Concepts of Genetics (11th Edition)
Concepts of Genetics (11th Edition)
11th Edition
ISBN: 9780321948915
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 6, Problem 23ESP

In studies of recombination between mutants 1 and 2 from Problem 21, the results shown in the following table were obtained.

Chapter 6, Problem 23ESP, In studies of recombination between mutants 1 and 2 from Problem 21, the results shown in the

  1. (a) Calculate the recombination frequency.
  2. (b) When mutant 6 was tested for recombination with mutant 1, the data were the same as those shown above for strain B, but not for K12. The researcher lost the K12 data, but remembered that recombination was ten times more frequent than when mutants 1 and 2 were tested. What were the lost values (dilution and plaque numbers)?
  3. (c) Mutant 7 (Problem 21) failed to complement any of the other mutants (16). Define the nature of mutant 7.
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An Hfr strain that is hisE + and pheA + was conjugated to a strain that is hisE − and pheA −. The conjugation was interrupted at different times, and the percentage of recombinants for each gene was determined by streaking on media that lacked either histidine or phenylalanine. The following results were obtained:   A. Determine the map distance (in minutes) between these twogenes.B. In a previous experiment, it was found that hisE is 4 minutesaway from pabB and that PheA is 17 minutes from pabB. Drawa genetic map showing the locations of all three genes.
A yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies.These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic,25% adenine-requiring auxotrophsa. Explain the difference between these two types ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?
Is )Given the experimental set up below, and knowing the genes for stalked eyes (st) and black bodies (b) are on the same chromosome and recessive to wild type (un-stalked eyes and light body), answer the questions below the diagram. 1. st st F1 F2 A = 489 B = 56 C= 497 D= 49 A) ) How many recombinant animals are in the F2 generation? B) woH) How far apart, (as both a % recombination and in map units) are the stalk eyed and black bodied genes on this chromosome?

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Concepts of Genetics (11th Edition)

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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license