AP BIO LABS REVIEW PPT - Manatee School for the Arts
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Transcript AP BIO LABS REVIEW PPT - Manatee School for the Arts
Animations and Videos
Bozeman - AP BIO Labs Review
AP Biology
Lab 1: Diffusion & Osmosis
AP Biology
2004-2005
Lab 1: Diffusion & Osmosis
Description
dialysis tubing filled with starchglucose solution in beaker filled with
KI solution
potato cores in
sucrose solutions
AP Biology
2004-2005
Lab 1: Diffusion & Osmosis
Concepts
semi-permeable membrane
diffusion
osmosis
solutions
hypotonic
hypertonic
isotonic
AP Biology
water potential
Lab 1: Diffusion & Osmosis
Conclusions
water moves from high concentration of
water (hypotonic=low solute) to low
concentration of water (hypertonic=high
solute)
solute concentration &
size of molecule
affect movement
through
semi-permeable
membrane
AP Biology
2004-2005
When a solution such as that inside dialysis
tubing is separated from pure water by a
selectively-permeable membrane water will move
by osmosis from the surrounding area where the
water potential is higher into the cell where water
potential is lower due to the presence of solute.
The movement of water into the cell causes the
cell to swell and the cell membrane pushes
against the cell wall to produce an increase in
pressure (turgor). This process will continue until
the water potential of the cell equals the water
potential of the pure water outside the cell. At this
point, a dynamic equilibrium is reached and net
water movement will cease.
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 1 - Diffusion and
Osmosis
AP LAB 1 - Diffusion and Osmosis
AP WEB LAB - Osmosis and Diffusion
AP Biology
Lab 2: Enzyme Catalysis
Description
measured factors affecting enzyme
activity
H2O2 H2O + O2
measured rate of O2 production
AP Biology
catalase
2004-2005
Lab 2: Enzyme Catalysis
Concepts
substrate
enzyme
enzyme structure
product
denaturation of protein
experimental design
rate of reactivity
reaction with enzyme vs. reaction without enzyme
optimum pH or temperature
test at various pH or temperature values
AP Biology
Lab 2: Enzyme Catalysis
Conclusions
enzyme reaction rate is affected by:
pH
temperature
substrate concentration
enzyme concentration
calculate rate?
AP Biology
2004-2005
In the first few minutes of an enzymatic reaction, the
number of substrate molecules is usually so large
compared to the number of enzyme molecules that
changing the substrate concentration does not (for a short
period at least) affect the number of successful collisions
between substrate and enzyme. During this early period,
the enzyme is acting on substrate molecules at a constant
rate. The slope of the graph line during this early period is
called the initial velocity of the reaction. The initial velocity,
or rate, of any enzyme catalyzed reaction is determined by
the characteristics of the enzyme molecule. It is always the
same for an enzyme and its substrate as long as
temperature and pH are constant and substrate is present
in excess. Also, in this experiment the disappearance of the
substrate is essential in this reaction. Eventually, the rate of
the reaction levels off.
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 2 - Enzyme
Catalyst
AP LAB 2 - Enzyme Catalyst
AP WEB LAB - Enzyme Catalyst
AP Biology
Lab 3: Mitosis & Meiosis
AP Biology
2004-2005
Lab 3: Mitosis & Meiosis
Description
cell stages of mitosis
exam slide of onion root tip
count number of cells in each stage to
determine relative time spent in each stage
crossing over in meiosis
farther gene is from centromere the greater
number of crossovers
observed crossing over in
fungus, Sordaria
arrangement of ascospores
AP Biology
2004-2005
Lab 3: Mitosis & Meiosis
Concepts
mitosis
interphase
prophase
metaphase
anaphase
telophase
I
meiosis
meiosis 1
meiosis 2
crossing over
tetrad in prophase 1
AP Biology
P
M
A
T
Lab 3: Mitosis & Meiosis
Conclusions
Mitosis
longest phase = interphase
each subsequent phase is
shorter in duration
Meiosis
4:4 arrangement in
ascospores
no crossover
AP Biology
any other arrangement
crossover
2:2:2:2 or 2:4:2
2004-2005
Sordaria analysis
total crossover
% crossover =
total offspring
distance from
=
centromere
AP Biology
% crossover
2
2004-2005
The relative lengths of the mitotic stages
are: 53.4% prophase, 17.4% metaphase,
16.8% anaphase and 12.4% telophase.
Meiosis is important for sexual
reproduction because it reduces the
chromosome number by half and it also
results in new combinations of genes
through independent assortment and
crossing over, followed by the random
fertilization of eggs by sperm.
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 3 - Mitosis and
Meiosis
AP LAB 3 - Mitosis and Meiosis
Bozeman - The Sodaria Cross
AP WEB LAB – Mitosis
AP WEB LAB - Meiosis
AP Biology
Lab 4: Photosynthesis
AP Biology
2004-2005
Lab 4: Photosynthesis
Description
determine rate of photosynthesis under
different conditions
light vs. dark
boiled vs. unboiled chloroplasts
chloroplasts vs. no chloroplasts
use DPIP in place of NADP+
DPIPox = blue
DPIPred = clear
measure light transmittance
paper chromatography to
separate plant pigments
AP Biology
2004-2005
Lab 4: Photosynthesis
Concepts
photosynthesis
Photosystem 1
NADPH
chlorophylls & other
plant pigments
AP Biology
chlorophyll a
chlorophyll b
xanthophylls
carotenoids
experimental design
control vs. experimental
2004-2005
Lab 4: Photosynthesis
Conclusions
Pigments
pigments move at different rates based on
solubility in solvent
Photosynthesis
light & unboiled
chloroplasts
produced
highest rate of
photosynthesis
AP Biology
2004-2005
Lab 4: Photosynthesis
Time
(min)
AP Biology
Light, Unboiled
Dark, Unboiled
% transmittance % transmittance
Sample 1
Sample 2
Light, Boiled
% transmittance
Sample 3
0
28.8
29.2
28.8
5
48.7
30.1
29.2
10
57.8
31.2
29.4
15
62.5
32.4
28.7
20
66.7
31.8
28.5
Animations and Videos
Bozeman - AP BIO Lab 4 - Plant
Pigments and Photosynthesis
AP LAB 4 - Plant Pigments and
Photosynthesis
Using a Spectrophotometer
AP WEB LAB – Pigments
AP WEB LAB - Photosynthesis
AP Biology
The solvent moves up the paper by capillary
action, which occurs as a result of the attraction
of solvent molecules to the paper and the
attraction of solvent molecules to one another. As
the solvent moves up the paper, it carries along
any substances dissolved in it, in this case
pigments. The pigments are carried along at
different rates because they are not equally
soluble in the solvent and because they are
attracted, to different degrees, to the cellulose in
the paper through the formation of hydrogen
bonds. Also, as the DPIP is reduced and becomes
colorless, the resultant increase in light
transmittance is measured over a time course
using a spectrophotometer.
AP Biology
Lab 5: Cellular Respiration
AP Biology
Lab 5: Cellular Respiration
Description
using respirometer to measure rate of
O2 production by pea seeds
non-germinating peas
germinating peas
effect of temperature
control for changes in pressure &
temperature in room
AP Biology
Lab 5: Cellular Respiration
Concepts
respiration
experimental design
control vs. experimental
function of KOH
function of vial with only glass beads
AP Biology
Lab 5: Cellular Respiration
Conclusions
temp = respiration
germination = respiration
calculate rate?
AP Biology
Germinating peas respire and need to consume
oxygen in order to continue the growing process.
Pea seeds are non-germinating and do not
respire actively. These seeds are no longer the
site of growth and thus do not need oxygen for
growth. In consideration to temperature, at higher
temperatures more oxygen is consumed which
means more respiration is occurring. 686
kilocalories are released during respiration. When
temperature decreases molecular motion slows
down and respiration decreases because less
energy is made available.
AP Biology
Animations and Videos
Bozeman - Lab 5 - Cell Respiration
Bozeman - AP BIO Lab 5 - Cell
Respiration
AP LAB 5- Cell Respiration
AP WEB LAB - Cell Respiration
AP Biology
Lab 6: Molecular Biology
AP Biology
Lab 6: Molecular Biology
Description
Transformation
insert foreign gene in bacteria by using
engineered plasmid
also insert ampicillin resistant gene on same
plasmid as selectable marker
Gel electrophoresis
cut DNA with restriction enzyme
fragments separate on gel based
on size
AP Biology
2004-2005
Lab 6: Molecular Biology
Concepts
transformation
plasmid
selectable marker
ampicillin resistance
restriction enzyme
gel electrophoresis
DNA is negatively
charged
smaller fragments
travel faster
AP Biology
Lab 6: Transformation
Conclusions
can insert foreign DNA using vector
ampicillin becomes selecting agent
no transformation = no growth on amp+ plate
AP Biology
Lab 6: Gel Electrophoresis
Conclusions
DNA = negatively
charged
correlate distance
to size
smaller fragments
travel faster &
therefore farther
AP Biology
Bacterial Transformation-Ampicillin Resistance:
In this exercise, we will introduce competent E.
Coli cells to take up the plasmid pAMP, which
contains a gene for ampicillin resistance.
Normally, E. Coli cells are destroyed by the
antibiotic ampicillin, but E. Coli cells that have
been transformed will be able to grow on agar
plates containing ampicillin. Thus, we can select
for transformants; those cells that are not
transformed will be killed by ampicillin; those that
have been transformed will survive.
AP Biology
Restriction Enzyme Cleavage of DNA: Restriction
endonuclease recognizes specific DNA
sequences in double-stranded DNA and digests
the DNA at these sites. The result is the
production of fragments of DNA of various
lengths corresponding to the distance between
identical DNA sequences within the chromosome.
By taking DNA fragments and systematically
reinserting the fragments into an organism with
minimal genetic material, it is possible to
determine the function of particular gene
sequences
AP Biology
Electrophoresis: Fragments of DNA can
be separated by gel electrophoresis when
any molecule enters the electrical field,
the mobility or speed at which it will move
is influenced by the charge (negative
charges travel to positive/top pole of gel),
the density of the molecule, (the smaller
the molecule, the faster it travels), the
strength of the electrical field, and the
density of the medium (gel) which it is
migrating.
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 6 - Molecular
Biology
AP LAB 6 - Molecular Biology
AP WEB LAB - Molecular Biology
AP Biology
Lab 7: Genetics (Fly Lab)
AP Biology
Lab 7: Genetics (Fly Lab)
Description
AP Biology
given fly of unknown genotype use
crosses to determine mode of
inheritance of trait
Lab 7: Genetics (Fly Lab)
Concepts
phenotype vs. genotype
dominant vs. recessive
P, F1, F2 generations
sex-linked
monohybrid cross
dihybrid cross
test cross
chi square
AP Biology
2004-2005
Lab 7: Genetics (Fly Lab)
Conclusions: Can you solve these?
Case 1
Case 2
AP Biology
2004-2005
From this lab, you will be able to find genotypes
and phenotypic expression within a fruit fly. Also,
recessive genes and mutations will be revealed
as the student crosses a variety of Drosophila
alleles. For example, if a female carrier for an xlinked, recessive trait, was crossed with a male
without the recessive trait the results would be:
½ males with x-linked trait ½ males without
½ female carriers ½ females without
0 females express sex linked traits
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 7 -Genetics of
Organisms
AP LAB 7 - Genetics of Organisms
AP WEB LAB - Genetics of Organisms
Bozeman - Chi-squared Test
AP Biology
Lab 8: Population Genetics
Description
simulations were used to study effects
of different parameters on frequency of
alleles in a population
selection
heterozygous advantage
genetic drift
AP Biology
2004-2005
Lab 8: Population Genetics
Concepts
Hardy-Weinberg equilibrium
p+q=1
p2 + 2pq + q2 = 1
required conditions
large population
random mating
no mutations
no natural selection
no migration
AP Biology
gene pool
heterozygous advantage
genetic drift
founder effect
bottleneck
2004-2005
Lab 8: Population Genetics
Conclusions
recessive alleles remain hidden
in the pool of heterozygotes
even lethal recessive alleles are not
completely removed from population
know how to solve H-W problems!
to calculate allele frequencies, use p + q = 1
to calculate genotype frequencies or how
many individuals, use, p2 + 2pq + q2 = 1
AP Biology
Assuming that Hardy-Weinberg equilibrium is maintained
allele and genotype frequencies should remain constant
from generation to generation. For this to happen the five
following situations must all occur:
1. Population is very large. The effects of chance on
changes in allele frequencies is
thereby greatly reduced.
2. Individuals show no mating preference, i.e. random
mating.
3.There is no mutation of alleles.
4. No differential migration occurs, (no immigration or
emigration).
5. All genotypes have an equal chance of surviving and
reproducing, i.e. there is no natural selection.
AP Biology
In humans, several genetic diseases have
been well characterized. Some of these
diseases are controlled by a single allele
where the homozygous recessive
genotype has a high probability of not
reaching reproductive maturity. If this
were to occur both the homozygous
dominant and heterozygous individuals
will survive while the homozygous
recessive will become extinct.
AP Biology
Animations and Videos
Bozeman - Lab 8 - Population Genetics
and Evolution
AP LAB 8 -Population Genetics and
Evolution
AP WEB LAB - Population Genetics and
Evolution
AP Biology
Lab 9: Transpiration
AP Biology
Lab 9: Transpiration
Description
test the effects of environmental factors
on rate of transpiration
temperature
humidity
air flow (wind)
light intensity
AP Biology
2004-2005
Lab 9: Transpiration
Concepts
transpiration
stomates
guard cells
xylem
adhesion
cohesion
H bonding
AP Biology
Lab 9: Transpiration
Conclusions
transpiration
wind
light
transpiration
humidity
AP Biology
2004-2005
Conditions that cause a decreased rate of
water loss from leaves result in a
decreased water potential gradient from
stem to leaf and therefore in a decreased
rate of water movement up the stem to the
leaves. Conditions that cause an
increased rate of water loss from leaves
result in an increase in the water potential
gradient from stem to leaf and therefore in
an increase in the rate of water movement
up the stem to the leaves.
AP Biology
Normal Room Conditions (CONTROL)
When you expose a plant to room
conditions nothing is supposed to
happen. The reasoning for this is room
conditions don’t cause drastic changes in
the plants environment for major
transpiration or even water gain to occur.
The plant under room conditions is
considered to be your control.
AP Biology
Floodlight
When light is absorbed by the leaf, some of the light energy
is converted to heat and remember that transpiration rate
increases with temperature. We learned in Unit One of the
Campbell’s edition that when the temperature of liquid
water rises, kinetic energy of the water molecules
increases. As a result, the rate at which liquid water is
converted to water vapor increases. When the water is
turned into water vapor, it easily passes out through
stomata out into the outer atmosphere. The floodlight is an
example of a plant near the sun (which is why the plant is
one meter away from the light). Due to the aforementioned
properties of plants you should see a loss of water.
AP Biology
Fan (WIND)
An increase in wind speed results in an
increase in the rate of leaf water loss
because increased wind decreases the
boundary layer of still air at the leaf
surface. This boundary layer acts to slow
leaf water loss. Increased wind also
causes the rapid removal of evaporating
water molecules from the leaf surface.
This results in a low water potential in the
air immediately and the water level should
drop.
AP Biology
Mist (HIGH HUMIDITY)
Increased humidity in the air surrounding the leaf
decreases the water potential gradient between
the saturated air in the leaf air spaces and the air
surrounding the leaf, resulting in a decreased
rate of leaf water loss. However, when the
humidity of the air surrounding the leaf if very
low, the water potential of the air is low.
Therefore, the water potential gradient between
the air spaces of the leaf and the surrounding air
is high, and the rate of leaf water loss increases.
AP Biology
When there is a great amount of humidity,
transpiration decreases because of water
potential. When the humidity is at a low or
normal, the mesophyll cells in the plant are much
higher in water potential than the relatively drier
surrounding air. Due to the properties of water
potential, which states that water tends to
evaporate from the leaf surface moving from an
area of higher water potential to an area of lower
water potential, transpiration occurs. But,
because of the high humidity, the surrounding air
has a higher water potential than the mesophyll
cells and water loss is at a minimum.
AP Biology
Adaptations to reduce leaf water loss
include a reduced number of stomates, an
increase in the thickness of the leaf
cuticle, a decrease in leaf surface area,
and adaptations that decrease air
movements around stomates, such as
dense hairs and sunken stomates.
Because leaves are all different in size,
reporting the water loss without
considering a unit area would provide
non-comparable data.
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 9 –
Transpiration
AP LAB 9 – Transpiration
AP WEB LAB - Transpiration
AP Biology
Lab 10: Circulatory Physiology
AP Biology
2004-2005
Lab 10: Circulatory Physiology
Description
study factors that affect heart rate
body position
level of activity
determine whether an organism is an
endotherm or an ectotherm by
measuring change in pulse rate as
temperature changes
Daphnia
AP Biology
2004-2005
Lab 10: Circulatory Physiology
Concepts
thermoregulation
endotherm
ectotherm
Q10
measures increase in metabolic activity resulting from
increase in body temperature
Daphnia can adjust their temperature
to the environment, as temperature
in environment increases, their
body temperature also increases
which increases their heart rate
AP Biology
2004-2005
Lab 10: Circulatory Physiology
Conclusions
Activity increase heart rate
in a fit individual pulse & blood pressure are lower &
will return more quickly to resting condition after
exercise than in a less fit individual
AP Biology
Pulse rate changes in an ectotherm as external
temperature changes
2004-2005
The sphygmomanometer measures the blood pressure. The blood pressure cuff is
inflated so that blood flow stops to through the brachial artery in the upper arm. A
stethoscope is used to listen to blood flow entering the brachial artery. When blood
first enters the artery, snapping sounds called the sounds of Korotkoff are generated.
A. Blood pressure and heart rate increase when you move from a reclining to a
standing position counteracting gravitational pull on the blood
B. Elevated arterial blood pressure indicates increased arterial resistance to blood
flow
C. Fit individuals can pump a larger volume of blood with each contraction and
deliver more oxygen to muscle tissue than the hearts of unfit individuals. As a result,
blood pressure and heart rate increases are smaller for fit individuals, and the time
required to return to normal conditions is shorter for fit individuals than unfit
individuals.
D. For the Daphnia, remember that ectothermic animals use behavior to regulate their
body temperatures and that Q10 cannot be determined for endothermic animals
because body temperatures remain constant regardless of environmental
temperatures.
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 10 - Circulatory
Physiology
AP LAB 10 - Circulatory Physiology
AP Biology
Lab 11: Animal Behavior
AP Biology
2004-2005
Lab 11: Animal Behavior
Description
set up an experiment to study behavior
in an organism
Betta fish agonistic behavior
Drosophila mating behavior
pillbug kinesis
AP Biology
Lab 11: Animal Behavior
Concepts
innate vs. learned behavior
experimental design
control vs. experimental
hypothesis
choice chamber
AP Biology
temperature
humidity
light intensity
salinity
other factors
Lab 11: Animal Behavior
Hypothesis development
Poor:
I think pillbugs will move toward the wet
side of a choice chamber.
Better:
If pillbugs prefer a moist environment,
then when they are randomly placed on
both sides of a wet/dry choice chamber
and allowed to move about freely for
10 minutes, most will be found on the wet
side.
AP Biology
Lab 11: Animal Behavior
Experimental design
AP Biology
sample size
When conducting this experiment, a couple of
things should be understood.
A. Three variables are tested: light, temperature
and pH. The control is exposed to room light,
room temperature, and neutral pH is also
prepared. For each of the variables, a gradient is
established providing a continuous variation from
weak to strong intensities.
B. The histograms are prepared showing the
number of isopods in each of the four intensities.
A histogram is prepared for each of the three
variables and the control. From the data in the
histograms, conclusions cab be made describing
the habitat preferences of the isopod.
AP Biology
Animations and Videos
Bozeman - AP BIO Lab 11 - Animal
Behavior
AP LAB 11 - Animal Behavior
Bozeman - Q10: The Temperature
Coefficient
AP WEB LAB - Animal Behavior
AP Biology
Lab 12: Dissolved Oxygen
Dissolved O2 availability
AP Biology
Lab 12: Dissolved Oxygen
AP Biology
2004-2005
Lab 12: Dissolved Oxygen
Description
measure primary productivity by measuring O2
production
factors that affect amount of dissolved O2
temperature
as water temperature, its ability to hold O2 decreases
photosynthetic activity
in bright light, aquatic plants produce more O2
decomposition activity
as organic matter decays, microbial respiration consumes O2
mixing & turbulence
wave action, waterfalls & rapids aerate H2O & O2
salinity
as water becomes more salty, its ability to hold O2 decreases
AP Biology
Lab 12: Dissolved Oxygen
Concepts
dissolved O2
primary productivity
measured in 3 ways:
amount of CO2 used
rate of sugar (biomass) formation
rate of O2 production
AP Biology
net productivity vs. gross productivity
respiration
2004-2005
Lab 12: Dissolved Oxygen
Conclusions
temperature = dissolved O2
light = photosynthesis = O2 production
O2 loss from respiration
respiration = dissolved O2
(consumption of O2)
AP Biology
2004-2005
The amount of oxygen dissolved in natural water samples
is measured and analyzed to determine the primary
productivity of the sample. The amount of dissolved
oxygen is dependent upon many factors.
A. Temperature
B. Salinity
C. Photosynthesis
D. Respiration
Primary productivity is a measure of the amount of biomass
produced by autotrophs through photosynthesis per unit
time. It can be examined by the following factors
A. Gross Primary Productivity
B. Net Primary Productivity
C. Respiratory Rate
AP Biology
Animations and Videos
Bozeman - Lab 12 - Dissolved Oxygen
and Aquatic Primary Productivity
AP LAB 12 - Dissolved Oxygen and
Aquatic Primary Productivity
AP WEB LAB - Dissolved Oxygen
AP WEB LAB - Aquatic Primary
Productivity
AP Biology