LO 2.39: The student is able to justify scientific claims, using evidence, to describe how
timing and coordination of behavioral events in organisms are regulated by several
mechanisms.
SP 6.1: The student can justify claims with evidence.
Explanation: Many organisms behavioral events are not just sporadic, many of them are
in regulation with that of internal mechanisms. Things such as the mating period for
certain species is regulated by this; homeostasis is a prime example of this. Certain
animal behaviors are caused to keep the organism healthy and as close to a complete
equilibrium as possible. Things such as the timing of when a tree sheds its leaves to
prepare for winter, or when flowers pollinate in the spring time.
M.C. Question:
Within the endocrine systems negative feedback loop what does
insulin produce/form when entering the liver?
A) Increased glucose uptake within all cells
B) Glucagon breakdown
C) Glycogen formation
D) An unexplained increase in blood sugar
FRQ Question:
Describe in detail how the endocrine system keeps mammalian
organisms in a homeostatic state involving insulin and glucose?
(Hint: Describe the Steps)
Answer Key- L.O. 2.39
M.C. Question:
Within the endocrine systems negative feedback loop what does insulin produce/form when entering the liver?
A) Increased glucose uptake within all cells
B) Glucagon breakdown
C) Glycogen formation
D) An unexplained increase in blood sugar
FRQ Question:
Describe in detail how the endocrine system keeps mammalian organisms in a homeostatic state involving
insulin and glucose? (Hint: Describe the Steps)
Answer:
A negative feed back loop, although negative may increase or decrease a stimulus. A negative
feed back loop works with receptors, so if a stimulus has too high of a level the loop will work to
balance it out, the same goes for if the levels are too low. In the endocrine system glucose, insulin
from the pancreas, glycogen, glucagon and tissue cells are used in this process of homeostasis.
When an animal eats the blood glucose level rises, this is sensed by the nervous system, the
pancreas then receives a signal to release insulin into the blood. In response to the higher insulin
levels, glucose is then stored in cells and in liver cells as glycogen. This drops the glucose levels
back into homeostasis. The same goes for when the levels are too low. Instead of the pancreas
releasing insulin, glucagon is released. This breaks down glycogen stored within liver cells. Which
releases glycogen into the blood bringing the body back into homeostasis. This system is
essentially helping a person to determine when food is needed to support the body and when
eating is not required.
Learning Objective 1.3: The student is able to apply mathematical methods to data from a real or simulated population to
predict what will happen to the population in the future.
Science Practice 2.2: The student can apply mathematical routines to quantities that describe natural phenomena.
Explanation: A mathematical model can help make conclusions for how a population will change over time. For example, by
looking at a scatter plot of bacteria colonies over time an equation for the line of best fit can be derived. This equation will
help to predict how the population size will change in the years to come. Another example is the Hardy-Weinberg
equilibrium equation which helps to analyze allele frequencies in a population. The Hardy-Weinberg equation focuses on the
gene pool and describes genetic balance within a population. This mathematical method can be connected to Science
Practice 2.2 as students will be able to calculate changes in allele frequencies, which could then show evidence of evolution
in a population. Examples of various changes in allele frequencies that will disturb the Hardy-Weinberg equilibrium include
mutations, gene flow, and natural selection.
Multiple Choice Question: Which of the following would not have
a significant impact on allelic frequencies within a population?
A. A new specie of birds immigrate into an small, isolated island.
B. A large earthquake in Africa reduces the size of a meerkat
population.
C. A population of deer begins to choose a mate bases on the
size of the males antlers.
D. A population of iguanas adds an accessible local fruit to its diet
after its main food source is diminished by global warming.
FRQ: In a population of sheep black fleece (B) is dominant to a white
fleece(b). The only information known is that the percentage of the
homozygous recessive genotype(bb) is 36%. First, find the frequency
of each other genotype using the Hardy-Weinberg mathematical
method. Then, predict what will happen in the future if black sheep
are able to hide from predators. Explain your reasoning.
Each line represents the three
possible genotypes.
Answer Key: Learning Objective 1.3
Multiple Choice Question: Which of the following would not have a significant
impact on allelic frequencies within a population?
A. A new specie of birds immigrate into an small, isolated island.
B. A large earthquake in Africa reduces the size of a meerkat population.
C. A population of deer begins to choose a mate bases on the size of the males
antlers.
D. A population of iguanas adds an accessible local fruit to its diet after its main
food source is diminished by global warming.
FRQ: In a population of sheep black fleece (B) is dominant to a white fleece(b). The only information known is
that the percentage of the homozygous recessive genotype(bb) is 36%. First, find the frequency of each other
genotypes using the Hardy-Weinberg mathematical method. Then, explain how you could determine if the
population was evolving over time.
Solution:
1) The frequency of bb is 36%, which means that q2 = 0.36. If q2 = 0.36, then q = 0.6. Because p+q=1 according
to the basic Hardy-Weinberg formula we can conclude that p=0.4. The percentage of homozygous dominant
individuals is equal to 0.42 or 16%. The percentage of heterozygous individuals can be found by taking 2pq or
48%.
2) If Black sheep are better able to hide from predators like wolves then the percentage of heterozygous
individuals and homozygous dominant individuals would both increase while the percentage of homozygous
recessive would decrease proportionally. This is because individuals with more favorable phenotypes are
more likely to survive and pass their traits to subsequent generations.
SP 4.1: The student can justify the selection of the kind of data needed to answer a
particular scientific question.
Explanation: Chemical data which reveals Earth’s early conditions were suitable for creating organic compounds, such
as amino acids ,which are necessary for cellular life in Early Earth. Adenine was also discovered, which is biologically
significant because it is a key component of DNA and RNA, as well as ATP, which is a major source of energy. Geological
data shows that solid rock floated in magma and cooled to form gases such as carbon dioxide, and water vapor which
created an Early Earth which lacked oxygen. Single-celled cyanobacteria formed later which performed photosynthesis
and released oxygen and created biotic diversity amongst Early Earth. Geological data also shows that Earth’s earliest
rocks were created 3.8 billion years ago and the oldest fossils are 3.5 billion years old. Physical data proves that oceans
were formed when the temperatures plummeted and steam was turned into rain. However, larges rocks had a
tremendous impact on the Earth’s oceans and the oceans boiled which created condensation and the disappearance of
the oceans. Although oceans evaporated due to the formation of large rocks, liquid water returned to Early Earth
once temperatures began to rise again.
Multiple Choice Question: What single celled organism was
responsible for the production of oxygen, as well as isolating
carbon dioxide in organic molecules?
A.) Oxygen was not available at all in early Earth’s conditions.
B.) Amoeba
C.) Cyanobacteria
D.) Algae
Free Response Question: Before and after oxygen
became present on Earth, how did organisms
survive? Describe this process and how it is
affected by temperature increases. What are
three gases that were available in Earth’s early
atmosphere?
Answer Key LO-1.32:
What single celled organism was responsible for the
production of oxygen, as well as isolating carbon dioxide
in organic molecules?
A.)Oxygen was not available at all in early Earth’s
conditions.
B.)Amoeba
C.)Cyanobacteria
D.) Algae
Before and after oxygen became present on Earth, how did
organisms survive? Describe this process and how it is affected
by temperature increases. What are three gases that were
available in Earth’s early atmosphere?
Before oxygen was present on Earth, organisms survived by the
process of anaerobic respiration. Organisms used sunlight,
carbon dioxide, and water to stay alive. Oxygen was a byproduct
of this process and then became present in the Earth’s
atmosphere. Oxygen was hazardous to the anaerobic organisms
living on Earth, so the anaerobic organisms resorted to living in
the depths of the ocean, or they would live in the deep soil
where oxygen was not present. When temperature increased,
the rate of anaerobic respiration increased as well, due to the
increase in fermentation. However; If temperatures rose too
high, the organisms would die because their proteins would
denature. Three gases that were available in Earth’s early
atmosphere were water vapor, methane, and carbon dioxide.
Early Earth consisted of only prokaryotes because anaerobic
respiration occurs mostly in prokaryotes.
LO 2.38: The student is able to analyze data to support the claim that responses to information
and communication of information affect natural selection.
SP 5.1: The student can analyze data to identify patterns or relationships
Explanation: In natural selection, for example, the adaption of a population of organisms to their
environment is caused by differential reproductive success. This is caused by reproduction of the fittest,
meaning the one most fit for survival will be the one to reproduce and pass on its successful traits. Natural
selection occurs through interactions between the environment and the variability among individual
organisms in a population. In photoperiodism in plans, changes in the length of night regulate flowering and
preparation for winter. Through this process, different types of plants were developed that are adapted to
short period days or long period days. This is crucial for the development of the flowering process.
Responses to information and communication are vital to natural selection.
M.C. Question: In a population of giraffes longer necks are less common until a drought occurs and only tall
trees are left. The longer necked giraffes can more easily reach these tall tress. Which of the following types
of natural selection will occur during this drought season?
A) Disruptive Selection
B) Stability Selection
C) No natural selection will occur
D) Directional Selection
Learning Log/FRQ-style Question:
A population of gazelles begin to die off due to cheetahs hunting them. One side of the
population is really fast and can out run the other gazelles that will then be killed by the
cheetas.
A) What will occur in the population in order for the gazelles to gain a better survival rate?
B) Graph the change in the population.
C) Explain Darwin’s theory of natural selection and provide a second example.
ANSWER KEY– LO 2.38
In a population of giraffes longer necks are less common until a drought occurs and only tall trees are left.
The longer necked giraffes can more easily reach these tall tress. Which of the following types of natural
selection will occur during this drought season?
A) Disruptive Selection
B) Stability Selection
C) No natural selection will occur
D) Directional Selection
A population of gazelles begin to die off due to cheetahs hunting them.
One side of the population is really fast and can out run the other gazelles
that will then be killed by the cheetas.
A) What will occur in the population in order for the gazelles to gain a
better survival rate?
B) Graph the change in the population.
C) Explain Darwin’s theory of natural selection and provide a second
example.
A) Evolution will occur via natural selection in order to make the survival
rate of the gazelles increase by selecting for the faster trait.
B) See Graph to the Right
C) Natural selectio occurs through interactions between the environment
and the variability among indivisdual organisms in a population. The
product of natural selection is adaption of a population of organisms to
their environment. For an example, in a habitat there are red bugs and
green bugs. The birds prefer the taste of the red bugs, so soon there are
many green bugs and few red bugs. The green bugs reproduce and make
more green bugs and eventually there are no more red bugs.
(sample drawing)
Learning Objective 1.23: The student is able to justify the selection of data that address questions related to reproductive
isolation and speciation.
Science Practice 4.1: The student can plan and implement data collection strategies particular to a specific scientific question.
Explanation: The student be capable of analyzing data related to isolation and speciation in animal populations. The student
should be able to find significance and evidence of speciation and reproductive isolations through comparing and interpreting
data. This can be connected to the Science Practice because students must be able to make claims and predictions based off
scientific theories and models in order to understand the relevance of the data and how it represents the phenomenon of
speciation and reproductive isolation. For an example, the student should be able to explain how data representing biological
diversity can be effected by geographical isolation. Geographical isolation is a type of speciation that results in the emergence
of species. And example of geographical isolation would be a river. If a river emerges and separates a squirrel population, this
will create to separate gene pools. Over time, the phenotypes of the two populations will become more diverse and lead to
the creation of a new species.
Multiple Choice Question: Which of the following
would most likely cause speciation in a population of
chupacabras?
A. A severe drought depleted their food supply, causing
the population to migrate.
B. The population of males is drastically decreased due
to a specific pathogen.
C. An earthquake creates a ridge that divides the
population.
D. The chupacabras adapt to withstand various
ecosystems.
FRQ: A population of deer live in a forest. The ecosystem is change
through the mergence of a river, dividing up the population of deer
into two populations. After a long period of time, the river dries up
and unites the two species of deer. However, researchers observed
that the populations are interbreeding, but also observed that the
deer aren’t producing as much offspring as expected for their
population size. Provide a definition of geographical isolation and and
a possible explanation for how it may have resulted in the researchers
observations.
Answer Key: LO 1.23
Multiple Choice Question: Which of the following would most likely cause speciation in a
population of chupacabras?
A. A severe drought depleted their food supply, causing the population to migrate.
B. The population of males is drastically decreased due to a specific pathogen.
C. An earthquake creates a ridge that divides the population.
D. The chupacabras adapt to withstand various ecosystems.
FRQ: A population of deer live in a forest. The ecosystem is change through the mergence of a
river, dividing up the population of deer into two populations. After a long period of time, the
river dries up and unites the two species of deer. However, researchers observed that the
populations are interbreeding, but also observed that the deer aren’t producing as much
offspring as expected for their population size. Provide a definition of geographical isolation
and and a possible explanation for how it may have resulted in the researchers observations.
The population of deer is a single species. Once the river emerges and divides the
population, the population is divided. This creates two populations which are unable to
mate with each other due to the geographical barrier. Over time, the gene pool of the
populations become divergent from each other, resulting in differences in the two
populations genotypes and phenotypes. This eventually creates two species, which are
unable to have viable offspring with the other do to their genetic diversity. So, once they
are united, many of them interbred but only the deer that bred with the same species
produce offspring, which explains why the offspring numbers were lower than expected.
LO 2.25: The student can construct explanations based on scientific evidence that homeostatic mechanisms reflect
continuity due to common ancestry and/or divergence due to adaptation in different environments.
SP 6.2:The student can construct explanations of phenomena based on evidence produced
through scientific practices.
Explanation: Homeostatic mechanisms regulate an organism’s body through organs, glands, tissues, and cells. These mechanisms
contribute to homeostasis, the optimum physiological condition of the body. Homeotic mechanisms control body temperature,
blood sugar levels, blood pressure etc. A similarity in homeostatic mechanisms between two organisms can be evidence of
common ancestry. For example, the great white shark and the pacific bottlenose dolphin both have the blood vessels called
countercurrent heat exchanger which are involved in thermoregulation. These blood vessels trap heat in the body core and prevent
heat loss from extreme temperatures. However, honeybees adapted differently based on their environment to depend upon social
behaviors by clustering together in cold temperatures to retain heat. Honeybees adapted to rely on social behaviors as a
thermoregulatory mechanism.
M.C. Question:
If a scientist knows that the great white shark and the bluefin tuna share common ancestry, which of the following statements is
most likely true?
a) The bluefin tuna and the great white shark occupy the same ecological
role.
a) The two species live in the same exact environment.
b) The great white shark and the bluefin tuna share similar homeostatic
mechanisms.
a) Because they share common ancestry, both species are likely to suffer
from the same disease.
Learning Log/FRQ-style Question:
All organisms utilize homeostatic mechanisms to regulate internal body processes. Give an example of one type of
homeostatic mechanism and provide an example of an animal that uses this mechanism. Describe and explain how
this homeostatic mechanism helps maintain homeostasis. Finally, indicate how scientists can use homeostatic
mechanisms to trace common ancestry.
ANSWER KEY – LO 2.25
If a scientist knows that the great white shark and the bluefin tuna share common ancestry, which of the
following statements is most likely true?
a) The bluefin tuna and the great white shark occupy the same ecological role.
b) The two species live in the same exact environment.
c) The great white shark and the bluefin tuna share similar homeostatic mechanisms.
d) Because they share common ancestry, both species are likely to suffer from the same disease.
All organisms utilize homeostatic mechanisms to regulate internal body processes. Give an example of one type
of homeostatic mechanism and provide an example of an animal that uses this mechanism. Describe and explain
how this homeostatic mechanism helps maintain homeostasis. Finally, indicate how scientists can use
homeostatic mechanisms to trace common ancestry.
Mammals, like bears, use insulation to reduce the flow of heat between an animal and its environment. A bear’s
fur, which is part of the integumentary system, will raise in reaction to cold weather trapping a thicker layer of
air. This allows the bear’s body to retain heat and maintain optimum body temperature. Scientists can compare
homeostatic mechanisms between two organisms to test for common ancestry. Animals with common
homeostatic mechanisms are likely to share common ancestry. A difference in homeostatic mechanisms can
indicate a divergence due to environmental pressures.
LO 4.18: The student is able to use representations and models to analyze how
cooperative interactions within organisms promote efficiency in the use of energy and
matter.
SP 1.4: The student can use representation and models to analyze situations or solve
problems qualitatively and quantitatively.
Explanation: Multicellular organisms consist of cells and organs that have specialized
jobs. These organs use cooperative interactions in order to complete certain tasks. For
example, food digestion consists of many hormone signals and organs. They each have
a specific job. It’s like an assembly line. The digestion of food actually begins when your
eyes see the food that you are about to eat. Your brain will then tell your mouth to
begin to salivate. Amylase is an enzyme that is formed in your mouth that will begin to
break down sugar. The food then goes on through the stomach which is acidic to
further break down food. The next stop is the small intestine. Pancreatic amylase is
made to break down carbohydrates into glucose that the cells need.
M.C. Question: Which of the following statements concerning multicellular
organisms is false?
A) They have specialized cells.
B) They usually consist of plants, animals, and fungi.
C) They have the shortest life span.
D) They are usually found in eukaryotic lineages.
FRQ: Use the diagram to describe the specifics about the exchange of gases between
an organism and its environment. Why do organisms need to exchange gases with their
environment?
ANSWER KEY- LO 4.18
Which of the following statements concerning multicellular
organisms is false?
A) They have specialized cells.
B) They usually consist of plants, animals, and fungi.
C) They have the shortest life span.
D) They are usually found in eukaryotic lineages.
Use the diagram above to describe the exchange of gases between
an organism and its environment. Why do organisms need to
exchange gases with their environment?
Gases move through the plasma membrane via diffusion. These
molecules such as carbon dioxide and oxygen have no problem
getting through the membrane because of their size. Organisms
must exchange gases in order to get the energy they need. Cells
need to take in oxygen to complete respiration. Organisms release
carbon dioxide so that it doesn’t build up in the cell. This could
cause the cell to become acidic.
Learning Objective 2.22: The student is able to refine scientific models and questions about the effect of complex biotic and abiotic
interactions on all biological systems, from cells and organisms to populations, communities and ecosystems.
Science Practice 1.3: The student can refine representations and models of natural or man-made phenomena and systems in the
domain.
Explanation: The student should fully be able to understand, analyze and explain scientific models, as well as analyze the complex
biotic and abiotic interactions of every system. By doing this, the student should be able to refine and reconstruct scientific models of
laboratory experiments, as well as naturally occurring phenomena. When connecting the objective with the science practice, students
must realize that they need to be fully capable of dissecting how a variety of factors can help or hinder different systems. For
example, the student should be able to realize that the addition of certain ions like nitrate impact the growth of a waterborne species
in a sitting body of water. All biological systems are affected by the interactions of living and nonliving factors, which involve the
exchange of matter and free energy. If nitrate is added to a sitting pond instead of a phosphate, the algal population will bloom
initially before levelling off. While the nitrate and phosphate concentrations directly impact the pond, wind is the abiotic factor
responsible for “flagging,” when trees grow windward due to the wind pressure. Students should be able to extend their knowledge
and interpret how “flagged” trees impact the rest of a biological system. Since flagged trees are essentially injured, the wind can lead
to ineffective trees, which aren't able to contribute oxygen, shelter, and many other resources to its inhabitants.
A.
B.
C.
D.
Multiple Choice Question:
Which of the following is most likely the effect that one season of no forest fires would have on the chaparral biome?
Evergreen trees would adapt quickly to the lack of nutrients produced by fire.
All native mammals, such as deer and goats, would migrate to a new biome.
There would be no effect on the chaparral biome.
Nutrients found in soil and seeds that germinate after a hot fire would be lost, leading the inhabitants of the biome to starve.
FRQ: The west coast had long experienced a long drought, until this year. Rainfall has steadily increased to equate to the yearly average based on
yearly data between 1975-2015. Much of central Nevada, specifically the eastern side of the Sierra Nevada Mountains, is desert. This is due to the
“rain shadow effect.” Explain how biotic and abiotic factors facilitate this phenomena. Discuss at least 1 abiotic factor.
Answer Key: Learning Objective 2.22
Multiple Choice Question: Which of the following is most likely the effect that one season of no forest fires would have on the chaparral biome?
A.
B.
C.
D.
Evergreen trees would adapt quickly to the lack of nutrients produced by fire.
All native mammals, such as deer and goats, would migrate to a new biome.
There would be no effect on the chaparral biome.
Nutrients found in soil and seeds that germinate after a hot fire would be lost, leading the inhabitants of the biome to
starve.
FRQ: The west coast had long experienced a long drought, until this year. Rainfall has steadily increased to equate to the yearly
average based on yearly data between 1975-2015. Much of central Nevada, specifically the eastern side of the Sierra Nevada
Mountains, is desert. This is due to the “rain shadow effect.” Explain how biotic and abiotic factors facilitate this phenomena.
Discuss at least 1 abiotic factor.
As moist air moves away from the Pacific Ocean and hits the mountains, it flows to a higher altitude and rains along the coastal
range. The rain is an abiotic factor and facilitates the growth of coastal redwoods, which are the world’s tallest trees. The rainwater
provides ample hydration and fertilizes the soil, which supports tree growth. Past the coast range, precipitation increases as the
altitude of the mountains increase. While rain is an abiotic factor impacting the shadow effect, wind carries the precipitation to the
higher altitudes. The combination of rain and wind makes the peak of the mountain home to deeply packed snow. However, just over
the snowcapped peak is the part of the system impacted by the “rain shadow effect.” While the rain travels over peaks and
mountains, it can’t travel further than the highest peaks, so the mountain essentially blocks the rain and wind from continuing. This
leads to a desert in the central Nevada area. The biotic factors of the areas that receive high amounts of precipitation vs. no
precipitation vary in characteristics. The plants in the coast range likely have thin cuticles, because they are constantly being
drenched and hydrated. The plants in the central Nevada desert likely have thick cuticles, to retain as much water as possible.
Learning Objective 1.11- The student is able to design a plan to answer scientific questions
regarding how organisms have changed over time using information from morphology,
biochemistry and geology
S.P.- The student can design a plan for collecting data to answer a particular scientific question
•
Explanation- Species have changed throughout time through different mechanisms. At the
biochemistry level, natural selection of species for survival of the fittest has helped ensure
genetic variation. Mutations, and genetic drift have helped contribute to this process.
Evidence that some species have diverged from a common ancestor has been found with
homologous structures being present (which is where similar structures are found in different
species, but they have different functions). Genetic drift has contributed to the evolution of
species, as different environments cause a need for different adaptations. This is seen with
the difference in the Galapagos Island beak size and types with the Finches.
Multiple Choice Question
1.) Which of the following is an example of a homologous structure?
A.) The ability for desert animals to maintain their body temperature during cold
temperatures at night
B.) Insects in the Amazon having wings that blend in with their environment
C.) The front appendages of humans, cats, and bats
D.) The dorsal fin of dolphins and sharks
FRQ- style question- Scientists studying the Forest of Dean
and Forest of Mayberry believe that two species of bears
have evolved from a common ancestor. Propose a scientific
study to determine if species X and species Y come from
the same ancestor, and how speciation occurred.
Answer Key
Multiple Choice Question
1.) Which of the following is an example of a homologous structure?
A.) The ability for desert animals to maintain their body temperature during cold temperatures at
night
B.) Insects in the Amazon having wings that blend in with their environment
C.) The front appendages of humans, cats, and bats
D.) The dorsal fin of dolphins and sharks
FRQ- Scientists can first start the study by obtaining DNA from species X and Y. The
DNA can then be ran to see if there is any genetic sequencing. The DNA test will also
inform the scientists if any mutations seem to be present in certain sequences
between the two species. If the DNA sequences are similar, the next step would be to
determine how environmental factors effected speciation. For this, scientists could
look at the geographical locations of the Forest of Dean and the Forest of Mayberry
and determine if genetic drift could have occurred. Furthermore, abiotic
environmental factors such as temperature and natural resources could have caused
adaptations throughout the bears’ evolution. Finally, scientists can determine if there
was a common ancestor by observing the different structures of the bears. By
determining if they have things such as homologous or analogous structures, different
life expectancies, mating patterns, or different average heights and weights, scientists
can use morphology to see if species X and Y did come from a common ancestor, with
the addition of biochemistry and geology.
Learning Objective 3.25: The student can create a visual representation to illustrate
how changes in a DNA nucleotide sequence can result in a change in the polypeptide
produced.
Science Practice 1.1: The student can create representations and models of natural or
man-made phenomena and systems in the domain.
Explanation: Protein production begins when DNA is transcribed into mRNA. This mRNA
is modified, taken out of the nucleus, and translated into a protein by ribosomes.
Mutations can occur in DNA (not mRNA) which cause the mRNA to be different after
transcription. When this changed mRNA is translated, this often results in a different
protein, but sometimes the protein is unchanged. This is because different codon
sequences can sometimes code for the same amino acid. The polypeptide produced will
only be different if the mutation in the DNA causes the mRNA to change in such a way
that a different primary structure is created during translation.
Multiple Choice Question: Consider the following template strand of
DNA, found in an exon of a particular gene:
3’ TAC TCC CTA ATA GCT 5’
Which of the following diagrams illustrates how a mutation can result
in a different polypeptide?
A) DNA: 3’ TAC TCC CTA
ATA GCT 5’
mRNA: 5’ AUG GGG GAU
UAU CGA 3’
B) DNA: 3’ TAC TCC CTA
ATA GCT 5’
mRNA: 5’ AUG AGG GAU
UAU CGA 3’
C) DNA: 3’ TAC GCC CTA
ATA GCT 5’
mRNA: 5’ AUG CGG GAU
UAU CGA 3’
D) DNA: 3’ TAC TCG CTA
ATA GCT 5’
mRNA: 5’ AUG AGC GAU
UAU CGA 3’
Free Response Question:
Consider the codon TCT, which occurs in the DNA
sequence which codes for protein X. Explain how a
missense mutation could occur, and how a nonsense
mutation could occur. Draw and label a diagram showing
transcription and translation at this codon for one of these
mutations. Which mutation is more likely to drastically
change the function of protein X?
Answer Key – Learning Objective 3.25
Multiple Choice Answer: D
A)
B)
C)
D)
Incorrect because mutations occur in DNA, not mRNA
Incorrect because the polypeptide changed without any mutation in the DNA
Incorrect because there was a mutation in the DNA, but it was a silent mutation that did not change the amino acid sequence of the
protein
Correct because a mutation occurred in the DNA, causing a corresponding change in the mRNA, causing the second amino acid to be
different.
Free Response Sample Answer
The DNA sequence ACA would normally be transcribed into the mRNA sequence UGU, which would correspond to the amino acid Cysteine. A
missense mutation could occur if the third nucleotide was changed from Adenine to Cytosine. This sequence, ACC, would be transcribed as UGG,
and translation would result in the amino acid Tryptophan. A nonsense mutation could occur if the third nucleotide was changed from Adenine to
Thymine. This sequence, ACT, would be transcribed as UGA, a stop codon, which would stop transcription. The nonsense mutation would most
likely have a much more drastic effect on protein X, because it will end transcription too soon, making the protein incomplete, changing the
structure, thereby changing the function.
Missense Mutation
Transcriptio
n
Translation
LO 1.29-The student is able to describe the reasons for revisions of the scientific hypotheses of the origin
of life on earth
SP 6.3-The student can articulate the reasons that scientific explanations and theories are refined or
replaced
Explanation: Learning objective 1.29 relates directly to the hypothesis of Oparin and Haldane that said that early
earth’s atmosphere been a reducing environment where the energy could come form UV radiation and this is where
life was started from a “primitive soup”. This was then tested 30 years later by Miller and Urey using a closed system,
water, heat, hydrogen, methane, and, ammonia which were all thought to exist on earth at this time the only difference
is the theory that on earth there was also sparks of electric energy in the atmosphere not UV radiation from the sun
forming compounds that could collect on rocks like clay. From this experiment in this closed environment Miller and
Urey collected samples and those samples included amino acids, common proteins and, complex hydrocarbons. From
this these amino acids and proteins could have collected and dried on such material like clay which was cool enough to
solidify on and was present on early earth. This experiment has been repeated thousands of times and the results have
been every major building block of life on earth. This test of the theory of the “primitive soup” proved that with the
conditions and the gases thought to be on planet earth at the time life could be started from these basic components
but with the changes in the catalyst. These amino acids and proteins are theorized to come together and make such
basic compounds of life like DNA and RNA in Protobionts that have been discovered in some of the most harshest and
warmest places on planet earth.
Glucose-phosphate
20
m
Glucose-phosphate
M.C Question: What organic gas compounds
could be found in early earth’s atmosphere
Phosphorylase
that when combined to form amino acids and
Starch
proteins?
Phosphat
Maltos Amylase
A) CH4
e
e
B) H2
Mal
C) N2
(a) Simple reproduction. This
(b) Simple metabolism.
If enzymes—in
tos
lipothis case, phosphorylase and
D) A and B
e
amylase—are included in the solution
some is “giving birth” to smaller
E) A and C
from which the droplets selfliposomes (LM).
assemble,
some liposomes can carry out simple
metabolic
reactions and export the products.
Learning Log/FRQ-Style Question: On early earth there were only
organic compounds of gasses such as Methane, Ammonia, Hydrogen,
and, Water Vapor in the atmosphere due to the heat of the surface. How
could these gases come together to form complex compounds that could
build life in early earth’s atmosphere? What complex compounds would
these
reactions make to be able to become the building blocks of early life?
Answer Key
What organic gas compounds could be found in early earth’s atmosphere that when combined to form
amino acids and proteins?
A) CH4
B) H2
C) N2
D) A and B
A and C
On early earth there were only organic compounds of gasses such as Methane, Ammonia, Hydrogen, and,
Water Vapor in the atmosphere due to the heat of the surface. How could these gases come together to
form complex compounds that could build life in early earth’s atmosphere? What complex compounds
would these reactions make to be able to become the building blocks of early life? How could these
compounds be protected and turn into living cells such as prokaryotes?
These gases could come together with the catalyst of an electric current from the atmosphere that would
give the gases enough energy to break their bonds and form new bonds to create new compounds with the
compatibility of hydrogen, nitrogen, carbon and, oxygen. These compounds could come together to form
one of the most important compounds amino acids which would be able to make common proteins which
are necessary to every form of life especially early life which started with single cell prokaryotes. They
also could make hydrocarbons that are very important in life due to the amount of energy stored in there
covalent bonds that make up there complex structures which are mainly made up of early life’s large
amount of hydrogen and carbon atoms. All of these compounds could be protected by protobionts which
can perform basic reproduction and very basic metabolism all surrounded by a membrane. These amino
acids, proteins, and hydrocarbons could be protected by the protobionts and in this environment could start
to form RNA and DNA to form the first living prokaryote cell.
LO 1.22: The student is able to use data from a real or simulated population(s), based on graphs or models of types of
selection, to predict what will happen to the population in the future.
SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models.
Explanation: Inference of a population’s future change based on current data requires the student’s knowledge of natural
selection as part of evolution. Provided with data that indicates a trend or information that implies a factor which impacts
reproductive success of a certain trait or allele, students should be able to infer how the population will change in the
future. For example, given a table listing frequencies of the alleles on the sickle cell gene and assuming that in the
environment, use of pesticides reduces the rate of malaria infection, the student should be able to conclude that the
frequency of the sickle cell allele will decrease.
M.C. Question: The locus for a certain plant gene has two alleles, the dominant A which codes for red petals and the
recessive a which codes for white petals. Given that the frequency of A is 0.75 and the frequency of B is 0.25, what is the
percentage of individuals that are homozygous dominant, homozygous recessive, and heterozygous respectively? Round to
the nearest hundredth.
A)0.60,0.05,0.35
B)0.56, 0.06, 0.38
C)0.49, 0.09, 0.42
D)0.64, 0.04, 0.32
Learning Log/ FRQ Question:
Two types of three-spined stickleback fish have been observed. The open
water stickleback has longer spines to protect it from predators and the
bottom-dwelling stickleback has short spines to reduce predation by
dragonfly larvae. Measurements of the average spine length of two
populations of the fish are below. Assume that an invasive species has been
introduced that sharply reduces dragonfly numbers.
Using the data provided, explain the most likely change in these populations
in the future. What is the cause of this change? Would this change occur if
the number of dragonfly larvae had not decreased? Explain.
ANSWER KEY- LO 1.22
M.C. Question: The locus for a certain plant gene has two alleles, the dominant A which
codes for red petals and the recessive a which codes for white petals. Given that the
frequency of A is 0.75 and the frequency of B is 0.25, what is the percentage of individuals
that are homozygous dominant, homozygous recessive, and heterozygous respectively?
Round to the nearest hundredth.
A)0.60,0.05,0.35
B)0.56, 0.06, 0.38
C)0.49, 0.09, 0.42
D)0.64, 0.04, 0.32
Two variants of three-spined stickleback fish have been observed. The open
water stickleback has longer spines to protect it from predators and the
bottom-dwelling stickleback has short spines to reduce predation by
dragonfly larvae. Measurements of the average spine length of two
populations of the fish are below. Assume that an invasive species is
introduced that sharply reduces dragonfly numbers.
Using the data provided, explain the most likely change in these populations
in the future. What is the cause of this change? Would this change occur if
the number of dragonfly larvae had not decreased? Explain.
The spine length of the bottom-dwelling stickleback will most likely increase and become closer to the average spine
length of the open-water stickleback. Since dragonfly larvae are no longer an environmental factor that favors the
development of short spines, the bottom-dwelling stickleback’s average spine length will increase.
If the number of larvae did not decrease, the change would not occur. There would still be pressure and a reproductive
advantage for the bottom-dwelling stickleback to have short spines, so the phenotype would be more common in that
population.
L.O. 3.7 The student can make predictions about natural
phenomena occurring during the cell cycle.
S.P. The student can make claims and
predictions about natural phenomena based on
scientific theories and models.
Explanation: The cell cycle contains two different cycles of reproduction. Mitosis is
the production of somatic cells and Meiosis produces the productions of gametes.
Within both of these cycles they have their own phases that are very similar.
Mitosis has diploid cells and Meiosis has haploid cells. The mitosis cycle in the cell
can cause cancerous harm.
M.C. Question: Which of the following statements about mitosis is incorrect?
A)The daughter nuclei are not genetically identical to the parent nucleus.
B) Chromosomes separate during anaphase due to the interaction of polar
microtubules from opposite poles pushing against each other.
C)Chromosomes separate during anaphase when the kinetochore
microtubules shorten as they depolymerize.
D)Chromosomes move to the metaphase plate using motor proteins, a kind
of kinesin, attached to spindle fibers.
E)The centrosomes organize the microtubules of the spindle fibers.
Learning Log/ FRQ Question: Meiosis reduces chromosome number and
rearranges genetic information. Several human disorders occur as a result
of defects in the meiotic process. Identify ONE such chromosomal
abnormality ; what effects does it have on the phenotype of people with the
disorder? Describe how this abnormality could result from a defect in
meiosis.
Answer Key
M.C. Question: Which of the following statements about
mitosis is incorrect?
A)The daughter nuclei are not genetically identical to
the parent nucleus.
B) Chromosomes separate during anaphase due to
the interaction of polar microtubules from opposite
poles pushing against each other.
C)Chromosomes separate during anaphase when the
kinetochore microtubules shorten as they
depolymerize.
D)Chromosomes move to the metaphase plate using
motor proteins, a kind of kinesin, attached to spindle
fibers.
E)The centrosomes organize the microtubules of the
spindle fibers.
Meiosis reduces chromosome number and rearranges
genetic information. Several human disorders occur as
a result of defects in the meiotic process. Identify ONE
such chromosomal abnormality ; what effects does it
have on the phenotype of people with the disorder?
Describe how this abnormality could result from a
defect in meiosis.
One chromosomal abnormality is Turner Syndrome
which has one X missing or partial of an X missing.
This effects the mental state and may not be able to
be pregnant in the future. This abnormality could
result in a defect of meiosis from of nondisjunction.
LO 2.30: The student can create representations of models to describe nonspecific immune defenses in
plants and animals.
SP 1.1: The student can create representations and models of natural or man-made phenomena and systems
in the domain.
SP 1.2: The student can describe representations and models of natural or man-made phenomena and
systems in the domain
Explanation: Plants and animals have nonspecific immune defense to protect themselves from pathogens.
Animals have a 1st line of defense that is external defense, and a 2nd line of defense that is internal defense.
Animals use skin, mucous, and hair that physically blocks/traps invaders, coughing and sneezing that
eliminates invaders etc.(1st line of defense). Within the 2nd line of defense, patrolling cells/proteins are non
specific and don’t have memory. However, they’re effective at ridding the body of some invading
pathogens. Leukocytes are phagocytic white blood cells. Macrophages, neutrophils and natural killer cells
are examples of leukocyte that kills invading pathogens with different characteristics. Macrophages are
long-lived in the body, neutrophils are abundant, but only live for roughly 3 days. Natural killer cells kill
virus-infected cells and cancer cells. An additional response to pathogenic invaders is a fever. Activated
macrophages release signal molecule cytokines to trigger the brain to increase normal body temperature
(Inflammatory response) to inhibit bacterial growth. Plants don’t have specific immune defense like
animals do. Plants have cell walls that physically protect the plant from invaders. Plants also use
hypersensitive response; they can sense pathogens invading them. When the plant is infected, the plant
burst infected cells via oxidative burst.
M.C. Question: Which of the following statements concerning nonspecific immune defense in animals is
false?
A)Neutrophils are very abundant, but only live for roughly 3 days.
B)Inflammatory system increases body temperature to destroy invaders.
C)Hair works as 1st line of defense, and traps invaders.
ANSWER KEY – LO 2.30
M.C. Question: Which of the following statements concerning nonspecific immune defense in
animals is false?
A)Neutrophils are very abundant, but only live for roughly 3 days.
B)Inflammatory system increases body temperature to destroy invaders.
C)Hair works as 1st line of defense, and traps invaders.
D)Macrophages are long-lived in the body.
E)None of the above.
FRQ-style Question: Plants don’t have specific immune defense like animals
do. How do plants protect themselves from invaders? What are the
steps? Does it have memory?
-Plants use hypersensitive response to protect themselves from pathogens. When
bacteria give off proteins to infect the plant, the R gene senses the protein, and
then undergoes hypersensitive response. Then they set off an oxidative burst, free
up all the excess highly reactive oxygen they have inside the cell, and destroy
them. Then proteins will be released that are going to change the cell wall of all
the adjacent cells, so they’re not be able to infected. It doesn’t have memory, so
the next time it’s infected by that pathogen, it has to undergo hypersensitive
response again.
LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted
for
by Mendelian
SP 6.3
The studentgenetics.
can articulate the reasons that scientific explanations and
theories
are refined
or basic
replaced.
Mendel established
a few
discoveries for genetics that work in certain cases. Those
discoveries were that the hereditary determinant were genes that each had for each particular
trait. For example, a heterozygous individual carries a dominant allele and a recessive allele,
making up the gene pair. However, only one allele from the pair is passed on to the gamete.
Although these applied in some cases, they did not explain how codominance, incomplete
dominance, pleiotropy, etc. occurs. For example, with a cross between two homozygous
individuals, one recessive and one dominant, Mendelian genetics states that all offspring with
exhibit the dominant phenotype; however, if the gene exhibits incomplete dominance, then
the offspring will be a mix of the two original phenotypes. The student can see that while
MC
Question
Mendelian
Genetics can apply, revisions and new laws have to be made to account for
occurrences
likefollowing
codominance,
incomplete dominance, etc.
Which
of the
would
Learning Log/FRQ-style Question
follow the inheritance patterns
established by Mendelian
genetics?
a) Mitochondrial DNA inherited
from the maternal side.
b)Polygenic traits such as eye
color where multiple alleles
control the phenotype.
c) Seed homozygous dominant for
round shape crossing with
homozygous recessive for oval
shape and producing all round
offspring.
d)A brown cat and white cat
mating and producing white
Why is incomplete dominance not accounted
for by Mendelian genetics? Provide an
example of what would occur according to
Mendelian genetics and an example of what
actually occurs.
The answer is C because the cross
between both seeds perform exactly as
Which of the following would follow the inheritance
patterns established by Mendelian genetics?
Mendel predicts. Because both parents
a) Mitochondrial DNA inherited from the maternal side.
are homozygous, they have to pass down
b)Polygenic traits such as eye color where multiple alleles
the same exact allele to every offspring.
control the phenotype.
Therefore each offspring will be
c) Seed homozygous dominant for round shape crossing
heterozygous. This shows that many
with homozygous recessive for oval shape and producing
patterns can’t be explained by Mendelian
all round offspring.
genetics; new laws has to be introduced
d)A brown cat and white cat mating and producing white
to account for all of the strange
offspring with brown spots.
e)The facial hair gene present in both males and females,
inheritance patterns occurring. Strange
but expressed differently.
inheritance patterns: codominance,
Incomplete dominance is not accounted for by Mendelian
incomplete dominance, etc.
genetics because Mendelian predicts that a person, etc. that is
heterozygous will exhibit the dominant trait’s phenotype.
However with incomplete dominance, both of the traits are
exhibited. For example, a cross between a homozygous dominant
red flower crossed with a homozygous recessive white flower:
based on Mendel’s predictions, all of the flowers would be the
dominant red color. However, Mendelian genetics doesn’t account
R R occurs, meaning that all
for the fact that incomplete dominance
of the flowers are pink because Rr
a mix
Rr of the two phenotypes.
MC Answer
Answer Key- LO
3.16
r
r
Rr Rr
•
LO 1.30: The Student is able to evaluate scientific
hypotheses about
the origin of life on
Earth.
SP 6.5:
The student can evaluate alternative
RNA is the oldest genetic material, prior to the evolution of organic molecules. Initially, Earth only had inorganic molecules, lacking
any large magnitude of oxygen. The dominant
gases in the atmosphere
at the time were water vapor, carbon dioxide, and
scientific
explanations.
ammonia, spewed out by volcanoes after the sun erupted and crusted over. The inorganic molecules were monomers, building
blocks, which are amino acids and nucleotides. Once they joined together, they formed polymers, allowing molecules to replicate,
store, and transfer information. In the scientific world, there is skepticism around whether or not these combinations of
monomers occurred within a solution or a solid reactive surface. The first organisms were cyanobacteria, and they used the sun
for photosynthesis. Later on, they evolved into the multicellular organisms we see and are today. Protobionts are the product of
the combination of polymers, and they are small pieces of membrane that help regulate a stable internal environment. These
organisms had to ability to replicate, and cells, the basic units of life, passed on this genetic material, forming different species and
generations of evolution. Oparin’s hypothesis, well-known and respected, involves the primordial soup, which is the reactions of
many essential elements and compounds to form even more complex compounds. The Miller-Urey experiment stressed the idea
of atmospheric elements, which they believed to be abundant; hydrogen, methane, ammonia, and water vapor. They figured
oxygen wasn’t abundant due to the lack of living organisms at the time. Their theory claimed that lightning created the building
blocks of life, amino acids and sugars. Deviating from the original formation of reactions, Alexander Smith proposed that mineral
crystals in clay rearranged molecules into compounds and the building blocks of life. One of them most famous and well-known,
especially by children is the ice age. This theory suggests that ice was covering organic compounds, and once the Ice Age ended,
the organic compounds were released. Out of all these theories, it seems evident that RNA was the equivalent of DNA at the time
and was the first nucleic acid for coding genes and other materials. Now, it is used as an off-on switch for genes, but it’s
importance was unrivaled during the time of inorganic compounds. Lastly, Pansmeria is the claim that blasted material from some
other planet brought coding material for living organisms to Earth.
•
Multiple Choice Question:
–
•
Archaeabacteria were the first organisms on Earth. How do they differ from bacteria?
a. Archaea have intracellular organelles, while bacteria do not
b. Bacteria have intracellular organelles, while archaea do not
c. Bacteria has circular DNA, while archaea does not
d. They have different numbers of polymerases
FRQ:
– Scientists claim there is evidence of mitochondria and chloroplasts once being primitive bacterial cells. Identify
three pieces of evidence to justify this claim. Identify the theory behind these claims and explain how
mitochondria and chloroplasts arose from bacteria.
ANSWER KEY- LO 1.30
Archaeabacteria were the first organisms on Earth. How do they differ from bacteria?
a. Archaea have intracellular organelles, while bacteria do not
b. Bacteria have intracellular organelles, while archaea do not
c. Bacteria has circular DNA, while archaea does not
d. They have different numbers of polymerases
Scientists claim there is evidence of mitochondria and chloroplasts once
being primitive bacterial cells. Identify three pieces of evidence to justify this
claim. Identify the theory behind these claims and explain how mitochondria
and chloroplasts arose from bacteria.
-The size and structure of bacteria resemble that of the mitochondria and
chloroplasts. They all have ribosomes attached to them. Another piece of
evidence is that when engulfed, all three of these structures replicate by
binary fission once their host cell enters mitosis. As well, mitochondria and
chloroplasts have their own sets of DNA, proving they were once unicellular
aerobic bacteria with their own identities, until they were engulfed by
bacteria, through a process called endophagocytosis.
-This is the Endosymbiotic Theory.
-Mitochondria and chloroplasts arose from bacteria due to aerobic bacteria
being engulfed by the archaea. Once inside, they were not digested and the
aerobic bacteria began to grow, causing an overpower of energy to make
bacteria leak, allowing it to also breath. The two organisms could not live
independently of each other, so that’s where endosymbiosis comes in. The
structures depend on each other.
LO 3.9: The student is able to construct an explanation, using visual representations or narratives, as to how DNA in chromosomes is transmitted to the next
generation via mitosis, or meiosis followed by fertilization.
SP 6.2: The student can construct explanations of phenomena based on evidence produced through scientific practices.
Explanation: DNA in chromosomes is replicated during the processes of mitosis and meiosis. Mitosis creates a cell with the same amount of DNA that was found in the
original cell. Meiosis on the other hand creates four gametes by undergoing replication twice. The first replication will involve creating two daughter cells that don’t have
the same DNA due to crossing over that occurred in the first cycle of cell division. The cell will then undergo another round of cell division that further mixes the DNA and
creates four different gametes or sperm cells that have a complete haploid set of chromosomes from the original DNA, which makes each gamete unique. A picture that
shows the process of meiosis will show where the chromosomes in DNA cross over and the two cycles of replication that occur in meiosis. Once the gametes are made, the
DNA will then be transferred to the next generation through fertilization. Fertilization involves a sperm cell fertilizing an egg cell or two gametes combining, which causes
the DNA in both the sperm and egg cell to combine and create a zygote that will result in an offspring, which inherits DNA from both the male and female gamete. The
combination of two haploids will then create a diploid cell with genetic variability.
M.C. Question: During meiosis, between which phases is there a chance for the chromosomes to fail in replicating?
A) Prophase I and Anaphase I
B) Prophase II and Telophase II
C) Metaphase II and Telophase II
D) Telophase I and Prophase II
FRQ-style Question: Sexual reproduction involves two parents providing half of their chromosomes to create a zygote.
What is the process in humans beings that allow them to create cells that contain half of their chromosomes? Briefly
explain this process and the final results from it. What process allows humans to create a zygote? Explain this process
and draw a diagram of this process with proper labels.
Answer Key - LO 3.9
During meiosis, between which phases is there a chance for the chromosomes to fail in replicating?
A) Prophase I and Anaphase I
B) Prophase II and Telophase II
C) Metaphase II and Telophase II
D) Telophase I and Prophase II
Sexual reproduction involves two parents providing half of their chromosomes to create a zygote.
What is the process in humans beings that allow them to create cells that contain half of their chromosomes?
Briefly explain this process and the final results from it. What process allows humans to create a zygote?
Explain this process and draw a diagram of this process with proper labels.
Meiosis is the process in human beings that allow them to create haploid cells, which contain
half of their chromosomes. Meiosis involves two stages that are almost identical but result
in different amount of cells. The first stage involves making two daughter cells that have
genetic variation through crossing over. These daughter cells contain 46 chromosomes. The
second stage involves another cell division of the daughter cells that result in four haploid
cells or gametes that contain 23 chromosomes each. Each of the gamete cells are unique to
each other. Fertilization allows humans to create a zygote. This process involves a sperm
cell penetrating through an egg cell and fertilizing it. By fertilizing it, the haploid cells will
create a complete diploid cell that contains the 46 chromosomes, 23 chromosomes from
each parent.
LO 2.8: The student is able to justify the selection of data regarding the types of molecules that an animal,
plant, or bacterium will take up as necessary building blocks and excrete as waste products.
SP: 4.1: The student can justify the selection of the kind of data needed to answer a particular a scientific
question.
Explanation: The main macromolecules that organisms use are proteins, carbohydrates, lipids, and
nucleic acids. Every macromolecule is composed of smaller molecules and atoms. These building blocks
(monomers) are brought in from the environment and used to create macromolecules (polymers).
Carbon is used to synthesize carbohydrates, proteins, lipids, or nucleic acids. Nitrogen is used to make
proteins and nucleic acids. Phosphorus is used in nucleic acids and some lipids. The process of
combining molecules to create macromolecules is dehydration synthesis. In dehydration synthesis, the
hydrogen of a molecule binds with the hydroxyl group of another molecule to form water. The water
leaves (hence dehydration) and the molecules are bound together. The opposite of this reaction,
hydrolysis, involves adding water to split the molecules back up again. The splitting of these
macromolecules releases energy that the organism can use and the individual building blocks are
excreted from the organism.
M.C Question: What would happen if you heated a polymer to 130 degrees Celsius?
A) A protein would do the work faster as a result of the increased
temperature.
B) DNA would begin to replicate.
C) A polysaccharide would be synthesized from several monosaccharides
D) A protein’s tertiary structure would begin to unravel
E) The polar head of a lipid would fall off leaving just the nonpolar tail
Learning log/FRQ question:
Describe the structure and function of 2 of the 4 main macromolecules
used by organisms and give 2 examples for each.
Answer Key: LO 2.8
M.C Question: What would happen if you heated a polymer to 130 degrees Celsius?
A) A protein would do the work faster and more effectively as a result of the increased
temperature
B) DNA would begin to transcribe itself into RNA
C) A polysaccharide would be synthesized from several monosaccharides
D) A protein’s tertiary structure would begin to unravel
E) The polar head of a lipid would fall off leaving just the nonpolar tail
Desribe the structure and function of 2 of the 4 main macromolecules
used by organisms and give an example for each.
Carbohydrates, like Glucose or Sucrose, are made up of monosaccharides (simple sugars) joined together via
dehydration synthesis. Their function is to give immediate energy to the organism by their hydrolysis.
Proteins are made up of amino acids synthesized together by peptide bonds. Proteins are described in four
structures: primary, secondary, tertiary, and quaternary. Primary is just the amino acid sequence; the amino
acids bonded together by the peptide bonds. Secondary is when the chain of amino acids folds into a Alpha
helix or a Beta-pleated sheet. Tertiary is the overall 3D shape of the protein, which is based on the original
amino acid sequence and the properties of the amino acids (polar, nonpolar, etc). The quaternary structure is
when multiple Tertiary structures bind together to create a larger protein. Proteins have a variety of
functions from cell communication to catalyzing reactions. An example of a protein is ATP synthase which
phosphorylates ADP molecules, creating ATP. Also, Helicase is a protein that unwind the double-stranded
DNA molecules.
LO 3.11:The student is able to evaluate evidence provided by data sets to support the claim that heritable information is passed
from one generation to another generation through mitosis, or meiosis followed by fertilization.
SP 5.3: The student can evaluate the evidence provided by data sets in relation to a particular scientific question.
Explanation: Through the analysis of different data sets, students can evaluate whether heritable information is passed from one
generation to the next by fertilization. Analysis of Punnett squares can indicate the likelihood of having children with certain
genotypes and phenotypes as well as the chances of certain disease or condition from their parents. Pedigrees can indicate what
type of disorders are found in the family and whether they are inherited as a sex-linked disorder or possibly a mitochondrial
disorder. Due to random fertilization and the sixty-four trillion unique combinations for offspring, these data sets are widely used
to predict the chances of inheriting certain characteristics. Meiosis plays a large role in this process as there is also much
randomization in the process which leads to less clarity in which traits are being passed down.
M.C. Question: In corn, the trait for tall plants (T) is dominant to dwarf and colored kernels
(C) are dominant to white kernels. In a cross of corn, the probability of the offspring being
tall is 0.5 while the probability of colored kernels is 0.75. Which represents the parental
genotypes?
a.
TtCc X ttCc
b.
TtCc X TtCc
c.
TtCc X ttcc
d.
TTCc X ttCc
e.
TTCc X TtCC
Learning Log/FRQ-style Question: In a cross between two plants, one completely
heterozygous tall green (SsYy) and the other heterozygous tall yellow (Ssyy), the following
results were observed: 54 tall green, 50 tall yellow, 23 short green, 29 short yellow
a.
Explain the importance of random fertilization in this process.
b.
Calculate the expected phenotypic ratio of offspring. You may use Punnett squares
for your description but the results must be discussed.
c.
Use x2 calculations to determine whether or not the variations in the observed
results could be due to chance.
ANSWER KEY—LO 3.11
In corn, the trait for tall plants (T) is dominant to dwarf and colored kernels (C) are dominant to white kernels.
In a cross of corn, the probability of the offspring being tall is 0.5 while the probability of colored kernels is
0.75. Which represents the parental genotypes?
a.
b.
c.
d.
e.
TtCc X ttCc
TtCc X TtCc
TtCc X ttcc
TTCc X ttCc
TTCc X TtCC
In a cross between two plants, one completely heterozygous tall green (SsYy) and the other heterozygous tall
yellow (Ssyy), the following results were observed: 54 tall green, 56 tall yellow, 23 short green, 27 short yellow
a.
Explain the importance of random fertilization in this process.
b.
Calculate the expected phenotypic ratio of offspring. You may use Punnett squares for your description
but the results must be discussed.
c.
Use x2 calculations to determine whether or not the variations in the observed results could be due to
chance. ‘
a.
b.
c.
Random fertilization is important as there are 64 trillion unique combinations of an egg and a sperm
cell and every combination leads to different possibilities for the offspring.
The ratios indicate 3 Tall green, 3 tall yellow, 1 short green, 1 short yellow, meaning the ratio is
3:3:1:1. Since there are a total of 160 plants, the ratio is 60:60:20:20. This means the expected
phenotypes are 60 tall green, 60 tall yellow, 20 short green and 20 short yellow.
𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒅−𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅
𝟐
The equation for Chi-Square Test is
. The Chi-Square value is 0.6 + 0.266667 + 0.45
𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅
+ 2.45 = 3.7666. For the 3 degrees of freedom, the critical value is 7.815. Since the chi square value is
less than 7.815, we reject the null hypothesis and conclude that the observed results do fit the
expected ratio and are not due to chance.
LEARNING OBJECTIVE 2.24
ANALYZE AND IDENTIFY PATTERNS AND RELATIONSHIPS BETWEEN A BIOTIC AND ABIOTIC FACTOR AND A
BIOLOGICAL SYSTEM (CELL,ORGANISM,POPULATION,COMMUNITIES OR ECOSYSTEM)
SCIENCE PRACTICE
ANALYZE DATA TO IDENTIFY RELATIONSHIPS AND OR PATTERNS
THE ABILITY TO IDENTIFY PATTERNS AND RELATIONSHIPS BETWEEN BIOTIC AND ABIOTIC FACTORS IS VERY
BROAD, BUT ONE OF THE MOST IMPORTANT CONCEPTS IN BIOLOGY. WHEN EXAMINING AN ECOSYSTEM IT IS
IMPORTANT TO REALIZE THE INTERDEPENDENCY OF BIOTIC AND ABIOTIC FACTORS. THE CLIMATE OF AN
ECOSYSTEM , AN ABIOTIC FACTOR, GREATLY AFFECT A BIOTIC CELL IN THAT ECOSYSTEM. TEMPERATURE CAN
AFFECT HOW WELL A CELL FUNCTIONS, IF IT GETS TO HOT IT CAN DENATURE THE CELL. IN THE SAME WAY
BIOTIC FACTORS SUCH AS BIOMASS MASS CAN AFFECT AN ABIOTIC FACTOR LIKE AMOUNT OF CO2 IN THE
ATMOSPHERE. BIOTIC, LIVING ORGANISMS ALSO INTERACT WITH EACH OTHER, ON A SMALLER SCALE THE
RELATIONSHIPS AND PATTERNS BECOME EVIDENT. MUTUALISM IS AN EXAMPLE OF THIS INTERACTION, THE BEE
GATHERS A FOOD SOURCE FROM THE FLOWER, BRINGING THE FLOWERS POLLEN TO ITS NEXT FLOWER. ON A
LARGER SCALE THE AVAILABILITY OF WATER CAN HAVE A TREMENDOUS IMPACT ON A POPULATION OR THE
ENTIRE ECOSYSTEM, CAUSING A HUGE CHANGE IN POPULATION
ACCORDING TO THE GRAPH APPROXIMATELY WHAT IS THE
OPTIMAL TEMPERATURE FOR PHOTOSYNTHESES TO
OCCUR. WHY IS THIS A PARABOLIC GRAPH INSTEAD OF A
LINEAR GRAPH, JUSTIFY YOUR ANSWER. EXPLAIN HOW A
DROP IN TEMPERATURE WOULD AFFECT THE
CONCENTRATION OF CO2 IN THE PLANT, AND THE RATE OF
PHOTOSYNTHESIS.
THIS NUTRIENT IS NEEDED IN PLANTS TO MAKE DNA AND
PROTEINS.
a. Carbon
b. Magnesium
C. Nitrogen
D. Potassium
FRQ ANSWER
30-33˚C
THE GRAPH IS PARABOLIC BECAUSE AFTER 30˚C THE RATE OF PHOTOSYNTHESIS BEGINS TO
DECREASE. THIS IS DUE TO THE FACT THAT ENZYMES NEEDED FOR PHOTOSYNTHESIS BEGIN
TO BRAKE DOWN AND BECOME DENATURED.
A DROP IN THE TEMPERATURE WOULD CAUSE THE RATE OF PHOTOSYNTHESIS TO DECREASE,
THE DROP IN TEMPERATURE ALSO MEANS THAT THE CONCENTRATION OF CO2 WOULD
DECREASE. BECAUSE THE AMOUNT OF CO2 AVAILABLE IS DEPENDENT ON THE
TEMPERATURE.
THIS NUTRIENT IS NEEDED IN
PLANTS TO MAKE DNA AND
PROTEINS.
A. CARBON
B. MAGNESIUM
C. NITROGEN
D. POTASSIUM
LO 1.14: The student is able to pose scientific questions that correctly identify essential processes of shared, core life processes
that provide insights into the history of life on Earth.
SP 3.1: The student can pose scientific questions.
Explanation: What does every organism have in common that joins us all? What is the base molecular unit?
What cellular parts and systems signal that organisms share a common ancestor? All organisms have DNA,
RNA, genes and proteins, which points to a first organism that passed DNA to the next generation. These
genetic codes and information are essential for telling cells what process to carry out. Organisms carry out
processes, some that are shared with others. Animals don’t have cell walls like plants, but both perform
mitosis and have a nucleus as well as mitochondria. Single Celled organisms perform processes inside the
cell although both are made of cells and perform cellular respiration. The sharing of cellular processes and
genetic information points back to a common ancestor that passed down these shared processes to help life
get to where it is today. The relationship between identifying essential processes of shared, core life
processes and the ability to pose scientific questions, is that you have to be able to pose scientific questions
to identify essential processes of shared core life processes. Without posing questions, you can’t fully
identify or understand core processes of organisms.
Multiple Choice: There are processes that organisms have
to carry out to survive, and many organisms perform these
processes for certain cellular/bodily functions. Which of the
following does not correctly identify what the process
does?
A) Cellular Respiration helps create ATP, or the energy
necessary for other cellular functions.
B) Cellular Signaling helps transmit signals to other cells
or throughout the body that helps perform functions.
C) Mitosis is the division of the cell, and cytokinesis
division of the cytoskeleton
D) Phagocytosis in the immune system is the process of
phagocytes engulfing pathogens to rid the body of
them.
Free Response Question: Binary fission and mitosis are seen
In bacteria and most eukaryotes. Both have similarities, and it is
known that prokaryotes preceded eukaryotes on Earth. Being a
process that occurs, explain how this could point to a common
ancestor? Provide evidence for how this could show that all life
shares a common ancestor.
ANSWER KEY- LEARNING OBJECTIVE 1.14
There are processes that organisms have to carry out to survive, and many organisms perform these processes for
certain cellular/bodily functions. Which of the following does not correctly identify what the process does?
A) Cellular Respiration helps create ATP, or the energy necessary for other cellular functions.
B) Cellular Signaling helps transmit signals to other cells or throughout the body that helps
perform functions.
C) Mitosis is the division of the cell, and cytokinesis division of the cytoskeleton.
D) Phagocytosis in the immune system is the process of phagocytes engulfing pathogens to rid
the body of them.
Free Response Question(5x2=10 points): Binary fission and mitosis are seen in bacteria and most
eukaryotes. Both have similarities, and it is known that prokaryotes preceded eukaryotes on Earth. Being a
process that occurs, explain how this could point to a common ancestor? Provide evidence for how this
could show that all life shares a common ancestor.
Both processes are essential for organisms to produce more cells(1 point) in their body, although binary fission is
also a way for prokaryotes to asexually reproduce (1 point). This leads us to conclude that all organisms share a
common ancestor especially if all organisms use similar ways of cell division (1 point). This is believed because
prokaryotes arrived on earth first and binary fission existed first as the simplest cell division process, which may have
evolved over time (1 point for providing that prokaryotes and binary fission came first). Also, some characteristics are
shared with these processes. For example, during binary fission and mitosis, the chromosomes or sister chromatids
move to opposite ends of the cells, and proteins such as tubulin are related to proteins used in binary fission (1 point
for providing any justification or example of similar characteristics in both processes).
LO 3.2: The student is able to justify the selection of data from historical investigations that support the claim that DNA is the source of
heritable information.
SP 4.1: The student can justify the selection of the kind of data needed to answer a particular scientific question.
Explanation: The first step in determining DNA’s role as the source of heritable information was the confirmation that it could alter bacteria via a
process known as transformation: the external assimilation of DNA into a cell. British medical officer Frederick Griffith, while studying the
bacterium Streptococcus pneumoniae in 1928, found that dead pathogenic “S” cells mixed with nonpathogenic “R” cells led to the death of the
injected laboratory mouse. This observation, along with the presence of living S cells found in the dead mouse’s blood, suggested that the
nonpathogenic bacteria had been “transformed” by the DNA of the pathogenic cells, leading to their production in the mouse. The next step was in
1952 when Alfred Hershey and Martha Chase sought the source of viral reprogramming- either the protein or DNA of the bacteriophage was
responsible. They performed an experiment with two batches, one that contained phages marked by radioactive sulfur (tagging the protein) and
another with radioactive phosphorus (tagging the DNA). Both batches infected bacterial cells and were then separated in a blender and centrifuged.
Finally, the radioactivity of the pellet and the liquid products was tested. In the protein-tagged batch, no radioactivity was found in the bacterial cells,
indicating that phage protein was not responsible for heritable information. In the DNA-tagged batch, radioactivity was found inside the pellet
content (bacteria), indicating that DNA was truly the source of genetic material. A final confirmatory observation was made by Erwin Chargaff in
1947 when he observed molecular diversity between DNA pair ratios of different species, an indicator that DNA is a credible fount of genetic
information and diversity.
M.C. Question: How would the outcome of Hershey and Chase’s experiment have been different if the phage protein contained the heritable genetic
material?
A. Neither batch would contain any radioactivity in the pellet.
B. Both batches would contain radioactivity in the pellet.
C. The sulfur-tagged batch would contain radioactivity in the liquid.
D. The phosphorus-tagged batch would contain radioactivity in the liquid.
E. The phosphorus-tagged batch would contain radioactivity in the pellet.
Learning Log/FRQ-Style Question: For TWO of the three aforementioned
DNA investigations, a) identify one unique finding of the
experiment and b) describe how each solidified DNA’s role as the
source of heritable material in organisms.
ANSWER KEY- LO 3.2
How would the outcome of Hershey and Chase’s experiment have been different if the phage protein contained the heritable genetic material?
A. Neither batch would contain any radioactivity in the pellet.
B. Both batches would contain radioactivity in the pellet.
C. The sulfur-tagged batch would contain radioactivity in the liquid.
D. The phosphorus-tagged batch would contain radioactivity in the liquid.
E. The phosphorus-tagged batch would contain radioactivity in the pellet.
For TWO of the three DNA investigations, a) identify one unique finding of the experiment and b) describe how each solidified DNA’s role as the source
of heritable material in organisms.
a) One unique finding of Griffith’s investigation was that dead S cells were unable to infect a mouse on their own but capable with the help of
“transformable” R cells. One unique finding of Hershey and Chase’s investigation was that bacterial cells infected by phosphorus-tagged DNA displayed
radioactivity once cultured and reproduced. One unique finding of Chargaff’s investigation was that unlike previous perceptions of universal DNA, the
molecular base pair ratios from one species to another were different.
b) Griffith’s experiment showed that DNA could be transmitted heritably between bacteria cells. Hershey and Chase’s experiment showed that DNA was
the definite source of heritable information in phages (viruses) because this same DNA could be found in future generations of cultured bacteria.
Chargaff’s experiments solidified DNA’s stance as the source of heritable information by indicating that it showed diversity in its molecular makeup
across species.
LO 2.16: The student is able to connect how organisms use negative feedback to
maintain their internal environments
SP 7.2: The student can connect concepts in and across domain(s) to
generalize or extrapolate in and/or across enduring understandings and/or
big ideas.
Explanation: To maintain its internal environment, the human body uses feedback mechanisms,
which are responses that trigger other activities or processes. Negative feedback mechanisms are
the most common because they attempt to maintain a target level. An example could be body
temperature. The body has a set temperature that is required for homeostasis, 37 C. When the
temperature rises above that 37 C the hypothalamus sends signals to the brain causing the body to
sweat to cool down and when the temperature drops below 37 C, the hypothalamus send signals to
the brain causing shivering to warm up.
M.C. Question: Which of the following is not an
example of a negative feedback loop?
A. Hypothalamus regulates body temperature
B. ADH regulates osmolarity
C. Milk is made until a baby is done nursing
D. Insulin and glucagon regulate blood glucose level
Learning Log/ FRQ Question: Sarah hasn’t eaten
anything all day and her blood glucose levels are low. In
what ways will her body regulate her blood glucose
levels? What if she has consumed too much glucose in a
day?
Answer Key- LO 2.16
•
A.
B.
C.
D.
•
•
M.C. Question: Which of the following is not an example of a negative
feedback loop?
Hypothalamus regulates body temperature
ADH regulates osmolarity
Milk is made until a baby is done nursing
Insulin and glucagon regulate blood glucose level
Learning Log/ FRQ Question: Sarah hasn’t eaten anything all day and her
blood glucose levels are low. In what ways will her body regulate her blood
glucose levels? What if she has consumed too much glucose in a day?
Blood glucose levels are controlled mainly by two hormones secreted by the
pancreas, glucagon and insulin. As Sarah’s blood glucose levels decrease the
brain signals the pancreas to release glucagon which breaks down glycogen in
the liver cells that releases glucose into the blood. When blood glucose levels
rise the brain signals the pancreas to secrete insulin which causes the liver
cells to build polymers of glucose called glycogen.
LO 2.30 The student can create representations or models to describe nonspecific immune defenses in plants and animals
SP 1.1 The student can create representations and models of natural or man-made phenomena and systems in the domain
SP 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain
Explanation: Nonspecific immune defenses provide constant unchanging protection against any foreign invader, and make up the first and second innate lines of
defense. Barriers are a type of nonspecific immune defense that blocks foreign invaders from entering the body of an organism. In animals, skin and mucous
membranes are barriers that mechanically defend against invaders, while secretions like mucous, tears, and saliva are barriers that destroy invaders with lysozyme
enzymes. In plants, cell walls, epidermis, and tree bark are examples of nonspecific barriers. Leukocytes are white blood cells that patrol the body to remove foreign
invaders and there are 3 main types. Neutrophils are the most abundant leukocytes; they are very mobile but last only 3 days. Macrophages are long lived phagocytic
leukocytes that also signal other responses with cytokines. Natural killer cells are leukocytes that destroy infected human cells with the perforin enzyme that perforates
the infected cells. The complement system is a system of proteins that destroy dangerous cells by forming a cellular lesion on the plasma membrane of invading
bacterial cells. The inflammatory response is a nonspecific response triggered by mast cells releasing histamine and prostaglandin as chemical signals. Capillaries dilate
to become more permeable and deliver platelets for clotting and macrophages for defense. Temperature is increased to decrease bacterial growth, stimulate
macrophages, and repair the damaged area.
M.C. Question: If a cancer cell were produced in the human body, how would the
nonspecific immune system remove it?
A. A neutrophil would target it with phagocytosis
B.
A macrophage would target it with phagocytosis
C.
A natural killer cell would target it with perforin
D. The compliment system would produce a protein to target it
FRQ Question: A fever is another form of a nonspecific immune reaction. How would a
fever be triggered? How would a fever fight pathogens in the body?
LO 2.30 Answer Key
M.C. Question: If a cancer cell were produced in the human body, how would the
nonspecific immune system remove it?
A. A neutrophil would target it with phagocytosis
B.
A macrophage would target it with phagocytosis
C.
A natural killer cell would target it with perforin
D. The compliment system would produce a protein to target it
FRQ Question: A fever is another form of a nonspecific immune reaction. How would a
fever be triggered? How would a fever fight pathogens in the body?
A fever would be triggered when a macrophage is introduced to a pathogen, and
releases cytokines. The cytokines signal for the brain to increase body temperature.
One result would be dilated capillaries, increasing blood flow and supplying
macrophages. The increased temperature also stops bacteria growth by denaturing the
bacteria’s proteins, and speeds up repair of damaged areas. Fevers cause the liver and
spleen to remove iron from the blood supply and store it, which prevents bacteria using
the iron to grow.
LO 2.6: The student is able to use calculated surface area-to-volume ratios to predict which cell(s) might eliminate
wastes or procure nutrients faster by diffusion.
SP 2.2: The student can apply mathematical routines to quantities that describe natural phenomena.
Explanation: The surface area of an object is calculated by squaring its length (cm²). The volume of an object is
measured by cubing its length (cm³). This means that, as the length of an object such as a cell increases, the cell’s
surface area increases by a factor of 2 and its volume increases by a factor of 3. This causes the volume to grow much
faster than the surface area. The larger a cell gets, the smaller its ratio of surface area to volume becomes. The smaller
this ratio, the more energy a cell has to expend moving molecules around inside of it. A very small cell would have a
very high ratio and wouldn’t have to use much energy to move waste products outside of the cell or move nutrients
inside of the cell and to the correct organelles for processing. As a general rule, the smaller a cell is, the more efficient
it will be.
M.C. Question: Why would a single-celled organism with a side length of 5
µm have a competitive advantage to a similar organism with a side length of
20 µm when disposing of waste products?
A) Because it would be able to produce lysosomes more quickly
B) Because it would have to move the waste a shorter distance, so would
expend less energy
C) Because its Golgi body would be closer to its vacuole
D) Because it would be able to produce more ATP
Learning Log/FRQ-style question:
a) Cell A is cube-shaped and has a side length of 4.5cm. Cell B is also cube
shaped and has a side length of 7.9cm. Calculate the surface area to volume
ratio for both Cell A and Cell B. The surface area of a cube is equal to 6(side
length)², and the volume of a cube is equal to (side length)³ .
b) Name and describe two advantages cells with a high surface area to
volume ratio have over cells with a low ratio.
ANSWER KEY- LO 2.6
Why would a single-celled organism with a side length of 5 µm have a competitive advantage to a similar organism with
a side length of 20 µm when disposing of waste products?
A) Because it would be able to produce lysosomes more quickly
B) Because it would have to move the waste a shorter distance, so would expend less energy
C) Because its Golgi body would be closer to its vacuole
D) Because it would be able to produce more ATP
a) Cell A is cube-shaped and has a side length of 4.5cm. Cell B is also cube shaped and has a side length of 7.9cm.
Calculate the surface area to volume ratio for both Cell A and Cell B. The surface area of a cube is equal to 6(side
length)², and the volume of a cube is equal to (side length)³ .
b) Name and describe two advantages cells with a high surface area to volume ratio have over cells with a low ratio.
a)
b)
In image to right
Cells with a higher surface area to volume ratio will
expend less energy transporting waste from
organelles to outside of the cell because they will
have to move the waste a shorter distance. Cells with
a higher surface area to volume ratio will also be able
to move nutrients to the proper organelles faster
after the nutrients have diffused through the cell
membrane because the nutrients will have a shorter
distance to travel.