The typical human gene contains an average of 8 exons. The exons average about 145 nucleotides...
- The typical human gene contains an average of 8 exons. The exons average about 145 nucleotides in length, and the internal introns are usually about 10x that size. Using these values, calculate the following. Please show your work so that partial credit can be given when applicable.
a.(5 points) Assuming only internal introns, a 100 nucleotide 5’ untranslated region, and a 800 nucleotide 3’ untranslated region, how long is the typical pre-mRNA in humans?
b.(5 points) Assuming complete and perfect splicing, a one nucleotide 5’ cap, and a 200 nucleotide poly-A tail how long is the processed mRNA?
c.(EXTRA CREDIT 5 points) Assuming that the exons are cobbled together in the correct order, but that one or more exon might be missing, how many additional variants of the mRNA could be produced, functional or not? (To clarify, you could have E1E2E3E4E5E6E8 or even E1E8, but not E7E1E3 or E7E7.)
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The typical human gene contains an average of 8 exons. The
exons average about 145 nucleotides...
5. A eukaryotic protein-encoding gene contains two introns and three exons: exon 1–intron 1–exon 2–intron 2–exon...
5. A eukaryotic protein-encoding gene contains two introns and three exons: exon 1–intron 1–exon 2–intron 2–exon 3. The 5ʹ splice site at the boundary between exon 2 and intron 2 has been eliminated by a small deletion in the gene. Describe how the pre-mRNA encoded by this mutant gene would be spliced. Indicate which introns and exons would be found in the mRNA after splicing occurs
Question 8 (1 point) The primary RNA transcript of the chicken ovalbumin gene is 7700 nucleotides...
Question 8 (1 point) The primary RNA transcript of the chicken ovalbumin gene is 7700 nucleotides long, but the mature mRNA that is translated on the ribosome is 1872 nucleotides long. This size difference occurs primarily as a result of: O addition of the poly A tail O removal of exons O splicing O addition of the 5' cap Question 7 (1 point) Which of the following recognizes the promoter region of a gene in E. coli allowing RNA polymerase...
Hint 1. How is a complete initiation complex formed? Drag the labels to their correct locations...
Hint 1. How is a complete initiation complex formed? Drag the labels to their correct locations in the diagram to describe the roles of the various transcription factors in the formation of a complete initiation complex Reset Help part of initial committed complex TFIIA IIB added during formation of minimal initiation complex sa cosci.ccccccca RNA polymerase II added during formation of complete initiation complex Submit Request Answer Hint 2. How is pre-mRNA processed into mature mRNA? Select the two statements...
Hemoglobin is a protein that is found in red blood cells. It binds to oxygen in...
Hemoglobin is a protein that is found in red blood cells. It binds to oxygen in the lungs and it carries it to tissues and cells throughout the body. Hemoglobin is made of four polypeptide chains, two called “alpha-globins” (a) and two “beta-globins" (B). The B-globin polypeptide is produced in the cells based on the sequence of the HBB gene. The structure of the HBB gene that codes for beta-globin is represented below. The primary transcript is 1606 nucleotides long....
This problem will require you to think about gene expression (transcription and translation) as well as...
This problem will require you to think about gene expression (transcription and translation) as well as energy metabolism. 1. Consider a yeast cell which wants to express a 900 bp gene. Assuming this gene has no introns, ignoring any 5’ or 3’ untranslated regions, and assuming a 100 bp poly-A tail, how many nucleotides are required to transcribe this gene? For simplicity’s sake, let’s assume all the nucleotides are ATP. 2. Based on what you know about translation, how many...
Genes in eukaryotes are often organized into exons and intrans, which require splicing to produce an...
Genes in eukaryotes are often organized into exons and intrans, which require splicing to produce an mRNA that can be translated. The gene organization is the order of the DNA segments that comprise the gene starting with the promoter, the first exon, the first intron, the second exon, and so on. The interspersed intrans can make gene identification difficult in eukaryotesparticularly in higher eukaryotes with many introns and alternative spliced mRNAs. Prediction of many genes and their organization has been...
QUESTION 6 Assume you are studying a protein-coding gene, ACEX, which includes 4 exons as illustrated...
QUESTION 6 Assume you are studying a protein-coding gene, ACEX, which includes 4 exons as illustrated in the gene map below. The 5' UTR and 3' UTR segments are each 25 bp long. Exons 1 thru 4 are 100, 200, 300, 400 bp long, respectively. Each intron is 200 bp each. The locations of the relevant EcoRI sites within the ACEX locus are indicated, but the location of other restriction enzyme sites (like BamHI) are not shown." EcoRI probe EcoRI...
hello, two of these circled answers are incorrect. 1 6. The promoter sequences are the positions...
hello, two of these circled answers are incorrect.
1 6. The promoter sequences are the positions that: signal the initiation site of a gene (+1) B) bind the transcriptional factor that is associated with RNA polymerase e) attach the correct nucleotide triphosphate to the template DNA strand D) separate the two DNA strands CUA 7. A particular triplet of bases in the coding sequence of DNA is GAT. The anticodon on the tRNA that binds the mRNA codon is A...
Please help with 1-16!!! (two pictures are attached) Thanks! Transcription . Although both prokaryotes and eukaryotes...
Please help with 1-16!!! (two pictures are attached) Thanks!
Transcription . Although both prokaryotes and eukaryotes put a cap and a tail on the mRNA, only eukaryotes have introns that have to be spliced out. (T/F) 2. The poly A tail on cukaryotic mRNA protects the RNA from rapid degradation in the cytoplasm. (T/F) 3. The polyA tail is added to eukaryotic mRNA immediatel after transport of the message from the nucleus. (T/F) 4. is usually a single stranded molecule....
Humans and Chimps share a 9,826 nucleotides out of 10,000 in sequences. Therefore, the Human and...
Humans and Chimps share a 9,826 nucleotides out of 10,000 in
sequences. Therefore, the Human and Chimp nucleotide difference out
of 10,000 is 174. Gorillas and my new species differ by only 165
nucleotides. A) K=(-3/4)In(1-(4(0.0174)/3))=0.0176050171.
L=K/2t=0.0176050171/2(6600000)=1.33x10^-9. B)
K=(-3/4)In(1-(4(0.0165)/3))=0.0166842067. t=K/2L=0.0166842067. That
is 6.2x10^6 years ago. By using the molecular clock method to
estimate the two species, we can say that the species most likely
diverged nearly 6 million years ago. C) I am uncertain what to
write here but I...
Question 8 (1 point) The primary RNA transcript of the chicken ovalbumin gene is 7700 nucleotides long, but the mature mRNA that is translated on the ribosome is 1872 nucleotides long. This size difference occurs primarily as a result of: O addition of the poly A tail O removal of exons O splicing O addition of the 5' cap Question 7 (1 point) Which of the following recognizes the promoter region of a gene in E. coli allowing RNA polymerase...
Hint 1. How is a complete initiation complex formed? Drag the labels to their correct locations in the diagram to describe the roles of the various transcription factors in the formation of a complete initiation complex Reset Help part of initial committed complex TFIIA IIB added during formation of minimal initiation complex sa cosci.ccccccca RNA polymerase II added during formation of complete initiation complex Submit Request Answer Hint 2. How is pre-mRNA processed into mature mRNA? Select the two statements...
Hemoglobin is a protein that is found in red blood cells. It binds to oxygen in the lungs and it carries it to tissues and cells throughout the body. Hemoglobin is made of four polypeptide chains, two called “alpha-globins” (a) and two “beta-globins" (B). The B-globin polypeptide is produced in the cells based on the sequence of the HBB gene. The structure of the HBB gene that codes for beta-globin is represented below. The primary transcript is 1606 nucleotides long....
QUESTION 6 Assume you are studying a protein-coding gene, ACEX, which includes 4 exons as illustrated in the gene map below. The 5' UTR and 3' UTR segments are each 25 bp long. Exons 1 thru 4 are 100, 200, 300, 400 bp long, respectively. Each intron is 200 bp each. The locations of the relevant EcoRI sites within the ACEX locus are indicated, but the location of other restriction enzyme sites (like BamHI) are not shown." EcoRI probe EcoRI...
hello, two of these circled answers are incorrect.
1 6. The promoter sequences are the positions that: signal the initiation site of a gene (+1) B) bind the transcriptional factor that is associated with RNA polymerase e) attach the correct nucleotide triphosphate to the template DNA strand D) separate the two DNA strands CUA 7. A particular triplet of bases in the coding sequence of DNA is GAT. The anticodon on the tRNA that binds the mRNA codon is A...
Please help with 1-16!!! (two pictures are attached) Thanks!
Transcription . Although both prokaryotes and eukaryotes put a cap and a tail on the mRNA, only eukaryotes have introns that have to be spliced out. (T/F) 2. The poly A tail on cukaryotic mRNA protects the RNA from rapid degradation in the cytoplasm. (T/F) 3. The polyA tail is added to eukaryotic mRNA immediatel after transport of the message from the nucleus. (T/F) 4. is usually a single stranded molecule....
Humans and Chimps share a 9,826 nucleotides out of 10,000 in
sequences. Therefore, the Human and Chimp nucleotide difference out
of 10,000 is 174. Gorillas and my new species differ by only 165
nucleotides. A) K=(-3/4)In(1-(4(0.0174)/3))=0.0176050171.
L=K/2t=0.0176050171/2(6600000)=1.33x10^-9. B)
K=(-3/4)In(1-(4(0.0165)/3))=0.0166842067. t=K/2L=0.0166842067. That
is 6.2x10^6 years ago. By using the molecular clock method to
estimate the two species, we can say that the species most likely
diverged nearly 6 million years ago. C) I am uncertain what to
write here but I...
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