7.014 Quiz II Handout

Transcription

1 7.014 Quiz II Handout Quiz II: Wednesday, March 17 12:05-12: **This will be a closed book exam** Quiz Review Session: Friday, March 12 7:00-9:00 pm room Open Tutoring Session: Tuesday, March 16 4:00-6:00 pm room

2 Question 1 Indicate whether the following statements are true or false. If a statement is false, provide a brief explanation for why it is false. a) Semi-conservative replication means that both of the parental DN strands serve as templates for the synthesis of the new progeny strands, so the new double stranded DN molecules are composed of one new and one old strand. b) Nucleic acid primers are required for both the initiation of DN replication and for the initiation of transcription. c) The 5' to 3' direction of DN synthesis implies that growth occurs by addition of new dntps to the exposed 5'-OH group of the growing strand. d) During DN replication, DN synthesis occurs in the 5' to 3' direction in the leading strand, and in the 3' to 5' direction on the lagging strand. e) Loss of the 3' to 5' exonuclease activity of DN polymerase in E.coli should increase the rate of DN synthesis, but not affect the fidelity of DN synthesis. f) If you heated DN with a high percentage of /T base pairs (bps) compared to / bps, you would expect this DN to be more resistant to thermal melting (separation of the two strands) than if you heated DN with a high percentage of / bps compared to /T bps. g) Shown below is a diagram of a replication fork in a double-stranded DN sequence found in a prokaryotic cell. This region of the DN will be replicated by the enzyme DN polymerase. The replication fork will proceed to the right ( >). 5' strand 3' 5' 3' strand B i) Which strand ( or B) of the parental DN duplex will serve as the template for the newly synthesized leading strand and which strand ( or B) will serve as the template for the newly synthesized lagging strand (the one synthesized from Okazaki fragments). ii) Briefly describe the reason that the lagging strand must be synthesized from Okazaki fragments. 2

3 Question 2 onsider the following hypothetical short mrn found in a bacterium: UUUUUUUUU 3 a) What are the first two and last two amino acids of the protein produced from this mrn? ( table of the genetic code can be found on page 10). Be sure to indicate the amino and carboxyl termini. b) single base pair mutation (substitution, insertion, or deletion) results in the production of a protein only 3 amino acids long. What nucleotide was altered (give the #) and in what way was it altered in this mutant? c) Shown below is the double-stranded genomic DN which encodes the above mrn as it would appear within the chromosomal DN of the bacterium. The remaining DN of the organism is indicated by xxx. i) Which of the following would be required for transcription of the mrn encoded by this DN. ircle all that are appropriate. intron start codon promoter origin of replication exon stop codon terminator ii) Indicate on the figure below approximately where each of the items you circled in (i) above would be located. 5 xxxxxxxxxxxxxxtttttttttxxxxxxxxxxxxxx3 3 xxxxxxxxxxxxxxttttttttttttxxxxxxxxxxxxx5 3

4 Question 3 The fictitious bacterium E. kellogg requires two enzymes for the synthesis of the sugar krispiose. Enzyme converts cheeriose to wheatiose; enzyme B converts wheatiose to krispiose. enzyme enzyme B cheeriose wheatiose krispiose The krispiose operon is regulated by repression. The repressor, encoded by gene R, is constitutively expressed. In the presence of krispiose, the repressor binds the operator and prevents the transcription of genes and B. In the absence of krispiose, enzymes and B are synthesized from a single mrn. PR gene R P O gene gene B You have isolated several strains that have mutations in the krispiose biosynthesis pathway. Their genotypes are listed in the table below. For example, the " " in the genotype R indicates a loss-of-function of that component. The R s indicates a superrepressor that can bind the operator in the absence of krispiose. ssume that genes that are not listed are wild-type. a) In the table below, predict the level of activity ("H" for high or "L" for low) of the enzymes, and B, for the following E. kellogg mutants when grown in medium with or without krispiose. The activity levels for the wild-type strain is given. enotype without krispiose with krispiose enzyme enzyme B enzyme enzyme B wild type H H L L B R O P O R s O 4

5 Question 3, continued b) You make partial diploids of E. kellogg strains using a single copy F' plasmid introduced into the E. kellogg cell. The genotypes of two partial diploids are indicated below. For example, the " " in the genotype R indicates a loss-of-function of that component. You then determine the levels of activity of enzymes and B in strains grown with and without krispiose in the medium. The results are shown in the table below. Provide an explanation for the phenotype observed in these two partial diploids. i) enotype: hromosome F' Plasmid without krispiose with krispiose ii) P R + R + P + O + B + P R + R + P + O + B + enotype: hromosome F' Plasmid P R + R P + O + + B + P R + R + P + O + B + enzyme enzyme B enzyme enzyme B H H H H without krispiose with krispiose enzyme enzyme B enzyme enzyme B H H L L Question 4 In order to investigate the biosynthesis of arginine in the yeast, S. cerevisiae, you mutagenize haploid yeast and screen for arginine auxotrophs. You isolate 6 such mutants, mate them to create diploids with pairwise combinations of mutations. You obtain the following results ("+" indicates that the diploid grows on minimal media and " " indicates no growth on minimal media): m1 m2 m3 m4 m5 m6 wild type m m m m4 + + m5 + + m6 + wild type + a) Sort the 6 haploid mutants into complementation groups. 5

6 Question 4, continued b) What does the number of complementation groups tell you about the number of genes in the arginine biosynthetic pathway? c) Briefly explain why the diploid formed by mating m1 and m2 is able to grow on minimal medium while the diploid formed by mating m1 and m4 is unable to grow on minimal medium. When you grow the following yeast strains on minimal media, you observe that certain chemical intermediates accumulate in the medium. The results are shown below for various mutant strains (" + " indicates the wild-type allele and " " indicates the mutant allele): Strain enotype Intermediate 1 2 +, 3 + ornithine B 1 +, 2, 3 + citrulline 1 +, 2 +, 3 argino-succinate D 1, 2, 3 + ornithine E 1 +, 2, 3 citrulline F 1, 2 +, 3 ornithine d) Based on the data for strain, does the enzyme mutated catalyze the synthesis or the breakdown of ornithine? e) Why does citrulline, but not argino-succinate, accumulate in the medium in which strain E is grown? f) Based on these data, draw the biosynthetic pathway for arginine. 6

7 U The enetic ode: U UUU phe (F) UU ser (S) UU tyr (Y) UU cys () UU phe U ser U tyr U cys UU leu (L) U ser U STOP U STOP UU leu U ser U STOP U trp (W) UU leu U pro (P) U his (H) U arg (R) U leu pro his arg U leu pro gln (Q) arg U leu pro gln arg UU ile (I) U thr (T) U asn (N) U ser (S) U ile thr asn ser U ile thr lys (K) arg U met (M) thr lys arg UU val (V) U ala () U asp (D) U gly () U val ala asp gly U val ala glu (E) gly U val ala glu gly U U U U Question 1 Solutions a) True. b) False. lthough DN polymerase requires a short RN primer for DN replication, a primer is not required by RN polymerase for transcription (RN synthesis). c) False. DN synthesis occurs by addition of dntps to the exposed 3' OH group of the growing strand. d) False. ll of the DN is synthesized in the 5' to 3' direction. DN on the lagging strand is made in short pieces which are joined together later, so that the lagging strand lengthens in the 3' to 5' direction. e) False. Loss of the 3' to 5' proofreading exonuclease activity of DN polymerase means that the fidelity of DN synthesis is lowered and, therefore, DN synthesis is much more error-prone. 7

8 f) False. Because each / base pair is held together by three hydrogen bonds and each /T base pair is held together by two hydrogen bonds, higher temperatures are required to separate /- rich DN than are necessary to separate /T-rich DN. Therefore, the /rich DN is more resistant to thermal melting compared to the /T-rich DN. g i) The template for lagging strand synthesis will be strand of the parental DN duplex because to synthesize in a 5' to 3' direction on this strand you must start at the fork and move towards the 5' end of the template strand. Strand B will serve as the template for the leading strand because 5' to 3' synthesis can start at the 3' end of strand B and proceed in the same direction as the replication fork. ii) Since all polynucleotides must be synthesized in a 5' to 3' direction, the direction of synthesis of the newly synthesized lagging strand is opposite to the movement of the replication fork. In order for DN polymerase to synthesize the lagging strand, the synthesis is done in a discontinuous manner. Question 2 a) H 3 N+ - met - val asp - lys - OO b) several possible: 16U --> ; delete 15; or delete 16. c) Shown below is the double-stranded genomic DN which encodes the above mrn as it would appear within the chromosomal DN of the bacterium. The remaining DN of the organism is indicated by xxx. i) Which of the following would be required for transcription of the mrn encoded by this DN. intron start codon promoter origin of replication exon stop codon terminator ii) The promoter would be in the left set of xxxx & terminator would be in the right set of xxxx. 8

9 Question 3 a) enotype without krispiose with krispiose enzyme enzyme B enzyme enzyme B wild type H H L L B H L L L R H H H H O L H L H P O L L L L R s O H H H H b) i) Operators only act in cis. The O mutation cannot affect the expression of genes and B from the F' plasmid. The O + on the F' plasmid does not affect expression from the chromosome. The result of the O mutation is constitutive expression of the and B genes from the chromosome. ii) Repressor can act in trans; therefore, the wild-type repressor on the F' plasmid can affect expression of genes and B from both the chromosome and the F' plasmid. s a result, the wild-type phenotype is restored in this partial diploid. Question 4 a) roup 1: m1, m4, and m6 roup 2: m2 roup 3: m3 and m5 b) There are most likely three genes in the arginine pathway. c) There are several genes involved in arginine biosynthesis. lthough each haploid parent has a mutation that causes a block at a specific point in the pathway, a diploid may have one functional allele for each gene in the pathway. The two mutants, m1 and m2, have mutations in different genes and are thus able to complement each other, while the m1 and m4 mutants have mutations in the same gene and are thus unable to complement each other. d) The enzyme mutated catalyzes the step in the pathway that breaks down ornithine. e) The enzyme encoded by gene 2 is functions before of the enzyme encoded by gene 3; therefore, there is little or no argino-succinate produced in strain E. The intermediate that accumulates is before the first affected enzyme. f) ornithine citrulline argino-succinate argini 9