Unit 3: Homeostasis Practice Test			Name:__________________________________

____	1)	Charles and Francis Darwin concluded from their experiments on phototropism by grass seedlings that the part of the seedling that detects the direction of light is the
A)	tip of the coleoptile.	
B)	part of the coleoptile that bends during the response.	
C)	base of the coleoptile.	
D)	cotyledon.	
E)	phytochrome in the leaves.	

____	2)	Which of the following conclusions is supported by the research of both Went and Charles and Francis Darwin on shoot responses to light?
A)	When shoots are exposed to light, a chemical substance migrates toward the light.	
B)	Agar contains a chemical substance that mimics a plant hormone.	
C)	A chemical substance involved in shoot bending is produced in shoot tips.	
D)	Once shoot tips have been cut, normal growth cannot be induced.	
E)	Light stimulates the synthesis of a plant hormone that responds to light.	

____	3)	According to modern ideas about phototropism in plants,
A)	light causes auxin to accumulate on the shaded side of a plant stem.	
B)	auxin indirectly inhibits elongation of plant stem cells.	
C)	auxin is produced by the apical meristem of the coleoptile and moves downward.	
D)	all hormones move downward via the xylem.	
E)	cytokinins are more directly involved than auxins.	

____	4)	Auxin triggers the acidification of cell walls, which results in rapid growth, but also stimulates sustained, long–term cell elongation. What best explains how auxin brings about this dual growth response?
A)	Auxin binds to different receptors in different cells.	
B)	Different concentrations of auxin have different effects.	
C)	Auxin causes second messengers to activate both proton pumps on the plasma membrane and certain genes within the same cells.	
D)	The dual effects are due to two different types of auxins that are produced by different genes.	
E)	Other antagonistic hormones modify auxin's effects.	

____	5)	According to the acid growth hypothesis, auxin works by
A)	dissolving sieve plates, permitting more rapid transport of nutrients.	
B)	dissolving the cell membranes temporarily, permitting cells that were on the verge of dividing to divide more rapidly.	
C)	changing the pH within the cell, which would permit the electron transport chain to operate more efficiently.	
D)	increasing wall plasticity and allowing the affected cell walls to elongate.	
E)	greatly increasing the rate of deposition of cell wall material.	

____	6)	Many plants flower in response to day–length cues. Which of the following statements best summarizes this phenomenon?
A)	As a rule, short–day plants flower in the summer.	
B)	As a rule, long–day plants flower in the spring or fall.	
C)	Long–day plants flower in response to long days, not short nights.	
D)	Flowering in day–neutral plants is only influenced by day length if there is an exceptionally warm spring.	
E)	Flowering in short–day and long–day plants is controlled by phytochrome.	

____	7)	Plants often use changes in day length (photoperiod) to trigger events such as dormancy and flowering. It is logical that plants have evolved this mechanism because photoperiod changes
A)	are more predictable than air temperature changes.	
B)	alter the amount of energy available to the plant.	
C)	are modified by soil temperature changes.	
D)	can reset the biological clock.	
E)	are correlated with moisture availability.	

____	8)	What does a short–day plant require in order to flower?
A)	a burst of red light in the middle of the night	
B)	a burst of far–red light in the middle of the night	
C)	a day that is longer than a certain length	
D)	a night that is longer than a certain length	
E)	a higher ratio of Pr to Pfr	

____	9)	Plants that have their flowering inhibited by being exposed to bright lights at night are
A)	day–neutral plants.	
B)	short–night plants.	
C)	devoid of phytochrome.	
D)	short–day plants.	
E)	long–day plants.	

____	10)	A long–day plant will flower
A)	in the late fall.	
B)	when the night is shorter than a critical value.	
C)	only under artificial light in the summer.	
D)	during short days with proper fertilization.	
E)	regardless of the photoperiod imposed.	

____	11)	An example of a properly functioning homeostatic control system is seen when
A)	the core body temperature of a runner rises gradually from 37°C to 45°C.	
B)	the kidneys excrete salt into the urine when dietary salt levels rise.	
C)	a blood cell shrinks when placed in a solution of salt and water.	
D)	the blood pressure increases in response to an increase in blood volume.	
E)	the level of glucose in the blood is abnormally high whether or not a meal has been eaten.	

____	12)	Positive feedback differs from negative feedback in that
A)	positive feedback benefits the organism, whereas negative feedback is detrimental.	
B)	the positive feedback's effector responses are in the same direction as the initiating stimulus rather than opposite to it.	
C)	the effector's response increases some parameter (such as body temperature), whereas in negative feedback it can only decrease the parameter.	
D)	positive feedback systems have only effectors, whereas negative feedback systems have only receptors.	
E)	positive feedback systems have control centers that are lacking in negative feedback systems.	

____	13)	In a survivably cold environment, an ectotherm is more likely to survive an extended period of food deprivation than would an equally sized endotherm because the ectotherm
A)	maintains a higher basal metabolic rate.	
B)	expends more energy per kg of body mass than does the endotherm.	
C)	invests little energy in temperature regulation.	
D)	metabolizes its stored energy more readily than can the endotherm.	
E)	has greater insulation on its body surface.	

____	14)	An example of an ectothermic organism that has few or no behavioral options when it comes to its ability to adjust its body temperature is a
A)	terrestrial lizard.	
B)	sea star, a marine invertebrate.	
C)	bluefin tuna, a predatory fish.	
D)	hummingbird.	
E)	honeybee in a hive.	

____	15)	Endothermy
A)	is a characteristic of most animals found in tropical zones.	
B)	is a characteristic of animals that have a fairly constant body temperature.	
C)	is a term equivalent to cold–blooded.	
D)	is a characteristic of mammals but not of birds.	
E)	is seen only in insects and in certain predatory fishes.	

____	16)	Seasonal changes in snake activity are due to the fact that the snake
A)	is less active in winter because the food supply is decreased.	
B)	is less active in winter because it does not need to avoid predators.	
C)	is more active in summer because that is the period for mating.	
D)	is more active in summer because it can gain body heat by conduction.	
E)	is more active in summer as a result of being disturbed by other animals.	

____	17)	Innate immunity
A)	is activated immediately upon infection.	
B)	depends on a newly infected animal's previous exposure to the same pathogen.	
C)	is based on recognition of antigens that are specific to different pathogens.	
D)	is found only in vertebrate animals.	
E)	utilizes highly specific antigen receptors on B cells.	

____	18)	A key part of the humoral immune response is
A)	the attack of cytotoxic T cells on infected host cells.	
B)	the production of antibodies by plasma cells.	
C)	perforation of infected host cells by perforin.	
D)	the attack of phagocytes on living pathogens.	
E)	the initiation of programmed cell death in infected host cells.	

____	19)	Antigens are
A)	proteins found in the blood that cause foreign blood cells to clump.	
B)	proteins embedded in B cell membranes.	
C)	proteins that consist of two light and two heavy polypeptide chains.	
D)	foreign molecules that trigger the generation of antibodies.	
E)	proteins released during an inflammatory response.	

____	20)	The function of antibodies is to
A)	inject toxins into living pathogens.	
B)	secrete cytokines that attract macrophages to infection sites.	
C)	release perforins to disrupt infected cells.	
D)	act as Toll-like receptors.	
E)	mark pathogenic cells for destruction.	

____	21)	Jenner's successful use of cowpox virus as a vaccine against the smallpox virus is due to the fact that
A)	the immune system responds nonspecifically to antigens.	
B)	the cowpox virus made antibodies in response to the presence of smallpox.	
C)	cowpox and smallpox are antibodies with similar immunizing properties.	
D)	there are some antigenic determinants common to both pox viruses.	
E)	cowpox and smallpox are caused by the same virus.	

____	22)	Which of the following statements is true about the images above?
A)	The beet core in beaker A is at equilibrium with the surrounding water.	
B)	The beet core in beaker B will lose water to the surrounding environment.	
C)	The beet core in beaker B would be more turgid (rigid) than the beet core in beaker A.	
D)	The beet core in beaker A is likely to gain so much water that its cells will rupture.	
E)	The cells in beet core B are likely to undergo plasmolysis.	

____	23)	A plant cell with a solute potential of -6.5 bars maintains a constant volume when immersed in an open beaker containing a solution that has a solute potential of -3 bars.  What is the pressure potential of the cell?
A)	-6.5 bars	D)	3.5 bars	
B)	-3.5 bars	E)	6.5 bars	
C)	-0.54 bars	

____	24)	An open beaker of pure water has a water potential (Ψ) of
A)	–0.23 MPa.	
B)	+0.23 MPa.	
C)	+0.07 MPa.	
D)	–0.0000001 MPa.	
E)	0.0 (zero).	

____	25)	If ΨP = 0.3 MPa and ΨS = –0.45 MPa, the resulting Ψ is
A)	+0.75 MPa.	
B)	–0.75 MPa.	
C)	–0.15 MPa.	
D)	+0.15 MPa.	
E)	–0.42 MPa.	

____	26)	The value for Ψ in root tissue was found to be –0.15 MPa. If you take the root tissue and place it in a 0.1  M solution of sucrose (Ψ = –0.23 MPa), the net water flow would
A)	be from the tissue into the sucrose solution.	
B)	be from the sucrose solution into the tissue.	
C)	be in both directions and the concentrations would remain equal.	
D)	occur only as ATP was hydrolyzed in the tissue.	
E)	be impossible to determine from the values given here.	

____	27)	If isolated plant cells with a water potential averaging –0.5 MPa are placed into a solution with a water potential of –0.3 MPa, which of the following would be the most likely outcome?
A)	The pressure potential of the cells would increase.	
B)	Water would move out of the cells.	
C)	The cell walls would rupture, killing the cells.	
D)	Solutes would move out of the cells.	
E)	The osmotic pressure of the cells would decrease.	

____	28)	Transpiration in plants requires all of the following except
A)	adhesion of water molecules to cellulose.	
B)	cohesion between water molecules.	
C)	evaporation of water molecules.	
D)	active transport through xylem cells.	
E)	transport through tracheids.	

____	29)	What would enhance water uptake by a plant cell?
A)	decreasing the Ψ of the surrounding solution	
B)	increasing the pressure exerted by the cell wall	
C)	the loss of solutes from the cell	
D)	increasing the Ψ of the cytoplasm	
E)	positive pressure on the surrounding solution	

____	30)	Inflammatory responses typically include
A)	clotting proteins migrating away from the site of infection.	
B)	increased temperature and swelling in an inflamed area.	
C)	reduced permeability of blood vessels to conserve plasma.	
D)	release of substances to decrease the blood supply to an inflamed area.	
E)	inhibiting the release of white blood cells from bone marrow.	

Unit 3: Homeostasis Practice Test
Answer Section

MULTIPLE CHOICE

	1)	ANS:	A	PTS:	1	TOP:	Concept 39.2		

	2)	ANS:	C	PTS:	1	TOP:	Concept 39.2		

	3)	ANS:	A	PTS:	1	TOP:	Concept 39.2		

	4)	ANS:	C	PTS:	1	TOP:	Concept 39.2		

	5)	ANS:	D	PTS:	1	TOP:	Concept 39.2		

	6)	ANS:	E	PTS:	1	TOP:	Concept 39.3		

	7)	ANS:	A	PTS:	1	TOP:	Concept 39.3		

	8)	ANS:	D	PTS:	1	TOP:	Concept 39.3		

	9)	ANS:	D	PTS:	1	TOP:	Concept 39.3		

	10)	ANS:	B	PTS:	1	TOP:	Concept 39.3		

	11)	ANS:	B	PTS:	1	TOP:	Concept 40.2		

	12)	ANS:	B	PTS:	1	TOP:	Concept 40.2		

	13)	ANS:	C	PTS:	1	TOP:	Concept 40.3		

	14)	ANS:	B	PTS:	1	TOP:	Concept 40.3		

	15)	ANS:	B	PTS:	1	TOP:	Concept 40.3		

	16)	ANS:	D	PTS:	1	TOP:	Concept 40.4		

	17)	ANS:	A	PTS:	1	TOP:	Concept 43.1		

	18)	ANS:	B	PTS:	1	TOP:	Concept 43.2		

	19)	ANS:	D	PTS:	1	TOP:	Concept 43.2		

	20)	ANS:	E	PTS:	1	TOP:	Concept 43.3		

	21)	ANS:	D	PTS:	1	TOP:	Concept 43.3		

	22)	ANS:	A	PTS:	1	TOP:	Concept 36.2

	23)	ANS:	D	PTS:	1	TOP:	Concept 36.2

	24)	ANS:	E	PTS:	1	TOP:	Concept 36.2		

	25)	ANS:	C	PTS:	1	TOP:	Concept 36.2		

	26)	ANS:	A	PTS:	1	TOP:	Concept 36.2		
	27)	ANS:	A	PTS:	1	TOP:	Concept 36.2		

	28)	ANS:	D	PTS:	1	TOP:	Concept 36.3		

	29)	ANS:	E	PTS:	1	TOP:	Concept 36.2

	30)	ANS:	B	PTS:	1	TOP:	Concept 43.1