A. With just this information, how many genes does it take to make DNA polymerase III? 8 assuming d' and d are different polypeptides

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1 NAME Gene 603 Exam 1 October 4, 2,002 I. According to recent work in the lab of Charles McHenry, The E. coli DNA polymerase III holoenzyme contains polypeptide subunits in the arrangement (aeq) 2 -t 2 gd'd-(b 2 ) 2. A. With just this information, how many genes does it take to make DNA polymerase III? 8 assuming d' and d are different polypeptides B) Polypetide t is 643 amino acids long and g is only 430, but the sequence of these 430 are exactly the same as the first 430 amino acids of polypeptide t. Give two possible models to account for this observation. 1) editing of some messages changing codon 431 to a stop 2) differential splicing to eliminate last exons past codon 430 3) post-translational protein cleavage 4) duplicate genes, one of which is truncated or has nonsense C) In a slow growing mutant amino acid 27 in both polypeptides changed from histidine to valine. i. Does this information help decide between your models? Explain not unless #4 was a choice; in that case it would be eliminated; in all others both polypeptides would be expected to show this change ii. Give the specific base-pair change that occurred in the DNA. A:T changed to T:A iii. List 5 different terms that describe this mutation, telling how each applies. Be as brief a possible. missense (changed amino acid) transversion (purine replaced by pyrimidine and vice versa) point (single base pair change in DNA) forward (the mutation produced a new allele) leaky (the new protein functions less efficiently than normal) base pair substitution (one base pair "replaced" with another) iv. Name a chemical mutagen that might cause the mutation. alkylating agents such as mustard gas and EMS are known to cause transversions

2 2) E. coli glycine auxotrophs 1, 2 & 3 were streaked across each other on a minimal plate with just enough glycine to allow a few cell divisions. However, in some intersections of the streaks, it was clear that one of the strains was able to grow. (1 grows near 3, etc. as shown. A. Which mutant is blocked in the last step of G synthesis? #3 (it cannot grow without glycine) B) If X and Y are precursors of G, show a pathway, indicating where each mutant strain is blocked X Y glycine (3 spills Y allowing 2 & 1 to grow a little more etc) 3) How were the following isotopes useful in establishing DNA as the genetic material or the mode of replication? (Group as appropriate.) 14 N, 15 N, 32 P, 35 S Separate batches of phage, grown in 32 P to label DNA and in 35 S to label protein, were allowed to infect E. coli. Once the phage ghosts were sheared off, only 32 P was fond inside bacteria and in phage progeny, showing DNA contained the genetic information for reporduction. After many generations of growing E. coli in 15N medium almost all the DNA bases contained 15 N, giving dense DNA. When the bacteria were washed, stavared for N, and returned to 14N medium, samples collected at the first division showed all DNA of intermediate density, meaning one strand contained 14 N and the other 15 N. Density in succeeding generations agreed with semi conservative model of DNA replication.

3 4) As strange as it may now seem, the ability to determine amino acid sequences in proteins came earlier than sequencing DNA. The effects of some mutations in the phage T4 lysozyme gene could be determined by the changes in amino acid sequence. The normal sequence of one fragment is: lys-ser-pro-ser-leu-asn-ala. One treatment with acridine gave a mutant strain that lacked lysozyme. Treatment of this strain with?? gave a revertant that had the amino acid the following sequence for the same fragment: lys-val-his-his-leu-met-ala. a) What mutagen was used to induce the revertant? another acridine dye b) How can these changes be explained? the first treatment deleted one A in first codon causing a frameshift mutant and no enzyme detected; the reversion added a base (G) before/in the ala codon restoring the reading frame. c) As well as possible, give the mrna sequence that codes for the normal lysozyme peptide fragment. AAA/G-AGU-CCA-UCA-CUU-AAU-GCX (take out an early A and the new reading frame accounts for the revertant once a G is added to restore the ala codon) 5) Kits that can be purchased for in vitro (test tube) translation for messages from bacteria differ from those sold for translating eukaryotic mrna. a) What are the general components that would be found in both kits? ribosomes, all trnas, activating enzymes & ATP, (or charged trnas) initiation and elongation factors, GTP- ie everything needed to translate except for mrna b) Given that the same genetic code applies to both systems, explain why different kits are required. The leaders of the mrnas use different methods for initiation; CAP recognition in eukaryotes versus Shine-Delgarno match in prokaryotes. Once translation is initiated, either message will be translated into the same polypeptide by both systems.

4 6) In E. coli, mutations selected to suppress UAG nonsense mutations also often suppress UAA mutations. However, the reverse is never seen. Explain why this observation is true. Selection for UAG suppressor will get sense trnas with anticodons changed to AUC or AUU which can also pair due to wobble. Selection or UAA suppressors should always give AUU anticodons. (In my view these should suppress both too? anybody know why/if they don t?) 7) Describe the "ClB" technique and tell how/why it was first used. These are 3 X chromosome markers ion Drosophila that were used in expreiments to prove that X-rays cause mutations. C= crossover suppressor, l is a recessive lethal and B is a partly dominant Bar eye mutation. Wide Bar females were mated to irradiated males and their wide bar daughters were then mated to normal males. If the irradiated X also has a (different) recessive lethal, no sons will be produced. Muller found that the mutation rate was a function of the dose of X-ray (3% of the test females had no sons for each kiloroentgen of X-ray to the irradiated males). 8) Beadle and Tatum chose Neuospora for very specific attributes. What were the useful attributes and tell the significance of each. Haploid (phenotype = genotype; mutations not masked) Grows on defined minimal/complete media (allows identification of auxotrophic mutants) Controlled crosses possible (permit genetic segregation analysis) Millions of genetically identical spores (conidia) available (can increase odds of finding rare mutations) Rapid life cycle and inexpensive to grow

5 9) A canavanine sensitive (can S ), serine prototroph (ser + ), strain of S. typhimurium that can grow on the sugar xylose (xyl + ) was infected with P22. Progeny phage were collected and added to cells of a can R, ser -, xyl - strain. xyl + progeny were identified as colonies on "plate 1" without selection for the other markers. These colonies were then tested via replica plating to plates 2 and 3 to identify double transductants with the following results: 100 can s 50 ser + 0 [Donor DNA] a) Give the relevant media components used in : 1. plate 1 used to select the xyl transductants add xylose as sole carbon source and add serine so not require ser+ to grow 2. plates 2 and 3 used to identify double transductants: one has serine & canvanine to test for canavanine resistance; the other has xylose and no serine (or canavanine) to test for ser+ b) Are any of the markers linked? How do you know? xylose and canvanine must be close together since more than half of the recipients receive both markers, even when low numbers of donor phage are used for infection of the recipient

6 10. Amino acid # 245 in a human protein called p53 is known to differ in 3 naturally occurring mutant alleles. A) Fill in the missing information in the chart below concerning this site. LEVEL Mutant 1 Mutant 2 Mutant 3 Normal ds DNA AGC GAC TGC GGC Sequence TCG CTG ACG CCG mrna codon Sequence AGC GAC UGC GGC Amino acid # 245 Ser Asp Cys Gly B) Based on your "normal" DNA sequence, which mutations: i) are missense mutations? all 3 ii) resulted from a transversion? #3 C) If only the amino acid substitutions were known and not one of the mrna codons, the "normal" column could have a different answer. What would it be as far as the codon and amino acid are concerned? UAC and Tyr 11. What is "telomerase"? Why is it important? The ribozyme that replaces the ends of linear chromosomes after DNA replication. Without it, the RNA primers used to start DNA replication could not be replaced and the chromosomes would shorten every cell division

7 12. The following picture is from a 1962 Genetics text. Label the 4 bases as appropriate. C G A T Do you notice anything "unusual" about this drawing? If so, what? It seems to have a 3'phosphate end, rather than a normal 5' phosphate and 3'OH we would expect to see 13. Aflatoxin B 1 removes guanine bases from the DNA backbone. What is the general name for this type of defect. Outline a potential repair mechanism. "AP" site Several models - one is exonuclease recognizes the AP site, cleaves the sugar phosphate backbone (or removes a number of bases in the strand) followed by an exonuclease to lengthen the gap and a polymerase and finally a ligase to seal the remaining nick.