AP Biology Fall Semester Review 2014

Unit 0: Science Skills/Experimental Design

 Graphing o Graph should have: . Title – should be descriptive of the x axis, y axis, and key (if applicable) . X-axis – labeled with units . Y-axis – labeled with units . Key (if applicable) o Rate = y/x (x = time) Ex: photosynthesis rate = # of disks floating/time o Bar graphs – noncontinuous data, distinct components (ex: colors) o Line Graphs – continuous data (ex: time, pH level)  Components of a controlled experiment o Tests only ONE variable . Ex: in photosynthesis lab, if testing light, one group uses light, the other group does not, but EVERY OTHER PART OF THE EXPERIMENT IS THE SAME o Must have a control (baseline for comparison) . Ex: How does exercise affect heart rate? – your control would be the heart rate at rest o How do you increase validity of data? . Large sample size – ex: 100 pillbugs . Many number of trials – ex: 5 trials . The experiment should be repeatable and obtain the same/similar results  Chi-squared analysis o Using pillbug lab as an example – after experiment you find that 47 pillbugs are on the side with sugar water and 26 pillbugs are on the side with water o Null Hypothesis: There is no significant difference between observed and expected values OR Pillbugs have no preference for sugar water vs. regular water OR Our results/data was due to random chance. o Observed value for sugar water = 47, Observed value for regular water = 26. Total # pillbugs = 73 o Expected value for sugar water = 36.5, Expected value for regular water = 36.5 o Chi-square value = 6.02 o 1 degree of freedom (df) o Using the chart, our chi-square value is more than the critical value of 3.84 at p = 0.05, therefore we reject our null hypothesis. Our chi-square value means that there is <5% chance that our data was due to random chance. According to the data, pillbugs DO have a preference for sugar water over regular water. o Example #2: There are 3 doors for a building and 1 door (Door #3) is always breaking because we think that more people use it because it is an automatic door. Data for how many people go through the doors is taken for a day. The results are Door #1: 60 people, Door #2: 66 people, Door #3: 80 people. Find the chi square value to determine if Door #3 is more popular than the other two doors. . Null hypothesis: There is no significant difference between observed and expected values OR There is no difference between the 3 doors (Door #3 is not more popular than the other two doors.) . Expected values for each door will be 206/3 = 68.667 . Chi-square = 3.06 . 2 degrees of freedom . using the chart, our chi-square value is less than the critical value of 3.84 at p = 0.05, therefore we accept our null hypothesis. This means that there is a 10-25% probability that we got our data by random chance and door #3 is actually not more popular than the other two doors

Unit 1: Biochemistry

 Chemical bonds – from weakest to strongest: van der waals/hydrophobic interactions, Hydrogen bonds, ionic bonds, covalent bonds  Nonpolar vs. Polar functional groups o Non-polar (hydrophobic) . Hydrocarbon chains, methyl groups (-CH3) o Polar (hydrophilic) – mostly because they contain oxygen which is HIGHLY electronegative (this also is important to know with the ETC of respiration – oxygen draws the electrons through the ETC because it is SO electronegative.) . Hydroxyl (-OH), Carboxyl (-COOH), Carbonyls (C=O)  Properties of Water o All properties of water are because of its polarity, which allows water to form hydrogen bonds (make sure you know which part of water is slightly negative and which is slightly positive. Also make sure you know how to draw a hydrogen bond between two water molecules.) o Properties: . Universal solvent – many things can dissolve in water . Cohesion (water binding with water) and adhesion (water binding to other molecules) – relates to why water has high surface tension (it is hard to break the surface of water) . High specific heat – takes a lot of energy to raise the temperature of water . High heat of vaporization – takes a lot of energy to break the hydrogen bonds holding water together, “hard to evaporate”, important for living things because water keeps organisms cool by evaporative cooling . Ice floats – the density of solid water is higher than the density of liquid water  pH o Acids donate H+ (acidic solutions have a high concentration of H+) . Ex: carboxyl group (study the structure – why is it an acid?) o Bases accept H+ (basic solutions have a lower concentration of H+) . Ex: amino group (study the structure – why is it a base?)  Functional groups – you need to be able to recognize which ones are polar/nonpolar, acid/base  Macromolecules are many monomers that are combined together to form polymers o Dehydration (condensation, synthesis) reaction – anabolic and endergonic o Hydrolysis (digestion) reaction – catabolic and exergonic a. Carbohydrates – CHO 1:2:1 ratio, monomer=monosaccharides, 2=disaccharides, 3 or more=polysaccharides  Used for energy (cell respiration)  Examples (1) glucose- immediate energy to make ATP (2) starch- stored energy in plants (3) glycogen- stored energy in animals (stored in liver) (4) cellulose- plant cell wall

b. Lipids – C, H, O (not a 1:2:1 ratio) *P only in phospholipids (1) fats, waxes, oils and sterols (2) Saturated fats have single bonds between carbons, unsaturated fats have at least one double bond between carbons (kinky); plants make polyunsaturated; animals make monounsaturated (3) Phospholipids make up cell membranes (double layer) and are amphipathic- hydrophilic and hydrophobic (4) Uses- in all membranes; stored energy, protection, insulation, myelin sheath of nerves

c. Proteins - C, H, O, N (may have other elements in R group) (1) Monomer- amino acids (20 total types), 2=dipeptide, 3 or more= polypeptide (2) Parts of amino acid= carboxyl group (COOH) on one end, amino group on the other end (NH2), central carbon and variable R group (can be hydrophobic or hydrophilic) which determines chemical properties. (3) Protein Folding- shape determines function; primary= a.a. chain; secondary= beta pleated sheet or alpha helix( hydrogen bonds); tertiary=globular; folds in on itself (disulfide bridges, hydrogen bonds, hydrophobic interactions; ionic bonding); quartenary= more than one polypeptide. (4) Uses- protein carriers in cell membrane, antibodies, hemoglobin, enzymes, most hormones

d. Nucleic acids – C, H, O, N (1) Monomer= nucleotide, 2 = dinucleotide, 2 or more polynucleotide (2) Nucleotide made up of sugar, phosphate and base (3) Used to store genetic information (4) DNA is double stranded, has deoxyribose, A, G, C, T (5) RNA is single stranded, has ribose, A, G, C, U (6) mRNA- copies genetic message; rRNA- attaches mRNA and makes up ribosomes (most common);tRNA- carries amino acids; DNA- carries genetic code

Unit 2: Cells Review the notes from the review session before Unit 2 exam – make sure you review water potential (what it means and how you calculate it)

Unit 3: Viruses, Immune System, Cell Signaling  Viruses o Structure: capsid (made of protein) + nucleic acid (either DNA or RNA) o Animal viruses may also have a membrane (made of phospholipids and proteins – derived from the host cell) surrounding the capsid o Bacterial viruses are called barteriophages – or just phages . Lytic cycle – destroy host cell . Lysogenic cycle – nucleic acid of virus is integrated with host genome and replicates along with the host genome. Then, environmental signals could cause the viral genome to come out of the host genome and enter the lytic cycle (which will then destroy host cell)  Immune System – READ THE PDF POSTED IN RESOURCES FOR FINAL o Non-specific/Innate . 1st line of defense – skin, mucous, stomach acid . 2nd line of defense – inflammatory response – white blood cells (phagocytes) and fever o Specific/Adaptive . 3rd line of defense – B cells and T cells  B cells – make antibodies  T cells – attack infected cells o Make sure you know additional vocabulary – antibody, antigen, leukocytes, helper T cells  Cell Signaling – READ THE CELL COMMUNCATION OVERVIEW HAND OUT o Reception – signal/ligand binds to receptor protein (very specific), causes conformational change that will start transduction o Transduction – proteins or second messengers (small molecules, not proteins) pass signal along, usually there are conformational changes in protein, phosphorylation is a way that these proteins become activated. One key advantage – amplification o Response – there is a cellular response – usually a transcription factor is activated that turns transcription of a certain gene on