two or more solutes in the same direction across the membrane.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

 

101) Compare and contrast primary active transport and secondary active transport, noting any special differences.

Answer:  Both ultimately depend on the energy of ATP, but dependence is indirect in secondary, direct in primary.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

102) Explain the process of secondary active transport and how it uses ATP.

Answer:  Secondary active transport uses the energy released from moving one molecule down its concentration gradient to push other molecules against their concentration gradient. ATP is used to create the chemical disequilibrium (or concentration gradient) for the first molecule.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

 

103) Explain the purpose of having both the reversible GLUT transporters as well as the SGLT transporters in the body.

Answer:  See Figure 5.16 in the chapter.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

 

104) Compare and contrast penetrating solutes and non-penetrating solutes.

Answer:  This is discussed in the “Osmosis and Tonicity” section of the chapter.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Application

 

105) Explain the term dynamic steady state.

Answer:  Dynamic indicates that materials are constantly moving from compartment to compartment, but steady state implies there is no net movement between the compartments.

Section:  Osmosis and Tonicity

Learning Outcome:  5.1

Bloom’s Taxonomy:  Application

 

106) Describe the three major roles of structural proteins.

Answer:  1. connect membrane to the cytoskeleton to maintain cell shape

2. create cell junctions that hold tissues together
3. attach cells to the extracellular matrix by linking cytoskeleton fibers to extracellular collagen and other protein fibers

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

107) Draw a membrane channel protein from two different perspectives, clearly indicating the pore in each.

Answer:  See Figure 5.11 in the chapter.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Comprehension

 

108) How is a carrier protein like a ship canal?

Answer:  See Figure 5.12 in the chapter.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Application

109) What property of some cell membranes is associated with impermeability to water molecules?

Answer:  Higher concentrations of cholesterol in the cell membrane reduce membrane permeability to water.

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

 

 

110) Explain the process of receptor-mediated endocytosis and exocytosis.

Answer:  See Figure 5.19 in the chapter.

Section:  Vesicular Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Comprehension

 

111) Explain the five rules for diffusion and the two rules for simple diffusion across a membrane.

Answer:  See Table 5.6 in the chapter.

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Comprehension

 

112) Write the five rules for osmolarity and tonicity. Are the relative osmolarity and tonicity of an extracellular solution compared to intracellular fluid always the same? If they are, explain why. If they are not, give specific examples of when they are different.

Answer:  See Table 5.4 in the chapter.

Section:  Osmosis and Tonicity

Learning Outcome:  5.4

Bloom’s Taxonomy:  Comprehension

 

113) Explain and distinguish between the following:

1. chemical reaction equilibrium
2. chemical equilibrium
3. osmotic equilibrium

Answer:

1. Chemical reaction equilibrium is achieved in reversible reactions when the rate of the forward reaction equals the rate of the reverse reaction. At this state there is no net change in the concentration of reactants and products in the system. This does not mean that concentrations are equal.
2. Chemical equilibrium occurs when the concentration of a particular solute in one location equals that in another. Typically the locations compared are intracellular vs. extracellular.
3. Osmotic equilibrium occurs when total solute concentration is the same, though chemical disequilibrium may exist.

Section:  Osmosis and Tonicity

Learning Outcome:  5.1

Bloom’s Taxonomy:  Comprehension

 

114) You are a server in a restaurant, always interested in going the extra mile for your customers. Patrick, a regular customer in your section, has ordered sweet iced tea and has an appointment in 10 minutes, so he must drink quickly then leave. The kitchen staff makes only unsweetened tea, but there are sugar packets on the tables. What should you do for Patrick to provide the best sweet tea, and what general principle of diffusion does this illustrate? (Hint: Will sugar dissolve quickly in an iced drink?)

Answer:  Make sugar syrup for Patrick by heating a small amount of water with a generous amount of sugar. The sugar diffuses through the water as it dissolves, and this will occur much quicker in warm water (diffusion rate increases with increasing temperature). Then add the syrup to his iced tea and serve.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Application

 

115) Provide the basic physics definition of the term fluid. What is bulk flow relative to body compartments? What types of matter move by bulk flow? What is fundamentally different in the behavior of these types of matter?

Answer:  A fluid is a substance that flows. Bulk flow is movement of a fluid, usually within a body compartment. Liquids and gases are fluids, and they flow. Gases are compressible, but liquids are not.

Section:  Transport Processes

Learning Outcome:  5.5

Bloom’s Taxonomy:  Comprehension

 

116) When you eat a large meal and your body absorbs a lot of glucose and that makes its way to the interstitial fluid before going into the cell. 100% of the glucose should be absorbed into the cell from the interstitial fluid. Why does nearly all of the glucose enter the cell, rather than only half of it?

1. A) It is moved by active transport.
2. B) It is modified by the cell, so there is still more glucose on the outside of the cell than inside it.
3. C) Insulin forces glucose into the cell against a concentration gradient.
4. D) The cells make ATP so fast, they use up all the glucose as soon as it enters the cell.

Answer:  B

Section:  Integrated Membrane Processes: Insulin Secretion

Learning Outcome:  5.14

Bloom’s Taxonomy:  Application

 

 

117) If a 10% sucrose solution is separated from a 20% sucrose solution by a membrane impermeable to sucrose, in which direction will net movement of water occur?

1. A) from the 10% sucrose solution to the 20% sucrose solution only
2. B) from the 20% sucrose solution to the 10% sucrose solution only
3. C) from the 10% sucrose solution to the 20% sucrose solution and from the 20% sucrose solution to the 10% sucrose solution
4. D) There will be no net movement of water in this case.
5. E) None of the answers are correct.

Answer:  A

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Application

118) If a 10% sucrose solution is separated from a 20% sucrose solution by a membrane permeable to sucrose, in which direction will net diffusion of sucrose take place?

1. A) from the 10% sucrose solution to the 20% sucrose solution
2. B) from the 20% sucrose solution to the 10% sucrose solution
3. C) from the 10% sucrose solution to the 20% sucrose solution and from the 20% sucrose solution
4. D) neither from the 10% sucrose solution to the 20% sucrose solution nor from the 20% sucrose solution to the 10% sucrose solution
5. E) There will be no diffusion in this case.

Answer:  B

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Application

 

119) The concentration of calcium inside a cell is 0.3%. The concentration of calcium outside the cell is 0.1%. How could the cell transport even more calcium to the inside?

1. A) passive transport
2. B) active transport
3. C) osmosis
4. D) exocytosis
5. E) All of the answers are correct.

Answer:  B

Section:  Transport Processes

Learning Outcome:  5.6

Bloom’s Taxonomy:  Application

 

 

120) When the ions move across the cell membrane, an electrical potential change results. As you are probably aware, nerves trigger muscle movement. When you place your finger on a sharp object (such as a tack), you quickly draw your hand away and develop a sensation of pain. Which types of channels are operating under these conditions to allow nerve conduction and subsequent muscle movement?

Answer:  Mechanically gated (in sensory cell receptor potentials), voltage-gated (in nerve and muscle action potentials), and chemically gated (in neuromuscular synaptic transmission).

Section:  The Resting Membrane Potential

Learning Outcome:  5.12

Bloom’s Taxonomy:  Application

 

121) Design an experiment to test whether molecular weight does indeed influence the rate of diffusion. Be sure to list all controlled variables.

Answer:  Answers will vary. Controlled variables may include temperature, pH, composition and volume of solvent or medium, amount of solute added. Solutes could be soluble dyes of different molecular weight, and rate of diffusion could be estimated by observing the extent of coloration around a dye crystal at specified intervals.

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Evaluation

122) Cells of the intestine are very permeable to water while some cells of the kidney tubule are not at all permeable to water. Can you suggest some ways these two types of cells might be structurally different from each other?

Answer:  Kidney cells may have fewer open channels through which water can pass, and/or they may have more cholesterol in their membranes.

Section:  Diffusion

Learning Outcome:  5.7

Bloom’s Taxonomy:  Analysis

 

123) How are molarity and osmolarity different? What property of salts necessitates this distinction? How does this property affect the behavior of water?

Answer:  Molarity is the number of molecules per liter of solution, while osmolarity is the number of independent particles per liter. The ionization of salt in water illustrates the importance of this distinction: one mole of sodium chloride dissociates to produce a total of two moles of particles (one mole Na+ and one mole Cl-), or two osmoles. Osmosis is diffusion of water. A one molar solution of sodium chloride (two osmolar) produces higher osmotic pressure than a one molar solution of glucose, which does not dissociate.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Analysis

 

 

124) Define osmolarity and tonicity. How are they similar? How are they different?

Answer:  Osmolarity refers to the concentration of individual particles in solution. Tonicity refers to the behavior of a cell in a solution. They are similar in that both are related to particles in solution. They are different in that osmolarity depends only on the total concentration of particles in solution, whereas tonicity depends on nature of the particles (i.e., are they penetrating or nonpenetrating) as well as on the concentration of the different particles.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Analysis

 

125) Explain the steps that occur in a pancreatic beta cell so that insulin is secreted.

Answer:  See Figure 5.26b in the chapter.

Section:  Integrated Membrane Processes: Insulin Secretion

Learning Outcome:  5.14

Bloom’s Taxonomy:  Application

126) Indicate the relative osmolarities of the following solutions.

Solution a:      2 osmolar NaCl

Solution b:      1 molar NaCl

Solution c:      900 milliosmolar glucose

 

1. a is ________ osmotic to b. D.  a is ________ osmotic to c.
2. b is ________ osmotic to a. E.   b is ________ osmotic to c.
3. c is ________ osmotic to a. F.   c is ________ osmotic to b.

Answer:  This is easier to answer if all three solutions are described in equivalent terms. Solution b is 2 osmolar, because of the dissociation of sodium and chloride. Solution c is 0.9 osmolar.

1. iso D. hyper
2. iso E. hyper
3. hypo F. hypo

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Application

 

127) When the kidney goes into failure, one of the signs that doctors will see is that red blood cells will crenate (shrivel up). Why does this happen in kidney failure? What process is occurring to the blood cells?

Answer:  Because the kidney is failing to filter particles out of the blood effectively, the plasma becomes hypertonic or hyperosmotic in comparison to the intracellular compartment of the blood cell. Since the cell membrane is impermeable to the ions, but permeable to water, water will leave the cell to try to balance the tonicity and osmolarity with the plasma and in the process the cell will shrink.

Section:  Osmosis and Tonicity

Learning Outcome:  5.4

Bloom’s Taxonomy:  Analysis

 

128) Red blood cells with an internal osmolarity of 300 mOsM are placed in the following solutions. Designate each solution according to its osmolarity and tonicity, and explain what happens to the cells and why.

1. 200 mOsM NaCl
2. 400 mOsM urea
3. 100 mOsM urea plus 200 mOsM NaCl
4. 300 mOsM urea
5. 300 mOsM NaCl
6. 200 mOsM urea plus 300 mOsM NaCl
7. 400 mOsM NaCl

Answer:

5. Hyposmotic, hypotonic. The cell swells. By Rule 5 in Table 5.8 in the chapter, hyposmotic solutions are always hypotonic, because the intracellular solutes are mainly nonpenetrating (Rule 1) thus there will be a net flow of water into the cell.
6. Hyperosmotic, hypotonic. The cell swells. Urea is a penetrating solute, so some urea will move into the cell down its concentration gradient. This will increase the osmolarity inside the cell, causing a net flow of water into the cell.
7. Isosmotic, hypotonic. The cell swells. Urea is a penetrating solute, so there will be a net movement of urea into the cell, raising the osmolarity and causing a net flow of water into the cell.
8. Isosmotic, hypotonic. The cell swells. Urea will penetrate the cell, raising the osmolarity and causing a net flow of water into the cell.
9. Isosmotic, isotonic. No change in cell size. Sodium and chloride are nonpenetrating solutes, so there will be no net ion flow across the membrane. Because there is no osmotic pressure, there will also be no net flow of water.
10. Hyperosmotic, isotonic. No change in cell size at equilibrium. Initially water leaves the cell due to the higher osmolarity outside the cell. Then, because there is a concentration gradient for urea, urea will enter the cell, increasing its osmolarity, and bringing some water into the cell. The nonpenetrating solute concentrations in cell and solution initially are equal, therefore there will be no net movement of water at equilibrium.
11. Hyperosmotic, hypertonic. The cell shrinks. There are no penetrating solutes, and water exits due to the higher osmolarity.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Analysis

 

129) There are two solutions separated by a semipermeable membrane. Solution a is 0.3 M glucose, and solution b is 0.15 M NaCl. Will there be a net flow of water across this membrane? Why or why not?

Answer:  In osmoles, solution a is 0.3 OsM and solution b is 0.3 OsM (because sodium and chloride dissociate into separate particles). The solutions are isosmotic, and there is no net water flow.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Analysis

 

130) There are two solutions separated by a semipermeable membrane. Solution a is 0.2 M NaCl and solution b is 0.1 M CaCl2. Will there be a net flow of water across the membrane? Why or why not?

Answer:  In osmoles, solution a is 0.4 OsM and solution b is 0.3 OsM. Solution a is hyperosmotic, so there will be a net flow of water into solution a until equilibrium is established.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Analysis

 

131) Define the term transport maximum and describe a way cells can increase their transport capacity.

Answer:  The transport maximum occurs when all carrier binding sites are filled with substrate. At this point adding more substrate will no longer increase the rate of transport. In order to increase the capacity and raise the maximum rate of transport, some cells can increase the number of carrier proteins in the membrane.

Section:  Protein-Mediated Transport

Learning Outcome:  5.9

Bloom’s Taxonomy:  Application

 

132) Diuretics cause the kidneys to produce large amounts of urine. Unfortunately, they can also cause the loss of large quantities of K+ in the urine. What effect might prolonged use of diuretics have on nerve or muscle cells?

Answer:  Low concentrations of potassium in the blood is a condition called hypokalemia. Interstitial fluids would similarly become low in K+. As the resting potential of nerve and muscle cells depends primarily on extracellular K+ concentration, the potential would be altered. Decreased extracellular K+ would increase the concentration gradient for movement of K+ out of the cells, which would gradually hyperpolarize the potential as positive ions exit and make the cells less excitable (farther from threshold).

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Analysis

 

133) If someone has a muscle cramp or spasm, a commonly discussed treatment is to consume a banana (or another food high in potassium). Why would this be a possible treatment for muscle cramps or spasms?

Answer:  Potassium is a cation that leaves the cell during an action potential or depolarization of a muscle or nerve cell, and the muscle or nerve cell needs to move potassium back into the cell to allow the cell to repolarize and relax. If there is a deficiency in potassium, muscle and nerve cells may take longer to repolarize and therefore relax, so increasing your dietary intake of potassium may help cells repolarize quicker.

Section:  The Resting Membrane Potential

Learning Outcome:  5.13

Bloom’s Taxonomy:  Analysis

 

134) Which membrane transport process(es) is/are abnormal in people with cystic fibrosis? What is the consequence of the abnormal transport? Which organ systems are affected? How is this disease treated? How long do cystic fibrosis patients normally live? What are some of the possible causes of death related to this disease? What is the cause of cystic fibrosis?

Answer:  Active transport of chloride is impaired, in the airways, sweat glands, and pancreas. The affected epithelia are involved in production of sweat and mucus. Thus, the respiratory, integumentary, and digestive systems are affected. Treatments include replacement of pancreatic digestive enzymes, which are blocked from secretion by the mucus buildup in secretory ducts, and respiratory therapies to loosen mucus in the airways and treat recurring infections. Gene therapy is being explored as well. Median survival is 37 years as of the publication date of the textbook. Causes of death can be related to malnutrition and respiratory illness. This is a genetic disease, in which the gene coding for the chloride transporter is abnormal.

Section:  Epithelial Transport

Learning Outcome:  5.11

Bloom’s Taxonomy:  Analysis

 

You are walking to school one day when you notice an alien spaceship that has crashed in a nearby field. You and some other physiology students collect samples from the beings inside of the spaceship. First, you collect a liquid sample of what appears to be alien blood so that you can test for the concentration of solutes inside the alien blood cells. You then extract some of these cells, place them in various concentrations of glucose in water, then look at them under the microscope. Below is what happens to the cells when they are placed in various concentrations of glucose:

 

Percent Glucose in Water	Condition of Alien Blood Cells
0.02%	cells lyse
0.05%	cells lyse
0.10%	cells lyse
0.12%	cells crenate (shrivel up)
0.20%	cells crenate

 

Table 5.1

 

135) Refer to Table 5.1. From cells located in another part of the alien’s body, you find that the protein-to-lipid ratio of the cell membrane is about 20% protein, 78% lipid, and 2% carbohydrate. Assuming the aliens use their cells as Earthlings do, and have the same terrestrial physiology, what is the most likely function of these cells?

Answer:  Similar to the myelin membrane around nerve cells—good insulators.

Section:  Protein-Mediated Transport

Learning Outcome:  5.8

Bloom’s Taxonomy:  Analysis

 

 

136) Refer to Table 5.1. How can you determine the osmolarity and tonicity of the alien blood and alien cells?

Answer:  Quantitative chemical analysis would determine osmolarity. To determine tonicity, drop the cells into various solutions and observe the cells’ response under the microscope.

Section:  Osmosis and Tonicity

Learning Outcome:  5.4

Bloom’s Taxonomy:  Analysis

137) How does the beta cell in the pancreas react to release insulin when glucose levels are low?

Answer:  ATP dependent potassium gates keep the insulin releasing channels closed when there is enough glucose getting into the cell. When glucose levels drop and the amount of ATP the beta cell is making drops, it eventually loses the energy to keep the gates closed, therefore opening and allowing insulin to be released into the blood.

Section:  Integrated Membrane Processes: Insulin Secretion

Learning Outcome:  5.14

Bloom’s Taxonomy:  Analysis

 

138) What is the approximate concentration of solute present in alien blood cells?

Answer:  Equivalent osmolality to a 0.11% solution of glucose

Section:  Osmosis and Tonicity

Learning Outcome:  5.4

Bloom’s Taxonomy:  Analysis

 

139) A patient is given an intravenous infusion of glucose solution that has a concentration of 25 grams of glucose per liter. If the infusion is given at a rate of 4 milliliters per minute, what is the mass flow of glucose into the body?

Answer:  25 g glucose/1000 mL solution × 4 mL solution/min = 0.1 g glucose/min

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Analysis

 

140) A 70-year-old man weighs 145 pounds and has 48% of his body weight in the form of water. How many liters of water is that?

Answer:  0.48 ×  ×  × 1 L = 31.64 L

Section:  Osmosis and Tonicity

Learning Outcome:  5.3

Bloom’s Taxonomy:  Analysis

 

 

141) A dehydrated patient needs a saline solution IV in order to be rehydrated. Unfortunately the hospital is poorly stocked with saline solutions. The nurse was asked by the doctor to mix up 1 L of a 0.45% saline solution using NaCl and distilled water. How would she do this? What is the osmolarity of this solution?

Answer:  4.5 g NaCl into 950 mL of distilled water.

Adjust the pH if necessary to 7.4, then add more distilled water, up to 1000 mL.

 

×  ×  ×  = 0.154 osmoles/L

Section:  Osmosis and Tonicity

Learning Outcome:  5.4

Bloom’s Taxonomy:  Evaluation

142) Based on what you know about the characteristics of membrane transport, explain the results shown on the graph. Each mixture consists of equal parts of glucose and fructose at the indicated concentration.

 

Answer:  The cell has two separate membrane transport molecules, one specific for glucose and the other for fructose. The transporters become saturated somewhere between the concentrations of 0.5 mM and 1.0 mM, thus any higher concentration of these sugars produces no further increase in transport.

Section:  Protein-Mediated Transport

Learning Outcome:  5.9

Bloom’s Taxonomy:  Analysis

 

 

143) You mix one liter of 300 mOsM NaCl with two liters 450 mOsM glucose.

1. What is the osmolarity of the new solution?
2. What is the final osmolarity of the NaCl in the new solution?
3. What is the final osmolarity of glucose in the new solution?
4. What is the tonicity of this new solution compared to a red blood cell with 300 mOsM nonpenetrating solute?

Answer:

1. 300 mOsM NaCl × 1 L = 300 mOsmoles NaCl.

450 mOsM glucose × 2 L = 900 mOsmoles glucose.

Total solutes = 900 mOsmoles + 300 mOsmoles = 1200 mOsmoles.

Total volume = 1 L + 2 L = 3 L. 1200 mOsmoles/3 L = 400 mOsM solution.

1. 300 mOsmoles/3 L = 100 mOsM NaCl.
2. 900 mOsmoles/3 L = 300 mOsM glucose.
3. Solution is hyperosmotic. Glucose is a penetrating solute, so glucose diffuses into the cell, raising the osmolarity, causing water to diffuse into the cell. The cell swells. Thus, the solution is hypotonic.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3, 5.4

Bloom’s Taxonomy:  Analysis

 

144) Nurse Cameron has been asked to mix an isotonic intravenous solution for an emergency room patient who has lost a lot of blood. The available solutes include glucose (m.w. 180), NaCl (m.w. 58.5), and urea (m.w. 60). How should she make up 10 L of IV solution with an osmolarity of 290 mOsm (isosmotic), making sure that it will also be isotonic?

Answer:  Nurse Cameron should make her solution contain only nonpenetrating solutes, i.e., she should use NaCl but not glucose or urea.

10 L (0.290 Osmoles/L) (1 mole NaCl/2 Osmoles) (58.5 g/1 mole NaCl) = 84.8 g NaCl.

She should add 84.8 g NaCl to about 9.5 L distilled water, mix until dissolved, adjust the pH if necessary to 7.4, then add more water to a final volume of 10 L.

Section:  Osmosis and Tonicity

Learning Outcome:  5.3, 5.4

Bloom’s Taxonomy:  Analysis