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

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

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

Students have asked these similar questions

12. When glutamic acid is heated to 180°C., it loses a molecule of water to form a lactam. The bond that creates the lactam is an amide bond that forms between the amino nitrogen and the carbon of the side chain -COOH group. What is the most reasonable structure for this lactam? (NOTE: The structure of glutamic acid is shown in Question 1.) COOH COOH ا کچھ جو کچھ C D COOH B COOH E COOH

10. Which one of the following compounds is the major organic product of the series of reactions shown here? Ph A OH Ph B CO₂Et Br -H Ν ΚΑ CO₂Et 1. NaOEt 1. NaOH, H₂O 2. H3O+ 2. PhCH2CH2Br 3. heat NH2 Ph OH NH2 0 OH Ph OH NH2 Ph D NH2 E OH

1. What is the isoelectric point of glutamic acid? (Glutamic acid has pKa1 2.10, pKa2 4.07, pKaз 9.47) A) pH 2.1 D) pH 6.8 B) pH 3.1 C) pH 4.1 E) pH 9.5 HO NH2 Glutamic acid (shown without charges) OH

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.

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

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

Concept explainers

Videos

Explanation of Solution

Explanation of Solution

Want to see more full solutions like this?

Chapter 26 Solutions