AMH 2010 Valencia College Federalist Paper by James Madison Discussion

Los Angeles Valley College DNA The Foundation of Life Biology Lab Report

Los Angeles Valley College DNA The Foundation of Life Biology Lab Report.

12 DNA: The Foundation of Life BackgroundDNA is the foundation of life. All living organisms including humans, monkeys, dogs, plants, molds and bacteria have DNA. If it’s alive, it has DNA in its cells. What is so important about DNA? First, DNA functions as the hereditary molecule that is passed on from one generation of cells or organisms to the next. DNA can do this because it has the ability to self-replicate through DNA replication. Second, DNA functions as the source of information for synthesizing each and every type of protein the cell may need. In this process called gene expression, information is said to “flow” from a gene in the DNA to a similar molecule called mRNA which is then used as a template to make very specific types of proteins. Proteins make up all the tissues which give an organism its structure (e.g. shape, texture) and function. They also serve as enzymes to enable most necessary chemical reactions. Proteins are thus extremely critical to life. DNA has a unique structure that allows it to function as the hereditary molecule and the foundation for protein synthesis. Briefly, it consists of a double helix structure. Each of the two strands of the double helix consists of a long string of nucleotides. There are four varieties of nucleotides. Each nucleotide consists of a deoxyribose sugar, phosphate group and base. There are four different types of bases: adenine (A), thymine (T), guanine (G) and cytosine (C) (see Fig. 1 below). In the double helix, the two strands have complementary base-pairs, which means that each base on one strand pairs specifically to another base on the second strand. A pairs with T and G pairs with C. These base pairs and therefore the two strands are held together by hydrogen bonds. This base-pair pattern is common across all life. These nucleotides can be strung together in different sequences to form the DNA strand. It is this variation in DNA nucleotide sequence that leads to variation in proteins made by the cells which leads to the variation seen in different organisms. Figure 1. Four nucleotides with deoxyribose sugar, phosphate group and base (adenine, thymine, guanine and cytosine)PurposeModel and describe the structure of DNAModel and explain the function of DNA in DNA replication and gene expressionDescribe how DNA contributes to the diversity of lifeMaterials1 DNA template strandDNA polymerase12 DNA nucleotides withBases: A (orange), T (yellow), G (green), C (blue)phosphate group (red bead)deoxyribose sugar (white bead)3 Hydrogen bonds: clear connectors RNA polymerase12 RNA nucleotides with bases A (orange), U (purple), G (green), C (blue)phosphate group (red bead)ribose sugar (pink bead) Large and small subunits of ribosome tRNAs with amino acids and release factorProcedure Activity 1: DNA ReplicationTo replicate DNA, the two original strands must first separate. These are your template strands. Then DNA polymerase, an enzyme, uses each strand as a template for building a new, complementary strand using the base-pairs (A to T, G to C). Each original (“parental” or “old”) strand is now paired with a new (“daughter”) strand, so there are now two identical DNA molecules instead of 1. DNA replication is called semi-conservative because each DNA molecule has 1 strand that is old and 1 strand that is new so half or “semi” of the molecule is “conserved” or old while the other half is new. Model the process of DNA replication using pop beads.Here, you will go through a model of the process of DNA replication using beads. This simplified bead model can help you visualize the process better. Watch DNA replication, transcription & translation by Manuel Robles (watch only the first 47 seconds) and follow along with the procedure and fill-in belowTake your pre-made template single strand of DNA. This is the “old” strand which is written below.Place the end of the template strand with the “T” (yellow bead) on the DNA polymerase.Now, working inside the DNA polymerase, start creating the new replicated strand by complementary pairing single nucleotides to the old strand (A to T, G to C, etc.). Write in the nucleotide sequence of the new DNA strand below. Finish until all DNA nucleotides are paired.Connect the two strands (old and new) using the clear connectors (hydrogen bonds) to link the bases. Each base pair should be connected, but for simplicity, use only 3 clear connectors for the entire DNA molecule.Twist your double stranded molecule to the left to demonstrate the “helix”. Fig 2. DNA Replication Note: You are only replicating one strand of DNA. In reality, both old strands are replicated.Fill in Old DNA strand T A C G C G A A G A T T New DNA strand ____________________________________ Activity 2: Gene ExpressionNow let’s look at DNA as the foundation for making proteins. DNA is organized into chromosomes. Each chromosome contains genes, which is a specific sequence of nucleotides that have the instructions for making one protein. In gene expression, information flows from the gene on the DNA to RNA to protein in two steps: Transcription: DNA is used to make mRNA (messenger RNA) in the nucleus. Since DNA and RNA are similar (both consists of nucleotides), this step is said to “transcribe” DNA into mRNA. For this to occur, first RNA polymerase, an enzyme, unwinds the DNA and uses one DNA strand as a template and makes a complimentary mRNA strand using base pairs. This base-pair is the same as for DNA replication, except that A pairs with U instead of T. After, mRNA is released, edited and exits the nucleus. *Note: mRNA also differs from DNA because it is single, not double, stranded and uses ribose, not deoxyribose, sugar.Translation: mRNA is used as a template to synthesize the protein in the cytoplasm. Since mRNA and proteins consist of entirely different building blocks (nucleotides vs. amino acids), this step is said to “translate” mRNA into the language of proteins. For this to occur, first a molecule called the ribosome binds to the mRNA. An mRNA codon (set of 3 nucleotides) within the ribosome binds to an anticodon (complimentary nucleotides to the codon) on the tRNA The tRNA brings a specific amino acid according to the genetic code (see Figure 3 below). Then the next codon on the mRNA goes through the same steps to bring in another amino acid which then attaches to the previous amino acid. Translation starts with the mRNA codon AUG. It ends with a release factor bind to a stop codon (UAA, UAG or UGA). After translation, the string of amino acids, the new protein, folds into its correct 3D structure.To review the process of gene expression, watch Protein Synthesis by Amoeba Sisters (8 min 46 sec).Model the process of gene expression using pop beads.Here, you will continue to model using beads. You will perform the process of gene expression with transcription and translation using beads. Watch “DNA replication, transcription & translation” by Manuel Robles (48 seconds to the end at 2 min 16 sec) and follow along with the procedure and fill-in below.Transcription: Unwind and disconnect your double stranded DNA molecule into 2 single strands.Take your original single strand of DNA. Place the end with the “T” on top of the “RNA polymerase” inside the nucleus. Make an mRNA strand using the DNA as the template strand and complimentary base pairs (recall: A–>U). Make sure you are using the nucleotides with the pink bead since RNA contains ribose sugars (pink bead) not deoxyribose sugar (white bead). Write your mRNA sequence below: Old DNA strand: T A C G C G A A G A T TmRNA sequence w/codons: _________________________________ anti-codon: _______ _______ _______ release factor amino acid: _______ _______ _______ Go until the entire DNA strand is transcribed into mRNA.Move your mRNA strand to the small subunit of the ribosome which is in the cytoplasm.Translation: Start at the beginning of your mRNA strand with the start codon: AUG, in the center of the ribosome.Write down the correct anti-codon sequence on the tRNA molecule that pairs with the codon using base-pairs. Bring in the complimentary tRNA molecule. Write the correct amino acid that it carries according to the genetic code (see Figure 3).Place the large unit of the ribosome on top of the small subunit of the ribosome, mRNA and tRNA molecules.Circle the next codon on the mRNA sequence above. Shift your next codon on the mRNA model to the center of the ribosome. The first tRNA is now shifted to the left end of the ribosome.Attach the new amino acid to the first amino acid. Remove the old tRNA molecule from the ribosome.Repeat steps 7, 9 and 10 for the third codon.Stop when you reach your stop codon, UAG. There is no tRNA for this stop codon. Instead there is a protein called release factor.Detach your amino acid chain of 3 amino acids. This is your new protein!Remove your mRNA strand from the ribosome.Separate the large and small units of the ribosome.Discussion/Conclusion1. Replication Where does DNA replication occur in the cell? _______________________________What molecule serves as template for DNA replication? ______________________What enzyme catalyzes the chemical reaction that adds nucleotides to a growing DNA strand?________________________________Gene Expression TranscriptionWhere does transcription occur? ______________________What section of the DNA serves as template for transcription? __________________What molecule is synthesized during transcription? _________________What is the substrate for synthesizing mRNA? _________________________What enzyme catalyzes the reaction that adds nucleotides to a growing RNA strand? ____________________________ TranslationWhere does translation occur? ______________________What molecule serves as template for translation? ____________________________What molecule is synthesized during translation? _______________________What is the substrate for synthesizing proteins? ________________________What organelle adds amino acids to the growing polypeptide (protein) strand? ______________________Summary Fill in the table below with theDNA sequence of your “gene.”sequence of the mRNA strand transcribed from your gene.amino acid sequence of your “protein” translated from your mRNA.name of the two processes of gene expression and their location.13 The Cell Cycle & MitosisBackgroundEvery hour, millions of your skin cells shed and millions of your red blood cells die. These cells need to be replaced with new ones to maintain your skin and blood. They are just a few examples of the many cells that compose a large organism. In addition, growth of organisms from a single cell to adulthood involves an increase in number of cells. The basic mechanism by which a body maintains itself, grows or repairs wounds is through cell division. The cell’s life cycle is an orderly series of events in which one parent cell gives rise to two new daughter cells, each which contain identical chromosomes to the parent cell. The phases of the cell’s life cycle are divided into interphase, which has two gap phases in which a cell enlarges as well as prepares to divide and an S phase, in which DNA is replicated. Mitosis is the division of the nucleus into two identical ones and is traditionally divided into 4 phases: prophase, metaphase, anaphase and telophase. Cytokinesis is the division of the cytoplasm and occurs during telophase. The diagram below helps illustrate the cell cycle”To review the cell cycle, watch The cell cycle (& cancer)” by Amoeba sisters (9 min 19 sec). Ignore the portion on CDK and cyclins (7:05 – 7:52 time). To review mitosis, watch The amazing cell process that uses division to multiply” by Amoeba Sisters (8 min 26 sec). The phases of mitosis begin at 5 min and 30 sec.You will be looking for cells undergoing mitosis in prepared slides of sections of fish blastula and onion (Allium) root tip. These phases of mitosis are continuous and there is no pause between them so it may be helpful to think of these slides as a snapshot in time. The fish blastula is a small ball of cells that is formed after the egg is fertilized and is actively growing so it contains many mitotic animal cells. The onion root tip is actively growing into the ground so it contains many mitotic plant cells.PurposeObserve and draw the 4 phases of mitosis in onion root tip and fish embryo.Describe events that occur in each phase of mitosis.MaterialsVirtual microscopeOnion (genus Allium) root tip slides White fish blastula (embryo) slides ProcedureGo to the virtual microscope.Load the slide for “Onion root”.Click Launch and then Explore.Click the “?” in the microscope slide box.In the Slide catalogue, select “Sample slides”. Inside, select “Plant slides”, then “Onion root”. The virtual microscope loads the onion root slide at the center of the stage. You will see a blurry image within the field of view.View the onion root at 4x, then 10x, then 40x.Click “4X” (scanning power objective lens). Use your mouse to adjust the Coarse Focusuntil the image is in focus. Fine tune it using the Fine Focus. Slowly slide the Light Adjustto the right to obtain optimum brightness. Click “10X” (low power objective lens). Adjust the Coarse Focus, Fine Focus & Light Adjustto obtain the clearest image with optimum light.Click “40X” (high power objective lens). This time, only adjust the Fine Focus & Light Adjust to obtain the clearest image with optimum light.Search for each of the 4 phases of mitosis and draw each phase.Click on the image in the field of view & move it. Search for each of the 4 phases of mitosis. Draw the image in the field of view for high power for one of the phases.Repeat steps until you have observed & drawn images of mitosis phases (prophase, metaphase, anaphase & telophase). Make sure to draw the image of chromosomes clearly for each phase.When finished, click “Remove slide”.Load the slide for “Whitefish interphase”Click the “?” in the microscope slide box.In the Slide catalogue, select “Sample slides”. Inside, select “Animal slides”, then “Whitefish interphase”.Follow the same procedure (#3-4) used to observe mitosis phases in the onion root. To find all stages, observe the “Whitefish late prophase” & “Whitefish late interphase” slides.Draw images of white fish mitosis phases (prophase, metaphase, anaphase & telophase) with chromosomes & plasma membrane clearly shown.Now that you have identified the stages of mitosis in stationary slides, watch Real Microscope Mitosis (1 min 28 sec) for a time lapse of cells undergoing cell division (mitosis & cytokinesis) in real time. Try to identify when the cell is interphase, prophase, metaphase, anaphase and telophase.Discussion/Conclusion Identify the cell cycle stages: A. _____________________________ B.______________________________C.______________________________D. ______________________________Stages between D & B. __________________________During what phase of the cell cycle does DNA replication occur?__________________During mitosis, at what phase do chromosomes first become visible?________________During what 2 phases of mitosis are chromosomes each composed of two chromatids?_____________________________ _______________________________During what 2 phases of mitosis are chromosomes each composed of one chromatid?_____________________________ _______________________________How does cytokinesis differ in plants and animals?What is the term used to refer to the “division of the cytoplasm”? _________________Are the two cells created from cell division & mitosis the genetically identical or different?Write the stages of the cell’s life cycle in order, starting with the stage immediately after cytokinesis.15 DNA Extraction and Gel ElectrophoresisBackgroundToday, DNA technology is used for various purposes such as identifying criminals, performing paternity tests, understanding genetic diseases, engineering food and drugs and much more. To understand all these applications of DNA technology, we first need to understand the basics of DNA. Scientists have known about DNA since the mid-1800’s. It wasn’t until the 1950’s, however, that Francis Crick and James Watson with the help of Rosalind Franklin and Maurice Wilkins understood its role as genetic material. DNA is present in almost every cell of an organism. For example, if DNA is used to identify individual suspects in a forensic case, scientists can get DNA from almost any cell such as hair, blood or semen. First, however, scientists need to extract the DNA from the cells. In this lab, we will perform a simple procedure to extract DNA from saliva and strawberries.After extracting the DNA, it needs to be analyzed. British geneticist, Dr. Alec Jeffreys, first developed DNA gel electrophoresis in 1984. By using this procedure, you can distinguish one individual from the other by the differences in their DNA sequences. Gel electrophoresis is a standard method used to separate DNA fragments of different sizes. In this method, DNA fragments are cut by restriction enzymes and then loaded on a gel made of agarose polymer, a chain of sugar molecules. An electric field is created with a negative and positive end. Since DNA has a negative charge due to its phosphate group, it will move from the negative to the positive end of the gel. The rate at which a DNA fragment moves depends on its size. The smaller the DNA fragment, the faster and further it travels down the gel. The larger the DNA fragment, the slower and less it travels down the gel. Thus, the DNA fragments are separated based on size alone. Each individual will have different size DNA fragments due to their different DNA sequences. So the pattern of DNA fragments on the gel will be unique for each individual. If DNA is obtained from the crime scene and the suspect, scientists can run both DNA samples using gel electrophoresis. If the DNA patterns match on the gel, then you found the right suspect! If not, then you need to keep searching. In this lab, we will investigate a hypothetical crime scenario.Materials (DNA extraction)For Human DNA Extraction:2 glassesWaterSpit2 teaspoons detergent (ie. dish soap, hand soap, shampoo, laundry detergent)½ teaspoon Salt3 drops contact solution or pineapple juice2 teaspoons rubbing alcohol (chilled in freezer)Stick (ie. toothpick or popsicle stick)For strawberry DNA Extraction:~1/4 cup frozen strawberriesPlastic bag that you can seal or close1 glass2 teaspoons detergent (ie. dish soap, hand soap, shampoo, laundry detergent)½ teaspoon Salt½ cup of waterRubbing alcohol (~2-3 tablespoons)Stick (ie. toothpick or popsicle stick)Filter (ie. coffee or tea filter) Procedure (DIY DNA extraction)WatchHuman DNA Extraction | DIY Crime Scene Investigator Activities | Whodunit?(2 min 9 sec), a video of DNA extraction from human cheek cells, and follow the procedure:Collect cells containing DNA in salivaCollect human cheek cells by taking a mouthful of water and move it around in your mouth for 2 minutes.Spit water into empty glass #1.Concentrate the DNA in the sampleAdd a pinch (about ¼ teaspoon) of kitchen salt (NaCl) to glass #1 with spit water and stir gently.Lyse (break open) the cells to release the DNAPour ½ cup of water into an empty glass #2. Add 2 teaspoons of detergent and stir.Add 2 teaspoons of detergent water to glass #2 your salty spit water with cheek cells.Let the sample sit for 10 minutes. Degrade proteins that were released from broken cells Transfer mixture in glass #2 with spit, salt and detergent to a narrow glass.Add 3 drops of contact lens fluid or pineapple juice and mix gently. Let the sample sit for 10 minutes to give the proteases time to do their work.Precipitate the DNA from the solutionSlowly drop 2 teaspoons of chilled alcohol down the side of your tube to precipitate DNA. Let sit for 1 minute. You should see strings of DNA forming on the detergent solution / alcohol boundary.Fill in observations in Table 1 belowExtract DNA from strawberriesWatch “How to extract DNA from strawberries”(9 min 45 sec) or “Science Mom extracts DNA from a strawberry” (4 min 36 sec) and follow the procedure:Let ~¼ cup of frozen strawberries defrost and become soft.Put strawberries in bag and seal. Smash strawberries with hand.Create lysis solution: Add to a cup, 2 teaspoons of detergent, ½ teaspoon of salt to cup and ½ cup of water to cup. Stir.Add lysis solution from cup into strawberries in bag.Don’t over mix and avoid bubbles.Filter strawberry solution back into cup.Add equal parts of rubbing alcohol to the strawberry solution. You should see strings of DNA forming on the strawberry mixture / alcohol boundary.Try wrapping the stringy DNA with a stick.Figure 1. DNA extracted from human cheek cells (left) and strawberries (right)Results (DNA extraction)Discussion/Conclusion (DNA extraction)What is the role of each of the following in extracting DNA?a. Cheek cells and strawberriesb. Detergentc. Saltd. Contact solution / pineapple juicee. Alcohol How does a temperature above 60ºC affect DNA?DNA in saliva and strawberries are only 50% similar. How are they 50% different? 4. Did this difference show up in your observations of your extracted DNA? Please explain.Materials (gel electrophoresis demo)Gel electrophoresis apparatus Agarose gel Tray for gelElectrophoresis buffer solution Power supply DNA samples MicropipetteProcedure (gel electrophoresis demo)Watch “Agarose gel electrophoresis” (4 min 6 sec) to view this gel electrophoresis procedure for the class.Set-up includes gel electrophoresis apparatus with tray containing agarose gel. Tray ends are matched with appropriate ends of the apparatus.Pour in electrophoresis buffer on both side of the tray to cover the gel completely.Load 10 µl of each DNA sample obtained from the different sources into the wells of the gel with micropipette. Go in order from left to right with DNA A-F. (Instructor may ask student volunteers for this step)Connect gel electrophoresis apparatus to power supply. The positive lead (red) should be at the end opposite from the wells. The negative lead (black) should be at the end with the wells.Set power supply at 80 V with the setting of “low” select. Turn on. Check for proper running of the apparatus. You should see gas bubbles forming in the buffer solution.Run gel for ~35 minutes. *Carefully remove the tray with the gel and slide gel into viewing tray.Visualize gel. Record the number and patterns of bands from each sample.Figure 2. Sample results of Gel electrophoresis of DNA fragmentsResults (gel electrophoresis)Figure 3. Results of Gel electrophoresis of DNA samples from the lab, from A to EDraw and label your DNA wells in Figure 4 from left to right at the top of the gel. Use Figure 3 results.Crime Scene DNA ASuspect #1 DNA BSuspect #2 DNA CSuspect #3 DNA DSuspect #4 DNA ELabel the charges (-) and (+) at the appropriate ends of the gel.Document the DNA pattern results on your gel.Discussion/Conclusion (gel electrophoresis)In the experimental data on the gel, what does each band represent?Why do different bands travel different lengths?Why does DNA travel down the gel towards the positive charge?Why do most individuals have different band patterns on a gel?According to the data from the gel, who committed the crime? How does the data from the gel support your answer? 16 Biotechnology Article and DebateBackgroundBiotechnology is relevant to our everyday lives. There are many benefits of biotechnology that advance our lives and the world that we live in. At the same time, there are negative consequences and ethical issues that we have to contend with. Regardless, biotechnology research will continue to progress and impact us. Where do you stand in this debate?PurposeResearch a current biotechnology topicUnderstand pros and cons to biotechnologyDo a literature searchDiscuss a scientific topicMaterialsComputerBiotechnology articleProcedureFind an article related to biotechnology that has been published within the past 1 year.- Go to reliable websites ending in .org, .edu, .gov to select an article. There are some websites that are reliable ending in .com, but check the source!- Please make sure that you are collaborating with your classmates.Read it carefully.Fill out the questions on the following page.Each student shouldSummarize your article for your group.Discuss your views on the topic (e.g. pros, cons, where you stand)Ask your peer questions and give your stance on their topic. Review and organize the pros and cons of biotechnology.Discuss and debate using peer review. Discuss if you would support your lab mates technology. Why or Why not?Biotechnology Article Information Title: ________________________________________________ Author(s) ________________________________________________ Publication Date: ________________________________________________ Publication source: ________________________________________________Summarize the article in 2-3 sentences:Discussion/Conclusion Discuss pros to using this technology.Discuss cons to using this technology.What was the most interesting aspect about this article for you?Do you support this technology? Why or why not?
Los Angeles Valley College DNA The Foundation of Life Biology Lab Report

Bond Pricing and Interest Rates

essay order Bond Pricing And The Term Structure Of Interest Rates: A New Methodology For Contingent Claims Valuation We read the paper Bond pricing and the term structure of interest rates by Heath, Jarrow, and Morton. Their paper presents a theory for valuing contingent claims under a stochastic term structure of interest rates. The methodology takes as given an initial forward rate curve and a family of potential stochastic processes for its subsequent movements. A no arbitrage condition restricts this family of processes yielding valuation formulae for interest rate sensitive contingent claims which do not explicitly depend on the market prices of risk. In relation to the term structure of interest rates, arbitrage pricing theory has two purposes. The first is to price all zero coupon (default free) bonds of varying maturities from a finite number of economic fundamentals, called state variables. The second, is to price all interest rate sensitive contingent claims, taking as given the prices of the zero coupon bonds. The primary contribution of this paper, however, is a new methodology for solving the second problem, i.e., the pricing of interest rate sensitive contingent claims given the prices of all zero coupon bonds. The methodology is new because (i) it imposes its stochastic structure directly on the evolution of the forward rate curve, (ii) it does not require an “inversion of the term structure” to eliminate the market prices of risk from contingent claim values, and (iii) it has a stochastic spot rate process with multiple stochastic factors influencing the term structure. The model can be used to consistently price (and hedge) all contingent claims (American or European) on the term structure, and it is derived from necessary and (more importantly) sufficient conditions for the absence of arbitrage. The Difference Between Duration And Maturity In Bonds – Based on the article: Bond Price Volatility and Term to Maturity: A generalized Re-specification Most investors and especially we who are taking the course Fixed Income Securities are familiar with the bonds maturity. The article wants to illustrate why we can’t just rely on the length of maturity when estimating how volatile the price for a certain bond is. As we have read before during the course, there is a common and accepted thumb rule that tells us that for a given change in yields, the price change for the bond will be greater the longer the term to maturity is. Therefore I first of all want to highlight the difference between the duration and maturity. Firstly I will have a short explanation of these two terms and further I will continue this paper by explain some important parts from the article. When it comes to maturity, we all know the maturity is the point in time when the investor receives back the principal. We also know that a bond will increase in value, that is, the price of the bond will increase, if the interest on the market decrease and vice versa. From this statement above, it may be clear that the longer maturity, the more changes in the interest rates can be waited and the more volatile the bond price will be. The duration of a bond will show how sensitive a bond’s price is to changes in the interest rate. It’s a measurement for how much the bond price will change due to a one percentage change in the interest rate on the market. Duration of 7 means for instance that if the interest rate raises by one percentage means that the price of the bond will fall 7 percentages. The duration is simply the weighted average amount of time that it takes for the investor to be repaid all cash that is both the coupon payments and the end, principal payment. Therefore the duration will always be less than the maturity, except for zero-coupon bonds where they will be equal. These two properties are important when it comes to duration: The first one is that the longer the maturity, the higher the duration. The second one is that the lower the coupon payment the higher the duration. With these facts above, I want to highlight the important aspect of this what the article goes through. Indeed, the price volatility is connected to the time structure of the bond, but it’s not direct mathematically related to the maturity in a pure simple way. Since there are evidence that duration is more accurate, the authors for the article wants to generalize the following: For a given basis point change in market yield, percentage changes in bond prices vary proportionally with the duration and are greater, the greater the duration of the bond. There is also true that there is an inverse relationship between duration and coupon. This means that a higher coupon bond will automatically be seen as a shorter-term bond than a bond that has a lower coupon payment, even if they in fact have the same maturity period. Furthermore, this means that a comparison of these bonds with equal maturity will underestimate the default risk premium in periods of upward sloping yield curves and also overestimate the premium in periods of downward sloping yield curves. For instance, referred to the above information, there is evidence that the duration varies inversely with coupon rates. A 50 year 8 percent coupon bond, yielding 6 percent, have approximately the same duration as a 20 year 2 percent coupon bond yielding the same amount of 6 percent. To sum up the article and this paper, I want to highlight the complexity of the relationship between the bond price volatility and the maturity, as well as the relation between then bond value and the duration, even if I believe that duration is a more accurate measurement of price volatility. Expectations, Bond Prices, And The Term Structure Of Interest Rates The term structure of interest rates is of great importance when dealing with bonds, since the interest rate significantly affects the bond price. Burton G. Malkiel examines the relationship between market interest rates and bond prices in his article Expectations, Bond Prices, and the Term Structure of Interest Rates, where he takes the position that Lutz theory of Basic Behavioural postulate is correct and important in understanding the behaviour of market interest rates of securities with different term to maturity. Lutz says that investors decide whether to invest in bond based on their expectation of future short rates, since they are not able to predict long term rates. Malkiel furthermore aims to ease the principle hypothesis by Hicks and Keynes, that future prises are biased expectations of future spot prices. A bond’s market price or value is determined by four factors: the face value of the bond; the coupon or interest paid periodically to the bondholder; the effective interest rate per period; and the number of years to maturity. The lower interest rate, the higher the bond price – hence the significant relationship between the two factors. The term structure in this sense is important since the investor wants to choose the term and bond that is most beneficial. According to Malkiel, the term structure is based on investors’ expectations, which is influenced by the normal range of interest rates. If interest rates appear to be very high relative to the normal range, investors may expect that interests will fall and vice versa. The term structure problem is furthermore analysed with a combination of spot and forward trading, resulting in longer term rates as combinations of relevant forward short rates: (1 R2)2 = (1 r1)(1 r2). When long-term average rates are below the current short rate future short-term rates are expected to fall, and conversely, long rates will exceed the current short rate if future short rates are expected to rise. Additionally, when interest rates are believed to be high in relation to historical averages, investors will prefer long-term bonds while issuers prefer to sell short-term securities, whilst low interest rates will encourage investors to buy shorts and issuers to sell longs. Supporting Lutz’s theory, Malkiel makes the conclusion that short and intermediate areas of the yield curve exhibit more dramatic responses to changes in expectations. This is due to the fact that investors cannot predict the long term rates; they only interpret the near past and current market conditions. What is also worth mentioning is that the term structure fluctuates more between e.g. one to two years, and three to six years, compared to a very long period of time, e.g. thirty-four and sixty-eight years. This is also due to the fact that it is difficult to predict changes in interest rates for such a far away future. This furthermore explains that the yield curve tends to flatten out the longer term to maturity. To conclude, investors will choose to purchase bonds depending on their expectations of how interest rates will change in the near and long-term future, the term-structure, and that presumably would be most beneficial in terms of bond price and returns. Does Duration Extension Enhance Long-Term Expected Returns? The article’s main purpose is to give investors important information regarding duration and if you can gain a profit or not in the long-term. The author is using empirical evidence mainly from the U.S. Treasury bond market over the past 25 years. All the results of the past returns depend on the interest rate trend in the period the authors are looking at. The focus in the article lies on the long-run expected return differentials across bonds with different maturities. The risk premium is defined as the long-term return exceeding short-term risk-less rate. The writer means that the one-year bill earns on average 150 basis points higher return than one-month bill and after two years the yield curve will remain a constant line. In other words the return of the bill will stay reasonably the same at two years but recall that long-term bonds are riskier than short-term bonds because it is difficult to predict the future. In other words it is uncertain how much the bonds are worth in the long-term because there are many different causes that affect the value in the future. The article discusses the bond risk premium using six theories. There are three classic term structure hypotheses. The first is called pure expectations which means assuming that there is no risk premium. The second is the liquidity/risk hypothesis explaining the compensation for return volatility. The third explains the increase and decrease with duration depending on time horizon called the preferred habitat theory. Ilmanen presents two modern asset pricing theories. One that explains the risk premium proportional to return volatility and the other one that clarifies CAPM. The latter explains that the risk of assets depends on the sensitivity to aggregate wealth as in stock market sensitivity (β correlation) and risk premium (β Market risk premium) which in turn depends on market volatility and risk aversion level. Equilibrium model means assets performing poorly in bad times should earn positive risk premium while assets performing well are accepted for low yields but other non-risk related factors are also mentioned. To sum up the article long-term bonds are riskier than short-term bonds and investors earn positive risk premium for bearing this risk. Various models specify that expected returns are linear in duration and return volatility but other factors may contribute. References Longstaff, F. A., and E. S. Schwarz (1992), “Interest Rate Volatility and Term Structure: A Two-Factor General Equilibrium Model”, Journal of Finance, Vol. 47(4), pp. 1259-1282. Heath, D., R. Jarrow, and A. Morton (1992), “Bond Pricing and Term Structure of Interest Rates: A New Methodology for Contingent Claims Valuation”, Econometrica, Vol. 60(1), pp. 77-105. Hopewell, M. H., and G. G. Kaufman (1973), “Bond Price Volatility and Term to Maturity: A Generalized Re-specification”, The American Economic Review, Vol. 63(4), pp. 749-753. Malkiel, B. G. (1962), “Expectations, Bond Prices, and Term Structure of Interest rates”, The Quarterly Journal of Economics, Vol. 76(2), pp. 197-218. Ilmanen, A., (1996), Does Duration Extension Enhance Long-term Expected Returns?” Journal of Fixed Income, September, pp. 23-36.

Critical Thinking Blocks

Critical Thinking Blocks. I’m trying to study for my English course and I need some help to understand this question.

Rationale to Critical Thinking
Think about the blocks to critical thinking you studied in this lesson. As the Critical Thinking Blocks presentation (attached) noted, it is normal to get caught up in common obstacles like egocentrism. Think of a time when you witnessed one of these blocks in action, either in yourself or in someone else. What characteristic of critical thinking mentioned earlier in this lesson could have been useful to avoid or overcome the block that you witnessed?
Critical Thinking Blocks

Walden Universit Week 5 Confidence Intervals & Reject the Null Hypothesis Discussion

Walden Universit Week 5 Confidence Intervals & Reject the Null Hypothesis Discussion.

Required ReadingsFrankfort-Nachmias, C., Leon-Guerrero, A., & Davis, G. (2020). Social statistics for a diverse society (9th ed.). Thousand Oaks, CA: Sage Publications.
Chapter 8, “Testing Hypothesis: Assumptions of Statistical Hypothesis Testing” (pp. 241-242)Wagner, III, W. E. (2020). Using IBM® SPSS® statistics for research methods and social science statistics (7th ed.). Thousand Oaks, CA: Sage Publications.
Chapter 6, “Testing Hypotheses Using Means and Cross-Tabulation”Warner, R. M. (2012). Applied statistics from bivariate through multivariate techniques (2nd ed.). Thousand Oaks, CA: Sage Publications.Applied Statistics From Bivariate Through Multivariate Techniques, 2nd Edition by Warner, R.M. Copyright 2012 by Sage College. Reprinted by permission of Sage College via the Copyright Clearance Center.
Chapter 3, “Statistical Significance Testing” (pp. 81–124)Applied Statistics From Bivariate Through Multivariate Techniques, 2nd Edition by Warner, R.M. Copyright 2012 by Sage College. Reprinted by permission of Sage College via the Copyright Clearance Center. Magnusson, K. (n.d.). Welcome to Kristoffer Magnusson’s blog about R, Statistics, Psychology, Open Science, Data Visualization [blog]. Retrieved from http://rpsychologist.com/index.htmlAs you review this web blog, select [Updated] Statistical Power and Significance Testing Visualization link, once you select the link, follow the instructions to view the interactive for statistical power. This interactive website will help you to visualize and understand statistical power and significance testing.Note: This is Kristoffer Magnusson’s personal blog and his views may not necessarily reflect the views of Walden University faculty.American Statistical Association (2016). American Statistical Association Releases Statement on Statistical Significance and P-Values. Retrieved from http://www.amstat.org/newsroom/pressreleases/P-ValueStatement.pdfAs you review this press release, consider the misconceptions and the misuse of p-values in quantitative research.To prepare for this Discussion:Review the Learning Resources related to hypothesis testing, meaningfulness, and statistical significance.Review Magnusson’s web blog found in the Learning Resources to further your visualization and understanding of statistical power and significance testing.Review the American Statistical Association’s press release and consider the misconceptions and misuse of p-values.Consider the scenario:A research paper claims a meaningful contribution to the literature based on finding statistically significant relationships between predictor and response variables. In the footnotes, you see the following statement, “given this research was exploratory in nature, traditional levels of significance to reject the null hypotheses were relaxed to the .10 level.”By Day 3Post your response to the scenario in which you critically evaluate this footnote. As a reader/reviewer, what response would you provide to the authors about this footnote?Be sure to support your Main Post and Response Post with reference to the week’s Learning Resources and other scholarly evidence in APA Style.By Day 5Respond to at least one of your colleagues’ posts and explain the benefits and consequences of the “relaxed” level of significance.Be sure to support your Main Post and Response Post with reference to the week’s Learning Resources and other scholarly evidence in APA Style.
Walden Universit Week 5 Confidence Intervals & Reject the Null Hypothesis Discussion