Final 03

Final 03 - Name: 12. Lac Operon (30 points) page 5 A. Given...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name: 12. Lac Operon (30 points) page 5 A. Given the genotypes, fill in the appropriate phenotypes for LacZ and LacY. The inducer is lactose or IPTG. Genotype 1 I+O+Z+Y+ 2 IS O+ Z+ Y3 IS O+ Z- Y+ / F’I-OcZ+Y4 I+OcZ+Y-/ F’ I-O+Z+Y5 I-OCZ+Y+ / F’ IS O+Z-Y- Phenotypes LacZ expression LacY expression no inducer inducer no inducer inducer + + + + + + + + + + 10 pt, 1/2 ea B. Name one strain that indicates that OC is epistatic to IS, and what data tells you that? #5, lacZ and lacY expression (credit given for either) #3 lacZ expression C. A new gene is identified that regulates the Lac operon, called Gene T. An allele Gene T-2 was analyzed. We know that Gene T+ acts in the glucose, ADC1, CRP pathway. The assays of Gene T+ function are done in a I- cell, with glucose present. (The phenotypes would be complicated by reporting LacZ expression with no glucose and with glucose column. However, you can figure this out using this data alone). 2pt i. Is Gene T-2 recessive or dominant (circle one) 2pt Strain LacZ expression + + + + + - ii. Draw the glucose, Gene T+,ADC1,CRP, LacZ, pathway, with gene order and sign of regulation (bar or arrow) Glucose • iii. ADC1+ GeneT+ CRP LacZ 1. GeneT+ 2. GeneT-2 3. GeneT-2/ Gene T+ 4. ADC1* 5. ADC1* GeneT-2 6. ADC17 . ADC1- Gene T-2 8. CRP 9. CRP- GeneT-2 4pt order, 2 pt each bar/arrow=12pt overall Why is the combination of strains 2, 4 and 5 not informative for trying to determine the order of function of GeneT+ and ADC1+? Not informative because both single mutants have same phenotype. 4pt NAME page 6 13. True/False short answer. (26 points-4 points each. 1 point T or F, 3 points short explanation. Note that F is worth 6 points.) Explanation must have some relevance to True False for full credit. Where statement is true, make one short comment that indicates you know something about the statement. A. Mendel was a funny-looking Italian hairdresser turned molecular biologist. False, geneticist, monk, Austrian--any answer will do for this joke question. I enjoyed some of your senses of humor! B. A complementation test requires recombination between two mutant genomes to test how far apart the mutations are. False - complementation test determines if two mutations are in same or different gene OR- recombination test uses recombination and determines distance between two genes. C. The definitive experiment in the phage test of genetic code used three suppressor mutations, two of which were “+” and one “-” as indicated by the sign convention. False, combination of 3 “+” or 3”-” was used in the definitive experiment. Credit was given for a variety of other answers. D. Recombination between an Hfr and an F- requires one SINGLE crossover to generate a viable, intact recombinant product in the F-. False requires 2 single crossovers or 1 double crossover. E. Most of the human genome encodes either a protein or an RNA. False- only about 3-5% of human genome encodes protein or RNA,most is junk. F. (6 points) Oligonucleotide length problem. A researcher says he needs an 6 base long oligonucleotide to uniquely detect a specific DNA sequence in the 10,000 base long HIV genome, and a 16 base long oligonucleotide to detect the same sequence when HIV is integrated into the human genome (which is 3x109 bases long). False - need at least a 7 base pair long oligonucleotide (4N>10,000 N=7, 4N=16,384) A 16 base pair long oligonucleotide is sufficient (416> 4 x 109) Name: A.. page 7 14. Genetic Recombination and Complementation Using Phage (14 points) Below are the data generated in analysis of the mutants. Mutant #1 is given with an “X”. On the map below, place where the other 4 mutations lie. If one of the mutations is a deletion, draw a bar to indicate the region deleted. Indicate the location of the genes consistent with the data (label genes A, B etc) State here which mutations are dominant, if any. None--2 pt B. C. HINT: Solve mutation position first, mapping mutations 1, 4 and 5 first. Cross 1x1 1x2 1x3 1x4 1x5 2x3 2x4 2x5 3xwt 3x4 3x5 4x5 Complementation + + + - Recombination (%) 0 0.25 0 0.5 1.5 0 0.75 1.25 not relevant 0.25 0.25 0.0 1.0 + 2.0 2.0 3.0 X#1 X#4 X#2 X#3. X#5 Gene A Gene B 2 pt each mutation, 2 pt each gene. Note that deletion 3 must overlap with gene A, and 2 pt part C Name 15. Yeast Genetics Questions (30 points) page 8 A.Mitotic recombination is needed to generate MATa/MATa cells from a MATa/MATa cell (used for complemention tests). In the diagram below, draw how a MATa/MATa cell can be formed by mitotic recmbination, starting with the G1 cell shown. In your drawing indicate the site of recombination and features of chromosome segregation if needed (10pt) Draw answer in space to right. MATa MATa 1 1 1&3 cM MATa MATa MATa Sly+ cosegregate 3 2 centromeres 3 Sly- MATa MATa MATa 4 10 pt: 4 pt G2 cell, 2 pt crossover, 2 pt 1 &3 coseg, 2pt final product (& 2 and 4, but that was not essential to show) B. From the map shown, what fraction of the MATa/MATa cells will by homozygous for one of the sly alleles, and which allele will it be (2pt)? 0- all products still heterozygous. “0” is enough. Genotype C. Order of function. Ste91* has one mutation, Ste92* has one mutation. Ste91* and ste92* are not linked. Phenotypes of diploids and haploids are shown. i. Is ste91* dominant or recessive (circle one). ii. Is ste92* dominant or recessive (circle one) iii. Draw the pathway of the order of function, from pheromone to mating. Include bars and arrows. 1. 2. 3. 4. 5. 6. Wt Mating response (schmoo) no pheromone pheromone Ste91* Ste91*/+ - Ste92* + Ste92*/+ + Ste91* ste92* - + + + - Pheromone Ste92+ Ste91+ mating 2p ea i and ii;10 pt pathway (4pt order, 6pt bars/arrows) D. Another two mutations are isolated, called ste93* and ste93**. Ste93* is dominant and fails to mate, while ste93** is recessive and always mates. State whether you can determine the mating phenotype of a MATa/MATa ste93*/ste93** cell, and if so what that phenotype would be. Yes..it fails to mate (phenotype of dominant mutant) 4 pt. ...
View Full Document

This note was uploaded on 04/02/2008 for the course MCB 320 taught by Professor Walsh during the Spring '08 term at Arizona.

Ask a homework question - tutors are online