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Gene Expression Discovery of DNA & Structure Classwork

1. Discuss the statement: “DNA is evolution”. 2. Describe the independent and dependent variables in Griffiths’ experiment. 3. Describe what Griffith concluded from this experiment. 4. Avery decides to use an ASSAY to perform his test. Describe the benefit of using this format over

Griffith’s experiment. 5. How did the structure of nucleic acids affect Hershey and Chase’s experimental design? 6. Describe the significance of X-ray diffraction being used in the search for the structure of DNA. 7. Compare and contrast characteristics of an RNA and a DNA molecule. 8. Chromosome size and structure can vary widely depending on the type of organism. Briefly describe

the typical eukaryotic and prokaryotic chromosome. Homework Read the following article “Isolating Hereditary Material: Frederick Griffith, Oswald Avery, Alfred Hershey, and Martha Chase” By: Clare O'Connor, Ph.D. (Biology Department, Boston College) © 2008 Nature Education and answer the questions that follow. http://www.nature.com/scitable/topicpage/isolating-hereditary-material-frederick-griffith-oswald-avery-336

9. Describe the relationship between polysaccharides found in the capsule and the virulence of the pneumococci.

10. Take a close look at figure 3, this provides a detailed of Avery’s experiment. Provide a hypothesis for Avery’s experiment.

11. Describe the conclusion Avery was able to ascertain after completing his experiment. 12. Alfred Hershey and Martha Chase choose to use bacteriophage for their experiment. Explain why they

choose this for their experiment, and provide a brief description of a bacteriophage. 13. Edwin Chargaff contributed to the discovery of DNA structure by noting that the amount of adenine in

any molecule of DNA was equal to the amount of thymine. The same held true of guanine and cytosine. Explain the significance of Chargaff’s rule to the discovery of DNA structure.

14. Describe the major chemical features of a DNA molecule. 15. Sketch and label a typical eukaryotic chromosome. Include the following in you drawing. Centromere,

histone, short arm and long arm. DNA Replication Classwork

16. Give the complimentary strand to the following DNA sequence: 5’ATTCGCATGCCAGTTA3’ 17. Discuss why DNA replication is known as “semi-conservative” replication. 18. Describe the direction that new nucleotides are added to the template side of the molecule. Which

carbon does the new nucleotide phosphate bond to, and what is the name of this bond? 19. Enzymes are important proteins used in chemical processes. Describe the role of topoisomerase,

helicase, and DNA polymerase in DNA replication. 20. Originally it was thought that DNA was copied from 3’ to 5’ and 5’ to 3’ in the same way on both

strands, however DNA polymerase only works in the 3’ to 5” direction. Describe how Okazaki fragments work.

21. Suppose there were a mutation and the enzyme primase could not be synthesized. What effect would this have on DNA replication?

22. PCR is a technique used to amplify small amounts of DNA. Describe the procedure used in this process.

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Homework 23. What is the purpose of DNA replication in the cell? 24. Give the complementary strand for the following DNA sequence: 3’GCATACGATATTCCGG5’ 25. Why does DNA base pairing occur? 26. DNA polymerase makes a mistake once in every 104 – 105 nucleotides. Discuss the enzyme

responsible for fixing these errors and what occurs when the mistakes are repaired. 27. The lagging strand of DNA requires additional enzymes to replicate than the leading strand. Explain

why this is so. 28. A single DNA molecule is placed in a PCR machine. How many DNA molecules will be present after

30 cycles? 29. PCR techniques require enzymes from thermophilic bacteria. Explain.

Transcription & Translation Classwork

30. Describe what is meant by the phrase “Universal Genetic Code”, and explain the significance of having 64 codons that code for only 20 amino acids.

31. Explain why RNA polymerase does not bind to the “non-template strand”. 32. Describe how the lac operon works. 33. Walter Fleming stated that chromatin appeared like threadlike structures in the cell’s nucleus. Describe

how chromatin can affect gene expression. 34. Transcribe the following gene (template strand) and show all steps of RNA processing (exons in bold)

5’TACCCGCCTAATCAGTGCTTCCTTTATCCGCATGCCCGGATCGGAATT3’ 35. After transcription, mRNA must be stripped of useless strands that were copied and intermingle the

target gene. Surprisingly, many biology students confuse the two because of the first two letters in each of their names. Provide a proper explanation for the coding and non-coding parts involved in RNA splicing.

36. Gene expression is the molecular process of reading the order of nucleotides in a DNA molecule and making the coded product. Describe the requirements needed for a gene to be expressed.

37. RNA plays a critical role in the process of translation. Describe the structure and role of the three types of RNA that are used during translation.

38. Describe the significance of methionine; please provide the mRNA codon. 39. Translate the processed RNA from #34 into the correct polypeptide using the codon chart. Draw the

polypeptide beneath the RNA and label it appropriately.

Homework 40. In 1958 Francis Crick stated what he called the “Central Dogma of Biology”. To what was he referring? 41. Describe the trp operon. 42. Describe the attachment site for RNA polymerase. Also describe what may inhibit it from attaching to

DNA. 43. Describe the role and mechanism of transcription factors in gene regulation. 44. Following transcription the mRNA has to under go what is known as mRNA processing. Provide a

description to mRNA processing, and an explanation as to why it is needed. 45. Provide a summary of Gene Regulation in Eukaryotic Cell. Along with you written summary provide a

15 base long of DNA, then use is as an example for each step. Be sure to point of the 3’ and 5’ end. 46. Translation is a crucial step in the assembly of a protein. Describe the three stages of translation

involved in protein synthesis. 47. Describe the difference between the P-site and the A-site on the ribosome. Which does mRNA enter

first? 48. Hypothesize why there is only one start codon yet there are 3 stop codons. 49. The follow gene can code for more than one polypeptide. This will be dictated by which parts of the

gene are inhibited and which are expressed. Given the following DNA strand, show pre mRNA,

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alternate splices, and translation to amino acids. The parts that are underlined and italics are part of the same polypeptide.

DNA Template - AAG AAT GGC CGG TTA AGA CCC mRNA - alternate splices amino acids

50. Given the following polypeptide, provide a possible DNA molecule, please label the 5’ and 3’ end. Glu – Ser – Arg – Phe – Leu - Val Recombinant DNA Classwork

51. In you own words describe “recombinant DNA”. What was the significance of development of Humulin? 52. Briefly outline the steps involved in making Recombinant DNA. 53. Splicing a red fluorescent gene from a jellyfish into the genome of a puppy has created transgenic

glowing puppies. These two species are not closely related. Explain how the puppy’s cells can utilize this gene.

Homework

54. Restriction enzymes play an important role in Recombinant DNA technology. Describe their role and explain how you decide which enzyme to use.

55. Some restriction enzymes makes cut that result in blunt ends while other create sticky ends. Describe the difference between blunt ends and sticky ends.

56. Describe the role of gel electrophoresis in recombinant DNA technology.

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Free Response

1. The above diagrams illustrate the experiments carried out by Griffith and Hershey and Chase- respectively.

a. Describe the hypothesis or conclusion that each set of experiments proved or set out to test. b. Why were the Griffith experiments inconclusive as to the source of the genetic material?

2. If dividing cells are treated with the pyrimidine analog bromo- deoxyuridine(BrdU), during DNA replication, the cells are fooled into incorporating it —instead of thymidine (thymine) — into their DNA. One of the properties of the resulting DNA is that it fails to take up stain in a normal way. When cells are allowed to duplicate their chromosomes once in BrdU, the chromosome that appear at the next metaphase stain normally. However, when the cells duplicate their chromosomes a second time in BrdU, one of

the sister chromatids that appears at the next metaphase stains normally, while its sister chromatid does not (Harlequin Chromosomes). Note that each chromatid is normally completely stained or not (circled). http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/B/Bodycote_L.jpg

a) Using your knowledge of DNA replication, describe a possible hypothesis of why only one sister chromatid gets stained darker than the other chromatid after the second round of cell division?

b) What will you expect to see in the chromosome staining pattern if cells are allow to divide for one more round of DNA replication and cell division after the initial 2 rounds of cell division and DNA replication?

Thymine

bromodeoxyuridine

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3. Lactose metabolism is controlled by the lac operon, which consists of the lacZ, lacY, and lacA genes encoding β-galactosidase, lactose permease, and transacetylase, respectively. Expression of the operon is negatively regulated by a transcription factor, the lac repressor (LacI), which dissociates from its specific binding sequences of DNA, the lac operators, in the presence of an inducer to allow transcription (Fig. A). The repressor LacI and permease LacY form a positive feedback loop. Expression of permease increases the intracellular concentration of the inducer lactose which causes dissociation of LacI (lactose repressor) from the promoter (Plac), leading to even more expression of permeases.

Cells with a sufficient number of permeases will quickly reach a state of full induction (Figure B), whereas cells with too few permeases will stay uninduced. (B) After 24 hours of growth in media containing 30 μM TMG, strain SX700 expressing a LacY-YFP fusion exhibits all-or-none fluorescence in a fluorescence-phase contrast overlay (bottom, image dimensions 31 μm × 31 μm). Fluorescence imaging with high sensitivity reveals single molecules of permease in the uninduced cells (top, image dimensions 8 μm × 13 μm). A Stochastic Single-Molecule Event Triggers Phenotype Switching of a Bacterial Cell Paul J. Choi, Long Cai, Kirsten Frieda, and X. Sunney Xie Science 17 October 2008: 322 (5900), 442-446.

a) If there is no lactose present, why will

the E. coli cells not produce the proteins involve in lactose metabolism?

b) The boxed cells in figure B shows cells will single molecules of permease present within the cells, why are these cells not fully fluorescent?

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4. The diagram shows 2 possible pathways of polymerase exchanges that are supported by the results of observing changes in the fluorescence and bleaching (removing fluorescence) of single molecular fusions of polymerase III, the major DNA polymerase in bacteria, within the replication fork of live E. coli cells.

Polymerase Exchange During Okazaki Fragment Synthesis Observed in Living Cells Giuseppe Lia, Bénédicte Michel, and Jean-François Allemand Science 20 January 2012: 335 (6066), 328-331

a) Describe why so many proteins are required to carry out the process of DNA replication?

b) Why does DNA need to get copied? c) Select the model that you think that

cells would favor and provide a rationale for your choice.

5. For more than a decade, circadian (daily rhythms) researchers have examined the RNA molecules produced cyclically in many tissues. Consistently, the rhythmically expressed genes exhibited expression peaks throughout the entire day, with an enrichment for transcripts peaking at the dawn and dusk transitions. The core transcriptional machinery of the mammalian circadian clock includes BMAL1, CLOCK, and NPAS2 proteins that positively regulate gene targets that contain the

circadian cis-regulatory sequence known as an enhancer box (E box), or a slight sequence variant the E′box, in their promoters or enhancers. The circadian regulators PER1, PER2, PER3, CRY1, and CRY2 have a repressive role on these same targets. The transcriptional activity takes place in three phases (see the figure). First, a transcriptionally poised state occurs as both the repressor, CRY1, and the activators, BMAL1 and CLOCK, occupy regulatory sequences. As the amount of CRY1 binding slowly decreases, the transition to

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activation (the second phase) occurs as the coactivator protein p300 and marks of active chromatin (DNA and protein) [histone acetylation (H3K9ac) and methylation (H3K4me1)] are established. In the final repressive state, BMAL1 and CLOCK occupancy decreases and accumulation of the repressive factors PER1, PER2, and CRY2 is observed on target genes. It will be interesting to see how this transition occurs at each individual promoter, whether the same DNA region can retain both repressors and activators in the poised state, or whether this observation reflects an average of multiple cells in different states of transition. Circadian Surprise—It's Not All About Transcription Colleen J. Doherty and Steve A. Kay Science 19 October 2012: 338 (6105), 338-340.

a) Select one activity either performed by a cell or an organism that may be regulated by the circadian clock described in the above reading. Describe the rationale by your choice.

b) Using the above graph, provide evidence to support the idea that BMAL1 and CLOCK regulate daytime activity.

c) Our day is 24 hours long; the length of day varies in other planets. How would a change in the length of the day affect the circadian clock? How would the information be relayed to the transcription pathway that is the circadian clock?

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Gene Expression-Answer Key

1. DNA is the genetic material passed on to future individuals. Small changes in the DNA, such as mutations can be passed onto future generations. These changes can add up over time and lead to evolution of the organism.

2. Independent variable - type of Bacteria injected Dependent variable – survival of the mice (alive or dead)

3. Griffith concluded that the R- strain had been “transformed into a deadly strain of bacteria. 4. An assay allows the scientist to analyze differences in test tubes after treating each differently with

more specific reactions. 5. Hershey and Chase were able to use the elemental structure of nucleic acids vs. proteins; that nucleic

acids contain phosphorus, while proteins contain sulfur to differentiate between the two compounds in vivo.

6. X-Ray diffraction enabled scientists to see DNA’s double helix structure. X-Ray diffraction involves shooting subatomic particles into a substance, and when they diffract off of a structure, a clear image is observed.

7. RNA – Uracil is used as a nitrogenous base in place of Thymine on a DNA molecule. RNA is single stranded and can therefore be folded into different shapes while DNA is double stranded and can only be double helix in shape.

8. Eukaryotic chromosomes are typically linear and larger than prokaryotic chromosomes. They can be found in scaffold arrangements. The DNA is wound around protein structures called histones. These beads on a string fold up making an arm called a chromatid. Prokaryotic cell have circular DNA that contains a significantly smaller amount of nucleotides. All the genes are arranged in a circle.

9. Avery found that the pneumococci that were encapsulated where more virulent than the non-encapsulated form.

10. If assays of virulent bacteria are treated with enzymes to kill protein, DNA, and RNA are mixed with non-virulent bacteria then transformation will occur in both assays that were treated with protein enzyme and RNA enzyme.

11. That DNA was the transforming agent. 12. A bacteriophage is a virus that infects a bacterium, it consists of a protein coat and DNA, it also

reproduces quickly and can easily be harvested. Hershey and Chase used radioactive P and S to track the differences in the protein and DNA, since only DNA contains P and only proteins contain S. The infected bacteria with the phage that had the radioactive S, and they infected another group of bacteria with the radioactive P. The results showed that the bacteria contained radioactive P entered the bacterium.

13. Chargaff’s rule suggest that the A pairs with T and G pairs with C. 14. DNA is double helix. The backbone or sides of the latter are alternating sugar- phosphate molecules.

The rungs of the latter are made up of the nitrogenous bases: Adenine, thymine, guanine and cytosine. Adenine and guanine are purines (double ring structures). Thymine and cytosine are pyrimidines (single ring structures). A-T always bonds together with 2 H bonds, likewise G-C are held together with 3 H bonds. The two strands run antiparellel.

15. Drawings should include all terms. 16. 3’ TAAGCGTACGGTCAAT 5’ 17. Each new DNA molecule contains half of the original strand and half of the new strand. 18. Each new nucleotide bonds in the 5’ to 3’ direction. Nucleotides are added to the –OH end of carbon 3

by a phosphodiester bond. 19. Helicase unzips the double strand by breaking the H bonds between nucleotide bases, topoisomerase

controls over winding or underwinding of the DNA. DNA polymerase follows behind helicase and copies the template as its being exposed.

20. Primase adds RNA nucleotides as a primer for DNA polymerase on the lagging strand. This happens a couple of nucleotides at a time. Ligase binds the fragments together.

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21. Primase adds primers to the lagging strand, without it you could not copy the lagging stand. 22. Heat the DNA to denature, separating the double helix. Cool the DNA and add primers and

polymerase to the mixture. Slightly increase the temperature to speed the rate of replication. Repeat the cycle until you have the desired amount of DNA.

23. DNA contains genetic information necessary to the development and functioning of all living organisms. Cell division occurs in order for an organism to live and grow. When this occurs, all of the cell’s genetic information from the DNA must be replicated and passed on to each of the daughter cells.

24. 5’ CGTATGCTATAAGGCC 3’ 25. Base pairing occurs between a pyrimidine and a purine because two purines would be too large and

two pyrimidines would be too small to reach each other. Adenine pairs with Thymine because there are two spots on each molecule to hydrogen bond. Guanine and Cytosine each have three spots to hydrogen bond. This base pairing forms the double helix structure which twists.

26. DNA polymerase repairs its own errors. When an incorrect nucleotide is recognized, polymerase reverses its direction by one base pair of DNA, excises the incorrect nucleotide and replaces it.

27. DNA Polymerase can only read in the 3’ to 5’ direction (leading strand). The new strand off of the lagging strand must be made in reverse, and this is why additional enzymes are used.

28. 1,073,741,824 29. These enzymes can withstand the high temperature it takes to denature the DNA strands. 30. All life uses the same 4 bases A, T, C, G in 3 letter combinations to code for amino acids is the

universal genetic code. Having 64 codons and only 20 amino acids provides room for errors that may occur at any point.

31. This is because the RNA makes a mirror image of the gene. Once RNA polymerase opens the replication bubble, corresponding RNA nucleotides will attach to the DNA nucleotides until the gene is transcribed.

32. Allolactose will bind to the repressor causing a change in its shape, making it fall off. Now RNA polymerase can bind to the promoter and transcribe the lac gene. When lactose is all broken down and no longer present, the repressor goes back on the promoter.

33. Chromatin inhibits the action of RNA polymerase. When DNA is packed in chromatin it is not accessible to RNA polymerase, therefore transcription cannot occur because the DNA is packed to tightly.

34. TACCCGCCTAATCAGTGCTTCCTTTTACCGCATGCCCGGATCGGAATT AUGGGCGGAUUAGUCACGAAGCAAAUAGGCCUACGGGCCUAGCCUUAA GAUGGGCGGAUUAGUCACGAAGGAAAUAGGCGUACGGGCCUAGCCUUAAAAAAAA AUGGGCGGAUUAGUCACGAAGGAAAUAGGCGUACGGGCCUAGCCUUAA mGAUGGGCGGAUUAGAAAUAGGCCCUUAAAAAAAA

35. The RNA transcript has regions that are noncoding called introns. The other regions are called exons because they are expressed.

36. The gene must be unpacked from chromatin. The right transcription factor must be present. Transcription occurs. Cap and Tail must be added to the mRNA. Pre-mRNA must be edited (spliced), Nuclear pores allow passage to the cytoplasm. mRNA comes in contact with a ribosome. Translation occurs. Protein is used within the cell or exported to the environment.

37. mRNA, tRNA an rRNA are all single stranded nucleic acids. mRNA carries the information as to which polypeptide is to be make. It travels from the DNA to the ribosome. tRNA carries the amino acids that correspond to the codons on the mRNA. rRNA makes up the ribosomal subunits.

38. Methionine is the first amino acid code for by mRNA, its codon is AUG. 39. Met - Gly – Gly – Leu – Glu – Ile – Gly – Pro 40. DNA makes RNA, which code for a protein. 41. This is a repressible operon that is usually on. It codes for genes that produce the enzyme tryptophan.

When tryptophan is present in the environment, it acts as a co-repressor and the trp operon is not used. 42. The promoter region is the attachment site for RNA polymerase, once attached the RNA polymerase

can unwind the DNA to create an initiation bubble. This bubble provides more space for the RNA

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polymerase to access the DNA molecule. Protein inhibitors may attach to or near the promoter region and block transcription.

43. Transcription factors regulate when RNA polymerase can bond to the region of DNA that is to be transcribed. The transcription factors can bind to either a promoter region or enhancer region on the DNA. Depending on the type of TF, the gene will either be regulated up or down.

44. The 5’ end receives a modified guanine cap and the 3’ end gets a poly-A-tail added. These modifications help mRNA get out of the nucleus . They also provide protection from hydrolytic enzymes in the cytoplasm. Finally, they help ribosomes attach to the mRNA.

45. Answers may vary. 46. The first step is initiation. The small ribosome subunit rRNA, mRNA and tRNA with Met aa. Next the

large subunit rRNA will join the small unit. Elongation – mRNA moves to the p-site in codon units (3 bases), At the p-site the tRNA corresponding anticodon and amino acid will bind to the matching codon. Next the amino acids are linked together by a peptide bond. Finally, termination is where the ribosome reaches the stop codon. The polypeptide is release and the ribosome breaks apart.

47. The P-site is the new protein will emerge, and the A-site is where the amino acids are delivered. The mRNA enters the A-site first.

48. There are a couple of hypotheses that can be accepted here. One case: there is a frameshift mutation, 3 codons provide more of an out. Another is that they could provide room for the evolution of future amino acids.

49. AAG AAT GGC CGG TTA AGA CCC UUC UUA CCG GCC AAU UCU GGG

UUC UUA GCC UCU and CCG AAU GGG Phe-Leu-Ala- Ser and Pro-Asn-Gly

50. 3’ CTTAGAGCAAAAAATCAA 5’ (Other possible strands are acceptable) 51. Recombinant DNA is when genes from one organism are spliced into the DNA of another organism.

Humulin was the first insulin produced by bacteria that contained human insulin genes. The bacteria manufactured human insulin.

52. 1. Choose the gene of interest. 2. Cut the gene of interest form the genome. 3. Isolate the gene of interest. 4. Make more of the gene of interest. 5. Paste the gene into the host’s DNA. 6. Collect the protein

53. Both a puppy and a jellyfish rely on the universal genetic code, RNA polymerase, and ribosomes to express genes.

54. A restriction enzyme is an enzyme that cuts the DNA at a specific nucleotide sequence. They recognize a specific sequence of nucleotides and make a cut across both strands of DNA.

55. Sticky ends are when the restriction enzymes cut the DNA in a stair step pattern. Usually this occurs when it cuts through an inverted palindrome. Blunt ends are straight cuts though both DNA strands. These are done by small restriction enzymes and done at a mirror-like palindrome site.

56. Gel electrophoresis separates the DNA fragments being worked based on length.

1. Question 1 a. The Griffith experiments showed that the rough bacteria can take a substance from its

environment, which the Avery experiments later proved to be a nucleic acid and not a protein, and acquired a trait or ability that it did not possess from an organism that was no longer living (the smooth bacteria). The Hershey and Chase experiments showed that virus injected their DNA into bacteria and not their protein coat, so DNA was sufficient to generate new viral particles within the bacteria and pass traits, the whole characteristics of a virus, to the next generation of virus.

b. The Griffith experiments did not isolate the different cellular products to pin point the causative agent of the transformation and the source of the new trait in the bacterium; therefore, the source of the new trait could be assumed to be the DNA and further experiments needed to carried out to be sure that DNA or nucleic acids were responsible for the observed phenomena.

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Learning Objectives: LO 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information. [See SP 6.5] LO 3.2 The student is able to justify the selection of data from historical investigations that support the claim that DNA is the source of heritable information. [See SP 4.1]

2. Question 2 a. The BrdU disrupts the chemical properties of the nitrogenous base stacking that occurs in the

inner core of the DNA molecule; if one strand contains BrdU, which is the case after one replication round because of the semi-conservative nature of DNA replication, the chemical properties are not disturb sufficiently enough to interfere with the staining of the chromosomes. But after the second round of DNA replication, one chromosome will have both strands of DNA with BrdU instead of thymine that will stain less dark than the other chromosome that will only have one strand with BrdU instead of thymine and stain darker.

b. There will be 2 possibilities. If the chromosomes segregate with one dark and light chromosome going to each new cell, then the staining pattern at the next metaphase will show sister chromatids with dark and light chromosomes and another sister chromatids with only light chromosomes. But if the chromosomes segregate with two dark chromatids going to one new cell and two light chromatids going to the other new cell, then the cell with 2 dark chromatids will have chromatids showing one dark and one light sister chromatids at metaphase- and the cell with 2 light chromatids will have both sister chromatids showing light staining patterns.

LO 3.3 The student is able to describe representations and models that illustrate how genetic information is copied for transmission between generations. [See SP 1.2] LO 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information. [See SP 6.5] LO 2.37 The student is able to connect concepts that describe mechanisms that regulate the timing and coordination of physiological events. [See SP 7.2]

3. Question 3 a. Without lactose to bind allosterically to the repressor protein on the lac operator, RNA

polymerase is blocked from attachment to the DNA sequence. b. The concentration of lactose is not high enough to activate transcription of the LacY gene.

Learning Objectives: LO 2.15 The student can justify a claim made about the effect(s) on a biological system at the molecular, physiological or organismal level when given a scenario in which one or more components within a negative regulatory system is altered. [See SP 6.1] LO 2.16 The student is able to connect how organisms use negative feedback to maintain their internal environments. [See SP 7.2] LO 2.17 The student is able to evaluate data that show the effect(s) of changes in concentrations of key molecules on negative feedback mechanisms. [See SP 5.3] LO 2.18 The student can make predictions about how organisms use negative feedback mechanisms to maintain their internal environments. [See SP 6.4] LO 2.19 The student is able to make predictions about how positive feedback mechanisms amplify activities and processes in organisms based on scientific theories and models. [See SP 6.4]

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LO 2.20 The student is able to justify that positive feedback mechanisms amplify responses in organisms. [See SP 6.1]

4. Question 4 a. The DNA molecule is a large biomolecule with the information to make proteins and to regulate

their activity; its integrity and ability to be replicated is of prime importance to the survival of cells. Evolution selected for the complex set of proteins that is required to open the double-stranded molecule, protect the single stranded DNA molecules, polymerases to generate the new strands, RNA primases to replicate the opposite strand, the ligases to join DNA fragments, and helicases to twist the DNA and relieve torsion in the large molecule. The slew of proteins are require to maintain the information in competent enough to produce functional proteins at the right time.

b. DNA needs to get copied to have an extra set of genetic information for a new cell when a cell divides. Sometimes DNA gets copied to amplify the protein production capacity of the cells or tissues the DNA resides in.

c. Both pathways are plausible. Pathway A requires the least number of steps, making the cell more efficient. Pathway B has more steps but the speed of the steps may be faster and may be ultimately more efficient.

Learning Objectives: LO 3.3 The student is able to describe representations and models that illustrate how genetic information is copied for transmission between generations. [See SP 1.2] LO 2.39 The student is able to justify scientific claims, using evidence, to describe how timing and coordination of behavioral events in organisms are regulated by several mechanisms. [See SP 6.1] LO 2.40 The student is able to connect concepts in and across domain(s) to predict how environmental factors affect responses to information and change behavior. [See SP 7.2]

5. Question 5 a. Any activity or behavior that shows fluctuations or changes during the 24 hour cycle is a

candidate activity to be controlled or influenced by the circadian clock. Some examples of circadian regulated systems are the foraging activity of animals (diurnal vs nocturnal), mitochondrial respiration, and cell division.

b. BMAL1 and CLOCK are assisted by P300 to help RNA polymerase transcribe many mRNAs that are involved in regulating the efficiency of the cell. By aligning the activity of these transcription factors with the time of day, the cell can carry cellular activities that it can do or coordinated more productively and/or with less energy during the day.

c. The cell must recognize that a change in the length of the day has occurred; this recognition most likely involves photoreceptors or other light-sensing cells that can relay the information to other cells via hormones. The circadian clock will try to adjust as much as evolutionarily possible to make the cellular activity more efficient. If the cell cannot make the necessary adjustments, the cell will become extinct.

Learning Objectives: LO 2.35 The student is able to design a plan for collecting data to support the scientific claim that the timing and coordination of physiological events involve regulation. [See SP 4.2] LO 2.36 The student is able to justify scientific claims with

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evidence to show how timing and coordination of physiological events involve regulation. [See SP 6.1] LO 2.40 The student is able to connect concepts in and across domain(s) to predict how environmental factors affect responses to information and change behavior. [See SP 7.2]