BIS101 Winter 2007 Final Exam: Genetics and Protein Synthesis - Prof. Mark F. Sanders, Exams of Biology

Download BIS101 Winter 2007 Final Exam: Genetics and Protein Synthesis - Prof. Mark F. Sanders and more Exams Biology in PDF only on Docsity! NAME: ID# Last, First BIS101-2 Winter 2007 FINAL EXAM SHOW ALL WORK TO GET FULL CREDIT. I,________________________, ID#(6 last digits)_________________________ authorize posting of my grade in Briggs (1st floor) by the 6 last digits of my ID# Note: Final exams will not be returned and no re-grading petitions will be honored. Only claims for clerical errors (errors adding points or transferring grades to roster) will be accepted. The exams will be only available for inspection by appointment until April 15th at 116 Asmundson. 1 FINAL W’07 QUESTION VALUE SCORE 1 20 2 10 3 24 4 20 5 13 6 10 7 22 8 24 9 10 10 9 11 20 12 18 TOTAL 200 2 5) The following factors, structures, sequences or enzymes play a role in protein synthesis in prokaryotes and eukaryotes. Identify with a check mark the correct column in which process do the they primarily participate and whether it applies to prokaryotes, eukaryotes or both. (13) Item Transcription Translation (or after transcription) prokar eukar prokar eukar RNA polymerases x x Shine-Dalgarno sequence x G-cap x Sigma factor x Transcription factors x x mRNA splicing x “TATA”/Pribnow boxes x x poly A tail x Amino acyl tRNA synthetase x x 6. Write the expected progeny genotypes and phenotypes (male fertile or male sterile, including ratios when applicable) for the crosses shown in the table. S indicates sterile cytoplasm, F fertile cytoplasm (10) female parent male parent progeny of cross SMsMs Fmsms S Msms (fert) Fmsms Smsms none Smsms FMsms 1/2 S Msms (fert) 1/2S msms (ster) Fmsms FMsms 1/2 F Msms (fert) 1/2 F msms (fert) SMsms FMsms 1/4 S MsMs (fert) 1/2 S Msms (fert) 1/4 S msms (ster) 5 7) (22) A circular DNA molecule is cleaved with two individual restriction enzymes HaeIII and EcoRI and then with a combination of the two enzymes. The fragments obtained are: HaeIII 1400 bp, 600 bp EcoRI 1500 bp, 500 bp HaeIII and EcoRI 1100 bp, 400 bp, 300bp, 200bp a) Draw the restriction map. (18) b) HaeIII recognizes the 4 base pair sequence GATC CTAG Estimate the number of fragments yielded by this enzyme in a DNA molecule containing 50% GC and 3000 Kb in size. (4) 3000 x (1/4)4 = 11.7Kb or 11,719 bp 400 200 1100 300 H E H E 6 8) (24) In each of the merozygotes below indicate whether or not there would be expression in ß-gal or permease from the lac operon: Lactose present Lactose present Lactose absent Lactose absent Genotype ß-gal permease ß-gal permease a. F’i-p-o+z+y + /i+p+o+z -y - N N N N b. F’isp+o+z+y + / i+p+ocz+y + Y Y Y Y c. F’ i+p+ocz -y + /i+p+o+z -y - N Y N Y d. F’ i+p-ocz +y - /i+p+o+z -y + N Y N N e. F’ i+p-o+z -y + /i-p+o+z +y - Y N N N f. F’i+p-o+z+y - /i-p+ocz -y + N Y N Y 9) Tomato trisomics are useful for physically mapping genes on their respective chromosomes. Genes found on the chromosome pair affected will deviate from Mendelian ratios. Provide the expected phenotypic and genotypic ratios resulting from crossing a trisomic plant of genotype Aaa by a homozygous recessive diploid plant of genotype aa (A=purple stem, a=green stem). a) Use a Punnett square to figure out all possible gametic genotypes and ratios as well as genotypes and phenotypes of all the progeny. (8) a aa aaa 2 Aa 2Aaa A Aa 2a 2aa 3 purple ; 3 green (or 1:1) b) Now write the phenotypic ratio for the diploid portion of the progeny. (2) 1 purple: 2 green 7