Download Solutions: Chapter 4 and 5 Review Sheet

 REVIEW SOLUTIONS: Molecular Genetics Chapter 4 and 5 Test Chapter 4 questions: Pg. 230 -­‐ #1, 6, 8-­‐10, 13, 14, 16, 19 1. Erwin Chargaff determined the percentage composition of each of the nitrogenous bases in a DNA molecule. He illustrated that the proportion of adenine is equal to the proportion of thymine and that the proportion of guanine is equal to the proportion of cytosine. In addition, he determined that the sum of adenine plus guanine is equal to the sum of thymine plus cytosine. This information led to the incorporation of complementary base pairing in Watson and Crick’s model of DNA. Complementary base pairing explains the nitrogenous base proportions discovered by Chargaff. 6. TTAACGTAT 8. Rosalind Franklin produced an X-­‐ray diffraction pattern of DNA that suggested it was helical in nature. Her findings also indicated that the diameter of a DNA molecule was constant at 2 nm. The helical nature of DNA transpired into two antiparallel DNA strands. The constant diameter supports the complementary base pairing of a purine and pyrimidine throughout the molecule (if pairing included a purine-­‐purine or pyrimidine–pyrimidine combination, the diameter would vary from 2 nm), with a deoxyribose sugar and phosphate backbone. The constant width suggested a repeating pattern. We now know that DNA is a polymer of nucleotides. 9. Both strands of a DNA molecule act as a template during replication. If both new complementary strands were synthesized in the same direction (toward the replication fork), one strand would built in the 5' to 3' direction (the leading strand), while the other complementary strand would have to be built in the 3' to 5' direction (because of the antiparallel nature of DNA strands). However, DNA polymerase III, the enzyme responsible for the synthesis of new complementary strands, can only build in the 5' to 3' direction. To deal with this problem a mechanism has evolved where one strand, the lagging strand, is built in small segments away from the replication fork. Initially, small segments of RNA, known as RNA primers, are laid by primase along the lagging strand template. DNA polymerase III uses these segments to start synthesizing small sections of DNA away from the replication fork. These small sections of DNA sequence are called Okazaki fragments. DNA polymerase I will then excise the RNA primers and replace them with the appropriate nucleotides. DNA ligase joins the gaps in the Okazaki fragments by creating a phosphodiester bond. The result is discontinuous replication of the 3' to 5' parental strand. 10. (a) If single-­‐stranded binding proteins were malfunctioning during DNA replication, then the two strands of the unwound DNA molecule would rehybridize. Template strands would not be available for primase to lay down primers and for DNA polymerase III to build complementary strands. DNA replication would not proceed. (b) If primase was inactivated, DNA polymerase III would not be able to build a complementary strand. Primase lays down primers (short sequences of RNA) that DNA polymerase III uses as starting points to start building a complementary strand. DNA replication would not proceed. (c) If DNA polymerase I was malfunctioning, the RNA primers that were laid down by primase as starting points for elongation would remain in the DNA molecule. The result would be a DNA molecule embedded with stretches of RNA sequence 13. DNA can be classified as a polymer because consists of repeating units known as nucleotides. A nucleotide comprises a deoxyribose sugar, a nitrogenous base, and a phosphate group. 14. A 3' end is represented by the letter b. The 5' ends are represented by the letters a, c, and d. 16. The following proportions would be found in the complementary strand: 15% thymine, 30% adenine, 20% cytosine, and 35% guanine. The proportions in the whole molecule would be as follows: adenine = (15% + 30%)/2 = 22.5% thymine = (30% + 15%)/2 = 22.5% cytosine = (35% + 20%)/2 = 27.5% guanine = (20% + 35%)/2 = 27.5% 9. (a) Length in metres of average E .coli chromosome: 1.02 x 10-­‐3 m long. (b) There are 300 000 turns in the structure. Chapter 5 questions: Pg. 274 -­‐ #1, 2, 5 -­‐ 8, 10, 11, 13, 14 and 18 1.(a) Eukaryotic genomes are divided into two areas. The noncoding regions are known as the introns and the coding regions are known as the exons. (b) The specific DNA regions where an RNA polymerase attaches and initiates RNA synthesis are called promoters. In prokaryotes, many genes may be under the control of the promoter and an operator. The whole system is known as an operon. (c) The transfer of genetic information from DNA to RNA is called transcription. The enzyme that facilitates the process is called RNA polymerase and can only synthesize messenger RNA in the 5' to 3' direction. Synthesis ceases when the termination sequence is reached. 2. Messenger RNA (mRNA) represents the product of transcription of a gene. Ribosomes found in the cytoplasm synthesize proteins using mRNA as the template. Since DNA is transcribed into mRNA, the DNA itself does not have to exit the nucleus and be subjected to degradation. Transfer RNA (tRNA) delivers amino acids to the ribosome. The amino acids are used to build protein during the process of translation. Ribosomal RNA (rRNA) binds with ribosomal proteins to form ribosomes. 5. (a) transcriptional: This type of control regulates which genes are transcribed (DNA to mRNA), or it can control the rate at which transcription occurs. Use of transcription factors. (b) posttranscriptional: -­‐ addition of tail and cap. Introns are removed and exons are spliced. (c) translational: It controls how often and how rapidly mRNA transcripts will be translated into proteins. This control affects the length of time it takes for mRNA to be activated and the speed at which cytoplasmic enzymes destroy mRNA. (d) posttranslational: -­‐ addition of various chemical groups; folding 6. (a) A substitution would be less harmful than a deletion, because it does not result in a shift of the reading frame. A substitution can result in a change in one amino acid, while a deletion can result in the translation of many new amino acids in a growing polypeptide chain. (b) An addition in an intron region is less harmful than an addition in an exon region. Intron regions are noncoding and therefore will not be transcribed and translated into protein. An addition in an exon region can result in frameshift mutations that can lead to different amino acid sequences in a protein. (c) do not complete (d) An inversion is more harmful than a substitution. A substitution can result in a change in one amino acid residue in a protein, whereas an inversion results in many amino acid residue changes in the protein sequence. 7. RNA polymerase transcribes in the 5' to 3' direction, therefore, the 3' to 5' strand will be the template strand. 3'–
TTCATGTCGTA–5': template strand 5'–AAGUACAGCAU–3': mRNA strand lys–tyr–ser: polypeptide chain 8. (a) If spliceosomes were inactivated, introns would not be excised from eukaryotic mRNA. The result would be dysfunctional proteins, as they would contain many extraneous amino acid residues encoded by the introns. (b) If RNA polymerase were inactivated, no proteins would be synthesized because none of the DNA would be transcribed. (c) If poly-­‐A polymerase were inactivated, the mRNA transcript would be subjected to possible degradation on exit into the cytoplasm. If the transcript were degraded, then the proteins synthesized would be missing the end portion of their sequences, depending how far along the degradation had proceeded. (d) If tRNA were inactivated, translation would not take place. Amino acids could not be delivered to the ribosomes that build polypeptide chains. (e) If ribosomes were inactivated, proteins would not be synthesized. 10. (a) Answers will vary. A possible mRNA sequence could be AUG GGU CCU GUU CGU. (b) More than one sequence may exist because more than one codon encodes for an amino acid. For example, the amino acid proline is encoded by the codons CCU, CCC, CCA, and CCG. (c) The variability found in the genetic code can possibly offset mutations. For example, it does not matter what the third ribonucleotide is in a codon that encodes for proline. If the two first ribonucleotides are intact CC, then proline will be placed in the polypeptide chain. The variability originates in the third ribonucleotide of a codon. 11. Adenine and thymine share two hydrogen bonds, while guanine and cytosine share three hydrogen bonds. Less energy is required to break two hydrogen bonds and unwind a DNA molecule as compared with three hydrogen bonds. 13. (a) If tryptophan levels are high, then the tryptophan molecule is bound to the tryptophan repressor. The repressor–tryptophan complex binds to the trp operon operator, blocking transcription of the tryptophan synthesizing genes. RNA polymerase is bound to the promoter but cannot get through. (b) If the tryptophan levels are low, the tryptophan repressor molecule cannot bind to the trp operon operator. RNA polymerase is free to transcribe the tryptophan synthesizing genes. 14. Every codon must consist of a triplet of base pairs because there are 20 amino acids. If doublets of base pairs were used to code for amino acids only 16 possibilities would exist (42 = 16), which would not be enough to code for all the amino acids. The triplet combination allows for 64 combinations (43 = 64), which are more than enough combinations to code for all amino acids, as well as the start and stop signals. 18.(a) 5'–GGC CAG AAA CAA GAA–3' gly–gln–lys–gln–glu (b) codon GGC for gly, anticodon CCG,CCA, CCC, CCU codon CAG for gln, anticodon GUC, GUU codon AAA for lys, anticodon UUU, UUC codon CAA for gln, anticodon GUU, GUC codon GAA for glu, anticodon CUU, CUC Chapter 4 and 5 questions: Pg. 324 -­‐ #13 -­‐ 16, 20, 23(a-­‐g), 24, 27, 28, 30, 31, 34 13. Hydrogen bonds, which are individually weak but collectively strong enough to ensure the stability of the DNA molecule, hold together the two DNA strands in the double helix. Constant “breaking” of these hydrogen bonds for replication and protein synthesis. 14. Ensuring the quality control of new DNA strands is the job of DNA polymerases I and III. During replication, they proofread the newly synthesized strands for missing or mismatched bases. If they find a mismatched nucleotide, they will backtrack over the strand, excise it, and replace it with the appropriate nucleotide.In some instances, mutations can be beneficial for organisms if they provide a competitive advantage; in fact, mutations are a major cause of evolution because they allow organisms to physically adapt to the changing environment in which they live. 15. DNA replicates in a semiconservative manner by having each strand of the double-­‐ stranded parent DNA molecule serve as a template for, and ultimately end up as part of, the newly formed daughter strand. The process occurs as the parent strand is uncoiled and unzipped by the enzymes gyrase and helicase so that it is only single stranded. SSBs are added to keep the strands from reannealing, and RNA primers are added by primase. The two new daughter strands are synthesized in opposite directions with respect to the replication fork. DNA polymerase III synthesizes the nucleotides on the parent templates in the 5' to 3' direction. One strand is synthesized continuously and is known as the leading strand. The second strand is replicated discontinuously and is known as the lagging strand. The lagging strand consists of Okazaki fragments that are eventually annealed. DNA polymerase I and III check and repair any mistakes in the two new daughter strands, which each contain a newly synthesized single strand and one parent single strand. 16. The seven stages, which break down the events of DNA replication, are listed below: i. The enzyme gyrase relieves any tension from the unwinding double helix. ii. The enzyme helicase breaks the hydrogen bonds holding the two complementary parent strands together, resulting in an unzipped helix that terminates at the replication fork. iii. Single-­‐stranded binding proteins anneal to the newly exposed template strands, preventing them from reannealing. iv. The enzyme primase lays down RNA primers that will be used by DNA polymerase III as a starting point to build the new complementary strands. v. DNA polymerase III adds the appropriate deoxyribonucleic triphosphates to the 3' end of the new strand, using the template strand as a guide. The energy in the phosphate bonds is used to drive the replication process. The leading strand is built continuously toward the replication fork. A lagging strand comprising short segments of DNA, known as Okazaki fragments, is built discontinuously, away from the replication fork. vi. DNA polymerase I excises the RNA primers and replaces them with the appropriate deoxyribonucleotides. DNA ligase joins the gaps in the Okazaki fragments by the creation of a phosphodiester bond. vii. DNA polymerase I and DNA polymerase III proofread by excising incorrectly paired nucleotides at the end of the complementary strand and adding the correct nucleotides. 20. An overview of protein synthesis beginning with the movement of the mRNA to the ribosome is outlined for eukaryotes below: Ribosome subunits (large and small) bind to the 5' cap of the mRNA transcript, sandwiching the mRNA between them, and translation commences. The large ribosome’s subunit contains two sites, the A (acceptor) site and the P (peptide) site. The ribosome moves along the mRNA, reading the code in triplets known as codons. Translation does not occur until it reads the triplet AUG, which is the start codon. The start codon corresponds to the amino acid methionine. When the start codon is in the P site, tRNA delivers the amino acid methionine. tRNA recognizes the codon because of the complementary anticodon. The second codon is now in the A site. The appropriate tRNA delivers the next amino acid in the protein sequence. A peptide bond is formed between methionine and the second amino acid. The ribosome shifts over one codon. The tRNA that delivered methionine is released to obtain another methionine amino acid. The second amino acid tRNA shifts over to the P site. The third amino acid, coded by the third codon, is brought intothe A site by the next tRNA. A peptide bond is formed between the second and the third amino acid. The process of elongation continues until a stop codon is read in the A site. The stop codons are UAG, UGA, and UAA. The ribosome stalls when it reaches a stop codon. A protein known as the release factor recognizes that the ribosome has stalled and causes the ribosome subunits to disassemble, releasing the mRNA and newly formed protein. The protein is folded and modified and then targeted to the area of the cell where it is required. 23. (a) Exons are segments of DNA that code for part of a specific gene, while introns are the noncoding region of the DNA. (b) A codon is a sequence of three bases in DNA or complementary mRNA that serves as a code for a specific amino acid, while an anticodon is a group of three complementary bases on tRNA that recognizes and pairs with a codon on mRNA. (c) A missense mutation is a mutation that results in the single substitution of one amino acid in the resulting polypeptide, while a nonsense mutation is a mutation that converts a codon from an amino acid into a termination codon. (d) RNA is a single stranded polymer of nucleotides consisting of a ribose sugar, a phosphate, and four nitrogenous bases (uracil substituted for thymine). Three types of RNA, each with varying functions, exist: mRNA is the end product of transcription and is translated into protein by ribosomes; tRNA is responsible for delivering amino acids to ribosomes for translation; rRNA binds with ribosomal protein to form ribosomes. DNA is a double-­‐stranded polymer of nucleotides consisting of a deoxyribose sugar, a phosphate, and four nitrogenous bases; it carries the genetic information of an organism. (e) An inducer is a molecule that binds to a repressor protein and causes a change in conformation, resulting in the repressor protein falling off the operator, while a corepressor is a molecule (usually the product of an operon) that binds to a repressor to activate it. (f) Promoters are regions upstream from the transcription site that contain binding sites for RNA polymerase and a number of proteins that regulate the rate of transcription in the adjacent gene. Operators are regulatory sequences of DNA to which a repressor protein binds. (g) Induced mutations are mutations caused by a chemical agent or radiation, while spontaneous mutations are mutations occurring without chemical change or radiation but because of errors made in DNA replication. 24. In a eukaryotic cell, an mRNA primary transcript must undergo numerous modifications before it can leave the nucleus for translation. A 5' cap, consisting of 7-­‐methyl guanosine that forms a modified guanine nucleoside triphosphate, must be added. This cap protects the mRNA from digestion by nucleases and phosphates as it exits the nucleus and enters the cytoplasm. A poly-­‐A tail, consisting of approximately 200 adenine ribonucleotides added by the enzyme poly-­‐A polymerase, must be added to the 3' end of the mRNA transcript. This tail is also essential in the protection of the mRNA from nucleases and phosphates. Finally, since the mRNA contains both coding regions, exons, and noncoding regions, introns, it must be modified to remove the introns. The introns are removed from the primary transcript by particles made from RNA and proteins called spliceosomes. It is necessary to remove the introns because they do not code for proteins; if they are translated, they will lead to the improper folding of the produced proteins, rendering them dysfunctional. Once these modifications have been made to the mRNA primary transcript, it is known as the mRNA transcript and is ready to leave the nucleus for translation. 27. If one strand contains the nucleotide proportions 15% A, 30% T, 20% G, and 35% C, the other strand in the double helix must contain the bases in the following proportions: 30% A, 15% T, 35% G, and 20% C, since bases are complementary. 28. The roles of the following enzymes in DNA replication are outlined below: DNA ligase: joins DNA fragments by catalyzing the formation of a bond between the 3' hydroxyl group and the 5' phosphate group on the sugar-­‐phosphate backbones. DNA gyrase: relieves tension produced by the unwinding of DNA. DNA helicase: unwinds double-­‐helical DNA by disrupting hydrogen bonds. DNA polymerase I: removes RNA primers and replaces them with the appropriate deoxyribonucleotides during DNA replication. 30. Silent mutations do not lead to deleterious effects in an organism because they either occur in introns, noncoding regions of DNA, or have no effect on the translation of proteins, because of the redundant nature of the genetic code. Introns are regions are cut out of the mRNA in the process of transcription, thus preventing the mutation from manifesting itself in the organism. Silent mutations that do occur in coding regions, exons, do not affect the translation process and thus the organism as a whole, because many codons can be coded for with multiple base sequences. Thus, even if one base is mutated, the codon may still code for the same amino acid. 31. The lac and trp operons are similar in their mode of action, because they both work in conjunction with a given molecule to determine when regulation should occur. The operons differ in their mode of action because with the lac operon, the LacI repressor protein binds to the operator when lactose levels are low, while with the trp operon, the corepressor tryptophan binds to the trp repressor protein, and the complex binds to the operator, when tryptophan levels are high. Additionally, with the lac operon, high levels of lactose induce the operon, while with the trp operon, high levels of tryptophan repress the operon. See Figure 2 (page 256) and Figure 3 (page 257) of the Student Text. 34. In a double-­‐stranded DNA molecule complementary bases (A–T, C–G) must exist in the same proportion. On this basis, the following results were determined: Sample A: Single stranded, the proportion of adenine is not equal to the proportion of thymine, and the proportion of guanine is not equal to the proportion of cytosine. Sample B: Double stranded, the proportion of adenine is equal to the proportion of thymine and the proportion of guanine is equal to the proportion of cytosine. Sample C: Single stranded, the proportion of guanine is not equal to the proportion of cytosine.