please draw in the answeres, thank you APromoter regiBCDELLF[Choose ][Choose ]3' poly A tailPromoter region of DNATerminatorStop codon5'CapC terminus of polypeptideExonSplice sitesIntronN terminus of polypeptideChoose ]G[Choose ] Match the letter labels in the figure below to theterms. Some terms will not be used.DAUGAB CFFFF000000000000000000000000000000 GE

Detailed explanation: A: Promoter region of DNAThe wavy arrow typically represents the start of…

Answered: A Promoter regi B C D E LL F [Choose ]…

Biology: The Unity and Diversity of Life (MindTap Course List)

15th Edition

ISBN:9781337408332

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter9: From Dna To Protein

Section: Chapter Questions

Problem 8SQ

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Codon-Anticodon Recognition: Base-Pairing Possibilities (Integrates with Chapter 11.) Draw base-pair structures for (a) a G:C base pair. (b) a C:G base pair. (C) a G:U base pair, and (d) a U:G base pair. Note how these various base pairs differ in the potential hydrogen-bonding patterns they present within the major groove and minor groove of a double-helical nucleic acid.
Figure 28.41 gives some examples of recombination in IgG codons 95 and 96, as specified by the Vkand Jkgenes. List the codon possibilities and the amino acids encoded if recombination occurred in codon 97. Which of these possibilities is less desirable?
Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.
Exploring the Structure of the 30S Ribosomal Subunit Go to www.pdh.org and bring up PDB file 1GIX, which shows the 30S ribosomal subunit, the three tRNAs, and mRNA. In the box on the right titled â€˜Biological Assembly.â€� click â€œMore Images.â€� and then scroll down to look at the Interactive Vic By moving your cursor over the image, you can rotate it to view it from any perspective. a. How are the ribosomal proteins represented in the image? b. How is the 16S rRNA portrayed? c. Rotate the image to see how the tRNAs stick out from the structure. Which end of the tRNA is sticking out? d. Where will these ends of the tRNAs lie when the 50S subunit binds to this complex?
Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also, write out the amino acid sequence of the protein encoded by this message. tRNA: UAC UCU CGA GGC mRNA: protein: How many hydrogen bonds would be present in the DNA segment?
Consequences of the Wobble Hypothesis Point out why Cricks wobble hypothesis would allow fewer than 61 anticodons to be used to translate the 61 sense codons. How might wobble tend to accelerate the rate of translation?
Using Figures 8.7 and 8.9 as a guide, draw a dinucleotide composed of C and A. Next to this, draw the complementary dinucleotide in an antiparallel fashion. Connect the dinucleotides with the appropriate hydrogen bonds. FIGURE 8.9 The two polynucleotide chains in DNA run in opposite directions. The left strand runs 5 to 3, and the right strand runs 3 to 5. The base sequences in each strand are complementary. An A in one strand pairs with a T in the other strand, and a C in one strand is paired with a G in the opposite strand. FIGURE 8.7 Nucleotides can be joined together to form chains caled polynucleotides. Polynucleotides are polar molecules with a 5 end (at the phosphate group) and a 3 end (at the sugar group). An RNA polynucleotide is shown at the left, and a DNA polynucleotide is shown at the right.
A fragment of bacterial DNA reads: 3’ -TACCTATAATCTCAATTGATAGAAGCACTCTAC- 5’ Assuming that this fragment is the template strand, what is the sequence of mRNA that would he transcribed? (Hint: Be sure to identify the initiation site.)
Imagine if there were 200 commonly occurring amino acids instead of 20. Given what you know about the genetic code, what would be the shortest possible codon length? Explain.
If mRNA is complementary to the DNA template strand and the DNA template strand is complementary to the DNA nontemplate strand, then why are base sequences of mRNA and the DNA nontemplate strand not identical? Could they ever be?
Alternative Splicing Possibilities Suppose exon 17 were deleted from the fast skeletal muscle troponin T gene (Figure 29.46). How many different mRNAs could now be generated by alternative splicing? Suppose that exon 7 in a wild-type troponin T gene were duplicated. How many different mRNAs might be generated from a transcript of this new gene by alternative splicing?
Functional Consequences of Y-Family DNA Polymerase Structure The eukaryotic translesion DNA polymerases fall into the Y family of DNA polymerases. Structural studies reveal that their fingers and thumb domains are small and stubby (see Figure 28.10). In addition, Y-family polymerase active sites are more open and less constrained where base pairing leads to selection of a dNTP substrate for the polymerase reaction. Discuss the relevance of these structural differences. Would you expect Y-family polymerases to have 3-exonuclease activity? Explain your answer.

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