AP Bio review. AP Bio Test he AP Biology Exam is approximately three hours long and has two sections. Section 1 is Multiple Choice and Grid-in questions

AP Bio review

AP Bio Test he AP Biology Exam is approximately three hours long and has two sections. Section 1 is Multiple Choice and Grid-in questions and Section 2 is Free Response Questions.. Each section is worth 50% of the final exam grade. Section 1 lasts 1 hour and 30 minutes and contains 63 multiple choice questions and 6 grid-in questions. Section 2 lasts 1 hour and 20 minutes (plus a 10 minute reading period), and contains 8 questions consisting of 2 long free response questions and 6 short free response questions. Due to the increased emphasis on quantitative skills and application of mathematical methods in the questions, students are allowed to use simple four-function calculators (with square root) on the entire exam. Students also receive a formula list as part of their testing materials.

QOD: Chi-squared Review

2 types of chi-squared Genetics Expected is figured out using punnet square, compare to given observed to see if this is likely or not If there is a sig difference, must be other factors affecting the genetics (mutation, different linkage, etc) Difference in observed and expected in experiment (pill bug preference) Null hypothesis is there is no difference in options (even number of specimen in each) If sig difference, means there is choice going on, may help support original hypothesis

Using a chi-squared analysis, what is the chi squared value and the probability that red-eyes is a sex linked in the cross below of a heterozygous female and red-eyed male.

Hardy Weinberg

The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in a biology class, and determined that 150 could detect the bitter taste of PTC and 65 could not. What is the predicted frequency of the recessive allele? t 0.55 What is the predicted frequency of dominant allele? T 0.45 In a population of 10,000 people, how many would be heterozygous (assuming Hardy-Weinberg equilibrium)? Homozygous dominant? Homozygous recessive? Calculate all of the potential frequencies. TT =.203 / Tt =.495 / tt =.302

The I A "allele" for the ABO blood groups actually consists of two subtypes, I A1 and I A2, either being considered "I A ". In Caucasians, about 3/4 of the IA alelles are I A1 and 1/4 are I A2 (Cavalli-Sforza and Edwards, 1967). What would be the expected proportions of I A1 I A1, I A1 I A2, and I A2 I A2 among I A I A individuals? I A1 I A1 = 0.563 / I A1 I A2 = 0.063 / I A2 I A2 = 0.375

How would there be potential and kinetic energy in bonds of molecules? Energy = the ability to do work Kinetic Energy (energy of motion) Potential Energy (stored energy) Potential energy is store in the bonds as they form Kinetic energy occurs as the bonds break and can be used to do work

What property of ATP makes it an energy carrier? The last phosphate bond is considered high energy bond, when hydrolyzed, it releases releases energy. P ADP 2 Phosphate groups

Dehydration- requires energy, releases water Hydrolysis- releases energy, requires water

Explain the Bicarbonate Buffer system: H 2 O + CO 2 H 2 CO 3 HCO 3 - + H + HCO 3 - = Bicarbonate (weak base) H 2 CO 3 = Carbonic acid (weak acid) Major buffer system in blood Maintains blood pH between 7.38 and 7.42 reservoir of H + donate H+ when [H + ] falls absorb H+ when [H + ] rises

Chemistry in Biology activation energy: the minimum amount of energy needed for a reaction. Enzymes lower the activation energy What is activation energy and what does it have to do with enzymes? Exothermic vs endothermic?

MATCHING a. carbohydrate b. lipids c. protein d. nucleic acids 1. contains adenine and thymine 2. lactose 3. chains of amino acids 4. long term energy storage in mammals 5. cholesterol 6. chains of fatty acids and glycerol 7. plant cell walls

Proteins in the Cell What parts of the cell are involved with protein production and what role does each part play? Most of the Endomembrane: Nucleolus Rough ER Golgi Vesicles Ribosomes

Figure 5.9

Movement across the plasma membrane Diffusion Osmosis Facilitated Diffusion Active Transport Passive Transport Requires energy Does not require energy Water High to low concentration gradient Low to high concentration gradient Requires a protein CO2 and O2 Glucose Na + /K + pump Add these to the tree map ( put some in more than one place):

How is plasmolysis different from just a cell in a hypertonic solution? Can salt diffuse through a plasma membrane?

Prokaryote Eukaryote Cells DNA Cytoplasm Nucleus Plasma Membrane Ribosome Cell wall Membrane Bound Organelles Specific in function Complex Multicellular Plants, Animals, fungi No Organelles Mostly single cells Simple Archaea and bacteria both

Plant CellAnimal Cell Eukaryote Nucleus Mitochondria Muscle cells Plasma Membrane Ribosome lysosome centrioles humans Cell wall Chloroplast vacuole Fern ER Cytoplasm Golgi apparatus

Channel Proteins - form small openings for molecules to diffuse through Transport Proteins - regulate movement of substances across membrane Carrier Proteins- binding site on protein surface "grabs" certain molecules and pulls them into the cell Gated Channels - similar to carrier proteins, not always "open"

Plasma Membrane Passive Diffusion H 2 O = Osmosis Facilitated Diffusion Active Transport need energy Noncharged, small particles, CO 2 & O 2, water High to low Low to high Remember lipids are nonpolar

QOD: Compare and contrast photosynthesis and cellular respiration

How are they connected? glucose + oxygen carbon + water + energy dioxide C 6 H 12 O 6 6O 2 6CO 2 6H 2 O ATP +++ Heterotrophs + water + energy glucose + oxygen carbon dioxide 6CO 2 6H 2 O C 6 H 12 O 6 6O 2 light energy +++ Autotrophs making energy & organic molecules from light energy making energy & organic molecules from ingesting organic molecules oxidation = exergonic reduction = endergonic

Pg 129b Light & H 2 O O2O2 CO 2 glucose ATP NADPH ADP NADP ProcessLight?LocationReactantProduct (photosynthesi s) 1. ETC Light dependent rxn Thylakoid membrane Light H 2 O ATP NADPH O 2 2. Calvin Cycle Light Independent rxn stromaATP NADPH CO 2 Glucose Photosynthesis

A = photosystem II B = photosystem I C = H20 D = Electron Transport Chain E = ATP Synthase AB = ATP AC = phospholipids AD = light (energy)

2 NADH 2 ATP net 2 ATP 2 NADH 6 NADH 8 NADH Cellular Respiration Summary 2 FADH 2 6 CO 2 2 ATP net 34 ATP possible - Each NADH produces 3 ATP - Each FADH 2 produces 2 ATP

Breakdown of One Glucose Molecule Calculations from each: NADH= 2 or 3 ATP can be made FADH 2 = 2 ATP can be made 1. Glycolysis Produces: 2 ATP2 NADH 2. Krebs Cycle (including pre-Kreb s ) - Produces: 2 ATP 8 NADH 2FADH 2 Total:4ATP10 NADH 2 FADH 2 3. ETC - Produces:x 3 x 2 30 ATP 4 ATP = 34 ATP Total: 4ATP + 34 ATP + a grand total of 38 ATP ! Theoretical Yield http://www.youtube.com/watch?v=j7g PtASv0SQ http://www.youtube.com/watch?v=0IJ MRsTcwcg For simplicity, however, we look at the theoretical maximum yield of ATP per glucose molecule oxidized by aerobic respiration. Cellular Respiration

Lactic Acid Fermentation Alcoholic Fermentation

Cell Cycle: Control of the Cell Cycle G1 Checkpoint - Check to see if DNA is damaged G2 Checkpoint - Check to see if DNA is replicated properly M Checkpoint - spindle assembly checkpoint, check for alignment of chromosomes G 2 Gap 2 Cell prepares for mitosis G 1 Gap 1 Cell growth and performs normal functions S synthesis DNA is replicated

Control of Cell Cycle Growth Factors: signal cell growth Cyclin: family of internal signaling proteins that increase or decrease the cell cycle (feedback/ homeostasis) P53: stop cell at G1 if there is damage RB: protein responsible for interpreting growth signals and available nutrients Somatic Cells: body cells Apoptosis: programed cell death

Interphase Prophase Metaphase Anaphase Telophase IPMAT

Meiosis is actually TWO divisions, this results in FOUR daughter cells, each with HALF the number of chromosomes. These cells are HAPLOID! - Involves crossing over and independent assortment to increase genetic variation

Goals Genetics practice questions Work on cheat sheet One side of a standard size paper (back of appendix B) Online review Textbook online post tests Lab bench online lab review

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Genetic Crosses Monohybrid cross Dihybrid cross Dominance and incomplete dominance Multiple alleles (blood types) Sex linked (colorblind, MD, hemophilia) Pedigrees Test cross (cross with homozygous recessive)

Key: R = round r = wrinkled P = RR x rr Genotype: 4/4 Rr Genotype ratio: 0:1:0 Phenotype: 4/4 round Phenotype ratio: 4:0 1:0 #1 R R r Rr Rr F1F1 If these two plants had 100 seeds, how many would you expect to be round? Wrinkled? 1000

Key: RR= Red flowers Rr = Pink Flowers rr = White Flowers P = RR x rr Genotype: 4/4 Rr Genotype ratio: 0:1:0 Phenotype: 100% pink Phenotype ratio: 0:1:0 #2 a) R R r Rr Rr F1F1

Key: In pigs, T = curly tailB =brown coat t = straight tailb = white coatComplete dominance P = TtBb x TtBb TB, Tb, tB, tb TB Tb tB tb G TB, Tb, tB, tb X TB, Tb, tB, tb Genotype: 1/16 TTBB 2/16 TTBb 2/16 TtBB 4/16 TtBb 1/16 TTbb 2/16 Ttbb 1/16 ttBB 2/16 ttBb 1/16 ttbb Phenotype 9/16 curly tail & brown coat 3/16 curly tail & white coat 3/16 straight tail & brown coat 1/16 straight tail & white coat Pheno ratio: 9 : 3 : 3 : 1 TTBBTTBbTtBBTtBb TTBbTTbbTtBbTtbb TtBBTtBbttBBttBb TtBbTtbbttBbttbb 1a.

P = AB x OO Genotype: AO : BO Phenotype: Type A; Type B #1 A B O AO BO F1F1 1. Cross: Type AB father with a type 0 mother. What are the possible blood types of the offspring? What are the possible blood types of the offspring? Type A, Type B

Key: X H X H = female normal blood clotting X H X h = female carrier (normal blood clotting) X h X h = female hemophiliac P = X h Y X X H X h Genotype: 1/4 XHXh; XHY; XhXh; XhY Phenotype: female carriers; female hemophiliac; male normal; 1/4 male hemophiliac #2 Xh Y XH XHXh XHY Xh XhXh XhY F1F1 X H Y = male normal X h Y = male hemophiliac What percentage of their girls will be expected to be hemophiliacs? 50% of the girls would be expected to be hemophiliacs.

1. XCY male normal 7. XeY male colorblind 2, XCXe female carrier 8. XCXe female carrier 3. XCY male normal 9. XCXC or XCXe 4. XCXe female carrier 10. XeY male colorblind 5. XCXe female carrier 11. XCY male normal 6, XCY male normal 12. XeXe female colorblind

Name of DisorderType (autosomal, sex linked, dominant, recessive) Description/ Symptoms Type of people group/ treatment / other Tay SacksRecessive Cystic Fibrosisrecessive MDSex linked Sickle Cell Diseaserecessive Achondroplasia Dominant Huntington's Disease Dominant HemophiliaSex linked Hereditary Genetic Disorders

Pg 192 Autosomal Recessive vs Autosomal Dominance aa Aa

There are 2 Types of Nucleic Acids: DNARNA 1. Monomers are nucleotides. 2. Each nucleotide has 3 parts: a. Phosphate group b. Deoxyribose, a 5-carbon sugar b. Ribose, a 5-carbon sugar c. Nitrogenous base: *adenine thymine *adenine uracil *cytosine - guanine 3.Double Stranded3. Single Stranded 4. Only in nucleus4. May leave nucleus Okazaki fragment: a string of nucleotides added to the lagging size at once RNA primer: locates the area where DNA polymerase bind and will start Telomeres: nonsense DNA at end of chromosome to protect from loss of genes

The Central Dogma DNA RNA Amino Acid (Protein)

Figure 14.5 Every 3 bases on mRNA (messenger RNA) is called a CODON Each CODON specifies one AMINO ACID Chains of amino acids are proteins (ex. hemoglobin)

The Process: 1.RNA polymerase attaches to the DNA molecule at the promoter the start signal. The DNA unwinds and unzips in one region, exposing the gene. (-Gene: a piece of DNA that codes for a protein.) 2.Complimentary bases form a new mRNA molecule. 3.RNA polymerase reaches the termination signal or mRNA transcript- the stop signal. 4.The DNA re-winds and re- zips The mRNA can now leave the nucleus.. Protein Synthesis: Transcription

Translation The process: 1. Initiation- mRNA attaches to a ribosome- tRNA attaches to the start codon 2. Elongation-A tRNA with an amino acid joins with mRNA according to complementary base pairing (codons to anticodons) -The amino acid joins the peptide chain by forming a peptide bond. Elongation (cont.)- The tRNA is released into the cytoplasm -The next tRNA is positioned, the polypeptide chain grows. 3. Termination- the process ends when a stop codon is reached (4). Disassembly- the ribosome falls apart and the protein is released.

Repressors versus Inducers repressor operon: turns transcription OFF it is normally on but can be turned off when tyrptophan is present ex: Tryp operon inducer operon: turns transcription ON is normally off but can be turned on when lactose is present ex: lac operon Grammar Time A woman is pregnant and the baby is late. The doctor says they will "INDUCE" labor tomorrow. What does he mean? http://highered.mcgraw- hill.com/sites/0072556781/student_view0/c hapter12/animation_quiz_3.html

Recombinant DNA Genetic Engineering

Generic Cell Signaling Pathway Reception Chemical message (ligand) docks at receptor on cell membrane and changes its shape Transduction switching message from chemical signal received on cell outside to chemical messages on interior of cell Response Signal transduction cascade occurs until end result is reached

Parts of Virus Nucleic acid - Double- or single-stranded DNA or Double- or single-stranded RNA capsid Is the protein shell that encloses the viral genome Envelopes Membranous coverings (derived from the membrane of the host cell) 18 250 mm 7090 nm (diameter) 20 nm50 nm (b) Adenoviruses RNA DNA Capsomere Glycoprotein

Steps of virus production (lytic cycle) 1.Attachment- capsid combines with receptor 2. Penetration - the virus is engulfed by the cell (Cell can enter Lysogenic or Lytic Cycle) 3. Biosynthesis - viral components are made (protein coat, capsid, DNA/RNA) 4. Maturation - assembly of viral components 5. Release - viruses leave host cell to infect new cells (often destroys host) Viruses multiply, or replicate using their own genetic material and the host cell's machinery to create more viruses. Viruses cannot reproduce on their own, and must infect a host cell in order to create more viruses. (See McGraw Hill animation)McGraw Hill animation http://highered.mcgraw-hill.com/sites/007352543x/student_view0/chapter20/entry_of_virus_into_host_cell.html

Bacteria Structure 4. Nucleoid region contains a circular loop of DNA 5. Plasmids are rings of DNA, used in reproduction 6. Ribosomes in cytoplasm synthesize proteins 2. Flagella is used for movement 3. Pilli (Fimbrae) help bacteria cling to surfaces (cilia) (Prokaryotes do not have organelles or a membrane bound nucleus!) 1. Outside the plasma membrane of most cells is a rigid cell wall that keeps the cell from bursting or collapsing

Prokaryotic Nutrition Bacteria differ in their need for, and tolerance of, oxygen (O 2 ). a. Obligate anaerobes: no O 2 are unable to grow in the presence of O 2 ; this includes anaerobic bacteria that cause botulism, gas gangrene, and tetanus. b. Facultative anaerobes: O 2 optional are able to grow in either the presence or absence of gaseous O 2. c. Aerobic organisms: need O 2 (including animals and most prokaryotes) require a constant supply of O 2 to carry out cellular respiration. staphylococcus is a gram-positive, facultative anaerobe

Autotrophic Prokaryotes a. Photoautotrophs are photosynthetic and use light energy to assemble the organic molecules they require. b. Chemoautotrophs make organic molecules by using energy derived from the oxidation of inorganic compounds in the environment. Reduce CO 2 by oxidizing ammonia, nitrites and nitrates (nitrogen fixing bacteria),

Natural selection is a mechanism of evolution! Natural Selection: organisms best adapted to their environment tend to survive and transmit their genetic characteristics in increasing numbers Evolution: gradual change in a species over time

Natural Selection in 4 Basic Steps 1. Heritable variation 2. More individuals are produced than can improve survival 3. Some individuals have traits (adaptations)that help them survive & reproduce 4. Survivors pass those traits to their offspring (increases the number that have the trait) * Results in a population adapted to its environment

Types of Selection

Evidence of Evolution The theory of evolution is supported by evidence that can found in 1. The Fossil Record: traces the history of life 2. Biogeography: study of range and distribution of plants and animals 3. Anatomy: homologous, vestigial structures 4. Embryology: all vertebrates have the same basic pattern of development 5. Biochemistry: DNA, amino acids are similar I related organisms, all life of same few elements

5 Agents of evolutionary change MutationGene Flow Genetic Drift Natural Selection Non-random mating

MACROEVOLUTION Speciation the splitting of one species into two or more species Allopatric Speciation Sympatric Speciation PREZYGOTIC Isolation: prevent a zygote from forming POSTZYGOTIC Isolation: prevent reproduction after zygote formation

Phylogeny TAXA Lancelet (outgroup) Lamprey Salamander Leopard Turtle Tuna Vertebral column (backbone) Hinged jaws Four walking legs Amniotic (shelled) egg CHARACTERS Hair (a) Character table Hair Hinged jaws Vertebral column Four walking legs Amniotic egg (b) Phylogenetic tree Salamander Leopard Turtle Lamprey Tuna Lancelet (outgroup) 0 00 0 0 0 00 0 0 00 000 1 11 1 1 1 1 11 1 1 11 11 An outgroup is a species or group of species that is closely related to the ingroup, the various species being studied Systematists compare each ingroup species with the outgroup to differentiate between shared derived and shared ancestral characteristics Homologies shared by the outgroup and ingroup are ancestral characters that predate the divergence of both groups from a common ancestor

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AP Bio review. AP Bio Test he AP Biology Exam is approximately three hours long and has two sections. Section 1 is Multiple Choice and Grid-in questions