L.O 4.20: The student is able to explain how the distribution of ecosystems changes over time
by identifying large-scale events that have resulted in these changes in the past.
SP 6.3: The student can articulate the reasons that scientific explanations and theories are
refined or replaced
Explanation: Before a student understands the learning objective he or she must first understand what
“ecosystem distribution” means. Ecosystem distribution is the area over which an ecosystem covers. It is
crucial to understand the role that abiotic and biotic factors play in the distribution of ecosystems resulting
in changed. Abiotic factors may be floods, earthquakes, tornadoes, or any other form of a natural disaster.
Biotic factors are those that are living; animals, and humans that alter the ecosystems.
M.C. Question:
In a Type I survivorship curve which are you most
likely to see?
A)
B)
C)
D)
Human Being
Benthos
Snake
B&D
FRQ Question:
Professor Q discovers a new species that many of his colleagues
have categorized as a k strategist. Professor Q believes this new species is an r strategist.
This species has two offspring and exhibits parental care. Between professor Q and his colleagues, who is
correct? Define both k and r strategists and provide examples of each.
4.20 Answer Key
In a Type I survivorship curve which are you most
likely to see?
-Humans. Type 1 curves are characterized by high age-specific survival probability in
early and middle life, followed by a rapid decline in survival in later life. They are typical
of species that produce few offspring but care for them well, including humans and
many other large mammals.
Professor Q discovers a new species that many of his colleagues
have categorized as a k strategist. Professor Q believes this new species is an r
strategist.
This species has two offspring and exhibits parental care. Between professor Q and his
colleagues, who is correct? Define both k and r strategists and provide examples of
each.
-Student must state that professor Q is incorrect. R strategists have many offspring and
do not exhibit any type of parental care.
-They must also define both r and k strategists to receive credit as well as provide
examples
LO 1.10: The student is able to refine evidence based on data from many scientific disciplines that
supports biological evolution.
SP 5.2: The student can refine observations and measurements based on data analysis.
Explanation: Evidence supporting biological evolution can be obtained from multiple scientific disciples and methodologies.
Direct observations, specifically through studying artificial selection could be used as justification. Utilization of fossil records
provides ample evidence as well. Analyzing homology and convergent evolution (evident in analogous structures), which can be
broken up into specifically looking at embryos, molecular details, homologous structures, vestigial structures, and the “similar
problem, similar solution” situation – provides support regarding evolution. Evidence from biogeography, specifically endemic
species, also provides evidence supporting biological evolution. Perhaps the most commonly used points of comparison are
molecular details; DNA comparisons between species connected via a common ancestor that share similar core DNA with
deviations based upon environmental specifications. Nucleotide BLAST analysis is used in such an instance, to determine the
percent similarity between any two DNA sequences.
M.C. Question: A bat’s wing, a whale’s flipper, and a horse’s forelimb contain essentially the same basic bone structures.
These structures are examples of _____, and how do you know?
A. Analogous structures; similar location and function
B. Homologous structures; similar location and function
C. Vestigial structures; similar location and function
D. Analogous structures; unrelated species
E. Homologous structures; unrelated species
Learning Log/FRQ-style Question: Decide whether
the following relationships represent homology or
analogy:
a. A hedgehog’s quills and an agave’s spines
b. A bobcat’s paw and a chimpanzee’s hand
c. An eagle’s wing and a fly’s wing
How would a data set of DNA or protein similarities, or
a cladogram allow analogy/homology to be determined?
ANSWER KEY– LO 1.10
A bat’s wing, a whale’s flipper, and a horse’s forelimb contain essentially the same basic bone structures.
These structures are examples of _____, and how do you know?
A) Analogous structures; similar location and function
B) Homologous structures; similar location and function
C) Vestigial structures; similar location and function
D) Analogous structures; unrelated species
E) Homologous structures; unrelated species
Decide whether the following relationships are examples of analogy or homology:
a. A hedgehog’s quill and an agave’s spine
b. A bobcat’s paw and a chimpanzee’s hand
c. An eagle’s wing and a fly’s wing
How would a data set of DNA/ protein similarities or a cladogram allow analogy/homology to be
determined?
a. Analogy; the two species are incredibly
unrelated.
b. Homology; the two species are both mammalian
and the structures have a similar location and
function
c. Analogy; one species is avian and the other
species is an insect – similar function, but
different structure/origin
Due to the fact that analogy is said to
be when structures have a similar
function but different structure or
origin, a set of DNA/protein values
would draw a low percent similarity as
opposed to if homology were present.
A cladogram for analogy would also not
necessarily reveal a common ancestor,
whereas in the case of homology a
common ancestor would be easily seen
on a cladogram.
LO 4.17 The student is able to analyze data to identify how molecular interactions affect structure
and function.
SP 5.1 The student can analyze data to identify patterns or
relationships.
Explanation: As has been stated many times throughout this course
shape, in this case structure, determines function. There are
signal transduction pathways are started by receptor proteins
binding to a ligand to change its shape and causes the entire
sequence of transduction to go off. The ligand attaching to this
receptor protein is like pushing down the first tile in a line of
dominos. For this to start correctly the ligand must be able the
have a matching tertiary structure as the receptor protein site
for binding to occur. If this correct structure isn't perfectly
made then the bind will never happen and neither will the cells
transduction. Proteins and other signal molecules are coded for
by usually thousands of DNA sequences to make that one
specific molecule that will twist and bend to be able to fit the
receptor protein, however if the DNA sequencing is messed up
in any way then the correct signal molecule will never be
made. The codons that compose the long genetic sequences
that are read for transcriptions can sometimes be effected by
insertion, deletion, or frame shift mutations. Any one of these
three mutations could potentially disrupt the entire code for a
certain signal and prevent different types of cell
communications and possibly prevent an important task from
being achieved. These mutations can occur due to splicing,
radiation, and interaction with toxic materials. If given a data
table with a before and after segment of DNA and notice two
base pairs are missing from the second then you know that a
deletion mutation has occurred and can usually assume that an
incorrect protein will be made because when the protein is
being assembled the tRNA will bring the incorrect amino acid
and, its new structure will give it a new function making it
useless for what it was originally supposed to do and the
receptor protein will remain vacant and un-signaled.
•
Multiple Choice: A receptor protein has the shape to
receive a U shaped ligand, but in the transcription
process parts of the genetic code for this ligand were
deleted and a J shaped ligand was produced instead. The
receptor protein will…
•
A) Successfully bind because enough of the receptor
space was filled to start transduction
•
B) Not bind to the ligand but the ligand will set off the
chain reaction by hovering around the area
•
C) Signal the immune system to remove the new ligand
because it doesn’t fit to its receptor site and therefore is
unnecessary
•
D) Not bind and the chain reaction will not be carried out
FRQ: Mutations happen all the time throughout cells and the
human body ATTAGTACC is the original genetic code
a) describe different kinds of mutations that can affect
this sequence. b) write out a new sequence that displays
these mutations, c) describe what these mutations can do
to the proteins structure.
Answer Key
•
•
•
•
•
Multiple choice answer.
Not A- since the ligand doesn’t have the correct
shape it wont bind at all
Not B- signal receptors can only be activated by
a ligand binding to their receptor site not by the
wrong ligand hovering around its area
Not C- if anything the didn’t fit for receptors was
immediately wiped out then there wouldn’t be
hardly any signal molecules floating around
because it would go past the wrong receptor and
immediately be destroyed
Answer is D because- the new signal molecule
has the wrong shape, therefore will not bind to
the receptor site and will not cause the
transduction or any other process to be carried
out.
•
a) Substitution- where one of the bases is
switched with another, completely changing the
letter that was first present
Insertion- where a brand new is base and would
add to the total number of bases making a
sequence 10 bases long when it was originally
only 9
Deletion- when bases of the sequences are gotten
rid of completely and turn the total number of
base pairs from 9 to 8 or less.
b) Original- ATTAGTACC
Substitution- ATAAGTACC
Insertion- ATTATGTACC
Deletion- ATTAACC
c) All of these mutations if done to the correct spots
in the sequence could cause newly shaped
proteins with no appropriate function since their
shape is slightly off and shape defines the
function of a molecule. This could prevent
processes like transduction and scaffolding from
occurring and halting cell to cell communication
Learning Objective: 3.14 - The student is able to apply mathematical routines to determine
Mendelian patterns of inheritance provided by data sets.
Science Practice: 2.2 - The student can apply mathematical routines to quantities that describe natural
phenomena.
General Statement: Mendelian patterns of inheritance can be determined using probability and Punnett squares,
which are mathematical processes based on given data. With clues to genotypic ratios and identities we are able
to develop mathematical routines to solve problems and fill in blanks relating to genotypes, phenotypes, and
inherited traits. With this quantitative data we can predict allelic trends in a gene pool for generations to come.
Multiple Choice: A child has a father with a recessive disease. The child's grandmother on his mom's side also has
the same disease but his grandfather on his mom's side is homozygous dominant. What is the probability that the
child carries the disease?
(a) 100%
(b) 50%
(c) 25%
(d) 0%
FRQ:
This family tree shows the family history of an inherited disease called mercerosis, a recessive disease that causes
the infected to be able to teach AP Biology at Career Center High School in North Carolina. There is no indication
that the disease is sex-linked.
(a) To what extent is it possible to determine if individual A is infected with the disease, a carrier, or homozygous
dominant? Explain.
(b) Individual B was tested and is revealed to have the genotype AA. Determine individual A’s genotype.
(c) Individual A breeds with someone who is a carrier for the disease. Using the information from part b, draw a
Punnett square demonstrating the potential genotypes for the offspring and compare the probabilities for each
genotype in a single child.
Answers:
Multiple Choice: The answer is (b) 50% because it can be determined that the mother’s genotype is Aa, this, along with the given
information that the father’s genotype is aa, allows you to make a simple Punnett square to determine that there is a 50% chance
the disease will be fully inherited.
FRQ:
(a) The data shows that individual A has a father with mercerosis and a mother who does not have the disease. However, it is
impossible to determine whether the mother is a carrier or homozygous dominant. By making Punnett squares for the 2
possibilities (shown below)
we can determine that individual A must be heterozygous or homozygous recessive.
(b) With new information that reveals the genotypes of both parents, a Punnett square can be made (shown below)
to determine that Individual A must be heterozygous (Aa)
(c) A Punnett square can be made using the new information (shown below). This reveals that the offspring must be heterozygous
or homozygous recessive. The Punnett square also reveals that there is a 50/50 chance that any given
offspring will be homozygous recessive, and thus inheriting the disease, or heterozygous, and thus
becoming a carrier. For both situations, the recessive allele will be passed on to the offspring.
LO 4.1 The student is able to explain the connection between the sequence and the subcomponents of a
biological polymer and its properties.
SP 7.1 The student can connect phenomena and models across spatial and temporal scales.
Explanation: Since every polymer is made of many monomers, each different monomer determines the property
of each different polymer. So that all polymers are different, variations in components within biological systems
provide a greater flexibility to respond to changes in the environment. As composed parts of an ecosystem
interact with each other, the resulting interactions allows characteristics not found in the individual part alone.
Information is coded into nucleic monomers that make up nucleic acids. The structures of each nucleotide are a
five-carbon sugar, a phosphate and a nitrogen base. DNA differs in structure, thus differing in function. In
proteins, the specific order of amino acids in a polypeptide interact with the environment to determine overall
shape, which involved the primary, secondary, tertiary and quaternary shape, thus creating different functions.
Examples of proteins whose shape determines their function are enzymes. If a protein loses its shape at any
time, it may no longer be functional.
M.C. Question: Animal cell membranes are made from a phospholipid bilayer.
This bilayer is made from hydrophilic heads and hydrophobic tails that
keep all unwanted compounds out of the cell, while allowing others to
enter. If a protein was outside of the cell wanting to enter through the
bilayer, could it?
a) Yes, because proteins are large enough to pass though
b) Yes, because proteins are polar molecules and can easily pass through
c) No, because proteins also have hydrophilic heads
d) No, because proteins are polar and too large
FRQ: Shape and components of polymers change the function of each polymer.
Explain the a) function of and b) what monomers make up peptide bonds.
Include a diagram to further explain your understanding.
ANSWER KEY – LO 4.1
Animal cell walls are made from a phospholipid bilayer. This bilayer is made from hydrophilic heads and
hydrophobic tails that keep all unwanted compounds out of the cell, while allowing others to enter. If a protein
was outside of the cell wanting to enter through the bilayer, could it?
a) Yes, because proteins are large enough to pass through
b) Yes, because proteins are polar molecules and can easily pass through
c) No, because proteins also have hydrophilic heads
d) No, because proteins are polar and too large
The phospholipid bilayer of cells are hydrophilic, allowing polar molecules to be passed through the wall.
Most compounds such as proteins and carbohydrates are nonpolar and cannot be passed. Also large
molecules such as macromolecules are too large to be passed through the layer. With these stipulations not
many things besides ions, lipids and a few other molecules can pass.
FRQ: Shape and components of polymers change the function of each polymer.
Explain the a) function of and b) what monomers make up the polymer protein.
Include a diagram to further explain your understanding.
a)Each protein carries a different function depending on the shape and
structure of the protein. Proteins carry out almost every task that needs
to be done in the human body such as ATP synthase and DNA replication.
Proteins are used as pumps for active transport, making up our DNA and
breaking down the food we eat.
They are macromolecules that are as diverse as the functions they serve.
b) Proteins are formed by many polymers of amino acids being
bonded together by peptide bonds. Amino acids are made from
a carboxyl group, an H group, an R group and an amino group
which bind to other amino acids groups though the peptide bonds,
creating a polypeptide thus creating a protein.
L.O. 2.23: The student is able to design a plan for collecting data to show that all biological systems (cells, organisms,
populations, communities and ecosystems) are affected by complex biotic and abiotic interactions.
S.P. 4.2: The student can design a plan for collecting data to answer a particular scientific question.
Explanation: Biotic factors are everything living, while abiotic factors are everything non living. For example, if you take an new
species into a new environment, that is an abiotic factor. If the carbon dioxide level in a specific place rises over 60 years, then
that is an abiotic factor. At the cellular level, bacteria
can form a biofilm, an abiotic factor. A biofilm is slime-like
material that bacteria form if they have something to grow on
and room to spread and grow. On the organism level,
the biotic factor would be the predator/prey
relationship. It’s an up and down cycle of population
between the two. For communities and ecosystems,
biotic and abiotic factors would matter, for example,
a food web.
Multiple Choice Question: Which answer is the best reason for
Why it’s necessary for the predator and prey population numbers to stay on a continuous cycle?
A) The predators won’t have anything to eat if they eat all of the prey.
B)The predator will eat the prey until the number of prey get too low, so then some of the predators die because there isn't
enough food for all of them to eat. While the numbers of predators is low, that allows for the prey to reproduce and their
population numbers rise.
C) The prey only reproduce at the certain time of the year, so the predator will eat them until they're almost gone. The predators
wont have any food, so some of them die off. When the prey start to reproduce, that causes their numbers to rise , allowing more
predators to eat, so therefore allowing them to reproduce at a much faster rate, thus allowing their numbers to rise.
D) The predator and prey die and reproduce at different times, causing the number to rise and fall at different times.
Learning Log/FRQ Style Question:
Every year for 40 years you measure the amount of carbon dioxide in the atmosphere in New York City, and you notice that the
level is rising very fast. Give and explain 4 factors, 2 biotic and 2 abiotic, that could be the cause of the fast rising levels of carbon
dioxide in New York City.
ANSWER KEY – L.O. 2.23
Multiple Choice Question: Which answer is the best reason for
Why it’s necessary for the predator and prey population numbers to stay on a continuous cycle?
A) The predators won’t have anything to eat if they eat all of the prey.
-This is true, but it is not specific enough to be the best answer out of the possibilities.
B)The predator will eat the prey until the number of prey get too low, so then some of the predators die because there isn't
enough food for all of them to eat. While the numbers of predators is low, that allows for the prey to reproduce and their
population numbers rise.
This is the correct answer because if the prey population rises, then the predator population will rise because
they have more food to eat and that makes more of them able to reproduce. Once the predators eat so much of the prey that
they don’t have enough food to sustain the whole population, the predators will die off, allowing for the prey to repopulate,
thus starting the cycle again.
C) The prey only reproduce at the certain time of the year, so the predator will eat them until they're almost gone. The predators
wont have any food, so some of them die off. When the prey start to reproduce, that causes their numbers to rise , allowing more
predators to eat, so therefore allowing them to reproduce at a much faster rate, thus allowing their numbers to rise.
-The pattern of rise and fall would not be anything like it should be if this were the case. It doesn’t make much
sense for this to be the best answer.
D) The predator and prey die and reproduce at different times, causing the number to rise and fall at different times.
-This isn’t the reason for the rise and fall of the population. Yes, this does happen, but this would not cause the
numbers to increase and decrease in a related pattern to each other.
Learning Log/FRQ Style Question: Every year for 40 years you measure the amount of carbon dioxide in the atmosphere in New
York City, and you notice that the level is rising very fast. Give and explain an abiotic or biotic factor that could be the cause of the
fast rising levels of carbon dioxide in New York City.
Anything living that could cause carbon dioxide to be released into the atmosphere. Many factories have been built around
there lately, so a large amount of factories would lead to more people in that area, and people give of carbon dioxide when we
breathe. If you combine the amount of carbon dioxide all of those factories are giving off, and the amount that all the people
are giving off, compared to how much they were 40 years ago, that would explain the high increase in carbon dioxide in New
York City in the past 40 years.
LO 2.9: The student is able to represent graphically or model quantitatively the exchange of molecules between an organism and its
environment, and the subsequent use of these molecules to build new molecules that facilitate dynamic homeostasis, growth, and
reproduction.
SP 1.4: The student can use representations and models to analyze situations or solve problems
qualitatively and quantitatively.
Explanation: Many organisms take in molecules from their environment, such as carbon, and utilize these molecules to build new molecules,
such as sugars, that aid in the growth, reproduction, and homeostasis of the organism. Then through biological processes, these molecules that
were formed are then broken down and the original molecules absorbed, such as carbon, are reemitted into the environment. For instance,
plants take in CO2 from the atmosphere through their stomata and utilize this CO2 during the Calvin cycle of photosynthesis in order to produce
the sugar G3P, which is later converted into glucose. This glucose is then broken down during respiration, providing the ATP energy needed by
the plant to grow and perform basic biological tasks. Once the glucose is used in respiration, the plant emits CO 2 back into the environment as a
waste product of this process. This basic exchange of molecules between organisms and their environment can be seen in various other
processes performed by various other organisms as well, such as animals, bacteria, etc. By being familiar with such processes, the student must
be able to quantitatively and/or graphically demonstrate his knowledge of such processes by performing tasks such as graphing the CO2
consumption and oxygen production of a plant and calculating the net gain of ATP of a plant after performing respiration with a given number of
glucose molecules. Through this graphical and quantitative understanding of the exchange of molecules between an organism and its
environment and the subsequent use of these molecules, the student must then be able to interpret representations and models, such as
graphs, in order to analyze a situation, such as the photosynthesis of a plant, and solve specific problems, such as calculating the photosynthetic
rate of a plant and determining the overall significance/effect of this rate.
MC Question: According to the graph on the right, which of the
following accounts for the behavior represented in the graph?
A) The rate of photosynthesis increases due to an increase in kinetic
energy, but eventually decreases due to excess transpiration
B) Increased temperature causes stomata to open wider, increasing
the amount of CO2 absorbed, but eventually denatures the enzyme
Rubisco, causing the rate of photosynthesis to decrease
C) Excessive transpiration causes a decrease in the photosynthetic
rate due to fewer water molecules splitting and give off their
electrons
D) A and C are correct
FRQ: Photosynthesis is a process performed by plants that is central to the production
of food and energy. Describe the specific mechanism in which plants obtain, utilize,
and emit the molecules used and produced during photosynthesis. Given a plant that
absorbs 18 CO2 molecules, calculate the total number of ATP and NADPH molecules
used in the Calvin cycle. Show the work that leads to your answer.
Answer Key: LO 2..9
MC Question: According to the graph on the right, which of the following accounts for the
behavior represented in the graph?
A) The rate of photosynthesis increases due to an increase in kinetic energy, but eventually
decreases due to excess transpiration
B) Increased temperature causes stomata to open wider, increasing the amount of CO2 absorbed,
but eventually denatures the enzyme Rubisco, causing the rate of photosynthesis to decrease
C) Excessive transpiration causes a decrease in the photosynthetic rate due to fewer water
molecules splitting and give off their electrons
D) A and C are correct
A is correct because as temperature increases, kinetic energy increases, causing the reaction rate (photosynthetic rate) to increase as
well. However, the photosynthetic rate eventually decreases because as the temperature reaches a certain point, the amount of water
that evaporates due to transpiration is significantly high, resulting in less water to function as a reactant during the light reactions. C is
also correct because as fewer water molecules are available as reactants, fewer electrons are being emitted from the splitting of these
molecules in order to replace the electron that was sent to the ETC, thus decreasing the photosynthetic rate. B cannot be correct because
although increasing the temperature does denature enzymes, it does not cause the stomata to open wider to allow more CO2 to enter.
FRQ: Photosynthesis is a process performed by plants that is central to the production of food and energy. A) Describe the specific
mechanism in which plants obtain, utilize, and emit the molecules used and produced during photosynthesis. B) Given a plant that
absorbs 18 CO2 molecules, calculate the total number of ATP and NADPH molecules used in the Calvin cycle. Show the work that leads
to your answer.
A) During the light reactions of photosynthesis, a photon of light hits a pigment molecule of Photosystem II and the energy emitted jumps from
one pigment molecule to another until it reaches chlorophyll a and excites and electron. This electron is captured by a primary electron acceptor,
which then sends the electron down an ETC, powering the formation of an H+ gradient, which subsequently powers chemiosmosis and ATP
synthase, producing ATP. This electron then arrives at photosystem I. In order to replace this lost electron, water must be absorbed through the
stomata so that it can be split, emitting O2. Light then hits a pigment molecule in Photosystem I, transferring the energy of the photon across
various pigment molecules until it excites an electron at the chlorophyll a molecule, which is then captured by another primary electron acceptor.
The electron is then sent down another ETC and is added to NADP+ by NADP+ reductase, forming NADPH. Then during carbon fixation of the
Calvin cycle, CO2 is absorbed through the stomata and is added to RUBP by the enzyme rubisco. This creates a six carbon molecule that is split in
half into a three carbon molecule. This process is powered by the hydrolysis of ATP. Then electrons from the NADPH formed during the light
reaction are gained by the carbon molecule, forming the sugar G3P, which is later formed into glucose. ATP is then hydrolysized into ADP to power
the regeneration of RUBP. The glucose formed is then reused by the light reactions during respiration in order to produce ATP, which emits CO2
back into the atmosphere.
B) 3 CO2 molecules used per Calvin cycle 6 cycles
6x9 ATP= 54 ATP molecules used
6x6 NADPH= 36 NADPH molecules used
LO 1.17: The student is able to pose scientific questions about a group of organisms whose relatedness is described by a
phylogenetic tree or cladogram in order to (1) identify shared characteristics, (2) make inferences about the evolutionary
history of the group, and (3) identify character data that could extend or improve the phylogenetic tree.
SP 3.1: The student can pose scientific questions.
For example, scientists can observe certain traits found in a group of differing organisms and compare them using a
cladogram to determine the species’ closeness, and determine possible common ancestors. From this, scientists can then
hypothesize how or why certain species may have diverged from each other and investigate possible methods of speciation
(i.e. allopatric, sympatric).
M.C. Question: Plants in arid regions have developed an adaptation to develop defense mechanisms to defer other
organisms from eating the plant. Consider the spines on a Saguaro cactus (scientific name: Carnegiea gigantea), the
thorns on the Prickly Acacia tree (Vachellia nilotica), and the prickles on the Wood’s rose (Rosa woodsii). All organisms
have defense mechanisms to resist herbivores, and live in similar climates. These defense structures are most likely
A) Homologous structures; because they are not closely related, but exemplify divergent evolution.
B) Analogous structures; because they are not closely related, but exemplify convergent evolution.
C) Vestigial structures; because the defense mechanisms no longer have any use for the organisms.
D) Homologous structures; because they are closely related, and exemplify divergent evolution.
Learning Log/FRQ: Phylogeny is the evolutionary history of a species.
a) Identify and explain two examples of evidence of how we can
determine the phylogeny of an organism.
b) Construct a cladogram from the chart given.
c) Identify and explain two mechanisms for genetic change and
explain how each lead to evolution
Lamprey Clown Fish Gecko Alligator Chicken
Jaw Bones
0
1
1
1
1
4-Legged Locomotion
0
0
1
1
1
Amniotic Eggs
0
0
0
1
1
Feathers
0
0
0
0
1
Example of a cladogram
Answer Key LO 1.17
M.C. Question: Plants in arid regions have developed an adaptation to develop defense mechanisms to defer other
organisms from eating the plant. Consider the spines on a saguaro cactus (scientific name: Carnegiea gigantea), the
thorns on the prickly acacia tree (Vachellia nilotica), and the prickles on the Wood’s rose (Rosa woodsii). All organisms
have defense mechanisms to resist herbivores, and live in similar climates. These defense structures are most likely
A) Homologous structures; because they are not closely related, but exemplify divergent evolution.
B) Analogous structures; because they are not closely related, but exemplify convergent evolution.
C) Vestigial structures; because the defense mechanisms no longer have any use for the organisms.
D) Homologous structures; because they are closely related, and exemplify divergent evolution.
Explanation: Analogous structures are not similar in structure, but are similar in function. Homologous structures can be
similar in structure and function. Since homologous structures in organisms are similar in structure, it is evident that
they came from a common ancestor. Analogous structures only serve the same function in the organisms like a
defense mechanism in plants, because the organisms undergo similar environmental pressures.
Learning Log/FRQ: Phylogeny is the evolutionary history of a species.
a) Identify and explain two examples of evidence of how we can determine the phylogeny of an organism.
b) Construct a cladogram from the chart given.
c) Identify and explain two mechanisms for genetic change and explain how each lead to evolution.
a)
We can compare fossils to observe similar structures in an organism and possible common ancestors, and follow
evolution from the common ancestor. Another example of evidence is DNA sequencing to see how closely related
the species are by sequencing their DNA. The more similar, the more closely related a species is to a common
ancestor.
c) Nondisjunction of chromosomes during meiosis can lead to an
b)
Chicken
increase in gamete variability, if livable, and the extra chromosomes
Alligator
will lead to a divergence of a new species that can survive and
reproduce successfully in its environment. Also, allopatric speciation
Gecko
of a population can lead to genetic change because when a
Carp
population is split by a geographic barrier, gene flow is disrupted and
only certain genes end up on either side of the barrier. By continual
Lamprey
reproduction, the genome of the populations will be different enough
to where interbreeding between the populations does not produce
viable offspring, hence creating new species.
LO 2.33: The student is able to justify scientific claims with scientific evidence to show that
timing and coordination of several events are necessary for normal development in an organism
and that these events are regulated by multiple mechanisms.
SP 6.1: The student can justify claims with evidence.
Explanation: Scientists have used model organisms for investigating the roles of cell signaling, induction, and apoptosis in
development. C. elegans is often used because scientists know its entire cell lineage and it is very reproducible. Cell signaling
is seen as early as the 4 cell stage in C. elegans as cell-surface proteins from cell 4 bind to cell 3. When cell 3 divides, the
posterior end will go on to make intestine and the anterior daughter cell goes on for a different fate. This was supported
when cell 4 was experimentally removed and no intestine formed. This told scientists of the importance of induction early in
development. Observed apoptosis in C. elegans revealed that it occurs the same amount of times and at precisely the same
points in the lineage each time. Based on experimental tampering (removing genes), scientists have discovered that
homeotic genes influence these developmental patterns and sequences as well.
M.C. Question: What would be the result of a mutation in the
cytoplasmic determinants and signal proteins of an early embryo?
A) The cells of the embryo would undergo apoptosis.
B) The cells would not differentiate and specialized tissues would
be absent.
C) The embryo would develop two heads.
D) There would be no effect because healthy cytoplasmic
determinants and signal proteins could be absorbed from its
surroundings.
Learning Log/FRQ-Style Question: Scientists have discovered many types
of genes that code for proteins at specific stages in the developmental process of a Drosophilia embryo.
These genes are proven to have close counterparts in the animal kingdom, supporting the evolution of
development. Identify and describe two types of these genes and predict what would happen if they had a
mutation.
What would be the result of a mutation in the cytoplasmic determinants and signal
proteins of an early embryo?
A)
B)
C)
D)
The cells of the embryo would undergo apoptosis.
The cell would not differentiate and specialized tissues would be absent.
The embryo would develop two heads.
There would be no effect because healthy cytoplasmic determinants and signal proteins
could be absorbed from its surroundings.
Scientists have discovered many types of genes that code for proteins at specific stages in the
developmental process of a Drosophilia embryo. These genes are proven to have close
counterparts in the animal kingdom, supporting the evolution of development. Identify and
describe two types of these genes and predict what would happen if they had a mutation.
The initial activity of developmental genes occur in maternal effect genes or egg-polarity genes.
These genes control the orientation of the egg and consequently the fly as anterior-posterior and
dorsal-ventral axes are formed. Mutations in these genes are often embryonic lethals. Another
important set of genes are homeotic genes. Homeotic genes are regulatory genes that specify
the types of appendages and other structures that form at each segment. Mutations would
cause an entire structure characteristic of a particular segment of an animal to arise in the wrong
segment.
LO 2.18 The student can make predictions about how organisms use negative
feedback mechanisms to maintain their internal environments.
SP 6.4 The student can make claims and predictions about natural phenomena
based on scientific theories and models.
Explanation: Organisms react to environmental conditions with behavioral and
physiological mechanisms. Negative feedback mechanisms maintain homeostasis in
specific internal conditions. They control physiological processes to keep the
condition at its set point. Negative feedback is when a change occurs in a system, a
corrective mechanism is triggered to bring the system back towards the set point.
The goal is stabilization of a certain condition. An example of a negative feedback
mechanism is temperature regulation in animals. If an animal’s body temperature
rises above normal, the animal will sweat to release heat and vasodilation will occur,
which will bring the blood to the surface of the skin to release heat through
convection. If the animal has fur, the fur will lay down to contain heat in the body. If
this causes the body temperature to drop below normal, these mechanisms will be
turned off. Vasoconstriction will occur which holds the blood towards the inside of
the body and goosebumps will form to raise hair and prevent convection. These
processes are continuous in maintaining homeostasis.
Multiple Choice Question: The hypothalamus is a key component in maintaining
homeostasis in humans. Where is it located and why is it located there?
A) In the brain stem so it can easily sensor the outside environment.
B) In the brain so it can be centrally located to act as the control center.
C)In the brain because it is heavily protected by the skull.
D) In the right hand so it can regulate bodily actions.
Free Response Question: A) Describe one negative feedback mechanism in plants or
animals. Explain what would occur if this mechanism failed.
B) List two problems that arise when a negative feedback mechanism does not work
properly. Describe one of them in detail.
Answer Key for LO 2.18
Multiple choice: The correct answer is B) In the brain so it can be centrally located to act as the
control center. This is the correct answer because by being located in the brain, the
hypothalamus can easily control regulation of many autonomic functions of the peripheral
nervous system.
FRQ: A) One negative feedback mechanism in plants is their response to water limitations. If
there is too much water in a plant: ethylene increases and can cause leaf abscission, stomata
close to ensure leaf turgor, and roots change to halt water intake from soil. If there is too little
water in plants: roots grow to reach deeper soil to find water, leaves are lost or wilting/curling of
leaves occur to decrease surface area, and metabolic paths are changed to resist drought. If
plants did not respond to water limitations, they would not survive due to either drought or
drowning.
B) One problem that can arise in response to a failed negative feedback mechanism is
dehydration, which is caused by decreased antidiuretic hormone. A disease that can arise as a
result of a negative feedback mechanism not working properly is Grave’s disease. In a person
with Grave’s disease, the thyroid stimulating immunoglobulin binds to the TSH receptor,
mimicking TSH. This activates secretion of T3 and T4 while the actual TSH blood levels are
decreasing. The TSH levels are decreasing because the hypothalamus → pituitary → thyroid
feedback is not working. This results in very high levels of thyroid hormones.
LO 3.40: The student is able to analyze data that indicate how organisms exchange information in response to internal
changes and external cues and which can change behavior.
SP 5.1: The student can analyze data to identify patterns or relationships.
Explanation: Many organisms have learned to observe the environment in which they live. They use all senses to determine if the environment in
which they live in is safe and fit for survival. For example, a plant’s behavior in a particular environment like a tall grass savanna can be attributed
to an environmental stimulus like the sun. Though many plants don’t have the readily available sunlight they require like those on the forest floor,
plants will behave in many different ways to get sunlight to perform photosynthesis and get the nutrients and minerals they need. Plants are
photoautotroph's meaning they use light to make energy from inorganic compounds, so a plants survival is dependent upon the amount of sunlight it
can obtain to make energy and starch during photosynthesis. Usually the more sunlight a plant receives the more it will show growth behavioral
characteristics. This provides an example of how plants behave because of their attraction to take in more light. The process a plant undergoes in
order to do this is called phototropism and is stimulated by the plant hormone auxin. In addition, in order to continue growing taller a plant also has
to take in more water, nutrients, and minerals so the plant has to constantly has to grow its roots deeper for stability and this purpose. This behavior
is called gravitropism and it’s in response with the plant growing taller.
M.C. Question: In which of the following environments would you expect a plant to the grow
tallest and have deep roots in the soil the best, based on what you know about plant nutrition?
A) On a tropical rain forest floor in a forest that lets
in moderate usable sunlight for 3-4 hours and
receives moderate rain.
B) In the African Sahara Desert that’s hot, receives
usable sunlight for 8-10 hours, is prone to
sandstorms, and receives very little rain.
C) Under a shaded place that receives little usable
sunlight about 1-2 hours and lots of rain.
D) In an open field that has a moderate climate that
receives usable sunlight on average for 4-8 hours
with average rain.
Free Response Question: Say you placed a long
leafy plant upside down in a dark room. You leave
the plant there for about a day. Would you expect the
Auxin levels of the plant to be high or low? Also,
based on what you know about phototropism and
gravitropism what would happen if an agar block
where placed on one side of the plant that was visible
to one light in the room was left on right above the
plant?
Answer Key-LO 3.40
M.C. Question: In which of the following environments would you best expect a plant to grow tallest, and have deep roots in
the soil the best, based on what you know about plant nutrition?
A) On a tropical rain forest floor in a forest that lets in moderate usable
sunlight for 3-4 hours and receives heavy rain.
B) In the African Sahara Desert that’s hot, receives usable sunlight for 8-10
hours, is prone to sandstorms, and receives very little rain.
C) In an shaded place that receives little usable sunlight about 1-2 hours and
lots of rain.
D) In an open field that has a moderate climate, receives usable sunlight on
average for 4-8 hours with average rain.
The correct answer is D, because although most would think a rainforest is
ideal for a plant, it can be good for the plant if the tall trees and taller plants
are letting in more sunlight so a smaller plant on the forest floor can thrive.
In this case, however, the rain forest is letting in less than the normal levels
of sunlight a rain forest plant on the floor should be receiving. As D states,
in an open field the plants will receive the average amount of sunlight
needed in order to be healthy and strong . Also, there isn't as much
competition because its an open field so all plants should be able to thrive
in this kind of environment with good sunlight exposure and rain to
perform photosynthesis well for survival. Also, they will be getting more
water, minerals, and nutrients from the good soil.
Free Response Question: Say you placed a long leafy plant upside down
in a dark room. You leave the plant there for about a day. Would you
expect the Auxin levels of the plant to be high or low? Also, based on what
you know about phototropism and gravitropism what would happen if an
agar block where placed on one side of the plant that was visible to one
light in the room that was left on right above the plant?
I would expect for the auxin levels to be low although there is no light,
because auxin levels rise in the presence of light on the side that isn't
facing the light to help the plant grow in the direction of the light.
Phototropism would drive the plants auxin levels very high in this case
to help the plants stems grow taller in order to help the leaves face the
light in the room to do photosynthesis. Gravitropism would assist in
allowing the plants stems to curve and grow tall towards the light as
well as dig the plants roots deeper for more nutrients. This would
allow for an otherwise upside down plant to grow upright like a
normal plant and orient itself in its own geospatial understanding in
accordance to its environment. This kind of environment the plant is in
causes the plant to behave in these ways to increase its chances of
survival in a less than desired environment.
LO 4.12 The student is able to apply mathematical routines to quantities that describe communities composed
of populations of organism that interact in complex ways.
SP 2.2 The student can apply mathematical routines to quantities that describe natural phenomena.
Explanation: With the use of the Chi-square test, scientists can compare observed population with population that we would
expect to obtain according to a specific hypothesis. For example, you expected a cross between two different colored
flowers would yield a 1:1:2 ratio. By comparing the observed data with the expected ratio, you can tell whether the
deviations between the observed and expected the result of chance, or were they due to other factors. If the null-hypothesis
is accepted, then there is no significant difference between the observed and expected data. Recombinants are offsprings
that have traits that do not match either parent in the parental generation. Scientists can calculate the recombination
frequency by using the equation:
M.C. Question: If the recombination frequency between A and D gene is 14.5%, between D and B is 5.2%, and between A and C is 25.3, What is the
recombination frequency between D and C? The order of the genes on a chromosome is A-D-B-C.
A) 19.4%
B) 4.3%
C) 5.6%
D) 10.8%
FRQ: Mr. Mercer was attempting to predict the outcome of his students’ exam results. He
predicted that the outcome of the results would be in the following ratio; 1 with a level five:
9 with a level three or four: and 2 with a level two or below. After the test was completed,
he counted the students and he found 15 with a level five, 90 with a level three or four, and
27 with a level two or below. Calculate the chi-squared value. Based on your result, state
whether Mr. Mercer’s prediction is accepted or rejected. Explain your reasoning.
Answer Key- LO 4.12
M.C. Question: If the recombination frequency between A and D gene is 14.5%, between D and B is 5.2%, and between A and C is 25.3, What is the
recombination frequency between D and C? The order of the genes on a chromosome is A-D-B-C?
A) 19.4%
B) 4.3%
C) 5.6%
D) 10.8%
FRQ Answer key:
Expected
ratio
1 level five
Observed
#
15
Expected #
11
O-E
4
(O-E)^2
16
(O-E)^2/E
1.45
1/12 * 132 = expected # of level five students = 11
9/12 * 132 = expected # of level three or four students = 99
2/12 * 132 = expected # of level two or one students = 22
Degrees of freedom = 3-1 = 2 (3 different characteristics - 1)
9 level
three or
four
90
99
-9
81
0.82
2 level two
or one
27
22
5
25
1.14
12 total
132
132
total
total
0
total
sum=
3.41
Since 3.41 is less than 5.991, Mr. Mercer’s null hypothesis is accepted.
Therefore, there is no significant difference between his observed and
expected data. Mr. Mercer was right.
LO 3.18: The student is able to describe the connection between the regulation of gene
expression and observed differences between different kinds of organisms.
SP 7.1: The student can connect phenomena and models across spatial and temporal scales.
Explanation: In both prokaryotes and eukaryotes, the process of gene regulation is controlled by
enhancers, teminators and promoters. In prokaryotes, the process of gene regulation is very simple,
because it is started by an inducer and inhibited by a repressor. Regulatory proteins demonstrate positive
or negative control when they either allow transcription or inhibit transcription. The process of gene
regulation in eukaryotes is different because rather than inducers and repressors, activators and
repressors increase or decrease regulation of genes. Regulation in eukaryotes leads to differentiation in all
sorts of body cells. It also causes differentiation of cells in one body, for example, an eye cell is different
from a liver cell because they both have different activator proteins present, which ultimately means
different functions of cells.
Multiple Choice Question: Which of the following
In both eukaryotes and prokaryotes, the process of gene regulation gives way to different phenotypic
statements is NOT true concerning gene regulation?
differences between organisms.
A.) In prokaryotes, the process of transcription and
translation occurs simultaneously.
B.) Eukaryotic gene expression, most notably the
process of translation, occurs only in the nucleus of the
cell.
C.) Gene regulation in both prokaryotic and eukaryotic
cells involves mRNA
D.) In eukaryotes, DNA unpacking involves histone
acetylation
DNA demethylation.
Explain theand
process
of lactose metabolism in
bacteria. Draw a picture to illustrate your answer.
Answer Key:
Multiple Choice Question: Which of the following
statements is NOT true concerning gene regulation?​
A.) In prokaryotes, the process of transcription and
translation occurs simultaneously​
B.) Eukaryotic gene expression, most notably the process
of translation, occurs only in the nucleus of the cell. ​
C.) Gene regulation in both prokaryotic and eukaryotic
cells involves mRNA​
D.) In eukaryotes, DNA unpacking involves histone
acetylation and DNA demethylation.
Explain the process of lactose metabolism. Draw a picture to
illustrate your answer.
Lactose metabolism usually occurs in bacteria, such as E. Coli. These bacterium have a certain
section of their genes that is inducible, meaning that it can be triggered to break down sugar
that is used in cellular metabolism. A repressor gene is located on the bacterial DNA on a
segment called the operon (in between the promoter and the genes) and prevents gene
expression. When lactose appears in the cell, it binds to the repressor molecules, which
causes the repressor to release itself from the DNA. Then, RNA polymerase will begin
transcription since it the DNA is not being blocked by the repressor in the controlling
region.The process begins, and the DNA is transcribed into mRNA and the mRNA is translated
into proteins. The proteins are enzymes that allow for the digestion of lactose sugar.
LO 2.3: The student is able to predict how changes in free energy availability affect organisms, populations, and
ecosystems.
SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and
models.
Explanation: Free energy is total amount of energy in an ecosystem. The energy is taken in by the producers directly from the
sun and thus the producers in a ecosystem have the most energy. Producers obtain this energy through photosynthesis and
chemosynthesis. Now, as other organisms consume the producers they obtain a portion of their energy and the rest of the
energy is converted into low quality heat. The theories that supports this change of energy is the First and Second laws of
Thermodynamics. The first law states the energy can neither be created nor destroyed, so the energy is just downgraded into
low quality heat. The second law states that when energy is transferred, there will be less energy available at the end of the
transfer process than at the beginning. This process continues up the food chain to the top predators, who obtain the least
amount of energy from what they consume. Now, if the amount of free energy availability decreases, the amount of
organisms that that ecosystem can support also decreases for there would not be enough energy to go around, and vice versa,
when the amount of free energy increases, then the amount of organisms that an ecosystem can provide for increases.
Therefor, changes in free energy availability can result in changes in population sizes. So lets say that the amount of sunlight
in an area decreases, then the amount of produces decreases, then the amount of consumers of those produces decreases and
everything that eats that consumer decreases, and this reduction happens to the top of the food chain.
Multiple Choice Question:
All of the following examples are true about free energy EXCEPT…
A. If the producers in an ecosystem decrease then the top predators will follow
B. If 10% of the free energy is transferred to the next tropic level, then 90% of the energy is turned into low quality heat
C. The free energy in the ecosystem comes from the ecosystem
D. The energy in an ecosystem comes from the sun
E. If an ecosystem is 5% efficient, then 90% of the energy is turned into low quality heat, 5% is transfers to the next tropic level, and 5% of the
energy is lost.
FRQ
An ecosystem is 10% efficient, and had 5 trophic levels what would be the amount of energy in each trophic level if you start with 50,000 joules?
How much energy would be converted into low quality heat at each level? What percentage of the original population of predators would not be
able to survive if there
were five top predators and the joules of energy received by the producers was only 10,000 joules and the same efficiency? What problems in the
ecosystem
would occur if this happened?
Answer Key- L.O. 2.3
Multiple Choice Question:
All of the following examples are true about free energy EXCEPT…
A. If the producers in an ecosystem decrease then the top predators will follow
B. If 10% of the free energy is transferred to the next tropic level, then 90% of the energy is turned into low quality heat
C. The free energy in the ecosystem comes from the ecosystem
D. The energy in an ecosystem comes from the sun
E. If an ecosystem is 5% efficient, then 90% of the energy is turned into low quality heat, 5% is transfers to the next tropic level, and 5% of the
energy is lost.
FRQ
An ecosystem is 10% efficient, and had 5 trophic levels what would be the amount of energy in each trophic level if you start with 50,000 joules?
How much energy would be converted into low quality heat at each level? What percentage of the original population of predators would not be
able to survive if there
were five top predators and the joules of energy received by the producers was only 10,000 joules and the same efficiency? What problems in the
ecosystem
would occur if this happened?
If an ecosystem is 10%, and had 5 trophic levels, then you would have 50,000 joules at the producers level( the first trophic level) and then
you would multiply 50,000 joules by .1 for the 10% efficiency and get 5000 joules for the second trophic level with 45000 joules being
degraded to low quality heat, then multiply the remain energy by .1 each energy level, getting 500 joules for the third trophic level and
4500 degraded to low quality heat, 50 joules for the fourth level and 450 joules degraded to low quality heat, and then 5 joules for the fifth
trophic level and 45 joules degraded into low quality heat. Now, when the joules of energy is reduced to 10000, you would repeat the
process in the first part by having 10000 joules in the first level and multiplying the remaining energy by .1, so the energy in the fifth level
would be 1 joules which would only support 1 predator instead of five, causing only 20% of the predators of the original population to
remain. This extreme decline in top predators would possible cause an increase in lower animals such as herbivores. This increase of
herbivores would decrease the number of producers and this could cause erosion and water shortage because herbivores would each
plants, riparian vegetation keeps erosion on the sides water such as streams and rivers from going into the river and blocking the water.
Such negative effects could destroy an ecosystem for an ecosystem needs top predators to keep the lower trophic levels in check, but in
order to do so they need enough energy. The other levels would also reduce in population size for there wouldn’t be enough energy to
support the original populations.
Befo
Afte
LO 1.16: The student is able to justify the scientific claim that organisms share many conserved core processes and features that evolved
and are widely distributed among organisms today.
SP 6.1 : The student can justify claims with evidence.
Explanation: Organisms share many conserved core processes and feature that are widely distributed among organisms today. These
processes provide evidence that all organisms (Archaea, Bacteria, and Eukarya, both extant and extinct) are linked by lines of descent from
common ancestry. Elements that are conserved across all domains of life are DNA and RNA as carriers of genetic information, a universal
genetic code, and many metabolic pathways. The existence of these properties implies that they were present in a universal ancestor and that
present life evolved from a universal ancestor. Phylogenetic trees graphically model evolutionary history and can represent both acquired traits
and those lost during evolution.
M.C. Question: Structural evidence supports the relatedness of
all eukaryotes. Which of the following is not an example of this?
A) Cytoskeleton
B) Membrane-Bound Organelles
C) Large size of the cell
D) Endomembrane system, including the nuclear envelope
Learning Log/FRQ Style Question: Two students are arguing over whether
they believe that the organisms today share the same conserved core processes
as those a long time ago. Student A says they do whereas student B says they
don’t. Explain who is right and give three examples of characteristics in
organisms that can prove this statement.
•
ANSWER KEY – LO 1.16
M.C. Question: Structural evidence supports the relatedness of all eukaryotes. Which of the following is not an example of this?
A)
B)
C)
D)
Cytoskeleton
Membrane-Bound Organelles
Large size of the cell
Endomembrane system, including the nuclear envelope
Learning Log/FRQ Style Question: Two students are arguing over whether they believe that the organisms today share the same conserved
core processes as those a long time ago. Student A says they do whereas student B says they don’t. Explain who is right and give three
examples of characteristics in organisms that can prove this statement and explain.
Student A is correct in proclaiming that organisms did indeed evolve from a single ancestor and that they share many conserved core
processes today. DNA and RNA are protein structures that are used by every living organism and they are the carriers of the genetic
information that make up the organism characteristics. That brings me to my second point, DNA and RNA have a universal genetic code
and this can be proven because we can take genes for insulin in humans and insert it into bacteria to make it at faster rates. This proves
that bacteria and humans have very similar DNA despite being so phenotypically different. The last and final point is metabolic
pathways. These pathways allowed organisms to be less dependent on outside sources of amino acids and other compounds. This
allows for rapid advancement of organisms and these processes are still shown throughout organisms present today.
Learning Obective 3.7 – The student can make predictions about natural phenomena occurring during the cell cycle.
Science Practice 6.4 – The student can make claims and predictions about natural phenomena based on scientific theories and models.
Explanation: One cell cycle, mitosis, which occurs in the body cells, involves 5 stages: Interphase, Prophase, Metaphase,
Anaphase, and Telophase/Cytokinesis. During Interphase, chromosomes duplicate and remain attached to one another.
Following this is Prophase, in which the chromosomes condense and become visible, spindle fibers form, and the
nuclear membrane dissolves. During Metaphase, the copied chromosomes align in the middle of the spindles. Then in
Anaphase, the chromosomes separate into two genetically identical groups and move to opposite ends of the cell.
Finally, in Telophase and Cytokinesis, nuclear membranes form around the two sets of chromosomes, and the cells
separate from one another. The other cell cycle, Meiosis, occurs in the gametes of an organism. Meiosis involves a
reduction in the amount of genetic material. It is divided into two steps: meiosis I and meiosis II. Meiosis I includes
Prophase I, Metaphase I, Anaphase I and Telophase I. Meiosis II (Second division) includes Prophase II, Metaphase II,
Anaphase II and Telophase II. Each of these phases are similar but not identical to the corresponding mitosis
phases. The goal of meiosis is to create four haploid cells, all with different
genotypes. Differences within the daughter cells genes result from:
crossing over (the exchange of genes between homologous chromosomes)
and random arrangement of the paternal and maternal chromosomes.
Multiple Choice Question: If a somatic cell has 24 chromosomes, how
many rounds of ________ cell division will result in 32 diploid cells?
a) Meiotic, 4
b) Mitotic, 4
c) Meiotic, 3
d) Mitotic, 5
Free Response Question: Meiosis and mitosis are very similar
processes. However, they do have their differences. Contrast the
cell cycles of mitosis and meiosis.
Learning Objective 3.7 – Answer Key
Multiple Choice Answer: If a somatic cell has 24 chromosomes, how many rounds of ________ cell division will result
in 32 diploid cells?
a) Meiotic, 4
b) Mitotic, 4
c) Meiotic, 3
d) Mitotic, 5
The correct answer is D because somatic cell division is a result of mitotic cell division. Also, cell division must occur 5
times in order to get 32 cells.
Math:
1
2
4
8
16
32
= 5 Rounds of Cell
x2
x2
x2
x2
x2
Division
Free Response Answer: Mitosis occurs in the somatic cells of the body, whereas meiosis occurs within the sex cells of
the organism. The two processes also vary slightly in the steps involved with the division of the cells. Because Meiosis
results in the formation of 2 gamete cells, two separate rounds of cell division, called Meiosis I and Meiosis II. During
Meiosis I, the tetrads are separated into two homologous pairs and are split up into two different daughter cells. In
Meiosis II, the chromatids are separated into two more daughter cells each, resulting in 4 total daughter cells. In
prophase of mitosis, the duplicated chromosomes form sister chromatids. In Prophase I of Meiosis I, a tetrad is formed
by synapsis of homologous chromosomes. In Mitosis the chromosomes are then lined up in the middle during
Metaphase, and are then broken apart during Anaphase and Telophase to form two diploid cells that are clones of the
parent cell. In Metaphase I, the tetrads are lined up in the middle, and then during Anaphase and Telophase they are
pulled into two cells (where the sister chromatids remain together). Then during Meiosis II, these processes are
repeated to turn the two cells into four unique haploid gamete cells.
Lindsey Hoots – 8th Period
LO 2.28 The student is able to use representations or models to analyze quantitatively and qualitatively the
effects of disruptions to dynamic homeostasis in biological systems
SP 1.4 The student can use representations and models to analyze situations or solve problems qualitatively
and quantitatively.
Explanation: Homeostasis may be disrupted at the ecosystem level as well as the molecular and cellular
level, which affects the health of the organisms involved. Disruption at the ecosystem level for example may
include invasive species, which are generally introduced by humans and exploit the resources and organisms
in a particular area outside of native range. The effects can be seen through graphs or other visuals
quantitatively by taking account for the numbers of both domestic and invasive species and graphing them
over time. Qualitative analysis can be useful with analyzing flow charts or similar models to track the events
that occur overtime when an invasive species is introduced such as an observable changes in colors of
components of the ecosystem or changes in appearances, textures or smells, which could possibly help solve
the initial problem. Any drastic changes in numbers of individuals (quantitative) or changes in appearance
(qualitative) would lead the person to believe that homeostasis has been disrupted and be able to observe
the effects of the imbalance.
M.C. Question: To the right is a diagram of a positive feedback mechanism for blood clotting. What would
happen if a genetic mutation caused an imbalance of homeostasis where a majority of the blood
clotting proteins are inhibited?
a) The platelet plug would be formed faster than normal.
b) The blood vessel would bleed for a much longer period of time.
c) The platelet plug would not be formed at all.
d) The blood vessel would bleed much less than normal.
FRQ: FRQ: The following graph depicts the levels of blood glucose and insulin in the blood of a healthy
person
after eating. Insulin is a hormone released by the pancreas to lower blood glucose levels. Diabetes is a
homeostatic imbalance of blood glucose levels. Explain how the graph would be different for a person with
Type I Diabetes who produces little to no
insulin. Then explain how the graph would be
different for a person with Type II Diabetes
who produces insufficient insulin.
Answer Key: LO 2.28
M.C. Question: To the right is a diagram of a positive feedback mechanism for blood clotting. What would happen if a
genetic mutation caused an imbalance of homeostasis where a majority of the blood clotting proteins are inhibited?
a) The platelet plug would be formed faster than normal.
b) The blood vessel would bleed for a much longer period of time.
c) The platelet plug would not be formed at all.
d) The blood vessel would bleed much less than normal.
FRQ: The following graph depicts the levels of blood glucose and insulin in the blood of a healthy person after eating.
Insulin
is a hormone released by the pancreas to lower blood glucose levels. Diabetes is a homeostatic imbalance of blood
glucose levels.
Explain how the graph would be different for a person with Type I Diabetes who produces little to no insulin. Then
explain how
the graph would be different for a person with Type II Diabetes who
produces insufficient insulin.
For a person with Type I Diabetes, the red line for blood insulin would
be extremely low and close to the bottom of the graph for the entire day
and would fluctuate in very small increments since the pancreas
produces little to no insulin. The black blood sugar line would spike to
much higher amounts and remain high for most of the day since there
would be little to no insulin to keep the levels in balance. The blood glucose would eventually
decrease but not to a sufficient extent and the process would take much longer. For a
person with Type II diabetes, the pancreas would detect high levels of blood glucose and
produce insulin. Since the insulin is inefficient, the insulin levels would follow the same
hill pattern as the diagram, but spike at much higher intervals since the blood glucose
would remain high. The black line for blood glucose would follow a similar pattern as a
person with Type I and remain high throughout the day , spike after eating, and only
decrease in small amounts over an extended period of time. In diseases, any “hills”
would be much higher and much wider than the diagram.
Diagram that further explains the LO to the right.
LO 1.1 The student is able to convert a data set from a table of numbers that reflect a change
in the genetic makeup of a population over time and to apply mathematical methods and
conceptual understandings to investigate the cause(s) and Effect(s) of this change.
SP 1.5 The student can re-express key elements of natural phenomenon across multiple
representations in the domain.
Explanation: Allelic frequency measures the proportion of a given allele in the pool for a
particular gene in a population. These proportions can be multiplied together (assuming they are
randomly assorted, as in a Hardy-Weinberg Equilibrium) to find the phenotypic proportions in
the population. p refers to the dominant allele, and q refers to the recessive; since there are no
other options, p+q=1. Squaring this results in an equation that gives genotypic frequencies,
pp+pq+qq=1; pp represents homozygous dominant, pq represents heterozygous, and qq
represents homozygous recessive.
Problem: A population of corn plants includes a single gene that determines height, and that
gene displays simple dominance, so there is a short allele (which is recessive) and a tall allele
(which is dominant). In the population, you note that 49% of the plants are short (they display
the recessive phenotype). What proportion are heterozygous?
a) .03 b) .21 c) .3 d) .51 e) .7
FRQ: Measures of allelic frequency are only stable when
all five conditions that cause microevolution fail. What are
those five conditions, and what is the name for the state
of stability.
Answer KEY- LO 1.1
Solution: The proportion of short corn plants=pp=.49. p = the square root
of pp = .7; 1-p=q=.3
pq=heterozygous=.7*.3=.21
FRQ: Hardy Weinberg equilibrium occurs when all of the causes of microevolution are not
occurring. If natural selection or mate selection are influencing who survives or reproduces
(respectively), differential migration is causing alleles to exit or enter the population in net
terms, mutation is effecting the allelic locus, or the population is small enough for genetic
drift to effect it, then the requirements for Hardy-Weinberg equilibrium fail.