Most fall flowering plants are
A) long-day plants
B) short-day plants
C) short-night plants
D) day-neutral plants
Answer: B) short-day plants
Biology Questions and Answers.
A) long-day plants
B) short-day plants
C) short-night plants
D) day-neutral plants
Answer: B) short-day plants
A) Spinach, a long-day plant, flowers in the tropics because the days are so long.
B) Ragweed, a short-day plant, can't flower in the northern US because of the frost in the late summer months.
C) Cockleburs, a short-day plant, can live where there are long flashes of red light at night.
D) Chrysantheums, a short-day plant, bloom in the summer when the days are long.
Answer: B) Ragweed, a short-day plant, can't flower in the northern US because of the frost in the late summer months.
A) A plant tastes bitter when a herbivore attempts to eat it
B) The leaves of a holly plant are slippery, making them difficult to eat
C) Thorns on the stems of pear bushes detract herbivores
D) The leaves of sensitive plants close up rapidly when they are touched
Answer: A) A plant tastes bitter when an herbivore attempts to eat it
A) aak tree
B) kudzu vine
C) sunflower
D) grass
Answer: B) kudzu vine
A) phototropism
B) gravitropism
C) thigmotropism
D) hydrotropism
Answer: C) thigmotropism
A) will flower if a period of continuous darkness was longer than a critical period.
B) flower in days where the day is short or the day is long.
C) flower when a period of darkness is shorter than the day.
D) flower if the night period is interrupted by a flash of light.
Answer: A) will flower if a period of continuous darkness was longer than a critical period.
A) occur in the morning and the evening
B) occur only in the presence of external stimuli
C) be able to reset if external cues are provided
D) affect seasonal changes
Answer: C) be able to reset if external cues are provided
A) A cocklebur is a long-day plant.
B) When the night is longer than a critical length, the cocklebur will flower.
C) The length of the light period controls flowering.
D) Interrupting the light period with darkness will prevent flowering.
Answer: C) The length of the light period controls flowering.
A) increased number of stomates
B) decreased thickness of the leaf cuticle
C) sunken stomates
D) increased leaf surface area
Answer: C) sunken stomates
A) close their stomata due to the low water potential outside the leaf
B) open their stomata temporarily then close due to environmental stress
C) close their stomata due to the high water potential outside the leaf
D) keep their stomata open due to the electrical potential of the guard cells
Answer: D) keep their stomata open due to the electrical potential of the guard cells
A) it is much quicker than conventional cleanup of sites
B) it can be effective at extremely deep sites by plants with taproots
C) plants have unique abilities to absorb and tolerate toxic substances
D) toxins absorbed by the plant increase reproductive rates of the plants
Answer: C) plants have unique abilities to absorb and tolerate toxic substances
A) engaging in a mutualistic relationship with fungi
B) sending out rootlike projections that tap into vessels of a host plant
C) possessing root nodules containing nitrogen-fixing bacteria
D) breaking down dead organisms extra-cellularly for sugars
Answer: D) breaking down dead organisms extra-cellularly for sugars
A) passive transport
B) active transport
C) phagocytosis
D) diffusion
Answer: B) active transport
A) provide proteins to sieve-tube members
B) transport organic nutrients to all parts of the plant
C) allow water to pass from cell to cell
D) provide internal support
Answer: A) provide proteins to sieve-tube members
A) higher water potential inside the cell
B) higher concentration of solutes outside of the cell
C) lower osmotic potential outside of the cell
D) lower pressure potential in the extracellular liquid
Answer: B) higher concentration of solutes outside of the cell
A) does not allow plants to increase in height
B) facilitates the dispersal of spores
C) makes it more difficult for animals to recognize flowers
D) promotes easier distribution of nutrients to fast-growing tissues
Answer: D) promotes easier distribution of nutrients to fast-growing tissues
A) the secretion of abscisic acid
B) Decrease in turgor pressure due to the exit of K+ ions
C) low humidity levels
D) high light intensity
Answer: D) high light intensity
A) Epidermis
B) Vascular bundle
C) Xylem
D) Sieve-tube member
Answer: C) Xylem
A) Parenchyma
B) Meristem
C) Epidermis
D) Schlerenchyma
Answer: C) Epidermis
A) Parenchyma
B) Meristem
C) Epidermis
D) Schlerenchyma
Answer: D) Schlerenchyma
A) Parenchyma
B) Meristem
C) Epidermis
D) Xylem
Answer: C) Epidermis
A) Parenchyma
B) Meristem
C) Epidermis
D) Xylem
Answer: D) Xylem
A) Strawberries
B) Potatoes
C) Sugar cane
D) Grapes
Answer: D) Grapes
A) mesophyll
B) endodermis
C) cork cambium
D) tracheids
Answer: A) mesophyll
A) spiny leaves
B) broad, wide leaves
C) fleshy leaves
D) sunken stomata
Answer: B) broad, wide leaves
A) Secondary phloem is not produced every year.
B) Phloem builds up from season to season.
C) Girdling is not harmful to a tree.
D) Cork cells have a waxy layer that makes them waterproof.
Answer: D) Cork cells have a waxy layer that makes them waterproof.
A) Taproot
B) Fibrous roots
C) Prop root
D) Aerial roots
Answer: D) Aerial roots
A) the microsporangium divides by meiosis
B) the pollen sacs produce microspores
C) meiosis produces four megaspores
D) one sperm fertilizes the egg, the other forms an endosperm nucleus
Answer: D) one sperm fertilizes the egg, the other forms an endosperm nucleus
A) large and showy
B) colorful
C) aromatic
D) light-colored
Answer: D) light-colored
A) produce microspores and megaspores
B) have male and female reproductive structures
C) have separate trees producing seeds and pollen
D) are pollinated by insects via wind
Answer: C) have separate trees producing seeds and pollen
A) are more broad
B) have a single strand of vascular tissue
C) are found in ferns
D) can gather more sunlight
Answer: B) have a single strand of vascular tissue
A) Alternation of generations
B) Covered by a cuticle
C) Possess stomata
D) Possess a dominant gametophyte stage
Answer: D) Possess a dominant gametophyte stage
A) Bryophytes
B) Charophytes
C) Gymnosperms
D) Angiosperms
Answer: D) Angiosperms
A) Bryophytes
B) Charophytes
C) Gymnosperms
D) Angiosperms
Answer: D) Angiosperms
A) Bryophytes
B) Charophytes
C) Gymnosperms
D) Angiosperms
Answer: C) Gymnosperms
A) Bryophytes
B) Charophytes
C) Gymnosperms
D) Angiosperms
Answer: A) Bryophytes
A) the type of fruit
B) mode of embryo protection
C) apical growth
D) the type of vascular tissue
Answer: B) mode of embryo protection
A) during asexual reproduction only
B) during sexual reproduction only
C) during asexual and sexual reproduction
D) because they are hermaphroditic
Answer: C) during asexual and sexual reproduction
A) Archaea
B) Bacteria
C) Green algae
D) Lichen
Answer: D) Lichen
A) Archaea
B) Bacteria
C) Green algae
D) Lichen
Answer: C) Green algae
A) Archaea
B) Bacteria
C) Green algae
D) Lichen
Answer: D) Lichen
A) mutualistic, in which the fungus receives nutrients from the algal cells, and the algal cells are protected from dessication by the fungus
B) controlled parasitism, with the fungus receiving nutrients, and the algae not benefiting at all from the interaction
C) commensal, with the fungi getting no benefits from the algae
D) a predator-prey relationship, as the fungi feeds on the algae
Answer: B) controlled parasitism, with the fungus receiving nutrients, and the algae not benefiting at all from the interaction
A) Fungi reproduce only asexually.
B) Fungi reproduce asexually through the use of runners.
C) Fungi produce nonmotile gametes during meiosis.
D) Fungi produce spores during both sexual and asexual reproduction.
Answer: D) Fungi produce spores during both sexual and asexual reproduction.
A) the secretion of toxins that suppress the immune system.
B) the production of lesions that produce penicillin.
C) the release of enzymes that degrade keratin and collagen in the skin.
D) the digestion of cellulose in the cell walls.
Answer: C) the release of enzymes that degrade keratin and collagen in the skin.
A) bacteria
B) fungus
C) yeast
D) lichen
Answer: B) fungus
A) Fungal cells lack chloroplasts
B) Plant cell walls are composed of peptidoglycan
C) Fungi lack motility at any stage in their life cycle
D) Many plant cells are multinucleate
Answer: A) Fungal cells lack chloroplasts
A) genetic recombination occurs in the two cells that begin the sexual process
B) it is a reproductive process
C) it is not a sexual process, but a reproductive process as four daughter cells are produced
D) clones are produced with an identical, equal amount of DNA
Answer: A) genetic recombination occurs in the two cells that begin the sexual process
A) Tubulin
B) Pepsin
C) Actin
D) Keratin
Answer: C) Actin
A) Carotenoid
B) Chlorophyll a
C) Phycoerythrin
D) Chlorophyll b
Answer: A) Carotenoid
A) haploid spores by meiosis
B) haploid gametes
C) diploid gametophytes by mitosis
D) diploid zygotes
Answer: C) diploid gametophytes by mitosis
A) form psuedopods
B) discharge excess water
C) are essential in reproduction
D) contain carotenoid pigments
Answer: B) discharge excess water
A) asexually by mitosis and sexually by meiosis
B) through binary fission
C) by sexual reproduction through conjugation
D) with the aid of sex pillus
Answer: A) asexually by mitosis and sexually by meiosis
A) Brown algae
B) Water molds
C) Red algae
D) Amoeboids
Answer: C) Red algae
A) Brown algae
B) Water molds
C) Red algae
D) Amoeboids
Answer: A) Brown algae
A) Brown algae
B) Water molds
C) Red algae
D) Euglenoids
Answer: B) Water molds
A) unicellular
B) colonial
C) filamentous
D) prokaryotic
Answer: D) prokaryotic
A) high levels of glucose will prevent expression from other operons that metabolize lactose.
B) low levels of glucose will cause the lactose mechanism to be turned off.
C) cAMP binds to activators to slow down transcription of the lac operon.
D) lactose will be synthesized only when there is ample glucose in the cell.
Answer: A) high levels of glucose will prevent expression from other operons that metabolize lactose.
A) when the substrate is in the environment and needs to be metabolized.
B) when a signal molecule is scarce.
C) when a metabolite activates the repressor.
D) when a repressor is in its active form.
Answer: B) when a signal molecule is scarce.
A) These molecules are nonprotein coding.
B) They are able to inhibit translation.
C) They can cause degredation of the mRNA.
D) They are unable to cause disease.
Answer: D) They are unable to cause disease.
A) frameshift
B) substitution in the first base
C) deletion
D) point
Answer: D) point
A) transcriptional
B) posttranscriptional
C) translational
D) posttranslational
Answer: A) transcriptional
A) adding extra nucleotides to the DNA.
B) the removal of introns from the nucleotide sequence by snRNP's.
C) inducers bind with the repressor, causing a shape change and release from the operator.
D) regulatory genes make the repressor proteins that bind with the operator and prohibit the function of RNA polymerase.
Answer: B) the removal of introns from the nucleotide sequence by snRNP's.
A) the repressor will code for an operator and change shape.
B) tryptophan will bind to the repressor.
C) the binding site for tryptophan will change shape.
D) RNA Polymerase will attach to the promoter, and trytophan is produced.
Answer: B) tryptophan will bind to the repressor.
A) A regulator gene codes for a repressor protein that controls the operon.
B) An operator is the sequence of DNA where the RNA polymerase attaches.
C) Promoters code for enzymes and are transcribed as a unit.
D) Structural genes are located outside the operon and code for enzymes.
Answer: A) A regulator gene codes for a repressor protein that controls the operon.
A) transcription and translation occur simultaneously in eukaryotes.
B) ribosomal subunits are larger in prokaryotes.
C) operons regulate gene expression in eukaryotes.
D) related prokaryotic genes are organized into groups called operons.
Answer: D) related prokaryotic genes are organized into groups called operons.
A) XXY
B) XXXY
C) XXYY
D) XYY
Answer: D) XYY
A) homozygous dominant
B) heterozygous
C) homozygous recessive
D) heterozygous dominant
Answer: B) heterozygous
A) 1/4
B) 1/8
C) 3/64
D) 10/64
Answer: D) 10/64
A) The F2 offspring are all tall.
B) Half of the F2 offspring are short.
C) All of the F1 offspring are short.
D) Only 1/4 of the F2 offspring are short.
Answer: D) Only 1/4 of the F2 offspring are short.
A) The father could be homozygous A.
B) The mother has homozygous B blood type.
C) The maternal grandmother is homozygous A.
D) The paternal grandmother could have type O blood.
Answer: D) The paternal grandmother could have type O blood.
A) AB+ or B+
B) AB- or O+
C) A+ or B+
D) AB- or AB+
Answer: A) AB+ or B+
A) A primrose develops white flowers when grown above 32C, and red flowers when grown at 24C
B) The more genes involved, the more continuous is the variation in phenotypes.
C) True breeding red and white flowered four o'clocks produce pink flowered offspring at any temperature.
D) A curly haired caucasian and a straight haired caucasian will have wavy haired offspring, regardless of the time of year.
Answer: A) A primrose develops white flowers when grown above 32C, and red flowers when grown at 24C
A) Cystic fibrosis is more common in males than females.
B) Parents who are phenotypically normal can have an afflicted child.
C) Individuals with cystic fibrosis are all heterozygous.
D) females have a decreased chance of inherting cystic fibrosis due to the inactivation of the second X chromosome.
Answer: B) Parents who are phenotypically normal can have an afflicted child.
A) 1/64
B) 1/8
C) 1/2
D) 0
Answer: D) 0
A) A and O
B) O and B
C) B and AB
D) A and B
Answer: D) A and B
A) 1/4
B) 1/2
C) 1/3
D) 1/6
Answer: A) 1/4
A) The unique separation of sister chromatids during the first meiotic division.
B) The presence of maternal DNA in one cell, and paternal DNA in another cell.
C) The physical exchange of regions of homologous chromosomes.
D) Two divisions create more diversity than one division.
Answer: C) The physical exchange of regions of homologous chromosomes.
A) Crossing over occurs between homologous chromosomes.
B) A tetrad or synaptonemal complex is formed at the equator.
C) The spindle apparatus extends outward from the centrosomes.
D) Four haploid daughter cells are formed after cytokinesis.
Answer: C) The spindle apparatus extends outward from the centrosomes.
A) skin cells
B) egg cells
C) muscle cells
D) nerve cells
Answer: B) egg cells
A) crossing over
B) chromatids
C) centromere
D) kinetochore
Answer: A) crossing over
A) Two sister chromatids did not separate into the proper daughter cells during anaphase.
B) Homologous chromosomes were sorted incorrectly during prophase.
C) The protein structure of nonkinetochore microtubules was defective, and incorrectly shortened during anaphase.
D) The cleavage furrow did not completely separate the two daughter cells.
Answer: A) Two sister chromatids did not separate into the proper daughter cells during anaphase.
A) the incorrect separation of this pair of sister chromatids during meiosis II
B) the failure of two sets of homologous chromosomes to line up properly during metaphase I
C) the nondisjunction of this set of homologues during meiosis II
D) a crossing over of genetic information of homologous chromosomes during prophase I
Answer: A) the incorrect separation of this pair of sister chromatids during meiosis II
A) depend on mutations to generate variation.
B) produce great numbers of offspring within a limited amount of time.
C) depend on crossing over during mitosis to generate genetic variation.
D) increase genetic variation due to an independent assortment of chromosomes.
Answer: D) increase genetic variation due to an independent assortment of chromosomes.
A) In animals, the haploid stage consists of only the gametes.
B) In plants, there is both a diploid and a haploid multicellular form.
C) In fungi, the dominant form is usually haploid.
D) In plants, the spore is a multicellular diploid form.
Answer: D) In plants, the spore is a multicellular diploid form.
A) the both produce haploid daughter cells.
B) sister chromatids move towards the poles in both.
C) homologous chromosomes pair and undergo crossing over in both.
D) they both contain two nuclear divisions.
Answer: B) sister chromatids move towards the poles in both.
A) crossing over between homologous chromosomes.
B) proper alignment of paternal chromosomes and maternal chromosomes on the proper side of the metaphase plate.
C) a differing number of homologous chromosomes in each of the daughter cells after meiosis I.
D) the production of genetically identical daughter cells after mitosis.
Answer: A) crossing over between homologous chromosomes.
A) Kinases become activated by cyclins, and inhibit mitosis.
B) Cyclins accumulate during the G1, S, and G2 phases and inhibit MPF.
C) MPF is turned off, which leads to the destruction of cyclin.
D) At the G2 checkpoint, MPF complexes form as aggregations of Cdk and cyclin.
Answer: D) At the G2 checkpoint, MPF complexes form as aggregations of Cdk and cyclin.
A) The duplicated DNA content of the reproducing parental cell is separated into two equal halves only in mitosis.
B) Four daughter cells consisting of a haploid compliment of chromosomes are produced in binary fission.
C) Binary fission does not separate sex chromosomes, but is able to separate autosomes.
D) Their is neither a nucleus nor centromeres in prokaryotic cells that undergo binary fission.
Answer: D) There is neither a nucleus nor centromeres in prokaryotic cells that undergo binary fission.
A) a decrease in the length of the cell cycle
B) density-dependent inhibition
C) apoptosis of cells without appropriate cyclins
D) cells passing through the M1 checkpoint
Answer: D) cells passing through the M1 checkpoint
A) haploid
B) gametes
C) somatic
D) homologous
Answer: C) somatic
A) the production of a cell plate in animals and a cleavage furrow in plants.
B) centriole formation in plant cells.
C) the fusing of vesicles to form a cell wall in plants.
D) the presence of asters during spindle formation in animal cells.
Answer: D) the presence of asters during spindle formation in animal cells.
A) the nucleolus disappears.
B) spindle formation is complete.
C) the centromeres divide.
D) centrosomes migrate away from one another.
Answer: C) the centromeres divide.
A) sister chromatids
B) homologous chromosomes
C) kinetochores
D) cell plates
Answer: A) sister chromatids
A) sister chromatids
B) homologous chromosome
C) kinetochore
D) chromatin
Answer: C) kinetochore
A) Flasks B and D will turn pink.
B) Only flasks A and B will turn pink.
C) Only flask D will turn pink.
D) None of the flasks will turn pink.
Answer: B) Only flasks A and B will turn pink.
A) CO2
B) O2
C) glucose
D) glycogen
Answer: B) O2
A) proteins
B) fats
C) simple carbohydrates
D) nucleic acids
Answer: B) fats
A) Anaerobic respiration requires oxygen in order to generate ATP.
B) The citric acid cycle produces NADPH, which releases energy to fuel chemiosmosis.
C) Oxygen is formed as water combines with electrons at the end of an electron transport chain.
D) Glucose is broken down into two pyruvate molecules during glycolysis.
Answer: D) Glucose is broken down into two pyruvate molecules during glycolysis.
A) integral proteins translocate H+ ions across the membrane, creating a pH gradient.
B) ATP synthase molecules facilitate the movement of H+ ions across the membrane.
C) ATP synthase phosphorylates ADP.
D) the electrons are taken up by oxygen to make water.
Answer: A) integral proteins translocate H+ ions across the membrane, creating a pH gradient.
A) nerve cells
B) fat-storage cells
C) liver cells
D) cardiac muscle cells
Answer: B) fat-storage cells
A) triglyceride
B) DNA
C) cholesterol
D) glucose
Answer: D) glucose
A) glycolysis.
B) fermentation.
C) substrate-level phosphorylation.
D) electron transport.
Answer: D) electron transport.
A) Both cause a decrease in acidity due to the accumulation of lactic acid.
B) Both lead to the lack of lactate dehydrogenase to break down pyruvate.
C) Trematol inhibits the conversion of pyruvate to glucose, leading to an insufficiency of ATP.
D) Lactic acid is produced anaerobically during strenuous exercise, and will accumulate in the muscles in both situations.
Answer: D) Lactic acid is produced anaerobically during strenuous exercise, and will accumulate in the muscles in both situations.
A) is aerobic and produces 2 ATP per glucose molecule
B) converts glucose to pyruvate
C) yields 34 ATP molecules through the process of chemiosmosis
D) produces O2 as a by-product
Answer: B) converts glucose to pyruvate
A) the rate of photosynthesis will vary with light intensity.
B) the higher the irradiance, the higher the rate of photosynthesis.
C) the rate of photosynthesis decreases with an increase in atmospheric carbon dioxide.
D) the rate of photosynthesis is unaffected by water availability.
Answer: A) the rate of photosynthesis will vary with light intensity.
A) green
B) red
C) orange
D) yellow
Answer: A) green
A) Leaves produce more chlorophyll as temperatures drop.
B) Carotenes and Xanthophylls are produced only in the fall, when leaves turn yellow or orange.
C) The chlorophyll breaks down, and the accessory pigments are more noticable.
D) Reduced amounts of sunlight reflect less green light, making it easier to see than other pigments.
Answer: C) The chlorophyll breaks down, and the accessory pigments are more noticable.
A) carbon dioxide is fixed into sugar.
B) the hydrolysis of water supplies electrons to photosystem II.
C) ATP is broken down into ADP and P.
D) NADPH+ is broken down into NADP and H.
Answer: B) the hydrolysis of water supplies electrons to photosystem II.
A) electrons are accepted by water at the end of photosystem II.
B) water is split at the initiation of photosystem II.
C) protons from water are used in the production of carbohydrates in the stroma.
D) CO2 is oxidized, and carbon is utilized in the production of carbohydrates.
Answer: B) water is split at the initiation of photosystem II.
A) captures CO2 from the atmosphere and converts it to two 3-carbon sugars.
B) transports H+ across the chloroplast membrane in chemiosmosis.
C) combines with CO2 to form oxaloacetate in C4 photosynthesis.
D) releases a molecule of CO2 as it is converted to pyruvate.
Answer: A) captures CO2 from the atmosphere and converts it to two 3-carbon sugars.
A) CO2 is obtained during the day and incorporated into a 3-carbon compound.
B) They require fewer enzymes, and have no specialized anatomy.
C) avoiding photorespiration by converting CO2 to an acid and storing it during the night, and converting it back into CO2 during the daytime.
D) avoidance of photorespiration by first incorporating CO2 into a 4-carbon compound and then translocating it into bundle sheath cells.
Answer: D) avoidance of photorespiration by first incorporating CO2 into a 4-carbon compound and then translocating it into bundle sheath cells.
A) occur in the stroma of the chloroplasts.
B) occur when hydrogen and carbon dioxide react to form a carbohydrate.
C) release oxygen as a by-product and split water in the process.
D) utilize NAD+ as an electron acceptor.
Answer: C) release oxygen as a by-product and split water in the process.
A) trypsin
B) peptin
Answer: A) trypsin
A) result in an increased rate of reaction until there is no substrate left.
B) not affect the rate of the reaction.
C) result in a decreased rate of reaction because there will be a higher percentage of substrate molecules.
D) initially result in a decreased rate of reaction, but will then increase toward the end of the reaction period.
Answer: A) result in an increased rate of reaction until there is no substrate left.
A) Endergonic reactions can be accelerated by coupling them to an exergonic reaction.
B) Reactants contain less energy than products in exergonic reactions.
C) Enzymes increase the activation energy necessary for a reaction to proceed.
D) Competitive inhibitors bond with an enzyme outside of the active site.
Answer: A) Endergonic reactions can be accelerated by coupling them to an exergonic reaction.
A) utilizes energy captured in photosynthesis.
B) occurs in the mitochondria.
C) is an exergonic reaction.
D) occurs during aerobic respiration.
Answer: A) utilizes energy captured in photosynthesis.
A) ATP
B) NAD+
C) NADP+
D) FAD
Answer: C) NADP+
A) ATP
B) NAD+
C)NADP+
D) ADP
Answer: D) ADP
A) ATP
B) NAD+
C) NADP+
D) FAD
Answer: A) ATP
A) ATP
B) NAD+
C) NADP+
D) FAD
Answer: C) NADP+
A) Facilitated diffusion transports molecules against the concentration gradient, and requires energy to do so.
B) Active transport transports small molecules with the concentration gradient in large quantities, but does not require energy.
C) Receptor-mediated endocytosis makes use of receptor proteins, and transports solids or liquids with the help of lysosomes into the cell.
D) Osmosis is an active transport process that moves water against the concentration gradient.
Answer: C) Receptor-mediated endocytosis makes use of receptor proteins, and transports solids or liquids with the help of lysosomes into the cell.
A) gap junctions
B) plasmodesmata
C) desmosomes
D) tight junctions
Answer: B) plasmodesmata
A) to become turgid due to the increase in solute in the surrounding environment.
B) to lose water to the surrounding hypotonic environment.
C) to grow excessively, due to the increased N-P-K fertilizer in the soil.
D) to plasmolysize, due to osmosis of water to a lower water potential.
Answer: D) to plasmolysize, due to osmosis of water to a lower water potential.
A) will lyse without special adaptations to combat water concentration fluctuations.
B) will remain flaccid, due to an ineffective central vacuole.
C) will remain turgid, due to the elastic cell wall.
D) will shrivel, due to the loss of water from the large central vacuole.
Answer: C) will remain turgid, due to the elastic cell wall.
A) K+
B) glucose
C) HCO3-
D) CO2
Answer: D) CO2
A) increasing the temperature.
B) decreasing pressure.
C) increasing the size of the molecules.
D) decreasing concentration.
Answer: A) increasing the temperature.
A) The net movement of water will be out of the cell.
B) Sucrose will move out of the cell, and glucose will move into the cell.
C) There will be no net movement of water.
D) Glucose will move out of the cell.
Answer: D ) Glucose will move out of the cell.
A) A cell is freeze-fractured and splits the phospholipid bilayer into two layers, and the membrane proteins go with one of the layers.
B) Human and mouse sauce labeled with protein markers are fused, and the hybrid cell contains a mixing of proteins on the surface.
C) There are two types of protein receptors in plants to transport CO2 and NH4-; however, in some microorganisms, proteins that transport NH4 can also transport CO2.
D) Integral proteins on the surface of stomach cells can transport either acids or bases.
Answer: B) Human and mouse sauce labeled with protein markers are fused, and the hybrid cell contains a mixing of proteins on the surface.
A) phagocytosis.
B) ion-mediated transport.
C) exocytosis.
D) cotransport.
Answer: D) cotransport.
A) A carrier proteins moves sucrose across the membrane with the concentration gradient.
B) cytoplasmic Na+ mines to the sodium potassium pump, ATP is used, and NA+ is released to the outside.
C) water moves from an area of high free-water concentration to an area of low free-water concentration.
D) Aquaporins facilitate the diffusion of water in the red blood cells of mammals.
Answer: B) cytoplasmic Na+ mines to the sodium potassium pump, ATP is used, and NA+ is released to the outside.
A) nucleus
B) DNA
C) chromosomes
D) nucleolus
Answer: A) nucleus
A) cell wall
B) cytoskeleton
C) chloroplast
D) lysosome
Answer: D) lysosome
A) Mitochondria
B) Smooth endoplasmic reticulum
C) Chloroplast
D) Lysosome
Answer: B) Smooth endoplasmic reticulum
A) Mitochondria
B) Smooth endoplasmic reticulum
C) Chloroplast
D) Lysosome
Answer: C) Chloroplast
A) Mitochondria
B) Smooth endoplasmic reticulum
C) Chloroplast
D) Lysosome
Answer: A) Mitochondria
A) Mitochondria
B) Smooth endoplasmic reticulum
C) Chloroplast
D) Lysosome
Answer: C) Chloroplast
A) Mitochondria contain many folds called cristae.
B) Chloroplasts contain stacks of thylakoids.
C) Microvilli (very small folds) are present on cells of the small intestine.
D) The large central vacuole maintains pH and allows enzymes to act.
Answer: D) The large central vacuole maintains pH and allows enzymes to act.
A) red pine
B) Streptococcus
C) C. amoeba
D) mushroom
Answer: A) red pine
A) the Golgi bodies do not transport fats out of the cell in vesicles, and they accumulate in the cell.
B) the lysosomes do not contain the appropriate enzyme to metabolize the fat.
C) the fats are not permeable to the cell wall and they accumulate in the cell.
D) hydrogen peroxide is not produced by the peroxisomes to destroy the fats.
Answer: B ) the lysosomes do not contain the appropriate enzyme to metabolize the fat.
A) water transport cells in plants
B) sperm cells in animals
C) sugar transport cells in animals
D) cells lining the oviduct in animals
Answer: D) cells lining the oviduct in animals
A) cells that transport water
B) cells that transport sugars
C) leaf or fruit tissues
D) storage cells in roots
Answer: C) leaf or fruit tissues
A) higher rate of combustion than saturated fats
B) no double bonds
C) more rigid at room temperature
D) lower boiling pount
Answer: D) lower boiling point
A) DNA is single-stranded, and RNA is double-stranded.
B) DNA contains uracil.
C) RNA bonds purines to purines, and DNA bonds pyrimidines to pyrimidines.
D) DNA contains deoxyribose, and RNA does not.
Answer: D) DNA contains deoxyribose, and RNA does not.
A) they specify the correct structure of a polypeptide
B) they are carbohydrates that bond with proteins to maintain proper shape
C) they are organic molecules that protect against disease
D) they are protein molecules that assist in the proper folding of other proteins
Answer: D) they are protein molecules that assist in the proper folding of other proteins
A) Primary
B) Helical
C) Secondary
D) Tertiary
Answer: D) Tertiary
A) amino group
B) a carbon
C) carboxyl group
D) R group
Answer: D) R group
A) creates glycostatic linkages between the glycerol and the fatty acids.
B) removes one molecule of water for each fatty acid joined to the glycerol.
C) bonds water to the glycerol molecule to form a triacylglycerol.
D) requires water to form peptide bonds.
Answer: B) removes one molecule of water for each fatty acid joined to the glycerol.
A) cellulose.
B) glycogen.
C) starch.
D) chitin.
Answer: C) starch.
A) length.
B) branching.
C) bonding patterns.
D) composition.
Answer: D) composition.
A) Carbon
B) Silicon
C) Hydrogen
D) Oxygen
Answer: A) Carbon
A) carbon
B) sodium
C) iron
D) potassium
Answer: A) carbon
A) Surface tension allows molecules to form a bubble within the fibers of the paper towel and be absorbed.
B) Cohesion between the polar ends of the water molecules and the carbons in the paper towels allow water to be absorbed.
C) Cohesion of the water molecules to other water molecules from a chain that is absorbed by the paper towel.
D) Ionic bonding between the water and the paper towel allow water to be absorbed.
Answer: C) Cohesion of the water molecules to other water molecules from a chain that is absorbed by the paper towel.
A) temperature changes would require a great deal of energy input.
B) the water in organisms would heat quickly and cause the cells to burst.
C) the water in lakes would evaporate on a hot day.
D) ectothermic animals would be more adapted to their surroundings.
Answer: B) the water in organisms would heat quickly and cause the cells to burst.
A) hydrogen atoms have a partial negative charge, and will be attracted to one another.
B) the slightly positive hydrogen of one molecule is attracted to the slightly negative oxygen of another molecule.
C) hydrogen is more electronegative than oxygen, and the electrons spend more time closer to the oxygen atom.
D) its tow hydrogen atoms are joined to the oxygen atom by single covalent bonds.
Answer: B) the slightly positive hydrogen of one molecule is attracted to the slightly negative oxygen of another molecule.
A) accepting hydrogen ions from the solution when they are in excess.
B) irreversibly combining with hydrogen ions to neutralize a base.
C) dissociating to yield OH- ions to neutralize an acid.
D) donating hydrogen ions when they are in excess.
Answer: A) accepting hydrogen ions from the solution when they are in excess.
A) tears.
B) stomach fuilds.
C) blood.
D) urine.
Answer: B) stomach fluids.
A) equal sharing of electrons.
B) transfer of electrons.
C) uneven sharing of electrons.
D) asymmetric distribution in molecules.
Answer: B) transfer of electrons.
A) their cumulative effects reinforce molecular structure, making them biologically significant.
B) they are not found in organic molecules.
C) they occur between molecules that are far apart.
D) they occur when two atoms are equally electronegative and are attracted to one another.
Answer: A) their cumulative effects reinforce molecular structure, making them biologically significant.
A) nonpolar covalent
B) ionic
C) polar covalent
D) hydrogen
Answer: D) hydrogen
A) Hydrogen bonds require less energy to break than do covalent bonds.
B) Water cools more slowly than other liquids.
C) The hydrogen bonds between water hold more heat energy than most other liquids.
D) Water's surface tension is due to hydrogen bonds between water molecules.
Answer: C) The hydrogen bonds between water hold more heat energy than most other liquids.
A student set up an experiment using Drosophila melanogaster. The student wished to determine the LC-50 dose of caffeine exposure to this particular species of fruit flies. The lethal concentration is reached when 50% (LC-50) or more of the flies die when exposed to a toxin. The student mixed various concentrations of caffeine with the fruit fly media, added 50 eggs to each vial, and waited one week to analyze the results. Below are the results of the study. PICTURE
A) The student should run additional trials at each level of exposure.
B) The student should place additional eggs in each vial.
C) The student should use higher concentrations of caffeine.
D) The student should compare results with eggs of another insect.
Answer: A) The student should run additional trials at each level of exposure.
A) genus
B) phylum
C) class
D) order
Answer: A) genus
A) Population
B) Ecosystem
C) Community
D) Species
Answer: A) Population
A) plant
B) animal
C) fungus
D) prokaryote
Answer: A) plant
A) Kingdom Plantae
B) Domain Bacteria
C) Kingdom Fungi
D) Domain Archaea
Answer: D) Domain Archaea
A) species
B) population
C) genus
D) biosphere
Answer: D) biosphere
a. nonadaptive traits
b. acclimation
c. convergent traits
d. developmental homology
e. adaptation
Answer: A
a. is created by the direct action of natural selection
b. tends to be reduced by when diploid organisms produce gametes
c. must be present in a population before natural selection can act upon the population
d. arises in response to changes in the environment
Answer: C
a. stabilizing selection
b. directional selection
c. sexual selection
d. disruptive selection
Answer: D
a. sexual selection
b. frequency-dependent selection
c. stabilizing selection
d. balancing selection
e. disruptive selection
Answer: B
a. stabilizing selection
b. disruptive selection
c. sexual selection
d. artificial selection
e. directional selection
Answer: B
a. heterozygote advantage and stabilizing selection
b. population bottleneck and Hardy-Weinberg equilibrium
c. mutation and natural selection
d. founder effect and genetic drift
e. sexual selection and inbreeding depression
Answer: D
a. increased genetic drift
b. lower average fitness in both populations
c. higher average fitness in both populations
d. decreased genetic difference between the two populations
e. increased genetic difference between the two populations
Answer: D
a. natural selection has not had sufficient time to create the optimal design in each case, but will do so given enough time
b. in many cases, phenotype is determined by genotype and the environment
c. though we may not consider the fit between the current skeletal arrangements and their functions excellent, we should not doubt that natural selection ultimately produces the best design
d. natural selection is generally limited to modifying structures that were present in previous generations and in previous species
Answer: D
a. You cannot predict a change in the courtship songs at the two lakes.
b. Genetic drift will cause the songs to differentiate even more.
c. The songs will become more similar to each other.
d. Males will become louder.
e. Disruptive selection will cause the songs to differentiate even more.
Answer: C
a. Genetic drift decreased the population's fitness.
b. Genetic drift increased the population's fitness.
c. The population was relatively small.
d. Any of the statements can be true.
e. The population experiences a decrease in genetic variation.
Answer: D
a. stabilizing selection
b. disruptive selection
c. sexual selection
d. directional selection
e. artificial selection
Answer: A
a. Natural selection cannot account for losses, but accounts only for new structures and functions.
b. Natural selection accounts for these losses by the principle of use and disuse.
c. Under particular circumstances that persisted for long periods, each of these structures presented greater costs than benefits.
d. The ancestors of these organisms experienced harmful mutations that forced them to lose these structures.
Answer: C
a. Hummingbirds are the best pollinators of certain flowers, but bees are the best pollinators for orchids.
b. The strong, thick beak of a woodpecker helps it find insects in trees.
c. In some hornbill species, the male helps seal the female in a tree with her nest until the young are ready to fledge.
d. Turtle shells provide protection but are heavy and burdensome when moving.
Answer: D
a. The Tamiflu-resistance gene will undergo mutations that convert it into a gene that has a useful function in this environment.
b. If the Tamiflu-resistance gene involves a cost, it will experience directional selection leading to reduction in its frequency.
c. If the Tamiflu-resistance gene confers no benefit in the current environment, and has no cost, the virus will increase in frequency.
d. The new virus will maintain its Tamiflu-resistance gene, in case of future exposure to Tamiflu.
Answer: B
a. f(A1) = 0.9700, f(A2) = 0.0300
b. f(A1) = 0.9997, f(A2) = 0.0003
c. f(A1) = 0.9600, f(A2) = 0.0400
d. f(A1) = 0.9800, f(A2) = 0.0200
e. f(A1) = 0.9604, f(A2) = 0.0392
Answer: A
a. is more important in eukaryotes than in prokaryotes
b. decreases fitness
c. happens in all populations
d. does little to change allele frequencies
e. has no effect on genetic variation
Answer: D
a. Estimate the frequency of alleles relevant genes for drought tolerance.
b. None of these. Gene flow does not contribute to evolution.
c. All of the above.
d. Determine if the population is in Hardy-Weinberg equilibrium.
e. Transplant pitcher plants from other populations into the bog.
Answer: D
a. can favor beneficial mutations
b. does not affect harmful mutations
c. is a random process
d. creates beneficial mutations
Answer: A
a. Mutation caused genetic drift and decreased fitness.
b. Mutation caused the fixation of new alleles.
c. Mutations caused major changes in rodent physiology over time.
d. Mutation led to increased genetic variation.
Answer: D
a. no genetic drift
b. random mating
c. no gene flow
d. no mutation
e. no selection
Answer: D
a. frequency of dominant allele
b. expected frequency of homozygous dominant genotypes
c. expected frequency of homozygous recessive genotypes
d. frequency of recessive allele
e. expected frequency of heterozygous genotypes
Answer: E
a. no genetic drift
b. no mutation
c. random mating
d. no selection
e. no gene flow
Answer: E
a. Natural selection
b. Gene flow
c. Genetic drift
d. All of the above
e. Both natural selection and genetic drift.
Answer: D
Answer: True
Answer: False
Answer: False
Answer: False
A. random mating
B. migration
C. genetic drift
D. selection
Answer: B
Answer: random mating, no genetic drift, no gene flow, no mutations, no natural selection
A. 0.81
B. 0.09
C. 0.18
D. 0.01
E. 0.198
Answer: C
A. Large population size
B. No mutations
C. No immigration or emigration
D. Dominant alleles more frequent than recessive alleles
E. No natural selection
Answer: D
A) always
B) often
C) rarely
D) never
Answer: C
A) duplicate an existing allele
B) lead to deletion and loss of an allele
C) guarantee that duplicated alleles are going to be different from each other
D) A & B
E) A, B & C
Answer: D
A) true
B) false
Answer: B
A) true
B) false
Answer: B
A) true
B) false
Answer: B
A) true
B) false
C) cannot be determined
Answer: B
A) DNA replication.
B) transcription.
C) translation.
D) protein degradation.
Answer: A
a. Hummingbirds are the best pollinators of certain flowers, but bees are the best pollinators for orchids.
b. The strong, thick beak of a woodpecker helps it find insects in trees.
c. In some hornbill species, the male helps seal the female in a tree with her nest until the young are ready to fledge.
d. Turtle shells provide protection but are heavy and burdensome when moving.
Answer: D
A. Direct sunlight
B. Bright indirect light
C. Medium light
D. Bright light
Answer: B. Bright indirect light
A. Two tentacles
B. Six legs
C. Two pairs of wings
D. Three body segments
Answer: A. Two tentacles
A. Cut flowers in the cooler
B. Insurance for the shop's building
C. Energy bills
D. Membership fee for FTD
Answer: A. Cut flowers in the cooler
A. Silica gel
B. Sand
C. Glycerin
D. All of the above
Answer: D. All of the above
A. Black
B. Gray
C. White
D. Any of the above
Answer: C. White
A. Hydrangea
B. Dahlia
C. Zinnia
D. Asiatic Lily
Answer: A. Hydrangea
A. Scarification
B. Stratification
C. Vernalization
D. Germination
Answer: D. Germination
A. Fill the spray tank
B. Read the label
C. Put on personal protective equipment
D. Rinse out the sprayer tank
Answer: B. Read the label
A. Allelopathy
B. Anthesis
C. Transcription
D. Translation
Answer: A. Allelopathy
A. Fungus gnat
B. Greenhouse whitefly
C. Western flower thrip
D. Shore fly
Answer: C. Western flower thrip
A. Transduction
B. Translocation
C. Transpiration
D. None of the above
Answer: C. Transpiration
A. To keep the greenhouse temperatures lower
B. To reduce light intensity
C. To reduce pest populations
D. A and B
Answer: D. A and B
A. Wax flower
B. Bird-of-Paradise
C. Miniature Carnation
D. Leatherleaf Fern
Answer: B. Bird-of-Paradise
A. Vermiculite
B. Perlite
C. Peat
D. Bark Medium
Answer: B. Perlite
A. Texture
B. Focal
C. Filler
D. Form
Answer: C. Filler
A. 1⁄2
B. 1 1⁄2
C. 3
D. 5
Answer: B. 1 1⁄2
A. American
B. Greek
C. European
D. Egyptian
Answer: D. Egyptian
A. Petiole
B. Stipule
C. Midrib
D. Apex
Answer: A. Petiole
A. Nitrogen
B. Phosphorus
C. Potassium
D. Sunlight
Answer: B. Phosphorus
A. Petiole
B. Stoma
C. Cuticle
D. Vein
Answer: C. Cuticle
A. Force
B. Fertilize
C. Prune
D. Divide
Answer: A. Force
A. Perlite
B. Vermiculite
C. Sand
D. Peat Moss
Answer: D. Peat Moss
A. Arrangement
B. Margins
C. Shape
D. Forms
Answer: A. Arrangement
A. Foot candles
B. Particles
C. Humidity
D. Dew point
Answer: C. Humidity
A. Air layering
B. Compound layering
C. Whip or tongue grafting
D. Cleft grafting
Answer: A. Air layering
A. Systemic
B. Fumigants
C. Contact
D. Repellents
Answer: B. Fumigants
A. Standard pot
B. Bulb pan
C. Azalea pot
D. Mum pot
Answer: B. Bulb pan
A. Fungus
B. Virus
C. Bacteria
D. Environmental Condition
Answer: A. Fungus
A. 500-1,000
B. 2,000-3,000
C. 4,000-6,000
D. 7,000-8,000
Answer: C. 4,000-6,000
A. 100%
B. 150%
C. 250%
D. 500%
Answer: C. 250%
A. 1 inch
B. 12-14 inches
C. At least 2 feet
D. 6-8 inches
Answer: D. 6-8 inches
A. Cell division
B. Cell enlargement
C. Cell specialization
D. Cell protection
Answer: B. Cell enlargement
A. 6.8-7.4
B. 6.2-6.8
C. 4.5-5.8
D. 5.8-6.2
Answer: C. 4.5-5.8
A. Emulsifiable concentrate
B. Granules
C. Dusts
D. Baits
Answer: A. Emulsifiable concentrate
A. Heartwood
B. Xylem
C. Phloem
D. Cambium
Answer: D. Cambium
A. Piercing
B. Clutch
C. Hook
D. Stitch
Answer: A. Piercing
A. Where disease organisms move into the flower from the air clogging the stem.
B. Where a bubble of air enters the xylem preventing water from reaching the flower.
C. Where the different between room and flower cooler temperature clog the stem.
D. Where the flower stem clogs when the flower cooler temperature drops below 36 degrees F.
Answer: B. Where a bubble of air enters the xylem preventing water from reaching the flower.
A. An "O" shape
B. A "C" shape
C. A "S" shape
D. A "H" shape
Answer: C. A "S" shape
A. Tussy Mussy
B. Cascading
C. Presentation
D. Round Cluster
Answer: B. Cascading