Ap Biology Formulas and Equations Sheet, Cheat Sheet of Biology

Download Ap Biology Formulas and Equations Sheet and more Cheat Sheet Biology in PDF only on Docsity! AP BIO EQUATIONS AND FORMULAS REVIEW SHEET #1 Formulas: Mode = value that occurs most frequently in a data set Median = middle value that separates the greater and lesser halves of a data set Mean = sum of all data points divided by the number of data points Range = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum) Standard Deviation = 1 )( 2   n xxi where x = mean and n = size of the sample Example problem: One of the lab groups collected the following data for the heights (in cm) of their Wisconsin Fast Plants: 5.4 7.2 4.9 9.3 7.2 8.1 8.5 5.4 7.8 10.2 Find the mode, median, mean, and range. Show your work where necessary. 4.9 5.4 5.4 7.2 7.2 7.8 8.1 8.5 9.3 10.2 Mode:__5.4, 7.2 ___ Median = average of middle 2 values = (7.2 + 7.8) / 2 = 7.5 Median:____7.5_____ Mean = 4.7 10 74 10 2.103.95.81.88.72.72.74.54.59.4   Mean:______7.4____ Range = 10.2 – 4.9 = 5.3 Range:_____5.3_____ Find the standard deviation by filling in the following table. Heights (x) Mean ( x ) xx  2)( xx  5.4 7.4 -2 4 7.2 7.4 -.2 .04 4.9 7.4 -2.5 6.25 9.3 7.4 1.9 3.61 7.8 7.4 -.2 .04 8.1 7.4 .7 .49 8.5 7.4 1.1 1.21 5.4 7.4 -2 4 7.8 7.4 .4 .16 10.2 7.4 2.8 7.84 27.64  2)( xx  Standard deviation: 9 64.27 = 1.75 Interpret the standard deviation in the context of the problem. The heights of Wisconsin Fast Plants in this sample are, on average, is about 1.75 cm away from the mean height of 7.4 cm. AP BIO EQUATIONS AND FORMULAS REVIEW SHEET #2 Formulas: Chi Square e eo 2 2 )(   o = observed individuals with observed genotype e = expected individuals with observed genotype Degrees of freedom equals the number of distinct possible outcomes minus one Degrees of Freedom p 1 2 3 4 5 6 7 8 0.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51 0.01 6.64 9.32 11.34 13.28 15.09 16.81 18.48 20.09 Example problem: Wisconsin Fast Plants have two very distinctive visible traits (stems and leaves). Each plant will either have a purple (P) or green (p) stem and also have either have green (G) or yellow (g) leaves. Suppose that we cross a dihybrid heterozygous plant with another plant that is homozygous purple stem and heterozygous for the leaf trait. Make a Punnett square to figure out the expected ratios for the phenotypes of the offspring. PpGg x PPGg PG Pg PG Pg PG PPGG PPGg PPGG PPGg Pg PPGg PPgg PPGg PPgg pG PpGG PpGg PpGG PpGg pg PpGg Ppgg PpGg Ppgg Purple stem/Green leaves 12 Purple stem/Yellow leaves 4 so the expected ratio of the phenotypes is 12:4 or 3:1. Suppose a class observed that there were 234 plants that were purple stem/green leaves and 42 that were purple stem/yellow leaves. Does this provide good evidence against the predicted phenotype ratio? Find expected values. 234 + 42 = 276 total. 207 4 3 *276  69 4 1 *276  df= n – 1 = 1 o e o – e e eo 2)(  234 207 27 3.52 42 69 -27 10.57 so X 2 = 3.52 + 10.57 = 14.09 Because 14.09 > 3.84, the data did not occur purely by chance. Using your understanding of genetics, what might be one reason why the class got these results? One of the conditions must have been violated, probably the random or independent condition. This could lead you to a test cross for linkage, of which you could check map units. But most likely, sample size is the culprit. AP BIO EQUATIONS AND FORMULAS REVIEW SHEET #4 – Answer Key Formulas: Rate Population Growth Exponential Growth Logistic Growth dY/dt dN/dt = B – D Nr dt dN max         K NK Nr dt dN max dY = amount of change B = birth rate D = death rate N = population size K = carrying capacity rmax = maximum per capita growth rate of population Notes dt dN = t N   = timeinchange sizepopulationinchange = population growth rate Example 1: There are 300 falcons living in a certain forest at the beginning of 2013. Suppose that every year there are 50 falcons born and 30 falcons that die. a. What is the population growth rate (include units)? Interpret the value. dN/dt = B – D = 50 – 30 = 20 falcons/year. Each year the falcon population will increase by 20 falcons. b. What is the per capita growth rate of the falcons over a year? Interpret the value. Nr dt dN max so 20 = rmax 300 so rmax = .0667 The falcon population will increase by 7% each year. c. Fill in the table and construct a graph. Year Population 2013 300 2014 330 2015 363 2016 399.3 2017 439.23 2018 483.153 d. Find the average rate of change for the falcon population from 2013 to 2018 (include units). Interpret the value. We have two data points: (2013, 300) and (2018, 483). Average rate of change = slope 6.36 20132018 300483        x y slope falcons/year. Over the past five years, the falcon population has increased by 36.6 per year on average. Example 2: Kentwood, Michigan had a population of 49,000 in the year 2013. The infrastructure of the city allows for a carrying capacity of 60,000 people. rmax = .9 for Kentwood. a. Is the current population above or below the carrying capacity? Will the population increase or decrease in the next year? Current population is below the carrying capacity so we would expect the population to increase. b. What will be the population growth rate for 2013 (include units)?         K NK Nr dt dN max =        000,60 000,49000,60 )000,49)(9.0( 8085 people/year c. What will be the population size at the start of 2014. 49,000 + 8085 = 57,085 people d. Fill in the following table: Year Population size Population growth rate 2013 49,000 8085 2014 57,085 2496.041625 2015 59,581 374.4296457 2016 59,955 40.04611425 2017 59,996 4.034052134 e. What happened to the population size over the years? What happened to the population growth rate over the years? Population size got closer and closer to the carrying capacity of 60,000. Population growth rate got smaller and smaller. f. Explain your answer from part (e) using what you know about carrying capacity. As the population gets closer and closer to the carrying capacity, the resources of the environment start to deplete and the growth rate slows. Limiting factors take over (especially density-dependent ones, space, food, water, shelter, etc) g. Explain your answer from part (e) using the formula:         K NK Nr dt dN max Look at this part of the formula:        K NK As the size of the population (N) gets closer and closer to the carrying capacity (K), then this part of the formula gets smaller and smaller, thus reducing population growth rate. Q10, Dilution, pH Review Key 1. The rate of metabolism of a certain animal at 10ºC, is 27 lO2 g -1h-1. What are its rates of metabolism at 20, 30, and 40 ºC if the Q10 is 2? If it is 2.5? ((T2T1)/10)) R2 R1 x Q10 R2 = 27 * 2((20 - 10)/10) = (27 x 21) = 54 lO2 g-1h-1 Temperature ºC Rate2 if Q10 = 2 20 R2 = (27 x 21) = 54 lO2 g -1h-1. 30 R2 = (27 x 22) = 108 lO2 g -1h-1. 40 R2 = (27 x 23) = 216 lO2 g -1h-1. Temperature ºC Rate2 if Q10 = 2.5 20 R2 = (27 x 2.51) = 67.5 lO2 g -1h-1. 30 R2 = (27 x 2.52) = 168.75 lO2 g -1h-1. 40 R2 = (27 x 2.53) = 421.875 lO2 g -1h-1. graph showing the effect of Temp on Rx rate 2. The following table reports the rates of metabolism of a species at a series of ambient temperatures: Temperature (ºC) Rate of Metabolism (lO2 g -1h-1.) 15 10 20 13.42 30 21.22 (a) Calculate the Q10 values for each temperature interval (10/(T2T1)) Q10 {R2/R1} Interval (15-20) = (13.42/10)10/(20-15) = 1.3422 = 1.8 Interval (20-30) = (21.22/13.42)10/(30-20) = 1.581 = 1.58 Interval (20-30) = (21.22/10)10/(30-15) = 2.122(2/3) = 1.6 Surface area to Volume and Water Potential Review 1) Cells throughout the world have variable shapes and sizes. Because of this, and because structure is designed around function, certain shapes are optimal for certain processes. Analyze the following cells (units not to scale), and determine the following… Cell 1 (spherical) where the radius is 3 mm Cell 2 (flat and rectangular) where the height is 0.5mm, length is 4mm, width is 2mm A) What is the surface area to volume ratio of both cells? How to calculate Surface Area? Surface area How to calculate Volume? Volume Surface area to Volume Ratio Cell 1 = 4π r 2 4π(3) 2 = 113 mm 2 4/3 π r 3 4/3 π (3) 3 = 113mm 3 133/133 = 1:1 Cell 2 = Σa 16 + 4 + 2 = 22 mm 2 L x W x H 0.5 x 2 x4 = 4 mm 3 22:4 = 11:2 B) Conclusion: Compare the ratios and explain why one cell would be more efficient than another. Cell 2 has a much higher surface area to volume ratio, hence it would be more efficient as substances could diffuse in and out faster, allowing for quicker chemical processes. C) Are you made of lots of large cells or lots of small cells? Why? How do you actually grow in height? Made of small cells. They have a high surface area to volume ratio, making communication between them, and processes within them much more efficient. You grow in height by making MORE cells. While cells will grow in size, they will reach a limit of efficiency, and then perform mitosis. 2) Water potential in potato cells was determined in the following manner. The initial masses of six groups of potato cores were measured. The potato cores were placed in sucrose solutions of various molarities. The masses of the cores were measured again after 24 hours. Percent changes in mass were calculated. The results are shown below. Molarity of Sucrose in Beaker Percent Change in Mass 0.0 M 18.0 0.2 5.0 0.4 -8.0 0.6 -16.0 0.8 -23.5 1.0 -24.0 Graph these data to the right of the table. From your graph, label where the cells were hypotonic and the solution was hypertonic, and vice versa. Determine the apparent molar concentration (osmolarity) of the potato core cells. Osmolarity = 0.3M cell and beaker are isotonic Above 0.3M cells are hypotonic while beaker is hypertonic (as cells are gaining mass = water is moving in) Under 0.3M cells are hypertonic and beaker is hypotonic (as cells are losing mass = water is moving out) Looking at the water potential equation, Pressure potential is always (positive/negative), while solute potential is always (positive/negative). When Solution potential goes down (gets more negative), water potential DECREASES When Pressure potential goes down (gets smaller), water potential DECREASES When would the pressure in a cell rise? (Under what conditions?) When a cell is in a hypotonic environment, where water is entering the cell (hence building up pressure) What would happen to the solute potential when Concentration is increased (justify with equation)? WHY? When Concentration is INCREASED Solute potential would go DOWN. –iCRT (when C is increased, that is a bigger multiplier to the negative equation) What would happen to the solute potential when Temperature is increased (justify with equation)? WHY? When Temperature is INCREASED Solute potential would go DOWN. –iCRT (when T is increased, that is a bigger multiplier to the negative equation) What would happen to the solute potential when the dissolved substance is glucose vs. salt (justify with equation)? WHY? When Ionization constant is INCREASED Solute potential would go DOWN. –iCRT (when i is increased, that is a bigger multiplier to the negative equation) Salt (i=2) glucose (i=1) Why is water potential important for plants? What are they lacking? Allows for the movement of materials through the organism. Drives water up the plant through xylem by transpiration and cohesion-tension theory. Also builds up pressure for translocation of sugar through phloem. While plants have a vascular system, they lack the muscular pump (heart) to move materials and create pressure for them. Predict what would happen to animal cells placed in 0.0M and 1.0M concentration solutions. 0.0M solution = the cell would swell and possibly lyse from the pressure. 1.0M solution = the cell would shrivel and become very inefficient. ΔG = ΔH - T ΔS What is Entropy? = a measurement of Disorder/randomness When ΔS is positive this means there is MORE Disorder/randomness When ΔS is negative this means there is LESS Disorder/randomness What is ΔH? = a measurement of ENTHALPY When ΔH is positive this means the reaction is ENDOTHERMIC (taking in energy) When ΔH is negative this means the reaction is EXOTHERMIC (releasing energy) What is Gibbs Free energy? = a measurement of SPONTANEITY When ΔG is positive this means the reaction will happen IF ENERGY IS ADDED When ΔG is negative this means the reaction will happen SPONTANEOUSLY ΔG (Joules) ΔH (Joules) T (Kelvin) ΔS (J/K) -500 1000 300 5 -400 1100 300 5 -300 1200 300 5 -200 1300 300 5 -100 1400 300 5 0 1500 300 5 100 1600 300 5 200 1700 300 5 300 1800 300 5 400 1900 300 5 What happens to ΔG when ΔH goes up ? WHY? ΔG INCREASES when ΔH INCREASES. This happens because the reaction needs more and more energy to occur (more endothermic), and therefore will not happen as spontaneously. What happens to ΔG when ΔH goes down ? WHY? ΔG DECREASES when ΔH DECREASES. This happens because the reaction needs less and less energy to occur (more exothermic), and therefore becomes more and more likely to happen spontaneously. Figure 1 Coupling doesn't occur all by itself. In this example, if this experiment were set up so that the ATP would have to be hydrolyzed in one tube and the glucose phosphorylated in another, no coupling would be possible. Coupling can occur only when the partial reactions are part of a larger system. In this example, coupling occurs because both partial reactions are carried out by the enzyme hexokinase. In other cases, coupling can involve membrane transport, transfer of electrons by a common intermediate, or other processes. Another way of stating this principle is that coupled reactions must have some component in common. The “orderliness” of your body is not favored by free energy. Explain (in terms of free energy and disorder) why you need to perform digestion? Energy coupling. You digest food (natural tendency toward disorder) to harness that energy to build ATP (ordered), of which the breaking of ATP (natural tendency toward disorder), can be used to order your body. Why does decomposition of a dead animal happen in terms of energy? What would happen if we increase temperature? Why do we freeze food? Decomposition is the breakdown of ordered cells into disorder. It should happen naturally ΔG is negative. If you increase temperature, ΔG becomes more negative. So increasing temperature will speed up decomposition. We freeze our food to LOWER TEMPERATURE, which also increases ΔG. This slows down decomposition. Explain why plant cells need light to build sugar (in terms of energy). Building sugar (ordering molecules) would not happen spontaneously. Energy must be added (aka sunlight). Primary Productivity – The rate at which organic materials are stored 6CO2 + 6H2O → C6H12O6 + 6O2 One can determine Primary Productivity by measuring dissolved oxygen in the water (as it is hard to measure it in the air) Conversion Factors: 1 ml of O2 = .536 mg of Carbon assimilated To convert: ppm O2 = mg O2/L mg O2/L x 0.698 = ml O2/L ml O2/L x 0.536 = mg carbon fixed/L Fill in the table and Graph Net and Gross Productivity vs % of light Using your data table, what seems to be the trend as the % of light decreases? WHY? As % of light decreases, productivity decreases resulting in less carbon assimilated. Less light = Less photosynthesis = Less carbon dioxide used = Less oxygen released into the water Using your data table, what seems to be the trend as the % of light increases? WHY? As % of light increases, productivity increases resulting in more carbon assimilated. More light = more photosynthesis = more carbon dioxide used = more oxygen released into the water Where would you say this organism is using as much energy as they are making? WHY? When Net = 0 (about 9%) This is where photosynthesis and respiration of the plant seem to be equal. No net oxygen is being released into the water, what is being released is equally being used by the plant in respiration. Using your table and graph, explain why most of the time there are bigger plants on land than in the sea? Explain this in terms of evolution. Plants evolved to live on land because it is easier to acquire carbon dioxide (more abundant), and there is more light available as it doesn’t have to pass through water. % light DO (mg O2/L) Gross PP = Bottle – Dark (mg O2/L) Net PP = Bottle – Light (mg O2/L) Gross Carbon fixed in mgC/L Gross PP x 0.698 x 0.536 Initial 8.4 -- -- -- Dark 6.2 -- -- -- 100% 10.2 4 1.8 1.50 65% 9.7 3.5 1.3 1.31 25% 9.0 2.8 0.6 1.05 10% 8.5 2.3 0.1 0.86 2% 7.1 0.9 -1.3 0.34