biology ii lab practical review part i last updated 11-29-07
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Review ContentReview Content
This review is intended to reflect the This review is intended to reflect the general material as presented in the lab general material as presented in the lab documents and does not take into account documents and does not take into account specific emphases that individual professors specific emphases that individual professors may or may not stress. Always check with may or may not stress. Always check with your professor for any clarifications. your professor for any clarifications.
Geologic Time-lineGeologic Time-line What are the current hierarchical set of divisions for What are the current hierarchical set of divisions for
describing geologic time, going from larger to smaller units.describing geologic time, going from larger to smaller units. a. Period, Era, Epoch, Eona. Period, Era, Epoch, Eon b. Era, Epoch, Eon, Periodb. Era, Epoch, Eon, Period c. Eon, Era, Period, Epochc. Eon, Era, Period, Epoch d. Epoch, Period, Era, Eond. Epoch, Period, Era, Eon
Answer: cAnswer: c In the Paleozoic Era is the Carboniferous Period before or In the Paleozoic Era is the Carboniferous Period before or
after the Permian Period?after the Permian Period? Answer: Before: Cambrian, Ordovician, Silurian, Devonian, Answer: Before: Cambrian, Ordovician, Silurian, Devonian,
Carboniferous, PermianCarboniferous, Permian
Geologic Time Chart
Martin, R. Aidan. 2003. Copyright and Usage Policy.World Wide Web Publication, www.elasmo-research.org/copyright.htm
Half LifeHalf Life
Assume a half life is 5,730 years.Assume a half life is 5,730 years. A fossil has 1/16 the Carbon 14 compared to A fossil has 1/16 the Carbon 14 compared to
when it was living.when it was living. How many half lifes have occurred?How many half lifes have occurred? Answer: (½)Answer: (½)n n where n = the number of half where n = the number of half
lifes. ½ x ½ x ½ x ½ = 1/16 = four half lifes.lifes. ½ x ½ x ½ x ½ = 1/16 = four half lifes. What is the age of the fossil?What is the age of the fossil? Answer: 5,730 yrs./half life x 4 half lifes = Answer: 5,730 yrs./half life x 4 half lifes =
22,920 yrs.22,920 yrs.
Baculites compressus (Ammonite)
Crinoid Palaeocoma egertoni Brittlestar
White Shark Tooth Carcharodon sp. Bear Claw (Ursa sp.) Amber
Fossil Identification
Archaeopteryx Eurypterid Fossil Lobster
Smilodon
Homo neandertalensis(skull on left) Fossil Fern
Fossil Identification
Copralite CopraliteBryozoan Archimedes sp.
Horseshoe Crab Priscacara oxyprion Bony Fish
Ammonite
Fossil Identification
Methods for fossil datingMethods for fossil dating
Which are methods for determing fossil dating?Which are methods for determing fossil dating? a. Paleomagnetic datinga. Paleomagnetic dating b. Rubidium-strontium dating b. Rubidium-strontium dating c. Flourine datingc. Flourine dating d. Carbon 14 datingd. Carbon 14 dating e. Relative datinge. Relative dating f. d. and e.f. d. and e. g. All the aboveg. All the above
Answer: g.Answer: g.
Fossilization ProcessesFossilization Processes
Otzi was a complete human preserved in a Otzi was a complete human preserved in a glacier for over 5300 yrs. What type of glacier for over 5300 yrs. What type of fossilization was this?fossilization was this?
Answer: FreezingAnswer: Freezing Name five other types of fossilization.Name five other types of fossilization. Answer: Amber, Casts or Molds, Imprints, Answer: Amber, Casts or Molds, Imprints,
Petrification, Tar PitsPetrification, Tar Pits
Mechanisms of EvolutionMechanisms of Evolution Name the five conditions necessary for Hardy-Weinberg to Name the five conditions necessary for Hardy-Weinberg to
occur.occur. Answer: Random mating, no natural selection, large Answer: Random mating, no natural selection, large
population sample, no mutations, no immigration or population sample, no mutations, no immigration or emigrationemigration
125 students out of 500 have an unattached ear which is 125 students out of 500 have an unattached ear which is recessive. Determine p, q, precessive. Determine p, q, p22, q, q22 and 2pq. and 2pq.
Answer: 125/500 = Answer: 125/500 = .25.25 = q = q22
(.25)(.25)1/21/2 = = .5.5 = q = q .5 + p = 1 1- .5 = .5 + p = 1 1- .5 = .5.5 = p = p .5.522 = = .25.25 = p = p22
2 x .5 x .5 = 2 x .5 x .5 = .5.5 = 2pq = 2pq
Genotype vs. PhenotypeGenotype vs. Phenotype
Attached ears is a dominant trait. The Attached ears is a dominant trait. The genotype for this is:genotype for this is: a. EEa. EE b. Eeb. Ee c. eec. ee d. a or b d. a or b e. All of the abovee. All of the above
Answer: d. Answer: d.
TaxonomyTaxonomy
The Five Kingdom system, Monera, Protista, The Five Kingdom system, Monera, Protista, Plantae, Fungi and Animalia, was changed to Plantae, Fungi and Animalia, was changed to the current Three Domain System. Name the the current Three Domain System. Name the domains.domains.
Answer: Bacteria, Archaea, EukaryaAnswer: Bacteria, Archaea, Eukarya Methanogens, extreme halophiles and extreme Methanogens, extreme halophiles and extreme
thermophiles belong to which domain?thermophiles belong to which domain? Answer: ArchaeaAnswer: Archaea
Which is gram negative?Which is gram negative?
Answer: One on the rightAnswer: One on the right Which has more peptidoglycan in the cell wall?Which has more peptidoglycan in the cell wall? Answer: One on the leftAnswer: One on the left Peptidoglycan is made up of what?Peptidoglycan is made up of what? Answer: Polymers of sugars and amino acidsAnswer: Polymers of sugars and amino acids
Prokaryotic Cell Nutritional ModesProkaryotic Cell Nutritional Modes
What are the energy and carbon sources for a What are the energy and carbon sources for a photoheterotroph?photoheterotroph?
Answer: Light (energy) and organic compounds Answer: Light (energy) and organic compounds (carbon)(carbon)
Merismopedia Gleocapsa cocci - yogurt
bacilli, cocci, spirochete Anabaena Oscillatoria
Bacteria Identification
MicroscopesMicroscopes
If the ocular lens is 10X and the objective lens If the ocular lens is 10X and the objective lens is 40X, then what is the total magnification?is 40X, then what is the total magnification?
Answer: (10X) x (40X) = 400XAnswer: (10X) x (40X) = 400X The diameter of the FOV at 400X is what?The diameter of the FOV at 400X is what? Answer: .45 mmAnswer: .45 mm What is the size of a single cell in mm if it What is the size of a single cell in mm if it
takes 5 cells to fit across the diameter of the takes 5 cells to fit across the diameter of the FOV?FOV?
Answer: .45mm / 5 cells = .09mm/cellAnswer: .45mm / 5 cells = .09mm/cell
Protista Tentative CladesProtista Tentative Clades D P E A S C A R CD P E A S C A R C Diplomonadida - Diplomonadida - Giardia lambliaGiardia lamblia
Parabasala - Parabasala - Trichomonas vaginalisTrichomonas vaginalis (STD) (STD)
Euglenazoa - Euglenazoa - Kinetoplastids (Kinetoplastids (TrypanosomaTrypanosoma/sleeping sickness); Euglinids (/sleeping sickness); Euglinids (EuglenaEuglena))
Alveolata - Alveolata - Dinoflagellata (Dinoflagellata (CeratiumCeratium/red tide); Apicomplexa (/red tide); Apicomplexa (PlasmodiumPlasmodium/malaria); /malaria); Ciliates ( Ciliates (ParameciumParamecium) )
Stramenopila - Stramenopila - Oomycetes (water molds, rusts, mildews); Bacillariophtes (diatoms); Oomycetes (water molds, rusts, mildews); Bacillariophtes (diatoms); Chrysophytes (golden algae); Phaeophytes (brown algae) Chrysophytes (golden algae); Phaeophytes (brown algae)
Radiolarians – Radiolarians – test (skeleton) made of silica with axopodia extending throughtest (skeleton) made of silica with axopodia extending through
Cercozoa - Cercozoa - diverse amoebas: Chlorarachniophytes, Foraminifera, slender plasmadiverse amoebas: Chlorarachniophytes, Foraminifera, slender plasma strandsstrands
Amoebazoa -Amoebazoa - lobed pseudopodia: Gymnamoeba, Entamoebas, cellular slime molds, lobed pseudopodia: Gymnamoeba, Entamoebas, cellular slime molds, plasmodial slime molds plasmodial slime molds
Rhodophyta - Rhodophyta - red algae (red algae (PorphyraPorphyra)) Chlorophyta - Chlorophyta - green algae (green algae (Ulva, Chlamydomonas, VolvoxUlva, Chlamydomonas, Volvox))
Giardia - Diplomonadida
Trypanosoma - Euglenazoa
Diatom - Stramenopila
Diatom – Stramenopila
Volvox - Chlorophyta
Stentor - Alveolata
Vorticella - Alveolata
Protista ID
Paramecium - Alveolata
Cercozoa - Radiolarians
Euglena - Euglenazoa
Trichomonas vaginalis Parabasala
Plasmodium - Alveolata
Amoeba - Amoebazoa
Protista ID
Protista ID - AlgaeProtista ID - Algae
Porphyra - Rhodophyta
Fucus – Brown (Phaeophytes)Stramenopila
HoldfastStipeFloatLamina/Blade
Ulva sp. - Chlorophyta
Kelp – Brown(Phaeophytes)Stramenopila
Plants IPlants I
Seedless Non-Vascular Phylums - Name ThemSeedless Non-Vascular Phylums - Name Them Answer: BryophytaAnswer: Bryophyta
Hepatophyta Hepatophyta Anthocerophyta Anthocerophyta
Seedless Vascular Phylums – Name ThemSeedless Vascular Phylums – Name Them Answer: LycophytaAnswer: Lycophyta
Pterophyta, which includes ferns, Pterophyta, which includes ferns, PsilotumPsilotum, horsetails (, horsetails (Equisetum)Equisetum)
Bryophyte Life Cycle
What are the two gametangia below?
Answer: Archegonium with egg (on left) and Antheridium with sperm (on right)
Are they gametophyte or sporophyte?
Answer: gametophyte
Pterophyta Pterophyta PterophytaPsilotum
AnthocerophytaBryophytaPterophytaEquisetum
Hepatophyta
Lycophyta
Name the Phylum of each example.
Sporophyte Sporophyte Sporophyte
GametophyteGametophyteSporophyte
Gametophyte
Sporophyte
What are the dominant generations of each example?
Cycadophyta - male Cycadophyta - female Ginkgophyta
Gnetophyta Coniferophyta ConiferophytaPlants IIWhat are the Phylums of each example?
Identify which is the female ovulate cone and the male Identify which is the female ovulate cone and the male staminate cone?staminate cone?
Answer: Ovulate cone on left; staminate cone on rightAnswer: Ovulate cone on left; staminate cone on right Are the pictures below sporophyte (2N) or gametophyte (1N)?Are the pictures below sporophyte (2N) or gametophyte (1N)? Answer: Sporophyte (2N)Answer: Sporophyte (2N)