Answer Key. Unit 3 Molecular Genetics. Answers to Unit 3 Preparation Questions Assessing Student Readiness. (Student textbook pages )

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1 Answer Key Unit 3 Molecular Genetics Answers to Unit 3 Preparation Questions Assessing Student Readiness (Student textbook pages ) 1. Characteristic Prokaryotes Eukaryotes Relative cell size small large Cell number in typical single multiple organism Location of genetic material cytoplasm nucleus Membrane-bond genetic no yes material Number of chromosomes one several 2. c 3. Any three of: they are not cells; they have no cytoplasm, cell membrane, or organelles; they cannot reproduce outside of host cells; they are dormant outside of host cells. 4. c 5. d 6. d 7. a. endoplasmic reticulum (ER) b. Synthesis of proteins and synthesis of lipids and lipid-containing molecules. For example, in the liver, the ER helps detoxify the blood of drugs and alcohol. In the testes and ovaries, the ER produces testosterone and estrogen. 8. c 9. e 10. d 11. Enzymes help facilitate chemical reactions by acting as protein catalysts that increase the rate of the reaction. 12. These indentations (active sites) facilitate substrateenzyme binding as the active site changes in shape to accommodate the substrate in what is called induced fit. 13. Enzymes are classified according to the type of reactions they catalyze. 14. nucleotide nucleic acid gene protein 15. d Nucleotides are the building blocks of nucleic acids, which make up genes, which code for proteins. 16. a. Hydrogen bonds link nucleotides together between the two strands that make up a DNA molecule. These bonds occur between complementary bases. b. The covalent bonds that link adjacent nucleotides together within each strand are called phosphodiester bonds. These bonds occur between the phosphate group on one nucleotide and a hydroxyl group on the sugar of the next nucleotide in the same strand. 17. e 18. An allele is a different form of the same gene. Homologous chromosomes carry two alleles of the same gene. Differences in these alleles account for differences in hair colour. 19. a 20. a 21. e 22. Deletion a piece of a chromosome is cut out Duplication a piece of a chromosome appears twice or more Inversion a piece of a chromosome is flipped Translocation a segment of a chromosome is attached to another chromosome 23. a. prophase I b. crossing over c. During crossing over, the chemical bonds that hold the DNA together in the chromosome are broken and reformed. In some cases, the chromosomes do not reform correctly. 24. The chromosome errors described result in genetic disorders when they delete, alter, or duplicate genes. In some case, however, the errors affect regions of the genome that lack genes or are within non-coding regions of a gene. These errors are not harmful. 25. d Biology 12 Answer Key Unit 3 MHR TR 1

2 26. Sample answer: Genetic engineering can create transgenic organisms that secrete a human protein for medical use. First, the human gene that codes for the protein is injected into an egg from a donor goat. Then the egg is placed in a host goat where a transgenic goat develops. Finally, this transgenic goat produces milk containing the human hormone. 27. c 28. Viruses enter host cells and direct the activity of their host cell s DNA. This makes viruses useful tools for producing a copy of a gene. Researchers can insert the gene of interest into the virus genome. The virus then directs the host cell to make multiple copies of the virus; each new virus that the cell produces will contain the gene of interest as well. 29. a. transgenic or genetically modified organism b. It has a different genotype, as it now contains four new genes c. It has a different phenotype, as it now displays higher levels of iron, vitamin A, sulfur, and a new enzyme. Chapter 5 The Structure and Function of DNA Answers to Learning Check Questions (Student textbook page 207) 1. Genetic material must contain information that regulates the production of proteins. It also must be able to accurately replicate itself to maintain continuity in future generations. Genetic material must allow for some mutations so that there is variation within a species. 2. Griffith used two forms of S. pneumoniae: a pathogenic S-strain and a non-pathogenic R-strain. After injecting mice with a mixture of heat-killed S-strain and live R-strain, the mice died. Griffith concluded that something from the heat-killed S-strain transferred to the R-strain to transform it into a pathogenic form. 3. When they treated heat-killed pathogenic bacteria with a protein-destroying enzyme, transformation still occurred. When they treated heat-killed pathogenic bacteria with a DNA-destroying enzyme, transformation did not occur. These results provided strong evidence for DNA s role in transformation. 4. Two different radioactive isotopes were used to trace each type of molecule. One sample of T2 virus was tagged with radioactive phosphorus ( 32 P), since phosphorus is present in DNA and not protein. The other sample of T2 virus was tagged with radioactive sulfur ( 35 S), since sulfur is only found in the protein coat of the capsid. 5. The independent variable in the experiment was the type of radioactive isotope used to tag the virus. The dependent variable in the experiment was the presence of radioactivity inside the infected bacterial cells. Controls include the usage of the same type of virus in both experiments and the usage of the same protocol for infecting bacterial cells in both experiments. 6. Bacterial cells that are infected by viruses with 32 P-labelled DNA would not be radioactive. Bacterial cells infected by viruses with 35 S-labelled capsid proteins would be radioactive. (Student textbook page 212) 7. Answers should resemble Figure 5.4 on page 208 of the student textbook, with labels for phosphate group, sugar group, and nitrogen-containing base. 8. Nucleotides in DNA have a deoxyribose sugar, while nucleotides in RNA have a ribose sugar with a hydroxyl group at carbon 2. In addition to the sugar group, each nucleotide is attached to a phosphate group and a base. The bases are adenine, cytosine, guanine, and thymine in the case of DNA, and adenine, cytosine, guanine, and uracil in the case of RNA. 9. Chargaff s rule states that, in the DNA nucleotides, the amount of adenine will be more or less equal to the amount of thymine, and the amount of guanine will be equal to the amount of cytosine. The number of A-T nucleotides will not necessarily equal the number of C-G nucleotides. This overturned Levene s earlier hypothesis that the nucleotides occurred in equal amounts and were present in a constant and repeated sequence. 10. Franklin used X-ray photography to analyze the structure of DNA. Her observations provided evidence that DNA has a helical structure with two regularly repeating patterns. She also concluded that the nitrogenous bases were located on the inside of the helical structure, and the sugar-phosphate backbone was located on the outside, facing toward the watery nucleus of the cell. Pauling s methods of assembling three-dimensional models of compounds led to the discovery that many proteins had a helical structure. Watson and Crick also used this information to propose that DNA had a helix shape. 2 MHR TR Biology 12 Answer Key Unit 3

3 11. Diagrams should resemble Figure 5.7B on page 213 of the student textbook. Base pairing and directionality of strands should be shown. 12. Nucleic acids are soluble in water. Therefore, the nitrogenous bases, which are somewhat hydrophobic, must be positioned away from the water found in the nucleoplasm, and the polar phosphate groups (which are hydrophilic) must be on the outside of the molecule, interacting with the water. (Student textbook page 222) 13. The main objective of DNA replication is to produce two identical DNA molecules from a parent DNA molecule. 14. DNA replication occurs during the S phase of interphase, prior to cell division, ensuring that there is a copy available for each new daughter cell. 15. Conservative model Two new daughter strands form to create a new double helix, and the original DNA strands re-form into the parent molecule. Semi-conservative model Each new DNA molecule contains one strand of the original DNA and one newly synthesized strand. Dispersive model Parental DNA is broken into fragments. Therefore, the daughter DNA contains a mix of parental and newly synthesized DNA. 16. Nitrogen is a component of DNA and is incorporated into newly synthesized daughter strands. Having a light form ( 14 N) and a heavy form ( 15 N) allowed the separation of different DNA strands based on the amount of isotope present in the newly synthesized DNA. DNA with more 15 N would be denser than DNA with 14 N, and could therefore be separated by centrifugation. 17. Meselson and Stahl concluded that DNA replication is semi-conservative. After one round of replication, DNA appeared as a single band, midway between the expected positions of 15 N-labelled DNA and 14 N-labelled DNA. After the second round of replication, DNA appeared as two bands, with one band corresponding to 14 N-labelled DNA and the other band in the position of hybrid DNA (half 14 N and half 15 N). In additional rounds of replication, the same two bands were observed, therefore supporting the semi-conservative model. 18. Each new cell that is produced must have an exact copy of parental DNA. The daughter strands of DNA are part of a DNA molecule that will be in the daughter cells. This ensures that newly born cells are similar to parents and maintain their genetic identity. (Student textbook page 227) 19. Initiation Helicase enzymes unwind DNA to separate it into two strands. A replication bubble is formed when single-strand binding proteins stabilize the separated strands. Elongation New DNA strands are synthesized by joining free nucleotides together. This is catalyzed by DNA polymerase, which synthesizes the new strands that are complementary to the parental strand. Termination The two new DNA molecules separate from one another. 20. Replication takes place in a slightly different way on each DNA strand because DNA polymerase can only catalyze elongation in the 5 to 3 direction. In order for both strands of DNA to be synthesized simultaneously, the method of replication must differ. 21. On the leading strand, DNA synthesis takes place along the DNA molecule in the same direction as the movement of the replication fork. On the lagging strand, DNA synthesis proceeds in the opposite direction to the movement of the replication fork. The lagging strand is synthesized in short fragments called Okazaki fragments. 22. DNA replication requires the use of many enzymes that have specific roles. The presence of numerous specialized enzymes may reflect the importance of having accurate DNA replication, since mutations in DNA can change the genetic makeup of an organism. 23. Answers may include: DNA polymerases have a proofreading function during which they excise incorrect bases and add the correct bases. Mismatch repair involves a group of enzymes that identify, remove, and replace incorrect bases. 24. Many tissues and organs require continuous cell regeneration. Therefore, DNA replication must be quick and accurate so that new daughter cells receive exact copies of DNA from the parent cell. Answers to Caption Questions Figure 5.2 (Student textbook page 205): If a live strain had been transferred, the effects would have been due to that strain, not due to the transfer of a substance form it to the R-strain, which makes it pathogenic. Figure 5.3 (Student textbook page 207): The results would have also shown that protein was not the hereditary material. However, it would not have directly demonstrated the role of DNA as the hereditary material since RNA also contains phosphorus. Biology 12 Answer Key Unit 3 MHR TR 3

4 Figure 5.10 (Student textbook page 215): Twisting a rubber band around itself mimics how DNA supercoiling. The rubber band becomes compacted due to the coils that twisting forms. This model is also useful since it demonstrates the tension that is created by supercoiling. A rubber band may become linearized, where supercoiling in bacterial DNA occurs because it is a circular. The rubber band model also does not reflect the double-stranded nature of DNA. Figure 5.16 (Student textbook page 221): If DNA had not been uniformly labelled with 15 N, the banding patterns would not accurately reflect the presence of parental DNA. Answers to Section 5.1 Review Questions (Student textbook page 218) 1. Griffith s experiments showed the existence of a transforming principle. That is, something in the heatkilled pathogenic bacteria (S-strain) could transform the non-pathogenic bacteria (R-strain) into a pathogenic form. This result led to Avery s experiments on Streptococcus pneumoniae to identify the molecules that caused this transformation. Avery s research concluded that DNA was the transforming principle. 2. a. Graphic organizers should include information about experimental setup (i.e., the use of two radioactive isotopes to differentially label DNA and capsid protein), experimental procedure (i.e., the use of agitation in a blender to dislodge viruses, and subsequent centrifugation), and results. A summary of the experiment is found in Figure 5.3 on page 207 of the student textbook. b. The results showed that DNA is the hereditary material. 3. Miescher isolated nuclein from the nucleus of white blood cells. He found that this material was present only in the nuclei of cells. Further experimentation showed that nuclein was a weakly acidic phosphoruscontaining substance. Nuclein would later be known as nucleic acid or, more specifically, DNA (deoxyribonucleic acid). 4. Diagrams should resemble the marginal portion of Figure 5.4 on page 208 of the student textbook. 5. The nucleotide composition of the human would be different from the nucleotide composition of the mouse because the composition of DNA is unique to each species. However, the percentage of adenine will remain approximately the same as the percentage of thymine, and the percentage of cytosine will remain approximately equal to the percentage of guanine in each species. 6. C = 26%; G = 26%; T = 24% 7. Diagrams should include labels for sugar-phosphate molecules ( handrails ), nucleotide base pairing ( rungs ), and directionality of both strands and resemble the close up of Figure 5.7 on page 213 of the student textbook. 8. a. Levene proposed that DNA was composed of nucleotides, and that each of the four types of nucleotides contained one of four nitrogen-containing bases, a sugar molecule, and a phosphate group. b. Chargaff showed that DNA is composed of repeating units of nucleotides in fixed proportions (i.e., the percent composition is of adenine is the same as thymine, and the percent composition of cytosine is the same as guanine). Chargaff s rule helped Watson and Crick infer that adenine paired with thymine, and cytosine paired with guanine. c. Franklin determined that DNA had a helical structure, with nitrogenous bases located on the inside of the structure, and the sugar-phosphate backbone located on the outside. This information led to Watson and Crick s ladder-like double helix model of DNA, with the sugar-phosphate molecules acting as handrails and the bases making up the rungs. d. Pauling discovered that proteins have a helical structure. This discovery influenced Watson and Crick to propose that DNA was shaped like a helix. 9. a. 5 -ATTGAACAT-3 b. 5 -GATTAACGG-3 c. 5 -CGGAGCTAA A gene is a functional unit of DNA. It is a specific sequence that encodes for proteins or RNA molecules. A genome is an organism s complete genetic makeup. It is composed of an organism s total DNA sequence. 11. Venn diagrams should include: Prokaryotes Only double-stranded, circular DNA packed in the nucleoid; DNA is compacted via supercoiling; most are haploid; genomes contain very little non-essential DNA; contain plasmids Prokaryotes and Eukaryotes chromosomal DNA is much larger than their cells, and therefore must be compacted Eukaryotes Only total amount of DNA is much greater than in prokaryotes, and they therefore have greater compacting and levels of organization (i.e., nucleosomes, chromatin, chromosomes); doublestranded linear DNA is contained in the nucleus; most are diploid; genomes can vary widely in size and complexity (i.e., some have large non-coding regions) 4 MHR TR Biology 12 Answer Key Unit 3

5 12. Diagrams should include the DNA molecule winding around histones to form nucleosomes, which are connected to each other by DNA and may resemble Figure 5.12 on page 216 of the student textbook. 13. a. There is no set relationship between the complexity of an organism (number of genes in an organism) and the total size of its genome. An organism may have an enormous number of base pairs in its genome and very few genes if the bulk of its genome consists of non-coding DNA. b. Comparing the genomes of the two organisms would show what genes they have in common, and would indicate their evolutionary relationship how closely or distantly related they are. 14. A mutation in a protein-coding region would not necessarily be more detrimental than a mutation in a non-coding region since the latter may contain regulatory sequences (i.e., regions that can influence the production of proteins and RNA molecules). In addition, since multiple codons exist for a given amino acid, a mutation in the protein-coding region of DNA may not alter the protein sequence. Answers to Section 5.2 Review Questions (Student textbook page 229) 1. Daughter cells must have the same genetic information as parent cells. 2. a. Flowcharts should include a summary of experimental steps and results shown in Figure 5.16 on page 221 of the student textbook. b. If DNA replication was conservative, only two bands would appear following one round of replication: one band of 14 N-only DNA (newly synthesized DNA), and one band of 15 N-only DNA (old parental DNA). Diagrams should show two distinct bands, one labelled light ( 14 N) and the other labelled heavy ( 15 N). 3. Initiation Replication begins at the replication origin. Helicases bind to the DNA at each replication origin. The helicases cleave and unravel a section of the original double helix, creating Y-shaped areas (replication forks) at the end of the unwound areas, which form a replication bubble. Single-strand binding proteins stabilize the separated strands. These single strands serve as templates for the semiconservative replication of DNA. Elongation New DNA strands are produced when DNA polymerase inserts into the replication bubble. A primase synthesizes an RNA primer that serves as the starting point of new nucleotide attachment by DNA polymerase. DNA polymerase can only synthesize the new nucleotide chain in the 5 to 3 direction. As a result, one strand (the leading strand) is replicated continuously in the 5 to 3 direction, in the same direction that the replication fork is moving. The other strand, known as the lagging strand, is replicated in short segments, still in the 5 to 3 direction, but away from the replication fork. These fragments, called Okazaki fragments, are joined together by DNA ligase. Termination When replication is complete, the two new DNA molecules separate from one another and the replication machine is dismantled. Each new molecule of DNA contains one parent strand and one new strand. 4. Early development is a very rapid process, and many key molecules are produced during this time. Therefore, it is expected that more replication origins would be present in developing embryo cells. 5. A replication bubble is formed as the DNA double helix unwinds during initiation. The replication forks are the Y-shaped regions of the replication bubble, and move along the DNA in opposite directions as replication proceeds. Diagrams should resemble the third portion of Figure 5.17 on page 223 of the student textbook, with labels on the replication bubble, replication fork, and the double-headed arrow showing the direction(s) of unwinding. 6. Diagrams should illustrate continuous DNA synthesis on the leading strand and discontinuous DNA synthesis on the lagging strand. Diagrams may resemble a simplified version of Figure 5.19 on page 224 of the student textbook with labels for leading strand, lagging strand, Okazaki fragments, RNA primer, DNA polymerase, DNA ligase, parent DNA, and directionality of strands. 7. An RNA primer is necessary for DNA synthesis on the lagging strand. The primer provides a free 3 -hydroxyl end, which DNA polymerase can extend by adding new nucleotides. 8. DNA polymerase adds new nucleotides to the 3 end of a growing chain during replication. DNA polymerase also proofreads newly formed base pairs and replaces any nucleotides that have been incorrectly added. 9. a. No RNA primer would be synthesized. Therefore, synthesis of the lagging strand cannot be initiated. b. Okazaki fragments on the lagging strand cannot be joined together. c. Unwinding of the DNA double helix during initiation would not occur. Biology 12 Answer Key Unit 3 MHR TR 5

6 10. A The semi-conservative model states that each new molecule of DNA would contain one strand of original parent DNA and one new strand of daughter DNA. B The dispersive model states that new molecules of DNA would be hybrids containing a mixture of old parent DNA and new daughter DNA strands. C The conservative model states that one molecule of DNA would contain two new daughter DNA strands, and the other molecule of DNA would contain the original parent DNA strands. 11. Graphic organizers should include: Prokaryotes Only rate of replication is faster compared to eukaryotes; five DNA polymerases have been identified in prokaryotes; circular chromosome of prokaryotes have a single replication origin Prokaryotes and Eukaryotes require replication origins; have 5 3 elongation; have continuous synthesis on the leading strand, and discontinuous synthesis on the lagging strand; require a primer for Okazaki fragments on the lagging strand; require the use of DNA polymerase enzymes Eukaryotes Only rate of replication is slower compared to prokaryotes due to complicated enzymes complexes and proofreading mechanisms; 13 DNA polymerase enzymes have been identified in eukaryotes; linear chromosome has multiple replication origins; presence of telomeres due to linear nature of eukaryotic chromosome Answers to Chapter 5 Review Questions (Student textbook pages 235 9) 1. d 2. b 3. e 4. b 5. c 6. c 7. b 8. d 9. b 10. a 11. d 12. d 13. a 14. a 15. a. While studying DNA in the early 1900s, Phoebus Levene reported that the nucleotides were present in equal amounts, and that they appeared in chains in a constant and repeated sequence of nitrogen bases. Therefore, most scientists thought that the great variety of proteins was an important factor, and must be the hereditary material. Scientists assumed that the molecular structure of DNA was just too simple to provide the great variation in inherited traits. b. Oswald Avery, Colin MacLeod, and Maclyn McCarty conducted a series of experiments and discovered: When they treated heat-killed pathogenic bacteria with a protein-destroying enzyme, transformation still occurred. When they treated heat-killed pathogenic bacteria with a DNA-destroying enzyme, transformation did not occur. These results provided evidence that genetic information was carried on DNA. 16. Franklin s X-ray diffraction images showed that DNA had a helical structure, with two regularly repeating patterns. She also concluded that the sugar-phosphate backbone was located on the outside, and the nitrogencontaining bases protruded inward. Watson and Crick used these observations to construct the threedimensional model of DNA. 17. a. Each body cell produces two daughter cells from itself, and one of the two strands could go to each daughter cell. b. With bases on the outside, the DNA would not be uniform width throughout, which all evidence indicated. Also, the weak hydrogen bonds between the nitrogen bases could be broken easily. The bonds between the sugar and phosphate portions of the nucleotides are much stronger. c. From Franklin s X-ray photographs, they reasoned that DNA was twisted into a spiral, or helix. Since the spiral consisted of two strands wound around each other, they called it a double helix. 18. Eukaryotic DNA is compacted in the nucleus through different levels of organization. DNA associates with histones to form nucleosomes. It can be further compacted by the coiling of nucleosomes to produce 30 nm fibres. Additional compacting is achieved through the formation of loop domains of the 30 nm fibre on a protein scaffold. This scaffold can condense further through folding. 19. Purines, such as adenine and guanine, have a doublering structure. Pyrimidines, such as thymine and cystosine, have a single-ring structure. 6 MHR TR Biology 12 Answer Key Unit 3

7 20. Hydrogen bonds between the bases hold the two strands of DNA together. 21. The two strands of a DNA molecule are antiparallel since each strand has directionality. At each end of the DNA molecule, the 5 end of one strand is across from the 3 end of the complementary strand. 22. Since nitrogen is a component of DNA, it would be incorporated into newly synthesized strands of DNA. Two different isotopes of nitrogen were used to distinguish between the original parental strand and the newly synthesized daughter strand. Furthermore, having a light form ( 14 N) and a heavy form ( 15 N) of nitrogen allowed the separation of different DNA strands based on the amount of isotope present in the newly synthesized DNA. DNA with more 15 N would be denser than DNA with 14 N, and therefore could be separated by centrifuge and visualized. 23. Producing exact copies ensures that when a cell divides, the offspring cells will receive the same genetic information as the parent cell. 24. An RNA primer is required for discontinuous synthesis of DNA on the lagging strand. The RNA primer provides a free 3 hydroxyl end from which DNA polymerase can add nucleotides. 25. The replication machine consists of the complex of proteins and DNA that interact at the replication fork. These proteins include DNA polymerase, an enzyme that joins nucleotides together to create a complementary strand of DNA (elongation); DNA ligase, an enzyme that joins Okazaki fragments together; primase, an enzyme that constructs the RNA primer needed for replication to begin; helicases, a group of enzymes that cleave and unravel a segment of the double helix to enable replication; and single-strand binding proteins, which help stabilize the unwound strands. 26. Having multiple origins of replication increases the speed of replication. Instead of starting at one end and finish at another, having multiple start points increases the efficiency of replication. 27. DNA polymerase has a proofreading mechanism, which identifies and excises an incorrect nucleotide, and then inserts the correct nucleotide. Another mechanism of error correction is mismatch repair, where a group of enzymes can detect deformities in the newly synthesized strand of DNA caused by mispairing of nucleotides. This group of enzymes excises the mispaired nucleotides and inserts the correctly paired nucleotide. Errors that are not corrected before cell division become mutations in the genome, which are passed on to daughter cells once cell division occurs. 28. Telomeres are present to ensure that important genetic information is not lost during replication of linear eukaryotic DNA. 29. Nucleotides can come fully-formed from a variety of food sources. Nucleotides can also be formed by our bodies using components that come from our diet (i.e., sugar, phosphates, nitrogen). 30. Different tissues all develop from the same fertilized egg cell (zygote). While the tissues have the same genes, only those genes necessary for a specific tissue s functions are active. 31. Sample answer: There are some similarities. However, DNA words are limited to sequences of amino acids. Each section of code has only one meaning, resulting in one specific protein. This does not compare to the arrangement of letters in a language, which results in words that can have a great variety of meanings. 32. a. 5 -GATGTACAG-3 b. 5 -ATCAGCGAT-3 c. 5 -AATACGCCG G = 16%, T = 34% 34. Sample B is the viral DNA because the percentages of adenine and thymine are not the same. Similarly, the percentages of guanine and cytosine are not the same, as they are in sample A, which shows complementary base pairing of these respective bases. Complementary base pairing does not occur in a single-stranded DNA virus. 35. a. Sample answer: The small fragments could be Okazaki fragments that were not joined together properly. This could be due to a lack of ligase, a mutant ligase, or a mutation in the ligase gene that produces a mutant ligase enzyme. b. Experimental design should include the experimental setup, proper controls (i.e., one reaction tube with cells cultured without ligase, and another with ligase), and expected results. 36. A similar base composition does not necessarily mean that the DNA sequences are similar (i.e., the order of the nucleotides in the DNA sequence are not necessarily the same in both organisms). 37. Linker DNA is responsible for joining nucleosomes together. Micrococcal nuclease preferentially cuts linker DNA, which would lead to disrupted formation of 30 nm fibres. Diagrams should include an illustration of the beads on a string appearance Biology 12 Answer Key Unit 3 MHR TR 7

8 of eukaryotic DNA organization with labels for DNA, histones, nucleosomes, and linker DNA (refer to Figure 5.12 on page 216 of the student textbook). An accompanying diagram should illustrate micrococcal nuclease cutting linker DNA between nucleosomes. 38. Answers and diagrams should include: If replication was continuous in a 5 to 3 direction, the two DNA strands would not be antiparallel and would instead be parallel to each other. If continuous replication was able to occur in both 5 to 3 and 3 to 5 directions, then primase, RNA primers, Okazaki fragments, and DNA ligase would not be necessary for synthesis on the lagging strand. Additionally, telomeres would not be necessary since there would be complete synthesis of the lagging strand of eukaryotic DNA. 39. The weak hydrogen bonds in DNA break easily, making it easier for the two strands in the molecule to separate during replication. The strong covalent bonds ensure that the sequence of nucleotides remains fixed in each strand. 40. Radioactive phosphorus ( 32 P) would label newlysynthesized DNA strands. Diagrams should resemble the semi-conservative model of replication shown in Figure 5.15 on page 220 of the student textbook, where the 32 P-labelled strands would correspond to the newly synthesized daughter strands. After two rounds of replication, both strands of new DNA double helix should incorporate 32 P UGACU DNA polymerases are expected to be more active in continuously dividing skin cells, since DNA replication would occur more often. Heart muscle cells do not divide as frequently as skin cells, and therefore DNA replication (and DNA polymerase activity) occurs less often. 43. Answers could include targeting bacterial-specific topoisomerases. These enzymes are essential for supercoiling and DNA replication, which are both required for the survival of bacterial cells. Since this is a bacteria-specific target, the side effects of this drug on eukaryotic cells could be reduced. 44. A defect in DNA helicase could result in delayed unwinding of the DNA double helix. Therefore, the initiation of DNA replication would not occur efficiently, which could lead to DNA instability and cell death. 45. Primase would no longer be needed to synthesize a RNA primer on the lagging strand, since a 3 hydroxyl end would already exist. Ligase would not be necessary since Okazaki fragments would not be synthesized on the lagging strand. 46. Concept maps should include information on two or more experiments that addressed a similar hypothesis (i.e., the transforming principle experiments performed by Griffith and Avery). The experimental approaches and results for each experiment should be outlined in the concept map, in addition to how the different experiments are related to one another (i.e., did the result from one experiment help direct subsequent experiments?). Using a variety of experimental approaches provides confidence and confirmation of results. 47. Timelines should be a chronological order of contributions from the scientists presented in the chapter: Miescher; Levene; Griffith; Avery, MacLeod, and McCarty; Hershey and Chase; Chargaff; Pauling; Franklin; Watson and Crick. 48. Sample diagram: 3 end S P 5 end S G C P P S S G C P P S S A T P P S S C G P P S S T A P P S S A T P 5 end hydrogen bonding 3 end 49. Graphic organizers should include: Prokaryotes Only rate of replication is faster compared to eukaryotes; five DNA polymerases have been identified in prokaryotes; circular chromosome of prokaryotes have a single replication origin. Prokaryotes and Eukaryotes require replication origins; have 5-3 elongation; have continuous synthesis on the leading strand, and discontinuous synthesis on the lagging strand; require a primer for Okazaki fragments on the lagging strand; require the use of DNA polymerase enzymes Eukaryotes Only rate of replication is slower compared to prokaryotes due to complicated enzymes complexes and proofreading mechanisms; 13 DNA polymerase enzymes have been identified in eukaryotes; linear chromosome has multiple replication origins; presence of telomeres due to linear nature of eukaryotic chromosome 8 MHR TR Biology 12 Answer Key Unit 3

9 50. Diagrams should resemble the bottom illustration in Figure 5.16 on page 221 of the student textbook, with the first set of parental and new strands labelled first round, and the next set labelled second round. 51. Flowcharts should include: Initiation Helicase enzymes unwind the DNA double helix to separate it into two strands. A replication bubble and replication forks are formed when single-strand binding proteins stabilize the separated strands. Topoisomerase enzymes help to relieve the strain on DNA caused by unwinding. Elongation New DNA strands are synthesized by joining free nucleotides together. This is catalyzed by DNA polymerase, which synthesizes the new strands that are complementary to the parental strand. Synthesis of the new DNA strands occurs continuously on the leading strand, and discontinuously on the lagging strand. Termination The two new DNA molecules, each composed of one parental strand and one new daughter strand, separate from one another. 52. Diagrams should resemble Figure 5.17 on page 223 of the student textbook, with labels for replication bubble, replication forks (each end of the bubble), and the double-headed arrow showing the direction(s) of unwinding. 53. DNA can only be synthesized in the 5 to 3 direction. Diagram should therefore illustrate continuous DNA synthesis on the leading strand and discontinuous DNA synthesis on the lagging strand. Diagram may resemble a simplified version of Figure 5.19 on page 224 of the student textbook with labels for leading strand, lagging strand, Okazaki fragments, RNA primer, DNA polymerase, DNA ligase, parent DNA, and directionality of strands. 54. See Table below. 55. Journal entries should be written in the first person and summarize the knowledge of heredity related to his or her research, using scientific terminology. Challenges could include a lack of research facilities, lack of financial support, the shortcomings of the current technology, lack of consensus among research colleagues, or lack of support for the work in the broader scientific community. Thoughts of future significance of the research may reflect later discoveries. 56. Articles should summarize Watson and Crick s findings, including their model of DNA and include how this discovery would affect the general public (i.e., advances in medicine and genetics, ethical issues and concerns). The article should be written in language that is aimed at the general public and is appropriate for the time period. 57. Concept maps should illustrate the different levels of organization of eukaryotic DNA. Answers may resemble Figure 5.12 on student textbook page 216, with the addition of a nucleotide label above the DNA molecule, nucleosome at the structure illustrating DNA wrapped around histones, and chromatin on all of the non-condensed forms of genetic material. Question 54 Enzyme Function Absence of Enzyme Helicase Unwinds the double stranded DNA at the replication fork. Other enzymes (below) cannot bind to the DNA because the DNA would remain double-stranded. Topoisomerase II Relieves strain on DNA that is generated from unwinding of the double helix. Unwinding may not occur efficiently. Primase DNA polymerase I, II, and III DNA ligase Synthesizes an RNA primer used to generate Okazaki fragments. A group of enzymes that: Adds new nucleotides to the 3 end of a growing chain. Proofreads the newly formed base pairs and cleaves out any nucleotides that do not fit. Removes ribonucleotides at the 5 end (removes the RNA primer). Joins Okazaki fragments together on the lagging strand. The DNA strands would be open, but synthesis could not begin because DNA polymerase has to have an existing chain with a 3 end to add new nucleotides. Only primer strands would exist on the opened DNA strands. No new DNA would be synthesized. The leading strand would be normal. However, the Okazaki fragments making up the lagging strand would never be joined, and therefore the new DNA would never be complete and functional. Biology 12 Answer Key Unit 3 MHR TR 9

10 58. This would leave replication errors (i.e., mispairing of bases) uncorrected. These errors would then become mutations in the genome, which are then passed onto daughter cells once cell division occurs. 59. The graphic organizer should effectively represent the points outlined in the Chapter 5 summary on page 234 of the student textbook. 60. Answers may include: Human disease Many genes that are implicated in human disease have parallel versions in model organisms (i.e., yeast, mice, fruit flies), where they can be studied easily in various experimental settings. Gene function Studying parallel human genes in model organisms may also provide insight into gene function. Evolutionary biology Comparative genomics also allows researchers to study which regions of a genome have been conserved amongst different species. These conserved regions are thought to be essential and important regions of the genome. Likewise, divergent regions may confer speciesspecific function and contribute to morphological and functional changes. 61. Answers should include specific examples of how the chosen scientist s research would benefit the scientific research community and the general public. Social, legal, and ethical implications may be addressed. The paragraph should be convincing, as this is an example of writing that would be included in a grant application. 62. a. When the last RNA primer from the lagging strand is degraded in linear DNA, the gap that remains is unable to be filled since there is no adjacent fragment where nucleotides can be added. Therefore, the new DNA molecule will be shorter than the parent DNA molecule. Since telomeres are at the ends of eukaryotic chromosomes, they are the sequences which become shorter after each round of replication. b. Telomerase is an enzyme that synthesizes telomeres and replaces sequences that have been lost. In childhood, telomerase activity in cells is high. As people age, the activity of telomerase decreases, which can result in shorter telomeres and therefore shorter chromosomes. This may lead to loss of important coding information. Certain lifestyle factors may influence telomere length based on its effects on telomerase activity. For example, smoking may cause accelerated symptoms of aging due to decreasing telomerase activity. Exercise, on the other hand, may delay the symptoms of aging since it increases telomerase activity. 63. Answers may include: If there was a method to increase telomerase activity, who would have access to fountain of youth technology? Should we be interfering with the natural aging process? Why would certain individuals want to delay aging? What are the possible benefits and risks of being able to delay aging? 64. a. Telomerase activity may be higher in cancer cells since these cells divide rapidly. b. Telomerase is present and active in normal cells. Inhibiting telomerase activity as a potential cancer therapy could therefore cause damage to normal cells (i.e., shortened or lack of telomeres, which would lead to more rapid shortening of chromosomes). 65. a. Students opinions should be supported with examples. b. Issues may addressed include: How would appropriate credit be determined? Should contributions be based on the percentage of work performed for the study or for the analysis? Should primary credit go to the scientist who proposed the hypothesis, or should it go to the scientist who actually carried out the experiment (i.e., did the bench work )? 66. Answers may be based on cultural, religious, or family values. Accept any reasoned argument. 67. Answers may include: DNA sequencing Genetic screening for diseases Therapeutic gene targets Gene expression (RNA and protein) DNA mutations 68. This required a rethinking of many ideas that formed the basis of other ideas. The lower number of genes may mean that the structures function differently than anticipated the number of number genes does not determine an organism s complexity there are a large number and sizes of introns non-coding regions act as regulatory sequences genes work in combinations (groups) to perform diverse function 10 MHR TR Biology 12 Answer Key Unit 3

11 69. In Linus Pauling s model, DNA replication would have to occur without nitrogen base pairing. Accept any well-reasoned answer. 70. Anti-viral drugs specifically target viral DNA polymerase to interfere with viral replication. Specificity is also required to ensure eukaryotic DNA polymerase activity in not adversely affected. 71. a. Because of the shared features/structure of their DNA. b. Graphic organizers (such as a table) should include the advantages and disadvantages of using the chosen organism and specific examples of significant research findings that were obtained using that organism. c. Ethical issues will vary but may include the harm done to the animal either by being kept in captivity or by the experimental process. 72. a. Mispairing of bases. Thymine should be paired with adenine, and cytosine should be paired with guanine. b. Mispairing of bases occurs during replication and may be due to flexibility in the structure of DNA. c. Mismatch repair can correct this error. A group of mismatch repair enzymes recognizes deformities that are caused by mispairing of bases. These enzymes then excise the incorrect nucleotide and insert the correct nucleotide. DNA polymerases can also correct this error by excising the incorrect nucleotide in a newly synthesized strand, and adding the correctly paired nucleotide. Answers to Chapter 5 Self-Assessment Questions (Student textbook pages 240 1) 1. b 2. c 3. e 4. c 5. a 6. e 7. c 8. a 9. e 10. b 11. C = 19%, G = 19%, T = 31% 12. a. Two different radioactive isotopes were used to trace each type of molecule. One sample of T2 virus was tagged with radioactive phosphorus ( 32 P), since phosphorus is present in DNA but not in protein. The other sample of T2 virus was tagged with radioactive sulfur ( 35 S), since sulfur is only found in the protein coat of the capsid. b. In one experiment, Hershey and Chase observed that most of the radioactively labelled viral DNA was in bacteria and not in the liquid medium. In a second experiment, they observed that the radioactively labelled viral capsid protein was in the liquid medium and not in the bacteria. These results demonstrated that viral DNA, not viral protein, enters the bacterial cell. Therefore, DNA is the hereditary material. 13. Franklin s X-ray diffraction images showed that DNA has a helical structure with two regularly repeating patterns. She also concluded that the nitrogenous bases were located on the inside of the helical structure, and the sugar-phosphate backbone was located on the outside, facing toward the watery nucleus of the cell. 14. Diagrams should include labels for sugar-phosphate molecules ( handrails ), nucleotide base pairing ( rungs ), and directionality of both strands and resemble the close up part of Figure 5.7 on page 213 of the student textbook. 15. Prokaryotic DNA is double-stranded, circular, and packed in the nucleoid. It is compacted via supercoiling. Linear eukaryotic DNA is compacted in the nucleus through different levels of organization. DNA associates with histones to form nucleosomes. It can be further compacted by the coiling of nucleosomes to produce 30 nm fibres. Additional compacting is achieved through the formation of loop domains of the 30 nm fibre on a protein scaffold. This scaffold can condense further through folding. The differences are due to structure (circular prokaryotic DNA versus linear eukaryotic DNA) and size. Since the total amount of DNA in eukaryotes is much greater than in prokaryotes, they have greater compacting and levels of organization. 16. Answers should include support for the student s interpretation of the statement. While protein-coding regions include genes which code for proteins, the noncoding regions have regulatory sequences which can influence and regulate the production of proteins and RNA molecules. 17. a. Answers could include identity theft or the identification of an individual s traits (e.g., disease) without consent (such as for job or insurance screening purposes). Biology 12 Answer Key Unit 3 MHR TR 11

12 b. Answers could include forensics, comparative genomics, screening for diseases, or tracing the origin or source of an illness. c. Opinions should be supported with an explanation that includes evidence of scientific understanding of the nature of a DNA sequence. Answers may also consider consent under certain circumstances (i.e., genetic screening for disease or identifying lineage). 18. a. Arrow should indicate movement to the left. b. Okazaki fragment C was made first. Primase synthesizes an RNA primer, which binds to the parental strand of DNA. DNA polymerase III adds new nucleotides to the free 3 -hydroxyl end of the primer. This newly synthesized fragment is an Okazaki fragment. c. Okazaki fragments are necessary for the discontinuous synthesis of the lagging strand during DNA replication. Once the Okazaki fragments are made, DNA polymerase I removes the RNA primer and the Okazaki fragments are joined together by DNA ligase to form a complete lagging strand. 19. DNA has weak hydrogen bonds, which exist between the nucleotides on opposite strands. These weak hydrogen bonds break easily, making it easier for the two strands to separate during replication. The strong covalent bonds that exist on the sugar-phosphate backbone ensure that the sequence of nucleotides remains fixed in each strand, since these bonds are not easily broken. 20. Diagrams should resemble Figure 5.17 on page 223 of the student textbook, with labels identifying the replication bubble, replication forks at each end of the bubble, and the double-headed arrow showing the direction of unwinding. 21. Answers should include the information summarized in Table 5.2 on page 224 of the student textbook, which lists the important proteins involved in DNA replication and their functions (helicase, primase, single-strand-binding protein, topoisomerase II, DNA polymerase I, II, and III, and DNA ligase). 22. A person without DNA polymerase would not exist, because DNA replication could not occur and cells would not be able to divide and survive. 23. Mutations in Mut genes could lead to an inability to correct errors in replication, such as mispairing of bases. Uncorrected errors become mutations in the genome, which are then passed on to daughter cells once cell division occurs. 24. Telomerase activity decreases as we get older, which results in the shortening of telomeres in somatic cells. This means that the age of an organism is reflected in the length of telomeres. 25. Venn diagrams should include: Prokaryotes Only rate of replication is faster; five DNA polymerases identified; circular chromosome of prokaryotes have a single replication origin Prokayotes and Eukaryotes require replication origins; have 5-3 elongation; have continuous synthesis on the leading strand, and discontinuous synthesis on the lagging strand; require a primer for Okazaki fragments on the lagging strand; require the use of DNA polymerase enzymes Eukaryotes Only rate of replication is slower due to complicated enzymes complexes and proofreading mechanisms; 13 DNA polymerase enzymes identified; linear chromosome has multiple replication origins; presence of telomeres due to linear nature of eukaryotic chromosome Chapter 6 Gene Expression Answers to Learning Check Questions (Student textbook page 246) 1. The black urine phenotype shown to be caused by a recessive inheritance factor (gene) that caused that production of a defective enzyme (protein). 2. Answers could use Figure 6.1 on page 245 of the student textbook as a guideline. The results of the Beadle and Tatum experiment showed that a single gene produces one enzyme (one-gene/one-enzyme hypothesis). This was later modified to the one-gene/one-polypeptide hypothesis since not all proteins are enzymes. 3. RNA is found in the nucleus and cytoplasm; the concentration of RNA in the cytoplasm is correlated with protein production; RNA is synthesized in the nucleus and transported to the cytoplasm. 4. Jacob and colleagues saw that bacteria infected with a virus had a newly synthesized virus-specific RNA molecule. This RNA molecule associated with bacterial ribosomes, which are the sites of protein production. Therefore, the RNA molecule carried the genetic information for the production of a viral protein. 5. Having DNA continually transport itself could increase the likelihood of damage to the DNA. Multiple steps in gene expression provide many opportunities for regulation. This allows the cell to have increased control over protein synthesis GAUUAACGG-3 12 MHR TR Biology 12 Answer Key Unit 3