An E. coli replication fork is shown in Figure 2.2. III 3' IV 5 II Figure 2.2 (i) Identify the lagging strand template and the leading strand template in Figure 2.2. Identify the strand that is made up of Okazaki fragments.
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- You are working in a research lab studying human Peptide Z, which is a short protein that plays a key role in responding to SARS-CoV-2 virus infection. Your advisor gives you the double-stranded DNA sequence encoding Peptide Z (below), but she neglects to tell you which are the coding and non-coding strands of this DNA before leaving on a six-month sabbatical. 5'-AGA CCC ATG CCT CAG CAG TTC TTT GGA TTA ATG TAA-3' 3'-TCT GGG TẠC GGA GTC GTC AAG AAA CCT AAT TẠC ATT-5' Using your understand of transcription, translation, and the codon table below, determine if the top strand or bottom strand is the template strand of the Peptide Z gene. (Enter either top strand or bottom strand.) What is the sequence of the RNA that encodes Peptide Z? Note that the entire DNA sequence is transcribed. 5' - - 3' What is the amino acid sequence of Peptide Z? Make sure you indicate the carboxyl-terminus and amino-terminus with C and N, respectively.This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' (i) Draw the structure of hairpin loop that will be formed during the end of transcription. (ii) Describe the function of the hairpin loop during transcription.1)give 3 differences between replication in prokaryotes and replication in Eukaryotes 2)For each item in the following table, decide whether it is related or involved in transcription, translation or replication. 1. Splicing 2. Stop codon 3. Lagging strand 4. RNA polymerase 5. DNA polymerase 6. Telomerase 3) Give the mRNA and the polypeptide (amino acid sequence) that results from the following DNA template strand: DNA template T A C A C G G G C G T A mRNA Amino acid sequence
- 40 Shown below is the antisense strand of DNA. What is the amino acid sequence that corresponds to this code? 5' AAAGCATACCGG 3' Second letter G. UUU Phe UCU UAU) Tyr UGC Cys UGU UUC UCC UAC Ser UAA Stop UGA Stop A UAG Stop UGG Trp UUA UCA UUG Leu UG CAU His CGU CGC CGA CGG CU CUU CUC CUA CUG CAC Pro Leu Arg CCA CAA) Gin CCG CAG AAU AUU AUC lle AUA AUG Met ACG ACU AGU AAC Asn AGC Ser ACC ACA Thr AAA1 AGA AAG Lys AGG Arg GAU) Asp GGC GAC Ala GAA GGU GUU GUC GUA Val GCU GCC Gly GCA GGA Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a PRO-VAL-SER-PHE PHE-ARG-MET-ALA GLY-HIS THR-LYS d LYS-ALA-TYR-ARG First letter Third letterii) Assume that the following polynucleotide is part of a longer DNA molecule. Write out the product that would be obtained from its replication.3 AAGCTTTCCGThe beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 5'.TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3' 3'.AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCATTATACCCCTACGTGATAG...5' promoter 2a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. d Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 -41 5' 3' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?
- Which of the following statements about DNA polymerase III holoenzyme from E. coli are correct. 1. It elongates a growing DNA chain approximately 100 times faster than does DNA polymerase I. 2. It associates with the parental template, adds a few nucleotides to the growing chain, and then dissociates prior to initiating another synthesis cycle. 3. It maintains a high fidelity of replication, in part by acting in conjunction with a subunit containing a 3' -> 5' exonuclease activity. 4. When replicating DNA, it is a molecular assembly composed of at least eight different kinds of subunits.Look at the double-stranded segment of DNA shown below. Imagine that the two strands have already been denatured, and the temperature has been decreased to an appropriate annealing temperature. Show where the two primers would anneal to the strands, then indicate the direction of extension on each new strand with an arrow. 5’--T C A G G A C G T A A G C T T G C A T A T C T C G A T G C T A A A T C A T—3’ 3’--A G T C C T G C A T T C G A A C G T A T A G A G C T A C G A T T T A G T A—5’ Primer #1: 3’ A C G A T T T 5’ Primer #2: 5’ G G A C G T A 3’This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: bottom strand is the noncoding strand). 5'-ААCGCATGAGAAAGCCCCCCGGAAGATCACСТТСCGGGGGCТТТАТАТААТТАGC-3' 3'-тTGCGTACтстттCGGGGGGCCTTCTAGTGGAAGGCCCCCGАААТАТАТТААТтCG-5' (i) Draw the structure of hairpin loop that will be formed during transcription. (ii) Illustrate how the hairpin loop structure initiates the termination of transcription.
- This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.DNA is made of two strands that are antiparallel. If one strand runs from 3’ to 5’ direction the other one will go from 5’ to 3’ direction. During replication or transcription, whatever the process is, it will always follow the 5’ to 3’ direction using the 3’ to 5’ directed strand as the template strand. Therefore, if following is the DNA sequence5’-CCG ATC GCA CAA-3’a) Using this sequence as template after transcription no protein can be translated. Why? I. Presence of start codonII. Absence of start codonIII. Due to mutationb) If you want to start the translation, what change you need in the second codon (from 5’ to 3’ direction)?I. Substitution of C with GII. No changeIII. Deletion of CIV. Both I & IIITo test patients for COVID19, lab workers will first convert all the RNA molecules extracted from a nasal swab to a double-stranded DNA copy (dsDNA). If the virus is present, its genomic sequence should be in some of the new dsDNA molecules. Part 1) A region of COVID genomic DNA sequence is shown below. Following convention, only the top strand is shown. Copy/paste the sequence into the text box and create the second strand. Be sure to label its ends. (You may need to reduce the font size so that it doesn't wrap around) AAGATCACATTGGCACCCGCAATCCTGCTAACAATGCTGCAATCGTGCTACAACTTCCTC Part 2) To test for the presence of COVID DNA sequence, lab workers use single-stranded DNA oligonucleotides as probes (short pieces of DNA that do not have a partner strand). If the two strands of DNA that you drew were separated from each other, where would the shorter DNA strand shown below be able to form continuous base pairs? Highlight that region in your dsDNA model. TGTAGCACGATTGCAGCATTG Note: If you…