The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon. Introduction : Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene. RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

Question

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Answer 1

Question

Chapter 26, Problem 1P

(a)

Summary Introduction

To determine: The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon.

Introduction:

Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene.

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

(a)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

Time needed by primary transcript is:

T=5300 nucleotides50−90 nucleotides per second=59−106 sec

The time can be rounded to 60-100 sec.

Conclusion

Conclusion:

Thus, the time required for primary transcript is 60-100 sec.

(b)

Summary Introduction

To determine: The length of the transcriptional bubble formed by RNA polymerase.

Introduction:

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template. RNA polymerase is majorly found in all living organism that takes parts in the central dogma process.

(b)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

The distance travelled by the replication bubble in 10 sec is:

=(10 sec)×(50−90nuceotide/sec)=500−900 nucleotides

Conclusion

Conclusion:

Distance travelled by the replication bubble in 10 sec is 500-900nucleotides.

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You’ve isolated a protein and determined that the Native molecular weight of the holoenzyme is 160 kD using size exclusion chromatography. Analysis of this protein using SDS-PAGE revealed 2 bands, one at 100 kD and one at 30 kD. The enzyme was found to be 0.829% NAD (by weight). What further can be said regarding the structure of the polypeptide?

Answer 2

Question

Chapter 26, Problem 1P

(a)

Summary Introduction

To determine: The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon.

Introduction:

Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene.

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

(a)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

Time needed by primary transcript is:

T=5300 nucleotides50−90 nucleotides per second=59−106 sec

The time can be rounded to 60-100 sec.

Conclusion

Conclusion:

Thus, the time required for primary transcript is 60-100 sec.

(b)

Summary Introduction

To determine: The length of the transcriptional bubble formed by RNA polymerase.

Introduction:

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template. RNA polymerase is majorly found in all living organism that takes parts in the central dogma process.

(b)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

The distance travelled by the replication bubble in 10 sec is:

=(10 sec)×(50−90nuceotide/sec)=500−900 nucleotides

Conclusion

Conclusion:

Distance travelled by the replication bubble in 10 sec is 500-900nucleotides.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 26, Problem 1P

(a)

Summary Introduction

To determine: The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon.

Introduction:

Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene.

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

(a)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

Time needed by primary transcript is:

T=5300 nucleotides50−90 nucleotides per second=59−106 sec

The time can be rounded to 60-100 sec.

Conclusion

Conclusion:

Thus, the time required for primary transcript is 60-100 sec.

(b)

Summary Introduction

To determine: The length of the transcriptional bubble formed by RNA polymerase.

Introduction:

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template. RNA polymerase is majorly found in all living organism that takes parts in the central dogma process.

(b)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

The distance travelled by the replication bubble in 10 sec is:

=(10 sec)×(50−90nuceotide/sec)=500−900 nucleotides

Conclusion

Conclusion:

Distance travelled by the replication bubble in 10 sec is 500-900nucleotides.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 26, Problem 1P

(a)

Summary Introduction

To determine: The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon.

Introduction:

Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene.

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

(a)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

Time needed by primary transcript is:

T=5300 nucleotides50−90 nucleotides per second=59−106 sec

The time can be rounded to 60-100 sec.

Conclusion

Conclusion:

Thus, the time required for primary transcript is 60-100 sec.

(b)

Summary Introduction

To determine: The length of the transcriptional bubble formed by RNA polymerase.

Introduction:

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template. RNA polymerase is majorly found in all living organism that takes parts in the central dogma process.

(b)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

The distance travelled by the replication bubble in 10 sec is:

=(10 sec)×(50−90nuceotide/sec)=500−900 nucleotides

Conclusion

Conclusion:

Distance travelled by the replication bubble in 10 sec is 500-900nucleotides.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 26, Problem 1P

(a)

Summary Introduction

To determine: The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon.

Introduction:

Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene.

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

(a)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

Time needed by primary transcript is:

T=5300 nucleotides50−90 nucleotides per second=59−106 sec

The time can be rounded to 60-100 sec.

Conclusion

Conclusion:

Thus, the time required for primary transcript is 60-100 sec.

(b)

Summary Introduction

To determine: The length of the transcriptional bubble formed by RNA polymerase.

Introduction:

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template. RNA polymerase is majorly found in all living organism that takes parts in the central dogma process.

(b)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

The distance travelled by the replication bubble in 10 sec is:

=(10 sec)×(50−90nuceotide/sec)=500−900 nucleotides

Conclusion

Conclusion:

Distance travelled by the replication bubble in 10 sec is 500-900nucleotides.

Answer 6

Chapter 26, Problem 1P

(a)

Summary Introduction

To determine: The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon.

Introduction:

Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene.

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

(a)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

Time needed by primary transcript is:

T=5300 nucleotides50−90 nucleotides per second=59−106 sec

The time can be rounded to 60-100 sec.

Conclusion

Conclusion:

Thus, the time required for primary transcript is 60-100 sec.

Answer 7

Chapter 26, Problem 1P

(a)

Summary Introduction

To determine: The time duration that is required by E.coli RNA polymerase to synthesize the primary transcript for the E.coli gene that encode for lactose metabolism have 5300 bp Lac operon.

Introduction:

Lac operon is a cell regulatory mechanism that is used to operate the metabolism of lactose in E. coli. The lactose is catabolized by a set of genes known as lac operon. The set of genes present in the lac operon are regulated by regulator gene.

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template

Answer 8

(a)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

Time needed by primary transcript is:

T=5300 nucleotides50−90 nucleotides per second=59−106 sec

The time can be rounded to 60-100 sec.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Conclusion

Conclusion:

Thus, the time required for primary transcript is 60-100 sec.

Answer 13

(b)

Summary Introduction

To determine: The length of the transcriptional bubble formed by RNA polymerase.

Introduction:

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template. RNA polymerase is majorly found in all living organism that takes parts in the central dogma process.

(b)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

The distance travelled by the replication bubble in 10 sec is:

=(10 sec)×(50−90nuceotide/sec)=500−900 nucleotides

Conclusion

Conclusion:

Distance travelled by the replication bubble in 10 sec is 500-900nucleotides.

Answer 14

(b)

Summary Introduction

To determine: The length of the transcriptional bubble formed by RNA polymerase.

Introduction:

RNA polymerase is a subunit that will catalyze the transcription process in which RNA polymer is synthesize from the DNA template. RNA polymerase is majorly found in all living organism that takes parts in the central dogma process.

Answer 15

(b)

Expert Solution

Explanation of Solution

Explanation:

The elongation of the RNA transcript in E.coli will run at the rate of about 50 to 90 nucleotide per second.

The distance travelled by the replication bubble in 10 sec is:

=(10 sec)×(50−90nuceotide/sec)=500−900 nucleotides

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Conclusion

Conclusion:

Distance travelled by the replication bubble in 10 sec is 500-900nucleotides.

Answer 20

Ch. 26 - Prob. 1P Ch. 26 - Prob. 2P Ch. 26 - Prob. 3P Ch. 26 - Prob. 4P Ch. 26 - Prob. 5P Ch. 26 - Prob. 6P Ch. 26 - Prob. 7P Ch. 26 - Prob. 8P Ch. 26 - Prob. 9P Ch. 26 - Prob. 10P

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