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MATH-A-PALOOZA!
Finding Your Way around the Formula Sheet
STANDARD DEVIATION
¢  How spread out
is the data?
¢  Is it consistent
or sporadic?
STANDARD DEVIATION
¢  Which
running back would you want on
your fantasy football team?
STANDARD DEVIATION
¢  Calculate
the mean of the data
¢  Make a data chart…step-by-step
¢  Subtract and square it…list it
¢  Count up sample size (-1)
xi
xi – xmean
(xi - xmean)2
Length of Beetles (cm)
2.0
2.4
2.1
2.5
2.2
--------------------------
Sum = ______
Sum = __ n= __
STANDARD ERROR OF THE MEAN
A measure of how likely your mean value is to represent
the overall set of data points
¢  Numerical values likely to be provided…just plug in!
¢ 
¢ Which
set of data has
more variability?
STANDARD ERROR OF THE MEAN
Ex: What is the SEM for James (from the fantasy football
example)?
¢ 
SEM
= 41.1 / 4
= 10.275
CHI-SQUARE
¢  What
is this even for???
¢  To determine if your data is in fact close enough to the
expected result that you can just say they are the same
¢  What is a NULL hypothesis?
— 
¢  If
Statement that the observed and expected
should be the same (IV does not affect DV)
chi-square # is LARGER than
the “cut-off”, then there is a LARGE
difference between observed & expected.
Should you ACCEPT or REJECT your null hypothesis?
CHI-SQUARE
¢  Roll
the dice twice!
¢  21 possible rolls…(11 possible values)
¢  Null hypothesis is…?
— 
Each combination should be equally
likely
(the dice are not fixed!)
¢  Chi-square
= 9.63
¢  What is the degrees of freedom?
¢  Which probability to use?
¢  What is critical value? (the “cut-off”)
¢  Is chi-square value higher or lower?
CHI-SQUARE
¢  A
cellular biologist wants to verify that cells spend 90% of their time in
interphase and only 10% in mitosis. Her null hypothesis states that 90% of
the observed cells should be in interphase, while only 10% should be in
mitosis. She prepares cells from onion root tips. Using a microscope, she
counted 1000 cells from her preserved specimen. Her data are shown
below. Calculate the X2 to the nearest hundredth.
Stage of the Cell Number of Cells Number of Cells
Cycle Observed Expected Interphase 872 Mitosis 128 CHI-SQUARE
p
0.05
0.01
1
3.84
6.64
Degrees of Freedom
2
5.99
9.32
3
7.82
11.34
4
9.49
13.28
5
11.07
15.09
¢  What
6
12.59
16.81
are your expected values? Why?
¢  Calculate (O-E)… and then go ahead and square it…record each
¢  Divide that by E… record each
¢  Take the sum of last column…this is your chi-square value
¢  What is the df? What is the p? What is your critical value (cut-off)?
¢  Should you accept or reject the null hypothesis? Why?
Stage of the Cell Number of Cells Number of Cells
Cycle Observed Expected Interphase 872 900 Mitosis 128 100 7
14.07
18.48
8
15.51
20.09
HARDY-WEINBERG
¢  Only
works for a non-evolving populations
¢  Frequencies of alleles & genotypes in gene pool remain constant
¢  Must
meet these requirements:
—  Very
large population size
—  No migration
—  No net mutations
—  Random mating
—  No natural selection
HARDY-WEINBERG
¢  “p”
is frequency of one allele; “q” is other
p+q=1
¢  When
gametes combine, 2 alleles join…
These are the organisms:
p2 + 2 pq + q2 = 1
ALWAYS make a key
(p, q, p2, 2pq, q2)
ALWAYS start by finding q or q2
…why?
ALWAYS re-check the question!
HARDY-WEINBERG
¢ In
a population in H-W equilibrium, 36% of the
individuals show the recessive trait. What is the
frequency of the dominant allele?
¢ Where to start? What is “given”?
—  q2 =
.36
—  So q = 0.6
—  So 1-q=p
…p=0.4
—  Frequency of dominant allele = 0.4
HARDY-WEINBERG
¢ In
a population with two alleles for a particular locus, B
and b, the allele frequency of B is 0.7. What would be
the frequency of heterozygotes if the population were in
H-W equilibrium?
—  What
is “given”?
—  p = 0.7
—  so, q = 0.3
—  Heterozygotes = 2pq
—  2pq = 2(0.7)(0.3) = 0.42
RATE & SLOPE
¢  To
measure rate/slope…
Y2 – Y1
X2 – X1 …just pick to points on the line if not specified
What is the rate of growth in the
pea plant between weeks 1 & 3?
60 – 10
3–1
= 25 cm/wk
RATE & SLOPE
What is the cost per production?
90,000-60,000
25,000-15,000…
= $3 per product
TEMPERATURE COEFFICIENT – Q10
¢ Why
is regulating temperature so important?
¢ Denaturing proteins – too HOT = proteins unfold
¢ Metabolism rates (Q10) - too COLD = reactions too slow
TEMPERATURE COEFFICIENT
¢ Many
ectothermic organisms have a Q10 value of 2.0,
but that may not always be the case. Always follow the
information given.
¢ When
given a choice…always choose two values that are
10 degrees apart… why?
¢ Use
the higher temperature values as your “k2” and “t2”
…why?
TEMPERATURE COEFFICIENT
If the metabolism rate of a certain ectothermic organism
is 28 units at 25oC, what will the rate be at 15oC?
¢ 
¢ 
What is the exponent value?
—  1
¢ 
…because 25-15 = 10
…and 10/10 = 1
How to find k1?
—  Has
to be half the value of k2, if Q10 = 2.0
—  So, k1 = 14 units of metabolism
TEMPERATURE COEFFICIENT
What is the Q10 value for an organism that has a
metabolic rate with a value of 39 units at 27oC and a
metabolic rate of 41 units at 37oC?
¢ 
¢ 
What is the exponent value?
—  1
¢ 
…because 37-27 = 10
How to find Q10?
—  Divide
… 41/39 = 1.05
…and 10/10 = 1
BIOMASS
¢ Total
dry mass (take out the water and measure the
¢ Must
get uniform values (exponents may NOT match)
mass)
¢ Write
out the value of each number without the
exponents
¢ Look
for the total biomass – don’t care about other
numbers …this is our total “budget”.
BIOMASS
¢ What
percent of biomass in
this community is within
the grasses?
¢ Look for total biomass –
don’t care about other
numbers above it!
¢ What
is that total?
¢ 1.8 x 10,000 = 18,000
BIOMASS
¢  Total
Biomass = 18,000
¢  Subtract each listed value from
total biomass. Whatever is
leftover is the total for the
grasses.
18,000 – 12,000 – 4400 = 1600
¢  But, why is this NOT the answer?
— 
¢  Calculate
percentage:
1600/18000 = 0.0888888
—  How to record?
—  = 9% or 8.8% (however they ask!)
GIBBS FREE ENERGY
How much energy is available
¢  Important to determine if chemical reactions will take place
¢ 
ΔG = ΔH – TΔS
¢  G = Free Energy
H = Enthalpy
¢  T = Temperature in Kelvin
¢ 
S = Entropy
Don’t forget to change the temperature to Kelvin!
(add 273 to Celsius value…is listed on formula sheet!)
¢ 
GIBBS FREE ENERGY
¢  An
experiment determined that when a protein unfolds to its
denatured (D) state from the original folded (F) state, the
change in Enthalpy is ΔH = H(D) – H(F) = 46,000 joules/mol.
Also the change in Entropy is ΔS = S(D) – S(F) = 178 joules/
mol. At a temperature of 20⁰C, calculate the change in Free
Energy ΔG, in j/mol, when the protein unfolds from its folded
state.
= (46,000) – (293) (178)
¢  ΔG = 46,000 – 52,154
¢  = - 6154 joules/mol
¢  ΔG
WATER POTENTIAL
Water wants to move from high to low osmotic pressure
Water potential = pressure + solute…if open to the air, pressure
=0
¢  Solute potential will be: zero or LOWER (negative)
¢  Pressure potential will be zero or HIGHER (positive)
¢  the
“i” will be 1 for sugars, 2 for salt
¢  ALWAYS draw the cells!
WATER POTENTIAL
In plants, this typicaly means, water moves from soil à roots à
shoots à air (transpiration!)
Be careful with the negatives!
(always moves from high to low!)
ALWAYS
DRAW
the
CELLS!!!
If water potential is greater inside the cell than outside, then water
will exit the cell
Plasmolysis results (plasma membrane separates from cell wall)
ALWAYS
DRAW
the
CELLS!!!
If water potential is lower inside the cell than outside, then water
will enter the cell
Cell becomes turgid
WATER POTENTIAL
¢  What
is the water potential of a cell with a solute potential of -0.67
MPa and a pressure potential of 0.43 MPa?
—  -0.67 + 0.43 = -.24 MPa
¢  Cell
A has a solute potential of -2.0 MPA and a pressure potential of
0.5 MPA. Cell B has a solute potential of -4.0 MPa and a pressure
potential of 0.9 MPA. Which way will water flow when the two cells
are placed against each other?
—  Cell A à -2.0 + 0.5 = -1.5
—  Cell B à -4.0 + 0.9 = -3.1
—  Which is more negative?
Water moves from cell A to cell B
WATER POTENTIAL
¢  Calculate
the water potential of a 0.15 M sucrose solution. Assume
a temperature of 27C.
¢  Note: Solute potential = - iCRT
Pressure potential in open air is 0.
i = 1 for sugars!
add 273 to Celcius number!
¢ 
= - (1) (0.15) (0.0831) (300)
= - 3.74
WATER POTENTIAL
¢  A
wind borne pollen grain with solute potential of -3.0 MPa dried
out somewhat after being blown about; this caused its turgor
pressure (pressure potential) to go to 0. It lands on a flower
stigma whose cells have a solute potential of -3.0 MPa and a
pressure potential of 1 MPa. Which way will water flow? From
the pollen grain to the stigma, or the stigma to the pollen grain?
¢  Pollen
à -3.0 + 0 = -3.0
¢  Stigma à -3.0 + 1 = -2.0
¢  Stigma is more negative…water leaves stigma & enters pollen
POPULATION GROWTH
¢  How
much is this population growing? Or declining?
¢  Rarely have ACTUAL birth (B) and death (D) numbers, but this
is what is provided on your formula chart!
¢ Growth
Rate = dN/dt = B – D
POPULATION GROWTH
¢  The
growth rate may be calculated (or provided)… labeled as “r”
¢  Populations
¢  Growth
that are not limited in growth are growing exponentially
= rN
¢  But,
of course, eventually that carrying capacity (K) is reached
because there are always limits!
¢  Growth
= rN x [(K-N)/K]
POPULATION GROWTH
¢  A
population of rats produces 224 new babies in a year. In the
same time period, 124 of the rabbits died. What is the growth
rate of this population?
¢  Growth = B – D = 224 – 124 = 100 rats/year
¢  A
population of 1000 roaches invades a new habitat with
virtually unlimited resources. They have a growth rate of 6.12.
What will their population total be after one year?
¢  dN/dt
= rN = 6.12 x 1000 = 6120
POPULATION GROWTH
¢ If
the carrying capacity is 2400 and r=2.1, what is the
growth rate for a population of 1000 individuals?
¢ Growth
= (2.1)(1000) x [(2400 – 1000)/2400]
= 2100 x 1400/2400
= 2100 x 0.5833
= 1224
SURFACE AREA & VOLUME
In a diffusion lab, agar cubes are used to illustrate the relationship
of surface area/volume/diffusion rate. Fill in the values missing in
the table.
Block cm x cm A 2 x 2 B 3 x 3 C 4 x 4 Surface Area
cm Volume Surface Area:Volume Ratio ml SURFACE AREA & VOLUME
If you put each of the blocks into a solution, into which block would that
solution diffuse throughout the entire block fastest? Slowest? How do you
explain the difference?
Describe the relationship between the surface area: volume ratio and the
percentage of the cube not affected by diffusion.
Block cm x cm Surface Area
cm Volume Surface Area:Volume Ratio ml A 2 x 2 24 8 3:1 B 3 x 3 54 27 2:1 C 4 x 4 96 64 1.5:1 OTHER MATH - PYRAMIDS
¢  How
much energy is in each level of this ecosystem?
¢  Use 10% Rule unless otherwise specified!
¢  Simply move the decimal place one place to the left for every level
you move up
OTHER MATH – MARK & RECAPTURE
¢  This
is used to estimate population sizes of animals where
individuals are highly mobile. It is of no value where animals do
not move or move very little. The number of animals caught in
each sample must be large enough to be valid.
¢  First
— 
capture à each animal is marked in a distinctive way
Then the marked animals are released into the natural habitat and left
for a period of time.
¢  Second
capture à only a portion of the second capture sample
will have marked animals
OTHER MATH – MARK & RECAPTURE
¢  Total
population =
¢  30
# in first sample (all marked) x total # in 2nd sample
# of marked in the second sample (recaptured)
turtles are captured in 1 km2, they are marked and released back into
the wild. Two weeks later 30 more turtles are caught. 6 had the marking of
the original population. Based on this information, what is the best
estimation of the turtle population in the area?
¢ 30
x 30 = 900
¢ 900/6 = 150
OTHER MATH – CROSSOVER FREQUENCY
¢  In
fruit flies, long wings (A) and gray bodies (B) are dominant to
vestigial wings and black bodies. In a cross of AaBb x aabb.
These are your results-
¢  123
long wing, gray body
¢  21 long wing, black body
¢  27 vestigial wing, gray body
¢  129 vestigial wing, black body
¢  Calculate
the cross over value (recombination frequency) for the
offspring of the test cross.
OTHER MATH – CROSSOVER FREQUENCY
¢  123
long wing, gray body
¢  21 long wing, black body
¢  27 vestigial wing, gray body
¢  129 vestigial wing, black body
¢  You
recognize that this is recombination because you don’t get
the expected 1:1:1:1 ratio.
¢  To calculate Rf =
number of recombinants/Total
= 48/300 = 16% Rf
¢  This also tells you that the gene for wing length and color are
16 map units apart
¢  Could you complete a chi-square for this?!?
OTHER MATH – CROSSOVER FREQUENCY
¢  There
are 4 genes on a single chromosome: A, B, C and D. They exhibit the
following crossing over frequencies:
A-B = 35%
B=25%
B-C = 10%
C-D = 15%
¢  Determine
C-A =25%
D-
the order of the genes on the chromosome.
¢  Start with the genes the farthest apart, and everything else
usually falls into place.
¢  ADCB
or BCDA (either is acceptable)
RESOURCES
Resources used include:
Campbell Biology Student Study Guide
Campbell Biology - diagrams & pictures
AP College Board – Biology Formula Sheet
AP College Board – released questions
Har-Ber HS - Math Review document
COMBINE H-W WITH CHI-SQUARE!
¢  In
a certain species of flowering plant, the red allele R is dominant
to the white allele, r. A student carried out a cross between a red
flowered plant and a white flowered plant. When planted, the 158
seeds that were produced from the cross matured into 92 plants
with red flowers and 66 plants with white flowers. Calculate the
chi squared value for the null hypothesis that the red flowered
parent was a hybrid for the flower color gene. Give your answer to
the nearest tenth.
COMBINE H-W WITH CHI-SQUARE!
¢  In
this cross Rr x rr you would expect a 1:1 ratio
¢  Observed
92 red
66 white
¢  Expected
79 red
79 white
¢  O-E
13
-13
¢  O-E2
169
169
¢  O-E2/ E
2.14
2.14
¢  Chi squared = 4.3
¢  What
is the df? The p? The critical value?
¢  What was the null hypothesis? Should you accept or reject it?
Why?