Review Projects – 2013 Big Idea 3 - answers Mr. Bennett

1. Review Projects – 2013 Big Idea 3 - answers Mr. Bennett

2. Answer Key--LO 3.1 M.C. Question: Which dyed-chemical would help recognize that DNA is the primary sources of heritable information, basing your prediction on the Hershey and Chase Experiment? A) Nitrogen B) Sulfur C) Phosphorus D) A and C E) A, B, and C 1) Compare and contrast the structures of DNA and RNA. 2) How are genes expressed? If there was a point mutation on the DNA, how would it be affected? • DNA is a double stranded molecule with a long chain of nucleotides of adenine, thymine, guanine and cytosine. It is self-replicating and uses deoxyribose sugar and a phosphate backbone. On the other hand, RNA is a single-stranded molecule and has a shorter chain of nucleotides, comprising of Adenine, uracil, guanine and cytosine. RNA uses ribose sugar, and a phosphate backbone and is more reactive because of the hydroxyl bonds. • Gene expressions follow from DNA to transcription which is made into RNA and then the RNA is translated into proteins which then stimulate other responses to express for that specific gene. If there was a point mutation, then there is a possibility that that gene would not be expressed correctly, because the point mutation in the DNA would result in a new RNA sequence and then could translate into other proteins that do not correspond with the gene. Therefore, genotypes and phenotypes would be altered due to the mutations in the genetic material. A mutation could also cause the production of a protein to stop sooner than normal because of a terminated anticodon.

3. Answer LO 3.2Multiple Choice Question:If bacteriophages infected cells by inserting their entire construction (proteins and DNA) into the target bacteria rather than just the molecule of inheritance, but only the molecule of inheritance transforms it, the Hershey Chase experiment: I: Would always produce a radioactive pellet II: Show radioactive molecules in all areas of the mixture after it is centrifuged III: Still would prove the same results as the molecule of inheritance IV: Have inconclusive results as to the molecule of inheritanceA) IB) I & IIIC) II & IIID) I & IVE) II & IV Free Response Answer:Both DNA and proteins incorporate carbon into their structure. Because of this, if phages are grown in a mixture with radioactive carbon, both the protein coat and the DNA would be radioactive. In terms of the experiment, DNA is the molecule of inheritance so the radioactive DNA would get inserted into the cell and make the pellet radioactive. Additionally, the lighter radioactive protein coat is not inserted into the cell so after centrifugation it remains suspended in the supernatant, causing it to be radioactive as well. Thus, the entire mixture is radioactive in this trial. These results suggest that one molecule, proteins or DNA is the molecule has to be the molecule of inheritance, because if neither are, the transformed cells in the pellet would not be radioactive, and if both are then the supernatant would not contain radioactive particles. But, these results do not decisively point to one or the other. Because they are both tagged within the same phage, one piece transforms the cells while the other stays suspended, giving off radioactivity in all parts of the mixture without determining which piece is where.

4. ANSWER KEY– LO 3.3 Which of the following stages of Meiosis is NOT responsible for a source of genetic variation? A) Prophase I B) Metaphase I C) Anaphase I D) Fertilization E) None of the Above Identify TWO of the processes in sexual reproduction that result in genetic variation and explain how each contributes to this. Crossing over, which occurs in Prophase I, is a vital source of genetic variation in Meiosis. Homologous chromosomes form pairs with each other to form tetrads. The non-sister chromatids then break at the same location and swap DNA, rejoining after. This creates genetic variation by generating hybrid chromatids which are unique from the rest in the cell as well as those of other cells. The process of fertilization also accounts for genetic variation. Since fertilization is usually random by nature, it also further helps to generate genetic variation within organisms. Each male and female gamete is unique, and since each gamete can represent around 8 million different chromosome combinations, a single fertilization could create upwards of 64 million unique zygotes.

5. ANSWER KEY– LO 3.4 M.C. Question: Why is the third anticodon usually irrelevant in the process of translation? A) All anti codons have the same third anticodon B) The proofreading mechanism of eukaryotic cells uses the third anticodon as an reference point for checking the rest of the polypeptide C) The anticodon sits on a kink in the tRNA, therefore the third anticodon doesn’t usually make complete contact D) Because tRNA carries 2 anticodons, not three. Ala- GCU, GCC, GCA, GCG Arg-CGU, CGC, CGA, CGG, AGA, AGG Asn- AAU,AAC Asp- GAU, GAC Cys-UGU, UGC, Gln- CAA,CAG Glu-GAA,GAG Gly-GGU, GGC, GGA, GGG His- CAU, CAC Ile-AUU, AUC,AUA Leu-CUU,CUC,CUA,CUG,UUG,UUA Lys-AAA, AAG Phe-UU,UUC Pro- CCU,CCC,CCA,CCG Ser-AGU,AGC,UCU,UCC,UCA,UCG Thr-ACU,ACC,ACA,ACG Trp-UGG Tyr-UAU, UAC Val-GUU,GUC,GUA,GUG Start- AUG Stop-UGA, UAG, UAA Learning Log/FRQ-style Question: 3’- …ATGTCTCGTGTTCTCTGA…-5’ is a strand of DNA whose complimentary strand has been transcribed into mRNA and then translated into a polypeptide chain. A) What is the complimentary strand? B) What is the mRNA transcription? C) What is the polypeptide chain? Use the table provided (on the next page.) D)What would happen to the polypeptide if there were a point mutation that changed the given DNA strand to 5’-… ATGTCTCGTGTTCTCTTA…-3’? A) The complimentary chain of DNA would be 3’- TACAGAGCACAAGAGACT-5’ Remember DNA is antiparallel and that T matches up with A and that C matches up with G. B) the mRNA would read 5’-AUGUCUCGUUCUCUGA-3’ all Ts turn to Us in RNA C) the polypeptide chain would be Met-Ser-Arg-Val-Leu-STOP, since mRNA is transcribed 5’ to 3’. D) the ribosome would keep translating the mRNA into the next coding region, since the stop codon would be interrupted, the polypeptide would be elongated past its necessary length and therefore the polypeptide would also fail to function.

6. LO 3.5 Answers MC Question LL/FRQ-style Question A) The scientists could find out what gene is responsible for the swelling of the worm by isolating genes that relate and respond to heat signals in the organism. They could then use plasmids to produce different samples recombinant DNA. Once this new DNA is inside of a bacterial cell, they can expose the different cells to heat and observe which ones swell and which don’t The use of cDNA microarry could also determine which gene is functioning in this response as it matches samples of known and unknown DNA based on base pairing rules. cDNA is a fluorescent probe that is used to observe the intensity of complementary base pairings around specific genes.. The more fluorescent cDNA is seen around a gene site, the more base paring activity is occurring and therefore, they are complimentary genes. B) First the gene of the donor DNA must be isolated. Using restriction enzymes to cut the DNA and the plasmid, ligase to seal them together and then placing it inside of a bacterial cell. The resulting bacteria must be tested so the appropriate strain can be selected. The selected recombinant DNA will then be inserted into a bacterial cell and natural replication processes will follow. • B- PCR can produce billions of copies of DNA from even a microgram of the sample. There is no point at which the enzymes involved in the chain reaction become denatured and no longer function, so far more than 1,000 copies of the DNA can be produced.

7. LO 3.6 Answer Key Multiple Choice: The best answer is D, because the substitution of U for C changes the second codon to a stop codon, terminating transcription early Free Response: a. Identification of both enzymes (DNA and RNA polymerase) (2 points) Similarities (1 point): -3’ to 5’ direction -adds base pairs to a phosphate backbone -Exhibits quaternary protein structure b. The student should describe RNA Polymerase’s inability to proofread, and how this can lead to changes in base pairs during the process of transciption from DNA to mRNA. (2 points) c. Adequate descriptions of effect on gene expression should include mention of how changes in base pairs which will alter the amino acid sequence and ultimately the protein, which is what will alter gene expression (2 points). Mutations (one point each) Frameshift Mutations: -base pair deletions -base pair insertions Differences (1 point): -RNA polymerase doesn’t need primers -DNA polymerase does need primers -RNA polymerase does not proofread -DNA polymerase does proofread -DNA polymerase needs helicase to unzip genetic material -RNA polymerase builds code with uracil -DNA polymerase builds code with thymine -RNA polymerase uses ribose nucleic acids -DNA polymerase uses deoxyribose nucleic acids Point Mutations: -base pair substitution

8. ANSWER KEY L.O. 3.7 • M.C. Question: Embryonic cells in humans cycle rapidly. This rapid occurrence of cell cycling is controlled by cyclin concentration and maturation promoting factor (MPF) activity. What is the effect of cyclin concentration on MPF activity? • As cyclin concentration increases, MPF switches itself off and destroys its own cyclin. • As cyclin concentration increases, cyclins associate with Cdk molecules, the MPF complex initiates mitosis. • As cyclin concentration decreases, the MPF will cause phosphorylation of various proteins of the nuclear lamina. • As cyclin concentration decreases, enough molecules of MPF are produced to pass the G2 phase checkpoint. • Cyclin concentration has no effect on MPF activity. Explanation: As cyclin concentration increases, cyclins associate with Cdk (Cyclin dependent kinase) molecules which results in a MPF complex that phosphorylates a variety of proteins of the nuclear lamina which promotes the fragmentation of the nuclear envelope during prometaphase of mitosis. As cyclin molecules accumulate and combine with Cdk molecules, enough MPF has then been produced to for the cycle to pass the G2 checkpoint, initiating mitosis. Learning Log/FRQ-Styled Question: A fern and a bird seem to be showing signs of similar types of disease. In order to get a better idea of how quickly the disease is multiplying, biologists observe the cell cycle of both the fern and the bird. They do this by observing how often cytokinesis, the final division of a cell into two daughter cells, occurs. The figure to the right shows two types of cells undergoing cytokinesis. Identify which cell is an animal cell and which cell is a plant cell. How do you know this? Compare and contrast cytokinesis in animal cells and plant cells. Answer: Cell A is a plant cell and cell B is an animal cell. You know this because Cell B is beginning to separate (cleavage furrow), clearly indicating that cleavage is about to occur. Cleavage only occurs in animal cells. In animal cells, cytokinesis occurs by a process called cleavage. The first step of cleavage is the formation of a cleavage furrow, or a shallow groove in the cell surface near the former metaphase plate. The furrow is a contractile ring of actin microfilaments which interact with myosin molecules, causing the ring to contract which causes the cleavage furrow to deepen and eventually pinch the parent cell in two. Unlike in cytokinesis in animal cells, in plant cells there is no cleavage furrow since plants have a cell wall. Instead of cleavage, Golgi apparatus-derived vesicles move along microtubules to the middle of the cell where they come together to produce a cell plate. Materials to make a cell wall are carried on the vesicles and collect in the cell plate as it grows. The cell plate enlarges until it’s surrounding membrane reaches the plasma membrane on the perimeter of the cell, then they fuse. This results in two daughter cells with their own plasma membranes; the cell wall forms from the materials in the cell plate. Cytokinesis in both plant and animal cells produces two daughter cells, but the main difference is the process by which this occurs (cleavage in animal cells, and cell plate/plasma membrane fusion in plant cells).

9. ANSWER KEY– LO 3.8 A culture of mammalian cells normally grows in one even layer on a substratum. The cells receive DNA from a virus that prevents density-dependent inhibition, but the cells do not grow and divide out of control. This is because the cells began apoptosis in response to the virus exhibit anchorage dependence produce MPF in response to the new DNA have undergone metastasis have too much cyclin Suppose give a tissue sample a drug that prevents degradation of cyclin. Explain why the cell grows and divides out of control. What would cause this in a natural environment? Explain the roll of metastasis in determining how much of a threat a tumor poses to the life of an organism. Why? Cyclin is the molecule which binds to Cdk and MPF to trigger the sub phases of the cell cycle including mitosis. Each sub phase is dependent upon a specific concentration of cyclin. If cyclin is not degraded, the cell will enter G1, S, G2, and M without regulation. Transformation would cause this phenotype in nature. Transformed tissue (cancerous tissue) that has undergone metastasis extremely dangerous because it is not contained by the body’s immune system and may proliferate into more tumors throughout the body. This can cause impaired function of multiple organs and organ systems instead of just one.

10. Key Maternal Possibility 1 Possibility 2 Paternal Two equally probable arrangements of chromosomes at metaphase I Metaphase II Daughter cells Combination 1 Combination 2 Combination 3 Combination 4 FRQ Question: Explain three ways in which meiosis contributes to genetic diversity within a species. • Independent assortment contributes to genetic variation because the way that homologous chromosomes line up at the metaphase plate is completely random. Each pair of homologous chromosomes is independent from the others in the way it is positioned on the metaphase plate, so each gamete is genetically different from each other. In humans, this makes it so that 223 combinations of chromosomes are possible in the resulting haploid gametes. • Crossing over is the process in which homologous chromosomes switch genes at a certain location. This creates recombinant chromosomes which are hybrids of both maternal and paternal chromosomes. Where crossing over occurs is completely random, and it happens approximately 2-3 times per chromosome. This creates more combinations that can occur in addition to the 223combinations caused by independent assortment. • Random fertilization is the product of independent assortment. Since there are 223 possible combinations of chromosomes caused by independent assortment, and two gametes, the amount of possible combinations of gametes is 223x 223. Since 223 = 8 million, and 8 million x 8 million = 64 trillion, there are 64 trillion unique combinations of chromosomes that the zygote can have. In adittion to this, crossing over creates recombinant chromosomes, so there are many more possible combinations than 64 trillion. LO 3.9 Multiple Choice Question: In what order do these steps occur ? 1. Sister chromatids separate 2. Homologous chromosomes separate 3. DNA replicates 4. Crossing over occurs 5. Tetrads are positioned at the metaphase plate A. 3, 5, 4, 2, 1 B. 3, 5, 4, 1, 2 C. 3, 1, 4, 2, 5 D. 3, 4, 5, 2, 1 E. 3, 4, 5, 1, 2

11. Answer Key 3.10 • What is the importance of crossing over? • It provide extra genetic materials for daughter cells. • It increases the likelihood that daughter cells contain different genetic material. • It produces the protein that associated with DNA in chromosomes. • It separate homologous chromosomes. • Learning Log Explanation. • In meiosis, chromosomes replicate (homologous) during meiosis I. Crossing over occur, non-sister chromatids swap their genetic information with each other and result in recombinant chromosomes. Some genes could get pass from the parental gene to the haploids. These genes could be favored by the environment, and by passing on this ensure the next generation’s ability to survive and reproduce. If crossing over does not occur then there would not be a genetic switch between non-sister chromatids; therefore, there will be a variation of species. Variation causes biological diversity. Without crossing over, all organisms would be the same forever with no change of DNA nor mutation. And there would not be evolution, and eventually the species would all die.

12. LO 3.12 Answers M.C. Question: A) Spindle fibers are made of microtubules, which extend from the chromosomes and attach to the outside of the cell. Learning Log: The idea of inheritable variation is what drives natural selection which is the basis of Evolution. Inheritable variation increases the chance that a trait which is most beneficial for reproductive success will be produced. Then adaptation will occur after the most beneficial trait is accumulated within a population. So in the face of a changing environment, that variation increases the likelihood that some individuals in a population will have inheritable variations that help them cope with the new environment.

13. LO 3.13 M.C. Question: Nondisjunction in sexual reproduction is caused by all of the following EXCEPT: A. When a pair of homologous chromosomes fail to move apart properly B. When an error in meiosis II occurs C. When an inactive Bar Body in a zygote becomes active, taking the place of the Y chromosome D. When one gamete receives two of the same type of chromosome and another receives no copy E. Both A and C Genetic disorders bring up many controversial topics. Explain why Sickle Cell Anemia is a genetic disorder and what are the symptoms of the disorder. What reproductive issues does this genetic disorder cause? What social, ethical, and medical issues does Sickle Cell Anemia pose? Sickle Cell Anemia is a genetic disorder because it is a gene that is inherited from the parents of the genetically disabled person. In order for the disorder to occur, both parents have to be a carrier of the disease and pass down the gene to their child. Because this disease affect the RBCs of the cells, it causes the RBCs to become sickle-shaped and unable to carry oxygen which creates the following symptoms in the affected: pain, physical weakness, pain, organ damage and paralysis. This in turn makes causes reproduction issues because a person with the disorder is more likely to have a child with the disorder. From this, arises ethical issues because a mother with sickle cell Anemia, may become pregnant and decide she doesn’t want to have a child with her which causes her to abort the fetus. This would create ethical issues for people who are anti-abortion and pro-abortion. The social issues that arise are that the mother may decide to create a “designer baby” that doesn’t have any health issues, which brings about the ethical question; “Only God can create life?” The Medical issues that come along, if the mother were to have the child; the child may need blood transfusions, pain medication, and other medical expenses that the family can’t provide.

14. ANSWER KEY L.O. 3.14M.C. Question: Huntington’s disease is caused by a dominant allele. If one of your parents is heterozygous (Dd) and has the disease and the other is homozygous recessive (dd), what is the probability that you will not have the disease?a) 1b) 3/4c) 1/2d) 1/4e) 0Answer: c) 1/2Dd x dd (parents)probability of child’s genotype being dd = 1/2 x 1 = 1/2Learning Log/FRQ-Style Question: Two characteristics that Mendel studied were stem length (tall=T, short=t) and seed shape (R=round, r=wrinkled). If two plants that are heterozygous for both traits (TtRr x TtRr), what proportion of the offspring would be expected in the following situations:a) homozygous dominant for both traits TTRR = 1/4 x 1/4 = 1/16b) homozygous recessive for both traits ttrr = 1/4 x 1/4 = 1/16c) heterozygous for both traitsTtRr = 1/2 x 1/2 = 1/4d) homozygous dominant for stem length and heterozygous for seed shapeTTRr = 1/4 x 1/2 = 1/8

15. LO 3.15 Answers • Multiple Choice: ¼ * ¼ = 1/16. A • FRQ: A cows pelt can either be black, white, or a mixture of both. Rather than one color taking dominance of the other, both genes are expressed in a heterozygote cow. B B This cross will bring rise to ¾ of the time black cows and ¼ of the time mixed cows. W WB BB BB B BB

16. ANSWER KEY LO- 3.16 In which of the following situations would Mendel’s laws accurately predict offspring genotypes and phenotypes? A) The determining of blood type by multiple alleles B) Flower color based on dominance between two alleles at one locus C) A plant whose flower color is influenced by soil acidity D) A rat’s fur color derived from two genes Polygenetic Inheritance Mendel’s ideas are correct in theory, but not universal. Upon discovering a new species of plant in which flower color varied, how could it be determined if the trait could be accounted for by Mendelian genetics or if different patterns applied? Before determining if genetics could determine phenotype, first comparing the plants environments and situations could help to decide if other factors influenced the plants. Next, a count of flower colors could show complete dominance if there were only two possibilities, and incomplete if there were three or more. Mendelian genetics can easily account for complete dominance in a test cross. Performing test crosses and checking for expected offspring ratios could show that the trait was accounted for by two alleles on one gene, or if it is determined by multiple genes or is sex-linked.

17. ANSWER KEY --- L.O. 3.17 Multiple Choice Question: Using the figure to the right, which of the following methods of inheritance are being expressed in the phenotypes by the set of alleles in the punnett square. Incomplete dominance Codominance Complete dominance Recessive trait • FRQ-Style Question: • Many phenotypes cannot be explained by Mendel’s Laws of inheritance. There are several phenotypic traits that are established based on other rules • Explain the pattern of inheritance that allows for there to be people born with blood type AB and list four parent blood type combinations that could result in an offspring with the AB blood type. • Colorblindness is a genetic disorder commonly found among men. It is rare to find a color blind woman. • Explain the genetic basis of how the gene causing colorblindness is transmitted . • Explain why mostly men are colorblind and why it is rare to find colorblind women. • Codominance allows for people to be born with type AB blood type because both genes are equally expressed which results in a person with type AB blood. To produce offspring with type AB blood the parent combinations could be IAIA×IBIB, IAIA×IAIB, IBIB×IAIB, IAIB×IAIB, IAIB×ii • b) i. The gene that causes colorblindness is a sex-linked recessive gene which means that it is found only the X chromosome. Sex-linked traits can be passed on through reproduction and it can only exhibit itself when there is two X chromosomes with the recessive allele for females or when the X chromosome for a male carries the recessive allele. • ii. Most men are colorblind because males only have one X chromosome in contrast to females who have two X chromosomes. This makes men more susceptible to colorblindness because if their X chromosome receives the recessive allele then the Y chromosome cannot compensate since the gene for that trait is not present on the X chromosome, only the Y chromosome. Women normally do not have colorblindness because even though women have two X chromosomes it is no common that both contain the recessive allele. Typically women are carriers of the allele that causes colorblindness and they have a dominant allele that causes the colorblindness to go unexpressed

18. LO 3.18 Answers M.C Question: Both unicellular and multicellular organisms must turn genes on and off continually in response to the signals given off from their internal and external environments, during the development of a multicellular organism which process must the organism go through so that each cell recieves its propper characteristic and function? Transcriptional Regulation Differentiation Cell Communication Variation FRQ Answer: Any step of gene expression may be controlled, from the DNA-RNA transcription step to post-translational modification of a protein. Gene expression is regulated through changes in the number and type of interactions between molecules that collectively influence transcription of DNA and translation of RNA. Transcription determines the flow of genetic DNA to mRNA therefore without transcription the genetic DNA wouldn’t be translated into proteins needed so that each gene can be expressed differently in organisms that are phenotypically different.

19. Signal NUCLEUS Chromatin Chromatin modification: DNA unpacking involving histone acetylation and DNA demethlation DNA Gene available for transcription Gene Transcription Exon RNA Primary transcript Intron RNA processing Tail mRNA in nucleus Cap Transport to cytoplasm CYTOPLASM mRNA in cytoplasm Degradation of mRNA Translation Polypetide Cleavage Chemical modification Transport to cellular destination Active protein Degradation of protein Degraded protein ANSWER KEY– LO 3.19 • M.C. Question: In eukaryotes, many genes may have to interact with each other, requiring more interacting elements than can fit around a single promoter. This physical limitation is overcome by: • alternating promoters and operators • placing promoters on both sides of each gene • the use of very long promoters • distant sites in a chromosome controlling transcription of a gene • E) having factors on one chromosome control genes on another gene (sample drawing) Learning Log/FRQ-style Question : What are the differences between gene expression that can be regulated in eukaryotic cells and in prokaryotic cells? Why? What is a control point in gene expression? How can gene expression be applied to a population? Draw and label this hypothetical situation in a eukaryotic cell. Gene expression in eukaryotes have the nuclear envelope that separates transcription from translation which offers an opportunity for post-transcriptional control in the form of RNA processing that is absent in prokaryotes. Eukaryotes have a greater variety of control mechanisms operating before transcription and after translation. A control point in gene expression is a point where gene expression can be turned on or off, accelerated or slowed down. This can apply to a population where organisms with similar genes have a phenotypic difference due to gene regulation and the expressed gene that causes that difference.

20. LO 3.20 M.C. Question: Which of the following is false of gene expression? A. The poly-A tail helps ribosomes attach to the 5' end of the mRNA B. The TATA box is a promoter DNA sequence crucial in forming the transcription initiation complex C. tRNA molecules transfer rRNA codon into a particular amino acid D. The enzyme aminoacyl-tRNAsynthetase joins each amino acid to the tRNA E. Template strand provides the template for ordering the sequence of nucleotides in an RNA transcript Learning Log/FRQ-style Question: Explain the elongation cycle of translation. What is the role of the hydrolysis of GTP in this process? The anticodon of the aminoacyltRNA base pairs with complementary mRNA codon at the A site. Then an rRNA molecule catalyzes the formation of a peptide bond between the new amino acid that is in the A site and the carboxyl end in the P site which has the growing polypeptide. This causes the polypeptide to attach to the tRNA at the A site. The ribosome moves the tRNA in the A site to the P site. The tRNA that is empty gets moved to the the E site and it is released. mRNA moves with the tRNA while bringing the next codon to be translated into the A site. Hydrolysis of GTP increases the efficiency and accuracy it acts like an allosteric effector.

21. LO 3.21 Answers Multiple Choice Answer: Letter B is the correct answer because prokaryotic cells rely on small-protein molecule binding for gene regulation. Enhancers are common in eukaryotes along with the lack of an operon. Eukaryotic genes escape silencing (genes being turned off permanently) by the histone code which modifies the positively charged amino acids. Methylation is used to regulate a gene by modifying the DNA strand after it has been replicated with adding a methyl group to any cytosine molecule that comes right before guannine; and methylation of cytosines can cause gene suppression. Acetylation is the adding of an acetyl group into an organic molecule which can cause gene supression as well. Barr bodies are the product of X inactivation of the X chromosome which is done through suppression of genes for this specific chromosome. Learning Log/FRQ Answer: Cells come to synthesize different molecules because specific cells produce specific signals. Then, these signals are released and bind to a signal receptor that fits the signal’s shape. This then starts the signal transduction pathway to regulate gene expression. Gene regulation’s effectiveness overall is controlled by the cell’s ability to express or suppress a specific gene based on the information passed through the cell to the enzymes who are making the proteins that will either suppress or express a gene. The synthesized molecules will signal to the enzyme what its action needs to be in relation to the specific gene at times when it is needed and then times when it is unnecessary.

22. Answer Key LO 3.22 M.C. Question: All of the following correctly describe the role of protein kinase, except? • Protein kinases regulate cell reproduction and growth. • Protein kinases transfers a phosphate group from ATP to a protein. • Protein kinase plays the largest role in the transduction step of cell communication • Protein kinasesdephosphorylate another protein. FRQ: Second messengers are small, nonprotein, water-soluble molecules or ions. They can readily spread throughout the cell by diffusion. Some examples of second messengers are cyclic AMP (cAMP), calcium ions, IP3, and DAG. The role of the second messengers are to rely signals received at the receptor to target molecules inside the cell– in either the cytoplasm or the nucleus. For example, calcium ions are released from the endoplasmic reticulum by two other 2nd messengers, IP3 and DAG, which open calcium channels.

23. ANSWER KEY– LO 3.23 Transcription of the structural genes in an inducible operon A) Always happens. B) begins when the pathway's substrate is present. C) starts when the pathway's product is present. D) stops when the pathway's product is present. E) doesn’t happen at all. Describe how gene regulation occurs in bacteria cells through the operon theory. The operon is a unit of DNA that controls the transcription of a gene. Its components are the promoter region- where polymerase attaches, the operator region- where RNA polymerase can be blocked if occupied by a repressor protein, the structural genes- that contain DNA sequences that code for enzymes that direct the production of a product and regulatory genes that lie outside the operon region and produce repressor proteins. They also produce activator proteins that assist the attachment of RNA polymerase to the promoter region. Two kinds of operons in bacterium are the lac and trp operons. The lac controls the breakdown of lactose. A regulatory gene produces a repressor that binds to the operator region. When in bind, RNA polymerase can not transcribe structural genes that code for enzymes that breakdown lactose. When lactose is available, some of it combines with the repressor to make it inactive. RNA polymerase can then transcribe the genes to break down lactose. The trp operon produces enzymes for the synthesis of tryptophan. A regulatory gene produces a repressor that does not bind to the operator. So RNA polymerase proceeds to transcribe the genes to synthesize tryptophan. When tryptophan is avaliable from the environment, the bacterium no longer makes it on its own. The rising levels of tryptophan induce some of it to react with the inactive repressor which in turn prevents the transcription of the structural genes. Figure 18.21 (campbell reece 7) Figure 18.21 (campbell reece 7)

24. Once the genotype is changed it leads to a different phenotypic expression and that specific organism will be different. Natural selection will either select for it or against it. The organism could have a beneficial expression and survive to reproduce and contribute the mutated gene to the gene pool and thus leading to other favored phenotypic expressions. Cryptic Coloration: The color of the organism allows it to blend in with its environment and evade predators. A change in genotype led to the phenotypic expression of the camouflage which allows the organism to survive to reproduce. Batesian Mimicry: An organism mimics or looks similar to an organism that can harm or poison a predator. The organisms that have the mutated gene that allows them to resemble another dangerous species fools predators and keeps them alive, those without the gene will die and be less prominent.Mechanical Plant Defenses: Thorns, hooks, or spiny leaves on a plant that serve as defense to predators. The mutation expressed weird and sharp growths which momentarily injure and/or deter predators from eating, or stepping on such plant. Answer Key: LO 3.24 During the evolutionary process, which of the following is the correct sequence of events? A.)Change in phenotype—change in genotype—speciation—selection B.)Speciation—change in genotype—selection—change in phenotype C.) Speciation—selection—change in phenotype—change in genotype D.) Change in genotype—change in phenotype—selection—speciation E.)Selection—speciation—change in phenotype—change in genotype

25. Answer LO 3.25 Multiple Choice Answer: A- Arg Ser Stop Learning Log/FRQ: • Frameshift: • By deleting the “G” In the sequence, all bases following that are shifted back one place. This is called a frameshift mutation. This mutation changes all the codons in the sequence following the mutation, therefore changing the amino acid sequence, and the polypeptide itself. Both insertion and deletion are examples of frameshift mutations. The earlier in the sequence the mutation occurs, the more alterred the protein is. b) Point: By substituting another nucleotide for “G”, it only changes one codon. This is a point mutation; the mutation only manipulates one nucleotide. This only changes one codon, therefore only changing one amino acid. Point mutations effect the outcome much less than the frameshift mutations .

26. ANSWER KEY– LO 3.26 Errors in which of the following processes can lead to an altered phenotype? • A) Transcription • B) mRNA processing • C) Translation • D) All of the above Name and describe 3 mechanisms that can introduce genetic variation in a population of bacteria. Conjugation: The transfer of genetic material between bacterial cells by direct cell-to-cell contact or by a bridge-like connection between two cells. Transduction: the process by which DNA is transferred from one bacterium to another by a virus. Transformation: The genetic alteration of a cell resulting from the direct uptake, incorporation and expression of exogenous genetic material (exogenous DNA) from its surroundings and taken up through the cell membrane(s)

27. Answer LO 3.27M.C. Answer: The multiple choice question answer is C. In general genetic variations in both prokaryotes and eukaryotes are completely different. In bacteria the genes are haploid, while in eukaryotes they are diploid. Also, bacteria have plasmids, while eukaryotes just have linear chromosomes. Crossing over takes place in both bacteria and eukaryotes; however, crossing over in bacteria is different than in eukaryotes. In bacteria crossing over involves a chromosome segment entering the cell and aligning with its homologous segment on the bacterial chromosome. The two break at corresponding point, switch fragments and rejoin, but crossing over in bacteria does not occur as often as in eukaryotes . Bacteria does asexual recombination, while eukaryotes do sexual recombination. Since C is correct answer E is automatically incorrect. FRQ Answer: a. 1. Crossing over occurs during prophase I in meiosis at the chiasmata between two homologous chromosomes. During this process the two chromosomes attach at the chiasmata and genes cross over, or switch chromosomes. This increases genetic variation because crossing over can occur multiple times per chromosome, so that the given genes for each chromosome are always different. 2. Independent assortment is the independent alignment and separation of chromosomes during metaphase I. Then in metaphase II this process occurs again, which can lead to 4 different genetic combinations. This increases genetic variation because each alignment and separation leads to different genes in each daughter cell. This process also increases genetic variation because there are 223 combinations in humans (2^23 because we have 23 chromosomes) or 8 million sperm and 8 million eggs. 3. Random fertilization is the probability that any one sperm will fertilize any particular egg. This randomized fertilization increases genetic variation because it contributes to evolution and gives over 64 billion different combinations (8 million sperm X 8 million eggs). b. No two people can be exactly alike because random fertilization states that since there are 223 combinations in humans (2^23 because we have 23 chromosomes) or 8 million sperm and 8 million eggs; therefore, there are roughly 64 trillion unique combinations and there are only about 7 billion living on earth. In addition, when genes swap chromosomes through crossing over it helps to mix up the genes on each chromosome, so that no chromosomes have the same genes; therefore, no two people can be exactly alike. Not only that, but the independent alignment and separation of chromosomes through independent assortment helps the mix up the genes that will be distributed to a daughter cell; therefore, there are many different combinations and no two people can look exactly alike.

28. LO 3.28 Answers M.C. Question: Which of the following statements concerning about independent assortment is true? E) B and C are correct. Learning Log/FRQ-style Question: Order independent assortment, crossing over and random fertilization in sequential order. Explain each one thoroughly by telling how it contributes to genetic variation within a population. First, random fertilization occurs during sexual reproduction. It is the random combination of chromosomes due to the different 8.4 million possibilities of sperm and 8.4 million different possibilities of the egg. This will allow the zygote to have any of 70 trillion diploid combinations. Then, crossing over occurs in meiosis I during prophase I. Crossing over is the exchange of genes between homologous chromosomes. This aides in genetic variation because it combines DNA inherited from two parents into a single chromosome. The last one is independent assortment. This happens in meiosis I of metaphase I. Independent assortment is the random assortment of chromosomes during gamete production; each chromosome is chosen random. It contributes to genetic variation because it helps maternal and paternal sets of chromosomes to come together to create more variation.

29. LO 3.29 Answers • Viruses may alter the genetics of host organisms through all of the following EXCEPT: • A) the lysogenic cycle of viral reproduction • B) the lytic cycle of viral production • C) Transduction of DNA from bacterium to bacterium • D) Provirus incorporation of viral DNA (synthesized from RNA) into the host cell’s DNA • E) None of the above • Why are viruses the primary agent for gene therapy, and how is this possible? Use an example of an important experiment to support your answer. Additionally, explain why the location of this practice in reference to the body is important. In gene therapy, a retrovirus that has been rendered harmless has a correct or desired gene inserted into the capsid, and the virus is allowed to infect bone marrow cells that have been extracted from the patient and cultured in a lab. Viral DNA then inserts into the bone marrow’s genetic information. Since these cells divide throughout an individual’s lifetime and contain stem cells that give rise to all the cells in the blood and immune system, bone marrow cells are the ideal cells to use. This technique was first tested in 1990 as part of a gene therapy trial for SCID (Severe Combined Immunodeficiency), and then again in 2000. Unfortunately, some patients developed leukemia in response to the retrovirus activity, causing retrovirus therapy to be suspended for further research.

30. Multiple Choice: The flu is wide spread and you had it and gave it to your best friend. You have the memory cells for when you recover. So why did you get the flu again in a year? A) The memory cells have died and so the flu can attack the immune system again. B) A new strain has been formed through mutation because of no proof reading with reverse transcriptase. C) A new strain has been formed through mutation because of DNA polymerase with no proof reading. D) When your body came in contact with the flu again, the flu cells produced faster than the memory cells could replicate. ANSWER KEY– LO 3.30 Free Response: a) Describe two types of mutations that can take place in a viruses genetic code. b) If a scientist discovers a substance that affects the HIV infected cell shown to the right and gives it the ability to proofread, how would that affect the genetic variation of the HIV infected cells? Why? • Mutations can happen with base substitution and gene rearrangement. Base substitution, or point mutation, is when one base is substituted for another. It can cause a nonsense mutation, a missense mutation, and a silent mutation. Gene rearrangement happens through four major categories. Deletion, with a loss in a single base pair or a loss of a larger portion of RNA. Duplications, where an extra copy of genes usually caused by unequal crossing over. Inversions, where the gene is reversed. Last there are translocations, when a portion of two different chromosomes breaks and rejoins in a different way. • This would decrease the amount of genetic variation in the HIV population. Many mutations occur during reverse transcriptase, which transcribes an RNA template into DNA during virus replication. Reverse transcriptase does not proofread when it translates the DNA. So without the proofreading many mistakes can be made and mutations occur easily in the HIV strand, because the DNA can easily be copied wrong and the mistakes are not found. So if a substance allows it proofread, then mistakes are more likely to be found and corrected on the strands. This will cause less mutations in the HIV viruses and their strands, and less genetic variation.

31. ANSWER KEY– LO 3.31 • A scientist is conducting a field study on the effects of a new kind of soil on Alternantheraphiloxeroides, commonly known as Alligator weed. The soil is supposed to amplify the effects of auxin, in a beneficial way, to the Alligator weed. Which of the following are not related to auxin? • Polar Transport • Apical Meristem • Proton Pumps • Promotes germination of seeds • Promotes double fertilization i) Plants and animals both evolved from single cell organisms, such as archaebacteria and eubacteria. Describe the difference between archaebacteria and eubacteria. ii) With evolution came cell communication, but even as time has passed the molecular processes of cell communication has remained the same in both plants and animals. Describe how plants and animals carry out cell communication. iii) The figure to the right, Figure 39.4, shows the process of de-etiolation in plants. A harmful pesticide that binds to the second messenger is introduced to the plant. What effect would it have on the de-etiolation process. • Archaebacteria are a kingdom in the Archaea domain, whereas Eubacteria is a kingdom in the Bacteria domain. Archaebacteria are known as the most primitive bacteria on earth and are simple single celled prokaryotes. They can live in extreme environments such as hot springs. Eubacteria are more complex than archaebacertia. They have complex DNA subunit patterns making them closer to eukaryotes. Eubacteria, being more complex, contain introns and archaebacteria do not. Introns are sections of nucleotide sequences that are removed by RNA splicing. Eubacteria are found in more neutral environments; therefore, eubacteria are present in greater numbers on earth than archaebacteria. Because of this, eubacteria have been studied more by humans. • Cell communication has three steps: reception, transduction, and response. During reception chemical signals bind to a specific cellular protein. This protein must be complimentary to the signal molecule. The signal molecule acts as a ligand because it binds to a specific, larger protein and causes a change in shape of the receptor. Transduction begins once the receptor changes shape. This change in shape triggers a series of changes of different molecules along the signal transduction pathway. Not all signal transduction pathways are the same types of relay molecules, although most relay molecules are proteins. For example, one major mechanism of transduction is protein phosphorylization. Protein phosphorylization is when a phosphate group is added to the protein, this is carried out by a protein kinase. Once the signal completes transduction a response is initiated. Responses are specific cell activities based on what signal was sent, and therefore, what receptor was activated. An example of a response is the opening or closing of ion channels, or cell elongation. • De-etiolation causes the plant to produce more chloroplasts and is a response to a stimulus to light. It causes greener and more plentiful leaves. The second messenger is part of the signal transduction pathway and activates the specific protein kinase which phosphorylizes a specific substrate. Once the change is caused in the substrate de-etiolation is carried out. If something is bout to the second messenger, cGMP, then it will change shape. The protein kinase will, therefore, not be activated because the kinase only binds to a specific shape. If the kinase does not get activated then de-etiolation cannot occur. If de-etiolation doesn’t occur the no chloroplasts will be produced which, in the long run will cause no new leaves to be produced and for current leaves to become a yellowish color and eventually die.

32. LO 3.33 Answer to multiple choice: B Answer to discussion question: The phosphorylation cascade is a chain reaction of enzymes phosphorylating one another until a response is caused in the cell. During this process, the signal is continuously amplified and changed until it reaches the end of the cascade, where it is received and interpreted by the cell (which then responds). The cascade is necessary to the cell signaling pathway because at each phosphorylation, the signal molecule has a conformational change that ultimately ends up in the right shape to be accepted by the receptor protein at the end of the chain. Protein kinase A cannot effectively activate the response because the signal would not yet be in its receptive shape, thus the receptor protein would not accept the signal. A phosphorylation cascade is also necessary to produce a cellular response to a signal because with each phosphorylation, the signal amplifies and can be sent out to many more receptor molecules than if only one protein kinase sent out a signal. A real world example of this is if one person told another person a joke, who then told 5 other people, who then told 20 other people. The strength in numbers (here people, in a real phosphorylation cascade protein kinases) is beneficial to producing a response to a larger number of receptors.

33. ANSWER KEY- LO 3.34 Which of the following is not a way that immune cells communicate? A. Killer T-cells B. Antigen-presenting cells C. Helper T-cells D. Neurotransmitters E. Cell-to-cell contact a. Explain the three stages of cell signaling. b. Discuss the following and explain what type of signaling takes place. i. paracrine signaling ii. synaptic signaling iii. hormonal signaling • The three stages of cell signaling are reception, transduction, and response. Reception is when a signal has caused a target cell to become detected. Transduction turns the cell signal into a response by binding a signal molecule. This causes a change in the receptor protein. Response is the result of the signal that has undergone transduction. • Paracrine signaling involves a secreting cell that sends molecules into the extracellular fluid. Synaptic signaling is when a nerve cell releases neurotransmitters into the synapse to communicate with neighboring cells. Hormonal signaling is the long distance communication where they can reach and body cells throughout the body.

34. LO 3.35 Answers Multiple Choice: D; The answer is D because first, there must be a stimulus to have a response. After the stimulus the body will response with the releasing of epinephrine which will continuously flow throughout the body until it finds a receptor on the plasma membrane that it will bind to. Following this binding the receptor will now change shape and activate the signal to be transmitted and amplified over the pathway through the activation of a G-protein. This will cause second messengers like cAMP to become activated and ultimately carry through with the message. The protein hormone is too polar to be able to cross the plasma membrane to reach the nucleus. FRQ:Plant- Plant: direct communication between cells must involve cell-to-cell contact. An example of this communication is the plasmodesmata in plants which are essentially holes in the plant's cell wall that allow for the cytoplasm between two neighboring cells to be shared. Neuron- Neuron: short distance communication, the cells are near each other, but to do touch in this communication. This process is highlighted with neurotransmitters in the nervous system that are released into a synaptic cleft that is quickly taken up by another cell and the message is transmitted. Also cell apoptosis is another example of short distance communication because as one cell releases the death hormone into the extra cellular matrix around it, other cells will receive the signal and begin their own cell death. Hormone- Target Cell: long distance cell communication deals exclusive with the signal being transmitted from one area of the body to a completely different area. An excellent example of this process is hormone signaling in the body through which a hormone such as human growth hormone is released by the pituitary gland and is then entered into the body's bloodstream. It will travel throughout the body searching for a cell that has receptors for it. Once it finds a cell in a completely different area of the body the receptor will begin to change in response to the signal and activate second messengers.

35. ANSWER KEY- LO 3.36 MC Question~ After viewing the figure to the left and your knowledge of the subject, which of the following is FALSE in regards to the signal transduction pathways of insulin? The pathway promotes the usage or storage of glucose There are negative and positive feedback mechanisms to regulate the signal transduction pathways The process of insulin production is an example of a trigger mechanism in a signal transduction pathway The activation of the IP3 (seen in figure as PI-3 K) causes an automatic halt of the production of lipids to prevent disruption of the signal transduction pathway. Free Response/LL Question~ Your Aunt Enid goes to see a doctor for her yearly check-up and is told that several of her protein kinases are not functioning properly and he is seeing abnormal cell growth in her thyroid. Explain what the doctor believes to be causing this abnormal cell growth and what this could lead to for Enid. Enid goes to a specialist who tells her that her protein kinases are working fine but something else is causing this same problem the doctor saw. What other factor related directly to the functioning of protein kinases might cause Enid’s demise? Free Response/LL Answer~ Aunt Enid’s original diagnosis is that her protein kinases specifically in her thyroid are not functioning properly. 2% of our genes code for protein kinases and they are extremely important in not only for transducing all sorts of signals but also for regulating cell reproduction, as Enid’s supposedly malfunctioning one does. If protein kinases are unable to regulate cell reproduction this can cause abnormal cell growth and contribute to the development of cancer, in this case thyroid cancer. When Enid goes to a specialist who finds that her protein kinases are functioning but she is still experiencing abnormal cell growth the specialist has likely determined that some of her protein phosphatases, or enzymes that can rapidly remove phosphate groups from proteins, are malfunctioning and unable to dephosphorylate. This means they are unable to perform one of their main jobs in regulating cell reproduction: making the protein kinases available for reuse which enables the cell to respond again to an extracellular signal. In Enid’s case this extracellular signal would be to halt cell reproduction but it cannot cause a response if her protein phosphatases are not working properly.

36. LO 3.37 Answer Key Multiple Choice: a) b) c) d) Learning Log Question:a) Signal transduction pathways require 3 basic steps. Reception, during which a ligand binds to a receptor, transduction, during which a signal is converted to a specific cellular response, and finally the response itself, during which a cell responds to the signal. If any of these stages is disrupted, the pathway is inhibited. For example, the hormone epinephrine acts G-protein-linked receptors to stimulate glycogen breakdown. If for some reason, G proteins are denatured, or the cell fails to hydrolyze GTP, then reception of the epinephrine fails to occur. Even if reception takes place, the pathway can be inhibited during transduction if any molecules in the phosphorylation cascade does not get activated. Even if reception and transduction go smoothly, the signal to break down glycogen can still be inhibited during the response stage if a cell has the wrong collection of proteins, because signal transduction requires specificity. b) Signaling can also be inhibited by drugs, neurotoxins, and other poisons. For example, bacteria that cause botulism and cholera act by disrupting G-protein linked receptors, which in turn prevents transduction and a response from taking place. Picture sources: http://www.cartage.org.lb/en/themes/sciences/botanicalsciences/plantreproduction/PlantBehavior/PlantBehavior.htm http://www.biology.arizona.edu/cell_bio/problem_sets/signaling/graphics/Pro_Kin_Activation.gif

37. ANSWER KEY– LO 3.38 A rare disease leads to an absence of DAG and IP3 in somatic cells. Which of the following effects would it induce in intracellular signaling? A) Membrane-bound receptors would not be able to receive signal molecules B) The second messenger system between the membrane receptor and the cellular response would be deactivated. C) The nucleus would receive the signal, but would not be able to transcribe the correct gene. D) All of these results would occur if DAG was absent. Learning Log/FRQ-style Question: Describe the basic elements that differentiate a gut’s response to epinephrine and a bicep’s response to epinephrine and why are these differences important? The gut has a different receptor for epinephrine than the bicep’s receptor for epinephrine. The gut utilizes α-type epinephrine receptors, which allows for a response that stimulates decreased blood flow to the digestive tract in order to focus on the flight or fight response. The bicep has β-type epinephrine receptors that cause the cellular response to allow for a quick reaction, or fight or flight, as there is increased blood flow to critical skeletal muscles. These different responses are important because when the body is producing adrenaline, often times there is a specific stimulus causing this and there therefore needs to be a response like fight or flight, rather than increased digestion, in order for the body to respond advantageously to the situation.

38. LO 3.39 M.C. Answer: C. receptor tyrosine kinase activity Learning Log/FRQ-Style Answer: Due to the fact that cell division of mammalian cells is driven by protein kinases that regulate progression through the various phases of the cell cycle, I would research ways to alter protein kinases in cell cycle regulation in order to slow cancer growth. The pyruvate kinase enzyme is involved in a critical energy-producing process known as glycolysis ― a process common in all cell types where glucose is broken down to produce adenosine triphosphate (ATP), the cell’s main energy source. Healthy cells and cancer cells use this process and enzyme differently. This finding of how cancer cells and normal cells use this kinase provides strategies to use cell cycle kinases as targets for therapeutic intervention.

39. ANSWER KEY– LO 3.40 MC Question Answer: B. posterior pituitary. This endocrine gland secretes oxytocin and antidiuretic hormone (ADH), which are not involved in the flight or fight response.The adrenal medulla produces epinephrine and norepinephrine, which causes reactions such as increased heart rate and blood pressure to occur once they have been entered into the bloodstream. The sympathetic nervous system is what stimulates muscles and activates glands in the body to prepare for the flight or fight response. An increased heart rate causes blood to be pumped faster through the body, so oxygen will travel faster and be more available for muscles to use for attack. The hypothalamus in the brain will activate the anterior pituitary which will release ACTH (adrenocorticotropic hormone) into the bloodstream, as well as the adrenal medulla which will release adrenaline. Free Response: When a mother eagle becomes threatened, the hypothalamus in her brain will activate the sympathetic nervous system. Her breathing will become deeper and her muscles tense up, in addition to her heartbeat speeding up and her pupils dilating. Decreased digestion will give her more energy for attack, and she will have adrenaline and other hormones in her blood that excite her. The adrenal glands will release the hormones. After the attack, the parasympathetic nervous system will calm the body down. The pupils will contract, heartbeat slow down and other systems as well.

40. Answer Key LO 3.41 • M.C. Question: Which of the following statements about epinephrine and norepinephrine is false? • They are members of the class compounds, the catecholamines • They are released by the adrenal medulla via nerve impulses • They respond to endocrine signals which may cause suppression of the nervous system • They cause a decrease in blood supply from the digestive system • and skin • E. They cause an increase in blood supply to the brain, heart and • skeletal muscles. • Suppose a male fish’s territory is invaded by the same species of fish. What may be his • physiological and behavioral responses be? If the invader had a • yellow underside, how might this affect the male’s future • responses to threats? Design an experiment to test this. • If the male fish’s territory is invaded, the fish’s muscles would contract causing its fins to raise. The fish might also demonstrate aggressive behavior towards the invader to protect his territory. If the fish detected the yellow underside as a threat , he might behave aggressively toward any thing with that same yellow color. To test this, one could create five different models. One model would resemble the physiology of the fish being tested, but with no yellow underside. The other four models would not resemble that fish, but they would have the yellow underside. The control would be the model fish with no yellow underside. The independent variable is the underside color of the fish models. The dependent variable is the actual fish’s response to the models. The results can be measured by the degree of behavioral response from the fish.

41. Answer Key-LO 3.42 Multiple Choice Question: Which of the following statements is incorrect regarding how organisms communicate and the resulting responses? • Most terrestrial mammals, as they are nocturnal, do not use visual displays as their main form of communication. • The act of communication between organisms increases the population’s likelihood of survival. • Many of the forms of signaling are controlled by an organism’s genetics. • Most signals are very inefficient when it comes to energy usage. • Chemical signaling involves animals emitting odors. Learning Log/FRQ-style Question: The figure on the right provides examples of animals exchanging information. Using the figure on the right, hypothesize what behavior the blue-footed booby is exemplifying as the male shows off his feet, as seen In the picture on the left. Be sure to include the type of signal with a justification, the reason for the behavior, the animal on the receiving end of the behavior, and the result of the behavior. The blue-footed booby’s behavior would most likely correlate to the peacock’s visual behavior of using their tails during courting rituals. With that in mind, the booby showing off its feet represents a visual behavior with the male bird attempting to court a female. Often times, visual signals used to court a mate include vibrant colors and patterns as well as dances. The organism, often the males, with the distinct coloration will compete to attract a mate. Just as in humans, the ones with the most appealing qualities will be chosen. In preforming this dance, the bird will likely find a female to reproduce with ensuring that his genes can be passed on to the next generation.

42. ANSWER KEY– LO 3.43 M.C. Question: Cocaine is a drug that floods the synapse with dopamine and blocks the reuptake of neurotransmitters how is this different from when dopamine is released normally? A) Normally the neurotransmitters are sent back through the synapse for reuse, which is why with cocaine, the synapse is flooded. B) This is the same as what normally occurs. C) Normally the neurotransmitters are used in the next neurotransmitter they get sent to, but cocaine blocks the sending of them, making them become flooded in the neurons. D) The cocaine blocks the ligand-gated ion channels sending the neurotransmitters into the space outside the neurons. Learning Log/FRQ-style Question: Discuss the steps that the nervous system goes through to react to an outside stimulus like the hitting of knee by the doctor. First the doctor taps on the knee which makes a tendon stretch the quadriceps. The sensors detect the sudden stretch of the quadriceps which convey information to the spinal cord. The motor neurons convey signals communicated from the sensory neurons to the quadriceps causing it to contract, jerking the leg forward. The interneurons in the spinal cord are also communicated with in the spinal cord. These interneurons stop the contracting of the hamstring by inhibiting the hamstrings motor neurons. If this toxin was injected, the victim would have decreased response to any stimulus because the extra neurotransmitters would decrease the ability of the signal sending ones to get through, drastically slowing the response time. http://chemistrybook2011.blogspot.com/2011/05/nervous-system-figure.html

43. ANSWER KEY– LO 3.44 MS (multiple sclerosis) is a disease that scars the nervous system (by degrading the myelin sheath covering of the axon). Given what you know about the function of the myelin sheath, which of following would be true in regarding the axon and how signals will travel down the axon? A)The axon would not be as insulated so signals will travel down the axon faster. B)The neurotransmitters would not travel down the axon at full speed. C)Signals will not travel down the axon at all. D)The axon will function normally. (sample drawing) Calcium is used to help signals get from one neuron to the next which helps a response triggered by an external stimulus. Describe a mechanism that would alter calcium concentration in the synapses of a neuron so that neurotransmitters would not be able to diffuse across the synaptic cleft and bind to the ligand channels on the postsynaptic membrane of a neighboring neuron. A mechanism that could be used to alter the concentration levels of calcium in the synapse of a neuron could be an inhibitor that binds onto the voltage-gated calcium ion channels on the synapse, which would close the channels altogether. Since the channels are closed then the calcium levels in the synapse will not rise, therefore the neurotransmitters will not be able to diffuse across the synaptic cleft, which also means that they will not bind to the ligand-gated ion channels in the neighboring neuron's postsynaptic membrane. Which would cause a response that was initiated to stop since neurons work on a “all or none” principle.

44. Answer Key – LO 3.45Which of the following is true regarding nerve impulses?A) They are transmitted rapidlyB) They are electrochemical in natureC) They are self-propagatingD) A and B are trueE) All of the above are true Describe, in detail, the process of direct synaptic transmission as well as the result of this process. A neurotransmitter binds to ligand-gated ion channels in the postsynaptic membrane, producing an excitatory or inhibitory postsynaptic potential. After release, the neurotransmitter diffuses out of the synaptic cleft, is taken up by surrounding cells, or is degraded by enzymes. A single neuron has many synapses on its dendrites and cell body. Whether or not it generates an action potential depends on the temporal and spatial summation of excitatory and inhibitory postsynaptic potentials at the axon hillock.

45. ANSWER KEY-LO 3.46 A spinal reflex differs from a normal sensory and motor reaction in that • Spinal reflexes are part of the central nervous system response, while normal reactions are part of the peripheral nervous system. • In a normal reaction, the spine transmits information through afferent nerve fibers, while reflex reactions are transmitted along efferent nerves. • Spinal reflexes can never be consciously controlled, while normal reactions can be controlled. • In a spinal reflex, the spine moves the muscles in response as soon as the sensory information reaches the spine while usually the impulse must reach the brain before a response. You were cooking dinner one night and cut your finger while slicing tomatoes. Describe how the nervous system detects this external signal, transmits and integrates the information, and then produces a response. Then draw a neuron, label and describe the parts. Skin receptors in the fingertip detect the cut, then send messages to the sensory neurons. The sensory neurons then relay information to the spinal cord through interneurons. Interneurons send information through the motor neurons, which cause your hand to withdraw before the brain senses the pain. Then, sensory neurons send messages through the spinal cord to the brain. The thalamus identifies sensory messages as pain and relays the information of pain to the somatosensory cortex in the brain. For a response to take place in a neuron, the stimulus must be strong enough to trigger a change in membrane potential, depolarizing the membrane to reach action potential. For depolarization to occur, a stimulus must open the activation gates on some Na+ channels. Na+ influx through those channels depolarizes the membrane, and can trigger an action potential if the depolarization reaches threshold. The action potential jumps from node to node as it travels along the axon, transmitting messages to target locations in the body.

46. What occurs sequentially when the nerve impulse is transmitted from the synapse of one neuron to the postsynaptic neuron? B) The synaptic vesicles release neurotransmitters into the synaptic cleft by exocytosis. Would severing a neuron’s axon stop the neuron from receiving or from transmitting information? Explain. Draw a diagram of a neuron and label it. Severing a neuron’s axon would stop the transmission of information, because the axon transmits away from the cell body. ANSWER KEY– LO 3.47

47. LO 3.48 Answer Key M.C Question: The correct answer is B) Na+ influx makes the inside of the membrane less negative due to the positive charge of the ion, allowing the cell to get closer to reaching its thereshold and action potential. Learning Logs/FRQ-style Question: Once the reflex is initiated by the tapping of the muscle, the sensors detect this sudden stretch in the muscle. The sensory neurons transmit the information through the spinal cord to communicate with the motor neurons. The sensory neurons of the quadriceps interact with interneuron in the spinal cord which causes the flexing of muscles. This shows how many neurons are involved in a reflex and how there are positive feedback responses to some stimuli.

48. LO 3.49 Answer Key (1 of 2) • A mutation in the Schwann cells on the axon causes the secretion of myelin for the myelin sheath to stop. • An inhibitory molecule renders the sodium and potassium channels of the neuron unusable after it fires a first time. • The dendrites are mutated in such a way that they are unable to experience inhibitory postsynaptic potentials. • The voltage-gated calcium ion channels on neuron are mutated in such a way that they stay open longer. Explanation: After the release of the neurotransmitters into the synapse, the neurotransmitters bind to the receptors of the ligand-gated ion channels in the postsynaptic membrane. The binding action changes the shape of the channel to be open and allows sodium and potassium to diffuse through. The diffusion causes depolarization of the postsynaptic membrane. However, the depolarization isn’t enough to let the neuron reach the potential threshold; this is why such depolarizations are known as excitatory postsynaptic potentials (EPSPs). For the postsynaptic neuron to reach the threshold, multiple EPSPs must occur in rapid succession. For this to happen, the presynaptic neuron must fire again (or multiple times) to create more EPSPs. The temporal summation of the EPSPs will allow the postsynaptic to reach action potential. Therefore, if the presynaptic neuron is unable to produce a 2nd or 3rd firing, then the postsynaptic neuron will not be able to achieve action potential. The myelin sheath only serves to increase conduction speed of an electrical impulse, so without it the impulse is only slower. EPSPs can still accumulate. The dendrites of the presynaptic neuron have no role in the action potential of the postsynaptic neuron, since the presynaptic neuron is receiving no inputs in this scenario. If the calcium ion channels were to stay open longer, then merely more calcium would come into the neuron. If anything the increased amount of calcium would increase the fusing action of synaptic vesicles carrying neurotransmitters to the presynaptic membrane, increasing the rate at which neurotransmitters flow into the synapse.

49. LO 3.49 Answer Key (2 of 2) • Sensors take in stimulus of intense heat on hand. Sensory neurons then convey that information to the spinal cord (part of the central nervous system). Once that information is in the spinal cord, sensory neurons interact with motor neurons. The motor neurons convey signals to the muscles in the arms, forcing them to contract and the arm to jerk backwards away from the stove top. During this pathway, the brain is not communicated with, even though it usually is the “headquarters” for neural activity. It was an evolutionary adaptation to not involve the brain in such a pathway, because the pathway occurs much faster without the brain in the mix. Going only to the spinal cord and back, information can be relayed much faster to elicit the reflex response. A faster reflex response increases an organism’s chances for survival and reproduction (for example, responding to a pouncing predator and fleeing). • For a neuron to achieve an action potential, the neuron must reach a threshold potential that is much more positive than its resting potential. When a neuron enters the process, no sodium or potassium channels are open. Then sodium channels start to open. Sodium begins to enter the cytosol and make the cell more positive. More and more sodium channels open depolarizing the neuron until the threshold is reached. After the threshold is reached, an electrical charge is created that originates at the axon and travels down to the synaptic terminals. The charge moves in one direction only because the portions of the axon behind the newly depolarized portions are becoming repolarized as potassium moves out of the neuron. A neuron’s primary form of communicating and interacting with another cell is through the use of neurotransmitters. The neuron packages these neurotransmitters into synaptic vesicles which flow down with the charge of an action potential. As the action potential moves towards the terminal, voltage-gated calcium ion channels are induced to open. Calcium ions then flow into the neuron which causes the synaptic vesicles to fuse with the presynaptic membrane and release the neurotransmitters into the synaptic cleft. Traveling through the synapse, neurons then bind to ligand-gated ion channels on the postsynaptic membrane. The channels open allowing sodium and potassium to diffuse. When these types of channels open, the membrane of the postsynaptic neuron depolarizes and are thus called excitatory postsynaptic potentials (EPSPs). The depolarization is not enough for the second neuron to reach a threshold potential. Therefore, either successive action potentials from the first neuron must occur, and EPSPs add up temporally, or the postsynaptic neuron receives EPSPs from multiple synapses of multiple neurons interaction with its dendrites. Some neurotransmitters bind to potassium-only channels, which allow potassium out of the membrane. This involves a hyperpolarization (neuron becomes more negative) and thus is called an inhibitory postsynaptic potential. An EPSP and an IPSP can cancel each other out. • Acetylcholine plays a major role in the contracting of muscles. When released into a synapse, it generally is excitatory to vertebrate skeletal muscles. GABA (gamma aminobutyric acid) is important in the “calming” of the body, which explains why it is inhibitory at most (if not all) chemical synapses). Other neurotransmitters include norepinephrine, dopamine, serotonin, and glutamate.

50. Answer LO 3.50 Multiple Choice Which of the following concerning a signal response is not true? A.) Sodium-gated channels are opened to allow NA+ into the neuron to produce an action potential. B.) Once integrated, motor neurons communicate with effector cells to create a response to an input. C.) Voltage gated Ca2+ channels are opened to allow Ca2+ ions into the presynaptic membrane. D.) Motor neurons travel through the autonomic nervous system to create a voluntary response to a signal. E.) Approximately -70 mV are required to reach an action potential. FRQ Response: A.)Motor neurons are fired to muscle cells or the endocrine system to produce a physiological response to a stimuli. A.)Mention and describe two examples of a physiological response, along with the neurotransmitter secreted, that would occur If you were to be awoken in the middle of the night due to a startling noise. If awoken during the night due to a startling noise, one of the first responses to most likely occur would be the acceleration of the heart. The acceleration of the heart rate is controlled by the sympathetic division of the autonomic nervous system. Acetylcholine would be released by neurons to excite the muscles of the heart, causing an increased heart rate. Another physiological response that would occur within the sympathetic nervous system would be the stimulation of the adrenal medulla. This stimulation would secrete norepinephrine which would increase the amount of adrenaline in the body. B.) During integration, the Central Nervous System processes the sensory input to create a response. Describe the process of the reception and transduction of an signal in a neuron. The signal is first brought into the neuron through the dendrite. The signal is brought into the axon hillock; it is here that the action potential is created and the impulse is sent down the axons to be released by the terminal buttons and passed through the synapse into the postsynaptic cell. Na+ channels are opened to allow the Na+ to travel against the concentration gradient K+ channels then open to repolarize the neuron and move the action potential down the axons.