ANSWER KEY– LO 3.1

1. ANSWER KEY– LO 3.1 Synthesis of a new DNA strand usually begins with? A thymine dimer DNA Primer DNA Ligase RNA Primer Describe the process of DNA replication and how sexual reproduction doesn’t produce identical offspring from the parent. Explain why changes occur in the DNA. An enzyme called helicase unwinds the double helix of DNA and several binding proteins bind to the sides to make sure the strands stay separated. Then DNA polymerase goes down the strand and adds new nucleotides and they pair with their complementary base on the same stand. Then the DNA is proof reads and ligase seals the fragments together creating two new strands. Sexual reproduction doesn’t produce identical offspring because there are multiple variations in the gametes the mother and father give to the zygote. Each made up of different genes, therefore different DNA sequences are made for the offspring. Other changes in DNA occur when there are mutations in the amino acid sequences. This can come from deletions or mutations, causing the DNA to be replicated incorrectly and for it to function differently. This mutation can then be passed on to the next generation. Picture Representation

2. Answer Key – LO 3.2 DNA’s use as heritable information relies on: A) Its nucleotide pairs B) The phosphate backbone C) Introns and exons D) The codon order E) Ribosomes DNA is often transferred between bacterial cells. Name three ways in which DNA can be transferred between two cells and explain how each works as a method of transfer of genetic information. Bacteria cells have several methods of transferring DNA between themselves. The first is a method called transformation. Transformation happens when a bacteria cell incorporates foreign DNA that has been excreted from another cell. It then uses this DNA with its own and begins displaying that traits that it codes for, thus showing that the DNA taken up is heritable information. An example of this is Griffith’s bacteria experiment where the live cells transformed to show traits of the dead cells. A second method of transfer is transduction. This method is the rarest and involves phages taking up pieces of one bacterium’s DNA accidently as they leave the infected cell and inserting it/them into another bacterium along with its own genetic material. Again the recipient cell begins to display traits from the new DNA showing that the DNA is heritable information. Finally, the last method of transfer of DNA from one bacterium to the next is called conjugation. Conjugation happens when one bacterium extends a sex pilus to another and reels it in until a mating bridge can be formed between them. The donor then transfers DNA into the recipient cell which it incorporates. The recipient cell than becomes a donor cell that can extend a sex pilus of its own, showing that the DNA transferred made genetic changes in the recipient cell.

3. ANSWER KEY- LO 3.3 Suppose there is a disorder which prevents synapsis from taking place during meiosis. This would cause of all the following EXCEPT • Four haploid cells containing identical genetic information will be produced. • A decrease in genetic variation due to the inability of a cleavage furrow to form. • The inability of chiasmata to hold the homologues together in meiosis I. • Genetically identical sister chromatids of each chromosomes in meiosis II. Describe the role of the following in DNA replication: Helicase, Primase, DNA Ligase, DNA polymerase. Explain what happens to the chromosomes during the three stages of meiosis. Explain how crossing over causes genetic variation. The first step in DNA replication is the unwinding of the double helix structure at replication forks by the enzyme helicase. Primase joins RNA nucleotides together one at a time to create a primer complementary to the template strand on the leading and lagging strand. DNA polymerase adds individual nucleotides with complementary nucleotides to the growing end of the new DNA strand. DNA ligase joins the 3’ end of the DNA that replaces the primer to the rest of the leading strand on the leading strand. DNA ligase also joins the Okazaki fragments on the lagging strand. The first stage of meiosis is interphase which involves the replication of chromosomes. Meiosis I begins with prophase I as the chromosomes begin to condense and synapsis occurs causing chromosomes to cross over and form tetrads. The tetrads line up in metaphase I and homologous chromosomes move toward opposite poles in anaphase I. Meiosis I ends with telophase I and cytokinesis which involves the formation of two haploid cells and the chromosomes remaining doubled. In prophase II, a spindle apparatus forms and in metaphase II, chromosomes are once again in the middle of each cell. The centromeres of each chromosome separate, causing the sister chromatids to come apart in anaphase I. The nuclei form and chromosomes condense in telophase II. Four genetically distinct daughter cells are produced. Crossing over begins early in prophase I as homologous chromosomes pair loosely along their lengths. The DNA molecules of two nonsister chromatids (one maternal and one paternal chromatid) are broken at the same place and then rejoined to each other’s DNA. Two homologous segments trade places producing chromosomes with new combinations of maternal and paternal genes which causes genetic variation.

4. L.O. 3.4 M.C. Question: Suppose a protein were to be built incorrectly, thus creating a nonfunctional polypeptide. All of the following could be a cause EXCEPT: There was an error as the GTP hydrolyzed anticodon and codon mismatched the wrong tRNA thus bonding the incorrect amino acid. The mRNA strand had an error in the nucleotide sequence after transcription and RNA processing. In the P site, the ATP failed in binding the amino acid from the incoming tRNA in the A site. The mRNA strand had an error which caused the stop codon to be out of place, causing the protein to be longer than required. FRQ: Draw and label a ribosome and its respective parts during translation. Describe the function of each element in the gene-to-polypeptide process. (Sample drawing) The large and small subunits of the ribosome serve as the main polypeptide synthesizer where the mRNA (which contains the nucleotide sequence and complementary codon for the tRNA) and the other units may function. The tRNA contains the amino acid (which is the building block for the polypeptide), bonds to the growing amino acid chain (polypeptide in the proses), and attaches to the A site. The A site is where the codon of mRNA and anticodon of tRNA match then moves the amino acid chain from the P site to the A site. The P site holds the tRNA that just accepted the amino acid chain after translocation. It then is in place to transfer the chain over and bond it to the new amino acid. Finally, the E site is the place in which the empty tRNA is translocated to, to be removed from the ribosome.

5. Answer Key-LO 3.5LO 3.5: The student can justify the claim that humans can manipulate heritableinformation by identifying at least two commonly used technologies. SP 6.4: The Student can make claims and predictions about natural phenomenabased on scientific theories and models. Which of the following is not true about restriction enzymes?A. They protect bacterial cells against the introduction of DNA from other organismsby cutting up foreign DNAB. They create a site for a eukaryotic gene to be inserted into a bacterial plasmidC. They bind bacterial DNA fragments togetherD. They the cut sugar phosphate backbone of bacterial DNA to create sticky endsHow could someone test to know if the specific gene is cloned correctly into the plasmid? Also, how could someone prepare large quantities of a particular gene or other DNA sequence using PCR and how would this be useful?This could be tested using Nucleic Acid Probe Hybridization. Where each colony known to contain recombinant plasmids are transferred to separate locations on a new ager plate and allowed to grow. Then some of each colony is transferred to a piece of filter paper. Then the filter is treated to break open the cells and denature the DNA and stick to the filter. Then a radioactive single-strand DNA probe is incubated with the filter and excess DNA is rinsed off. The filter is laid under photographic film, allowing any radioactive areas to expose the film. The black spots on the film correspond to the colonies containing the gene of interest. PCR ( the polymerase chain reaction) is a way to quickly amplify DNA molecules in vitro, producing billions of target DNA in a few hours, circumventing gene replication in host cells. PCR requires a double-stranded target nucleotide sequence to be copied with many nucleotides, heat-resistance DNA polymerase, and primers that are complementary to one of the two strands. First the target sequence is denatured to separate DNA strands. Then Annealing occurs which is cooling, to allow primers to form hydrogen bonds with the ends of target sequences. Then extension occurs, where DNA polymerase adds nucleotides to the 3’ end of each primer. This occurs many times, to have a large majority of target sequences.

6. L.O. 3.6 M. C. Which of the following would cause the most change in gene expression? • A codon deletion in the middle of a coding sequence. • A base pair substitution at the beginning of a coding sequence. c. A deletion of a single nucleotide in a coding sequence. • An insertion of a codon at the beginning of a sequence. Learning Log/FRQ-style Question: Sickle cell anemia is the result of a mutation of a nucleotide in a DNA sequence. Describe the process of the specific mutation causing the disease and how it leads to the expression of the disease. A point mutation is a change in only one or only a few bases in a DNA sequence. In the case of sickle cell anemia, it is a change in only one nucleotide, however, it is replaced with another nucleotide in its place. This is base pair substitution. The change in nucleotides causes the codon to change what protein it codes for. If it makes a different protein than the average person’s, this causes the protein to fold differently. The protein, glutamine, would fold and contribute to the disk shaped blood cells most people have. Sickle cell anemia’s codon doesn’t translate into glutamine, therefore, the protein folds differently and contributes to the sickle shaped blood.

7. Answer Key- L.O. 3.7 Increases in the enzymatic activity of some protein kinases important for the regulation of the cell cycle are due to a.       Kinase synthesis by ribosomes b.      Activation of inactive kinases by binding to cyclins c.       Conversion of inactive cyclins  to active kinases by means of phosphorylation d.      Cleavage of the inactive kinase molecules by cytoplasmic proteases e.      A decline in external growth factors to a concentration below the inhibitory threshold The cell cycle is fundamental to the reproduction of eukaryotic cells. Explain the role of THREE of the following in mitosis or cytokinesis: Kinetochores, Microtubules, Motor proteins, Actin filaments. Kinetochores are located in centromeres of condensed chromosomes and microtubules attach to kinetochores for chromosome positioning and movement. Microtubules attach to the to kinetochores on the sister chromatids. Throwing the chromosomes into an agitated motion moving them towards the center of the cell. Then during anaphase they will help with the migration of chromosomes to opposite ends of the cell. Motor proteins play a crucial role in the development of mitotic spindle as well as separating the chromosomes. Actin filaments interact with myosin motor proteins to generate force to pinch a cell in two and help the microtubules to position the spindle apparatus in the cell.

8. Answer Key: L.O. 3.8 What would be the effect on a cell if the synthesis of cyclin and was inhibited? a. The cell would enter the G0 phase. b. The cell would be unable to enter the M phase and would be unable to divide. c. Following division, Cdk would be unable to be recycled. d. Apoptosis would occur immediately. e. Both A and B. The interactions between cyclin dependent kinase (cdk) molecules and cyclins are an important aspect of the cell cycle. Discuss the interactions between these two molecules and their effects on the regulation of the cell cycle. Include an explanation of the effects on the cell if cyclins are produced in excess. The binding of cyclins and cdks is essential in producing the “go ahead” signal to override the checkpoints present in the cell cycle. These two molecules can combine to form MPF, which is important in signaling the cell to continue into M phase and undergo mitosis. Cyclin concentrations generally increase as the cell cyle goes on, reaching a peak at the peak of MPF activity. After M phase is terminated, cyclins are degraded, while cdks are recycled for use in the daughter cells. If cyclin production is excessive, there will also be excessive cdk-cyclin binding. The superfluous presence of these “go ahead” signals will cause the cell to contiunually grow and divide relatively unchecked, which could eventually lead to cancerous cell growth.

9. Answer Key – L.O. 3.9 Multiple Choice: Which of the following would result after meiosis II of a germ line cell with 10 pairs of homologous chromosomes? A.) 4 Gametes, each with 10 pairs of homologous chromosomes. B.) 4 identical copies of the original germ line cell. C.) 4 gametes, each with 10 chromosomes, 1 from each homologous pair. D.) 2 gametes, each with 10 sister chromatids. FRQ: Describe 2 ways by which a newly fertilized zygote will have genetic variability. This offspring will not be genetically identical to any other offspring of the same parents-why? *The Law of Independent Assortment says that it is completely random as to how tetrads line up during prophase I of meiosis. This causes some maternal chromosomes to be on the same side of the germ line cell as some of the paternal chromosomes, allowing potential gametes to have a combination of both maternal and paternal chromosomes. *Crossing Over occurs during prophase I of meiosis as well. The 2 inside chromatids, 1 from the maternal pair and and 1 from the paternal pair exchange genetic material so that the 2 inside chromosomes contain DNA from both the mother and father. This makes it so that each of the four chromosomes of a tetrad are genetically different from each other. *Random Fertilization refers to the act of the sperm gamete choosing an egg gamete to fertilize being totally and completely random. This is another way that the genes being passed from parent to offspring will be randomized.

10. Answer Key- LO 3.10 M.C. Question: In what way does the process of Meiosis contribute to genetic diversity? A) It splits the sister chromatids unevenly, allowing for some gametes to have more genetic material than others B) Homologous chromosomes exchange DNA in a process called crossing over in Meiosis 1 C) Gametes quickly exchange chromatids before they split apart in meiosis 2 D) Gametes form tight junctions between each other and exchange sections of chromosomes A) Explain how duplication, Rearrangement, and reduction of genetic material during meiosis contributes to genetic diversity within an organisim. B) Give an example of how an error in meiosis can lead to damaging traits in an organism. A) Replication within a cell during Meiosis 1 is necessary because the cell must split twice during the process of meiosis an their needs to be enough chromosomes so that there is exactly half of the genetic material in the gametes(haploid cells) than there is the parent cell in interphone(diploid cell). There are two instances of rearrangement that occur during meiosis. One is that the sister chromosomes cross over and randomly exchange genetic material during meiosis one. In Meiosis 2 even more genetic diversity is created by the random arrangement of chromatids into the gametes. Reduction occurs when the chromosomes of the daughter cells created by meiosis 1 split into sister chromatids that go into different gametes that have half the original amount of DNA. All of these processes insure that the same genetic material in a fertilized egg is not an exact replica of the mother or father. B) One example of a problem with meiosis is the disorder known as down syndrome or trisomy 21. This results from an error known as non-disjunction in meiosis 2 when a chromosome does not split into sister chromatids in the gamete. The result is to many copies of chromosome 21 in the offspring when the gametes fuse in reproduction. This defect is survivable but is known to cause cognitive and physical defects in the offspring.

11. ANSWER KEY-L.O. 3.11Based on what you know about mitosis and meiosis, which of the following is false concerning these two processes?A) Meiosis is the process of dividing a nucleus of the cell into two cells who each have half the number of chromosomes as the 1st phase.B) Mitosis is a way for the body to repair tissues and fibers.C) Meiosis is a process where cells are split two separate times, with chances of chromosomal mixing.D) Mitosis is an asexual, six step process that replicates cell information to make a cell of equal standards. If you have chromosomal deficiencies in your cells, is that a problem with the meiosis or mitosis process? Why? In what ways could a negative altercation of mitosis become worse than a negative altercation of meiosis? If you have deficiencies with your chromosomes then that would be a problem with the meiosis process. Meiosis is the first process involving chromosomal mixing, and development. Mitosis is the division of the cells that the body is provided. When you have an altercation in your chromosomes, one of the steps when the chromosomes separate to different sides of the cell, during meiosis, failed to work properly and you have an inefficient number of chromosomes on either sides of the cell. Then once this is complete, mitosis just replicates this process to where all your bodies cells consist of this deficiency. One way that mitosis could become a more serious issue than meiosis is when mitosis replicates the cells, they reproduce at a less than normal rate so internal tissues and fibers cannot be repaired efficiently. There is no problem with the cells that are being replicating in the body, just the rate of which it reproduces is inefficient with the bodies maturation.

12. L.O. 3.12 M.C. Question: Which of the following statements is true regarding meiosis? A) Meiosis involves the duplication and separation of genetically identical cells. B) Meiosis allows for increased genetic diversity by producing four gametes that have differing genetic material from the parent cell, thus enabling the offspring to differ from the two parent organisms. C) Epidemics can move through a population with greater ease if the species’ reproductive system uses meiosis, due to the resulting decrease in genetic variability. D) Meiosis is a more efficient way to reproduce as it requires less steps and energy than simple mitosis. Free Response Question:Suppose an organism has an unknown adaption due to the allele AaBb. Create a labeled diagram to trace the passage of the allele from the parent to its offspring. Why is the use of meiosis more beneficial to the species as a whole? What is one potential problem that the process of meiosis could cause for the offspring? The chromosomes of the diploid parent cell duplicate during interphase and then the cell enters meiosis I where the nonsister chromosomes will be able to exchange DNA. Then the cell will separate into two diploid cells. The sister chromosomes of each cell will then be pulled apart in anaphase II and the second round of cytokinesis will create four haploid cells. Each will have half the combination of the AaBb allele, such as Ab or ab. The other half will be provided by the other parent’s gametes. When the two gametes of opposite sex fuse, the resulting offspring will have a mix of the parents’ alleles. The benefits of meiosis include increases the genetic variability of the population, which can prevent diseases from devastating the species and can increase the species’ ability to adapt to a changing environment, which comes about from the genetic mixing that occurs from meiosis. The potential dangers of meiosis can include nondisjunction of a chromosome, which occurs when chromosomes do not separate before cytokinesis. This can cause the resulting gametes to have extra chromosomes or missing ones. The resulting offspring from these gametes may not survive to birth or may be very disabled due to being either trisomic or monosomic.

13. Answer Key L.O. 3.13 MC: A) Because all of the other ways would cause harm to the patient. FRQ Tourette Syndrome is caused by there being a extra 23 chromosome. The body doesn't know what to do the extra chromosome. This causes tics and learning disabilities. Colorblindness is the absence of a gene on the X chromosome that codes for the making of rods and cones in the human eye. The disorder causes the person to not be able to see certain colors such as Red or Green. Huntington's disease is a genetic disorder that caused the death of nerve cells around the age of 30. The victim would then slowly lose their motor function. Sickle cell anemia is a disorder that is a replacement gene that causes the blood cells from forming naturally round. The disease caused a change in the protein hemoglobin. The victims blood will be sickle in shape and be more likely to clot.

14. L.O. 3.14 Multiple Choice Ans: B (1/4) FRQ Ans: The three types of genetic dominance are complete dominance, incomplete dominance, and codominance. Complete dominance can be thought of as “traditional dominance,” it expresses one gene and not the other in a heterozygous genotype. An example of this is eye color. Say the allele for brown eyes is dominant to the allele for green eyes. In an offspring with the genotype Bb, the phenotype expressed would be completely brown. Incomplete dominance is when the phenotype of a heterozygous individual is a mix of the dominant and recessive allele. If the allele for straight hair is dominant to the allele for curly hair, a heterozygous individual would express the phenotype of wavy hair, a hybrid between curly and straight. Finally, codominance is when both the recessive and dominant genes are expressed side by side. A very common example for this is fur colors in cows. A cow with the genotype Bb (B being brown and b being white) would show the phonotype of brown with white spots.

15. L.O. 3.15 M.C. Question: Albinism is a lack of pigmentation in humans caused by an autosomal recessive gene. What is the probability that two parents with normal pigment have 3 children in a row that all suffer from Albinism? • 1/8 • 1/4 • 1/16 • 1/64 Learning Log/FRQ-Style Question: State the conclusions reached by Mendel in his work on the inheritance of characteristics. Explain how each of the following deviates from these conclusions: A. Autosomal linkage B. Sex-linked (X-linked) inheritance C. Polygenic (multiple-gene) inheritance • a) autosomal linkage are genes on the same chromosomes (not sex chromosomes); this is contrary to the idea of independent assortment. Examples are eye color and hair color • b) sex-linked inheritance (in humans) occurs on the X chromosome this means males only have one copy of the specific gene; Mendel assumed that organisms have 2 copies of each gene • c) polygenic inheritance means that some traits are controlled by more than one gene; this means that some traits show a range of phenotypes.

16. ANSWER KEY- LO 3.16 Colorblindness is sex-linked and recessive in humans. A woman with normal vision but whose father was color blind marries a man with normal vision whose father was also color blind. What type of allele can be expected of their four offspring? A) Both sons being colorblind, both daughters carriersB) One daughter and one son colorblind, and one daughter and one son carriersC) One son normal, one daughter normal, one son colorblind, and one daughter a carrierD) One son colorblind, one son carrier, and both daughter carriers Draw and label a pedigree showing the passage of the sex-linked allele of colorblindness. Make sure to describe all factors contributing to the outcome and why it happened. Males contain the sex chromosomes XY where as women have XX. For sex-linked traits carried on the X chromosome, it is more likely for a man to receive the disease than a woman. This is because a man only contains one X raising the chances of receiving the trait. In the pedigree, a colorblind male was crossed with a non carrier female, their results were as shown. When the heterozygous female crossed with the normal male, the received offspring, one of which being colorblind. When their female offspring was crossed with a colorblind male, their chances of having colorblind offspring raised. This is why they received one colorblind son, one normal son, one carrier daughter and one colorblind daughter. Having a colorblind daughter is rare, but in this case since the parents were both dominant for the trait, her chances rose, being colorblind.

17. ANSWER KEY 3.17 Which of the following statements concerning sex-linked traits is true? A female offspring's expression of a sex-linked trait is determined by the genotype of her mother. A female offspring's expression of a sex-linked trait is determined by the genotype of her father. Male and female offspring have the same probability for the expression of a sex-linked trait. d. A male offspring’s expression of a sex-linked trait is determined by the father. B and D X XX signifies an afflicted sex-linked trait X XX XX Results: 1 of 4 normal daughter 1 of 4 carrier daughter Y XY XY 1 of 4 normal son 1 of 4 afflicted male A female can be a carrier of a sex-linked trait but would not express it because females contain Barr bodies, and therefore the genotype of the female offspring’s mother has no effect on the offspring’s expression of a sex-linked trait (rule out A). In addition, because females have Barr bodies, male and female offspring do not have the same probability for the expression of a sex-linked trait (rule out C). Also, Male offspring's are given there X’s by their mothers, therefore a male’s offspring’s expression of a sex-linked trait is not determined by the father (rule out D and E). As shown by the chi square, a female offspring’s expression of a sex-linked trait is determined by the genotype of her father, so the answer is B. Look at the picture below and explain the phenotypic ratio of the first and second generations of offspring. Explain the inheritance principle that allows pink flowers to be made. Use punnett squares in your explanation. R R According to the punnett square, all of the first generation offspring (100%) have a heterozygous phenotype. Incomplete r RrRrdominace is the inheritance principle represented in this offspring. Incomplete dominance is when parent alleles are r RrRr blended, the combination of parental red and white phenotype will produce a pink phenotype for all of the 1st generation offspring. R r When two pink flowers mate, one out of four is proposed to be homozygous dominant (red), two out of four are proposed R RR Rr to be heterozygous (pink), and one out of four are proposed to be homozygous recessive (white). Therefore the r Rrrr phenotypic ratio of plant color when two pink flowers mate (2nd generation of offspring) is 1:2:1.

18. ANSWER KEY – LO 3.18 Human Cat Bat Whale In prokaryotes, gene expression involves: A) regulatory genes B) transcription factors C) Regulatory elements D) Polyribosomes Organisms share a high quantity of DNA. Give phenotypic evidence of different organisms sharing similar DNA.Homologous structures are similar structures in different organisms that suggest similar DNA sequences. Some examples of homologous structures include a human’s arm, a cat’s arm, a whale’s fin, and a bat’s wing. Organisms that have similar embryos are also organisms that share similar DNA sequences. An example of different organisms having similar embryos is the chicken and human embryo. The most likely reason that organisms have these similar DNA sequences is evolution from a common ancestor. Pharyngeal pouches Post-anal tail Chick embryo Human embryo

19. L.O. 3.19 MC Question: What would not be an instance when gene regulation would be used? • Stopping the growth of a tumor • Genetic changes in the body color of a fruit fly • The shift in a population toward larger animals • The building of certain proteins used in RNA The answer would be C because the other answers are all components of regulatory gene functions. The repressor regulatory genes aid in stopping the expansion of tumors. Activators can cause gene expressions that change color in organisms and also code for the proteins that build up RNA. FRQ Question: Regulatory genes regulate protein activity. Discuss TWO specific mechanisms of protein regulation in eukaryotic cells. The negative feedback loop created by the build up of a certain molecule can either repress or activate the gene through the promoter, which initiates the gene to be made or stopped depending on if it is repressed or activated.

20. Answers L.O. 3.20 • Why may a set of identical twins each have a feature that slightly appears to differ from his/her sibling?A )They aren’t really identical.B) There has been a mutation of genes in one of the twins.C) The code for that particular gene may be expressed in a variety of ways depending on how the body translates/reads it.D) There has been a deletion of insertion.E) Both a and d are correct. Gene expression regulates efficient cell function in many ways. Explain how the following contributes to this:a) cell differentiationb) role of proteinsc) DNA methylation • A) Cell differentiation is important to gene expression/cell function because it allows us to use the same DNA to create cells with different roles in the body. Less specialized cells become more specialized cells. Cellular differentiation almost never involves a change in DNA sequence and different cells can have very different physical characteristics but still have the same genome. This process allows for repair of damaged cells/tissues as well. • B) The role of proteins in gene expression and function of the cell is that the sequence of amino acids dictate the structure of the protein and the structure of the protein determines where the gene will act and what its role will be. Proteins are involved in transcription and histones. • C) DNA methylation can slide genes into their “off” stage and allow them to remain silent when necessary, and also assists with the process of cell differentiation.

21. ANSWER KEY- LO 3.21 Which of the following statements correctly describes promoters in E. coli?A) A promoter may be present on either side of a gene or in the middle of it. B)Every promoter has a different sequence, with little or no resemblance to other promoters. C)Many promoters are similar and resemble a consensus sequence, which has the highest affinityfor RNA polymerase holoenzyme. D) Promoters are not essential for gene transcription, but can increase its rate by two- to three-fold.Positive regulation involves an activator protein that, when bound in the vicinity of the regulated gene, facilitates the binding of RNA polymerase to its promoter. The affinity of the activator for the DNA may be either increased or decreased by a signal molecule. Negative regulation involves a repressor protein that, when bound near the regulated gene, hinders access of RNA polymerase to its promoter or inhibits its activity. Repressor affinity for its binding site (the operator) may be increased or decreased by a signal molecule

22. Answer Key—LO 3.22 M.C. Question: Explain why a steroid hormone binds to receptors inside the cell when protein ligands bind to receptors in the extra-cellular matrix. • Steroid hormones mediate gene expression, so they have to bind to their receptors while they are in the nucleus. • Protein ligands only bind to ligand-gated-ion channels, which must be embedded within the plasma membrane for ions to pass through them. • Steroid hormones can travel through protein channels across the plasma membrane to enter the cell and protein ligands cannot. • Steroid hormones are hydrophobic, so can travel through the plasma membrane and protein ligands are hydrophilic, so they cannot. Explain how a signal can be transmitted from one cell to another cell resulting in a physical response by the cell and the alteration of gene expression in the receptor cell. Include in your explanation two types of receptor proteins and describe how they function. A signal can betransmittedusing a ligand, a signalmolecule. Thisligand can eithertravelthroughthelymphsystemorthecirculatorysystem in endocrine, longdistance, signaling. Theligand can travelthrough gap junctions, tightjunctions, orplasmodesmataforparacrine, local, signaling. Once theligandreachesthecell, itbindsto a receptor proteineitherinsidethecellorembeddedwithinthe plasma membrane. Onetype of membrane-bound receptor proteinis a tyrosine-kinase receptor. When a ligand binds to a Tyrosine-Kinase receptor, the two separate halves form a dimer, which phosphorylates itself using ATP and then phosphorylates several separate relay proteins. G-protein-linked receptors are also membrane-bound receptor proteins. Once a ligand binds to a G-Protein-Linked receptor, the receptor phosphorylates a G-protein using GTP instead of ATP. The G-protein then phosphorylates an enzyme which activates other relay proteins. These relay proteins activated by the receptor proteins are part of a signal transduction pathway, which then causes physical changes within the cell, usually multiple changes per signal. One possible change is the activation of a gene possibly by moving necessary transcription factors to the nucleus. The gene transcripted into RNA could then be processed and translated into specific proteins coded for by that gene. This causes that gene to be expressed through the proteins it codes for.

23. Answer Key LO 3.23 M.C. Question: Which of the following is not involved in gene regulation? a) RNA processing b) Apoptosis c) Degredation of mRNA d) Cleavage and Transport FRQ: Cancer is caused by mutations and errors in regulation. Describe the factors necessary for cancerous cells to develop? Answer: Cancer occurs when RAS is hyperactive, causing signal pathways to always remain open. Having these signal pathways open call for constant cell division which leads to the formation of a tumor. P53 usually acts as a tumor supressor but in the case of a mutation of the p53 gene, the tumor is allowed to grow and develop into cancer.

24. ANSWER KEY– LO 3.24 M.C. Question: Which of the following is true? A) Nondisjunction of chromosomes is always fatal to an organism. B) A chromosomal point mutation generally causes more drastic changes in genotype than a chromosomal deletion. C) Females with trisomy X are phenotypically indistinguishable from normal XX females. D) Evolutionary forces always select against mutated organisms. The antibiotic-resistance gene developed by the bacteria described above is one example of a beneficial mutation. Provide another. Also, describe some of the mechanisms by which mutations are preventedduring or after DNA replication. Sickle-cell disease is caused by a point mutation in the hemoglobin protein of red blood cells. The disease causes the red blood cells of an individual to elongate into sickle-shaped rods that clump and clog blood vessels, resulting in pain and even paralysis. When an individual is heterozygous for this trait, he/she can still be affected by symptoms, but in certain environments it is better to be heterozygous than to be free of the disease completely: sickle-cell hemoglobin reduces malaria symptoms. In environments where malaria parasites are common, the mutation that causes sickle-cell can actually aid survival. During DNA replication, DNA polymerases proofread each nucleotide against its template as soon as it is added to the growing strand of new DNA. If an incorrect nucleotide is discovered, it is removed by the polymerase before synthesis of the strand continues. Mismatched nucleotides that evade proofreading are fixed by special enzymes during mismatch repair after synthesis is completed. If DNA is damaged—by ultraviolet rays of sunlight, for example—then, during nucleotide excision repair, the damaged segment is cut out by a nuclease and replaced with proper nucleotides by DNA polymerase and ligase.

25. ANSWER KEY - LO 3.25 Which of the following changes to the DNA sequence, AGC-GCT-TAA-CGC-AGT-ATT-CGA-GCT, would have the greatest effect when being translated into amino acids? a. The guanine in the first triplet is replaced with a thymine nitrogenous base. b. The guanine in the fourth triplet is replaced with a cytosine nitrogenous base. c. The guanine of the eighth triplet is replaced with a adenine nitrogenous base. d. The guanine of the seventh triplet is replaced with a thymine nitrogenous base. Information flow can be altered by mutation. Describe three different types of mutations and their effect on protein synthesis. The most common types of mutations are single base-pair mutations. Silent, missense, and nonsense are all mutations that occur due to the change of one nucleotide. In silent mutations, a single nitrogenous base is replaced with another. Such a change, however, does not effect the overall amino acid the codon represents. Therefore, no change in the protein/ peptide chain is made. A missense mutation is when a nucleotide is changed and causes a significant change in the codon. As a result, a different amino acid is brought by tRNA and the anticodon. The final protein is incorrect. Nonsense mutations are the most risky because, the resulting codon may code for “stop”. Therefore, the protein may not be formed at all, or the peptide chain will be terminated prematurely.

26. ANSWER KEY – LO 3.26 Multiple crosses involving genes known to occur on the same chromosome produce frequencies of phenotypes that suggest there is a high rate of crossover between these two genes. Which of the following is the most likely explanation for the phenotypic frequencies observed due to crossing over? (A) The two genes are far apart from one another. (B) The two genes are both recessive. (C) The two genes have incomplete dominance. (D) The two genes are both located far from the centromere. A change in an individual’s DNA may result in a phenotype that is affected by natural selection. Describe and explain a situation in which a change in an organism’s DNA may be beneficial for that organism’s survival and one in which the mutation would be detrimental. One mutation that proves to be beneficial for organisms is the heterozygote advantage, which occurs when heterozygous individuals have a better fitness than homozygotes. This change in DNA is particularly important in tropical regions where malaria is present. Homozygous recessive individuals carry an allele that causes sickle-cell disease, so being homozygous dominant or heterozygous would prevent the organism from contracting sickle-cell disease. However, heterozygotes are also protected against malaria. Due to this, heterozygotes for this allele are dominant in tropical areas affected by malaria because they are protected from sickle-cell disease and malaria, thus allowing them to reproduce and carry on their genes. In contrast, the point mutation that causes sickle-cell disease is a detrimental mutation. In a single substitution mutation, AU is placed in the DNA sequence instead of TA, producing valine as opposed to glutamic acid. This change causes different folding of the amino acids in the resulting protein, creating a sickle-shaped cell. This blood disorder severely restricts the length of an individual’s life, keeping them from procreating and passing on the unfavorable gene.

27. ANSWER KEY: LO 3.27 What is the role of exon shuffling in providing genetic diversity to a population? It allows transposons on a section of RNA to copy and paste itself elsewhere, thus creating different proteins. It causes bacteria to have a recombinant plasmid. It causes new genes to evolve as sections of DNA are moved to different places. It randomly assorts chromosomes along the Metaphase plate during Metphase I of Meiosis. Suppose you were an expert in DNA technology and were given the task to insert a gene of interest into the plasmid of a bacterial cell. Describe how the transformation process works. Describe three additional ways in which bacterial cells become genetically recombinant. When a gene of interest is inserted into a bacterial cell, the gene is first cut out of a chromosome by the help of restriction enzymes that have the ability to recognize a specific palindrome sequence. The gene of interest that is cut out of the chromosome is then placed inside the plasmid. DNA ligase is also added in order to seal the DNA fragments. After the gene of interest is inserted into the plasmid, the plasmid is then placed inside the bacterial cell. The host cell is then grown in a culture to form a clone of cells containing the “cloned” gene of interest. Genetic recombination can occur via transduction, which is when a virus takes DNA from one bacterial cell to another. Another process is conjugation, which involves the direct transfer of genetic material. The third process is transposition, which involves exon shuffling and the movement of sections of DNA from one place to the next.

28. L.O. 3.28 MCQ: A- diseases don't transfer into your DNA like viruses do. FRQ:Harry contracted a virus as a child which altered DNA, please explain in detail how this virus altered DNA and explain how, if so, this would increase diversity if he mated. When viruses infect a host they do a process of Transduction, in which they exchange or even give DNA to the host. Then this DNA is replicated and sent out to other cells to infect them. This DNA passes on when the male gamete meets the female gamete in fertilization. This is why HIV, and herpes can be passed along genetically.

29. L.O. 3.29 Which one of these statements is false about Transduction?A) Transduction can only occur in bacterial cells.B) Only one cell needs to be infected in order for genetic variation to happen.C) Specialized transduction happens during the lysogenic cycle.D) Transduction is not a 100% way to get a certain gene transferred. While many see viruses as bad, they; through transduction can be used to spread genetic material from one organism to another. Suppose you are trying to find out if a certain bacteria is resistant to virus, using your knowledge of transduction design an experiment to find out. In order to test which ones are resistant, you would first need to create a control group environment without the virus. Using a set of bacteria with the gene for glowing in the dark. These will act as the donor. Allow the virus to conduct transduction within the environment for 1 day to allow growth. On day 2, in the set of bacteria you want to test for resistance, place a small amount of the bacteria from the set that glows and was infected with the virus. After one day if the majority of bacteria in the dark glow, then transduction uncured, meaning that it is not resistant. If there is only a small portion glowing, then it most likely did not occur.

30. ANSWER KEY –LO 3.30 M.C. Question: Lysogenic bacteriophages have the capacity to transfer a piece of cell DNA adjacent to the prophage to another cell. This is known as specialized: A) Invagination B) Endocytosis C) Transduction D) Transformation Learning Log/FRQ-style Question: Consider the case of an RNA virus: Is genetic variation any more or less likely to occur than in a similar DNA virus? Explain. Draw and label a model of the RNA viral reproductive cycle and indicate where genetic variation is likely to occur. RNA viruses lack replication error-checking mechanisms, and thus have higher rates of mutation than DNA viruses. RNA polymerase doesn’t proofread newly formed strands like DNA polymerase does. Therefore, mutations are more likely to occur during the synthesis of new genetic material in RNA viruses than in their DNA-possessing counterparts. Due to the lack of a proofreading mechanism by RNA polymerase, a large number of mutations are likely to occur during transcription.

31. ANSWER KEY- LO 3.31 M.C. Question: Which of the following statements concerning cell communication is/are true? A)The use of pheromones can trigger reproduction and developmental pathways through cell communication. B)Apoptosis, programmed cell death, is achieved through a signal cascade that alters active genes. C)A signal cascade is initiated when a ligand binds to DNA. D)Both A and B Why is it important for cell-to-cell communication to function quickly and effectively? Why do you think cell communication in many species functions in very similar ways, and what do you think would happen to an organism that has dysfunctional signal cascade pathways? Cell-to-cell communication must function quickly and effectively because if they do not do so, they will not be able to correctly respond to the external environment or stimulus. Cell-to-cell communication must be functioning properly in order for an organism to carry out essential processes such as organ differentiation and whole organism physiological responses and behaviors. Many species share the same basic cell-to-cell communication mechanisms because they share evolutionary lines of descent, and communication schemes are the products of evolution. Cell-to-cell communication is mandatory for cell life and growth, so cells that had a selective advantage for adjusting to the environment more efficiently had their genes passed down over the ages. An organism that has dysfunctional signal cascade pathways would most likely die due to an inability to adjust to the environment. Defects in any part of the signal pathway often lead to severe or detrimental conditions such as faulty development, metabolic diseases, cancer or death.

32. Answer Key-LO 3.33 Phosphorylation cascades involving a series of protein kinases are useful for cellular signal transduction because A) they are species specific B) they always lead to the same cellular response because of specificity C) they amplify the original signal many-fold D) they counter the harmful effects of phosphatases by dephosphorylating proteins E) the number of molecules is small and fixed Explain how the binding of a signal molecule to a ligand gated ion channel affects the distribution of anions and cations in the membrane. For a cell signaling pathway such as this the signal molecule (ligand) binds to the target cell (active site of the cell) initiating the reception and transduction. Transduction occurs and the ion gate is now open. The response allows ions to flow freely, but when the the ligand dissociates from the receptor the ion gate closes and ions are not allowed to enter or leave the cell. Therefor when a ligand is binded anions and cations move freely and achieve equilibrium, but when it is not binded it is closes and ions do not move.

33. ANSWER KEY– LO 3.34 M.C. Question: Which of the following is not an example of endocrine signaling? A: Parathyroid Glands and Calcium B: Thyroid glands and iodine C: Glucagon and Insulin D: Growth Factors and Clotting factors E: None of the above FRQ Style Question: Blood glucose levels decrease in the bloodstream. Explain the process the body goes through in order to increase the level back to normal. If the blood glucose levels would have increased, what would be the difference? When the blood glucose level in the bloodstream is lowered, glucagon is release by the alpha cells in the pancreas. This immediate release causes the liver to turn the glucagon into sugar, and following with the release of the glucose into the bloodstream. On the other hand, if blood glucose levels are raised in the bloodstream, the pancreas is informed to release insulin through it’s beta cells. The presence insulin triggers the immediate response to cells(red blood cells, muscle cells and fat cells) to take the glucose from the bloodstream to allow it to achieve the normal blood glucose level. **Beta Cells are responsible for making and releasing Insulin. **Alpha Cells secrete Glucagon

34. Answer Key: LO. 3.35 • M.C. Question: What is the correct order in which a helper T communicates with a killer T cell? • 1. Killer T cells secrete perforin proteins. • 2. Killer T cells are activated • 3. Killer T cells initiate clone killer T cells • 4. Helper T cells activate & secrete cytokines • 5.Helper T cell binds to a molecule on a dendritic cell • 6. Helper T cells initiate clone helper T cells • A. 4,5,6,1,2,3 B 2,1,3,4,6,5 • C. 5,4,6,2,3,1 D 5,6,4,2,3,1 • E. 6,5,4,2,1,3 Learning Log/FRQ:

35. Answer Key- LO 3.36 M.C. Question: Which of the following statements concerning signal receptors is true? An intracellular receptor may receive polar hormones. ATP displaces ADP to activate the G-protein after reception. Tyrosine kinase receptors form a phosphorylateddimer to activate relay proteins. Ligand-gated ion channels allow ions to flow against the concentration gradient. Describe in detail the process by which epinephrine triggers a cell response. Be sure to include reception, transduction, and 1 example of a cell response. Epinephrine is a hormone (signal molecule) that attaches to a G-protein linked receptor. This causes a change in receptor shape, attracting an inactive G-protein. GTP displaces GDP to activate the G-protein which converts ATP to cAMP to activate protein kinase which starts a phosphorylation cascade. In this cascade, protein kinase uses an ATP molecule to provide a phosphate ion for the next protein kinase for its ATP; at the end of the cascade, the 3rd protein kinase will activate a final protein that triggers the cell response. This final protein will vary from cell to cell. In some cells, it will be glycogen phosphorylase which will produce glucose to provide energy for a fight or flight response. Epinephrine (visual approximation)

36. Answer Key LO 3.38 Multi Choice: FRQ: A) Describe the basic principles of signal transduction. B) Think of 3 examples of signal transduction receptors. C) Compare and contrast 2 of the examples of signal transduction receptors. A. In the signal transduction pathway, there are three stages. They are reception, cell’s detection of signal molecule; transduction, the binding of the signal molecule changing the receptor protein; and response, the cellular response finally triggered by the transduced signal. B. One example is G-protein linked receptors. Another is receptor tyrosine kinases. The last one is ion channel receptors. C. In G-protein linked receptors, a signal molecule activates a G protein linked receptor and a G protein binds to the receptor activating it. The GDP in the G protein becomes GTP and the G protein binds to an enzyme which triggers a cellular response. One phosphate is lost from the GTP to turn it back into GDP and its original state. In Ion Channel Receptors, the gate is closed at first. When a ligand binds to a receptor, the gate opens and specific ions can flow through the channel and rapidly change the concentration of that ion in the cell. That rapid change in concentration triggers cellular response. When the ligand dissociates from the receptor, gate closes and ions no longer enter the cell. • In signal transduction, if the receptor is altered, which of the following will happen? a) the cellular response will go on as normal b) there will be a different cellular response c) the cell will stop all functions d) the cell will burst

37. Answer Key- LO 3.39 Suppose a healthy individual who has never used a morphine based drug decided to inject themselves with heroin. How would the body react to the initial injection. A.) The individual would receive the instant gratification of a high with few residual symptoms. B.) The individual would experience nausea and vomiting due to the sudden chemical imbalance in the nervous system. C.) The individual would experience cross tolerance and would feel little to no effect from the drug. D.) The morphine ligand would not be able to bind to the receptor site and would have no effect on the individual. E.) The drug would immediately show inhibitory effects causing the individual to feel sluggish and uncaring. • In the body natural endorphin acts as a natural stimulant and pain reliever. Endorphins can inhibit neurons from firing and create a natural analgesic affect or excite the neurons which gives the organism a feeling of euphoria. In a recently recovering heroin addict, how would the body be affected if the person went through a vigorous workout? • Because endorphins are often released when an organism encounters various stressors (i.e. exercise, childbirth) the individual would lack the necessary levels to keep them energized. The lack of this stimulus would likely cause the person to become tired and sluggish faster than a completely healthy individual would. The absence of natural endorphin would also leave little relief for various strains to the body and result in large amounts of pain from muscle fatigue and other strains.

38. L.O. 3.40 Which of the following is a behavioral pattern resulting from an ultimate cause?A) A male robin attacks a red tennis ball because it resembles the breast of another male.B) A male robin attacks a red tennis ball because hormonal changes in spring increase its aggression.C) A male robin attacks a red tennis ball because a part of its brain is stimulated by red objects.D) A male robin attacks a red tennis ball because several times in the past red tennis balls have been thrown at it, and it has learned that they are dangerous.E) A male robin attacks a red tennis ball because it confuses it with an encroaching male who will steal his territory. FRQ Response A) Innate behavior is naturally inherited behavior. Learned behavior is acquired by the method of trial and error. Innate behavior has little to no influence from the environment, instead reflecting evolutionarily preferred behaviors embedded into organisms overtime. Learned behavior can be separated into two categories: classically conditioned (involuntary responses)and operantly conditioned (voluntary responses). Innate behaviors are adaptive in that characteristics present in previous ancestors allow for survival and therefore prevalence of the behavior in organisms today. For example, baby geese upon hatching imprint on the first figure walking away from them in order to increase survivorship by immediately relying on their mother’s leadership. This behavior was favored over isolated behavior and thus eventually became an innate, unconscious inclination in geese. Learned behaviors are adaptive in other respects, since they deal directly with environmental stimuli. For example, in Pavlov’s experiment, a dog learned an association between the ringing of a bell and the presentation of food. After multiple happenstances, the dog’s subconsciously salivated to the ring without the presentation of food, involuntarily preparing his mouth for possible food intake and increasing his adaptive chances of survivorship in digestion. B) First, a hypothesis would state that the male garter snake would only recognize the pheromones of female garter snakes in a controlled setting, seeing as how it would be reproductively more efficient. The independent variables would include equal numbers of female garter snakes and other snake species of similar physiognomy. The dependent variable would be the amount of times the male garter snake followed either of the two trails, as done by human observation and recording. The environment would generally be kept the same, in order to limit further variables that could push the results towards either of the two species. The experiment would have thirty male garter snakes, ten females garter snakes, and ten female non-garter snakes. The study would be conducted five times to heighten reliability and the results would be graphed for comparison. A control group of ten out of the twenty male garter snakes would be left to roam without the leading of pheromones i.e no females. Also, the mating season, and food would have to be accounted for, as circadian rhythms and consumption are often known to influence libido.

39. L.O. 3.41 How is the distance to a food source communicated by a dancing honeybee?A) by the direction it waggles its abdomenB) by how far it moves during the straight run portion of the danceC) by which direction it turns after making the straight runD) by the tempo or degree of vigor of the danceE) none of the above - bees can't communicate the distance • Description: Auditory- animals communicate through vocalizations. Communication through vocalization is essential for many tasks including mating rituals, warning calls, conveying location of food sources, and social learning. Examples: In Drosophila species, males produce a song by vibrating their wings, and females are able to recognize the songs of males through intervals between pulses, rhythm, and length of song. Birds also transfer information through their songs. • Description: Chemical-animals that communicate through odor emits a chemical called pheromones. The signal can be involved in courtship, warning, ect. Examples: Moths emit pheromones that can attract mates from several kilometers away. Minnows or catfish when in danger can disperse an alarm substance into the water to warn other members.

40. ANSWER KEY LO 3.42 After the action potential of a neuron, which of the following restores the cell to -70 mV? a) decrease in Na+ concentration b) increase in K+ concentration c) Aquaporins d) Sodium potassium pumps e) None of the above Suppose you place a neuron in an aqueous solution containing free Na+ and K+ ions. Thresholds are reached and action potentials are created. Why does the action potential only travel in one direction? After the cell is restored to normal charge, how does a single neuron interpret multiple inputs? Action potentials only travel in one direction because the Na+ ions travel through and make the cell become less positive. While the Na+ travels through the K+ ions are following behind to restore the negative charge. This is so that another action potential can be reached at another impulse. Most synapses on a neuron are located on its dendrites or cell body, whereas action potentials are generally initiated at the axon hillock. A single EPSP is usually too small to trigger an action potential in a postsynaptic neuron. However, if two EPSPs occur in rapid succession at a single synapse, the second EPSP may begin before the postsynaptic neurons membrane potential has returned to the resting potential. The EPSPs add together, an effect called temporal summation. EPSPs produced nearly simultaneously by different synapses on the same postsynaptic neuron can also add together, an effect called spatial summation. Through these summations several EPSPs can depolarize the membrane at the axon hillock to the threshold, causing the postsynaptic neuron to produce an action potential.

41. Answer Key-L.O. 3.43 (D)Multiple choice: •A women gets in a car accident and suffers from nerve damage in her right arm. After analyzing the injuries, the doctor concluded that a dendrite in her nerve cells  has been severed. What will  most likely be the  effect of this damage? •A. There will  be no interruption of her nerve signaling and she will have full range of motion in her arm. •B. There will be no longer term effects; the dendrite will  self-heal  after 1-2 weeks and she will regain full motion •C. She will  permanently lose motion in her entire upper body; damage to one dendrite will interrupt signaling in her upper body region •D. She will lose motion in her right arm, the damage to the dendrite will make it so that signals will be unable to travel through the right arm’s nervous system •E. She will lose motion in her right arm; the damage to the dendrite will have a direct effect on the axon of nerve cells  in her right arm causing her nerve cells to be able to signal to other nerve cells temporarily FRQ- When the you see the ball coming towards you, this signal is picked up by sensory cells, and turned into a nerve impulse. The nerve impulse is picked up through the dendrite of a neuron and travels through neuron and then is passed to the next neuron through the axon of the neuron and then travels through the axon terminal to the dendrite of another neuron. The signal continues to pass through the nervous system until it reaches the brain where it’s processed. From there the brain signals for the appropriate response, which in this case would be to catch the ball.

42. ANSWER KEY –LO 3.44 M.C. Question: Action potentials are normally carried in one direction from the axon hillock to the axon terminals. By using an electron probe, you experimentally depolarize the middle of the axon to threshold. What do you expect? • No action potential will be initiated. • An action potential will be initiated and proceed in the normal direction toward the axon terminal. • An action potential will be initiated and proceed back toward the axon hillock. • Two action potentials will be initiated, one going toward the axon terminal and one going back toward the hillock. • An action potential will be initiated, but it will die out before it reaches the axon terminal. Explanation of MC Question: By starting the signal in the middle of resting potential, you can make it move in either direction, as opposed to moving from threshold potential to resting potential, which is what naturally occurs. Learning Log/FRQ-Style Question:Ouabain, a plant substance used in some cultures to poison hunting arrows, disables the sodium-potassium pump. What change in the resting membrane potential would you expect to see if you treated a neuron with ouabain? Why? If the potassium-sodium pump is not activated, the poisoned neuron would be unable to pump out Na+, so the neuron would remain positive and never repolarize to -70 mV. The gradients of sodium and potassium would gradually disappear, resulting in a greatly reduced resting potential. Without the flow of these particles, action potential cannon be reached.

43. ANSWER KEY – LO 3.45 Which of the following is the correct order of sequence? A) Resting state – depolarization – Na+ influx – K+ efflux B) Na+ influx – depolarization – K+ efflux – resting state C) Resting state – K+ efflux – Na+ influx – deploraization D) K+ efflux – resting state – Na+ influx – depolarization Describe in details what happens between a presynaptic cell and a postsynaptic cell using the following terms: voltage-gated Ca2+ channel, Ca2+, synaptic vesicles, ligand-gated ion channels, and neurotransmitters. When an action potential depolarizes the plasma membrane of the end of the synaptic cell, it opens voltage-gated Ca2+ channels in the membrane, which causes the Ca2+ to enter the cell. Therefore, the Ca2+ concentration rises, causing synaptic vesicles to fuse with the presynaptic cell. These vesicles release neurotransmitter into the synaptic cleft, which bind to the ligand-gated ion channels in the post synaptic membrane and open the channels. Na+ and K+ diffuse through the channels , and the neurotransmitters are released.

44. Answer Key – LO 3.46 If the axon of a neuron was to be severed, what would most likely be the effect? A) The neuron would be unable to receive information from another neuron. B) The neuron would be unable to transmit information to another neuron. C) It would have no effect, depending on which side of the axon that was severed. D) The myelin sheath would break and the transmittance would slow down. It’s Halloween night and you go to a haunted house. While you are there, a masked character runs out and scares you. Without thinking, you punch the character in the face. This is considered a type of reflex. How does your brain integrate the information of fright and produce a response? How is this response created through the pathway of neurons and neurotransmitters? Be specific. When being frightened, the information is immediately sent to the brain to be processed. The signal of fright, or flight-or-fight in the sympathetic nervous system is activated by the hypothalamus and is first received by dendrites extending from the cell body of the neuron. The signal is then generated at the axon hillock and transmits down the axon. At the synaptic terminals, the information is passed from the presynaptic neurons to the postsynaptic neurons. This happens by means of chemical messengers called neurotransmitters. The neurotransmitter epinephrine (adrenaline), in this case, is released by the medulla and goes from the presynaptic neurons and binds to receptors in the postsynaptic neurons. The sensory neurons convey the information to the spinal cord. They then communicate with interneurons that signal your motor neurons to move your arm and hit the character.

45. ANSWER KEY – LO 3.47 Why are action potentials usually conducted in only one direction along an axon? The nodes of Ranvier can conduct potentials in only one direction. The brief refractory period prevents reopening of voltage-gated Na+ channels. The axon hillock has a higher membrane potential than the terminals of the axon. Ions can flow in only one direction Answer: If the calcium ions that surround a neuron were removed, the there would cease to be any communication between neurons. This is because when action potential depolarizes, the Ca+2 ions flood the membrane to cause the synaptic vessels that hold on to neurotransmitters to rise with the membrane potential, thus releasing the signal to bind with the ligand-gated ion channels in a postsynaptic membrane. Without Ca+2, this would not happen. Question: If all the Ca+2 in the fluid surrounding a neuron were removed, how would this affect the transmission of information within and between neurons? Why?

46. ANSWER KEY- LO 3.48 M.C. Question: Suppose you were to draw an image of a neuron that had just received an electrical impulse. Which of the following would not be included in your drawing? Why? Neurotransmitters, because they are not involved in nerve signaling. Nothing. It is a chemical signal that travels down a neuron, not an electrical one. ATP synthase, because neurons are not directly responsible for energy production. Mitochondrion, because neurons do not have the organelles that regular cells have. Explain how a gazelle is able to run when being chased by a cheetah in terms of the nervous system response, and draw and label a visual representation of a nerve cell involved in the response. (sample drawing) When the gazelle sees the cheetah, signals received by the Eye will travel to the brain via sensory neurons. Electrical impulses will travel from neuron to neuron, starting at the dendrites, moving down the myelinated axon as an action potential occurs (the depolarization of an axon as Na+ and K+ gates are opened), and ending at the synapse. Once at the synapse, neurotransmitters are moved by vesicles across the synaptic gap to the next neuron. Once processed and integrated within the brain, a motor output occurs via the activation of muscle cells by the brain, allowing the gazelle to run. In addition to the running, the parasympathetic nervous system will be activated, causing the gazelle to be ready for flight, triggering response such as dilated pupils, an increased heart rate, and slowed digestion.

47. Answer key – LO 3.49 Transmission across a synapse is achieved mainly by adiffusion of neurotransmitter across the cleft which is triggered specifically by?a)The depolarization of membraneb)The influx of Ca+ into the axon of sending neuronc)The initial resting potential of the neurond)The insulation of axons by Schwann cells Determine the function(s) of CA2+ in the transmission of signals between neurons. How would the removal of CA+ from the synaptic cleft affect the way the signals are transmitted? • CA2+ is a vital element that continues the transmission of signals by entering the axon of the sending neuron, and causing the vesicles in the axon to fuse with the plasma membrane. The release of the vesicles that contain neurotransmitters allows the neurotransmitters to bind to receptors on the receiving dendrite. Finally, the receptor sites open the ion channels. If Ca2+ channels are blocked, neurotransmitter release is inhibited . If the neurotransmitters are inhibited, then information cant be sent from one neuron to the other

48. ANSWER KEY- LO 3.50 M.C. Question: Which cells are involved after the information is transferred to the CNS? A) The sensory cells are sent by the sensory neurons to transfer the information to be integrated by the interneurons B) The sensory cells are sent by the sensory neurons to stimulate muscles after information is processed by interneurons C) The effector cells are sent by the motor neurons to transfer the information to be integrated by the interneurons D) The effector cells are sent by the motor neurons to stimulate muscles after the information has been processed by the interneurons Suppose that a doctor’s patient got into an accident in which her sensory neurons were damaged. How would the entire integration process be effected? Would a response still be elicited from a stimulus? Sensory neurons are the first step in the three part nervous system process in integrating information to prompt a response. The sensory neurons are stimulated by the body’s sensors such as light, sound, touch, hear, or smell. Sensory neurons are also triggered by internal changes such as blood pressure or muscle tension. After the sensory neurons are activated, the information is transmitted to the interneurons to be integrated and sent to the motor neurons to elicit a response. Therefore, without the sensory neurons, the interneurons will not be activated by the stimulus and no response would occur.