LIFE:SCIENCE OF BIOL.(LL) >CUSTOM<
11th Edition
ISBN: 9781319209957
Author: Sadava
Publisher: MAC HIGHER
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Chapter 16.5, Problem 4R
Summary Introduction
To review:
The involvement of mRNAs (messenger ribonucleic acids) in explaining the discrepancy between the assumed (80,000 to 150,000) and actual amount (21,000) of protein-coding genes in humans.
Introduction:
The mRNAs can produce different types of proteins. This protein production depends on the requirement of the body. The number of proteins and protein-coding genes varies in the body and accordingly the gene expression changes.
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A 2500 bp region of the human genome encodes two genes. One of the genes encodes a protein of 600 amino acids and the other gene encodes a protein of 280 amino acids. The mRNA sequences of the two genes do not contain any of the same nucleotide sequences (i.e. they do not overlap). How is this possible? Fully explain your answer.
The human genome contains thousands of sequences known as small open reading frames, some of which encode proteins of about 30 amino acids. What is the minimum number of nucleotides required to encode such a protein?
Consider a stretch of DNA (a hypothetical gene) that has the sequence 5’ ATG-CTA-TCA-TGG-TTC-TAA 3’
A) Transcribe and translate this gene using the genetic code table. Be sure to label the mRNA 3’ and 5’ ends. Write the amino acid sequence using 1 letter abbreviations.
B) Now, our hypothetical gene has undergone a mutation. The mutant sequence is....3’ TAC-GAT-AGT-ACC-AAT-ATT 5’5’ ATG-CTA-TCA-TGG-TTA-TAA 3’
Transcribe and translate the mutant sequence. Be sure to label the mRNA 3’ and 5’ ends. Write the amino acid sequence using 1 letter abbreviations.
C) Indicate the type of mutation (nonsense, missense, silent, or frame shift) present.
D) How severe of a consequence will this mutation likely be in terms of protein function (none, mild, moderate or severe)? Why?
Chapter 16 Solutions
LIFE:SCIENCE OF BIOL.(LL) >CUSTOM<
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- The following is a portion of an mRNA sequence: 3’ –AUCGUCAUGCAGA-5’ a)During transcription, was the adenine at the left-hand side of the sequence the first or the last nucleotide used to build the portion of the mRNA shown? Explain how you know. b)Write out the sequence and polarity of the DNA duplex that encodes this mRNA segment. Label the template and coding DNA strands. c)Identify the direction in which the promoter region for this gene will be located.arrow_forwardThe length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the MRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this difference? Explain in detail. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). A Ix BIUS Paragraph Arial 14pxarrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 867 nucleotides in length, and the other is 685 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 867 and 685 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 150 nucleotides.arrow_forward
- 2b) Prokaryotic cells can and do produce "polycistronic" mRNAs, which have multiple independent coding sequences coding for separate proteins on the same mRNA strand. Eukaryotic cells don't have polycistronic mRNAs, because only the first (most 5') coding sequence on such an mRNA would ever be translated in a eukaryotic cell. Explain why this is the case - why wouldn't a second, more 3' coding sequence on an mRNA be translated in a eukaryotic cell?arrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides. 99 62 120 84 102 27 117 Gene X E1 11 E2 12 ЕЗ 13 E4 Exon (E) Intron (I)arrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature MRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides. 99 62 120 84 102 27 117 Gene X E1 в в 11 E2 12 E4 Exon (E) Intron (1)arrow_forward
- Here is a eukaryotic gene. The numbers given are base pairs of exon and intron. How long in bases will the pre mRNA transcript be? Explain briefly. What is the maximum number of amino acids that could make up the protein product from the final mRNA? Explain briefly.arrow_forwardGiven the following DNA sequence of the template strand for a given gene: 5' TTTCCGTCTCAGGGCTGAAAATGTTTGCTCATCGAACGC3' Part A ) Write the mRNA that will be transcribed from the DNA sequence above (be sure to label the 5' and 3' ends). Part B ) Use the genetic code to write the peptide sequence translated in a cell from the mRNA in part A. Please use the 3 letter abbreviation for each amino acid. Part C: How would the peptide synthesized in a cell be different if the mRNA was translated in vitro (i.e. not in the cell)?arrow_forwardin the human gene for the beta chain of hemoglobin, the first 30 nucleotides in the amino acid coding region is represented by the sequence 3'TACCACGTGGACTGAGGACTCCTCTTCAGA-5'. What is the sequence of the partner strand? If the DNA duplex for the beta chain of hemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region.arrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY