Description

 

 

INSTANT DOWNLOAD COMPLETE TEST BANK WITH ANSWERS
Biology 12th Edition by Sylvia S. Mader – Test Bank
Sample  Questions

 

Chapter 01

A View of Life

 

 

Multiple Choice Questions

  1. A university biology department wishes to hire a scientist to work on the relationships among the wolves, moose, trees and physical features on an island. If you were charged with writing the job description, you should title the position
    A. population geneticist.
    B.  molecular biologist.
    C.  community ecologist.
    D.  organismal physiologist.
    E.  island zoologist.

A community ecologist studies the interactions and relationships that occur among groups of species coexisting within a region.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: Ecology

 

 

 

  1. Which of the following terms best describes a conceptual scheme in science that is strongly supported, has not yet been found incorrect, and is based on the results of many observations?
    A. a scientific model
    B.  an experiment
    C.  descriptive research
    D.  a scientific theory or principle
    E.  experimental results

A scientific theory or principle is a conceptual scheme in science that is strongly supported, has not yet been found incorrect, and is based on the results of many observations.
Bloom’s Level: 2. Understand
Learning Outcome: 01.03.02 Distinguish between a theory and a hypothesis.
Section: 01.03
Topic: General

  1. Choose the correct order of classification from most inclusive to exclusive.
    A. Domain-Kingdom-Phylum-Class-Order-Family-Genus-Species
    B.  Kingdom-Domain-Class-Phylum-Order-Family-Genus-Species
    C.  Kingdom-Domain-Class-Phylum-Order-Genus-Species-Family
    D.  Kingdom-Class-Phylum-Domain-Genus-Order-Family-Species

The correct order of classification from most inclusive to exclusive is: Domain-Kingdom-Phylum-Class-Order-Family-Genus-Species.
Bloom’s Level: 1. Remember
Learning Outcome: 01.02.02 Distinguish among the three domains of life.
Section: 01.02
Topic: General

 

 

 

  1. Which listing correctly indicates a sequence of increasing biological organization?
    A. molecule, cell, organelle, atom
    B.  organelle, tissue, cell, molecule
    C.  organ, tissue, atom, molecule
    D.  atom, molecule, organelle, cell

The correct sequence of increasing biological organization is: atom, molecule, organelle, cell.
Bloom’s Level: 2. Understand
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: General

 

True / False Questions

  1. The classification system most commonly used by biologists today contains five domains.
    FALSE

The classification most commonly used by biologists today contains three domains.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.02.02 Distinguish among the three domains of life.
Section: 01.02
Topic: General

 

 

 

 

Multiple Choice Questions

  1. Single-celled prokaryotes
    A. lack a membrane bounded nucleus.
    B.  are classified in the domains Bacteria and Archaea.
    C.  are found in almost all habitats.
    D.  All of the choices are correct.

Single-celled prokaryotes lack a membrane bounded nucleus, are classified in the domains Bacteria and Archaea and are found in almost all habitats. All of the choices are correct.
Bloom’s Level: 2. Understand
Learning Outcome: 01.02.02 Distinguish among the three domains of life.
Section: 01.02
Topic: General

 

Matching Questions

  1. Match the items below with their descriptions:
1.  Plantae       multicellular, photosynthetic organisms   1
2.  Animalia       multicellular organisms that ingest their food   2
3.  Fungi       obtain their food by absorption through filaments called hyphae   3

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.02.02 Distinguish among the three domains of life.
Section: 01.02
Topic: General

 

 

 

 

True / False Questions

  1. Humans have an exaggerated impact on the ecosystem through their use of extra sources of energy and raw materials, and the production of extra wastes that must be handled.
    TRUE

Humans have an exaggerated impact on the ecosystem through their use of extra sources of energy and raw materials, and the production of extra wastes that must be handled.

 

Bloom’s Level: 2. Understand
Learning Outcome: 01.04.02 Summarize the major challenges facing science and society.
Section: 01.04
Topic: General

Living organisms on Earth share many common characteristics. Which statements are TRUE and which are FALSE about nearly all living things?

 

  1. Living organisms are made up of cells.
    TRUE

The Cell Theory states that livings organisms are made of cells. This is a true statement.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

  1. Living things must obey the laws of chemistry and physics.
    TRUE

It is true that living organisms must obey the laws of chemistry and physics.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

 

 

 

  1. Living organisms show biological organization and other common characteristics of life.
    TRUE

In addition to being organized at the cellular level, living organisms exhibit other common characteristics of life.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: General

  1. Emergent properties can be used to distinguish living organisms from nonliving things.
    TRUE

The emergent properties of life can be used to distinguish living organisms from nonliving things.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

  1. Living organisms are composed only of organic elements, whereas nonliving things are made up of inorganic elements.
    FALSE

Living organisms are composed of both organic and inorganic molecules. Nonliving things may be composed or both organic and/or inorganic molecules.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

 

 

 

 

Multiple Choice Questions

  1. A hypothesis is tested by
    A. a prediction.
    B.  experimentation.
    C.  analysis of results.
    D.  formulation of a theory.

An hypothesis is tested by experimentation.
Bloom’s Level: 1. Remember
Learning Outcome: 01.03.02 Distinguish between a theory and a hypothesis.
Section: 01.03
Topic: General

 

True / False Questions

  1. The control group in an experiment receives all the same treatments as the experimental group(s), except for the one variable being tested.
    TRUE

The control group in an experiment receives all the same treatments as the experimental group(s), except for the one variable being tested.

 

Bloom’s Level: 2. Understand
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. Extinction can occur if a species is unable to adapt to a changing environment.TRUE

Extinction can occur if a species is unable to adapt to a changing environment.

 

Bloom’s Level: 2. Understand
Learning Outcome: 01.02.01 Explain the relationship between the process of natural selection and evolutionary change.
Section: 01.02
Topic: Evolution

 

Multiple Choice Questions

  1. Living organisms are characterized by
    A. adapting to the environment.
    B.  evolving over time.
    C.  displaying homeostatic controls.
    D.  all of the choices pertain to living organisms.

Living organisms are characterized by adaptation to the environment, evolving over time and displaying homeostatic controls. All of the choices are correct.
Bloom’s Level: 1. Remember
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

 

True / False Questions
Living and nonliving entities share some characteristics. Which statements are TRUE and which are FALSE about both living and nonliving entities?

 

 

  1. Both living and nonliving entities are organized at the cellular level.
    FALSE

Life is organized at the cellular level. Nonliving entities are not composed of cells.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: General

  1. Both living and nonliving entities exhibit homeostatic controls.
    FALSE

Living entities exhibit homeostatic controls in order to maintain a stable internal environment.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

  1. Both living and nonliving entities are composed of chemical elements.
    TRUE

Both living and nonliving entities are composed of chemical elements.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: General

 

 

 

  1. Both living and nonliving entities adapt to the environment.
    FALSE

A characteristic of life is adaptation to the environment.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

 

Multiple Choice Questions

  1. Which of the following domains contains the most primitive prokaryotes that live in extreme environments?
    A. Archaea
    B.  Bacteria
    C.  Plantae
    D.  Fungi
    E.  Eukarya

 

Bloom’s Level: 2. Understand
Learning Outcome: 01.02.02 Distinguish among the three domains of life.
Section: 01.02
Topic: General

 

 

 

  1. Which of the following concepts is NOT one of the unifying theories of biology?
    A. All organisms are composed of cells.
    B.  Life may arise through spontaneous generation.
    C.  Life comes only from life.
    D.  The internal environment of an organism stays relatively constant.
    E.  All living things have a common ancestor and are adapted to a particular way of life.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.03.02 Distinguish between a theory and a hypothesis.
Section: 01.03
Topic: General

  1. Which of the following does NOT represent homeostasis?
    A.Sensors detect CO2 levels in the blood and trigger an increase or decrease in the rate of breathing.
    B. When body temperature drops, you shiver to generate heat; when your body heats up, you sweat and the evaporation cools you.
    C. Feelings of hunger and then fullness affect the length of time and quantity of food you eat, keeping your weight near a “set point.”
    D. Energy is captured by plants, then transferred to consumers and decomposers, and eventually lost as heat.
    E. Cells adjusting the openings on the bottom of leaves respond to differences in water stress in order to maintain moisture inside the leaf.

 

Bloom’s Level: 3. Apply
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

 

 

 

  1. From Kansas to central Indiana to Pennsylvania, many people believe that they have water moccasins (also called “cottonmouths”) in their ponds. Some assert they have seen them, although they are certainly not so foolish as to try to capture one alive. Meanwhile, the fish and game offices and the range maps in the herpetology books indicate that this poisonous snake does not breed this far north. What is the most scientific attitude to assume on this issue?
    A. Observations by both the public and the fish and game officers are subjective so this is not easily resolved objectively.
    B.  A simple field trip to the pond locations-that resulted in the capture and confirmation of the identity of the snakes-would settle the matter.
    C.  Scientific books with range maps are based on field research and, therefore, determine the truth in this case.
    D.  Because living organisms are active, scientific theories in biology always change.

The very nature of science is one of change. New observations that contradict prevailing theories require further testing, with replication of results. Simply visiting the ponds and confirming the identity of the snakes would settle the matter.
Bloom’s Level: 4. Analyze
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

  1. Studying a brick does not predict the design of a skyscraper. Intense examination of muscle tissues does not allow you to predict the design of a kangaroo or clam. The structure of chlorophyll does not dictate the unique structure of a tree. These cases demonstrate
    A. the essential properties of life.
    B.  the levels of organization from atom to biosphere.
    C.  determinism, or how all phenomena are predictable effects of causes.
    D.  emergent properties that are easily predicted by examining their parts.
    E.  emergent properties that cannot be predicted by examining their parts.

These cases demonstrate emergent properties that cannot be predicted by examining their parts.
Bloom’s Level: 4. Analyze
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. One classic definition of life is “a self-replicating molecular assemblage.” However, clay particles (in clay soil) contain layered aluminum and iron compounds that determine the pattern of the adjacent layers of sediment. This is technically a self-replicating molecular assemblage.
    A. Therefore, it is living.
    B.  It is not living because it cannot think.
    C.  It is not living because there were no molecular changes (or chemistry) involved.
    D.  It is not living because there is no carbon involved; otherwise, such duplication would be living.
    E.  It is not living because it is a simple repetitive process without the ability to evolve or respond to the environment.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

  1. The expected outcome of an experiment is known as
    A. a scientific model.
    B.  an experiment.
    C.  prediction.
    D.  a scientific theory or principle.
    E.  experimental results.

A prediction is the expected outcome of an experiment based upon the knowledge of the factors in the experiment.
Bloom’s Level: 3. Apply
Learning Outcome: 01.03.01 Identify the components of the scientific method.
Section: 01.03
Topic: General

 

 

 

  1. Which of the following organisms is NOT ultimately dependent on the sun as a source of energy?
    A. A night-blooming flower is pollinated by night-flying bats.
    B.  An underground earthworm avoids the sun.
    C.  A cave fish feeds on debris that washes down to it.
    D.  All of the choices ARE ultimately dependent on the sun.
    E.  All of the choices are NOT ultimately dependent on the sun.

All of the choices ARE ultimately dependent on the sun.
Bloom’s Level: 3. Apply
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

  1. A study is conducted to determine the average length and weight of loblolly pine tree needles in the southeast United States. Is this data obtained through observation or experimentation?
    A. observation
    B.  experimentation
    C.  neither observation nor experimentation
    D.  both observation and experimentation

Some scientific data are obtained through observation rather than experimental results. This is true in descriptive research, which describes characteristics about a population.
Bloom’s Level: 3. Apply
Learning Outcome: 01.03.01 Identify the components of the scientific method.
Section: 01.03
Topic: General

 

 

 

  1. A doctor is testing the effectiveness of a new antibiotic.  He gives the first group of patients a placebo, a second group receives antibiotic A while the third group receives antibiotic B.  Which of the groups is considered the control group?
    A. the group that received antibiotic A
    B.  the group that received the placebo
    C.  the group that received antibiotic B
    D.  both groups receiving antibiotic A and B

 

Bloom’s Level: 3. Apply
Learning Outcome: 01.03.01 Identify the components of the scientific method.
Section: 01.03
Topic: General

  1. An earlier classification grouped organisms by whether they inhabited the air, land or sea. However, the five-kingdoms-of-life and three-domains system divided into class-order-family-genus-species as described in this chapter is superior because it
    A. better represents the origin of features held in common-the unity of life in DNA, etc.
    B.  better reflects the origin of adaptations-the diversity of life for differing environments.
    C.  allows the organization of over 900,000 different species.
    D.  groups organisms based on similarities related to their structure and evolution.
    E.  All of the choices are correct.

All of the choices are correct. The later classification systems better represent the unity and diversity of life, allow for better organization of ever-increasing numbers of species, and group organisms based on similarities related to their structure and evolution.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.02.02 Distinguish among the three domains of life.
Section: 01.02
Topic: General

 

 

 

  1. A cell is to a tissue as an atom is to a
    A. molecule.
    B.  subatomic particle.
    C.  electron.
    D.  population.

Tissues are made of cells and molecules are made of atoms.
Bloom’s Level: 3. Apply
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: General

  1. You are interested in the effect of increased carbon dioxide versus normal air, and also in the effect of green light versus full sunlight on the growth of corn plants in a greenhouse. Although you can set up your experiment inside a greenhouse, it is possible that there will be plant growth effects due to effects that you do not know and may never know. Which of the following are important to ensure control of unknown variables?
    A. An increase in carbon dioxide does not result in a substantial decrease of other necessary gases.
    B.  All seedlings are from one uniform strain.
    C.  The intensity or brightness of the green light equals the intensity of the full sunlight.
    D.  All temperatures and available water remain the same.
    E.  All of the choices are important.

All of the choices are important to conduct a valid, controlled experiment.
Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. Tropical rainforests have many species that are found in great abundance. A study in the Brazilian rainforest found 487 tree species growing on a single hectare (2.5 acres). In the US and Canada together, there are only 700 species of trees on millions of acres. In one park in a Peruvian rainforest, scientists have identified over 1300 species of butterflies, while in all of Europe there are approximately 320 butterfly species. These findings suggest that
    A. rainforests are biologically less diverse than other ecosystems on earth.
    B.  the number of tree species and butterfly species are about the same throughout the ecosystems of the world.
    C.  rainforests are biologically more diverse than other ecosystems.
    D.  as many as 400 species a day are lost due to human activity.
    E.  rainforests do not have any type of value to humans.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: Ecology

  1. Biodiversity in a particular ecosystem
    A. is the total number of species in that ecosystem.
    B.  includes the variability of the individual genes.
    C.  impacts the function of the ecosystem in which the species live.
    D.  All of the choices are correct.

All of the choices are correct. Biodiversity involves the total number of species and the variability of individual genes, and it impacts the function of the ecosystem in which the species live.
Bloom’s Level: 1. Remember
Learning Outcome: 01.04.02 Summarize the major challenges facing science and society.
Section: 01.04
Topic: Ecology

 

 

 

  1. Some members of Daphnia, a water flea, have a genetic mutation that causes them to prefer warmer environments. These members reproduce and pass these genetic changes to their offspring. The next generation will occupy warmer environments not previously occupied by this species. This is an example of
    A. adaptation.
    B.  homeostasis.
    C.  irritability.
    D.  All of the choices are correct.

This situation describes adaptation to the environment.

 

Bloom’s Level: 3. Apply
Learning Outcome: 01.01.02 Identify the basic characteristics of life.
Section: 01.01
Topic: General

  1. Some biologists study the complex interactions of animals and plants in forests or prairies. Such ecology field research often produces slightly different results for different researchers. In contrast, ecology experiments that are run indoors with one organism in a terrarium usually produce results that are repeatable. What is the most likely explanation?
    A. The scientific method is only useful in laboratory settings.
    B.  It is not possible to establish a control group outside of a laboratory.
    C.  It is easier to hold all but one variable constant in a laboratory.
    D.  Field research is only descriptive, and descriptive research is not strictly “science.”
    E.  Fieldwork is inductive; lab work is deductive.

It is easier to hold all but one variable constant in a laboratory.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

 

Short Answer Questions
Dr. James isolated Staphylococcus aureus, a type of bacteria, from the leg wound of a ten-year-old boy. He suspected these bacteria would grow better at body temperature than room temperature (72°F), but thought that he should collect data to support his thinking. Dr. James introduced the same number of Staphylococcus bacteria into each of six test tubes containing the same type and amount of nutrient broth. Three test tubes were incubated at 98.6°F (Group 1), while three test tubes (Group 2) sat at 72°F. After 24 hours, Dr. James compared the turbidity (indicative of growth) of all six tubes and rated each on a scale of 0 – 4. 0 indicates no turbidity (no growth), while 4 indicates high turbidity (high growth). The following data was collected:

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

  1. After reading the scenario, write the hypothesis that was being tested in Dr. James’ experiment.

The null hypothesis is:
Ho: There is no difference in the growth of Staphylococcus at 98.6°F and 72°F.
The alternate hypothesis may be written as follows:
HA: Staphylococcus will grow better at 98.6°F than at 72°F in a 24 hour period.

 

Bloom’s Level: 6. Create
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. Would you consider this to be a controlled experiment?

Yes, this may be considered a controlled experiment. Where possible, extraneous variables, such the type and amount of nutrient broth, same number of bacteria in the inoculums, same incubation time, etc. are held constant.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

  1. Evaluate the data presented here to reach a conclusion. Would you reject or accept the null hypothesis.

After analysis of the data, one should reject the null hypothesis. The growth of Staphylococcus at 98.6°F was greater than at 72°F.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

Multiple Choice Questions

  1. The dependent (responding) variable in this experiment is
    A. the temperature.
    B.  growth of bacteria, as indicated by the turbidity in the test tubes.
    C.  the time that the test tubes were allowed to sit.
    D.  amount of initial inoculum, or number of bacteria introduced into each test tube.

The dependent, or responding variable, is the variable that may or may not change, depending on the treatment introduced by the researcher.
Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. The independent (experimental) variable is
    A. temperature.
    B.  growth of bacteria.
    C.  incubation period.
    D.  amount of initial inoculum.

The temperature is the experimental, or independent variable, which is directly manipulated by the researcher.

 

Bloom’s Level: 3. Apply
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

  1. To ensure a controlled experiment all of the following conditions (variables) should be identical in Group 1 and Group 2:
    A. type of bacteria, temperature, and incubation period
    B.  temperature and amount of initial inoculum (bacteria used)
    C.  type of bacteria, incubation period, amount of bacteria used
    D.  degree of turbidity, incubation period, and amount of bacteria

The controlled variables should include the type of bacteria, incubation period, and the amount of bacteria introduced into each test tube.
Bloom’s Level: 4. Analyze
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. Choose which of the the following statements is a hypothesis for an experiment.
    A.Based on the data collected during an experiment, S. aureus grows better at body temperature than room temperature.
    B. Based on previous experience, it is predicted that S. aureus will grow better at body temperature than at room temperature.
    C. S. aureus grew equally well at room temperature and at body temperature.
    D. Based on the data collected during the experiment, it is confirmed that S. aureus grew better at room temperature.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

  1. Dr. James performed another experiment. Instead of inoculating the test tubes with Staphylococcus, he used the bacterium, Streptococcus. He found that Streptococcus grew better at body temperature than at room temperature. This is a replicate of the first experiment.
    A. True
    B.  False

This is not a replicate of the first experiment. Dr. James used Streptococcus instead of Staphylococcus.

 

Bloom’s Level: 3. Apply
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

 

Short Answer Questions

  1. Dr. James considers the bacteria grown at body temperature to be the control group and room temperature the experimental group. Do you agree with this reasoning?

One knows what to expect concerning the dependent variable in a control group. Since body temperature is 98.6°F and the bacteria were isolated from a wound on a leg, one would surmise that the bacteria would grow well at body temperature.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

Multiple Choice Questions

  1. A researcher reports he has conducted many experiments where a galvanometer (lie detector) was hooked up to detect a plant’s responses. He reports that when a spider was released near the plant, the spider’s decision to escape was picked up by the plant, “causing a reaction in the leaf.” When other researchers repeated the experiment, they could not get any galvanometer responses. The researcher then concluded that plants could be put into a faint by humans.
    A. This is a justified conclusion from a research design that appears to follow the scientific method.
    B.  The only problem with this general research plan is that it lacks a control.
    C.  Results must be somewhat repeatable and these results that only work for this researcher do not qualify as science.
    D.  The design is scientific; it just lacks a hypothesis.
    E.  If the researcher has actual numerical counts, this must be accepted as valid science.

Results must be somewhat repeatable and these results that only work for this researcher do not qualify as science.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. An elementary student decides to conduct an experiment comparing the effectiveness of two commercial soaps as cleaning agents, using each in paired tests of different types of stains and colors of cloth, where the only variable was the soap used. The student will use the judgment of classmates to decide if the stains remain equal or if soap one cleans better than the other in each test run. However, the student makes no prediction of which soap is expected to perform best.
    A.This lacks a hypothesis and is therefore not a scientific test.
    B. This lacks any control group (no-soap treatment) and therefore will provide no meaningful results.
    C. Because this is based on the subjective judgments of students, it is not objective and therefore not scientific.
    D. This is a scientific procedure, although it does lack a stated hypothesis describing an anticipated outcome.
    E. This experimental design has all the components and procedures of the scientific method.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

 

 

 

  1. For five years, you wake up before the alarm is set to ring each morning. This leads you to conclude that people have a built-in “alarm clock” capable of waking them up. From a science viewpoint, this conclusion
    A. is science because it is based on real observations.
    B.  is science because it is predictive of what will happen tomorrow morning.
    C.  is scientifically valid because 5 years x 365 days is a large number of trials.
    D.  may not be valid because it generalizes about all people, and there may have been other variables that could awaken you without a built-in clock.
    E.  cannot be scientifically treated because it involves human behavior.

From a science viewpoint, this conclusion may not be valid because it generalizes about all people, and there may have been other variables that could awaken you without a built-in clock.
Bloom’s Level: 5. Evaluate
Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions.
Section: 01.03
Topic: General

  1. The manner in which a scientist intends to conduct an experiment is called
    A. inductive reasoning.
    B.  the experimental design.
    C.  data collection and analysis.
    D.  the conclusion.

The manner in which a scientist intends to conduct an experiment is called the experimental design.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.03.01 Identify the components of the scientific method.
Section: 01.03
Topic: General

 

 

 

 

Short Answer Questions

  1. There is debate as to whether religious creationism should receive ‘equal time’ in the science classroom. Many firmly disagree with creationism being taught in public schools, maintaining that science does not address religious and supernatural concepts. How would you support this statement?

Although answers will vary, a discussion of the nature of science and scientific processes would be appropriate. Distinguishing between science and pseudoscience may also be included.

 

Bloom’s Level: 3. Apply
Learning Outcome: 01.03.01 Identify the components of the scientific method.
Section: 01.03
Topic: General

 

Multiple Choice Questions

  1. Arrange in order, the levels of ecological study from most inclusive to most exclusive:
    A. biosphere, ecosystem, community, population, individual organism
    B.  ecosystem, biosphere, population, community, individual organism
    C.  individual organism, community, population, ecosystem, biosphere
    D.  individual organism, population, community, ecosystem, biosphere

The levels of ecological study from most inclusive to most exclusive are: biosphere, ecosystem, community, population, individual organism.
Bloom’s Level: 2. Understand
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: Ecology

 

 

 

  1. Which definition best describes a population?
    A. the members of a species in a given area
    B.  the interaction between the organisms and their environment
    C.  the region of Earth that contains living organisms
    D.  the interaction between various groups of organisms in a given environment
    E.  all of the females of a given species in a particular area

A population consists of all of the individuals of a given species in a particular region

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: Ecology

  1. Which of the following features are present at the ecosystem level?
    A. chemical cycling through the food chain
    B.  energy flow that begins at the producer level
    C.  input of solar energy
    D.  complex interactions between a variety of populations
    E.  all are features of an ecosystem

All are features of an ecosystem
Bloom’s Level: 5. Evaluate
Learning Outcome: 01.01.01 Distinguish among the levels of biological organization.
Section: 01.01
Topic: Ecology

 

 

 

  1. Organisms belonging to the same _____ would be the most closely related.
    A. kingdom
    B.  phylum
    C.  family
    D.  class
    E.  order

Organisms belonging to the same family would be the most closely related.
Bloom’s Level: 1. Remember
Learning Outcome: 01.02.02 Distinguish among the three domains of life.
Section: 01.02
Topic: General

  1. What type of information does science provide for society?
    A. information about the natural world
    B.  information about the supernatural world
    C.  information about religious beliefs
    D.  information about religious beliefs and the natural world

Science is a systematic way of acquiring information about the natural world.

 

Bloom’s Level: 1. Remember
Learning Outcome: 01.04.01 Distinguish between science and technology.
Section: 01.04
Topic: General

  1. What is the best description of technology?
    A. Technology is the application of scientific knowledge to the interests of humans.
    B.  Technology is the development of new tools.
    C.  Technology is the use of power to make human life easier.
    D.  Technology is the advancement of the functionality of computers.

Technology is the application of scientific knowledge to the interests of humans.

 

Bloom’s Level: 2. Understand
Learning Outcome: 01.04.01 Distinguish between science and technology.
Section: 01.04
Topic: General

Chapter 03

The Chemistry of Organic Molecules

 

 

Multiple Choice Questions

  1. Saturated fatty acids and unsaturated fatty acids differ in
    A. the number of carbon-to-carbon bonds.
    B.  the consistency at room temperature.
    C.  the number of hydrogen atoms present.
    D.  all of the choices are differences between saturated and unsaturated fatty acid.

All of the choices are differences between saturated and unsaturated fatty acid.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.03.02 Distinguish between saturated and unsaturated fatty acids.
Section: 03.03
Topic: Lipids

  1. Which of the following carbohydrates would NOT be a molecule used for energy storage?
    A. starch
    B.  cellulose
    C.  glycogen
    D.  All of the above are used for energy storage

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.02.03 Compare the energy and structural uses of starch, glycogen, and cellulose.
Section: 03.02
Topic: Carbohydrates

 

 

 

  1. The lipids of the cell membrane and the lipids found in butter and vegetable oil differ in which of the following?
    A. the number of fatty acids attached to the glycerol molecule
    B.  the glycerol molecule
    C.  the carbon to carbon bonds
    D.  lipids of the cell membrane do not have hydrophobic sections of the molecule

The lipids of the cell membrane and the lipids found in butter and vegetable oil differ in the number of fatty acids attached to the glycerol molecule.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.03.03 Contrast the structures of fats, phospholipids, and steroids.
Section: 03.03
Topic: Lipids

  1. Which functional group will attach to a hydrocarbon chain to form alcohol?
    A. hydroxyl
    B.  carbonyl
    C.  carboxyl
    D.  amino
    E.  phosphate

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.01.02 Explain the relationship between a functional group and the chemical reactivity of an organic molecule.
Section: 03.01
Topic: Chemical Reactions

  1. Organic molecules are those that contain at least
    A. carbon.
    B.  carbon and oxygen.
    C.  carbon and hydrogen.
    D.  carbon, oxygen, and hydrogen.

Organic molecules are those that contain at least carbon and hydrogen.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.01.01 Explain how the properties of carbon enable it to produce diverse organic molecules.
Section: 03.01
Topic: Chemical Reactions

 

 

 

  1. The differences between organic and inorganic molecules do not follow simple absolute rules. However, most organic molecules are associated with living organisms. Which of the following statements does NOT correspond to the general distinctions between these types of molecules?
    A.Carbon dioxide (CO2) lacks hydrogen atoms found in organic molecules.
    B. Formaldehyde (CH2O) is a small molecule compared to most organic molecules.
    C. Salt (Na+Cl) is not an organic molecule but is important to the life of many organisms.
    D. Because they are in living organisms, organic carbon atoms are different from the inorganic carbon atoms forming the molecular structure of soot or a diamond.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.01.01 Explain how the properties of carbon enable it to produce diverse organic molecules.
Section: 03.01
Topic: Chemical Reactions

  1. A hydrocarbon is hydrophobic
    A. at all times.
    B.  only in the living cell environment.
    C.  except when it has an attached ionized functional group.
    D.  in carbohydrates but not in lipids.

A hydrocarbon is hydrophobic except when it has an attached ionized functional group.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.01.01 Explain how the properties of carbon enable it to produce diverse organic molecules.
Section: 03.01
Topic: Chemical Reactions

 

 

 

  1. What term is used for molecules that have identical molecular formulas but the atoms in each molecule are arranged differently?
    A. isotope
    B.  isomer
    C.  homomolecules
    D.  organic
    E.  balanced

Molecules that have identical molecular formulas but different molecular configurations are isomers.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.01.01 Explain how the properties of carbon enable it to produce diverse organic molecules.
Section: 03.01
Topic: Chemical Reactions

 

Short Answer Questions

  1. What is the molecular formula of these molecules? How do these molecules differ? Are these the same molecule?

These are different molecules, glyceraldehyde on the left and dihydroxyacetone on the right. They are structural isomers, having the same molecular formulas but different structural formulas.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 03.01.01 Explain how the properties of carbon enable it to produce diverse organic molecules.
Section: 03.01
Topic: Chemical Reactions

 

 

 

 

Multiple Choice Questions

  1. Identify the following molecule:A.  amino acid
    B.  hydrocarbon
    C.  carbohydrate
    D.  alcohol
    E.  cholesterol

This molecule is cholesterol.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.03.03 Contrast the structures of fats, phospholipids, and steroids.
Section: 03.03
Topic: Lipids

 

 

 

  1. Glucose (C6H12O6) can exist as both an open-chain form and a closed-ring form. Before 1900, glucose was only thought to occur as an open chain. Now we know that over 99 percent of the time, glucose occurs in the closed-ring form. What possible difference between these forms would give chemists a clue that the open-chain form was not present?
    A. Open-chain molecules can form polymers and glucose does not.
    B.  Only open-chain forms can undergo condensation, which does not occur with glucose.
    C.  An open chain presents ends with functional groups (in this case aldehyde), and glucose failed to undergo typical aldehyde reactions; a phenomenon that could be explained by having no end functional group in a ring structure.
    D.  Because glucose is solid at room temperature, it must have saturated hydrocarbon chains.
    E.  Glucose could not be “denatured” so it must be a tight chain.

An open chain presents ends with functional groups (in this case aldehyde), and glucose failed to undergo typical aldehyde reactions; a phenomenon that could be explained by having no end functional group in a ring structure.

 

Bloom’s Level: 3. Apply
Learning Outcome: 03.02.02 Distinguish among the forms of carbohydrates.
Section: 03.02
Topic: Carbohydrates

  1. What is the molecular formula for 5 glucose molecules?
    A. C30H50O25
    B.  C30H60O30
    C.  C30H52O26
    D.  C6H24O12

C30H52O26 is the molecular formula for 5 glucose molecules. C6H12O6 is multiplied by 5 to equal. C30H60O30. There are four bonds formed through dehydration synthesis to form the chain of 5 glucoses. One water is removed with the formation of each bond to equal: H8O4. Subtract H8O4 from C30H60O30 to equal C30H52O26.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 03.02.02 Distinguish among the forms of carbohydrates.
Section: 03.02
Topic: Carbohydrates

 

 

 

  1. A polysaccharide is a polymer made up of which kind of monomers?
    A. simple sugars
    B.  amino acids
    C.  nucleotides
    D.  alternating sugar and phosphate groups
    E.  fatty acids and glycerol

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.02.03 Compare the energy and structural uses of starch, glycogen, and cellulose.
Section: 03.02
Topic: Carbohydrates

  1. A lipid is a polymer made up of which kind of monomers?
    A. glucose or modified glucose molecules
    B.  amino acids
    C.  nucleotides
    D.  alternating sugar and phosphate groups
    E.  fatty acids and glycerol

A lipid is a polymer made of monomers of fatty acids and glycerol.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.03.03 Contrast the structures of fats, phospholipids, and steroids.
Section: 03.03
Topic: Lipids

  1. A dehydration reaction can also be called a (an) _________ reaction since it forms water.
    A. condensation
    B.  hydrolysis
    C.  isomeric
    D.  an energy-releasing
    E.  monomer formation

A dehydration reaction can also be called a condensation reaction since it forms water.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.01.03 Compare the role of dehydration synthesis and hydrolytic reactions in organic chemistry.
Section: 03.01
Topic: Chemical Reactions

 

 

 

  1. Which of the following are structural carbohydrate molecules?
    A. starch and glycogen
    B.  starch and cellulose
    C.  glycogen and cellulose
    D.  cellulose and chitin
    E.  glycogen and chitin

Cellulose and chitin are structural carbohydrate molecules.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.02.03 Compare the energy and structural uses of starch, glycogen, and cellulose.
Section: 03.02
Topic: Carbohydrates

  1. Which carbohydrate is found in the cell walls of plants?
    A. starch
    B.  chitin
    C.  cellulose
    D.  glycogen
    E.  glycerol

Cellulose is the carbohydrate found in the cell walls of plants.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.02.02 Distinguish among the forms of carbohydrates.
Section: 03.02
Topic: Carbohydrates

 

 

 

  1. Which carbohydrate is used in the liver for energy storage?
    A. starch
    B.  chitin
    C.  cellulose
    D.  glycogen
    E.  glycerol

Glycogen is used in the liver for energy storage.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.02.03 Compare the energy and structural uses of starch, glycogen, and cellulose.
Section: 03.02
Topic: Carbohydrates

  1. Which carbohydrate is found in the exoskeleton of insects and crabs?
    A. starch
    B.  chitin
    C.  cellulose
    D.  glycogen
    E.  glycerol

Chitin is found in the exoskeleton of insects and crabs.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.02.02 Distinguish among the forms of carbohydrates.
Section: 03.02
Topic: Carbohydrates

 

 

 

  1. Identify this molecule:A.  amino acid
    B.  hydrocarbon
    C.  carbohydrate
    D.  alcohol
    E.  lipid

This molecule is a carbohydrate. It may easily be identified by confirming the C:H:O ratio as 1:2:1.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 03.02.02 Distinguish among the forms of carbohydrates.
Section: 03.02
Topic: Carbohydrates

  1. If an animal needed to store energy for long-term use, but not be encumbered with the weight of extra tissue, which is the best molecule for storage?
    A. fructose and glucose in the form of honey
    B.  high-calorie fat molecules
    C.  complex cellulose molecules
    D.  starch
    E.  glycogen with extensive side branches of glucose

If an animal needed to store energy for long-term use, but not be encumbered with the weight of extra tissue, the best molecules for storage are high calorie fat molecules.

 

Bloom’s Level: 3. Apply
Learning Outcome: 03.03.01 Describe why lipids are essential to living organisms.
Section: 03.03
Topic: Lipids

 

 

 

  1. A peptide bond is found in which type of biological molecule?
    A. carbohydrate
    B.  lipid
    C.  nucleic acid
    D.  protein

A peptide bond is found in protein molecules.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.04.02 Explain how a polypeptide is constructed from amino acids.
Section: 03.04
Topic: Proteins

  1. The alpha helix and beta sheet are found at which level of protein organization?
    A. primary structure
    B.  secondary structure
    C.  tertiary structure
    D.  quaternary structure

The alpha helix and beta sheet are the secondary structure of a protein.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.04.03 Compare the four levels of protein structure.
Section: 03.04
Topic: Proteins

 

 

 

  1. After eating eggs for breakfast, you return in the evening, dunk the dirty dishes in water, and notice the yellow streaks remain “dried on.” However, after soaking awhile, the complex of various egg yolk molecules easily “washes off.” What has happened?
    A. Heating denatured the egg protein molecules, hydrolysis reactions then formed bonds in the dried egg, and soaking in water eventually resulted in condensation reactions where water broke these bonds.
    B.  Heating denatured the egg protein molecules, unorganized condensation reactions then formed bonds in the drying egg, and soaking in water eventually resulted in hydrolysis reactions where water broke these bonds.
    C.  The egg monomers were fused to become one polymer, which was easily dissolved by water back into monomers.
    D.  The presence or absence of water changes the molecules from hydrophilic to hydrophobic respectively.
    E.  The addition of water converted organic molecules into inorganic molecules.

Heating denatured the egg protein molecules, unorganized condensation reactions then formed bonds in the drying egg, and soaking in water eventually resulted in hydrolysis reactions where water broke these bonds.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 03.04.04 Understand the factors that affect protein structure and function.
Section: 03.04
Topic: Proteins

 

 

 

  1. Below freezing and above boiling, cells are unable to function as “liquid machinery.” However, most organisms’ cells are still limited from functioning throughout this full range of liquid temperatures. At the molecular level in different organisms, cells’ ability to vary in their tolerance to temperature, etc., is most closely related to variation in
    A. enzyme activity and protein denaturation.
    B.  ATP efficiency.
    C.  ability to form glucose polymers.
    D.  replication of nucleic acids.
    E.  extent of saturation of fatty acids.

At the molecular level in different organisms, cells’ ability to vary in their tolerance to temperature, etc., is most closely related to variation in enzyme activity and protein denaturation.

 

Bloom’s Level: 3. Apply
Learning Outcome: 03.04.04 Understand the factors that affect protein structure and function.
Section: 03.04
Topic: Proteins

  1. Which of these statements is NOT true about DNA?
    A.It is the genetic material of the cell.
    B. It forms a double helix.
    C. Adenine pairs with thymine and guanine pairs with cytosine.
    D. It contains the sugar ribose.
    E. The sugar and phosphate groups form the backbone of the molecule.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.05.02 Compare the structure and function of DNA and RNA nucleic acids.
Section: 03.05
Topic: Nucleic Acids

 

 

 

  1. Fish sperm is composed mostly of the male fish’s DNA. If we tested a sample chemically, we should find relatively high amounts of
    A. nitrogenous bases, sugar, and phosphate groups.
    B.  phospholipids and steroids.
    C.  amino acids and unsaturated fats.
    D.  triglycerides and ATP.
    E.  globular proteins and stored fats.

DNA contains high amounts of nitrogenous bases, sugar, and phosphate groups.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.05.02 Compare the structure and function of DNA and RNA nucleic acids.
Section: 03.05
Topic: Nucleic Acids

  1. Which statement is true about RNA?
    A. It contains adenine paired to thymine.
    B.  One of the bases from DNA is replaced by uracil.
    C.  It contains the sugar deoxyribose.
    D.  Its nucleotides contain twice as many phosphate groups as DNA’s nucleotides.
    E.  It is a double-stranded molecule.

In RNA, one of the bases from DNA is replaced by uracil.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.05.02 Compare the structure and function of DNA and RNA nucleic acids.
Section: 03.05
Topic: Nucleic Acids

 

 

 

  1. The reactivity of an organic molecule is primarily dependent upon ____________ of the molecule.
    A. the carbon skeleton
    B.  the attached functional groups such as a hydroxyl group
    C.  the isomer
    D.  All of the choices are correct.

The reactivity of an organic molecule is primarily dependent on the attached functional group of the molecule.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.01.02 Explain the relationship between a functional group and the chemical reactivity of an organic molecule.
Section: 03.01
Topic: Chemical Reactions

  1. Which Figure is that of an unsaturated fat?A.  Figure A
    B.  Figure B

Figure A is an unsaturated fat. It has double bonds between the some of the carbon atoms. Figure B is a saturated fat with all single bonds between the carbon atoms.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 03.03.02 Distinguish between saturated and unsaturated fatty acids.
Section: 03.03
Topic: Lipids

 

 

 

  1. A saturated fat
    A. is often solid at room temperature.
    B.  has fatty acids with no double bonds between the carbon atoms.
    C.  is of animal origin.
    D.  All of the choices are correct.

A saturated fat is often solid at room temperature, has fatty acids with no double bonds between the carbon atoms, and is of animal origin.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.03.02 Distinguish between saturated and unsaturated fatty acids.
Section: 03.03
Topic: Lipids

  1. A polypeptide has an amino acid sequence of: alanine-leucine-tryptophane-glycien-valine-alanine.
    This chain of amino acids is further organized into a helix that in-turn, folds in and around itself to form a globular structure. The primary structure of this polypeptide is
    A.  globular.
    B.  pleated-sheet.
    C.  alpha helix.
    D.  alanine-leucine-tryptophane-glycien-valine-alanine.

 

Bloom’s Level: 3. Apply
Learning Outcome: 03.04.02 Explain how a polypeptide is constructed from amino acids.
Section: 03.04
Topic: Proteins

 

 

 

  1. How many molecules of water are used to degrade this polypeptide, using hydrolysis reactions, into its constituent amino acids: alanine-leucine-tryptophane-glycine-valine-alanine?
    A. six
    B.  five
    C.  one
    D.  seven

Five molecules of water are used in hydrolysis reactions to degrade this polypeptide. One water for each broken bond.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 03.01.03 Compare the role of dehydration synthesis and hydrolytic reactions in organic chemistry.
Section: 03.01
Topic: Proteins

 

Essay Questions

  1. A strand of DNA has the following base sequence (genetic code): ATTGCGAATGGCA. Construct the complementary strand of DNA.

Adenine and thymine and cytosine and guanine are complementary base pairs. The complementary strand has the sequence: TAACGCTTACCGT.

 

Bloom’s Level: 6. Create
Learning Outcome: 03.05.02 Compare the structure and function of DNA and RNA nucleic acids.
Section: 03.05
Topic: Nucleic Acids

 

 

 

  1. Construct a portion of cell membrane using the typical phospholipid symbol. Indicate the location of the cytoplasm and outside of the cell. Explain how the chemical characteristics of phospholipids molecules dictate membrane structure and cause it to be fluid in nature.

A phospholipids bilayer forms due to the chemical nature of phospholipids molecules – both polar and nonpolar regions. The hydrophilic head are attracted to aqueous regions of the cytoplasm and extracellular fluid (or outside the cell). The hydrophobic tails cluster away from the aqueous regions. Kinks in the tails due to unsaturated bonds increase the fluidity of the bilayer.

 

Bloom’s Level: 6. Create
Learning Outcome: 03.03.04 Compare the functions of phospholipids and steroids in cells.
Section: 03.03
Topic: Lipids

  1. The feet of penguins and reindeer contain large amounts of unsaturated triglycerides. Why would these animals have these fats in their feet rather than saturated fats?

Triglycerides containing fatty acids with unsaturated bonds melt at a lower temperature than those containing only saturated fatty acids. An abundance of polyunsaturated fats helps to protect the organism from freezing in colder climates.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 03.03.02 Distinguish between saturated and unsaturated fatty acids.
Section: 03.03
Topic: Lipids

 

 

 

  1. Chitin and cellulose are molecules that do not spontaneously break down but can be digested by bacteria and some other microorganisms. Since carbon is not among the most common elements in the earth’s crust, what would happen if all of the chitin-digesting and cellulose-digesting organisms on the earth were destroyed?

Answers may vary. When chitin and cellulose containing organisms are decomposed the carbon may be recycled as carbon dioxide if oxygen is present or methane if there is no oxygen available. Inorganic elements are returned to the soil. If there are no decomposers, the carbon would remain as undigested chitin and cellulose. This would alter the carbon cycle and would leave undigested, dead organisms everywhere. In addition, the carbon available for the building of new life would decrease.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 03.02.02 Distinguish among the forms of carbohydrates.
Section: 03.02
Topic: Carbohydrates

  1. Do disaccharide isomers exist? If so, describe how they are formed.

Two glucose molecules (each C6H12O6) undergo a dehydration reaction to form maltose (C12H22O11).  A glucose molecule and a fructose molecule (each C6H12O6) undergo a dehydration reaction to form sucrose (C12H22O11).  A glucose and a galactose molecule (each C6H12O6) undergo a dehydration reaction to form lactose (C12H22O11).  The three disaccharides are isomers, having the same molecular formulas but different structures.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 03.01.03 Compare the role of dehydration synthesis and hydrolytic reactions in organic chemistry.
Section: 03.01
Topic: Carbohydrates

 

 

 

 

Multiple Choice Questions

  1. Choose the Figure that depicts polymer synthesis.A.  Figure 1
    B.  Figure 2

Figure 2 depicts a polymer being hydrolyzed. In degradation, the monomers in a polymer separate during a hydrolysis reaction. A bond is broken as water is added.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 03.01.03 Compare the role of dehydration synthesis and hydrolytic reactions in organic chemistry.
Section: 03.01
Topic: Chemical Reactions

  1. If a segment of DNA has 20% adenine in its base composition, what percent thymine is there?
    A. 20%
    B.  80%
    C.  60%
    D.  30%

Adenine and thymine are complementary base pairs. Since adenine bonds to thymine in DNA, the amount of thymine is the same, 20%.

 

Bloom’s Level: 3. Apply
Learning Outcome: 03.05.02 Compare the structure and function of DNA and RNA nucleic acids.
Section: 03.05
Topic: Nucleic Acids

 

 

 

 

Short Answer Questions

  1. If a certain segment of DNA has 30% adenine in its base composition, what percentage of guanine will there be? ev: 10_30_2013_QC_37868

Adenine and thymine are complementary base pairs. Since adenine bonds to thymine in DNA, the amount of thymine is the same, 30% also. This means the pyrimidines, adenine and thymine would be 60% of the base composition. Therefore, the purines, guanine and cytosine make up 40% of the base composition. Since guanine and cytosine are complementary base pairs, their individual concentrations should be equal, and half of the total purine concentration. Divide 40% by 2 to get 20%. Guanine is 20% of the base pair composition.

 

Bloom’s Level: 3. Apply
Learning Outcome: 03.05.02 Compare the structure and function of DNA and RNA nucleic acids.
Section: 03.05
Topic: Nucleic Acids

 

True / False Questions

  1. Carbon can form covalent bonds with as many as four other atoms.
    TRUE

It is true that carbon can form covalent bonds with as many as four other atoms.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.01.01 Explain how the properties of carbon enable it to produce diverse organic molecules.
Section: 03.01
Topic: Carbohydrates

 

 

 

  1. Waxes consist of a glycerol bonded to three long-chain fatty acids.
    FALSE

Triglycerides are composed of glycerol and three fatty acids. Waxes are long-chain fatty acids bonded with long-chained alcohols.

 

Bloom’s Level: 1. Remember
Learning Outcome: 03.03.03 Contrast the structures of fats, phospholipids, and steroids.
Section: 03.03
Topic: Lipids

  1. ATP is a protein that supplies energy to the cell.
    FALSE

ATP is a nucleotide-based molecule that supplies energy to the cell.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.05.03 Explain how ATP is able to store energy.
Section: 03.05
Topic: Nucleic Acids

  1. Starch is a protein that serves in energy storage in plant cells.
    FALSE

Starch is a polysaccharide that serves in energy storage in plant cells.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.02.03 Compare the energy and structural uses of starch, glycogen, and cellulose.
Section: 03.02
Topic: Carbohydrates

 

  1. Migratory birds store energy as glycogen which is lighter than fat.
    FALSE

Migratory birds store energy as fat, since gram per gram, fat stores more energy than glycogen.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.03.01 Describe why lipids are essential to living organisms.
Section: 03.03
Topic: Lipids

 

Multiple Choice Questions

  1. Which of the following functional groups represents sulfhydryl?
    A. SH
    B.  H-N-H
    C.  OH
    D.  0=C-OH
    E.  C=O

Sulfhydryl is represented by the functional group SH.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.01.02 Explain the relationship between a functional group and the chemical reactivity of an organic molecule.
Section: 03.01
Topic: Chemical Reactions

 

 

 

  1. Which of the following is a function of a steroid?
    A. sex hormone
    B.  transmission of genetic information
    C.  long term energy storage
    D.  insulation against cold
    E.  protective layer

Steroids function as sex hormones and as a component of the cell membrane (cholesterol)

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.03.04 Compare the functions of phospholipids and steroids in cells.
Section: 03.03
Topic: Lipids

  1. Due to an increased risk of heart disease many doctors have suggested shifting our diet to include more oils instead of fats.  As you cook dinner tonight the recipe calls for 2 tbs of oil.  What is the most likely source of the oil you should use in preparing your meal?
    A. canola or olives
    B.  bees wax distilled down to an oil form
    C.  melted butter
    D.  lard that was left out at room temperature
    E.  none of these choices are oils

The most likely source of oil would be canola and olive oils.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.03.03 Contrast the structures of fats, phospholipids, and steroids.
Section: 03.03
Topic: Lipids

 

 

 

  1. Which of the following is NOT a protein function?
    A. support
    B.  transportation
    C.  defense
    D.  motion
    E.  insulation

Proteins function in all of the following except for insulation.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.04.01 Describe the functions of proteins in cells.
Section: 03.04
Topic: Proteins

  1. Which of the following protein is correctly matched with its function?
    A. hemoglobin is found within the red blood cells
    B.  myosin is found within the ligaments
    C.  actin is found within the tendons
    D.  collagen is found within the muscle cells
    E.  hemoglobin is found within the cell membrane

Hemoglobin is found within the red blood cells.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.04.01 Describe the functions of proteins in cells.
Section: 03.04
Topic: Proteins

 

 

 

  1. Which of the following protein structures is exemplified by globular proteins?
    A. tertiary
    B.  secondary
    C.  primary
    D.  quaternary
    E.  all of the protein structures are exemplified by globular proteins

the tertiary protein structure is exemplified by globular proteins.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.04.03 Compare the four levels of protein structure.
Section: 03.04
Topic: Proteins

  1. What are the basic structures that make up a nucleotide?
    A. pentose sugar, phosphate and nitrogen-containing base
    B.  pentose sugar, nitrate and phosphorus-containing base
    C.  ribose sugar, phosphate and nitrogen-containing base
    D.  hydroxide group, phosphate and nitrogen-containing base
    E.  pentose sugar, sodium and nitrogen-containing base

A pentose sugar, phosphate and nitrogen-containing base are the components that make up a nucleotide.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.05.01 Distinguish between a nucleotide and nucleic acid.
Section: 03.05
Topic: Nucleic Acids

 

 

 

  1. Which of the following structures is classified as a nucleic acid?
    A. adenine
    B.  RNA
    C.  thymine
    D.  guanine
    E.  cytosine

RNA is a nucleic acid, the rest of the choices are nucleotides, the subunits of nucleic acids.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.05.01 Distinguish between a nucleotide and nucleic acid.
Section: 03.05
Topic: Nucleic Acids

  1. Which statement below correctly describes why ATP is a high energy structure?
    A. ATP contains 2 phosphate bonds that contain high levels of energy.
    B.  ATP contains hydro-carbon chains that hold energy.
    C.  ATP is composed of thymine which is a high energy molecule.
    D.  The last two phosphate bonds are unstable and easily broken.
    E.  The first phosphate bond is unstable and easily broken.

 

Bloom’s Level: 2. Understand
Learning Outcome: 03.05.03 Explain how ATP is able to store energy.
Section: 03.05
Topic: Nucleic Acids

Chapter 11

Mendelian Patterns of Inheritance

 

 

Multiple Choice Questions

  1. Which of the following reasons helped make Mendel successful with his genetic experiments?
    A. He had a strong background in mathematics.
    B.  He was very deliberate and followed the scientific method closely while doing his research.
    C.  He kept very detailed records of his research.
    D.  He was basing his research off of preexisting research.
    E.  All are reasons Mendel was successful with his genetic experiments.

All are reasons Mendel was successful with his genetic experiments.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.01.01 Describe how Mendel’s scientific approach enabled his genetic experiments to be successful.
Section: 11.01
Topic: Mendelian Genetics

  1. Which of the following is NOT a trait that is the result of, or is affected by, the interaction of more than one gene?
    A. human skin color
    B.  cleft palate
    C.  height
    D.  sickle cell anemia

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.04.04 Explain the concept of polygenic and multifactorial traits.
Section: 11.04
Topic: Non-Mendelian Inheritance

 

 

 

  1. Which of the following crosses would always result in offspring that only display the dominant phenotype?
    A. TT x tt
    B.  Tt x Tt
    C.  TT x TT
    D.  Tt x tt
    E.  Two of the crosses will always display the dominant phenotype

Both TT x TT and TT x tt will always result in offspring that only display the dominant phenotype. On average, one-fourth of the offspring of Tt x Tt will be tt and will not display the dominant phenotype. On average, one-half of the offspring of Tt x tt will be tt and not display the dominant phenotype.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. What aspect of Mendel’s background gave him the necessary tools to discover the laws of inheritance?
    A. He was a monk.
    B.  He was a teacher.
    C.  He lived in Austria.
    D.  He had studied mathematics and probability.
    E.  He corresponded with Charles Darwin.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.01.01 Describe how Mendel’s scientific approach enabled his genetic experiments to be successful.
Section: 11.01
Topic: Mendelian Genetics

 

 

 

  1. What is the blending theory of inheritance?
    A. Mendel’s theory of how the traits of parents are passed to offspring through the gametes
    B.  Darwin’s theory of how traits are passed from all parts of the parent’s body into the gamete to be transmitted to the offspring
    C.  The modern theory of how genetic information is passed from parents to offspring
    D.  An old theory that said that offspring show traits intermediate between those of the parents

The blending theory of inheritance was an old theory that said that offspring show traits intermediate between those of the parents. In Mendel’s theory, traits do not blend but remain discreet.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.01.02 Contrast blending and the particulate concept of inheritance.
Section: 11.01
Topic: Mendelian Genetics

  1. Which characteristic of pea plants were important in their selection as Mendel’s research organism?
    A. Peas are easy to cultivate.
    B.  Pea plants have a short generation time.
    C.  Pea plants are self-pollinating but can be cross-fertilized easily.
    D.  Many true-breeding varieties were available.
    E.  All of these were important characteristics in Mendel’s selection.

All of the factors listed were important in Mendel’s selection of the pea plant.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.01.01 Describe how Mendel’s scientific approach enabled his genetic experiments to be successful.
Section: 11.01
Topic: Mendelian Genetics

 

 

 

  1. In a classic Mendelian monohybrid cross between a homozygous dominant parent and a homozygous recessive parent, which generation is always completely heterozygous?
    A. F1 generation
    B.  F2 generation
    C.  F3 generation
    D.  P generation

The F1 generation is always heterozygous in a Mendelian monohybrid cross.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

  1. If a pea plant shows a recessive phenotype,
    A. the genotype may be TT or Tt.
    B.  the genotype may be Tt or tt.
    C.  the genotype can only be TT.
    D.  the genotype can only be tt.
    E.  the genotype may be TT, Tt, or tt.

If a pea plant shows a recessive phenotype, the genotype is tt. TT and Tt both express the dominant trait.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. Women with X-linked disorders always pass the genes for the disorder to ______, while men with X-linked disorders always pass the genes for the disorder to _______.
    A. only their daughters; only their daughters
    B.  both their daughters and sons; only their sons
    C.  both their daughters and sons; only their daughters
    D.  both their daughters and sons; their daughters and sons

Women with X-linked disorders always pass the genes for the disorder to both their daughters and sons, while men with X-linked disorders always pass the genes for the disorder to only their daughters.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.04.05 Understand how X-linked inheritance differs from autosomal inheritance.
Section: 11.04
Topic: Sex-Linked Inheritance

  1. The F2 offspring of a classic Mendelian monohybrid cross between homozygous dominant and homozygous recessive parents would produce the genotype(s)
    A. AA and Aa.
    B.  Aa and aa.
    C.  AA, Aa, and aa.
    D.  AA only.
    E.  Aa only.

The F2 offspring of a monohybrid cross would show the genotypes, AA, Aa, and aa.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. The offspring of a monohybrid testcross would have what possible genotype(s)?
    A. AA and Aa
    B.  Aa and aa
    C.  AA, Aa, and aa
    D.  AA only
    E.  aa only

The offspring of a monohybrid testcross could have either Aa or aa.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

  1. What are alleles?
    A. genes for different traits, such as hair color or eye color
    B.  alternative forms of a gene for a single trait, such as blue eyes or brown eyes
    C.  the locations of genes on a chromosome
    D.  recessive forms of a kind of characteristic carried by genes
    E.  dominant forms of a kind of characteristic carried by genes

Alleles are alternative forms of a gene for a single trait, such as blue eyes or brown eyes.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. If an individual with a dominant phenotype is crossed with an individual with a recessive phenotype, 4 of their 9 offspring show the recessive phenotype. What is the genotype of the first parent?
    A. AA
    B.  Aa
    C.  aa
    D.  The answer cannot be determined from this information.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. Which is NOT true according to Mendel’s law of segregation?
    A. Each individual contains two factors for each trait.
    B.  One factor must be dominant and one factor recessive in each individual.
    C.  Factors separate from each other during gamete formation.
    D.  Each gamete contains one copy of each factor.
    E.  Fertilization restores the presence of two factors.

All of the choices are true except the suggestion that one factor must be dominant and one factor recessive in each individual.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.02.01 Explain Mendel’s law of segregation and law of independent assortment.
Section: 11.02
Topic: Mendelian Genetics

  1. Some plants fail to produce chlorophyll, and this trait appears to be recessive. Many plants also self-pollinate. If we locate a pea plant that is heterozygous for this trait, self-pollinate it and harvest seeds, what are the likely phenotypes of these seeds when they germinate?
    A. All will be green with chlorophyll since that is the dominant trait.
    B.  All will be white and lack chlorophyll since this is self-pollinated.
    C.  About one-half will be green and one-half white since that is the distribution of the genes in the parents.
    D.  About one-fourth will be white and three-fourths green since it is similar to a monohybrid cross between heterozygotes.
    E.  About one-fourth will be green and three-fourths white since it is similar to a monohybrid cross between heterozygotes.

If we locate a pea plant that is heterozygous for this trait, self-pollinate it and harvest seeds, the likely phenotypes of these seeds when they germinate will be about one-fourth white and three-fourths green since it is similar to a monohybrid cross.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02

 

 

 

  1. The most common lethal genetic disease among Caucasians is
    A. neurofibromatosis.
    B.  Tay-Sachs disease.
    C.  phenylketonuria.
    D.  albinism.
    E.  cystic fibrosis.

The most common lethal genetic disease among Caucasians is cystic fibrosis.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.03.02 Identify the pattern of inheritance for selected single-gene human disorders.
Section: 11.03
Topic: Pedigrees

  1. The pedigree chart depicts the inheritance pattern of ____. Circles depict females and squares depict males. Colored shapes represent affected individuals (expressing a trait) and uncolored shapes are unaffected (do not express a trait).A.  an autosomal recessive characteristic with both parents being heterozygous
    B.  an autosomal dominant characteristic with both parents being homozygous dominant
    C.  an autosomal recessive characteristic with both parents being homozygous recessive
    D.  none.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 11.03.01 Distinguish between an autosomal dominant and an autosomal recessive pattern of inheritance.
Section: 11.03
Topic: Pedigrees

 

 

 

 

Short Answer Questions

  1. Create a pedigree chart that depicts an inheritance pattern of an autosomal dominant characteristic. Use circles to depict females and squares to depict males. Colored shapes represent affected individuals (expressing a trait) and uncolored shapes are unaffected (do not express a trait).

Answers may vary. This figure depicts the pattern of an autosomal dominant characteristic. The parents are heterozygous, and are affected because of the dominance of the trait. The offspring has received the recessive trait from both parents and is therefore not affected.

 

Bloom’s Level: 6. Create
Learning Outcome: 11.03.01 Distinguish between an autosomal dominant and an autosomal recessive pattern of inheritance.
Section: 11.03
Topic: Pedigrees

 

 

 

  1. A range of genotypes and phenotypes occur in polygenic inheritance. Draw a graph that depicts this pattern of continuous variation in polygenic inheritance.

The pattern takes the shape of a bell-shaped curve, or a normal distribution pattern. The graph should resemble the one depicted.

 

Bloom’s Level: 6. Create
Learning Outcome: 11.04.04 Explain the concept of polygenic and multifactorial traits.
Section: 11.04
Topic: Non-Mendelian Inheritance

  1. Where does independent assortment occur in meiosis (what stage)? Where is the law of segregation evident in meiosis? What is the result of independent assortment and segregation? Explain your answer.

In metaphase I of meiosis, homologous chromosomes align in many different orientations along the metaphase plate. The law of independent assortment relates to this random alignment and separation of homologous chromosomes. At metaphase II each daughter cell has only one member of each homologous pair in keeping with the laws of segregation. The result is gametes of all possible combinations of chromosomes and alleles.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 11.02.01 Explain Mendel’s law of segregation and law of independent assortment.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

 

Multiple Choice Questions

  1. If the parents are AO and BO genotypes for the ABO blood group, their children could include which of the following genotypes?
    A. AO and BO only
    B.  AO, BO, and AB only
    C.  AA, BB, and AB only
    D.  AO, BO, and OO only
    E.  AO, BO, AB, and OO only

If the parents are AO and BO genotypes for the ABO blood group, their children could include the following genotypes: AO, BO, AB, and OO.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.04.01 Explain the inheritance pattern of traits when more than two alleles for the trait exist.
Section: 11.03

  1. Haiti is settled by peoples of both African and European ancestry. A young couple, both with mixed ancestry, marry and have several children. The children vary widely in the amount of skin melanin production, with one child being lighter than both parents, and one being darker. The simple explanation for this is
    A. epistasis.
    B.  multiple alleles are available for the one chromosomal locus that governs skin color.
    C.  the environment affected the phenotype that developed.
    D.  polygenic inheritance.
    E.  gene linkage.

The simple explanation for this is polygenic inheritance.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.04.01 Explain the inheritance pattern of traits when more than two alleles for the trait exist.
Section: 11.04
Topic: Non-Mendelian Inheritance

 

 

 

  1. Lethal genes (genes that result in the failure to develop a vital organ or metabolic pathway) are nearly always recessive. Animal breeders who discover a unique trait and selectively breed to increase the occurrence of that trait often encounter a noticeable increase in lethal genes. Why?
    A. The lethal recessive gene may be incompletely dominant.
    B.  Spreading the gene among offspring of both sexes will increase the likelihood it will be sex-linked and expressed.
    C.  The selective-mating of closely related individuals, or inbreeding, increases chances that two recessive genes will “meet” in offspring.
    D.  “Pleiotropy” – the gene that is being selected for this trait may have the second effect of being lethal.
    E.  “Epistasis” – selection for the desired trait may result in “uncovering” the lethal gene.

Animal breeders who discover a unique trait and selectively breed to increase the occurrence of that trait often encounter a noticeable increase in lethal genes because the selective-mating of closely related individuals, or inbreeding, increases chances that two recessive genes will “meet” in offspring.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. Unattached earlobes (EE or Ee) are described in the textbook as dominant over attached earlobes (ee). A couple both have unattached earlobes. Both notice that one of their parents on both sides has attached earlobes (ee). Therefore, they correctly assume that they are carriers for attached earlobes (Ee). The couple proceeds to have four children.
    A. They can be certain that three will be heterozygous and one homozygous recessive.
    B.  If the first three are heterozygous, the fourth must be homozygous recessive.
    C.  The children must repeat the grandparents’ genotype (Ee).
    D.  All children must have unattached earlobes since both parents possess the dominant gene for it.
    E.  Two heterozygous, one homozygous recessive and one homozygous dominant is a likely outcome, but all heterozygous, or two, three or all four homozygous are also possible.

Two heterozygous, one homozygous recessive and one homozygous dominant is a likely outcome, but all heterozygous, or two, three or all four homozygous are also possible.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. In 1940, two researchers named Weiner and Landsteiner discovered that about 85 percent of the human population sampled possessed a blood cell protein that had been previously detected in Rhesus monkeys. This blood type was labeled Rh positive, and Rh+ was found to be dominant over the absence of the blood factor (Rh). Under normal Mendelian inheritance, which of the following statements is FALSE?
    A. Two Rh+ parents could have an Rh child.
    B.  Two Rh parents could have an Rh+ child.
    C.  An Rh child would require that both parents be carriers of at least one Rh gene.
    D.  It is possible with just one pair of parents to have children where some siblings are Rh and some are Rh+.
    E.  All of the choices are false.

Under normal Mendelian inheritance, two Rh- parents could NOT have an Rh+ child.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. Since each child of two heterozygous parents has a 50% chance of receiving a recessive trait from each parent,
    A. if the first child is phenotypically recessive, then the next child must be phenotypically dominant.
    B.  if the first child is phenotypically recessive, then the next child has a 3/4 chance of being phenotypically recessive.
    C.  if the first child is phenotypically recessive, then the next child has a 1/2 chance of being phenotypically recessive.
    D.  no matter what the first child’s phenotype, the next child will have a 1/4 chance of being phenotypically recessive.

Each individual offspring has the same chance of being phenotypically recessive. The phenotype of the first child has no effect on the phenotype of any other children.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. The ability to roll the edges of the tongue upward in a U-shape has been considered to be an inherited ability. The standard assumption is that tongue-rolling is a dominant allele at a single gene locus. Which of the following would cast doubt on this assumption?
    A. A teacher reports that after testing her class on the ability to roll their tongue, with very little effort the non-tongue-rollers can learn to also roll their tongues.
    B.  A student who can roll his tongue has a mother and father, both of whom cannot.
    C.  A student who cannot roll his tongue has a mother and father, both of whom can.
    D.  Two of the above are situations that would cast doubt on this assumption.

The non-tongue rollers who learn to roll their tongues and the student who can roll his tongue while his parents cannot are two situations that cast doubt on the heritability issue. The situation where a student cannot roll his tongue while his parents can is inconclusive.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 11.03.01 Distinguish between an autosomal dominant and an autosomal recessive pattern of inheritance.
Section: 11.02
Topic: Mendelian Genetics

  1. If the probability of event A is 3/4 and the probability of event B is 1/4, then the probability of both A and B occurring at the same time is
    A. 3/4.
    B.  1/4.
    C.  1 or absolute certainty.
    D.  1/2.
    E.  3/16.

If the probability of event A is 3/4 and the probability of event B is 1/4, then the probability of both A and B occurring at the same time is 3/16. The product rule of probability clarifies that you must multiply the chances of independent events to calculate the probability of inheriting a specific set of two alleles.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. The reason why some individuals who inherit polydactyly (having an extra digit on the hand or feet) but do not express the trait is due to ________.
    A. incomplete penetrance
    B.  incomplete dominance
    C.  gene linkage
    D.  pleiotropy
    E.  none of the answers are correct

Incomplete penetrance is when a dominant allele doesn’t fully express itself.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.04.02 Contrast incomplete dominance and incomplete penetrance.
Section: 11.04
Topic: Non-Mendelian Inheritance

  1. Computer simulations are sometimes used to demonstrate the outcome of monohybrid fruit fly crosses, where a student can run generation after generation of fruit flies with 100 offspring produced each generation, half male and half female, and a 3-to-1 phenotype ratio (or 75 to 25) in the F1 generation. Compared with real genetics results,
    A. rarely would exactly 100 fly offspring be produced or survive.
    B.  an exact balance between males and females would be rare.
    C.  a precise 3-to-1 ratio would be uncommon.
    D.  All of the choices are true.

In reality, one would rarely have exactly 100 fly offspring produced or survive, an exact balance between males and females or a precise 3-to-1 phenotype ratio.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. In what kind of classic Mendelian cross would you expect to find a ratio of 9:3:3:1 among the F2 offspring?
    A. monohybrid cross
    B.  dihybrid cross
    C.  testcross
    D.  None of the choices is correct.

In a classic Mendelian dihybrid cross, one would you expect to find a ratio of 9:3:3:1 among the F2 offspring.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

  1. If a woman is a carrier for the color-blind recessive allele and her husband is normal, what are their chances that a son will be color-blind?
    A. None since the father is normal.
    B.  50% since the mother is the only carrier.
    C.  100% because the mother has the gene.
    D.  25% because the mother is a hybrid.
    E.  None since he will also be just a carrier.

If a woman is a carrier for the color-blind recessive allele and her husband is normal, their chances that a son will be color-blind is 50%, since the mother is the only carrier.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.04.05 Understand how X-linked inheritance differs from autosomal inheritance.
Section: 11.04
Topic: Sex-Linked Inheritance

 

 

 

  1. In pea plants, the gene for round seed (R) is dominant, and wrinkled seeds (r) are recessive. The endosperm of the pea is also either starchy, a dominant gene (S), or waxy (s). What can be said of a fully heterozygous, dihybrid cross?
    A. It is impossible to secure offspring that are homozygous for both dominant genes.
    B.  It is impossible to secure offspring that are homozygous for both recessive genes.
    C.  It is impossible to secure offspring that are homozygous for one dominant gene such as round seed and homozygous recessive for the other recessive waxy gene.
    D.  All of these choices are impossible combinations in a dihybrid cross.
    E.  All of these choices are possible combinations in a dihybrid cross.

All of these choices are possible combinations in a fully heterozygous, dihybrid cross. Homozygous dominant for both genes, homozygous recessive for both genes, and homozygous for one gene and homozygous recessive for another are all possible combinations.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

  1. A testcross involves an individual exhibiting the dominant phenotype but an unknown genotype being crossed with an individual that has a(n) ___________ genotype.
    A. homozygous dominant
    B.  heterozygous dominant
    C.  homozygous recessive
    D.  any of the choices

A testcross involves an individual exhibiting the dominant phenotype but an unknown genotype being crossed with an individual that has a(n) homozygous recessive genotype.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.02.01 Explain Mendel’s law of segregation and law of independent assortment.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. If individuals exhibiting a dominant phenotype are crossed and produce only offspring with the dominant phenotype, what would be the logical genotype of the parents?
    A. both are homozygous recessive
    B.  one is heterozygous and one is homozygous dominant
    C.  both are homozygous dominant
    D.  both are heterozygous
    E.  two of the options may be correct

If individuals exhibiting a dominant phenotype are crossed and produce only offspring with the dominant phenotype, the parents may both be homozygous dominant or one may be heterozygous and one homozygous dominant.

 

Bloom’s Level: 3. Apply
Learning Outcome: 11.03.01 Distinguish between an autosomal dominant and an autosomal recessive pattern of inheritance.
Section: 11.03
Topic: Mendelian Genetics

  1. In the use of a Punnett square for genetic results of crossing individuals
    A. all different kinds of sperm are lined up either horizontally or vertically.
    B.  all different kinds of eggs are lined up either horizontally or vertically.
    C.  the results show the offspring’s expected genotypes.
    D.  All of the choices are correct.
    E.  Only two of the choices are correct.

In the use of a Punnett square for genetic results of crossing individuals, all different kinds of sperm are lined up either horizontally or vertically, all different kinds of eggs are lined up either horizontally or vertically and the results show the offspring’s expected genotypes. All of the choices are correct.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. If a human who is a tongue roller (T) and has unattached ear lobes (E) marries a person who cannot roll their tongue and has attached earlobes, could they produce an offspring that was also a non-tongue roller with attached earlobes? What would be the genotype of the first parent? the second parent?
    A. yes; TtEe; ttee
    B.  yes; TtEE; ttEe
    C.  no; TTEE; ttee
    D.  unable to determine from the information given

If a human who is a tongue roller (T) and has unattached ear lobes (E) marries a person who cannot roll their tongue and has attached earlobes, they could produce an offspring that is also a non-tongue roller with attached earlobes. The genotypes of the parents would be TtEe and ttee.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 11.02.03 Use a Punnett square and the law of probability to predict the chances of producing gametes and offspring.
Section: 11.02
Topic: Mendelian Genetics

  1. The particulate theory of inheritance
    A. preceded Mendel’s research by a century.
    B.  was proposed by Mendel.
    C.  is based on particles or hereditary units we now call genes.
    D.  All of the choices are correct.
    E.  Two of the above.

The particulate theory of inheritance was proposed by Mendel and is based on particles or hereditary units we now call genes.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.01.02 Contrast blending and the particulate concept of inheritance.
Section: 11.01
Topic: Mendelian Genetics

 

 

 

  1. As many as 60% of people in malaria-infected regions of Africa have the sickle-cell allele, but only about 10% of the U.S. population of African ancestry carries the allele. Malaria remains a major disease in central Africa but has not been a serious problem in the U.S. for many generations. What is/are the reason(s) for the difference in the percentages and what is a reasonable statement about future percentages?
    A. The presence of malaria in Africa maintains the advantage of the heterozygous sickle-cell trait, and the prevalence of malaria will likely continue to preserve the 60% rate.
    B.  The U. S. percentage may have always been somewhat lower due to immigration from nonmalaria regions, but changes in sites and rates of immigration could occur.
    C.  Lack of widespread malaria in the United States would have made both homozygous and heterozygous carriers of sickle-cell undergo several generations of negative selection, and we should expect this to continue unless innovative therapies give all individuals an equal chance of surviving and reproducing.
    D.  All of the choices are reasonable.

All of the choices are reasonable explanations for current and future differences in the percentages of the sickle cell allele in the U.S. and Africa.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 11.04.04 Explain the concept of polygenic and multifactorial traits.
Section: 11.04
Topic: Non-Mendelian Inheritance

 

 

 

 

Essay Questions

  1. An individual with blood type A marries an individual with blood type B.
    A. What blood types could their offspring exhibit?
    B. Provide the possible genotypes of parents and offspring produced.
    C. What pattern of inheritance is this?
  2. The offspring could have Blood Types A, B, AB or O.
    B. Possible gentotypes of parents and their offspring are:C. This is an example of multiple allelic traits and codominance.

 

Bloom’s Level: 6. Create
Section: 11.04
Topic: Non-Mendelian Inheritance

 

Multiple Choice Questions

  1. Which is associated with the inability to produce factor VIII in the blood?
    A. Williams syndrome
    B.  trisomy 21
    C.  color-blindness
    D.  Hemophilia A
    E.  Duchenne muscular dystrophy

The inability to produce factor VIII in the blood is associated with Hemophilia A.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.04.05 Understand how X-linked inheritance differs from autosomal inheritance.
Section: 11.04
Topic: Sex-Linked Inheritance

 

 

 

  1. When crossing a true-breeding red snapdragon flower with a true-breeding white flower of the same species, we secure all pink offspring. This would seem to support the pre-Mendel view that inheritance is a blending of parental traits. However, Mendel and conventional wisdom agree that “blending” of parental traits is not correct and that particles of inheritance are actually involved because
    A. in the case of incomplete dominance, only radioactive isotope tracers can follow the actual hereditary particles.
    B.  under blending theory, over many generations only the average (or pink flowers) would remain; there would be no way to get back to pure red and white.
    C.  it is possible to cross the pink F-1 generation and secure a predictable proportion of pure red and white flowers again, which is not accounted for under the blending theory.
    D.  there is no way to directly prove incomplete dominance is not a case of blending, but we can be sure of genes because of the other cases of dominance, etc., where genes are the only logical explanation.
    E.  two of these are true.

If blending were true, over many generations only the average (or pink flowers) would remain, there would be no way to get back to pure red and white. However, it is possible to cross the pink F-1 generation and secure a predictable proportion of pure red and white flowers again, which is not accounted for under the blending theory.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 11.01.02 Contrast blending and the particulate concept of inheritance.
Section: 11.01
Topic: Mendelian Genetics

 

 

 

 

Essay Questions

  1. Black and brown guinea pigs are bred. Assuming their color is determined by one gene, and that there is a dominant and a recessive allele, how would you determine which allele is dominant? How could you determine the genotypes of the parents? Explain.

Answers may vary.
Situation 1: If both parents are homozygous and the black allele is dominant, all F1 offspring will have a black color. The F2 offspring will have an average phenotypic ratio of 3 black: 1 brown.
Situation 2: If both parents are homozygous and the brown allele is dominant, all F1 offspring will have a brown color. The F2 offspring will have an average phenotypic ratio of 3 brown: 1 black.
Situation 3: If one parent is homozygous recessive and one is heterozygous, then an average of half the F1 offspring will be black and half will be brown. An average of half will have a homozygous recessive genotype and half will have a heterozygous genotype. To determine which allele is dominant, I would breed F1 offspring of the same color (i.e. two browns and two blacks). F1 parents expressing the recessive trait would have a homozygous recessive genotype and produce offspring of all the same phenotype. Two brown colored guinea pigs would produce all browns or two blacks would produce all blacks. When two heterozygous parents are crossed, the dominant phenotype will be expressed in ¾ of the offspring and only ¼ of the offspring will express the recessive trait. Therefore, I would study the F2 phenotypes of both crosses to determine the dominant and recessive traits.
Situation 4: I might also breed black colored guinea pigs from the parent generation with other black guinea pigs and brown colored guinea pigs with other brown guinea pigs. If they are homozygous, the color of offspring will remain the same as the parent color through the generations.

 

Bloom’s Level: 6. Create
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Pedigrees

 

 

 

  1. What is the blending theory of inheritance? How did Mendel disprove this theory?

The blending theory of inheritance stated that an individual’s genetic makeup is intermediate of that of the parents. In one of his experiments that disproved theory, Mendel used true breeding pea plants and performed reciprocal crosses: Pollen from tall plants with eggs from short plants was one cross and pollen from short plants crossed with eggs from tall plants was another cross. All offspring (F1 generation) were tall and resembled one parent. This did not support the blending theory, which suggests that all offspring would have been of an intermediate height.

 

Bloom’s Level: 6. Create
Learning Outcome: 11.01.02 Contrast blending and the particulate concept of inheritance.
Section: 11.01
Topic: Mendelian Genetics

 

True / False Questions

  1. Cystic fibrosis and Phenylketonuria are common autosomal dominant disorders.
    FALSE

Cystic fibrosis and Phenylketonuria are autosomal recessive disorders.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.03.02 Identify the pattern of inheritance for selected single-gene human disorders.
Section: 11.03
Topic: Non-Mendelian Inheritance

  1. In a case of incomplete dominance, the phenotypic ratio of the F2 generation is the same as the genotypic ratio.
    TRUE

It is true that in incomplete dominance, the phenotypic ratio of the F2 generation is the same as the genotypic ratio.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.04.02 Contrast incomplete dominance and incomplete penetrance.
Section: 11.04
Topic: Non-Mendelian Inheritance

 

 

 

  1. Each gamete carries one factor, now called an allele, for each inherited trait.
    TRUE

It is true that each gamete carries one factor, now called an allele, for each inherited trait.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.01 Explain Mendel’s law of segregation and law of independent assortment.
Section: 11.02
Topic: Mendelian Genetics

  1. If an individual is heterozygous for a particular trait, the gametes that individual produces will contain 3/4 dominant and 1/4 recessive alleles.
    FALSE

It is not true that if an individual is heterozygous for a particular trait, the gametes that individual produces will contain 3/4 dominant and 1/4 recessive alleles. They will theoretically produce an equal number of gametes that contain dominant alleles to those containing recessive alleles.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.01 Explain Mendel’s law of segregation and law of independent assortment.
Section: 11.02
Topic: Mendelian Genetics

  1. Polygenic traits such as height or weight are often influenced by the environment of the organism.
    TRUE

It is true that polygenic traits such as height or weight are often influenced by the environment of the organism.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.04.04 Explain the concept of polygenic and multifactorial traits.
Section: 11.04
Topic: Non-Mendelian Inheritance

 

 

 

 

Multiple Choice Questions

  1. If black fur is produced by a recessive allele, which genotype is most likely to produce a black individual?
    A.bb
    B. BB
    C. Bb
    D. Cb

Recessive phenotypes are produced by two recessive alleles (bb).

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. If white eyes are produced by a recessive allele, what is the likely genotype of a white-eyed individual?
    A. rr
    B.  Rr
    C.  RR
    D.  RB

Recessive phenotypes are expressed by two recessive alleles (rr).

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

 

 

 

  1. Which allele combination represents a homozygous dominant individual?
    A. AA
    B.  Aa
    C.  aa
    D.  none of these are homozygous dominant individuals

AA represents homozygous dominance.

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. Which allele combination represents a recessive individual?
    A. aa
    B.  AA
    C.  Aa
    D.  None of these represent a recessive individual

recessive genotypes are represented by two small letters (alleles).

 

Bloom’s Level: 1. Remember
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. Pleiotropy typically
    A. causes various symptoms to appear even though it only impacts a single gene.
    B.  causes a single symptom to appear even though it only impacts a single gene.
    C.  causes various symptoms to appear even though it only impacts multiple genes.
    D.  causes a single symptom to appear even though it only impacts multiple genes.

Pleiotropy typically causes various symptoms to appear even though it only impacts a single gene.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.04.03 Describe the effects of pleiotropy on phenotypic traits.
Section: 11.04
Topic: Non-Mendelian Inheritance

 

 

 

  1. Which occurrence is when an individual inherits a dominant gene but does not fully express the dominant phenotype?
    A. incomplete penetrance
    B.  incomplete dominance
    C.  polygenic inheritance
    D.  pleiotropy

Incomplete penetrance is when an individual inherits a dominant gene but does not fully express the dominant phenotype.

 

Bloom’s Level: 1. Remember
Section: 11.03
Topic: Non-Mendelian Inheritance

  1. Which of the following scenarios would not have been used by Mendel during his research?
    A. relied on his memory to remember his data
    B.  followed the scientific method when designing his experiments
    C.  used a statistic analysis to determine if his data was valid
    D.  researched multiple generations of pea plants to help derive his conclusions

Mendel was a very meticulous scientist who would have kept detailed notes of his research.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.01.01 Describe how Mendel’s scientific approach enabled his genetic experiments to be successful.
Section: 11.01
Topic: Mendelian Genetics

  1. Which factor led to Mendel’s success as a scientist?
    A. applied mathematics to the field of science
    B.  followed the scientific method very closely
    C.  kept detailed records of his research
    D.  all of these factors made Mendel successful

All of these factors made Mendel successful.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.01.01 Describe how Mendel’s scientific approach enabled his genetic experiments to be successful.
Section: 11.01
Topic: Mendelian Genetics

 

 

 

  1. Which statement about the blending concept of inheritance is correct?
    A. A cross between a tall man and a short woman would always produce children of medium height.
    B.  A cross between a tall man and a short woman can produce some children that are tall while others are short.
    C.  A cross between a tall man and a short woman produce tall children who can then produce short grandchildren.
    D.  Genetic material is unstable which accounts for such a wide diversity in our offspring.

The blending concept of inheritance believed that the offspring of contrasting parents are always of intermediate appearance.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.01.02 Contrast blending and the particulate concept of inheritance.
Section: 11.01
Topic: Mendelian Genetics

  1. Which of the following disorders is not a X-linked trait?
    A. hemophilia
    B.  Muscular dystrophy
    C.  Adrenoleukodystrophy
    D.  All of these are X-linked disorders

All of these are X-linked disorders.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.04.05 Understand how X-linked inheritance differs from autosomal inheritance.
Section: 11.03
Topic: Sex-Linked Inheritance

 

 

 

  1. During Mendel’s pea plants experiment he discovered that the trait for tallness is dominant to that of shortness.  Which of the following statements is correct?
    A. Tall = TT or Tt while short = tt
    B.  Tall = TT or tt while short = Tt
    C.  Tall = tt while short = TT or Tt
    D.  None of these statements are correct.

Tall = TT or Tt while short = tt

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.02.02 Compare and contrast dominant alleles with recessive alleles and their relation to genotype and phenotype.
Section: 11.02
Topic: Mendelian Genetics

  1. When an individual of African descent marries and has children with an individual of European descent their children often have a mid shade of skin color.  This can best be described as
    A. X-linked inheritance.
    B.  polygenic inheritance.
    C.  pleiotropic inheritance.
    D.  the dominance of the gene for dark skin tone.

Polygenic inheritance is when two or more alleles control the expression of a trait.  Human skin tone is an example of polygenic inheritance.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.04.04 Explain the concept of polygenic and multifactorial traits.
Section: 11.04
Topic: Non-Mendelian Inheritance

 

 

 

  1. When a round fruit plant is crossed with a long fruit plant their offspring are oval.  What is the best explanation for this scenario?
    A. pleiotropic inheritance
    B.  polygenic inheritance
    C.  incomplete dominance
    D.  X-linked inheritance

When two individuals of different phenotypes produce an offspring that has a third variation of the phenotype it is referred to as incomplete dominance.

 

Bloom’s Level: 2. Understand
Section: 11.04
Topic: Non-Mendelian Inheritance

  1. Which of the following is not part of Mendel’s law of segregation?
    A. Each individual has two factors for each trait.
    B.  The factors segregate during Meiosis.
    C.  Each gamete will contain two factors for a trait.
    D.  Fertilization gives the offspring two factors for each trait.

 

Bloom’s Level: 2. Understand
Learning Outcome: 11.02.01 Explain Mendel’s law of segregation and law of independent assortment.
Section: 11.02
Topic: Mendelian Genetics

Chapter 21

Protist Evolution and Diversity

 

 

Multiple Choice Questions

  1. The most widely accepted formal classification of protists assigns them to
    A. the kingdom Protista.
    B.  six different supergroups.
    C.  Domain Archaea.
    D.  Domain Eubacteria

The most widely accepted formal classification of protists assigns them to six different supergroups.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.01.03 Understand that protists represent multiple evolutionary lineages.
Section: 21.01
Topic: Protists

  1. Which of the characteristic(s) apply to the Supergroup Archaeplastida?
    A. They contain plastids that originated from endosymbiotic cyanobacteria.
    B.  They include multicellular and unicellular organisms.
    C.  They include land plants and the green algae, Chlorophyta.
    D.  All of these apply.
    E.  They include only multicellular organisms that contain plastids.

All of the characteristics apply. Archaeplastida contain plastids, are unicellular and multicellular, and include land plants and green algae.

 

Bloom’s Level: 1. Remember
Section: 21.02
Topic: Protists

 

 

 

  1. Which disease is mismatched with the incorrect causative agent?
    A. fish kills – dinoflagellate
    B.  fish ick – a ciliate
    C.  malaria – an amebozoan
    D.  African sleeping sickness – a trypanosome

The mismatched disease and causative agent is malaria and amebozoan. Malaria is caused by an Apicomplexan.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.03.01 Identify the characteristics of chromalveolates.
Section: 21.03
Topic: Protists

  1. _________ is a mixotrophic protozoan that is able to combine autotrophic and heterotrophic nutritional modes.
    A. Euglena
    B.  Plasmodium
    C.  Amoeba
    D.  Giardia

Euglena is a mixotrophic protozoan that is able to combine autotrophic and heterotrophic nutritional modes.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.04.01 Describe the distinguishing characteristics of excavates.
Section: 21.04
Topic: Protists

  1. Which protistan structure is not correctly matched with its function?
    A. pseudopodia – movement and feeding
    B.  cyst – a reproductive structure
    C.  trichocyst – defense and capture of prey
    D.  pyrenoid – synthesis of starch

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.01.03 Understand that protists represent multiple evolutionary lineages.
Section: 21.01
Topic: Protists

 

 

 

  1. Which is NOT a feature found in protozoans?
    A. heterotrophic
    B.  multicellular
    C.  usually motile
    D.  eukaryotic

Protozoans may be heterotrophic, motile, and eukaryotic. They are not multicellular.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.01.03 Understand that protists represent multiple evolutionary lineages.
Section: 21.01
Topic: Protists

  1. The various forms of algae (Archaeplastids) are NOT considered plants because they
    A. are not multicellular.
    B.  lack strengthened cell walls.
    C.  never have specialized tissues.
    D.  lack plant structures such as true roots, stems, and leaves.
    E.  All of the choices are differences that separate the algae from plants.

The various forms of algae (Archaeplastids) are NOT considered plants because they lack plant structures such as true roots, stems, and leaves.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.02.01 Identify the distinguishing characteristics of archaeplastida.
Section: 21.02
Topic: Protists

  1. Which organelles serve as the energy centers for most Protists?
    A. mitochondria and plastids
    B.  chloroplasts and Golgi Bodies
    C.  mitochondria and Endoplasmic reticulum
    D.  mitochondria and nucleus

The mitochondria and plastids serve as the energy organelles for most Protists.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.01.01 Explain the origin of eukaryotic organelles.
Section: 21.01
Topic: Protists

 

 

 

  1. Which protist is NOT correctly linked to the type of movement it shows?
    A. amoeboids—pseudopodia
    B.  ciliates-—cilia
    C.  zooflagellates—flagella
    D.  sporozoan—flexing the pellicle

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.01.01 Explain the origin of eukaryotic organelles.
Section: 21.01
Topic: Protists

  1. Protist reproduction
    A. is always asexual fission.
    B.  is always sexual with the adult haploid.
    C.  is always sexual with the adult diploid.
    D.  is always sexual with alternation of haploid and diploid generations.
    E.  may be asexual or any of these sexual cycles.

Protist reproduction may be asexual or any of these sexual cycles.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.01.01 Explain the origin of eukaryotic organelles.
Section: 21.01
Topic: Protists

 

 

 

  1. This name of the filamentous algae pictured here, belongs in the supergroup Archaeplastida, is  _______.A.  Ulva
    B.  Plasmodium
    C.  Chlamydomonas
    D.  Spirogyra
    E.  Volvox

This filamentous algae is Spirogyra.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.02.01 Identify the distinguishing characteristics of archaeplastida.
Section: 21.02
Topic: Protists

  1. Which of the following alga is mismatched with its description?
    A. Chlamydomonas—unicellular
    B.  Volvox—filamentous
    C.  Ulva—multicellular
    D.  Chara—multicellular

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.02.02 Describe the life cycles of archaeplastida.
Section: 21.02
Topic: Protists

 

 

 

 

Essay Questions

  1. Identify the Supergroup the organism in the picture below belongs in and describe the process that is occurring.© M.I. Walker/Science Source

The filamentous Charophyte, Spirogyra, belongs in the Supergroup Archaeplastida and it is sexually reproducing in this picture. The process, called conjugation, involves the temporary union of two filaments. The filaments line up parallel to each other and the cellular contents of one filament move into the cells of the other filament. A resistant, diploid zygospore is formed which can survive the winter. In the Spring, the zygospores undergo meiosis to produce new haploid filaments.

 

Bloom’s Level: 3. Apply
Learning Outcome: 21.02.02 Describe the life cycles of archaeplastida.
Section: 21.02
Topic: Protists

 

 

 

 

Multiple Choice Questions

  1. DNA sequencing suggests that among the green algae, the __________ are most closely related to land plants.
    A. Volvox
    B.  Spirogyra
    C.  Chara
    D.  Ulva
    E.  Chlamydomonas

DNA sequencing suggests that among the green algae, the Chara are most closely related to land plants.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.01.03 Understand that protists represent multiple evolutionary lineages.
Section: 21.01
Topic: Protists

  1. The members of the Supergroup Archaeplastida that have cell walls impregnated with calcium carbonate and help build coral reefs are the _________.
    A. diatoms
    B.  dinoflagellates
    C.  multicellular green algae
    D.  brown algae
    E.  red algae

A member of the Supergroup Archaeplastida has cell walls impregnated with calcium carbonate and helps to build coral reefs.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.02.01 Identify the distinguishing characteristics of archaeplastida.
Section: 21.02
Topic: Protists

 

 

 

  1. “Red tides” are produced by massive blooms of
    A. Archaeplastids.
    B.  Chromalveolates.
    C.  Excavates.
    D.  Amoebozoans.
    E.  Rhizarians.

“Red tides” are produced by massive blooms of Chromalveolates.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.03.01 Identify the characteristics of chromalveolates.
Section: 21.03
Topic: Protists

  1. Which feature has to be present in a protist in order to classify it as a photoautotroph?
    A. The presence of mitochondria within the cell.
    B.  The ability to synthesize organic compounds.
    C.  The presence of a light trapping pigment.
    D.  The presence of cellulose in the cell wall.
    E.  The ability to synthesize inorganic compounds.

The presence of a light trapping pigment is necessary to classify a protist as a photoautotroph.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 21.01.02 Assign protists into one of two groups based on mode of nutrition.
Section: 21.01
Topic: Protists

 

 

 

  1. The deposits of chalky fossils that built the White Cliffs of Dover were formed by
    A. Amoebozoans.
    B.  Rhizarians.
    C.  Excavates.
    D.  Opisthokonts.
    E.  Chromalveolates.

The deposits of chalky fossils that built the White Cliffs of Dover were formed by Rhizarians.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.05.03 Describe the unique properties of rhizaria tests.
Section: 21.05
Topic: Protists

 

 

 

  1. A distinctive form of this pictured organism is found in each geologic period. Which statement(s) about this organism is/are true? 8-26-2013© NHPA/SuperStock
    A.  The organisms pictured are foraminiferans, members of Supergroup Rhizarian and have a skeleton called a test.
    B.  These organisms may be used as index fossils to date sedimentary rock.
    C.  The Egyptian pyramids are built of this type of limestone.
    D.  All of the answer choices are correct.
    E.  The organisms pictured are foraminiferans, members of Supergroup Excavates.

All of the choices are true. These are foraminiferans, members of Supergroup Rhizarian with skeletons called tests. They are used as an index fossil to date other fossils and the pyramids are built of foraminferan limestone.

 

Bloom’s Level: 3. Apply
Gradable: automatic
Learning Outcome: 21.05.03 Describe the unique properties of rhizaria tests.
Section: 21.05
Topic: Protists

 

 

 

  1. Supergroup Opisthokinta includes all of the following organisms EXCEPT
    A. animals.
    B.  plants.
    C.  fungi.
    D.  choanoflagellates.

Supergroup Opisthokinta includes all of the following organisms except plants. It includes animals, fungi, and choanoflagellates.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.05.02 Identify the members of supergroup opisthokonta.
Section: 21.05
Topic: Protists

  1. Which definition would best describe the feeding mechanism of Euglena?
    A. carnivorous
    B.  decomposer
    C.  heterotrophic
    D.  photoautotrophic
    E.  mixotrophic

Mixotrophic refers to organisms that are both autotrophic and heterotrophic.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.04.01 Describe the distinguishing characteristics of excavates.
Section: 21.04
Topic: Protists

  1. Which of the following organisms belong to the supergroup Excavates?
    A. Euglenoids, Amoeboids & cellular slime molds
    B.  Diplomonads, euglenoids & parabasalids
    C.  Diplomonads, euglenoids & diatoms
    D.  Foraminiferans, radiolarians & fungi

Diplomonads, euglenoids & parabasalids all belong to the supergroup Excavates.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.04.01 Describe the distinguishing characteristics of excavates.
Section: 21.04
Topic: Protists

 

 

 

 

Essay Questions

  1. Describe the feeding mechanisms of members of the supergroup Archaeplastids.

The Archaeplastids consist of the green algae, red algae, land plants and charophytes.  All of these organisms have the ability to run photosynthesis due to the presence of pigments that can capture solar energy.  This would classify them as photoautotrophs.

 

Bloom’s Level: 6. Create
Learning Outcome: 21.02.01 Identify the distinguishing characteristics of archaeplastida.
Section: 21.02
Topic: Protists

  1. Describe the life cycle of Plasmodium vivax and explain how humans fit into its’ life cycle.

Answers may vary but should include:

1. The gametes will fuse in the digestive system of the female Anopheles mosquito.  The zygote will produce sporozoites which will migrate to the salivary glands.
2. When the mosquito bites the human host the sporozoites can be passed to the bloodstream of the host.
3.  Once in the bloodstream the sporozoites will migrate to the host’s liver and develop into the merozoites.
4. The merozoites will migrate into the bloodstream and infect the red blood cells as trophoblasts.
5.  The merozoites will reproduce asexually in the red blood cells of the host eventually causing the red blood cells to rupture.
6.  If bitten by another mosquito the merozoites can be taken up into the new mosquito starting the cycle all over again.
Bloom’s Level: 6. Create
Learning Outcome: 21.03.03 Describe the life cycle of Plasmodium.
Section: 21.03
Topic: Protists

 

 

 

 

Multiple Choice Questions

  1. A flagellated protist that is sexually transmitted is
    A. Plasmodium.
    B.  Giardia.
    C.  Entamoeba.
    D.  Trichomonas.
    E.  Paramecium.

A flagellated protist that is sexually transmitted is Trichomonas.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.04.02 Identify pathogenic excavate species.
Section: 21.04
Topic: Protists

  1. Which description best supports the endosymbiotic theory of organelles?
    A. Eukaryotic cells acquired mitochondria and plastids by engulfing free-living bacteria and developed a symbiotic relationship with them.  Mitochondria were derived from anaerobic bacterium while choloroplasts were derived from aerobic bacterium.
    B.  Eukaryotic cells acquired mitochondria and plastids by engulfing free-living bacteria and developed a symbiotic relationship with them.  Mitochondria were derived from aerobic bacterium while choloroplasts were derived from cyanobacterium.
    C.  Eukaryotic cells acquired nuclei and flagella by engulfing free-living bacteria and developed a symbiotic relationship with them.  Nuclei were derived from aerobic bacterium while flagella were derived from motile bacteria.
    D.  Eukaryotic cells acquired mitochondria and flagella by engulfing free-living bacteria and developed a symbiotic relationship with them.  Mitochondria were derived from aerobic bacterium while flagella were derived from motile bacteria.

Eukaryotic cells acquired mitochondria and plastids by engulfing free-living bacteria and developed a symbiotic relationship with them.  Mitochondria were derived from aerobic bacterium while choloroplasts were derived from cyanobacterium.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 21.01.01 Explain the origin of eukaryotic organelles.
Section: 21.01
Topic: Endosymbiosis

 

 

 

 

Essay Questions

  1. Describe the evidence that supports that protists have evolved into multiple evolutionary lineages.

Some of the main evidence available today is DNA sequencing that points to multiple protist lineages.  Protist lineages are very long and very old so it makes it difficult to determine the exact evolutionary lineages. The new DNA evidence has led scientists to classify protists and the other eukaryotes into six supergroups.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 21.01.03 Understand that protists represent multiple evolutionary lineages.
Section: 21.01
Topic: Protists

 

Multiple Choice Questions

  1. Which of the following organisms belong to the supergroup Amoebozoans?
    A. Amoeboids and euglenoids
    B.  Plasmodial and water molds
    C.  Plasmodial and cellular slime molds
    D.  Cellular slime molds and radiolarians
    E.  Choanoflagellates and diplomonads

Plasmodial and cellular slime molds belong to the supergroup Amoebozoans.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.05.01 Define the unique features of amoebozoans.
Section: 21.05
Topic: Protists

 

 

 

  1. Slime molds, found in the Supergroup Amoebozoa, were once classified as fungi. What characteristics of the slime molds distinguish them from fungi?
    A. Slime molds lack cell walls.
    B.  They are flagellated cells at certain stages of the life cycle.
    C.  The vegetative state of the slime mold is mobile and amoeboid.
    D.  All of these characteristics distinguish slime molds from fungi.

Slime molds lack cell walls, are flagellated cells at certain stages of the life cycle and the vegetative state of the slime mold is mobile and amoeboid. All of these characteristics distinguish slime molds from fungi.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.05.01 Define the unique features of amoebozoans.
Section: 21.05
Topic: Protists

  1. The endosymbiont hypothesis proposes all of the following EXCEPT
    A. mitochondria and chloroplasts were once free-living bacteria.
    B.  a nucleated cell engulfed an aerobic bacterium which evolved into mitochondria.
    C.  prokaryotes evolved when an aerobic bacterium engulfed a cyanobacterium.
    D.  chloroplasts originated when a nucleated cell engulfed a cyanobacterium.

The endosymbiont hypothesis proposes all of the following EXCEPT prokaryotes evolved when an aerobic bacterium engulfed a cyanobacterium.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.01.01 Explain the origin of eukaryotic organelles.
Section: 21.01
Topic: Endosymbiosis

 

 

 

  1. If a public health official was trying to reduce the potential for human exposure to a “red tide” which of the following actions would they take?
    A. Place a ban on eating fish that were exposed to the dinoflagellate Alexandrium catanella
    B.  Decrease the amount of nutrients that are washed into the aquatic ecosystem
    C.  Avoid eating shellfish that live in coastal areas that have a high influx of nutrients
    D.  All of these actions would help reduce the potential for human exposure to a “red tide”.

All of these actions would help reduce the potential for human exposure to a “red tide”.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 21.03.01 Identify the characteristics of chromalveolates.
Section: 21.03
Topic: Protists

 

Essay Questions

  1. Explain the features necessary for both Brown algae and Dinoflagellates to be classified in the supergroup Chromalveolata.

Brown algae will have chlorophyll a and c as well as an accessory pigment that gives them their brown color.  They will also store their food in the form of laminarin.  Sizes can range from small forms to large multicellular ones.
Dinoflagellates are unicellular, photoautotrophic algae that are encased by a shell of cellulose and silicate plates.  They also contain chlorophyll a and c and an accessory pigment that gives them their brown color.

 

Bloom’s Level: 6. Create
Learning Outcome: 21.03.01 Identify the characteristics of chromalveolates.
Section: 21.03
Topic: Protists

 

 

 

 

Multiple Choice Questions

  1. Which of the following protozoans is mismatched with the disease that it causes?
    A. Trypanosoma brucei—sleeping sickness
    B.  Trichomonas vaginalis—vaginitis and urethritis
    C.  Entamoeba histolytica—amoebic dysentery
    D.  Giardia lamblia—malaria

Trypansoma causes sleeping sickness, Trichomonas causes vaginitis and Entamoeba causes amoebic dysentery. Giardia causes intestinal disturbances, not malaria.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.04.02 Identify pathogenic excavate species.
Section: 21.04
Topic: Protists

  1. Imagine you are working for the Florida Public Health Department and you have been assigned the task of preventing malaria from spreading throughout the southern Everglades.  Which of the following scenarios would have the greatest chance of success?
    A. Spraying the edges of the Everglades with a pesticide that will kill mosquitos
    B.  Developing a vaccination that would prevent the ability of the merozoites to reproduce within the hosts bloodstream
    C.  Devise a insect repellant that would prevent the Anopheles mosquito from biting a human host
    D.  All of these scenarios would have a chance of success in preventing malaria from spreading throughout the southern Everglades.

All of these scenarios would have a chance of success in preventing malaria from spreading throughout the southern Everglades.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 21.03.03 Describe the life cycle of Plasmodium.
Section: 21.03
Topic: Protists

 

 

 

  1. A common cause of amoebic dysentery is
    A. Trichomonas vaginalis.
    B.  Entamoeba histolytica.
    C.  Plasmodium falciparum.
    D.  Giardia intestinalis.

A common cause of amebic dysentery is Entamoeba histolytica.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.05.01 Define the unique features of amoebozoans.
Section: 21.05
Topic: Protists

  1. Saprolegnia, a ________, is feeding on the dead goldfish pictured here.  This classifies it as a heterotrophic protist.© Noble Proctor/Science Source
    A.  slime mold
    B.  ciliate
    C.  water mold
    D.  green algae

Saprolegnia, a water mold, is feeding on the dead goldfish pictured here.  This classifies it as a heterotrophic protist.

 

Bloom’s Level: 3. Apply
Learning Outcome: 21.03.01 Identify the characteristics of chromalveolates.
Section: 21.03
Topic: Protists

 

  1. Amoebic dysentery is transmitted by (through)
    A. the bite of a mosquito.
    B.  the bite of a tsetse fly.
    C.  having sex with an infected partner.
    D.  ingesting contaminated water or food.

Amebic dysentery is transmitted by (through) ingesting contaminated water or food.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.05.01 Define the unique features of amoebozoans.
Section: 21.05
Topic: Protists

 

Essay Questions

  1. How are the Archaeplastids similar to and different from the plants?

Biologists have long thought that the Archaeplastids and plants are closely related. Plants are multicellular, eukaryotic, photosynthetic autotrophs. The Archaeplastids include single celled, filamentous, colonial and multicellular forms. Both plants and Archaeplastids have chlorophylls a and b and cells walls that contain cellulose. They both store their carbohydrate as starch. The charophytes of the Archaeplastids are thought to be more closely related to plants than the chlorophytes.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 21.02.01 Identify the distinguishing characteristics of archaeplastida.
Section: 21.02
Topic: Protists

 

 

 

  1. The human body generally forms antibodies (“immune chemicals”) to foreign substances detected in the blood plasma or in interstitial spaces within a short time. Explain why the malaria organism, unlike “short-lived” influenza and other infections, can survive in the human host to cause recurrent chills and fevers for years.

The malarial parasite has two hosts, both the mosquito and humans. In humans, the parasite spends most of its time inside liver and red blood cells, hiding from the immune system. Some species may remain dormant in liver cells for extended periods of time.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 21.03.03 Describe the life cycle of Plasmodium.
Section: 21.03
Topic: Protists

  1. Identify the organism in the picture and the Supergroup it belongs to. What characteristic(s) helped you to identify the organism?© Biophoto Associates/Science Source

The organism is a dinoflagellate that belongs to the Supergroup Chromalveolates. The cellulose plates (armor) and flagella help to identify the organism. Dinoflagellates are single celled organisms with protective cellulose plates. As in the picture, dinoflagellates normally have two flagella. One flagellum is in a transverse groove that encircles the organism and the other flagellum is in a longitudinal groove with its distal end free.

 

Bloom’s Level: 3. Apply
Learning Outcome: 21.03.02 Identify unique species of chromalveolates.
Section: 21.03
Topic: Protists

 

 

 

 

Multiple Choice Questions

  1. In Plasmodium, sexual reproduction occurs in the _________, while asexual reproduction occurs in ________.
    A. mosquito; humans
    B.  humans; mosquitoes
    C.  tsetse fly; humans
    D.  deer tick; deer

In Plasmodium, sexual reproduction occurs in the mosquito, while asexual reproduction occurs in humans.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.03.03 Describe the life cycle of Plasmodium.
Section: 21.03
Topic: Protists

 

 

 

 

True / False Questions

The evolutionary relationships between the eukaryotic supergroups are depicted in the diagram.

 

 

  1. Diatoms share a more recent common ancestor with dinoflagellates than with water molds.
    FALSE

Diatoms share a more recent common ancestor with water molds than with dinoflagellates.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 21.01.03 Understand that protists represent multiple evolutionary lineages.
Section: 21.01
Topic: Protists

 

Multiple Choice Questions

  1. Which of the following pairs of Protists would be most closely related due to both of them possessing plastids?
    A. Archaeplastids & Rhizaria
    B.  Archaeplastids & Chromalveolata
    C.  Rhizaria & Chromalveolata
    D.  Amoebozoa & Excavata

Archaeplastids & Chromalveolata both have plastids.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.02.01 Identify the distinguishing characteristics of archaeplastida.
Section: 21.02
Topic: Protists

 

 

 

 

True / False Questions

  1. The term “protists” is used to refer to eukaryotes that are not plants, animals or fungi.
    TRUE

It is true that the term “protists” is used to refer to eukaryotes that are not plants, animals, or fungi.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.01.03 Understand that protists represent multiple evolutionary lineages.
Section: 21.01
Topic: Protists

  1. The reason biologists believe that green algae and land plants, both members of the supergroup Archaeplastida, are related is because both of them possess chlorophyll a & b as well as cell walls made of starch.
    TRUE

It is true that both green algae and land plants possess chlorophyll a & b as well as cell walls made of starch.

 

Bloom’s Level: 3. Apply
Learning Outcome: 21.02.02 Describe the life cycles of archaeplastida.
Section: 21.02
Topic: Protists

  1. It is believed that eukaryotes evolved their mitochondria from a symbiotic relationship with a free-living aerobic bacteria.
    TRUE

It is true that scientists believe that eukaryotes evolved their mitochondria from a symbiotic relationship with a free-living aerobic bacteria.

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.01.01 Explain the origin of eukaryotic organelles.
Section: 21.01
Topic: Eukarya

 

 

 

  1. Chromalveolates that live along the rocky shores of the north temperate zones are able to withstand pounding tides and drying because of their holdfasts and mucilaginous cell walls.
    TRUE

It is true that Chromalveolates that live along the rocky shores of the north temperate zones are able to withstand pounding tides and drying because of their holdfasts and mucilaginous cell walls

 

Bloom’s Level: 2. Understand
Learning Outcome: 21.03.01 Identify the characteristics of chromalveolates.
Section: 21.03
Topic: Protists

  1. Most Chromalveolates have the alternation of generations life cycle, but some species of Fungus have a diplontic life cycle.
    TRUE

It is true that most Chromalveolates have the alternation of generations life cycle, but some species of Fungus have a diplontic life cycle.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.03.01 Identify the characteristics of chromalveolates.
Section: 21.03
Topic: Protists

  1. Water molds are heterotrophic because they are parasitic on fish, plants, and insects.
    TRUE

It is true that water molds are heterotrophic because they are parasitic on fish, plants, and insects.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.01.02 Assign protists into one of two groups based on mode of nutrition.
Section: 21.01
Topic: Protists

 

 

 

  1. The 1840s Irish potato famine was caused by a water mold parasite on potatoes.
    TRUE

It is true that the 1840s Irish potato famine was caused by a water mold parasite on potatoes.

 

Bloom’s Level: 1. Remember
Learning Outcome: 21.03.02 Identify unique species of chromalveolates.
Section: 21.03
Topic: Protists

Chapter 31

Animal Organization and Homeostasis

 

 

Multiple Choice Questions

  1. What type of fiber accounts for the strength of various connective tissues?
    A. elastic
    B.  collagen
    C.  actin
    D.  reticular

Collagen fibers account for the strength of various connective tissues. Collagen fibers contain collagen, a protein that gives them strength and flexibility.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

  1. What type of tissue is striated, contains cells with one nucleus, and is not under voluntary control?
    A. skeletal muscle
    B.  smooth muscle
    C.  cardiac muscle
    D.  nervous tissue

Cardiac muscle is striated, contains cells with one nucleus, and is involuntary.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. Which of the following is the most complex level of organization?
    A. cardiovascular system
    B.  brain
    C.  columnar epithelium
    D.  heart
    E.  lungs

The cardiovascular system is the most complex level of organization listed here.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.02.01 Distinguish among tissues, organs, and organ systems.
Section: 31.02
Topic: Levels of Biological Organization

  1. The correct sequence of increasing organizational complexity is
    A. organ, tissue, cell, organ system, organism.
    B.  cell, organ, organ system, tissue, organism.
    C.  cell, tissue, organ, organ system, organism.
    D.  organism, tissue, cell, organ system, organ.
    E.  tissue, cell, organ system, organism, organ.

The correct sequence of increasing organizational complexity is cell, tissue, organ, organ system, organism.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.02.01 Distinguish among tissues, organs, and organ systems.
Section: 31.02
Topic: Levels of Biological Organization

  1. What type of tissue lines body cavities and covers body surfaces?
    A. muscle tissue
    B.  nervous tissue
    C.  epithelial tissue
    D.  connective tissue

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.02 Identify the common locations of the various types of animal tissues.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. What type of tissue contracts and accounts for movements of organs or the entire body?
    A. muscle tissue
    B.  nervous tissue
    C.  epithelial tissue
    D.  connective tissue

Muscle tissue is responsible for contractions that account for movements of organs or the entire body.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

  1. _______ is responsible for receiving, interpreting, and producing a response to stimuli.
    A. Muscle tissue
    B.  Nervous tissue
    C.  Epithelial tissue
    D.  Connective tissue

Nervous tissue is responsible for receiving, interpreting, and producing a response to stimuli.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

  1. What type of tissue is bone and cartilage?
    A. muscle tissue
    B.  nervous tissue
    C.  epithelial tissue
    D.  connective tissue

Connective tissue includes bone and cartilage.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. The epidermis consists of
    A. stratified squamous epithelial cells.
    B.  collagen fibers.
    C.  elastic fibers.
    D.  adipose tissue.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.03.02 Identify the two main regions of skin and how these differ from the subcutaneous layer.
Section: 31.03
Topic: Integumetary System

  1. What tissue type is characterized by being a transport medium for nutrients and also includes blood and adipose tissues?
    A. muscle tissue
    B.  nervous tissue
    C.  epithelial tissue
    D.  connective tissue

Connective tissue includes blood and adipose tissue.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

  1. Which organ system does not play a role in acquiring materials and energy?
    A. cardiovascular
    B.  skeletal
    C.  muscular
    D.  digestive

The skeletal system does not play a role in acquiring materials or energy.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.02.02 List the major life processes carried out by each organ system in vertebrate animals.
Section: 31.02
Topic: Levels of Biological Organization

 

 

 

  1. Which of the following statements about epithelial tissue is NOT true?
    A. Squamous epithelium is composed of flattened cells.
    B.  Columnar epithelium is cube-shaped with the nucleus near the upper surface of the cells.
    C.  Simple epithelia have only a single layer of cells.
    D.  Pseudostratified epithelium looks like it has multiple layers, but all the cells are attached to the basement membrane.
    E.  Epithelium lining the respiratory tract contains cilia that move particles along its surface.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

  1. A thin layer of protein, within epithelial tissue, that anchors the epithelium to the extracellular matrix is the
    A. basement membrane.
    B.  plasma membrane.
    C.  compact bone.
    D.  hyaline cartilage.
    E.  neuroglia.

A thin layer of protein that anchors the epithelium to the extracellular matrix is the basement membrane.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

  1. An average person has about _____ neurons.
    A. 100,000
    B.  1 trillion
    C.  2 million
    D.  1 billion

An average person has about 1 trillion neurons.

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. All of the following are functions of connective tissue EXCEPT to
    A. line body cavities and cover surfaces.
    B.  bind and support body parts.
    C.  store energy as fat.
    D.  fill spaces.
    E.  produce blood cells.

Connective tissue binds and supports body parts, stores energy as fat, fills spaces, and produces blood cells. However, epithelial tissue lines body cavities and covers surfaces.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

  1. Which statement about connective tissue is NOT true?
    A. Loose connective tissue contains fibroblasts, different types of fiber, and a nonliving matrix.
    B.  Fibrous connective tissue includes bone and cartilage.
    C.  Blood is a connective tissue that contains a fluid matrix.
    D.  Adipose tissue provides insulation and padding in organs such as the skin.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. Which of the following features would be found in the epidermis layer of the skin?
    A. stratified squamous epithelium, cells produce keratin, contain melanocytes, offers a layer of protection against bacteria
    B.  stratified cuboidal epithelium, cells produce keratin, contain melanocytes, offers a layer of protection against bacteria
    C.  stratified squamous epithelium, cells produce keratin, contains a large amount o f adipose tissue, offers a layer of protection against bacteria
    D.  composed of collagen fibers, cells produce keratin, contain melanocytes, offers a layer of protection against bacteria
    E.  stratified squamous epithelium, contains specialized nerve endings, offers a layer of protection against injury

Stratified squamous epithelium, cells produce keratin, contain melanocytes, offers a layer of protection against bacteria are features of the epidermis.

Bloom’s Level: 4. Analyze
Learning Outcome: 31.03.02 Identify the two main regions of skin and how these differ from the subcutaneous layer.
Section: 31.03
Topic: Integumetary System

  1. Which is NOT a structure seen in compact bone?
    A. cylindrical structural units called osteons (Haversian systems)
    B.  central canals containing blood vessels and nerves
    C.  osteocytes located within lacunae
    D.  bony bars and plates with irregular spaces between them
    E.  thin extensions of bone cells within canaliculi

Compact bone contains cylindrical structural units called osteons (Haversian systems), central canals containing blood vessels and nerves, osteocytes located within lacunae, and thin extensions of bone cells within canaliculi. However, compact bone is not made of bony bars and plates with irregular spaces between them.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. Which statement is NOT true about cartilage?
    A. Cartilage cells are located in small spaces called lacunae.
    B.  The matrix of cartilage includes calcium salts.
    C.  The matrix of cartilage is both solid and flexible.
    D.  The human fetal skeleton is made up of cartilage.
    E.  Adult bodies possess structures that are made up of cartilage such as the nose, ear, and pads between the vertebrae in the backbone.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

  1. White blood cells fight infection by
    A. transporting large amounts of oxygen to the infected area.
    B.  engulfing infectious pathogens and making antibodies.
    C.  increasing platelet production and forming a plug to seal off the area.
    D.  producing antibiotics that will kill bacteria and fungi.

 

Bloom’s Level: 3. Apply
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

  1. Skeletal muscle
    A. is made of extremely small cells with one nucleus.
    B.  is not under voluntary control.
    C.  is found in the walls of the intestines and stomach.
    D.  has bands of actin and myosin filaments perpendicular to the length of the cell that produce contractions.

Skeletal muscle has bands of actin and myosin filaments perpendicular to the length of the cell that produce contractions.

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. What are the main features of melanocytes?
    A. Melanocytes are found in the dermis and they produce melanin when exposed to UV light.  The melanin is then distributed to epidermal cells.
    B.  Melanocytes are found in the epidermis and they produce melanin when exposed to UV light.  The melanin is then distributed to epidermal cells.
    C.  Melanocytes are found in the epidermis and they produce melanin when exposed to UV light.  The melanin is then distributed to dermal cells.
    D.  Melanocytes are found in the hypodermis and they produce melanin when exposed to UV light.  The melanin is then distributed to epidermal cells.

The main features of melanocytes are that they are found in the epidermis and they produce melanin when exposed to UV light.  The melanin is then distributed to epidermal cells.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 31.03.03 Explain the function of melanocytes in the skin and the effects of UV radiation.
Section: 31.03
Topic: Integumetary System

  1. What is most likely to occur when an individual is exposed to UV radiation?
    A. The melanocytes will produce melanin, the melanin is distributed into epidermal cells, a steroid is converted into vitamin D and stem cells can mutate leading to cancer.
    B.  The melanin will produce melanocytes, the melanocytes are then distributed into epidermal cells, a steroid is converted into vitamin D and stem cells can mutate leading to cancer.
    C.  The melanocytes will produce melanin, the melanin is distributed into dermal cells, a steroid is converted into vitamin D and stem cells can mutate leading to cancer.
    D.  The melanocytes will produce melanin, the melanin is distributed into epidermal cells, vitamin D is converted into a steroid and stem cells can mutate leading to cancer.

The melanocytes will produce melanin, the melanin is distributed into epidermal cells, a steroid is converted into vitamin D and stem cells can mutate leading to cancer.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 31.03.03 Explain the function of melanocytes in the skin and the effects of UV radiation.
Section: 31.03
Topic: Integumetary System

 

 

 

  1. Which statement is NOT true about nervous tissue?
    A. Cells in the nervous system that conduct electrical impulses are called neurons.
    B.  Neuroglia support and protect neurons.
    C.  The dendrites carry electrical impulses away from the cell body of the neuron.
    D.  Nerve fibers are long axons covered by a myelin sheath.
    E.  Neuroglia provide neurons with nutrients and keep them free of cellular debris.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.01.03 Explain how specialization of cells in tissues enhances tissue function.
Section: 31.01
Topic: Animal Tissues

  1. Which of the following is a function of the skin?
    A. protecting underlying body parts from physical trauma
    B.  preventing water loss
    C.  thermoregulation
    D.  monitor touch, pressure, temperature, and pain
    E.  All of the answer choices are functions of the skin

Functions of the skin include: protecting underlying body parts from physical trauma and water loss, thermoregulation, and monitoring touch, pressure, temperature, and pain.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.03.01 Distinguish between the functions of skin that are common to all animals and those that are unique to specific groups.
Section: 31.03
Topic: Integumetary System

 

 

 

  1. ______ make up more than half the volume of the brain.
    A. Neurons
    B.  Neuroglia
    C.  Nerves
    D.  Intercalated disks
    E.  Hyaline cartilage

Neuroglia make up more than half the volume of the brain.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.01.02 Identify the common locations of the various types of animal tissues.
Section: 31.01
Topic: Animal Tissues

  1. What is the body’s response to a rise in temperature above 37.0 degrees C?
    A. Blood vessels in the skin begin to dilate which increases blood flow to the surface of the skin.  Environmental variables will help decrease the temperature of the blood.  The nervous system also activates the sweat glands which help decrease temperature.
    B.  Blood vessels in the skin begin to constrict which increases blood flow to the surface of the skin.  Environmental variables will help decrease the temperature of the blood.  The nervous system also activates the sweat glands which help decrease temperature.
    C.  Blood vessels in the skin begin to dilate which increases blood flow to the surface of the skin.  Environmental variables will help decrease the temperature of the blood.  The heart rate decreases in an attempt to slow down blood flow.
    D.  Blood vessels in the skin begin to dilate which decreases blood flow to the surface of the skin.  Environmental variables will help decrease the temperature of the blood.  The heart rate increases in an attempt to slow down blood flow.
    E.  Blood vessels in the skin begin to dilate which increases blood flow to the surface of the skin.  Environmental variables will help decrease the temperature of the blood.  The nervous system will deactivate the sweat glands to help decrease body temperature.

Blood vessels in the skin begin to dilate which increases blood flow to the surface of the skin.  Environmental variables will help decrease the temperature of the blood.  The nervous system also activates the sweat glands which help decrease temperature.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 31.04.02 Evaluate the evolutionary benefits of regulating an internal variable, such as body temperature, versus the cost.
Section: 31.04
Topic: Homeostasis

 

 

 

  1. Which type of tissues would be damaged if someone was in a car accident and they had their kidney ruptured due to the force of the impact?
    A. squamous epithelium, cuboidal epithelium, smooth muscle and neural tissues
    B.  pseudostratified ciliated epithelium, cuboidal epithelium, smooth muscle and neural tissues
    C.  squamous epithelium, cuboidal epithelium, skeletal muscle and neural tissues
    D.  squamous epithelium, cuboidal epithelium, fibrocartilage and neural tissues

Squamous epithelium, cuboidal epithelium, smooth muscle and neural tissues would be damaged.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 31.01.02 Identify the common locations of the various types of animal tissues.
Section: 31.01
Topic: Animal Tissues

  1. Which of the following functions performed by the skin would be lost or seriously compromised if someone suffered a severe burn over most of the body?
    A. prevention of heat and water loss
    B.  prevention of massive and continuous infection from invasion by external bacteria, viruses, and parasites
    C.  production of sweat that evaporates and cools the body when it is overheating
    D.  conversion, in the presence of ultraviolet light, of precursor molecules to vitamin D
    E.  All of these are vital functions of the skin and would be compromised.

If someone suffered a severe burn over most of the body, all the vital functions of the skin would be compromised.

 

Bloom’s Level: 3. Apply
Learning Outcome: 31.02.02 List the major life processes carried out by each organ system in vertebrate animals.
Section: 31.02
Topic: Integumetary System

 

 

 

  1. If you wash your skin and hair several times a day, you will soon have dry skin that easily cracks and bleeds because the oils you have removed are necessary to keep the skin supple. This oil is
    A. pili secreted by arrectors.
    B.  lymph secreted by lymphatic ducts.
    C.  sebum secreted by sweat glands.
    D.  sebum secreted by sebaceous glands.
    E.  salty sweat secreted by sweat glands.

Sebaceous glands secrete an oil called sebum which lubricates the hair and skin.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.03.04 Describe the makeup and function of the accessory structures of human skin.
Section: 31.03
Topic: Integumetary System

  1. Which organs are located in the ventral cavity of the body?
    A. stomach, liver, heart and bladder
    B.  stomach, liver, heart and spinal cord
    C.  small intestine, brain, lungs and kidneys
    D.  All of the choices are correct.
    E.  None of the choices are correct.

The ventral cavity contains the stomach, liver, heart and bladder; E. The ventral cavity contains the stomach, liver, heart and bladder.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 31.02.03 Describe the two main cavities of the human body and the major organs found in each.
Section: 31.02
Topic: Levels of Biological Organization

 

 

 

  1. At one time it was thought impossible to operate within the body cavities. But then the Canadian doctor William Beaumont operated on a gunshot wound to the stomach. A few decades later we saw breakthroughs in heart operations and transplants. And the 1990s were called the decade of the brain because of the extensive research being done on how the brain functions. The order in which these discoveries were made, by cavity, are
    A. abdominal-thoracic-dorsal.
    B.  abdominal-dorsal-thoracic.
    C.  dorsal-thoracic-abdominal.
    D.  thoracic-abdominal-dorsal.
    E.  thoracic-dorsal-abdominal.

The stomach is located in the abdominal cavity, the heart in the thoracic cavity, and the brain in the dorsal cavity.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.02.03 Describe the two main cavities of the human body and the major organs found in each.
Section: 31.02
Topic: Levels of Biological Organization

  1. Usually it is not possible to penetrate inside a body cavity to reach interstitial spaces without cutting through some protective epidermal tissues. However, sperm reach the egg via the vagina, uterus, and fallopian tubes. Without crossing any epidermal barrier, the sperm finds the egg in what body cavity?
    A. abdominal cavity
    B.  pleural cavity
    C.  pelvic cavity
    D.  pericardial cavity
    E.  dorsal body cavity (cerebral-spinal)

Without crossing any epidermal barrier, the sperm finds the egg in the pelvic cavity.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.02.03 Describe the two main cavities of the human body and the major organs found in each.
Section: 31.02
Topic: Levels of Biological Organization

 

 

 

  1. The level of carbon dioxide in the blood is monitored by nerve tissue. If you exercise strenuously, the increase in cellular respiration consumes oxygen from the blood and produces surplus carbon dioxide removed in the blood. Since this physical exertion requires faster breathing to replenish oxygen supplies and carry away carbon dioxide, the carbon dioxide sensors result in signals from the brain to increase breathing rate. After you have rested, and carbon dioxide levels are back to normal, your breathing rate decreases too. This is an example of
    A. a response to a sensory impulse that causes a muscle or gland to react.
    B.  a negative feedback loop that stimulates a response similar to the stimulus.
    C.  a positive feedback loop that stimulates a response opposite to the stimulus.
    D.  the dynamic interplay between events that tend to change the internal environment and events that tend to keep it the same.
    E.  the dynamic interplay between events that tend to change the internal environment and events that tend to increase the changes.

This is an example of the dynamic interplay between events that tend to change the internal environment and events that tend to keep it the same.

 

Bloom’s Level: 3. Apply
Learning Outcome: 31.04.01 Define homeostasis and explain why it is an essential feature of all living organisms.
Section: 31.04
Topic: Homeostasis

  1. Homeostasis is the
    A. negative feedback loop that stimulates a response similar to the stimulus.
    B.  dynamic interplay between events that tend to change the external environment and events that tend to keep it the same.
    C.  the acquisition, transformation and use of energy.
    D.  maintenance of normal internal conditions by means of self-regulating mechanisms.

Homeostasis is the maintenance of normal internal conditions by means of self-regulating mechanisms.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.04.01 Define homeostasis and explain why it is an essential feature of all living organisms.
Section: 31.04
Topic: Homeostasis

 

 

 

  1. Which of the following statements concerning homeostasis in body temperature is NOT true? The body reacts to
    A. heat by stimulating the sweat glands.
    B.  cold by contracting skeletal muscles in shivering.
    C.  cold by rerouting blood away from the skin.
    D.  heat by constricting the blood vessels in the skin.
    E.  All of the statements are not true.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.04.02 Evaluate the evolutionary benefits of regulating an internal variable, such as body temperature, versus the cost.
Section: 31.04
Topic: Homeostasis

  1. Exocrine glands secrete products onto the surface of the skin and are classified as _______ tissue.
    A. epithelial
    B.  connective
    C.  muscle
    D.  nervous

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.03.04 Describe the makeup and function of the accessory structures of human skin.
Section: 31.03
Topic: Integumetary System

  1. Which of the following skin derivatives is mismatched with the organism on which it is found?
    A. mammals – hair and nails
    B.  amphibians – epidermal scales
    C.  fishes – bony scales
    D.  birds – feathers

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.03.01 Distinguish between the functions of skin that are common to all animals and those that are unique to specific groups.
Section: 31.03
Topic: Integumetary System

 

 

 

  1. Which of the following systems / organs are incorrectly matched with its role in maintaining homeostasis?
    A. Digestive system – provides nutrient molecules to the cells.
    B.  Respiratory system – transports oxygen and carbon dioxide to the cells throughout the body.
    C.  Liver – removes excess glycogen from the blood and stores it as glucose.
    D.  Pancreas secretes insulin when the body has an elevated blood glucose level.
    E.  All of the systems / organs listed are correctly matched with their role in maintaining homeostasis.

The liver does not remove excess glycogen from the blood and stores it as glucose.  It removes excess glucose and stores it as glycogen.

 

Bloom’s Level: 4. Analyze
Learning Outcome: 31.04.01 Define homeostasis and explain why it is an essential feature of all living organisms.
Section: 31.04
Topic: Homeostasis

  1. Which is most closely associated with negative feedback?
    A. As you continue to turn a faucet handle, more water flows out of it.
    B.  A child cuts his finger and the blood clotting process begins.
    C.  As the temperature falls in a house, the heater comes on.
    D.  As a woman goes into labor, hormones are secreted that stimulate the process.

An example of negative feedback is a heater coming on when the temperature in a house falls. All other scenarios are examples of positive feedback.

 

Bloom’s Level: 3. Apply
Learning Outcome: 31.04.03 Differentiate between positive and negative feedback mechanisms and list one specific example of each in animals.
Section: 31.04
Topic: Homeostasis

 

 

 

  1. The __________ body cavity will contain the heart, lungs and spleen.
    A. ventral
    B.  cranial
    C.  dorsal
    D.  abdominal
    E.  abdominopelvic

The ventral body cavity will contain the heart, lungs and spleen.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.02.03 Describe the two main cavities of the human body and the major organs found in each.
Section: 31.02
Topic: Levels of Biological Organization

  1. Which of the following structures begins in the dermis and extends to the surface of the epidermis?
    A. oil and sweat glands
    B.  hair follicles and nerve fibers
    C.  sweat glands and hair follicles
    D.  oil glands and hair follicles

Sweat glands and hair follicles both begin in the dermis and extend to the surface of the epidermis.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.03.04 Describe the makeup and function of the accessory structures of human skin.
Section: 31.03
Topic: Integumetary System

 

 

 

 

Matching Questions

  1. Match the terms in the word bank with the labeled structures on the diagram. Briefly define the function of each structure: axon, nucleus, cell body, dendrite, myelin sheath.
1.  A – dendrite       the part of a neuron that sends signals toward the cell body   1
2.  D – myelin sheath       the part of a neuron that contains the major concentration of the cytoplasm and the nucleus.   4
3.  C – nucleus       a membrane bound organelle within a neuron that contains chromosomes and controls the structure and function of the cell.   3
4.  B – cell body       white, fatty material that forms a covering over nerve fibers.   2
5.  E – axon       elongated portion of a neuron that conducts nerve impulses away from the cell body.   5

 

Bloom’s Level: 3. Apply
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

 

 

 

 

Short Answer Questions

  1. The production of red blood cells is regulated by a feedback loop involving erythropoietin. In non-disease states, the production of red blood cells is equal to the destruction of red blood cells, thus maintaining sufficient red blood cells to deliver adequate oxygen to the body’s tissues. However, when the number of red blood cells decrease and oxygen levels are low, erythropoietin is produced by the liver and kidneys to increase the production of red blood cells. Is this a positive or negative feedback loop? Explain your answer.

This is an example of a negative feedback loop, which keeps a variable close to a particular value or set point. A negative feedback system prevents change in the same direction. As the red blood cell count decreases, oxygen levels decrease. The low oxygen levels cause an increase in the production of erythropoietin, which in turn, increases red blood cell production. As red blood cells increase in number and oxygen levels become normal, erythropoietin production decreases.

 

Bloom’s Level: 5. Evaluate
Learning Outcome: 31.04.03 Differentiate between positive and negative feedback mechanisms and list one specific example of each in animals.
Section: 31.04
Topic: Feedback Mechanisms

 

Multiple Choice Questions

  1. In lactation, the more a baby suckles his mother’s breast to obtain milk, the more milk is produced. This is a
    A. positive feedback loop.
    B.  a negative feedback loop.

This is an example of a positive feedback loop, a mechanism that brings about an ever greater change in the same direction.

 

Bloom’s Level: 3. Apply
Learning Outcome: 31.04.03 Differentiate between positive and negative feedback mechanisms and list one specific example of each in animals.
Section: 31.04
Topic: Feedback Mechanisms

 

 

 

 

True / False Questions

  1. Stratified epithelium is composed of multiple layers of flattened cells that can often be found in the lining of blood vessels.
    TRUE

It is true that stratified epithelium is composed of multiple layers of flattened cells that can often be found in the lining of blood vessels.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.01.02 Identify the common locations of the various types of animal tissues.
Section: 31.01
Topic: Animal Tissues

 

 

 

 

Short Answer Questions

  1. Name the labeled cavities in the diagram at right and indicate an organ found in each.

A – Dorsal Cavity: contains the cranial and vertebral cavity
B – Ventral Cavity: contains the thoracic, abdominal and pelvic cavity
C – Cranial Cavity: contains the brain
D – Vertebral Cavity: contains the spinal cord
E – Thoracic Cavity: contains esophagus, heart and lungs
F – Abdominal Cavity: contains digestive organs, the liver, pancreas and other organs
G – Pelvic Cavity: contains reproductive organs, urinary bladder, and other organs
Bloom’s Level: 3. Apply
Learning Outcome: 31.02.03 Describe the two main cavities of the human body and the major organs found in each.
Section: 31.02
Topic: Levels of Biological Organization

 

 

 

 

True / False Questions

  1. Homeostasis is ultimately controlled by the nervous system.
    TRUE

It is true that homeostasis is ultimately controlled by the nervous system.

 

Bloom’s Level: 1. Remember
Learning Outcome: 31.04.01 Define homeostasis and explain why it is an essential feature of all living organisms.
Section: 31.04
Topic: Homeostasis

  1. Both the neural and endocrine systems play a role in coordinating body activities.
    TRUE

It is true that both the neural and endocrine systems play a role in coordinating body activities.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.02.02 List the major life processes carried out by each organ system in vertebrate animals.
Section: 31.02
Topic: Levels of Biological Organization

  1. Snakes, frogs, squirrels and humans are all classified as being homeothermic.
    FALSE

It is false that snakes, frogs, squirrels and humans are all classified as being homeothermic.

 

Bloom’s Level: 2. Understand
Learning Outcome: 31.04.01 Define homeostasis and explain why it is an essential feature of all living organisms.
Section: 31.04
Topic: Homeostasis

 

 

 

 

Essay Questions

  1. List the major life process that each of the following organ systems carries out: urinary, lymphatic, nervous, and cardiovascular.

Urinary system is responsible for eliminating wastes. Lymphatic system is responsible for protecting the body from pathogens. Nervous system is responsible for coordinating body activities. Cardiovascular system is responsible for transporting materials within the bloodstream.

 

Bloom’s Level: 6. Create
Learning Outcome: 31.02.02 List the major life processes carried out by each organ system in vertebrate animals.
Section: 31.02
Topic: Levels of Biological Organization

  1. List the four primary tissue types found in the human body and indicate their main function(s).
  2. Epithelial tissue: covers body surface, lines body cavities and forms glands.
    2. Connective tissue: binds and supports body parts.
    3. Muscular tissue: moves the body and its parts.
    4. Nervous tissue: receives stimuli and transmits impulses.

 

Bloom’s Level: 6. Create
Learning Outcome: 31.01.01 List and describe the four major types of tissues found in animals.
Section: 31.01
Topic: Animal Tissues

 

 

 

  1. Describe the problems that are associated with the loss of the epidermal layer of skin due to a burn.

The epidermis is the 1st line of defense against invading pathogens.  It will also help regulate body temperature as well as protect against UV radiation.  If this layer of skin is damaged due to a burn many of these functions will be lost or compromised.  An individual will need to be very cautious about infections as well as maintaining their body temperature.  Fortunately this layer can regenerate over time as long as the underlying dermis is not damaged.

 

Bloom’s Level: 6. Create
Learning Outcome: 31.03.02 Identify the two main regions of skin and how these differ from the subcutaneous layer.
Section: 31.03
Topic: Integumetary System