The original DNA sequence was CCGGAATT. It was replicated. Which of these newly replicated DNA molecules has a base-substitution error? 5'CCAGAATT3' 5'CCGGGAATT3. 5'CCGAATT3'
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- Below is a sample of a segment of DNA…(copy from left to right) 3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ 1.Assume the 6th amino acid is changed from T to G on the DNA template strand. What type of mutation is this? What effect would this have on the protein? Look up an example for this type of mutation. 2, Assume the 5th and 6th amino acids are removed from the DNA template strand. What type of mutation is this? How would this affect the protein? Look up an example of this type of mutation. 3.Which mutation changes the protein more...a point mutation or a frameshift mutation. Explain your reasoning. 4.What would be the problem if ATT was inserted into the DNA template strand after the second codon? (Be sure to consult the coding chart for amino acids). 5. What if the second amino acid was repeated over 5Ox. What amino acid is repeated? What type of mutation is this? If this is on chromosome 4, what genetic disorder is this?…The following DNA sequence was determined by Sanger sequencing, using a 20 nt long sequencing primer that ended ...AGTACAACAA-3'. 5'-agtacaacaa ctctcggtc tacggtacgc ctgcgggcgc gtagccaatc tagcacttcg-3' 3'-tcatgttgtt gagagccag atgccatgcg gacgcccgcg catcggttag atcgtgaagc-5′ A. If the technician forgot to add ddNTPs to the reaction, what would the sequencing chromatogram look like? Blank Many peaks, but only one at each position Overlapping peaks at every position All peaks are black There is only one peak, at 60 nt B.When the reaction is done correctly, ddCTP is labeld with a yellow fluorescent tag. When the Sanger sequencing reaction is complete, what will be the lengths, in nucleotides, of the three shortest products that have the yellow tag? C. Could you perform Illumina sequencing using ddNTPs? Why or why not? Explain.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.
- A linear piece of DNA was broken into random, overlapping fragments and each fragment was sequenced. The sequence of each fragment is shown below. Fragment 1: 5'-TAGTTAAAAC–3' Fragment 2: 5'-ACCGCAATACCCTAGTTAAA-3' Fragment 3: 5'-CCCTAGTTAAAAC-3' Fragment 4: 5'-ACCGCAATACCCTAGTT-3' Fragment 5: 5'-ACCGCAATACCCTAGTTAAA-3' Fragment 6: 5'-ATTTACCGCAAT-3' On the basis of overlap in sequence, create a contig sequence of the original piece of DNA.Given the following template DNA strand, what is the correct complementary DNA sequence? 3' CGC AGT GGA CAT TTC 5' O 5' GCG TCA CCT GTA AAG 3' O 5' GAA ATG TCC ACT GCG 3' O 5' GCG UCA CCU GUA AAG 3' O 5' GAA AUG UCC ACU GCG 3'A DNA synthesizer “machine” is used to create short single stranded DNA of any given sequence. You have used the machine to create the following the DNA molecules: (DNA #1) 5’- CTACTACGGATCGGG – 3’ (DNA #2) 5’- CCAGTCCCGATCCGT – 3’ (DNA #3) 5’-AGTAGCCAGTGGGGAAAAACCCCACTGG-3’ Now you add the DNA molecules either singly or in combination to reaction tubes containing DNA polymerase, dATP, dCTP, dGTP, and dTTP in a buffered solution that allows DNA polymerase to function. For each of the reaction tubes, indicate whether DNA polymerase will synthesize any new DNA molecules, and if so, write the sequence(s) of any such DNAs. DNA #1 plus DNA #3 DNA #2 plus DNA #3 DNA #1 plus DNA #2 DNA #3 only
- What will be the newly synthesized DNA from the template given? DNA Template 3 - CGGATGCCCGTATAC-5 O 3- GCCTACGGGCATATG -5 O 5-GCCTACGGGCATAAG -3 O 5- GCCTACGGGCATATG-3 O3-CGGATGCCCGTATAC -5The chromatogram shows fluorescent peak data from a dye-terminating nucleotide-sequencing reaction. The peaks are shown with shortest fragment on the left to longer fragments on the right. T •C A Select the DNA sequence that matches the data. 5-ТАТAСТТАСGAAGT-3' 5'-GTCCTACGGACGCG–3' 5'-ATATGAATGCTTCA–3' 5'-TGAAGCATTCATAT–3' 5-АСТТCGTAAGTATA-3'For the following short sequence of double stranded DNA, design primers (just ~ 3-4 bases) and show 2 copy cycles of PCR (refer to figure 13.25) for the amplification of this sequence of DNA (so that you have 4 double stranded DNA). 5’- GGTATTGGCTACTTACTGGCATCG- 3’ 3’- CCATAACCGATGAATGACCGTAGC- 5’
- The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all 5 groups and translate. Group A 5’-GGCAATGGGTTTGTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTTTCAAAAATTAAG-5’ Group B 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’ Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGFor the chromatogram below, what is the sequence of the template DNA from base 115 to 125? CTGTGTGAAA TTGT TA T C CGC T CACA A T TCCACA CA A CATACGAG CCGGAAG CA T AA 110 120 130 140 150 160 СТТТААСАAТА ТАTTCAATTТС ATAACAATTTC GAAATTGTTATA certain section of the coding (sense) strand of some DNA looks like this: 5'- ATGGGCCACTCATCTTAG-3' It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. I Don't Know mutant DNA 5'- ATG GGCCACAGTTCTTAG-3' 5'- ATG GG CTCATCTTAG - 3' 5'- ATG GGCCACGCATCTTAG-3' Submit type of mutation (check all that apply) ооооо O point O silent O noisy ооооо insertion deletion insertion O deletion Opoint Osilent noisy insertion O deletion ооооо Opoint silent O noisy X S Ⓒ2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibility