This preview shows page 1. Sign up to view the full content.
Unformatted text preview: WEEK 5 – LS4 DISCUSSION
Genevieve Kendall [email protected] Office Hours: 35pm Wednesday in BSRB 210C www.lsic.ucla.edu Computer Login: User name = [email protected] Password = Student ID # Open HP 3I Student on the desktop File > Connect > 18.104.22.168 MIDTERM II, November 5, 56:50pm, CS 50 Bring a calculator, no graphing calculators Bring your BRUIN CARD Write in blue/black ink if you might want a regrade Extra OH November 3 (Monday), 912am, BSRB 210C Normal OH time has been changed next week (November 5) from 2:304pm, BSRB 210C Next Week’s Discussion Section: I’m going to post slides early so come prepared with questions We will review complementation and recombination (2nd part of Lecture 8) 6 10/14 Tu Bacterial Genetics Chp 8 7 10/16 Th Bacterial Genetics/ Bacteriophage Genetics Chp 8 8 10/21 Tu Bacteriophage Genetics/ Benzer and the rII locus Chp 8, 14 9 10/23 Th Genetic code Chp 12 10 10/28 Tu Mutation and Gene Function Chp 13 11 10/30 Th Gene Regulation I Chp.20 11/5 Wednesday Midterm #2 57 PM (Material from Lectures 611) OUTLINE
Last week’s quiz question Lecture 7/8 Review Transduction Biosynthetic pathways Q&A Quiz 5 Pick up Quiz3/4 Practice questions QUIZ 4
Them r plateis rich m dia. There aste e plica plate all havesupple e e pt whe indicate (-arg s m nts xce re d or –le Mannosetake theplaceof glucosein thefirst plateand am u). s picillin is adde to these d cond plate .
1 2 Plate 1 3 2 4 + AMP Plate 2 1 3 4 3 4 Plate 3 1 2 ARG 3 4 2 LEU Plate 4 3 4 Master Plate + MANNOSE 1) What is these ctablem r for e plate(is it a drug, am acid, or sugar)? le arke ach ino 2) What is thege notypeof colony 1? 3) What is thege notypeof colony 2? From last week– How do we know that everything is lac+ in the replica plates?
lac+ lac+ lac+ lac+ bio+ bio+ bio bio cys+ cys cys+ cys 0 400 95 105 LECTURE 7 REVIEW Transduction – the process by which DNA is transferred from one bacterium to another by a virus (bacteriophage) Viral Reproduction Lytic – infects a bacterium, replicates and kills the host almost immediately Lysogenic (or lytic) – Inserted into the chromosome of Example: T4 the host bacterium and is replicated along with the bacterial chromosome
Example: Lambda Can go into a lytic cycle (ex: UV exposure) In the integrated state the virus is called a prophage LYTIC CYCLE TRANSDUCTION – 2 TYPES Generalized Transduction Random or nearly random bacterial DNA is packaged Specialized Transduction in the phage head in place of the phage chromosome Only bacterium (host) DNA, can be any segment of the host chromosome chromosome producing a phage with a piece of bacterial DNA (now a transducing particle) Characteristic of viruses that carry only certain genes between bacteria Recombination event occurs between host and phage Recombination between phage and host chromosome Only host genes adjacent to prophage integration site are transduced DETERMINING MOI GM02339 Multiplicity of infection
# Virus Particles # of Cells
GM03781 High MOI = almost certainty that all cells will be infected with GM03783 your viral vector GM05017 GM05112 COTRANSDUCTION
If two genes are close enough they are often packaged into the phage head on the same DNA fragment Example: consider a donor bacterium with genes a+b+ and a recipient bacterium with genotype ab If the genes are closely linked, then an a+b+ transductant (ie cotransductant) can be produced through infection by a single transducing phage If the genes are not closely linked then the a+b+ transductant can only be produced by the much rarer event of simultaneous infection of a recipient by two transducing phages, one that is a+ and one that is b+ MAPPING OF E. COLI GENES USING TRANSDUCTION WITH A TEMPERATE PHAGE
Donor: leu+thr+aziR Recipient: leuthraziS Data: Selected Marker Leu+ Leu+ Thr+ Thr+ Unselected Markers 50% = aziR 2% = thr+ 3% = leu+ 0% = aziR For the leu+ selected transductants 50% exhibit aziR and 2% thr+. This means that the leu and azi genes often are cotransduced on the same DNA molecule, and much less frequently the thr gene is on the same transducing DNA with the leu gene. For the thr+ transductants, 3% are leu+ and 0% are aziR. This confirms that the thr and leu genes can be on the same transducing DNA and also indicates that the azi gene is distant enough to never be on the same DNA molecule. EXAMPLE CONTINUED…
thr leu azi This is the order of the genes from the data on the last slide. The transductants are produced by crossingover between the piece of donor bacterial chromosome brought in by the infecting phage and the homologous region on the recipient bacterial chromosome. The infected donor DNA finds the region of the recipient chromosome to which it is homologous, and the exchange of parts is accomplished by double (or other even numbered) crossovers. TIPS
• The closer two genes are the higher the cotransduction frequency will be • If there is a high recombination frequency then there is a low cotransduction frequency so the genes must be far apart • If there is a low recombination frequency then there is a high cotransduction frequency so the genes must be close together Donor: asn+ile gal+ Re nt: asn-ile galcipie +
asn+ t ransductants 69%gal+ ile + 4%gal+ ile 27%gal- ile + 0%gal- ile gal+ t ransductants 70%asn+ ile + 3%asn+ ile 15%asn- ile + 12%asn- ile - I n onee rim nt asn+ t ransductants we se cte for and in theothe xpe e re le d r gal+ t ransductants we se cte for. Thepe ntageof e ge re le d rce ach notype was de rm d by re te ine plica plating and there sults areshown. What is theco-transduction fre ncy of asn-gal? que Donor: asn+ile gal+ Re nt: asn-ile galcipie +
asn+ t ransductants 69%gal+ ile + 4%gal+ ile 27%gal- ile + 0%gal- ile gal+ t ransductants 70%asn+ ile + 3%asn+ ile 15%asn- ile + 12%asn- ile asn+ile gal+ - I n onee rim nt asn+ t ransductants we se cte for and in theothe xpe e re le d r gal+ t ransductants we se cte for. Thepe ntageof e ge re le d rce ach notype was de rm d by re te ine plica plating and there sults areshown. What is theco-transduction fre ncy of asn-ile que ? Donor: asn+ile gal+ Re nt: asn-ile galcipie +
asn+ t ransductants 69%gal+ ile + 4%gal+ ile 27%gal- ile + 0%gal- ile gal+ t ransductants 70%asn+ ile + 3%asn+ ile 15%asn- ile + 12%asn- ile asn+ile gal+ - I n onee rim nt asn+ t ransductants we se cte for and in theothe xpe e re le d r gal+ t ransductants we se cte for. Thepe ntageof e ge re le d rce ach notype was de rm d by re te ine plica plating and there sults areshown be low. What is theco-transduction fre ncy of gal-ile que ? USING TRANSDUCTION TO DETERMINE GENE ORDER The higher the transduction frequency the closer the genes are together Example: Cotransduction frequencies are given for 3 different genes. Which 2 genes are closer together?
Genes Met+Arg+ Arg+Leu+ Leu+Met+ CoTransduction Freq. 11% 72% 18% C o-transduction fre ncie for 4 diffe nt ge s aregive The que s re ne n. com binations shown aretheonly one that could beco-transduce all s d, othe com r binations gave0%co-transduction. Usethis data to de rm theorde of thefour ge s. te ine r ne The data in the following table were obtained from threepoint transduction tests made to determine the order of mutant sites in the A gene encoding the a subunit of tryptophan synthetase in E. coli. Anth is a linked, unselected marker. In each cross, trp+ recombinants were selected and then scored for the anth marker (anth+ or anth). What is the linear order of anth and the three mutant alelles of the A gene indicated by the data in table? Cross 1 2 3 4 Donor Markers anth+ A34 anth+ A46 anth+ A223 anth+ A223 Recipient Markers anth A223 anth A223 anth A34 anth A46 Anth Allele in Recombinants 72 anth+ : 332 anth 196 anth+ : 180 anth 380 anth+ : 379 anth 60 anth+ : 280 anth % anth+ 18 52 50 20 Two additional mutations in the trp A gene of E. coli, trp A58 and trp A487, were ordered relative to trp A223 and the outside marker anth by threefactor transduction crosses as described in the previous problem. The results of these crosses are summarized in the following table. What is the linear order of the anth and the three mutant sites in the trp A gene? Cross 1 2 3 4 Donor Markers anth+ A487 anth+ A58 anth+ A223 anth+ A223 Recipient Markers anth A223 anth A223 anth A487 anth A58 Anth Allele in Recombinants 72 anth+ : 332 anth 196 anth+ : 180 anth 380 anth+ : 379 anth 60 anth+ : 280 anth % anth+ 82 48 50 80 LECTURE 8 (FIRST PART) REVIEW Gene A unit of genetic information that controls the synthesis of one polypeptide or one RNA molecule For our purposes consider the 1 gene – 1 polypeptide to be true (not always true in real science) DETERMINING BIOSYNTHETIC PATHWAYS FROM + AND 0 DATA precursor enzyme1 A enzyme2 B enzyme3 C A,B,C = Metabolites/intermediates Example: If a MutantX couldn’t grow when given intermediate A only, you would assume there was a defect, or it was mutant for enzyme 1
1. + means it can grow with that intermediate, 0 (or –) means it cannot grow with that intermediate
1. 2. 3. If the entire column is +, these metabolites must be at the end of the pathway, this represents your total number of endpoints If the entire row is +, this mutant most likely is defective in the enzyme that converts from the precursor to the first metabolite for the entire pathway If it cannot grow with that intermediate their must be a defect in the enzyme “upstream” Four mutant strains of Neurospora require one or more of the supplemented metabolites A through D to grow. From the data below, draw the metabolic pathway for the synthesis of these metabolites and show where mutant strains (1, 2, 3, and 4) are blocked. Metabolite Mutant 1 2 3 4 A 0 + 0 + B 0 + 0 0 C 0 + + + D + + + + Four mutant strains of Neurospora require one or more of the supplemented metabolites A through D to grow. From the data below, draw the metabolic pathway for the synthesis of these metabolites and show where mutant strains (1, 2, 3, and 4) are blocked. Metabolite Mutant 1 2 3 4 A 0 0 + 0 B + + + + C + 0 + 0 D + + + 0 Seven mutant strains of Neurospora were unable to grow on minimal medium unless it is supplemented with one or more of the metabolites A through G. On the basis of the data given below draw a biochemical pathway for the synthesis of these seven substances and show where the mutants are blocked in the pathway. Metabolite F 0 0 0 0 0 + 0 Mutant 1 2 3 4 5 6 7 A + 0 0 0 0 + 0 B 0 + 0 0 0 0 0 C 0 0 + 0 0 0 0 D 0 0 0 + 0 0 0 E 0 0 0 + + 0 0 G 0 0 0 0 0 0 + B+C 0 + + 0 0 0 + D+G + 0 0 + 0 + + B+C+ E + + + + + + + HOW DO YOU SOLVE THIS?
Mutant 1 2 3 4 5 6 7 A + 0 0 0 0 + 0 B 0 + 0 0 0 0 0 C 0 0 + 0 0 0 0 D 0 0 0 + 0 0 0 E 0 0 0 + + 0 0 F 0 0 0 0 0 + 0 G 0 0 0 0 0 0 + B+C 0 + + 0 0 0 + D+G + 0 0 + 0 + + B+C+ E + + + + + + + ANSWER
6 1 7 4 D G 2 3 B C Precursor F A 5 E To identify branch points in the pathway look for mutants that grow on all of the endpoints but not on any of the endpoints individually. For example, A grows on D+G and B+C+E but does not grow on any of these individually. TIP The best way to learn is to draw a somewhat complicated pathway and then work backwards and see if you can make the data table that represents your pathway – if you can do this you understand how to interpret data for biochemical pathways. QUIZ 5 Take out a piece of paper and write Your Name Student ID Number Quiz 3 Discussion Section (2B if at 10am, 2C if at 12) Show your work for partial credit Box your answer ANSWER Please write legibly or I will not be able to grade it Pick up Quiz 3 and 4 when you’re done QUIZ 5
1) Five mutants of Neurospora are unable to grow on minimal medium unless it is supplemented with one or more of the metabolites A through E. On the basis of the data given below (where + indicates growth, and 0 indicates no growth) draw a biochemical pathway for the synthesis of these five substances and show where the mutants are blocked in the pathway. Metabolite
Mutants A B C D E B+E 1 2 3 4 5 0 0 0 0 + 0 0 + + + + 0 + 0 + 0 0 + 0 + + + + 0 + + + + + + 2) What are you going to be for Halloween this year? OR Write a test question that you think would be good for Midterm II. ...
View Full Document