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- based on the pictures How many base substitutions are there between these sequences? Count ONLY the instances where there is ANY nucleotide difference between any of the sequences (leave out any indels in this count) Are there any indels in this alignment? (yes or no)1. Why do we need to check the isolated DNA for its quantity and quality? 2. The purity of DNA sample is below 1.8 A260/A280 so, where did the proteincontamination come from? Note: Cite the in-text citation and referencesHomologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesn’t the mismatch repair system of E. coli eliminate these mismatches?
- Heteroduplex DNA Formation in Recombination From the information in Figures 28.17 and 28.18, diagram the recombinational event leading to the formation of a heteroduplex DNA region within a bacteriophage chromosome.Design a pair of primers to amplify the entire length of the following 45 base pair sequence.Make each primer 14 bases long. Write the sequences of the primers in 5' to 3' order.(Hint: It will help for you to write out BOTH strands of the DNA sequence listed below.5'-GATGCCCGTTGGATAAATTGGGCGTCTAGAATCGGTCACACTTAG-3'Below are 9 possible primer pairs.● Determine which primer pair is the best choice.● Explain why the other primers are not good choices.● Calculate the Tm for each primer. Underline or highlight the region of DNA for the primer pair you chose as the best.Forward 1: 5’ gaaataattttgtttaactttaag 3’ Tm =Reverse 1: 5’ gtttaagacaaaatagtctgg 3’ Tm =Forward 2: 5’ gtaactcagctttcaggtcg 3’ Tm =Reverse 2: 5’ tctcggaatgttgcaacagc 3’ Tm =Forward 3: 5’ agattagcggatcctacctg 3’ Tm =Reverse 3: 5’ atgtgtaatcccagcagcag 3’ Tm =Forward 4: 5’ cattgattatttgcacggcg 3’ Tm =Reverse 4: 5’ aaaatcttctctcatccgcc 3’ Tm =Forward 5: 5’ tccataagattagcggatcc 3’ Tm =Reverse 5: 5’ tgcaagcttggctgttttgg 3’ Tm =Forward 6: 5’ gatcctacctgacgcttttta 3’ Tm=Reverse 6: 5’ aaataatgaattcgagctcggt 3’ Tm =Forward 7: 5’ataaaaaaatcgagataaccgtt 3’ Tm =Reverse 7: 5’aggtcgactctagaggatc 3’ Tm =Forward 8: 5’ctacctgttccatggccaac 3’ Tm=Reverse 8: 5’ ttcgggcatggcactcttg 3’ Tm=Forward 9: 5’ tccataagattagcggatcc 3’ Tm =Reverse 9: 5’…
- 8. Given the following two sequences, create a pairwise alignment by hand. Do not place any gaps in the sequences to optimize alignment. Use an identity matrix and BLOSUM62 substitution matrix to create the two sets of pairwise alignments. Report the identity, identity score, similarity, and similarity score.Sequence 1: ATPL MSequence 2: ATKI MA single strand of DNA, 24 nucleotides long, with the sequence 5'-TTTCCCgggAAAgggTTTAAAggg-3' is in a test tube. (Note that G's are shown in lowercase, so that your eye can better distinguish them from C's) Other than the appropriate buffer solution, what else needs to go in the test tube to so that we end up with a piece of double stranded DNA, 24 base pairs long, with the above sequence comprising one of the two strands?Are these answers correct? It's all for one problem where the E. coli chromosome contains 4,641,652 bp.
- The average of a number of closely related but nonidentical sequences is referred to as a(n) _____________.based on the picture What is the length in basepairs of these sequences? How many base substitutions are there between these sequences? Count ONLY the instances where there is ANY nucleotide difference between any of the sequences (leave out any indels in this count) Are there any indels in this alignment? (yes or no)Variable number tandem repeats (VNTRs) are repeating DNA sequences of about 15–100 bp in length, found both within and between genes. Why are they commonly used in forensics?