1.ProA wild-type strain of bacteria produces a protein with the amino acid proline (Pro) at one site. Treatment withnitrous acid causes a C to change to a U. Two different mutants are produced after nitrous acid treatment ofthe bacteria. One has a serine (Ser) in place of the proline, an the other has a leucine (Leu). Further treatmentof the two mutants with nitrous acid produces new mutant strains, both with phenylalanine (Phe) at the site.Treatment of the phenylalanine carrying mutants produces no change. USerLeuPhe→PheUсAบบบUUCUUA LOUUUGCUUCUCCUACUG-PheGUUGUCG GUAGUGLeuAUUAUC lleAUAAUG-MetValсUCUUCCUCAUCGCCUCCCCCACCGACUACCACAACGGCUGCCGCAGCGSerProThrAlaAUAUTyrCysUACUAA Stop UGA-StopUAGUGG-TrpCAU HisCACCGUCGCCAA Gin CGACAGCGGAAU AsnAACAAAAAG LysUGUUGCGAUGACGAAGAGAspGluAGUAGCAGAAGGGGUGGCGGAGGGGb. Explain why further treatment of the phenylalanine mutants with nitrous acid produces no change.ArgSerArgGlya. Use appropriate codons to show how these mutations occurred. Assume only a single nucleotide changeoccurs with each treatment of nitrous acid.

Name: 1.ProA wild-type strain of bacteria produces a protein with the amino
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Description: 1.ProA wild-type strain of bacteria produces a protein with the amino acid proline (Pro) at one site. Treatment withnitrous acid causes a C to change to a U. Two different mutants are produced after nitrous acid treatment ofthe bacteria. One has a s

From the genetic code we can find the codons of the following amino acids mentioned in the question.…

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1. Pro A wild-type strain of bacteria produces a protein with the amino acid proline (Pro) at one site. Treatment with nitrous acid causes a C to change to a U. Two different mutants are produced after nitrous acid treatment of the bacteria. One has a serine (Ser) in place of the proline, an the other has a leucine (Leu). Further treatment of the two mutants with nitrous acid produces new mutant strains, both with phenylalanine (Phe) at the site. Treatment of the phenylalanine carrying mutants produces no change. U Ser Leu Phe →Phe UUC UUUA UUG с UUU Phe Leu G CUU CUC CUA CUG AUU- AUC lle A AUA Leu AUG-Met GUU GUC GUA GUG Val с UCU UCC UCA UCG CCU CCC CCA CCG- ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala A UAU Tyr Cys UAC UAA Stop UGA-Stop UAG UGG Trp CAU His CAC CAA CAG GAU GAC GAA GAG Gin AAC Asn AAA AAG Lys Asp UGU UGC Glu CGU CGC CGA CGG AGU AGC AGA AGG GGU GGC GGA GGG G b. Explain why further treatment of the phenylalanine mutants with nitrous acid produces no change. Arg Ser Arg Gly a. Use appropriate codons to show how these mutations occurred. Assume only a single nucleotide change occurs with each treatment of nitrous acid.

1. Pro A wild-type strain of bacteria produces a protein with the amino acid proline (Pro) at one site. Treatment with nitrous acid causes a C to change to a U. Two different mutants are produced after nitrous acid treatment of the bacteria. One has a serine (Ser) in place of the proline, an the other has a leucine (Leu). Further treatment of the two mutants with nitrous acid produces new mutant strains, both with phenylalanine (Phe) at the site. Treatment of the phenylalanine carrying mutants produces no change. U Ser Leu Phe →Phe UUC UUUA UUG с UUU Phe Leu G CUU CUC CUA CUG AUU- AUC lle A AUA Leu AUG-Met GUU GUC GUA GUG Val с UCU UCC UCA UCG CCU CCC CCA CCG- ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala A UAU Tyr Cys UAC UAA Stop UGA-Stop UAG UGG Trp CAU His CAC CAA CAG GAU GAC GAA GAG Gin AAC Asn AAA AAG Lys Asp UGU UGC Glu CGU CGC CGA CGG AGU AGC AGA AGG GGU GGC GGA GGG G b. Explain why further treatment of the phenylalanine mutants with nitrous acid produces no change. Arg Ser Arg Gly a. Use appropriate codons to show how these mutations occurred. Assume only a single nucleotide change occurs with each treatment of nitrous acid.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter8: The Structure, Replication, And Chromosomal Organization Of Dna

Section: Chapter Questions

Problem 6QP: Recently, scientists discovered that a rare disorder called polkadotism is caused by a bacterial...

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Recently, scientists discovered that a rare disorder called polkadotism is caused by a bacterial strain, polkadotiae. Mice injected with this strain (P) develop polka dots on their skin. Heat-killed P bacteria and live D bacteria, a nonvirulent strain, do not produce polka dots when injected separately into mice. However, when a mixture of heat-killed P cells and live D cells were injected together, the mice developed polka dots. What process explains this result? Describe what is happening in the mouse to cause this outcome.
When Griffith injected mice with a combination of live rough-strain and heat-killed smooth-strain pneumococci, he discovered that (a) the mice were unharmed (b) the dead mice contained living rough-strain bacteria (c) the dead mice contained living smooth-strain bacteria (d) DNA had been transferred from the smooth-strain bacteria to the mice (e) DNA had been transferred from the rough-strain bacteria to the smooth-strain bacteria
1. The first base in the tRNA anticodon loop is also wobbling, that is one tRNA is able to pair with several codons of mRNA. If the firs base of the anticodon is U, its complement in the codon is: a.A or C b.A or G c.C or G d.A or U 2. The probability of an offspring having all the recessive traits from the cross AaBBCc X AaBbCC is: a.one-sixteenth b.three-fourth c.zero d.nine-sixteenth 3. The probability of an offspring having all the dominant traits from the cross AaBBCc X AaBbCC is: a.three-fourth b.one-sixteenth c.nine-sixteenth d.zero
A small section of a gene for a protein has the following nucleotide sequence:CTG GGA TCC TAA GGTWhich of the following mutations would cause a nonsense mutation in the sequence shown above? Select one: a. Replacement of second adenine base with a cytosine base b. Insertion of guanine base after the first thymine base c. Insertion of guanine base after the first adenine base d. Replacement of second cytosine base with a adenine base
5. You isolate ten new mutant yeast strains that are defective in synthesis of leucine, an amino acid. These Leu- mutants (numbered 1-10) were all isolated in a strain of mating type a (MAT a). S. cerevisiae yeast are either mating type a or α. As it turns out, your high school classmate, now at the University of Chicago, has independently isolated ten yeast Leu- mutants (numbered 11-20) in a strain of mating type α (MAT α). You and your ex-classmate decide to combine your resources and determine how many different genes are represented by your 20 mutant strains. You cross each of the MAT a strains to each of the MAT α strains. Your experimental observations are shown in the table below, where an empty square indicates that the diploid did not grow on minimal medium and a filled square indicates that the diploid did grow on minimal medium. Based on this experiment, what is the minimum number of genes required for leucine synthesis?
1. Below is the base sequence of protein for normal hemoglobin and the base sequence for the sickle cell hemoglobin:Normal: GGG CTT CTT TTTSickle: GGG CAT CTT TTTa. Transcribe and translate the normal and sickle cell DNA. b. Identify this as a point or frameshift mutation. Explain. c. If the base sequence read GGG CTT CTT AAA instead, would this result in sickle cell hemoglobin? Explain.
4. Some transposases use extensive DNA replication to leave one copy of the element behind in a process called ____________________.A. persistent transpositionB. chronic transpositionC. exchange transpositionD. replicative transpositionE. none of the above 5. Which of the following is not an example of a transposable element found in bacteria?A. Tn5B. Tn10C. IS1D. IS903E. Xis 6. Besides a transposase, many common composite transposons in bacteria encode genes for:A. Toxin productionB. Antibiotic resistanceC. DNA repairD. Homologous recombinationE. None of the above
When Griffith incubated heat-killed virulent S strain bacteria with live avirulent R strains, he found that R cells were transformed into lethal, disease-causing bacteria. What is the molecule responsible for the transformation of R cells into the S type? Why did the transformation occur? A. Proteins in the heat-killed S cell extract; they were able to synthesize the capsular polysaccharide in the R cells and make them virulent. B. The capsular polysaccharide in the heat-killed S cell extract; the polysaccharide was able to attach to R cells, thus making them virulent. C. DNA in the heat-killed S cell extract; the DNA altered the genetic makeup of R cells, allowing them to synthesize the polysaccharide capsule, thus making them virulent. D. Specialized lipids in the cell membrane of the S cells were able to integrate with the R cells, conferring the ability to evade host immunity.
A small section of a gene for a protein has the following nucleotide sequence: TAT AGG GAC CTA TGT Which of the following mutations would cause a missense mutation in the sequence shown above? a. Replacement of first cytosine base with guanine base b. Replacement of final thymine base with guanine base c. Replacement of second guanine base with cytosine base d. Replacement of first thymine base with adenine base
A small section of a gene for a protein has the following nucleotide sequence: GCT CTA GCT ATC TGA Which of the following mutations would cuase a silent mutation in the sequence shown above? a. Replacement of second adenine base with thymine base b. Replacement of first thymine base with adenine base c. Replacement of second guanine base with cytosine base d. Replacement of first cytosine base with guanine base
Explain how Benzer’s experimental results revealed thatthe rII region in bacteriophage T4 contains two genes,each composed of many nucleotide pairs.
A small section of bacterial DNA template (antisense) strand has the following nucleotide sequence: CGA AAA GAG AAT A mutation in the above sequence involved a substitution of a single base, resulting in an incomplete protein due to a nonsense mutation. Which of the following gen sequences exemplifies the mutation described above? a. CGA UAA GAG AAT b. CGA AAA AAG AAT c. CGA AAA GAG ACT d. UGA AAA GAG AAT