23 Ten mutant mouse strains with white coats were tested by crossing eachwith all the other strains. The results are shown in the table below. A "-"symbol indicates the offspring were white (failure to complement), and a"+" symbol indicates that the offspring showed wild-type brown coats(complementation).Which of the other strains have a mutation in the same gene as strain 2?abсdStrains 5, 6, and 10 have a mutation in the same gene as strain 2.Strains 1, 3, 4, 7, 8, and 9 have a mutation in the same gene as strain 2.All of the other strains have a mutation in the same gene as strain 2.MutantNone of the other strains have a mutation in the same gene as strain 2.12 +34 +5 +Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer.+++++6 +8 +910++1I++++++ + + + + ++++ I+++ I++++ I-+++ + +2 3 4 5 6 7 8 9 10Mutant+++-: offspring are mutant+: offspring are wild-typeI

Complementation usually occur if the mutations are in different genes. Complementation occurs when…

Answered: Ten mutant mouse strains with white…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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You've crossed two strains: (Hfr) pro+ lac+ arg+ with (F-) pro- lac- arg- . Interrupted mating studies indicated that arg+ enters the cell last, so you want to select for arg+ recombinants in order to map the genes by recombination. Which type of plates will you use for your initial selection? Containing proline, lactose, and arginine. Containing proline, glucose, and arginine. Containing arginine and glucose. Containing lactose and proline. Containing glucose and proline. Minimal media with nothing added.
An Hfr strain is used to map three genes in an interrupted mating experiment. The cross is Hfr/a+b+c+ rif x F-a-b-c- rifr (note that rifr is resistance to the antibiotic rifampicin). The a+ gene is required for the biosynthesis of nutrient A, the b+ gene for nutrient B, and the c+ gene for nutrient C. The cross is initiated at time=0. At various times, the mating mixture is plated out on three different types of medium. Each plate contains minimal medium (MM) plus rifampicin, plus specific supplements that are listed in the table below. The results for each time interval are shown as the number of colonies growing on each plate.An Hfr strain is used to map three genes in an interrupted mating experiment. The cross is Hfr/a+b+c+ rif x F-a-b-c- rifr (note that rifr is resistance to the antibiotic rifampicin). The a+ gene is required for the biosynthesis of nutrient A, the b+ gene for nutrient B, and the c+ gene for nutrient C. The cross is initiated at time=0. At various times, the…
Eight mutant bacteriophage strains cannot lyse a certain type of bacteria that can be lysed by wild-type bacteriophages. The mutant strains were allowed to infect the bacteria in a complementation test. A "+" indicates that lysis occurred with coinfection. A "-" indicates that lysis did not occur. OXWTMAO G K Т Т М А А Q C GKWTMAQC +++++ - - - - - - - + - + - + - - + - - - O +++ - + - - + + A cistron is defined by no complementation in the How many genes are controlling lysis in this bacteriophage? (Use a number not a word in the space) configuration.
DNA obtalned from the indicated donor strains is used to TRANSFORM the indicated recipient strains. The resulting progeny are plated on minimal medium so that only wild- type recombinants are scored. The number of wild-types for each cross is given in the chart below. What is the order of the genes? Donor Recipient wid type colonies a-b- c+ a+ b+ c- 273 a- b+ c- a+ b-c+ 462 a- b+ C+ a+ b-c- 2 a b- C+ a- b+ c+ O 4-b-c a-o-b b-a-c cannot be determined Question 18 of 26
E. coli strain B (permissive) were co-infected with rlla and rllb mutant phage T4. Progeny were plated in a dilution series with the following results: Lawn Dilution Number of plaques Strain B 10-6 10 Strain K 10-4 What is the recombination frequency? Remember that the readout is viability and you cannot see the inviable progeny. Answer:
A yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells 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 of results.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1auxotroph?
Using designations "Group A", “Group B," etc., organize the mutant strains into complementation groups. If the result of any cross does not fit your groupings, indicate which cross(es) you have excluded. Consider the consequence if one of the mutations was dominant. You should see that complementation tests could not be used to provide information about dominant mutations. a) If a mutation were dominant, how would the diploid be affected? What effect would this have on the determination of your complementation groups? b) How would you test if any of these trp¯ mutations were dominant?
A cross is made between two E. coli strains: Hfr arg + bio + leu+ × F− arg − bio − leu−. Interrupted mating studies show that arg+ enters the recipient last, and so arg+ recombinants are selected on a medium containing bio and leu only. These recombinants are tested for the presence of bio + and leu+. The following numbers of individuals are found for each genotype: arg+ bio+ leu+ 320 arg+ bio- leu+ 0 arg+ bio+ leu- 8 arg+ bio- leu- 48a. What is the gene order? b. What are the map distances in recombination percentages?
Homozygous wild-type male mice (AA BB CC) were crossed with triplemutant female mice (aa bb cc), forming an F1 generation with the followinggenotype (Aa Bb Cc). The F1 males were crossed with triple mutantfemales, forming the following F2 phenotypes”“a B c” 3“A b C” 3“a b c” 8“A B c” 5“a b C” 5“A B C” 8“a B C” 6“A b c” 6 44 Determine the sequence of the genes
With the technique of interrupted mating four Hfr strains were tested for the sequence in which they transmitted a number of different genes to an F strain. Each Hfr strain was found to transmit its genes is a unique sequence, as shown in the accompanying table. What is the gene sequence in the original strain from which these Hfr strains were derived?
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?
Below is a partially filled in complementation table. Please answer the following questions. 1 2 3 4 LO 5 6 1 2 I 3 4 5 4 + + 5 6 a. If you were to perform a complementation test with mutants 1 and 2, what would be the result? [Select] b. If you were to perform a complementation test with mutants 5 and 6, what would be the result? [Select] c. If you were to perform a complementation test with mutants 3 and 4, what would be the result? [Select] d. There is one complementation group that you can determine from the information above. If you performed a complementation test with mutant 1 and 6 and there was growth, would you say that the mutants were part of that complementation group? [Select] e. What is the minimum number of complementation groups that could exist based on the information above? [Select] f. What is the maximum number of complementation groups that could exist based on the information above? [Select]

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