Concept explainers
Videos
Figure 15.11 A scientist splices a eukaryotic promoter in front of a bacterial gene and inserts the gene in a bacterial chromosome. Would you expect the bacteria to transcribe the gene?
To analyze:
Whether the bacterial gene having eukaryotic promoter gets transcribed or not.
Introduction:
Eukaryotic promoters are different from prokaryotic promoters so the bacteria would not transcribe the gene.
Explanation of Solution
In case of transcription of genes, i.e. formation of RNA (mostly m-RNA) from DNA, promoters play an essential role. Promoters are the specific DNA sequences present near the starting point of transcription of a particular gene. Promoters help in the binding of RNA polymerase and transcription machinery to the specific point in the gene to initiate the process of transcription. They are mainly located upstream of a gene, towards 5’ end of the anti-sense strand. As these are specific recognition sequences, eukaryotic promoters are different from prokaryotic promoters.
Prokaryotic promoters have three essential elements:-10 elements,-35 element and UP (upstream promoter) element.
Eukaryotic promoters mainly contain a TATA box of six nucleotides (consensus sequence is 5’ TATAAA 3’) located around-30 (with reference to initiation site of transcription +1), while prokaryotic promoters have its counterpart-10 elements or Pribnow box of six nucleotides (consensus sequence is 5’ TATAAT 3’) located at-10. Both these have A-T rich elements. Prokaryotes also have one more promoter sequence of six nucleotides at-35 which is (5’ TTGACA 3’) called as-35 elements. Pribnow box or-10 element is essential in prokaryotes for the initiation of transcription and-35 element helps in increasing the transcription rate.UP element is A-T rich sequence located between-40 to-60.
Other conserved sequences in eukaryotic promoters are CAAT box (GGCCAATCT) around-80 and GC-rich boxes (GGCG), and octamer boxes (ATTTGCAT) located further upstream.Basically, eukaryotic promoters are much more complex and larger in structure than the prokaryotic promoters.Sigma factor (s factor) of prokaryotic RNA polymerase holoenzyme recognizes and binds to the promoter interaction site to initiate transcription.
So, a bacterial gene having eukaryotic promoter would not transcribe because eukaryotic promoter has different recognition sites/consensus sequences than the required prokaryotic recognition sites. RNA polymerase, transcription factors and other proteins of host bacterial chromosome would not be able to identify the eukaryotic promoter sequences and transcription of the gene would not occur.
So, the bacteria would not transcribe the gene with the eukaryotic promoter as they are different from the prokaryotic promoters.
Want to see more full solutions like this?
Chapter 15 Solutions
Biology 2e
Additional Science Textbook Solutions
Campbell Essential Biology (7th Edition)
College Physics: A Strategic Approach (3rd Edition)
Human Physiology: An Integrated Approach (8th Edition)
Genetic Analysis: An Integrated Approach (3rd Edition)
Chemistry: A Molecular Approach (4th Edition)
Biological Science (6th Edition)
- 13. When a smear containing cells is dried, the cells shrink due to the loss of water. What technique could you use to visualize and measure living cells without heat-fixing them? Hint: you did this technique in part I.arrow_forward10. Write a simple formula for converting μm to mm when the number of μm's are known. Use the variable X to represent the number of um's in your formula.arrow_forward8. How many μm² is in one cm²; express the result in scientific notation. Show your calculations. 1 cm = 10 mm; 1 mm = 1000 μmarrow_forward
- Find the dental formula and enter it in the following format: I3/3 C1/1 P4/4 M2/3 = 42 (this is not the correct number, just the correct format) Please be aware: the upper jaw is intact (all teeth are present). The bottom jaw/mandible is not intact. The front teeth should include 6 total rectangular teeth (3 on each side) and 2 total large triangular teeth (1 on each side).arrow_forwardAnswer iarrow_forwardAnswerarrow_forward
- calculate the questions showing the solution including variables,unit and equations all the questiosn below using the data a) B1, b) B2, c) hybrid rate constant (1) d) hybrid rate constant (2) e) t1/2,dist f) t1/2,elim g) k10 h) k12 i) k21 j) initial concentration (C0) k) central compartment volume (V1) l) steady-state volume (Vss) m) clearance (CL) AUC (0→10 min) using trapezoidal rule n) AUC (20→30 min) using trapezoidal rule o) AUCtail (AUC360→∞) p) total AUC (using short cut method) q) volume from AUC (VAUC)arrow_forwardQUESTION 8 For the following pedigree, assume that the mode of inheritance is X-linked recessive, and that the trait has full penetrance and expressivity and occurs at a very low frequency in the hum population. Using XA for the dominant allele and Xa for the recessive allele, assign genotypes for the following individuals (if it is not possible to figure out the second allele of a genotype, that with an underscore): 2 m 1 2 1 2 4 5 6 7 8 9 IV 1 2 3 5 6 7 8 CO 9 10 12 13 V 1, 2 3 4 5 6 7 8 9 10 11 12 13 a. Il-1: b. 11-2: c. III-3: d. III-4: e. If individuals IV-11 and IV-12 have another child, what is the probability that they will have a boy with the disorder?arrow_forwardAnswerrarrow_forward
- please,show workings ahd solutions calculate the questions showing the solution including variables,unit and equations all the questiosn below using the data a) B1, b) B2, c) hybrid rate constant (1) d) hybrid rate constant (2) e) t1/2,dist f) t1/2,elim g) k10 h) k12 i) k21 j) initial concentration (C0) k) central compartment volume (V1) l) steady-state volume (Vss) m) clearance (CL) AUC (0→10 min) using trapezoidal rule n) AUC (20→30 min) using trapezoidal rule o) AUCtail (AUC360→∞) p) total AUC (using short cut method) q) volume from AUC (VAUC)arrow_forwardQuestion about consensus and the relationship to transcriptional activityarrow_forwardQuestion about archaea and eukaryotic general transcription homologsarrow_forward
- Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning 
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
Biology Today and Tomorrow without Physiology (Mi...BiologyISBN:9781305117396Author:Cecie Starr, Christine Evers, Lisa StarrPublisher:Cengage Learning