202hw4KEY-1
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202hw4KEY-1

Course Number: BIOLOGY 202, Spring 2008

College/University: New Mexico

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Biology 202 Homework 4 Due: 10/2 or 10/3; in discussion Please type (not hand-write) ONLY the answers to question onto the question sheet itself (this sheet). Please highlight or otherwise indicate the parts that you have typed with italics, underlining or bolding. If a question requires additional working, please include all of your work pages. Do not put the work on this answer sheet. 1. The drawing below...

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202 Biology Homework 4 Due: 10/2 or 10/3; in discussion Please type (not hand-write) ONLY the answers to question onto the question sheet itself (this sheet). Please highlight or otherwise indicate the parts that you have typed with italics, underlining or bolding. If a question requires additional working, please include all of your work pages. Do not put the work on this answer sheet. 1. The drawing below depicts a replication fork undergoing replication of both the leading and lagging strand. (0.5 pts for each letter, total 2 pts) On the replication fork: a. Label all ends of template DNA and newly made DNA (in bold) with either 5 or 3. b. Identify the leading and lagging strands. c. Indicate the first Okazaki fragment made. d. Indicate the direction of synthesis (left to right or right to left) of each newly made DNA strand. The labels may be drawn in by hand and do not have to be typed. answer 2. In the Meselson-Stahl experiment, 15N-labeled cells were shifted to 14N medium at what we can designate as generation 0. a. For the semiconservative model of replication, what proportion of 15N-15N, 15N-14N, and 14N-14N would you expect to find after 1, 2, and 3 replication cycles? (1.5 pts) Generation 15N-15N 15N-14N 14N-14N 1 0: (0%) 2: (100%) 0 (0%) 2 0: (0%) 2: (50%) 2 (50%) 3 0: (0%) 2: (25%) 6 (75%) b. Register to View Answerin terms of the conservative model of replication. (1.5 pts) Generation 15N-15N 15N-14N 1 1: (50%) 0: (0%) 2 1: (25%) 0: (0%) 3 1: (12.5%) 0: (0%) 14N-14N 1 (50%) 3 (75%) 7 (87.5%) 3. Several different mutant strains of E. coli have problems in synthesizing DNA. Predict what enzyme or function is being affected in strains with the following problems. (0.5 pts each, total 2.5 pts) For the first 3, we can see different answers so please accept alternative answers below. a) Newly synthesized DNA contains many mismatched base pairs. Mismatch repair enzymes are not working correctly to remove incorrect bases and fill in correct bases. DNA polymerase could also be deficient in proofreading. b) Okazaki fragments accumulate and DNA synthesis is never completed. DNA ligase is not joining the deoxyribonucleotides to connect Okazaki fragments. DNA polymerase I might also be an answer and could be accepted. Its activity is actually a pre-req for ligase activity. c) No initiation of replication occurs. Helicase is not separating the 2 strands of DNA, or there is not enough ATP in the cells leading to no DnaA-ATP complexes which initiate DNA replication, or? initiator proteins that recognize DNA sequences specific to origins and that recruit other proteins such as helicase, are not working. They might answer DNA pol which is also acceptable. d) Strands remain in supercoiled state following initiation, which is never completed. Topoisomerase is not cutting and rejoining DNA downstream of the replication fork to relieve tension caused by unzipping. e) Final double-stranded nucleic acid contains both DNA and RNA DNA polymerase I is not removing RNA primers at the start of each Okazaki fragment. 4. The genome of the fruitfly Drosophila melanogaster contains roughly 170 million base pairs. DNA synthesis at proceeds a rate of 30 base pairs/second. In the early embryo, the genome is completely replicated in 5 minutes. What is the minimum number of bidirectional replication origins required in the genome to accomplish this? (2 pts) 18,889 origins (5 min=300 seconds. 170,000,000 base pairs/30 base pairs per second=5,666,666.67 base pair seconds. 5,666,666.67 base pair seconds/300 seconds=18,888.89 origins. Since there cannot be a partial origin we will round this up to 18,889 origins. Double check answer: 18,889 origins x 30 bp/sec x 300 sec = 170,001,000 base pairs. This is > 170 million so this was enough origins to complete replication in 5 minutes. However 18,888x30x300<170,000,000 so not enough) 5. The melting temperature, Tm, of genomic DNA is the temperature at which 50% of the DNA is unwound (in single stranded form rather than double stranded, often referred to as denatured). The table presents the melting temperature and the %GC content (the percentage of guanines and cytosines of the total nucleotide content of genomes) of several bacteriophage genomes (bacteriophages are viruses that infect bacteria). Graph the data in a way that presents the GC content on the horizontal axis and the melting temperature on the vertical axis. Describe the relationship between Tm and GC content. Suggest a molecular basis for this relationship. (2pts total- 1 pt for chart, 1 pt for reason) Phage 80 T3 Tm 86.5 91.5 89 90.5 92.1 90 %GC 44.0 53.8 49.2 53 57.4 49.6 hw 4 Q 5 93 92 91 90 Series1 89 88 melting temperature 87 86 0 10 20 30 %GC 40 50 60 70 As we can see from the chart, melting temperature increases with increasing %CG content. The molecular reason for this is that G and C nucleotides are joined with 3 hydrogen bonds while A and T are joined by only 2 H-bonds. Therefore, the more G and C nucleotides in a strand of DNA, the stronger that strand will be. 6. Nucleic acid molecules have been isolated from seven types of viruses. The base composition of these molecules has been analyzed and the findings are presented in the following table. In some instances, only a limited amount of information is available regarding the composition of the nucleic acid. Data are given in terms of percentage of total base composition. Virus Type BASE 1 2 3 4 5 6 7 A 12 20 18 26 23 T 12 18 0 26 0 18 C 38 30 26 24 28 G 38 30 30 26 U 0 20 0 17 24 Identify the type of nucleic acid found in each of the seven viruses. Also indicate, if possible, whether each molecule is single-stranded or double-stranded. In each case, indicate why you classified the molecule as you did. (0.5 pts each, 3.5 pts total) Virus 1: DNA because has T and no U, double stranded because #A=#T and #G=#C Virus 2: DNA because has T. Not enough info to tell if double or single stranded. Virus 3: RNA because has U and no T. Double stranded because #A=#U and #G=#C Virus 4: DNA because has T and no U. Single stranded because #A#T and #G#C Virus 5: RNA because has U. Not enough info to tell if double or single stranded. Virus 6: RNA because has U and no T. Single stranded because #A#U and #G#C. Virus 7: DNA because has T. Single stranded because #A#T.

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