Most eukaryotic cells transcribe a relatively high portion of their genome, but most of those RNA transcripts are likely not encoding for proteins. Elucidating the function(s) of that "junk" RNA is one of the next big challenges in molecular cell biology. You isolate one of those transcripts and determine that its sequence consists of 50% C, 30% U, and 20% A. What is the percentage of Gs in this transcript?
Most eukaryotic cells transcribe a relatively high portion of their genome, but most of those RNA transcripts are likely not encoding for proteins. Elucidating the function(s) of that "junk" RNA is one of the next big challenges in molecular cell biology. You isolate one of those transcripts and determine that its sequence consists of 50% C, 30% U, and 20% A. What is the percentage of Gs in this transcript?
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Its one problem with 7 questions
A) Most eukaryotic cells transcribe a relatively high portion of their genome, but most of those RNA transcripts are likely not encoding for proteins. Elucidating the function(s) of that "junk" RNA is one of the next big challenges in molecular cell biology. You isolate one of those transcripts and determine that its sequence consists of 50% C, 30% U, and 20% A. What is the percentage of Gs in this transcript? (don't write the unit)
B) Which stop codon(s) can be encoded by the RNA in the previous question ("How many times do we need to tell you to stop? 1").
List the codon(s) alphabetically. If you need to enter fewer codons than the alloted number of "blanks", enter N/A .
C) From what you can tell, the sequence of the transcript you isolated in the previous question* appears quite random. Assuming that it is random, you can expect to see the various combinations of any three nucleotides at a probability that depends of the frequency of each of the nucleotides. For example, you naturally should expect to see the triplet CCC more times than AAA. What is the expected percentage of triplets corresponding to stop codon(s) in this transcript (round to one decimal place and enter just the number without "%")? *(How many times do we need to tell you to stop? 1)
D) If the length of the transcript in the previous two questions is 2,500 nucleotides, how many stop codons do you expect to find in it, based on the frequency you calculated above?
E) On average, how many bases separate each of the occurrence of the stop codon(s) (round to the nearest integer).
F) The sequence between two potential stop codons is an "open reading frame" (ORF*). Upon closer examination of the RNA sequence from the previous questions** you determine the presence of an ORF that may encode for a putative protein which is 600 amino acid residues-long. How long is the ORF?
*Open reading frame: a stretch of DNA or RNA which is uninterrupted by a stop codon and may therefore encode for a protein.
**Questions "How many times do we need to tell you to stop" 1 through 5.
| 27 bases | ||
| 1,800 bases | ||
| 2,500 bases | ||
| 259 bases | ||
| 600 bases | ||
| 83 bases | ||
| 200 bases |
G) What may be the ramifications of your analyses of the RNA you isolated*? It is clearly containing an ORF which is longer than what you expected to find based on the sequence analysis you have conducted. What would be a logical conclusions based on the comparison of the lengths of the actual and predicted ORFs? Questions "How many times do we need to tell you to stop" 1 through 6. Pick 1, 2 or 3 below
| 1) The difference between the actual and predicted sizes of the ORF is not big enough to be significant and therefore your initial hypothesis that sequence is random is clearly random, is supported. | ||
| 2) Nature is random and your results, using the Bard's words, may be "full of sound and fury" but are really "signifying nothing". | ||
| 3) The difference between the actual and predicted sizes of the ORF is rather big and may suggest that the sequence of the RNA may not be as random as it first appeared. This raises the hypothesis that what started as piece of "junk" RNA may be actually encoding for a protein. |
