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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 14 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
- The intermediates A, B, C, D, E, and F all occur in the same biochemical pathway G is the product of the pathway, and mutations 1 through 7 are all G –, meaning that they cannot produce substance G. The following table shows which intermediates will promote growth in each of the mutants. Arrange the intermediates in order of their occurrence in the pathway at which each mutant strain is blocked. A “+” in the table indicates that the strain will grow if given that substance, an “o” means lack of growth.arrow_forwardSchizosaccharomyces 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_forward
- The pathway for cystiene synthesis in E.coli involves five precursor compounds, compound A-E. The requirements for each of the autotrophs are summarized in the table below. Use the data, define the biosynthetic pathway. Draw the pathway for cysteine synthesis. Fill in the blanks with A-E compounds as appropriate and then label the arrows with the auxotrophs C1-C8 that are unable to complete that step. There are may be more than one quxotroph that is defective at a particular step. that is - on C4 in Darrow_forwardWhich gene encodes the enzyme at the upper boxed step in the pathway? Which gene encodes the enzyme at the upper boxed step in the pathway? gene 1 gene 2 gene 3 gene 4arrow_forwardGive only typing answer with explanation and conclusion For the genotype shown below, which best describes the expression of the B-galactosidase gene. I+ Oc Z+ / F’ Is Constitutive Repressed Inducible None of the abovearrow_forward
- Indicate the presence of the the B galactosidase and permease. Please explain and follow the instructions in the picturearrow_forwardI have this strain of e coli. Is P+ o+ Z+ Y+ / I- P+ oC Z- Y+ Will beta-galactosidase and permease be expressed? If they are will they be inducible or constitutive?arrow_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_forward
- For 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_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_forwardFor the lac genotypes of Escherichia coli shown in the following Table 1, predict the expression of beta-galactosidase (Z) and permease (Y) is inducible or noninducible or constitutive. Explain your answer. Table 1 Genotype I-P+O+Z+Y+ (i) (ii) I+P+OCZ+Y+ (iii) ISP+O+Z+Y+ (iv) I+P+O+Z+Y-//I+P-O+Z+Y+ (v) ISP+OcZ+Y+//I-P+O+Z+Y- Condition No lactose No lactose lactose lactose No lactosearrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
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