1. Review of AP Formulas

2. REVIEW
We have covered most of the formulas this year and this is your ONE shot review of the entire list
In the next couple of days,we are going to review and practice using each of the formulas

3. Water Potential (Ψ)
Ψ = ΨP + ΨS
ΨS = -iCRT

4. Water Potential (Ψ) Ψ = ΨP + ΨS
Ψ = Free energy associated with water potential
ΨP = Pressure potential (force from water pressure, normally 0 in open container)
ΨS = Potential dependent on the solute concentration (how many particles of material are in solution

5. Water Potential (Ψ) What you need to know
Ψ = 0 (isotonic, there is no net movement of water into/out of system
Ψ > 1 (hypotonic to the solution, water has a net movement out of system)
Ψ < 1 (hypertonic to the solution, water has a net movement into the system)

6. Water Potential (Ψ) Solute potential (ΨS) has its own formula
ΨS = -iCRT
i = ionization constant (different based on the material used for the solute, usually one)
C = the molar concentration (molarity = moles/L)
R = pressure constant ( liters*bars/mole*K)
T = temperature in Kelvin (K = ̊C + 273)
Bar = 1 atm (at sea level)

7. Water Potential (Ψ)
A sample of 0.15M sucrose at atmospheric pressure (ΨP = 0) and 25 ̊C has what water potential? Since sucrose dose not break apart into ions (i = 1).

8. Water Potential (Ψ) Ψ = ? ΨP = 0 ΨS = -iCRT i = 1 C = 0.15M
Ψ = o +ΨS
ΨS = -iCRT
ΨS = - (1)(0.15M)(0.0831)(298K)
ΨS = -3.7 bars
Ψ = o -3.7bars
Ψ = -3.7bars
Ψ = ?
ΨP = 0
ΨS = -iCRT
i = 1
C = 0.15M
R = liters*bars/mole*K
T = 25 ̊C = 298K

9. Gibb’s Free Energy
ΔG = ΔH - T ΔS

10. Gibb’s Free Energy
ΔG = Gibb’s Free Energy, the amount of free energy available to a system
Unit = kJ (kiloJoules)
ΔH = heat of reaction (or Enthalpy), the amount of heat energy in a system
T = temperature
Unit (Kelvin  ̊C + 273)
ΔS = entropy, the amount of order/disorder in a system
Unit = J/K (Joules per Kelvin)

11. Gibb’s Free Energy ΔG – positive vs. negative
+ ΔG = reaction is endergonic (energy must be supplied to make the reaction occur, ANABOLISM)
- ΔG = reaction is exergonic (generally, this reaction is spontaneous and releases energy, CATABOLISM)
ΔH – positive vs. negative
+ ΔH = reaction is endothermic (heat energy is absorbed from the surrounding area)
- ΔH = reaction is exothermic (heat energy is released into the surrounding area)

12. Gibb’s Free Energy ΔS – positive vs. negative
+ ΔS = the entropy of the system is increasing (you have more free moving molecules, things are breaking apart, catabolism)
- ΔS = the entropy of the system is decreasing (you have more structured arrangement of molecules, tends to be fewer, more complex, molecules, anabolism)
EXAMPLE: HC2H3O2 has less entropy than 2 H+ (more number of molecules as opposed to just more atoms in molecule)

13. Gibb’s Free Energy
A biological reaction has an exothermic reaction (ΔH) that produces -55.8kJ of energy. If the reaction is decreasing entropy (ΔS) by -0.35kJ/K at a temperature of 25°C, what would be the free energy (ΔG)? Is this reaction exergonic (releasing energy) or endergonic (absorbing energy)?

14. Gibb’s Free Energy ΔG = ? ΔH = -55.8kJ ΔS = -0.35kJ/K T = 25°C = 298K
ΔG = ΔH – TΔS
ΔG = -55.8kJ – (298K)(-0.35kJ/K)
ΔG = -55.8kJ kJ
ΔG = +48.5kJ
NOTE: The reaction is endergonic (ANABOLIC)

15. Hardy-Weinberg
p2 +2pq + q2 = 1
p + q = 1

16. Hardy-Weinberg Hardy-Weinberg equation(s): p2 +2pq + q2 = 1 p + q = 1
ALLELE FREQUENCY
p = dominant allele
q = recessive allele
DETERMINES PHENOTYPE FREQUENCY
p2 = homozygous dominant
pq = heterozygous
q2 = homozygous recessive

17. Hardy-Weinberg Assumptions:
No net mutations occur, so alleles remain the same
Individuals neither leave nor come into the population (no gene flow)
The population is large (ideally infinite)
Natural Selection DOES NOT occur (so random mating and no environmental pressures)
Random mating (no sexual selection)

18. Hardy-Weinberg
A population of bunnies has the phenotype of 36% white bunnies(the recessive gene) and the rest are black. Based on this data, what are the frequency of each genotype?
Things to consider:
What are the possible genotypes?
Do you know any variables? 2pq, p2, or q2
How do you solve for p or q?

19. Hardy-Weinberg ANSWER: p2 = 0.16 or 16% (BB) 2 pq = 0.48 or 48% (Bb)
36 % of the bunnies are white
Genotype = bb
Therefore, 36% of the bunnies = q2
q2 = 0.36
q = 0.6
Since q = 0.6, then
p + q = 1
1 – q = p
p = 0.4 (40%)
ANSWER:
p2 = 0.16 or 16% (BB)
2 pq = 0.48 or 48% (Bb)
q2 = 0.36 or 36% (bb)

20. Population Formulas Rate: dY/dt Population Growth: dN/dt = (B – D)
Exponential Growth: dN/dt =rmax •N
Carrying Capacity: dN/dt =rmax •N • [(K-N)/K]

21. Population – Example 1
In a population of 1500, the number of births is 265 and the number of deaths is 100.
What is the population growth expressed as individuals/year?
What is the population growth expressed as a percent (rmax)?

22. Population – Example 1 Population growth as individuals/year
Population growth as a percentage
(165 / 1500) * 100
11% (the rmax value = 0.11)
To find the new population size
N = 1500 x 1.11

23. Population – Example 2
In a population of 2250, the number of births is 450 and the number of deaths is 125.
What is the population growth expressed as individuals/year?
What is the population growth expressed as a percent?
What is the population in 2 years?

24. Population – Example 2 Population growth as individuals/year
Population growth as a percentage
(325 / 2250) * 100
14.4% (the rmax value = 0.144)
To find the new population size for 2 years
N = 2250 x = 2574
N = 2574 x = 2944

25. Population – Example 3
You have a population of The rmax value is 30%/year. If the carrying capacity is 15,000, what is the population after 1 year?

26. Population – Example 3 N = 2500 rmax = 0.30 K = 15,000
N = 625 new individuals to the population
For total population:
N = N * [1 + rmax[(K-N)/K]]
N = 2500 * [ [(15, )/15,000]]
N = 2500 * 1.25
N = 3125

27. CHI-Square

28. CHI-Square x2: this is the chi-squared value
o: this is the value that was observed in the experiment
e: this is the expected value
Σ: this means that you take the sum of all the values

29. NULL HYPOTHESIS
The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result.
On the next page are the expected value for M&Ms by color.
Example of null hypothesis: The M&Ms will not vary from the values given by the company
NOTE: In the null hypothesis, it is a negative statement.

30. CHI-Square: Write Null Hypothesis
According to the Mars Company website, the colors of M&Ms are produced in certain percentages:
20% brown
20% yellow
20% red
20% blue
10% green
10 % orange

31. DATA Color Observed Brown 76 Yellow 65 Red 70 Blue 85 Green 28 Orange26
total
350

32. SOLVING FOR X2 Brown Yellow Red Blue Green Orange Observed (o) 76 6570 85 28 26
Expected (e) 35
o-e 6 -5 15 -7 -9
(o-e)2 36 25
225 49 81
(o-e)2/e
0.514
0.357
3.214
1.400
2.314 X2
7.799

33. CHECK YOUR ANSWER Degrees of freedom:
The sample size minus one
6 colors
Degrees of freedom: 5
Check the Chi-Square Table for 0.05
7.799 <11.07

34. NULL HYPOTHESIS
Since < 11.07, we will accept the null hypothesis:
The M&Ms do not vary significantly from the values given by the company. Therefore, they do not statistically vary from the expected values (NO OUTSIDE INFLUENCE)

35. Mean Standard Deviation Standard Error of the Mean
STATISTICS
Mean
Standard Deviation
Standard Error of the Mean

36. MEAN The formula for mean on the formula sheet can be simplified to:
Add all the values in a data set; and then
Divide the total by the number of data values in a sample
Example: Find the mean of the following
8, 12, 15, 10, 9
( )/5 = 10.8

37. STANDARD DEVIATION
The bell curve which represents a normal distribution of data shows what standard deviation represents.

38. STANDARD DEVIATION Explanation of the formula Find the variance.
a) Find the mean of the data.
b) Subtract the mean from each value.
c) Square each deviation of the mean.
d) Find the sum of the squares.
e) Divide the total by the number of items.
Take the square root of the variance.

39. PROBLEM 1
A math class took a test with these five test scores: 92,92,92,52,52.
Find the standard deviation for this class.

40. SOLUTION 1 Find the mean: (92+92+92+52+52)/5 = 76
Find the deviation from the mean:
92-76= = = = = -24
Square the deviation from the mean:

41. SOLUTION 1 Find the sum of the squares: 256+256+256+576+576= 1920
Divide the sum of the squares by the number of items :
1920/4 = 480 variance
Find the square root of the variance:
21.9

42. STANDARD ERROR OF THE MEAN
Definition: an indication of how well the mean of a sample estimates the mean of a population. It is measured by the standard deviation of the means of randomly drawn samples of the same size as the sample in question.

43. STANDARD ERROR OF THE MEAN PROBLEM
You have 2 standard deviations
Data Set 1: Standard deviation is 3.8 in a population size of 10,000
Data Set 2: Standard deviation of 1.3, in a population size of 25
Which sample is more indicative of the population

44. STANDARD ERROR OF THE MEAN SOLUTION
Data set 1: 3.8/√10,000 = 0.038
Data set 2: 1.3/ √25 = 0.26
What is means:
The data set with the smaller number, has the smaller deviation and is MORE indicative of the actual population mean

45. OTHER FORMULAS THAT MAY APPEAR ON AP EXAMpH
Dilution
Temperature Coefficient Q10
Primary Productivity
Surface Area/Volume

46. pH You don’t have to know how to use the formula Shortcut:
Whatever the exponent, that is the pH
Example 1: 1.0x10-3 = pH 3
Example 2: pH 8 = 1.0x10-8

47. Dilution
On the AP test, there may be times when they talk about diluting a sample
When they do this, you use the following formula
C1V1 = C2V2

48. Dilution Here is an example:
You have 250mL of a 3.5M solution, what will be the final volume necessary to create a 0.5M solution?

49. (250mL)(3.5M) = V2 (0.5M) 1750mL = V2 Dilution
NOTE: Be careful, sometimes they may ask you how much water you need to ADD to dilute. In that case, you would ADD 1500mL to bring it to the proper dilution

50. Temperature Coefficient Q10
This formula is for enzyme kinetics/reaction rates.
Sometimes they will have you calculate (usually in an essay) how much a reaction increases in speed due to temperature
The rate is standardized for an increase of 10°C

51. Temperature Coefficient Q10
Description of variables
k1 = the reaction rate at the first temperature
k2 = the reaction rate at the higher temperature
NOTE: Units are not important since they cancel out

52. Temperature Coefficient Q10
At a temperature of 20°C, an enzyme produces a product at the rate of 5mg/s. At a temperature of 25°C, the same enzyme produces product at the rate of 15mg/s. What is the Q10 for this enzyme?

53. Temperature Coefficient Q10
MEANING: Every 10°C increase, there is a 9-fold increase in reaction rate

54. Primary Productivity
Primary productivity is the rate at which photosynthetic organisms produce organic compounds in an organism
From the Light cycle, we know that plants produce O2 as waste
Furthermore, we can measure rates of O2 waste production to calculate rate of CO2 fixation in the Calvin cycle

55. Primary Productivity
The formula will allow you to either start with a mass density or a volume density:
mg O2/L, or
mL O2/L
If you start with mass, you can convert it to a volume by using the atmospheric density of O2

56. Primary Productivity Example:
An O2 detector is used to measure the mass of O2 produced by a plant during photosynthesis. If the sensor indicates that the plant produced 125mg O2/L, how much carbon was fixed in the Calvin cycle?

57. 46.8mg C fixed/L Primary Productivity 125mg O2 | 0.698mL =87.25mL O2/L
1L | mg
87.25mL O2 | mg C Fixed =
1L | mL O2
46.8mg C fixed/L

58. Primary Productivity
More commonly, water displacement is used to determine the mL O2/L
This method is used for water plants
When this value is discovered, there is a 1 step conversion to calculate mg C fixed/L

59. Primary Productivity Example:
It was determined that a water lily produced about 25mL O2/L in the course of the experiment. What was the amount of carbon fixed in the Calvin cycle

60. 25mL O2 | mg C Fixed =
1L | mL O2
13.4mg C fixed/L

61. Surface Area/Volume All of these are pretty straight forward
A lot of times these will be used to compare diffusion rates for volume vs. surface area
Reinforcing reasons for cell sizes