Big Idea 1
The Process of Evolution Drives the Diversity
and Unity of Life
Change in the genetic makeup
of a population over time is
evolution.
Organisms are linked by lines of
descent from common ancestry.
Life continues to evolve within a
changing environment.
The origin of living systems is
explained by natural processes.
Hardy-Weinberg Equations
Probabilities
Big Idea 2
Biological Systems Utilize Free Energy and Molecular Building Blocks
to Grow, Reproduce and Maintain Dynamic Homeostasis
Growth, reproduction & maintenance of the organization of living systems
require free energy & matter.
Growth, reproduction & dynamic homeostasis require that cells create & maintain
internal environments that are different from their external environments.
Organisms use feedback mechanisms to regulate growth & reproduction & to maintain
dynamic homeostasis.
Growth & dynamic homeostasis f a biological system
are influenced by changes in the system’s environment..
Many biological processes involved in growth,
reproduction & dynamic homeostasis include temporal
regulation & coordination.
Water Potential
Gibb’s Free Energy
Big Idea 3
Living Systems Store, Retrieve, Transmit and Respond to Information
Essential to Life Processes
Heritable information provides for continuity of life.
Expression of genetic information involves cellular & molecular
mechanisms.
The processing of genetic information is imperfect & is a source of genetic
variation.
Cells communicate by generating, transmitting
& receiving chemical signals.
Transmission of information results in changes
within and between biological systems.
Chi square
Gene Linkage
Big Idea 4
Biological Systems Interact and These Systems and Their Interactions
Possess Complex Properties
Interactions within biological systems lead to complex properties.
Competition & cooperation are important aspects of biological systems.
Naturally occurring diversity among & between components within biological
systems affect interactions within the systems.
1. On the AP Biology Exam, you probably
won’t have to do the actual count and
identification; you will be given a data table
and asked questions, likeFrom the data, calculate the percentage
of time a cell spends in interphase
Calculate the percentage of time a cell
spends in mitosis
Phase
Number
Interphase
52
Prophase/prometaphase
12
Metaphase
2
Anaphase
5
Telophase
1
Percent spent in each phase
2.
Step 1- count the total # of cells 72
Step 2- divide 52/72 x 100 to get % of time spent in interphase 72%
Step 3- Subtract 72% from 100% to get time spent in mitosis 28%
A Cellular Biologist wants to double check that statement that cells spend
90 percent of their time in Interphase as compared to the various stages of
Mitosis. She grows some Allium in her laboratory. She then takes one of the
plants, cuts off the root tips, stains the DNA in the cells so as to be able to
see the stages of the cell cycle. Her hypothesis states “If cells spend 90
percent of their time in Interphase, then she should be able to calculate the
relative time existing between Interphase and Mitosis based upon the cells
counted in her specimen.” She counted 1000 cells from her preserved
specimen under the microscope. Her data are shown below. Calculate the
X2 to the nearest hundredth.
Stage of the Cell Cycle
Number of Cells Observed
Number of Cells Expected
Interphase
872
900
Mitosis
128
100
3. To measure the population density of monarch butterflies occupying
a particular park, 100 butterflies are captured, marked with a small dot
on a wing and then released. The next day, another 100 butterflies
are captured, including the recapture of 20 marked butterflies. What
would you estimate the population to be? NOTE- this is not on your
formula sheet!
(100 x 100)/ 20 = 500
4.
Study this age structure diagram that shows the human population for
India. What percent of the population is under 20?
5.
2x2=4
6 x 4 = 24
6. The ability to taste PTC is due to a single
dominant allele (A). You sampled 215
individuals in biology and determined that 150
could taste PTC and 65 could not. How many
individuals in this population show the
following genotype?
AA, Aa, aa
65/216 = .3 = q2
q = .55
p = .45
(.45)(.45) = .2 = 20%
20% x 215 = 43 AA
2(.45)(.55) = .495 = 50%
50% x 215 = 107 Aa
7. The allele for the hair pattern called
“widow’s peak” is dominant over the allele
for no “widow’s peak”. In a population of
100 individuals, 64 show the dominant
phenotype. What is the frequency of the
recessive allele?
.6
64 show the dominant phenotype,
So 36 show the recessive phenotype.
Since this is a population of 100, 36%
Show the recessive phenotype
.36 = q2
.6 = q
8. Atmospheric pressure is the combined partial
pressures of all of the gases that make up the
atmosphere. At the summit of a high mountain, the
atmospheric pressure is 380mm/Hg. The partial
pressure of oxygen is 69mm/Hg. What percentage
Of the atmosphere is made up of oxygen at this
altitude?
18%
69/380 = .18 = 18%
(the average partial pressure of
oxygen at sea level is 21%)
9. Cytosine makes up 12% of the
nucleotides in a sample of DNA from
an organism. Approximately, what
percentage of the nucleotides in this
sample will be thymine?
38%
Remember Chargaff’s Law?
Amount of A = T and C=G
If 12% is cytosine, then 12% is guanine
76% is made up of adenine and thymine;
divided equally between them
10. How many oxygen atoms would be in the
polysaccharide that results from 100 glucose
molecules being linked together by dehydration
synthesis?
100 C6H1206 would give you 600 oxygen atoms
99 water molecules would be produced by dehydration synthesis
Each water molecule contains one oxygen atom
600 - 99 = 501
H 2O
HO
HO
H HO
H
H
11.
1st Law of
Thermodynamicsenergy cannot be
created or destroyed,
but it can change form.
18,000 energy
accumulated as
biomass; 12,000 going
to the tree layer; 4,400
going to the shrub
layer; 1,600 left, to go
to the grass layer.
1,600 is 9% of 18,000
(1,600/18,000 x 100)
12.
13. Geneticists working in an agriculture lab wanted to
develop a crop that combines the disease resistance of
rye grain with the high crop yielding capacity of wheat
grain. Rye grain has a diploid chromosome number (2n)
of 14 and wheat grain has a diploid chromosome
number of 42. The resulting grain is called triticale and is
an alloploidy plant. How many chromosomes are found
in the pollen grain of triticale?
Alloploidy results when two different plant species
combine their diploid genome to make new and
unique species. That would mean that this particular
species would have 56 chromosomes. The cells in a
pollen grain of would be haploid so the resulting
number is 28.
14. What is the water potential of a cell
with a solute potential of -0.67 MPa and a
pressure potential of 0.43 MPa?
-.24MPa
-0.67 + 0.43
How many naked ladies to you see in this picture?
15. You measure the total
water potential of a cell
and find it to be -0.24 MPa.
If the pressure potential of
the same cell is 0.46 MPa,
what is the solute potential
of that cell?
Since water potential is
equal to the solute potential
+ the pressure potential,
-0.24 MPa = 0.46 MPa +
X. Solve for x= -0.7
16. A wind blown pollen grain with a solute potential of -3.0 MPa has dried out
somewhat after blowing around in the wind. This has caused its turgor pressure
to go to zero. 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? Show how you deduced your answer.
17. From the Osmosis/Diffusion Lab. Once the molar concentration of a sucrose
solution has been determined, how would you determine the water potential of
that sucrose solution? Work one to show how. Assume room temperature (22C).
.4
-(1)(.4)(.0831)(295) = -9.8
Since this was done in an
Open container, your
pressure potential is 0, so
the water potential is the
same as the solute
potential
18. Use the Station 1 data to calculate the
Primary Productivity of a water sample.
Report your answer in units of
mg Carbon fixed/Liter
Station 1
4.2 mg O2/L  0.698 = 2.9 mL O2/L
2.9 mL O2/L  0.536= 1.6 mg Carbon fixed/L
19. What is the mean rate of growth per day between day 5
and day 25? Record your answer to the nearest hundredth of a
cm.
If this same rate of growth continues, how tall will the plant be
on day 50? Record your answer to the nearest hundredth
of a cm.
18-3 = 15
15/20 days = .75 cm
.75 x 25 = 18.75
18.75 + 18 = 36.75
20. A population of ground squirrels has an annual
per capita birth rate of 0.06 and an annual per capital death
rate of 0.02. Estimate the number of individuals added to
(or lost from) a population of 1,000 individuals in one year.
dN/dt = B-D
Change in population size/time = Birth rate – Death rate
0.06 – 0.02 = 0.04 x 1000 = 40 individuals added per year
21. In 2005, the United States had a population of approximately
295,000,000 people. If the birth rate was 13 births for every
1,000 people, approximately how many births occurred in the
United States in 2005?
295,000,000/1,000 = 295,000
295,000 x 13 = 3,835,000
22. A population of June beetles is estimated to
be 3,000. Over the course of a month, there are
400 births and 150 deaths. Estimate r and
calculate what the population size is predicted
to be at the end of the month. Predict the
population at the end of the second month.
Birth rate = 400/3000 = 0.1333 births
Death rate = 150/3000 = 0.0500 deaths
r = birth rate – death rate = 0.1333 –
0.0500 = 0.0833
3000 x 0.0833 = 250
250 + 3000 = 3250
3250 x 0.0833= 271
271 + 3250 = 3521
23. A population of frogs exhibits logistic growth. If the
carrying capacity is 500 frogs and r = 0.1 individuals (per
month), what is the maximum population growth rate
for this population? (Maximum population growth rate
occurs when N = K/2)
12.5
K/2 = 500/2 = 250= N
dN
dT
=
rN(K-N)
K
0.1 x 250 (0.5) = 12.5
individuals/month
24. What is the population standard deviation for
the numbers: 75, 83, 96, 100, 121 and 125?
1. Find the mean
2. Find the difference between the
mean and each number (some will be
+ and some will be –)
3. Square each difference (now they’re all +)
4. Add them together
5. Take the square root
6. Bam! That’s it.
Raise your hand
if you need help.
I hate standard deviations.
25. How many unique gametes could
be produced through independent
assortment by an individual with the
genotype AaBbCCDdEE?
Aa and Bb and Dd can each make 2. CC and EE can only make 1.
2x2x1x2x1=8
26. Assuming the genotype of F1 individuals in a
tetrahybrid cross is AaBbCcDd. independent
assortment of these four genes, what are the
probabilities that F2 offspring will have the following
genotypes?
AaBbCcDd x AaBbCcDd
¼ x ¼ x ¼ x ¼= 1/256
¼ AA ½ Aa ¼ aa
aabbccdd
¼ BB ½ Bb ¼ bb
¼ CC ½ Cc ¼ cc
AaBbCcDd ½ x ½ x ½ x ½ = 1/16
¼ DD ½ Dd ¼ dd
AABBCCDD ¼ x ¼ x ¼ x ¼= 1/256
AaBBccDd ½ x ¼ x ¼ x ½ = 1/64
AaBBCCdd ½ x ¼ x ¼ x ¼ = 1/128
Science Practices- The student can:
1. -use representations & models to communicate
scientific phenomenon & solve scientific problems.
2. -use mathematics appropriately.
3. -engage in scientific questioning to extend thinking
or to guide investigations within the context of the
AP course.
4. -plan & implement data collection strategies
appropriate
to a scientific question.
5. -perform data analysis & evaluation of evidence.
6. -work with scientific explanations & theories.
7. -is able to connect & relate knowledge across
various scales, concepts & representations in and
across domains.
Pose scientific questions
Refine scientific questions
Evaluate scientific questions
Put your enzyme stuff here-
Justify the selection of the kind of data
needed to answer a particular scientific
question.
Design a plan for collecting data to
answer a particular scientific question.
Collect data to answer a particular
scientific question.
Evaluate sources of data to answer a
particular scientific question.
Analyze data to identify patterns or
relationships.
Refine observations and measurements
based on data analysis.
Evaluate the evidence provided by data
sets in relation to a particular
scientific question.
Science Practices- The student can:
1. -use representations & models to communicate
scientific phenomenon & solve scientific problems.
2. -use mathematics appropriately.
3. -engage in scientific questioning to extend thinking
or to guide investigations within the context of the
AP course.
4. -plan & implement data collection strategies
appropriate
to a scientific question.
5. -perform data analysis & evaluation of evidence.
6. -work with scientific explanations & theories.
7. -is able to connect & relate knowledge across
various scales, concepts & representations in and
across domains.
1. Create a model
2. Describe what is taking place
Specifically, describe the relationship
between colony growth and sucrose
concentration.
3. Refine Model
Using the model, calculate the water
potential of the potato cores.
4. Use and Apply
The purpose of a particular investigation was to see the effects of
varying salt concentrations of nutrient agar and its effect on
colony formation. Below are the results Determine the mean for
each treatment and graph the results.
What might you be asked to do with this graph?
1. Describe the relationship between the salt concentration
and colony growth
2. Predict the salt concentration that would allow no
colony growth. Justify your answer.
5. Re-express
Signal transduction pathways are regulatory mechanisms in living things.
Identify a signaling molecule and the response brought about in
a) an animal
b) a plant
Forty flies were put into a choice chamber with two chambers. In one
chamber there was a cotton ball soak with vinegar. The other chamber
had nothing. After 20 minutes the number of flies were counted in both
chambers. This was repeated four more times. Perform a chi-square
analysis to determine if the difference between in the number of flies
found in the two chambers is significant.
Water
Vinegar
76
124
100
100
-24
24
576
576
5.76
5.76
11.52
In a certain population of deer on Fire Island,
NY, the allele for a black spot behind the eye is
dominant to the allele for no spot. After the
hunting season, the percent of deer with no
black spot is 15% and the population is in
Hardy-Weinberg Equilibrium. What is the
frequency for the allele for having no black
spot, to the hundredths?
15% = .15 = q2
q = .39
In a dihybrid cross between two heterozygotes,
if you have 200 offspring, how many should
show both dominant phenotypes?
112
This is a 9:3:3:1 ratio, with 9/16 showing
both dominant phenotypes.
9/16 = .56 = 56%
56% of 200 is 112
In this genetic cross, Aa x aa, there are 348
offspring. How many individuals are expected to
have the dominant phenotype?
174; this is a 1:1 ratio, so 50% are expected to
Have the dominant phenotype
In a typical Mendelian monohybrid cross, two
heterozygotes produce 400 offspring. How many
individuals are expected to have the recessive
phenotype?
100; this is a 3:1 phenotypic ratio
A study was conducted on the island of Daphne Major in the Galapagos Islands by Peter
and Rosemary Grant. This study lasted over 20 year s. The study investigated how the
type of seeds available to the finches impacted the depth of their beaks. In years when
rain and water were plentiful, the available seeds were smaller and easy to crack. In years
experiencing drought, fewer seeds were produced, and the finches had to eat the larger,
leftover seeds produced from previous years. During years of drought, birds with a
greater beak depth had a selective advantage.
Use the data above to determine the
increase in the mean of the depth of
the beak between the wet and dry
years. Give your answer to the nearest
hundredth of a millimeter.
In geckos, spots are dominant to the solid color.
If the frequency In a population of 700 geckos,
what percentage of the geckos would have
spots, if the frequency of the recessive allele
is 0.2, and the population is in Hardy-Weinberg
equilibrium?
96%
q= .2
p= .8
Homozygous dominant (p2) = .64 (64%)
Heterozygotes (2pq) = .32 (32%)
The formula is easy: it is the square root of the
Variance. So now you ask, "What is the
Variance?“The Variance is defined as: The average
of the squared differences from the Mean.
To calculate the variance follow these steps:
1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result
(the squared difference).
3. Then work out the average of those squared differences.
14,000 (a)
35 (c)
180
(b)
100 (d)
You are starting with 87,400 kJ and
simply subtracting to get the
answers.
Consider a field plot containing 200 kg of
plant material. Approximately how many kg
of carnivore production can be supported?
a.
200
b.
100
c.
20
d.
2