17.1 1. 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

17.1 1. 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

17.1 1. 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

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’

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

17.2 1. Compare and contrast the functioning of DNA polymerase and RNA polymerase.

DNA polymerase RNA polymerase

DNA polymerase Assembles chains from monomers RNA polymerase Assembles chains from monomers

DNA polymerase Assembles chains from monomers Complementary base pairing RNA polymerase Assembles chains from monomers Complementary base pairing

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’

DNA polymerase Needs a primer RNA polymerase Can start from scratch

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

17.2 2. Is the promoter at the upstream or downstream end of a transcription unit?

17.2 2. Is the promoter at the upstream or downstream end of a transcription unit? Upstream

17.2 3. In a prokaryote, how does RNA polymerase “know” where to start transcribing a gene? In a eukaryote?

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

17.2 4. How is the primary transcript produced by a prokaryotic cell different from that produced by a eukaryotic cell?

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

17.3 1. How does the alteration of the 5’ and 3’ ends of pre-mRNA affect the mRNA that exists in the nucleus?

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

17.3 2. Describe the role of snRNPs in RNA splicing.

17.3 2. Describe the role of snRNPs in RNA splicing. Joins with other proteins to form spliceosomes. Removes introns, splices exons together.

17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes?

17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes? THE CAT ATE THE RBAT

17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes? THE CAT ATE THE RBAT

17.3 3. How can alternative RNA splicing generate a greater number of polypeptide products than there are genes? THE CAT ATE THE RBAT

17.4 1. Which two processes ensure that the correct amino acid is added to a growing polypeptide chain?

17.4 1. Which two processes ensure that the correct amino acid is added to a growing polypeptide chain? Aminoacyl-tRNA synthase tRNA codon

17.4 2. Describe how the formation of polyribosomes can benefit the cell.

17.4 2. Describe how the formation of polyribosomes can benefit the cell. Multiple copies of a protein in a short time.

17.4 3. Describe how a polypeptide to be secreted is transported to the endomembrane system.

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.

17.5 1. Describe three properties of RNA that allow it to perform diverse roles in the cell.

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.

17.6 1. In figure 17.22 (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.

17.6 2. Would the arrangement shown in Figure 17.22 be found in a eukaryotic cell? Explain.

17.7 1. What happens when one nucleotide pair is lost from the middle of the coding sequence of a gene?

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

17.7 2. 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?

17.7 2. Template strand: 3’-TACTTGTCCGATATC-5’ 17.7 2. Template strand: 3’-TACTTGTCCGATATC-5’ Draw a double strand of DNA.

17.7 2. Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’ 17.7 2. Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’ Draw the resulting strand of mRNA, labeling all 5’ and 3’ ends

17.7 2. Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’ 17.7 2. Double strand: 3’-TACTTGTCCGATATC-5’ 5’-ATGAACAGGCTATAG-3’ mRNA: 5’-AUGAACAGGCUAUAG-3’

17.7 2. mRNA: 5’-AUGAACAGGCUAUAG-3’ Determine the amino acid sequence.

17.7 2. mRNA: AUG AAC AGG CUA UAG Determine the amino acid sequence.

17.7 2. mRNA: AUG AAC AGG CUA UAG Polypeptide: Met-Asn-Arg-Leu

17.7 2. Then show the same after a mutation changes the template DNA sequence to 3’-TACTTGTCCAATATC-5’.

17.7 2. Template strand: 3’-TACTTGTCCAATATC-5’ 17.7 2. Template strand: 3’-TACTTGTCCAATATC-5’ Draw a double strand of DNA.

17.7 2. Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’ 17.7 2. Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’ Draw the resulting strand of mRNA, labeling all 5’ and 3’ ends

17.7 2. Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’ 17.7 2. Double strand: 3’-TACTTGTCCAATATC-5’ 5’-ATGAACAGGTTATAG-3’ mRNA: 5’-AUGAACAGGUUAUAG-3’

17.7 2. mRNA: 5’-AUGAACAGGUUAUAG-3’ Determine the amino acid sequence.

17.7 2. mRNA: AUG AAC AGG UUA UAG Determine the amino acid sequence.

17.7 2. mRNA: AUG AAC AGG UUA UAG Polypeptide: Met-Asn-Arg-Leu

17.7 2. mRNA: AUG AAC AGG UUA UAG Polypeptide: Met-Asn-Arg-Leu 17.7 2. mRNA: AUG AAC AGG UUA UAG Polypeptide: Met-Asn-Arg-Leu The resulting polypeptide is the same. .