Download Ap Biology Formulas and Equations Sheet and more Cheat Sheet Biology in PDF only on Docsity!
AP BIO EQUATIONS AND FORMULAS REVIEW SHEET
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
xi x 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.
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 = 7. 4 10
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 ) x^ x ( x x )^2 5.4 7.4 - 2 4 7.2 7.4 - .2. 4.9 7.4 - 2.5 6. 9.3 7.4 1.9 3. 7.8 7.4 - .2. 8.1 7.4 .7. 8.5 7.4 1.1 1. 5.4 7.4 - 2 4 7.8 7.4 .4. 10.2 7.4 2.8 7. 27.64 ( x x )^2
Standard deviation: 9
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
Formulas:
Chi Square e
2 ( o^^ e )^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. 0.0 1 6.64 9.32 11.34 13.28 15.09 16.81 18.48 20.
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
276 * df= n – 1 = 1
o e o – e e
( o e )^2
42 69 -27 10.57 so X^2 = 3.52 + 10.57 = 14.
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
Formulas: p^2 + 2pq + q^2 = 1 p = frequency of the dominant allele in a population p + q = 1 q = frequency of the recessive allele in a population
Example problem: In 1990 the East Kentwood High School student body was made up of 90% right handed students. Being right handed (R) is the dominant trait over being left handed (r).
a. What is p and q for the population of 1990 East Kentwood High School students. Interpret each.
90% right handed means 10% are left handed. q^2 = .10 so q = .316. p + q = 1 so p = .684. p = .684 68.4% of all the alleles at EKHS in 1990 are R. q = .316 31.6% of all the alleles at EKHS in 1990 are r.
b. Find the percent of the student body in 1990 that are homozygous right handed, heterozygous right handed, and left handed.
p^2 = (.684)^2 = .468 so 46.8% of students are homozygous right handed (RR) 2pq = 2(.316)(.684) = .432 so 43.2% of students are heterozygous right handed (Rr) q^2 = (.316)^2 = .10 so 10% of students are left handed (rr)
Fast forward to today at East Kentwood. Mr. V. took a random sample of 100 East Kentwood students today and found that 18 of them were left handed.
c. What are the new p and q values? How do they compare with the values from 1990?
q^2 = 18/100 = .18 so q = .424. p + q = 1 so p =. p has decreased and q has increased. There are more recessive alleles in the population today.
There are many reasons why this apparent change could have occurred. Come up with an East Kentwood example for each:
Large Sample Size: Maybe Mr. V’s sample size is too small to accurately predict the proportion of the population that is left handed. Maybe his small sample just selected many lefties purely by chance.
Random Mating: Women are more naturally attracted to lefties and have had more babies with left handed males than right handed males since 1990.
Mutations: The chemicals in Mr. V’s room are modifying student’s brain structure, causing right handers to mutate and become lefties.
Gene Flow: Lefties are generally smarter than righties and love AP classes, so lefties from other school districts have been moving to Kentwood because of the strong AP program here.
Natural Selection: Left handed people are better drivers and are much less likely to die in a car accident. Car accidents have been killing off the right handed students since 1990.
AP BIO EQUATIONS AND FORMULAS REVIEW SHEET #4 – Answer Key
Formulas: Rate Population Growth Exponential Growth Logistic Growth
dY/dt dN/dt = B – D r N dt
dN max
^
K
K N
r N 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
changeintime
changein populationsize = 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.
r N 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. 2017 439. 2018 483.
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 2018 2013
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.
Q10, Dilution, pH Review Key
- The rate of metabolism of a certain animal at 10ºC, is 27 lO 2 g-1h-1. What are its rates of metabolism at 20, 30, and 40 ºC if the Q 10 is 2? If it is 2.5?
(( T (^2) T 1)/10))
R 2 R 1 x Q 10
R 2 = 27 * 2((20 - 10)/10)^ = (27 x 2^1 ) = 54 lO 2 g-1h-
Temperature ºC Rate2 if Q10 = 2 (^20) R 2 = (27 x 2^1 ) = 54 lO 2 g-^1 h-^1. (^30) R 2 = (27 x 2^2 ) = 108 lO 2 g-^1 h-^1. (^40) R 2 = (27 x 2^3 ) = 216 lO 2 g-^1 h-^1.
Temperature ºC Rate2 if Q10 = 2. (^20) R 2 = (27 x 2.5^1 ) = 67.5 lO 2 g-^1 h-^1.
30 R 2 = (27 x 2.5^2 ) = 168.75 lO 2 g-^1 h-^1. (^40) R 2 = (27 x 2.5^3 ) = 421.875 lO 2 g-^1 h-^1. graph showing the effect of Temp on Rx rate
- The following table reports the rates of metabolism of a species at a series of ambient temperatures: Temperature (ºC) (^) Rate of Metabolism (lO 2 g-^1 h-^1 .) 15 10 20 13. 30 21.
(a) Calculate the Q 10 values for each temperature interval
(10/( T 2 T 1))
Q 10 { R 2 / R 1 }
Interval (15-20) = (13.42/10)10/(20-15)^ = 1.342^2 = 1.
Interval (20-30) = (21.22/13.42)10/(30-20)^ = 1.58^1 = 1.
Interval (20-30) = (21.22/10)10/(30-15)^ = 2.122(2/3)^ = 1.
(b) Within which temperature interval (15-20 or 20-30) is the rate of metabolism most sensitive to temperature change?
SLIGHTLY MORE SO BETWEEN 15 AND 20 SINCE THE Q10 IS LARGER
(c) For this species, would a Q 10 calculated for 15 to 30 ºC be as useful as several for smaller temperature ranges? Calculate that Q 10 as part of your answer.
YES, IT AGREES REASONABLY CLOSELY WITH VALUES (I.E. THE VALUES DO NOT DIFFER MUCH FROM EACH OTHER
- The reaction rate for a certain process at 14 ºC is 15 units / time. (a) What would be the reaction rate at 20 ºC if the Q 10 = 1?
15 UNITS/TIME -- A FLAT Q10 BY DEFINITION IS TEMPERATURE INDEPENDENCE (1 to any power is still 1)
Surviving C 1 V 1 = C 2 V 2
C 1 = original concentration of the solution, before it gets watered down or diluted.
C 2 = final concentration of the solution, after dilution.
V 1 = volume about to be diluted
V 2 = final volume after dilution
By drawing the "X" through the equal sign and filling in the formula with letters of a size permitted by the borders of the "X", it reminds you that :
for all dilution problems C 1 > C2, and V 1 < V 2.
It makes sense because to dilute, we add water. This increases the volume but lowers concentration.
Examples by Type :
- Easiest: Joe has a 2 g/L solution. He dilutes it and creates 3 L of a 1 g/L solution.
How much of the original solution did he dilute?
Solution: We are looking for V 1 :
C 1 V 1 = C 2 V 2
2x = 1(3)
x = 1.5 L
Surface area to Volume and Water Potential Review
- 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 = 4 π r^2 4 π(3) 2 = 113 mm^2 4/3 π r^3 4/3 π (3)^3 = 113mm^3 133/133 = 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:
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.
- 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. 0.2 5. 0.4 -8. 0.6 -16. 0.8 -23. 1.0 -24.
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 200^1700 300 150 1700 310 5 100 1700 320 5 50 1700 330 5 0^1700 340
- 50 1700 350 5 - 100 1700 360 5 - 150^1700 370 - 200 1700 380 5 - 250 1700 390 5
What happens to ΔG when T goes up? WHY?
ΔG DECREASES when T INCREASES. AN INCREASE IN TEMPERATURE INCREASES KINETIC ENERGY OF MOLECULES MAKING IT MORE LIKELY THAT SPONTANEOUS COLLISIONS AND REACTIONS WOULD TAKE PLACE.
What happens to ΔG when T goes down? WHY?
ΔG INCREASES when T DECREASES. AN DECREASE IN TEMPERATURE DECREASES KINETIC ENERGY OF MOLECULES MAKING IT LESS LIKELY THAT SPONTANEOUS COLLISIONS AND REACTIONS WOULD TAKE PLACE.
ΔG ΔH T ΔS 6000 7500 300 5 4500 7500 300 10 3000^7500 300 1500 7500 300 20 0 7500 300 25
- 1500^7500 300 - 3000 7500 300 35 - 4500 7500 300 40 - 6000 7500 300 45 - 7500^7500 300
What happens to ΔG when ΔS goes up? WHY?
ΔG DECREASES AS ΔS INCREASES. AS DISORDER INCREASES (SINCE THAT IS THE NATURAL TENDANCY OF PROCESSES) SPONTANEITY WILL ALSO BECOME MORE LIKELY.
What happens to ΔG when ΔS goes down? WHY?
ΔG INCREASES AS ΔS DECREASES. AS DISORDER DECREASES (SINCE THAT IS NOT THE NATURAL TENDANCY OF PROCESSES) SPONTANEITY WILL BECOME LESS LIKELY.
Biochemical free energies are usually given as standard free energies of hydrolysis. For example, the hydrolysis of glucose-6-phosphate:
has ΔG° = -4.0 kcal/mole (-16.5 kJ/mole) under standard conditions. Therefore, the opposite reaction, the phosphorylation of glucose, is unfavored. However, the phosphorylation of glucose occurs readily in the cell, catalyzed by the enzyme hexokinase:
The other half of the phosphorylation reaction is the hydrolysis of ATP to yield ADP and inorganic phosphate (Pi):
under standard conditions has ΔG° = -7.3 kcal/mole (-31 kJ/mole).
The standard free energy change of the reaction can be determined by adding the two free energies of reaction:
Note that the reaction as written is unfavored; its free energy change is positive. Another way of stating this is that the reaction is endergonic, that is, the reaction involves a gain of free energy.
For the exergonic hydrolysis of ATP (the reaction involves a loss of free energy):
The two reactions are summed:
This is a simple example of energetic coupling, where an unfavorable reaction is driven by a favorable one, as shown in Figure 1.
Primary Productivity – The rate at which organic materials are stored 6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2
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 O 2 = .536 mg of Carbon assimilated To convert: ppm O 2 = mg O 2 /L mg O 2 /L x 0.698 = ml O 2 /L ml O 2 /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 O 2 /L)
Gross PP = Bottle – Dark (mg O 2 /L)
Net PP = Bottle – Light (mg O 2 /L)
Gross Carbon fixed in mgC/L Gross PP x 0.698 x
Initial 8.4 -- -- --
Dark 6.2 -- -- --
