Time mapping is performed in a cross involving the genes his,leu, mal, and xyl. The recipient cells are auxotrophic for all fourgenes. After 25 minutes, mating is interrupted, with the resultsin recipient cells shown below. Diagram the positions of thesegenes relative to the origin (O) of the F factor and to one another.(a) 90% are xyl+(b) 80% are mal+(c) 20% are his+(d) None are leu+

Time mapping is performed in a cross involving the genes his,leu, mal, and xyl. The recipient cells are auxotrophic for all fourgenes. After 25 minutes, mating is interrupted, with the resultsin recipient cells shown below. Diagram the positions of thesegenes relative to the origin (O) of the F factor and to one another.(a) 90% are xyl+(b) 80% are mal+(c) 20% are his+(d) None are leu+

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter15: From Dna To Protein

Section15.1: The Connection Between Dna, Rna, And Protein

Problem 1SB

See similar textbooks

Similar questions

The diploid that resulted from mating these two haploids would have which of the following auxotrophies? Select all that apply.Strain 1: MATa ura3-52 met15Δ0 leu2Δ0 Strain 2: MATalpha ura3-52 met15Δ0 trp1Δ63 a.) histidine b.) tryptophan c.) uracil d.) lysine e.) leucine f.) methionine
Time mapping is performed in a cross involving the genes his, leu, mal, and xyl. The recipient cells were auxotrophic for all four genes. After 25 minutes, mating was interrupted with the following results in recipient cells. Diagram the positions of these genes relative to the origin (O) of the F factor and to one another. (a) 90% were xyl+ (b) 80% were mal+ (c) 20% were his+ (d) none were leu+
A cross was performed between a yeast strain that requires methionine and lysine for growth (met− lys−)and another yeast strain, which is met+ lys+. One hundred asci were dissected, and colonies were grownfrom the four spores in each ascus. Cells from thesecolonies were tested for their ability to grow on petriplates containing either minimal medium (min), min+ lysine (lys), min + methionine (met), or min + lys+ met. The asci could be divided into two groupsbased on this analysis:Group 1: In 89 asci, cells from two of the four spore colonies couldgrow on all four kinds of media, while the other two spore coloniescould grow only on min + lys + met.Group 2: In 11 asci, cells from one of the four spore colonies couldgrow on all four kinds of petri plates. Cells from a second one ofthe four spore colonies could grow only on min + lys plates andon min + lys + met plates. Cells from a third of the four sporecolonies could only grow on min + met plates and on min + lys+ met. Cells from the…
The following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.
An Hfrstrain that is a *b*c*d* e*f* g *h* is mated with an F strain that is a b e d e f gh. The mating is interrupted at 5 minutes interval, and the genotypes of the F recombinants are determined. The results obtained are tabulated in Table 2. Draw the map of the Hfrchromosome and indicate the position of the origin of transfer, the direction of the transfer and the minutes between genes. Table 2:Entry time of Hfr chromosome into recipient cell. Time a d e f h 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
If an E. coli auxotroph X could grow only on a medium containing leucine, and an auxotroph Y could grow only on a medium containing threonine. (i) Write the genotype of both E. coli strains. (ii) How would you test whether DNA from X could transform Y?
During the analysis of seven rII mutations in phage T4, mutants1, 2, and 6 were in cistron A, while mutants 3, 4, and 5 were incistron B. Of these, mutant 4 was a deletion overlapping mutant5. The remainder were point mutations. Nothing was knownabout mutant 7. Predict the results of complementation (+ or -)between 1 and 2; 1 and 3; 2 and 4; and 4 and 5.
Schizosaccharomyces pombe, also known as "fission yeast," is a powerful model organism in molecular and cell biology. While performing a genetic screen, you discover an auxotrophic S. pombe strain that is unable to synthesize one or more vitamins. The following table represents the key experiments you performed during your genetic screen. Fill in the table with the outcome of each experiment for your mutant strain (using + for growth and - for no growth). Medium Rich media Minimal media Minimal media + all vitamins Minimal media + all amino acids Growth Wild-type + + + + Mutant + + + > > >
Have a look at the results in this SDS-PAGE gel. A MW ladder is shown. In the "control" lane, a sample of recombinant protein "LBT3" was loaded. Lane A shows proteins that have been prepared from leukocytes from a normal, healthy individual. Lanes B, C, D, E and F show protein prepared from leukocytes from individuals with lymphoma. What is the best explanation for the result shown for patient D? 200 150 100 75 50 37 20 Weight, kDa MW ladder Control A
From a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.
The following recombinants are recovered when conjugation occurs between an a*d*g+ donor and an adg recipient. at d+ g+ = 84% a d g+ = 6% at d g+ = 10% a dt g+ = less than 1% What is the map distance between the a and d genes? 10 map units 74 map units less than 1 map unit 84 map units 6 map units
b d f a + + + + d + e The normal flower color in garden peas is purple. A researcher interested in the biochemical pathway leading to purple flower color has isolated six white mutants, each of which is recessive to purple. The six white mutants have been designated a, b, c, d, e and f. Complementation tests have been carried out for each pair of mutants, and the results are indicated in the table above, where a “+" indicates complementation (purple flower color in the offspring of the complementation test) and a “-" indicates no complementation (white flower color). What is the number of genes (complementation groups) identified by these mutants? +| +|+|+|+