To analyze:
In the given table, the partial diploid genotypes are tested for the following aspescts:
Complete the following table for production of
Introduction:
In the lac operon, the partial diploids are produced by conjugation between
Based on the studies of structural gene mutations, Jacob, Monod, and Colleagues concluded that lac
These two wild type alleles are always dominant to the mutant type alleles.
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Chapter 12 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
- 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 + + + > > >arrow_forwardThe intermediates A, B, C, D, E, and F all occur inthe same biochemical pathway. G is the product of thepathway, and mutants 1 through 7 are all G−, meaningthat they cannot produce substance G. The followingtable shows which intermediates will promote growthin each of the mutants. Arrange the intermediates inorder of their occurrence in the pathway, and indicatethe step in the pathway at which each mutant strain isblocked. A + in the table indicates that the strain willgrow if given that substance, an O means lack of growth.SupplementsMutant A B C D E F G1 + + + + + O +2 O O O O O O +3 O + + O + O +4 O + O O + O +5 + + + O + O +6 + + + + + + +7 O O O O + O +arrow_forwardA unique aquatic plant was discovered from a lagoon in El Nido, Palawan. To determine the protein content of the plant, an adequate amount of the plant extract was acquired, and Bradford assay was performed. Determine the total protein concentration, in µg/mL, of the acquired plant extract. A bovine serum albumin (BSA) stock solution with a concentration of 250 µg/mL was used and mixtures with the following compositions and absorbance readings were prepared: Volume of BSA (mL) Volume of water (mL) Absorbance Tube # at 595 nm 1 0.00 2.00 0.000 2 0.20 1.80 0.112 3 0.40 1.60 0.225 4 0.60 1.40 0.318 5 0.80 1.20 0.432 6 1.00 1.00 0.551 The absorbance reading of the plant extract is 0.275.arrow_forward
- Three haploid fungal mutants that require compound W for growth were isolated. Each mutant contains a recessive allele in a single gene. Three compounds (A, B and C) in the biosynthetic pathway to W are known, but their order in the pathway is unknown. Each compound is tested for its ability to support the growth of each of the three mutants. Phenotypes of all of the three mutants are shown in the following table (“+" indicates growth, "-" indicates no growth). A C W Mutant 1 Mutant 2 Mutant 3 What would be the phenotype of a haploid mutant that contains both mutant alleles in mutant 2 and 3? Phenotype refers to growth or absence of growth on compounds A, B, C and WN. O Like mutant 1 O Like mutant 2 Like mutant 3 O Like wild typearrow_forwardFor each of the following mutant E. coli strains,plot a 30-minute time course of concentration ofβ-galactosidase, permease, and acetylase enzymesgrown under the following conditions: For the first10 minutes, no lactose is present; at 10 minutes, lactosebecomes the sole carbon source. Plot concentration onthe y-axis, time on the x-axis. (Don’t worry about theexact units for each protein on the y-axis.)a. I− P+ o+ Z+ Y+ A+ / I+ P+ o+ Z− Y+ A+b. I− P+ ocZ+ Y+ A− / I+ P+ o+ Z− Y+ A+c. IsP+ o+ Z+ Y+ A+ / I− P+ o+ Z− Y+ A+d. I− P− o+ Z+ Y+ A+ / I− P+ ocZ+ Y− A+e. I− P+ o+ Z− Y+ A+ / I− P− ocZ+ Y− A+arrow_forwardFor each of the following genotypes, explain how mutation (identified by a (-) will affect the organism grown in lactose medium. Indicate whether a) B-galactosidase will be synthesized or not, b) synthesis of B-galactosidase is inducible (1) or constitutive (C) and c) growth of the organism will occur or not. a. I-P+O+Z+Y+A+ b. I+P-O+Z+Y+A+ c. I+P+O-Z+Y+A+ d. I+P+0-Z+Y+A+/l+P+O+Z-Y-A- e. I-P+0+Z+Y+A+/l+P-O+Z+Y+A+arrow_forward
- Streptomycin resistance in Chlamydomonas may result from a mutation in either a chloroplast gene or a nuclear gene. What phenotypic results would occur in a cross between a member of an mt+ strain resistant in both genes and a member of a strain sensitive to the antibiotic? What results would occur in the reciprocal cross?arrow_forwardAssume that a series of compounds has been discovered in Neurospora. Compounds A–F appear to be intermediates in a biochemical pathway. Conversion of one intermediate to the next is controlled by enzymes that are encoded by genes. Several mutations in these genes have been identified and Neurospora strains 1–4 each contain a single mutation. Strains 1–4 are grown on minimal media supplemented with one of the compounds A–F. The ability of each strain to grow when supplemented with different compounds is shown in the table (+ = growth; o = no growth). Which biochemical pathway fits the data presented? Media Supplement Strain A B C D E F 1 o o o + + + 2 o o o o + + 3 o o o o + o 4 o o + + + + A) A → B → C → D → E → F B) A → B → C → F → D → E C) F → B → C → D → A → E D) A → B → C → D → F → E E) A → B → F → E → C → Darrow_forwardThe pathway for arginine biosynthesis in Neurospora crassa involves several enzymes that produce a series of intermediates as shown. O O O O ornithine citrulline ARG-E arginosuccinate arginine N-acetylornithine arginine You did a cross between ARG-E ARG-H* and ARG-E* ARG-H¯¯ Neurospora strains and identified an Arg- strain from an NPD tetrad. (Assume that Neurospora forms tetrads in the same way yeast do.) Which compound would rescue growth of this Arg- spore? N-aceltylornithine ARG-F ornithine citrulline ARG-G ARG-H → argininosuccinatearrow_forward
- In the Avery, McLeod, McCarty Experiment where supernatant from heat killed, virulent S Strain pneumonia solutions were added to non-virulent R Strain pneumonia cell cultures and allowed to grow in liquid media (i.e., broth). In tubes where Protease was added to the supernatant prior to cell culture, what was the observed effect when plating and growing the S. pneumonia cells to solid media?arrow_forward18) Based on your knowledge of lactose system in prokaryotes) and considering the following genotypes, complete the table below indicating a "+" where an enzyme is produced and a where no enzyme is produced. Note: Is is a mutant that cannot recognize allolactosearrow_forwardFour different uracil auxotrophs of Neurospora are tested for growth on uracil and uracil precursors. The data are shown in the table below. A plus sign (+) means growth. Mutant 1 is defective for which enzymatic step? Compound A B C D Uracil Mutant 1 - + - - + Mutant 2 - + - + + Mutant 3 + + - + + Mutant 4 - - - - + D to A A to D D to B C to A A to Carrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning