Failure To Thrive

Hello, I'm Milo and I'm doing some science.

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Essential Knowledge 4.A.3

Demonstrate your understanding of how interactions between external stimulie and gene expression result in differentiation of cells, tissues, and organs. 

Pluripotent stem cells have the potential to become any type of cell, since they all have the same exact genetic makeup. Differentiation of these stem cells is a result of transcription factors. Transcription is the process of converting DNA to RNA, therefore, transcription factors induce this process making it a sort of starting pistol for gene expression. Internally, stem cells become unipotent by only expressing genes queued to it by transcription factors. An external example of transcription factors at work is the differentiation of female and male embryos. At five weeks, all embryos are exactly the same. However, at week six, the testes of a male embryo release the SRY gene that is the transcription factor which signals the production of certain proteins, making an embryo male. 

Cells can also be effected by transcription factors triggered by their environment. Environmental changes, such the temperature being too high, can trigger transcription factors called heat shock factors. These heat shock factors can protect the cell from changes in temperature by altering it’s structure, making a protective coat around it, until it is no longer needed.

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1.B.1

Describe three conserved core biological processes mentioned above using specific examples

1. Chromosomes -  in the nucleus of most living things, chromosomes carry genetic information in the form of genes.

2. Cytoskeleton - a structure unique to eukaryotic cells. The cytoskeleton is a structure that helps cells maintain their shape and enables cells to carry out essential functions like division and movement.

3.  The Endomembrane System in Eukaryotic Cells - All eukaryotic cells contain an endomembrane system. The endomembrane system is comprised of nuclear envelope, ER, Golgi apparatus, the plasma membrane and more.

Justify how these conserved core processes/features support the idea of common ancestry for all organisms

These conserved core processes/features support the idea of common ancestry for all organisms because they are found in many organisms (in this case, all the ones I have listed are found in all eukaryotes). Since all organisms share these traits/processes it suggests that there is some common ancestry for all lifeforms. 

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Work for Peppered Moth Simulation Questions #21-24

Justify how an evolutionary change in a population is related to a change in the environment. What role do humans play in this change? What is the impact in the future of this change?

Using the peppered moth simulation as an example, if there were moths in a light forest (the environment) then light colored moths would be better suited for survival than black moths since they are less visible to predators because they can blend into their surroundings while the others cannot. Therefore (as proved in #21), the ratio of colors in an environment would change, most of the population being a part of the group that is more fit for survival in that environment. So the evolutionary change in this population of peppered moths (one color being more abundant than the other) is due to the environment they are in. Since humans have been steamrolling over habits of countless animals for quite some time, we know that the destruction of those environments can lead to monumental changes in the chances of survival for many species.

In the future, more species will most likely become extinct as a result of this because they are not fit to survive in an environment different than their original habitat. This is especially true since the destruction of entire habitats can happen within months while the rate of adaptation to be better suited within that environment can take many years. Because these creatures do not have time to gradually adapt, they die off.

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Essential Knowledge 2.A.1

An exergonic reaction is a spontaneous reaction that has a net release in free energy. Since cellular respiration releases free energy by breaking down carbohydrates, it is exergonic.

On the other hand, an endergonic reaction is a non-spontaneous reaction in which free energy is absorbed/captured from the environment in order to do work. Since photosynthesis captures light energy in order to form carbohydrates, it is considered endergonic. 

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Chapter 18-20 Portfolio

Chapter 18: Regulation of Gene Expression

Describe the structure of an operon. Include a model with your discussion.

[Source: Pearson Education, Inc.]

An operon is the entire stretch of DNA that includes the promoter, operator, and the genes that they control. 

Compare and contrast an inducible operon and a repressible operon. Include an example of each.

An inducible operon is usually “off” since it is, hence its name, able to be induced. An example of an inducible operon is the Lac operon that produces enzymes to break down lactose. In order to turn it “on” and make it active, a corepressor must be removed. (A corepressor is a moleule that cooperates with a repressor protein to switch an operon off). On the other hand, a repressible operon is usually “on”. In order to turn it “off”, a corepressor must be added. An example of a repressible operon is the Trp Operon. 

Explain the role of repressor genes in operons and why they are important.

Repressor genes prevent transcription/expression of the operon gene by binding to operator region on the operon. Additionally they prevent the RNA polymerase from synthesizing the operon gene.

Describe the impact of DNA methylation and histone acetylation on gene expression.

In histone acetylation, acetyl groups are attached to positively charged lysines in histone tails which loosens chromtin structure. This is useful since genes within highly packed heterochromatin are usually not expressed, and histone acetylation allows those to be demonstrated. In DNA methylation, the addition of methyl groups to certain bases in DNA is associated with reduced transcription in some species and causes long-term in activation of genes in cellular differentiation.

Compare and contrast the role of oncogenes, proto-oncogenes, and tumor suppressor genes in cancer.

Oncogenes are cancer causing agents, whereas Proto-oncogenes are the corresponding normal cellular genes that are responsible for normal cell growth and division. Tumor-suppressor genes help prevent uncontrolled cell growth, repair damaged DNA, control cell adhesion, and inhibit the cell cycle in the cell-signalling pathway. 

Chapter 19: Viruses

Develop a model and explain the components of a virus.

[Source: Untraceable]

Viruses are very small infectious particles consisting of nucleic acids enclosed in a protein coat and, in some cases, a membranous envelope. (Viral genomes may consist of either double or single stranded DNA or RNA). The shell built from protein subunits called capsomeres that covers the viral genome is called a capsid.

Compare and contrast the differences between lytic and lysogenic life cycles of a virus.

The lytic cycle is a phage reproductive cycle that results in the death of the host cell. It does this by producing new phages and digesting the host’s cell wall, releasing the progeny viruses. (A phage that only reproduces by the lytic cycle is a Virulent phage). The lysogenic cycle replicates the phage genome without killing the host. By this, the viral DNA molecule is incorporated into the host cell’s chromosome. Every time the host divides, it copies the phage DNA with it and passes the copies to daughter cells. (Phages that use both the lytic and lysogenic cycles are called temperate phages).

Chapter 20: DNA Technology and Genomics

Briefly discuss the following genetic engineering techniques - include a model of each and explain how each can be used to manipulate DNA

Gel Electrophoresis

[Source: Encyclopedia Britannica, Inc.]

An indirect method of rapidly analyzing and comparing genomes is gel electrophoresis. It uses a gel as a molecular sieve to separate nucleic acids or proteins by size. Current is applied across the gel and since DNA is negatively charged, it is attracted to the positive side of the gel. Molecules are sorted into “bands” by their size (shorter molecules move farther).

Plasmid-based Transformation

[Source: Universtiy of Walkato]

Plasmids can be used as vessels to carry foreign DNA into a cell. Once a DNA sequence of interest is cut by restriction enzymes, the two newly paired DNA sequences are joined by DNA ligase and the transformed plasmid is now ready for insertion into a host cell. Once inside the cell, the plasmid is copied by the host cell’s own DNA replication machinery and can be implemented in the cell.

Polymerase Chain Reaction (PCR)

[Source: Pearson Education, Inc.]

The polymerase chain reaction (PCR) can produce many copies of a specific target segment of DNA. A three-step cycle--heating, cooling, and replication--brings about a chain reaction that produces an exponentially growing population of identical DNA molecules.

Restriction Enzyme Analysis of DNA

[Source: Pearson Education, Inc.]

In restriction fragment analysis, DNA fragments produced by restriction enzyme digestion of a DNA molecule are sorted by gel electrophoresis. It is useful for comparing two different DNA molecules, such as two alleles for a gene. (Can also be used to prepare pure samples of individual fragments).

Develop a model involving the steps in gene cloning with special attention to the biotechnology tools that make cloning possible.

{CLICK TO BE DIRECTED TO IMAGE}

This is a diagram of molecular cloning using bacteria and plasmids. Even though it is not my own model, I feel like this model is so much more intricate than anything I could attempt to make by the due date of this portfolio.Therefore, I offer this as my model, since I really did learn a lot from it.

Explain the key ideas that make PCR possible.

The method relies on a thermal cycle, consisting of cycles of repeated heating and annealing of the reaction for enzymatic replication of the DNA. As the polymerase chain reaction progresses, the DNA generated is used as a template for replication, which begins a chain reaction in which the DNA template is increasingly amplified. 

Develop a model and explain how gel electrophoresis can be used to separate DNA fragments.

[Source: GBS IB Biology]

An indirect method of rapidly analyzing and comparing genomes is gel electrophoresis. It uses a gel as a molecular sieve to separate nucleic acids or proteins by size. Current is applied across the gel and since DNA is negatively charged, it is attracted to the positive side of the gel. Molecules are sorted into “bands” by their size (shorter molecules move farther than longer molecules). Thus, the DNA fragments are separated by size.

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Chapter 16 Remediation

I struggled most with the processes of DNA replication and repair in general. Specifically, the difference between the conservative and semi-conservative models of replication. The semi-conservative model of replication says that when DNA replicates, each of the daughter molecules has one old strand and one newly produced strand.   Whereas, in the conservative model, the two parental strands come back together after the procedure. (Conservative replication reverts back and conserves the original strand).  

I also struggled with the overall concept of replication. However, I now know that replication begins at sites called origins of replication that separate DNA strands, creating a replication bubble. This separation of DNA is called unwinding. Several proteins partake in this unraveling of parental strands of DNA, such as helicases that untwist the double helix, single-strand binding proteins that stabilize unpaired DNA strands by binding to them, and topoisomerase that helps relieve the strain put on the molecule by all of this twisting and untwisting. The single strands of DNA now act as templates for the production of new complementary DNA strands.

Chapter 17 Remediation

The topic I struggled the most with was the process of translation from RNA to a protein. In order for translation to occur, a correct match between a tRNA and an amino acid and a correct match between the tRNA anticodon and an mRNA codon must be present. Ribosomes facilitate coupling of tRNA anticodons with mRNA codons in protein synthesis. 

My main issue was understanding the structure of a ribosome. A ribosome has two subunits, a large and a small. On the small subunit, there is a mRNA binding site where mRNA is fed through in order to be translated. On the large subunit there are three additional binding sites: the P site (Peptidyl-tRNA binding site), the A site (Aminoacyl-tRNA binding site), and the E site (Exit site). The P site holds the tRNA that carries the growing polypeptide chain. The A site holds the tRNA that carries the next amino acid to be added to the chain. The E site discharges it when a stop codon in the mRNA reaches the A site, teminating the process.

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Chapter 16: The Molecular Basis of Inheritance

Explain the structure of DNA and nucleotides. Include a model of your explanation.

DNA is made up of a sugar-phosphate backbone (a 5 carbon sugar called Deoxyribose and a phosphate group) and a nitrogenous base (Adenine, Thymine, Guanine, Cytosine). This is then attached to another nitrogenous base (which is complementary to it) and sugar-phosphate backbone by a hydrogen bond. This creates the ladder-like rungs of the double helix.

Develop a model which explains the major steps to replication, specifically a replication bubble.

Compare and contrast the difference between replication, transcription, and translation. Replication in DNA occurs in the nucleus. It is the process of copying a double-stranded DNA molecule using a template DNA strand and DNA polymerase that produces new DNA. Transcription occurs in the nucleus and is the synthesis of mRNA from a DNA template using RNA polymerase. Translation is the synthesis of proteins from mRNA that occurs in the cytoplasm. 

Develop a model and explain how DNA is packaged into a chromosome.

In plant and animal cells, DNA is packaged into structures called chromosomes. A single strand of DNA is wrapped many times around various proteins (histones), to form structures called nucleosomes. These nucleosomes coil up tightly to create chromatin loops. The chromatin loops then wrap around each other to make a full chromosome. Each chromosome has two short arms (p arms), two longer arms (q arms), and a centromere that holds it together at the center.

Chapter 17: From Gene to Protein

Compare and contrast the key terms gene expression, transcription, and translation.

Gene expression is the process by which DNA directs protein synthesis which occurs in two stages: transcription and translation. Transcription is the synthesis of RNA that occurs in the nucleus using information in DNA that produces messenger RNA (mRNA). Translation is the synthesis of a polypeptide that occurs in the cytoplasm or ribosomes using information in the mRNA.

Develop a model and explain the process of transcription and translation.

Transcription is the synthesis of RNA that occurs in the nucleus using information in DNA that produces messenger RNA (mRNA). 

Translation is the synthesis of a polypeptide that occurs in the cytoplasm or ribosomes using information in the mRNA. It occurs in three stages: Initiation, Elongation and Termination.

Explain how eukaryotic cells modify RNA after transcription and why it is necessary. An example of a post-transcriptional modification is splicing. Splicing is the removal of non-coding regions (introns) of RNA in order to join the coding regions (exons). This process uses spliceosomes, which consist of a variety of proteins and several small nucleus ribonucleoproteins (snRNPs) that recognize the spice sites. Splicing is required in eukaryotes to generate mature coding RNAs out of previous RNAs still containing Introns, which need to be removed to produce the correct protein.

Develop a model which explains how point and frameshift mutations can impact a protein.

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AP Physics 1 Notesheets

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Enzyme Catalysis Lab

**NOTICE: All Procedures are paraphrased, or directly quoted from the “Lab 2 Enzyme Catalysis” Worksheet

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Exercise 2A: Test of Catalase Activity

This exercise investigates the effects of temperature and the presence of living tissue on the enzymatic activity of breaking down H2O2.

Procedure:

1. To observe the reaction to be studied, transfer 10 mL of 1.5% (0.44 M) H2O2 into a 50-mL glass beaker and add 1 mL of the freshly made catalase solution. The bubbles coming from the reaction mixture are O2, which results from the breakdown of H2O2 by catalase. Be sure to keep the freshly made catalase solution on ice at all times.

2. To demonstrate the effect of boiling on enzymatic activity, transfer 5 mL of purified catalase extract to a test tube and place it in a boiling water bath for five minutes. Transfer 10 mL of 1.5% H2O2 into a 50-mL glass beaker and add 1 mL of the cooled, boiled catalase solution. 

3. To demonstrate the presence of catalase in living tissue, cut 1 cm^3 of potato or liver, macerate it, and transfer it into a 50-mL glass beaker containing 10mL of 1.5% H2O2.

Analysis:

This experiment demonstrates how alterations in the environment, such as temperature, can effect an enzyme’s rate of reaction. Even though we did not physically perform this experiment, we can concur that putting an enzyme in any extreme environment (such as a temperature or pH that does not comply with that specific enzyme’s optimal condition needs), can cause the enzyme to refrain from working efficiently, unable to function, and ultimately denature it.  From this experiment, we can also conclude that living tissues contain their own specific enzymes. If catalase is put in the presence of living tissue, this reaction may take place more quickly.

Exercise 2B: The Base Line Assay

This exercise assists us in determining the amount of H2O2 that is initially present in a 1.5% solution in order to have a base line to compare other values to.

Procedure:

1. Put 10 mL of 1.5% H2O2 into a clean cup.

2. Add 1 mL of H2O (instead of enzyme solution)

3. Add 10 mL of H2SO4 (1.0 M)

4. Mix well.

5. Remove a 5-mL sample. Place this 5-mL sample into another cup and assay for the amount of H2O2. Use a burette to add KMnO4, a drop at a time, to the solution until a persistent pink or brown color is obtained. 

Analysis:

Establishing this baseline value is necessary to determine how much H2O2 is present in a 1.5% solution. By having this value, we are now able to calculate the efficiency of a catalase on the decomposition of H2O2 by measuring how much H2O2 is initially present before the introduction of an enzymatic solution.

Exercise 2C: The Uncatalyzed Rate of H2O2 Decomposition

The intent of this exercise was to observe the rate of decomposition in an uncatalyzed reaction, and to compare the results to the rate of decomposition of a catalyzed reaction.

Procedure:

1. Leave a small quantity of 1.5% H2O2 uncovered at room temperature for approximately 24 hours. 

2. Add 1 mL of H2O (instead of enzyme solution)

3. Add 10 mL of H2SO4 (1.0 M)

4. Mix well.

5. Remove a 5-mL sample. Place this 5-mL sample into another cup and assay for the amount of H2O2. Use a burette to add KMnO4, a drop at a time, to the solution until a persistent pink or brown color is obtained. 

6. Record your findings in the chart below.

Analysis:

From this exercise, we were able to observe that, when left out overnight and uncovered, the amount of H2O2 decomposed is about half of the baseline value. Since it was exposed to oxygen and light for a period of 24 hours, we can conclude that light causes H2O2 to spontaneously decompose, since most of the H2O2 was already decomposed spontaneously overnight before adding KMnO4. (This also explains why H2O2 is usually stored in a dark bottle in order to preserve the substance.) Since H2O2 is able to spontaneously decompose, we can conclude that the decomposition of H2O2 is an exergonic reaction.

Exercise 2D: An Enzyme Catalyzed Rate of H2O2 Decomposition

This exercise is intended use the amount of H2O2 used in the timed reaction to demonstrate the relationship between the rate of reaction over time.

Procedure:

1. 10 seconds

a. Put 10 mL of 1.5% H2O2 in a clean 50-mL glass beaker.

b. Add 1 mL of catalase extract.

c. Swirl gently for 10 seconds.

d. At 10 seconds, add 10 mL of H2SO4 (1.0 M)

2.  30, 60, 90, 120, 180, and 360 seconds

Each time, repeat Steps 1-4. However each time, allowing the reaction to occur for  30, 60, 90, 120, 180, and 360 seconds, respectively.

Note: Each time, remove a 5-mL sample and assay for the amount of H2O2 in the sample. Use a burette to add KMnO4.

3. Record your data in the table below (Table 2.1)

4. Graph the data for enzyme catalyzed H2O2 decomposition.

Analysis:

From the data we collected, we can observe how enzymatic activity impacts the decomposition of H2O2 over specific intervals of time. We can see, by looking at the graph above, that, initially, the H2O2 begins to decompose rapidly. While the amount of H2O2 used over time does continue to increase, our rate of reaction starts to slow down, eventually beginning to plateau. This is because, after a certain amount of time, all of the enzyme will be used up and there will be nothing left to keep reacting.

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Diffusion-Osmosis Lab

**NOTICE: All Procedures are paraphrased, or directly quoted from the “Lab 1 Diffusion and Osmosis” Worksheet

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Exercise 1A: Diffusion

     In this experiment,we will use dialysis tubing to demonstrate the passing of molecules through a selectively permeable membrane.

Procedure:

1. Take a 30-cm piece of 2.5-cm dialysis tubing that has been soaking in water. Tie off one end of the tubing to form a little candy-wrapper like bag. 

2. Test the 15% glucose/1% starch solution for the presence of glucose using Clinastix (or glucose strips). Record this color in Table 1.1.

3. Place 15 mL of the 15% glucose/1% starch solution in the bag. Tie the other end of the bag, leaving enough space for the expansion of the contents of the bag. Record the color of the solution in Table 1.1.

4. Fill a 250--mL beaker or cup two-thirds full with distilled water. Add approximately 4 mL of iodine to the distilled water and record the color of the solution in Table 1.1. Test this solution for glucose and record the results in Table 1.1.

5. Immerse the bag in the beaker of solution.

6. Allow your setup to stand for approximately 30 minutes or until you see a distinct color change in the bag or beaker. Record the final color of the solution in the bag, and of the solution in the beaker, in Table 1.1.

7. Test the liquid in the beaker and in the bag for the presence of glucose. Record the results in Table 1.1.

Analysis:

     This experiment demonstrates selective permeability wherein the dialysis tubing is the selectively permeable membrane. We can see that this membrane is selectively permeable because only certain substances are allowed to pass through it. Upon reviewing our data, we can determine that Glucose, Water, and Iodine were allowed through the membrane, while Starch was not. 

     We can see, by referring to Table 1.1, that there is a change in the presence of Glucose inside of the dialysis tubing bag after after soaking in Iodine (the glucose strip turned lime green when put in the bag afterwards, just like the H2O/Iodine solution). This indicates that Glucose left the bag through the selectively permeable membrane. By visually observing the data, we can also see that Iodine and Water (since they were both outside of the bag) did in fact pass through the membrane. After sitting for 30 minutes, the Iodine/H2O solution penetrated the dialysis tubing and entered into the bag, making the 15% glucose/1% starch solution that was once clear turn black. However, there was no indication that Starch was able to leave the dialysis bag. Therefore, we can conclude that the pores in the selectively permeable membrane were large enough to let some molecules (Glucose, Water, and Iodine) through, but were too small to let larger molecules, such as Starch, through. If we were to rank these all by size, they would be, from largest to smallest: Starch, Membrane Pores, Glucose, Water and Iodine.

Exercise 1B: Osmosis

     In this experiment we will use dialysis tubing to demonstrate the movement of water through a selectively permeable membrane because of solute concentration, by the process of osmosis.

Procedure:

1. Obtain six 30-cm strips of dialysis tubing.

2. Tie a knot in one end of each piece of dialysis tubing to form 6 bags. Pour about 15-24 mL of each of the following solutions into separate bags:

a) distilled water

b) 0.2 M sucrose

c) 0.4 M sucrose

d) 0.6 M sucrose

e) 0.8 M sucrose

f) 1.0 M sucrose

Remove most of the air from each bag by drawing the dialysis bag between two fingers. Tie off the other end of the bag. Leave sufficient space for the expansion of the contents in the bag. (The solution should fill only about one-third to one-half of the piece of tubing.)

3. Rinse each bag gently with distilled water to remove any sucrose spilled during filling.

4. Carefully blot the outside of each bag and record in Table 1.2 the initial mass of each bag, expressed in grams.

5. Label 6 cups indicating the molarity of the solution in the dialysis bag. Fill each cup two-thirds full with distilled water.

6. Submerge each bag in its designated cup.

7. Let them stand for 30 minutes.

8. At the end of 30 minutes remove the bags from the water. Carefully blot and determine the mass of each bag.

9. Record your group’s data in Table 1.2. 

Analysis:

     Unfortunately, our data was very skewed due to leakage in one of our dialysis bags. However, the graph below is the ideal outcome of this experiment where the percent change in mass and the molarity of each bag are directly and positively related. 

     Since the bags are put into 0.0 M distilled water, the 0.0 M distilled water inside the first bag does not move because it is in an isotonic environment. However, the rest of the bag’s masses do increase because of the fact that they are all in a hypotonic solution, making water rush into each of the bags.

     If all of the bags were put into a 0.4 M sucrose solution instead of water, each substance in the bag would react differently. 0.0 M distilled water and 0.2 M sucrose would rush out of the bag, making the mass of the bag less than the initial, since it is less concentrated that it’s environment. 0.4 M sucrose would not move, because it is in an isotonic solution. 0.8 M sucrose and 1 M sucrose mass’ would be more because it’s environmental concentration would diffuse into the bag, meaning that it is in a hypotonic solution.

Exercise 1C: Water Potential

     In this experiment, we will investigate water potential by observing potato cores placed in different molar concentrations of sucrose to determine the water potential of potato cells.

Procedure:

1. Pour 100 mL of each solution in its designated cup. (Label each of the six cups with the molarity of the solution.

2. Use a cork borer to cut potato cylinders. Do not include any skin on the cylinders. You will need four cylinders for each cup.

3. Keep your potato cylinders in a cup filled with water while waiting to use the scale so they do not dry out.

4. Determine the mass of the four cylinders together and record the mass in Table 1.4. Put four cylinders into each labeled cup. 

5. Cover the cups with plastic wrap to prevent evaporation.

6. Let it stand overnight.

7. Remove the cores from the cups, blot them gently on a paper towel, and determine their total mass.

8. Record the final mass in Table 1.4. Calculate the percentage change as you did in Exercise 1B.

9. Graph your data for the percent change in mass in Table 1.4.

     Since our data is extremely skewed by human error (and possibly because of the fact that we left our potato cores to soak for an entire weekend, instead of just 24 hours), I have made an example graph so we can actually determine the molar concentration of the potato core. 

     To determine the molar concentration of the potato core, we must determine the value at which the relationship line crosses the x-axis. In this ideal case, that value is 0.4 M. At this concentration, there is no net movement of water.

Exercise 1D: Calculation of Water Potential from Experimental Data

1. The solute potential of this sucrose solution can be calculated using the following formula:

Our Molar concentration (in an ideal situation) is 0.4 M and our temperature is 22 ° C, which is then converted to Kelvin, adding up to 295 K. We will use these to complete our calculations. 

2. Knowing the solute potential of the solution and knowing that the pressure potential of the solution is zero allows you to calculate the water potential of the solution. The water potential will be equal to the solute potential of the solution.

     Thus, the water potential of our ideal situation is -9.806 bars. 

Analysis:

     Water potential values are helpful because they help us in predicting the direction of the flow of water. Since water moves from high water potential to low water potential, we can predict, after seeing the water potential of the substance (in this case, a solution in a dialysis bag) and it’s surroundings, where water might flow. 

     If a potato core was allowed to dehydrate, it’s water potential decreases. The Molar concentration would become bigger, therefore making the absolute value of the answer larger. However, since water potential will be negative, that means that the number is technically smaller.

     If a plant cell has a lower water potential, it is hypertonic to it’s environment. The plant cell will gain water because water moves from high water potential to low water potential.

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Unit 2: Essential Knowledge Portfolio

Sections Covered (Click to see each section):

Chapter 6 Chapter 7 Chapter 8 Chapter 9 & 10 Chapter 11 Chapter 12

Due: TBD

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Chapter 6 Portfolio

Represent graphically and explain the relationship between surface area to volume ratios as it relates to the efficiency of cellular work.

The surface area of a cell determines the rate at which nutrients are taken in or eliminated through waste excretion. As a cell grows in size, its need for nutrients increases because more energy is needed to support the internal functions of the cell.  More nutrients means that more waste must be eliminated. However, in some cases, the cell is not able to keep up with the waste production because of the sheer amount of nutrients it must take in.

Compare and contrast the structure of Prokaryotic and Eukaryotic cells - be sure to include a discussion regarding the cellular organization of each.

Prokaryotic Cells 

[Source: Boundless]

No nucleus

DNA in an unbound region called the nucleoid

No membrane-bound organelles 

Eukaryotic Cells 

[Source: Pearson Education Inc.]

DNA in a nucleus that is bounded by a membranous nuclear envelope.

Have membrane-bound organelles

Cytoplasm in the region between the plasma membrane and nucleus.

Usually much larger than Prokaryotic Cells

Describe the basic structure and functions of key cell organelles (nucleus, Golgi, ER, mitochondria, chloroplast, vacuoles, plasma membrane).

Nucleus -  Command center for the cell. It is a membrane bound structure that contains the cell’s hereditary information and controls the cell’s growth and reproduction. It also regulates the synthesis of proteins in the cytoplasm through the use of messenger RNA (mRNA).  

Golgi -  The main function of the Golgi apparatus is to modify, sort and package the macromolecules that are synthesized by the cells. Golgi are composed of stacks of membrane-bound structures called cisternae. They possess special enzymes of the Golgi which help to modify and transport of the modified proteins to their destination.

Endoplasmic Reticulum -  Smooth Endoplasmic Reticulum acts as a storage organelle. Rough Endoplasmic Reticulum plays a very important role in the synthesis and packaging of proteins. Ribosomes are attached to the membrane of the ER, making it “rough.”

Mitochondria - The mitochondria acts as the powerhouse of a cell. It also acts as a digestive system by taking in nutrients, breaking them down and using them as energy (cellular respiration). What makes mitochondria so efficient  is the folding of the inner membrane. This increases the surface area inside the organelle. Since many of the chemical reactions happen on the inner membrane, the increased surface area creates more space for reactions to occur.

Chloroplasts - The process of photosynthesis which converts energy into sugars and the byproduct of the process is oxygen that all animals breathe happens in the chloroplasts of a prokaryotic cell. Chloroplasts make food by the process of photosynthesis using light energy, water, and carbon dioxide. The thylakoid system of a chloroplast carries out this process.

Vacuoles -  A vacuole is essentially a storage bubble. In a vacuole, there is a membrane that surrounds a mass of fluid. In that fluid is food or any variety of nutrients a cell might need to survive.

Plasma Membrane - The plasma membrane, or cell membrane, works as a barrier between the inner and outer surface of a cell. It is composed mainly of a phospholipid bilayer which we have covered in Section 4.A.1. But for a nice review, phospholipids are made of two fatty acids and a phosphate group attached to glycerol. When phospholipids are added to an aqueous solution, they self-assemble into a bilayer, with the hydrophobic tails pointing towards the interior and the hydrophillic heads pointing outwards towards the water, making a sealed off cell membrane.

Explain how several internal membrane-bound organelles and other structural features work together to provide a specific function for the cell and contribute to efficiency.

In the Endomembrane system, the Smooth Endoplasmic Reticulum, Rough Endoplasmic Reticulum, Golgi Apparatus, and Lysosomes work together to drive protein synthesis and transport, metabolism of lipids, and detoxification. Lipids are made in the Smooth ER. The ribosomes in the Rough ER create proteins which depart from the RER to the Golgi Apparatus. In the Golgi, the proteins are processed. They are then transported to Lysosomes which digest them.

Explain the structure and function relationships between chloroplasts and mitochondria

Both mitochondria and chloroplasts are enveloped by a double membrane, contain free ribosomes, have their own DNA, and grow/reproduce somewhat independently within a cell. 

Relate structural and functional evidence in chloroplasts and mitochondria to the endosymbiotic theory of their origins.

[milodoesscience]

Chapter 7 Portfolio

Describe passive transport and explain its role in cellular systems

Passive transport is the traveling of molecules through a plasma membrane without the assistance of ATP, the cells' main source of energy. Hydrophobic (nonpolar) molecules are able to pass through the membrane rapidly because they are able to dissolve in the lipid bilayer. Since hydrophopic molecules are able to get through a membrane with ease that means that they are also able to diffuse across it. Diffusion is the tendency of molecules to spread out through available space, moving down the concentration gradient which also requires no energy. In a cellular system, osmosis (the diffusion of water across a selectively permeable membrane) enables cells to gain or lose water based on the type of tonicity a solution has. For example, a plant cell thrives in a hypotonic solution where the solute concentration is less than that inside the cell, because this allows the cell to gain water, making it turgid. 

Explain how membrane proteins play a role in facilitated diffusion of charged and polar molecules in general and in relation to the specific molecules below.

Active transport proteins embedded in the plasma membrane speed up the transportation of specific charged and polar molecules across a membrane aided by ATP. One of these active transport systems is a Na+/Ka+ pump. The Na+/K+ pump is the major electrogenic pump in animal cells. An electrogenic pump is a transport protein that generates voltage across a membrane. A glucose transporter is a specific carrier protein in the plasma membrane of red blood cells transports glucose across the membrane much faster than it could travel on it’s own and rejects all other molecules.

Explain the terms: hypotonic, hypertonic or isotonic in relationship to the internal environments of cells. Illustrate each environment with a model and explain which environment is best for plant and animal cells. Justify and explain your model with evidence.

[Source: Pearson Education, Inc.]

A hypotonic solution is when the solute concentration is less than that inside of a cell, allowing the cell to gain water. This environment is best for plant cells. This is because plant cells have a cell wall, allowing the cell to take in a large amount of water without bursting (unlike blood cells that have no cell wall as pictured above). A plant cell in a hypotonic solution swells and becomes turgid, making the plant firm and crisp. If a cell and its surroundings are isotonic, the cell becomes flaccid because there is no net movement of water.

[Source: Pearson Education, Inc.]

A hypertonic solution is when the solute concentration is greater than that inside the cell, making the cell lose water.

[Source: Pearson Education, Inc.]

An isotonic solution is when the solute concentration is the same as that inside the cell which allows no net water movement. This environment is best for animal cells because too much water coming in (hypotonic) and losing too much water (hypertonic) is not good for the cell, as shown above.

Describe active transport. Propose a model illustrating this process.

[Source]

Active transport is the passing of molecules through a plasma membrane aided by ATP performed by specific proteins embedded in the membrane.

Describe the processes of endocytosis and exocytosis. Propose a model illustrating each process.

[Source]

In endocystosis, the brings in macromolecules by forming vesicles from the membrane.

In exocytosis, transport vesicles go to the membrane, fuse with it, and release their contents. It is used mainly to export products created by secretory cells.

Create a visual representation/model (ie. graph or diagram) to make predictions about the exchange of molecules between an organism and its environment, the use of these molecules, and consequences to the organism if these molecules cannot be obtained.

A plant cell is healthy and normal when in a hypotonic environment, taking in water until the cell wall opposes uptake. The cell body stiffens and stays in shape, making the cell turgid.

Water has many functions in plant cells. However, in this situation, water helps maintain the shape and functionality of the plant cell by either hydrating or dehydrating the cell body.

If water cannot be obtained or a plant cell is put in an isotonic or hypertonic environment, then the cell will become flaccid or plasmolyzed due to the inadequate amount of water being taken in. This makes the cell body either become limp or shrink away from the cell wall, respectively. 

[milodoesscience]

Chapter 8 Portfolio

Compare and contrast Endergonic and Exergonic reactions - include a model of each showing the change in free energy from reactants and products.

As shown above, the net change in free energy for an Endergonic Reaction is positive. Whereas, in an Exergonic Reaction, it is negative. When free energy decreases (Exergonic) the stability of a system increases. This occurs spontaneously and no input of energy is required. (This is the opposite for Endergonic Reactions)

Explain the key role of ATP in energy coupling reactions. Provide a model showing energy before and after phosphorlyation of the reactants.

To do work, cells manage energy resources by energy coupling, which is the use of an Exergonic process to drive an Enderognic one, mediated by ATP. ATP is made up of ribose (a sugar), adenine (a nitrogeneous base), and three phosphate groups. Energy is released when the terminal phosphate bond is broken down by the introduction of water (hydrolysis), sharing that phoshate with the reaction’s reactant (such as a monomer). The hydrolysis of ATP powers three types of cellular work, mechanical, transport, and chemical.

Analyze data showing how changes in enzyme structure, substrate concentration, and environmental conditions affect enzymatic activity - ie. include a graph of each example and predict the effects when one of the parameters is further changed. Justify and explain your predictions.

Source: Pearson Education, Inc.

Each enzyme has an ideal temperature and pH in which it can function. (These ideal conditions favor the most active shape for the enzyme, allowing it to work efficiently) If one of these conditions is altered (become too high or too low), such as the optimal pH for pepsin or temperature for a typical human enzyme,  the protein molecules will denature, and the enzyme will most likely become permanently inactivated.

Describe the relationship between the structure and function of enzymes.

An enzyme is a catalyst which is a chemical agent that speeds up a reaction without being consumed by the reaction. Enzymes are large proteins that bind small molecules. When bound to an enzyme, the bonds in the reactants can become more energetic thereby making it easier for them to achieve the transition state because, by bringing the molecules closer together, there doesn’t need to be as much energy for them to collide. This is one way that enzymes lower the activation energy of a reaction. 

Explain how enzymes accomplish biological catalysis. Provide examples.

Without enzymatic catalysis, most biochemical reactions are so slow that they would not occur under the mild conditions of temperature and pressure that are compatible with life. Enzymes speed up the rates at which reactions occur by working with a substrate that binds to the enzyme’s active site. Therefore, reactions that would an extremely long time without catalysis can occur in an instant if affected by the appropriate enzyme. 

Describe how enzyme-mediated reactions can be controlled through competitive and noncompetitive interactions.

Binding by inhibitors can prevent enzymes from catalyzing reactions. Some reversible inhibitors, or competitive inhibitors, resemble the substrate and compete for binding at the active site. Noncompetitive inhibitors hinder enzymatic reactions by binding to another part of the molecule.

Propose experimental designs by which the rate of enzyme function can be measured and studied.

Ways to measure the rate of enzyme function in an experiment could be monitoring the disappearance of substrate or the appearance of product. For this experiment to work properly, the initial rate of activity must be measured and the pH must be constant. Ways to measure either the disappearance or appearance of a product over time are called enzyme assays. There are many ways to go about this, however, enzyme assays always fall into two types: fixed-timed and continuous. Below are two ways to perform an Enzyme Assay, one is a Fixed Time Assay and the other, a Continuous Assay.

Source: ChemWiki

[milodoesscience]

Chapter 9 & 10 Portfolio

Compare the major features of chemoheterotrophic and photoautotrophic nutritional processes.

The major difference between these nutritional processes is that chemoheterotrophic (cellular respiration) utilizes chemical resources and photoautotrophic (photosynthesis) uses light. Photoautotropic processes use light energy to produce glucose, while chemoheterotrophic processes use this glucose to create actual energy. 

Explain the inputs, major processes, and outputs of glycolysis, fermentation, and aerobic cellular respiration.

Glycolysis:

Major processes: 

Breaks down glucose into two molecules of pyruvate. 

Can occur with or without the presence of oxygen.

Occurs in cytoplasm.

Input: 1 Molecule of Glucose

Output: 2 Molecules of Pyruvate (and 2 ATP and 2 NADH)

Fermentation:

Major processes:

In the absence of O2, Glycolysis teams up with Fermentation to produce ATP. This is because Fermentation generates NAD+ by oxidizing NADH, which is renewable energy that powers Glycolysis.

Prevents the buildup of NADH in the cytoplasm.

Less energy is yielded than in Aerobic Cellular Respiration.

Input: NADH

Output: NAD+, 2 ATP

Aerobic Cellular Respiration:

Major processes:

Oxygen is present which yields more energy than Anaerobic Cellular Respiration (Fermentation). 

Takes place in mainly the mitochondria of the cell

Glycolysis, Citric Acid Cycle, and Electron Transport Chain are the 3 main stages of Aerobic Cellular Respiration.

Input: 1 Molecule of Glucose (fed into the process of Glycolysis)

Output: Maximum Amount of 38 ATP produced

Trace the movement of energy and matter through all cellular respiratory processes.

A carbon atom enters into cellular respiration as a glucose molecule, which is broken down by glycolysis, forming CO2 from C6H12O6. This glucose molecule is ultimately broken down to form 2 molecules of pyruvate which are fed into the Citric Acid Cycle where carbon is present in each step and is renewed over and over (hence, from the name, cycle).

Match all cellular respiratory processes to their locations in a typical eukaryotic cell.

Glycolysis - Cytoplasm

Citric Acid Cycle - Within the Mitochondrial Matrix

Oxidative Phosphorylation - Inner Mitochondrial membrane

Electron Transport Chain - In the cristae of the mitochondrion (the plasma membrane)

Explain the inputs, major processes, and outputs of the light reactions and the Calvin Cycle.

Light Reactions:

Major Processes:

Creates ATP and NADPH for Calvin Cycle to use.

Input: Light Energy and H2O

Output: O2, ATP and NADPH

Calvin Cycle:

Major Processes:

Synthesizes simple sugars from CO2

Regeneration of RuPB

Oxidizes NADPH

Input: ATP, NADPH and CO2

Output: ADP, NADP+ and Sugar

Trace the movement of energy and matter through all photosynthetic processes.

A carbon atom enters into photosynthesis as CO2 and is oxidized, taking part in producing C6H12O2 and H2O.

6CO2 + 12H2O + Light Energy ---> C6H12O2 + 6H2O

Photosystem II and Photosystem I finish their reactions and pass the baton to the Calvin Cycle.

Carbon enters into the Calvin Cycle as CO2 and leaves as a sugar named glyceraldehyde 3-phosphate (G3P).

Match all photosynthetic processes to their locations in a typical eukaryotic, autotrophic cell.

Calvin Cycle - stroma of the chloroplasts

Light Reactions - in thykaloids

Describe the process of chemiosmosis and compare its function in photosynthetic and respiratory pathways.

Chemiosmosis is the process of moving ions across the concentration gradient. In respiratory pathways, the protons pass across the concentration gradient to generate ATP that can be utilized in more than one way. In photosynthetic pathways, it occurs in the light reactions where ATP is generated for the Calvin Cycle and building of glyceraldehyde 3-phosphate (the sugar produced by photosynthesis)

Explain the relationship between photosynthesis and cellular respiration at the molecular, organismal, and ecosystem levels of organization.

[Source: cK-12]

Cellular respiration is the reverse equation of photosynthesis, and vice versa. The products of one reaction are the reactants of the other. Photosynthesis being 6CO2 + 6H2O → C6H12O6+ 6O2 and cellular respiration being  C6H12O6 + 6O2 → 6CO2 + 6H2O. In an organism, such as a plant, photosynthesis generates the glucose that is used in cellular respiration to create ATP. The glucose is then turned back into carbon dioxide, which is used in the process of photosynthesis. In an ecosystem, trees use photosynthesis to produce oxygen for all of us to breath. We require this oxygen in our bodies to perform the process of cellular respiration. 

Explain how energetic requirements contribute to the adaptations of organisms.  Provide examples to support your statements.

Over time, organisms tend to adapt to the amount of energy available in order to thrive. For example, some species of yeast are aerobic (require oxygen for cellular respiration) and give off gas, while others can be anaerobic (do not require oxygen) and not give off gas. This is most likely due to the environments that these different species of yeast developed in having either too little or too much O2 for cellular respiration. 

Propose experimental designs by which the rate of photosynthesis and respiration can be measured and studied.

You can determine the rate of photosynthesis by measuring the uptake of CO2. You can put a plant in a ziplock bag and observe the concentration of CO2 using a CO2 monitor. To study the rate of photosynthesis more in depth, one could run trials with various plants (taking into account leaf size) and measure the uptake of CO2 for each. Additionally, you can also study the rate of cellular respiration by either measuring the amount of glucose or oxygen consumed.

Describing 2–3 different strategies that organisms employ to obtain free energy for cell processes (e.g., different metabolic rates, physiological changes, variations in reproductive and offspring-rearing strategies).

1. Organisms can obtain energy by eating other (once living) things, as shown in the transfer of energy between the levels of consumers in an ecosystem.

[Source: Field Studies Council]

We all get our energy from the sun whether or not we are eating plants that took part in photosynthesis, or we are eating animals that ate plants, or animals that ate animals that ate plants! If we don’t have food to break down, we don’t energy to drive cell processes, and therefore we die.

2. Additionally, based on an organism’s diet, it can have an extremely different metabolic rate than that of another.  For instance, the sloth. It feeds entirely on tree leaves which makes for a diet that is not very nutritious since most of the calories in the leaves are closed up by hard-to-digest cellulose. Additionally, the metabolic rate of the sloth is nearly 1/2 that of other mammals their size. This extremely slow metabolism assists sloths in dealing with the small amount of energy that they gain from their diet by taking longer to process the little calories they consume. 

Refine or revise a visual representation to more accurately depict the light-dependent and light-independent (i.e., Calvin cycle) reactions of photosynthesis and the dependency of the processes in the capture and storage of free energy.

Pose scientific questions about what mechanisms and structural features allow organisms to capture, store, and use free energy (e.g., autotrophs versus heterotrophs, photosynthesis, chemosynthesis, anaerobic versus aerobic respiration).

One question someone may ask is how, even without oxygen, cellular respiration can continue to take place and generate ATP. This is because, since Glycolysis can take place with or without oxygen, it can couple with Fermentation to resume the production of ATP. This is because Fermentation generates NAD+ which powers Glycolysis, rendering it able to continue even without the presence of oxygen. However, a downfall of this process is that it generates much less ATP than aerobic cellular respiration since Fermentation (anaerobic cellular respiration) produces only 2 ATP, while aerobic cellular respiration can produce up to 38 ATP.

Create a visual representation to describe the structure of cell membranes and how membrane structure leads to the establishment of electrochemical gradients and the formation of ATP.

[milodoesscience]

Chapter 11 Portfolio

Describe how signals are received by cells.

Signals are received by cells in three steps, Reception, Transduction, and Response. Receiving a signal depends on whether or not the cell has a receptor for that specific signal. A signaling molecule, or ligand, binds to a receptor protein (usually in the plasma membrane) and causes it to change shape. This change in shape triggers the next step of the process, Transduction.

Use an example to explain how a receptor protein recognizes signal molecules, causing the receptor protein’s shape to change, which initiates transduction of the signal.

[Source: Pearson Education, Inc.]

A G protein-coupled receptor works with the help of a G protein which acts as an on and off switch. If GDP (not to be confused with ADP) is bound to the G protein, the G protein is inactive. When the ligand binds to the G protein-coupled receptor, the G protein is activated by the formation of GTP and binds to the once inactive enzyme. This enzyme is then activated and omits a cellular response that leads to the next step of the process, transduction.

Describe the process of signal transduction - include a model in your discussion.

Transduction consists of cascades of molecular interactions that relay signals from receptors to target molecules in the cell. The receptor (or sometimes a secondary messenger) activates a relay molecule which activates another protein, which activates another protein, which activates another protein, etc. until it reaches the protein that activates the cellular response.

Explain the concept of signaling cascades - include a model in your discussion.

[Source: The Science Creative Quarterly]

When the relay molecule is activated by events that occur during Reception, it activates a protein, which activates another protein, which activates another protein, etc. until the whole cascade reaches the activating protein. This protein’s job is to trigger the cellular response, so when the energy from the cascade of proteins becoming active before it reaches it, it is able to commence the response. 

Use an example to explain how second messengers are often essential to the function of a signaling cascade.

Calcium is an important second messenger because cells can regulate its concentration using Ca pumps. Second messengers, such as Calcium Ions, are essential to the initial activation of the signalling cascade during Transduction, which drives cellular response. 

Explain the key features of a model that illustrates how changes in a signal pathway can alter cellular responses.

As seen in the figure above, alterations in a signal pathway, such as the introduction of another receptor, or the branching of a pathway can lead to different cellular responses or offer another way to get to the same response. Different kinds of cells have different collections of proteins which allow them to detect and respond to different signals. Even the same ligand can have different effects in cells depending on the composition of their pathways and proteins.

Provide an example and explain how it relates to cell communication over short distances. Neurotransmitters are chemical messengers that cross the synaptic gaps (very very tiny gaps) between neighboring neurons. When released by the sending neuron, neurotransmitters travel across the synapse and bind to the receptor sites on the receiving neuron, thereby influencing whether that neuron will generate a new impulse.

Provide an example to explain how endocrine signals are produced by endocrine cells that release signaling molecules, which are specific and can travel long distances through the blood to reach all parts of the body. Thyroid hormones found in animals are examples of hydrophobic chemical messengers. These messengers can readily cross the membrane of a cell in order to activate receptors.  They are often made inside of the cell and packaged into vesicles until the signal is received for them to be released.  When released, they will bind to cell-surface receptors (such as GPCR or tyrosine kinase receptors) and trigger a response inside of the target cell. 

Use any of the below examples to explain how, in multicellular organisms, signal transduction pathways coordinate the activities within individual cells that support the function of the organism as a whole.

DNA can be damaged by average physiological processes, exposure to environmental conditions, and many more factors. Once DNA is damaged, cells must trigger a response to repair it in order to prevent mutation and maintain stability among the genome.  Ataxia telangiectasia mutated (ATM) and Ataxia telangiectasia and Rad3 related (ATR) genes can prompt the DNA damage reaction which activates the specific signal transduction pathways that increase the expression of DNA repair genes as a cellular response.