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Published byKurtis Myers Modified over 9 years ago
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Draw the nontemplate strand of DNA for the template shown below. Compare and contrast its base sequence with the mRNA molecule. DNA A C C A A A C C G A G T mRNA U G G U U U G G C U C A
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Draw the nontemplate strand of DNA for the template shown below. Compare and contrast its base sequence with the mRNA molecule. DNA A C C A A A C C G A G T T G G T T T G G C T C A mRNA U G G U U U G G C U C A
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5’-GGGGGGGGGGGGGGGGGGGGGGGGGGGGGG-3’
17.1 2. What protein product would you expect from a poly-G mRNA that is 30 nucleotides long? 5’-GGGGGGGGGGGGGGGGGGGGGGGGGGGGGG-3’
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5’-GGGGGGGGGGGGGGGGGGGGGGGGGGGGGG-3’
17.1 2. What protein product would you expect from a poly-G mRNA that is 30 nucleotides long? 5’-GGGGGGGGGGGGGGGGGGGGGGGGGGGGGG-3’ Gly-Gly-Gly-Gly-Gly-Gly-Gly-Gly-Gly-Gly
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17.2 1. Compare and contrast the functioning of DNA polymerase and RNA polymerase.
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DNA polymerase RNA polymerase
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DNA polymerase Assembles chains from monomers RNA polymerase Assembles chains from monomers
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DNA polymerase Assembles chains from monomers Complementary base pairing RNA polymerase Assembles chains from monomers Complementary base pairing
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DNA polymerase Assembles chains from monomers Complementary base pairing Reads 3’→5’ Assembles 5’ →3’ RNA polymerase Assembles chains from monomers Complementary base pairing Reads 3’→5’ Assembles 5’ →3’
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DNA polymerase Needs a primer RNA polymerase Can start from scratch
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DNA polymerase Needs a primer to start Uses A, T, G & C Uses nucleotides containing deoxyribose RNA polymerase Can start from scratch Uses A, U, G & C Uses nucleotides containing ribose
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17.2 2. Is the promoter at the upstream or downstream end of a transcription unit?
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17.2 2. Is the promoter at the upstream or downstream end of a transcription unit? Upstream
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17.2 3. In a prokaryote, how does RNA polymerase “know” where to start transcribing a gene? In a eukaryote?
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17.2 3. In a prokaryote, how does RNA polymerase “know” where to start transcribing a gene? In a eukaryote? Prokaryote: RNA polymerase recognizes promoter Eukaryote: Transcription factors mediate binding
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17.2 4. How is the primary transcript produced by a prokaryotic cell different from that produced by a eukaryotic cell?
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17.2 4. How is the primary transcript produced by a prokaryotic cell different from that produced by a eukaryotic cell? Prokaryote: used immediately as mRNA Eukaryote: Must be modified before being used as mRNA
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17.3 1. How does the alteration of the 5’ and 3’ ends of pre-mRNA affect the mRNA that exists in the nucleus?
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17.3 1. How does the alteration of the 5’ and 3’ ends of pre-mRNA affect the mRNA that exists in the nucleus? Facilitates transportation Prevents degradation Facilitates ribosomal attachment
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17.3 2. Describe the role of snRNPs in RNA splicing.
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17.3 2. Describe the role of snRNPs in RNA splicing.
Joins with other proteins to form spliceosomes. Removes introns, splices exons together.
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17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes?
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17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes? THE CAT ATE THE RBAT
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17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes? THE CAT ATE THE RBAT
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17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes? THE CAT ATE THE RBAT
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17.4 1. Which two processes ensure that the correct amino acid is added to a growing polypeptide chain?
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17.4 1. Which two processes ensure that the correct amino acid is added to a growing polypeptide chain? Aminoacyl-tRNA synthase tRNA codon
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17.4 2. Describe how the formation of polyribosomes can benefit the cell.
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17.4 2. Describe how the formation of polyribosomes can benefit the cell. Multiple copies of a protein in a short time.
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17.4 3. Describe how a polypeptide to be secreted is transported to the endomembrane system. Signal peptide is recognized by SRP. SPR brings polypeptide to ER lumen.
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17.5 1. Describe three properties of RNA that allow it to perform diverse roles in the cell. Hydrogen bonds with DNA or RNA Specific 3-D shape Catalize chemical reactions.
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17.6 1. In figure (orange book and green book) number the RNA polymerases in order of their initiation of transcription. Then number each mRNA’s ribosomes in order of their initiation of translation.
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17.6 2. Would the arrangement shown in Figure be found in a eukaryotic cell? Explain.
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17.7 1. What happens when one nucleotide pair is lost from the middle of the coding sequence of a gene?
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17.7 1. What happens when one nucleotide pair is lost from the middle of the coding sequence of a gene? Frame shift mutation Nonfunctional protein
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The template strand of a gene contains the sequence 3’-TACTTGTCCGATATC-5’. Draw a double strand of DNA and the resulting strand of mRNA, labeling all 5’ and 3’ ends. Determine the amino acid sequence. Then show the same after a mutation changes the template DNA sequence to 3’-TACTTGTCCAATATC-5’. What is the effect on the amino acid sequence?
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17.7 2. Template strand: 3’-TACTTGTCCGATATC-5’
Template strand: 3’-TACTTGTCCGATATC-5’ Draw a double strand of DNA.
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17.7 2. Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’
Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’ Draw the resulting strand of mRNA, labeling all 5’ and 3’ ends
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17.7 2. Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’
Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’ mRNA: 5’-AUGAACAGGCUAUAG-3’
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mRNA: 5’-AUGAACAGGCUAUAG-3’ Determine the amino acid sequence.
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mRNA: AUG AAC AGG CUA UAG Determine the amino acid sequence.
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mRNA: AUG AAC AGG CUA UAG Polypeptide: Met-Asn-Arg-Leu
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Then show the same after a mutation changes the template DNA sequence to 3’-TACTTGTCCAATATC-5’.
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17.7 2. Template strand: 3’-TACTTGTCCAATATC-5’
Template strand: 3’-TACTTGTCCAATATC-5’ Draw a double strand of DNA.
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17.7 2. Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’
Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’ Draw the resulting strand of mRNA, labeling all 5’ and 3’ ends
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17.7 2. Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’
Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’ mRNA: 5’-AUGAACAGGUUAUAG-3’
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mRNA: 5’-AUGAACAGGUUAUAG-3’ Determine the amino acid sequence.
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mRNA: AUG AAC AGG UUA UAG Determine the amino acid sequence.
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mRNA: AUG AAC AGG UUA UAG Polypeptide: Met-Asn-Arg-Leu
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17.7 2. mRNA: AUG AAC AGG UUA UAG Polypeptide: Met-Asn-Arg-Leu
mRNA: AUG AAC AGG UUA UAG Polypeptide: Met-Asn-Arg-Leu The resulting polypeptide is the same. .
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