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.

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.

Name: The intermediates A, B, C, D, E, and F all occur in the same biochemic
Uploaded: 2024-03-20T06:42:55.000Z
Channel: Bartleby
Description: 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

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

Assume 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 → D
A number of auxotrophic mutant strains were isolated from wild-type haploid Neurospora crassa. These strains responded to the addition of certain nutritional supplements to minimal culture medium either by growth (+) or no growth (0) The data from this experiment are presented in the table below. Diagram a biochemical pathway, complete with positions of intermediates, that is consistent with the data. Indicate where in the pathway each mutant strain is blocked.
For E. coli strains with the lac genotypes show below, use a plus sign (+) to indicate the synthesis of β-galactosidase and permease and a minus sign (–) to indicate no synthesis of the proteins.
The 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 +
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. i. I+P+0-Z+Y+A+/l+P+O+Z-Y-A- ii. I-P+0+Z+Y+A+/I+P-O+Z+Y+A+
In this problem, just put the order of the intermediates on the pathway, starting with P and ending with Z and explain your reasoning. (So to get you started, notice the class 1 mutants—and there is only mutant in this class, which is mutant #5—won’t grow on minimal medium, but will grow if you give it substance Z. However, giving it substances, p, w, x, or y won’t help. This means that the gene that is mutated lies on the pathway in a place that the Z substances is to its right and all the other substances are to its left. The easiest next one to look at is the line with 2 pluses, and then the line with 3 pluses, and then the line with 4 pluses).
Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individual’s cell in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individual’s cells are given in the following table. A plus sign means that the protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. a) If an individual who is homozygous for the mutation found in individual 2 and heterozygous for the mutation found in individual 4 mates with an individual who is homozygous for the mutation found in individual 4 and heterozygous for the mutation found in individual 2, what could the…
Consider the following simple regulatory pathways. Assume the full pathway is shown. A- E- B- F- C- G- D- 1 A H- 2 B || L You identify several null mutations (a complete deletion of the gene). For each mutant (indicated with a - sign), determine whether the final product (I, J, K or L) is inducible, uninducible, or constitutive. 3 C 4 D- [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] E [Choose ] F G I H || J K
CTP synthetase catalyzes the glutamine-dependent conversion of UTP to CTP. The enzyme is allosterically inhibited by the product, CTP. Mamma- lian cells defective in this allosteric inhibition are found to have a complex phenotype: They require thymidine in the growth medium, they have unbal- anced nucleotide pools, and they have an elevated spontaneous mutation rate. Explain the likely basis for these observations.
The enzyme glucose oxidase isolated from the mold Penicillium notatum catalyzes the oxidation of B-D-glucose to D-glucono-6-lactone. The enzyme is highly specific for the B anomer of glucose and does not affect the a anomer. In spite of this specificity, the reaction catalyzed by glucose oxidase is commonly used in a clinical assay for total blood glucose -that is, for solutions consisting of a mixture of B- and a-D-glucose, as well as other sugars present in blood. The oxidation proceeds in the presence of oxygen and forms hydrogen peroxide, in addition to the lactone. A second enzyme, called peroxidase, catalyzes the reaction of hydrogen peroxide with colorless compounds to create a colored product, which is quantified with a simple spectrophotometer. What are the circumstances required to make this possible? Aside from allowing the detection of smaller quantities of glucose, what advantage does glucose oxidase offer over the Fehling's reagent for measuring blood glucose?
Sydney Brennen isolated Salmonella typhimurium mutants that were implicated in the biosynthesis of tryptophan and would not grow on minimal medium supplemented with intermediates in tryptophan biosynthesis, some mutants were able to grow while others remained unable to grow. Review the data attached to order the biosynthetic pathway by both enzymatic step and by intermediate biomolecule. Label the step impacted by each of the mutant cell lines.
The substrate which was used to investigate the enzymatic activity of wheat bran phosphatase was p-nitrophenyl phosphate. It is converted to p-nitrophenol and inorganic phosphate by this enzyme. P-nitrophenol has a pKa of 7.1 and is colorless, while its conjugate base, p-nitrophenolate has a yellow color. The enzyme assay was performed at pH 5.1 after the 0.5 N KOH is added at the end of each enzyme assay, the pH of the assay mixture is raised to 9.1. Why was the base added?