+ + d. + e + f 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?

+ + d. + e + f 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?

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

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 + + + > > >
The synthesis of flower pigments is known to be dependent on enzymatically controlled biosynthetic pathways. For the crosses shown here, postulate the role of mutant genes and their products in producing the observed phenotypes: (a) P1: white strain A * white strain B F1: all purple F2: 9/16 purple: 7/16 white (b) P1: white * pink F1: all purple F2: 9/16 purple: 3/16 pink: 4/16 white
A fluctuation test was carried out to determine the rate of mutation to Azetidine resistance (a toxic proline analog) in S. typhimurium. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures were grown to 10° cells / ml. A 0.1 ml sample of each culture was then plated on each plate (a total of 20 plates) to detect AztR mutants. The results are shown in the following table. Calculate the mutation rate of S. typhimurium to AztR. Culture # # AztR mutants Culture # # AztR mutants 11 12 3 4. 13 14 15 4. 30 303 97 69 14 16 17 18 19 20 10 19
In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?
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?
In an analysis of five rII mutants, complementation testing yielded the following results:
For each of the E. coli strains containing the lacoperon alleles listed, indicate whether the strain isinducible, constitutive, or unable to expressβ-galactosidase and permease.a. I+ o+ Z− Y+/ I+ ocZ+ Y+b. I+ o+ Z+ Y+/ I− ocZ+ Y−c. I+ o+ Z− Y+/ I− ocZ+ Y−d. I−P− o+ Z+ Y−/ I+ P+ ocZ− Y+e. Iso+ Z+ Y+/ I− o+ Z+ Y−
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.
For the following genotype, indicate if beta-galactosidase will be made in the presence of lactose and absence of lactose. P₁+1+PL+OcZ/ P₁1 P₁O+Z+ O O O beta-galactosidase will be made in both the presence and absence of lactose (+; +) beta-galactosidase will be made in the presence but not in the absence of lactose (+;-) beta-galactosidase will not be made in the presence but will be made in the absence of lactose (-; +) beta-galactosidase will not be made in either the presence or absence of lactose (-;-)
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]
Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6
Draw a hairpin structure like that shown in Figure 18.5 for the repeated sequence found in fragile-X syndrome (see Table 18.1).