ANSWERED QUESTION, I NEED EXPLANATION! DNA, mRNA FOR QUESTION 2, WHEN IT SAYS TH

Question

ANSWERED QUESTION, I NEED EXPLANATION! DNA, mRNA

FOR QUESTION 2, WHEN IT SAYS THE DNA SEQUENCE IS SHOWN ABOVE THE CODING STRAND, WHAT STRAND IS IT REFERING TO? I KNOW HOW TO TRASCRIBE (C-G/ G-C /A-U/ T-A) I AM JUST FINDING IT HARD TO FIRGURE OUT WHAT WE ARE SUPPOSED TO BE USE? PLEASE IGNORE ALL THE OTHER QUESTION AND ANSWER THIS.

PLEASE IGNORE ALL THE OTHER QUESTION AND ANSWER THIS.

Question:

COF CASE 2 Part 2 Cloning pMMO genes in E. coli will allow E. coli to produce the methane monooxidase that oxidizes methane to methanol... the first step to converting the methane to biomass. You want the heiress to understand that taking the DNA from M. capsulatus and cloning it in E. coli will result in E. coli producing the identical pMMO enzyme. It is all in the Central Dogma! The easiest way to show this is to look at the sequence of DNA and compare it to the sequence of amino acids in the protein. Again the college freshman helps you. You instruct the student to: get the gene sequence for pmoC1 from the National Center for BiotechnologyInformation site: http://www.ncbi.nlm.nih.gov/nuccore/14456718 and prepare a set of slides for you to show how the gene will be transcribed and translated to make a protein. Here is the sequence that she has prepared for your presentation:

1 GTCGACTGGG CACCAGCCGG ATGCGTCCGT CAACCCCGAC TGTTCCGCCA AGAACTCCGG

61 CCTTCTGTGG CGGGATGACG TTGAGCACAT CGGCACAGAA AGTGTCGAAC TCGGTTTCGA

121 CGGTTCGCCC GGCGGCATCG ATCCGCTCCA GCCGTCCCTC TTTTTCCGAG GAAACCCATT

181 CCACCATGTC TCCGTAGAGT TCGCGCCAGG CCTGCTCGAA TGCCGGCTGC TTGGAAAACC

241 GGGTCTTGGC ATCGAGTATC AGGATCTTGG AGCGGGGTTT GTGCTTTTTC ATGTAATAGG

301 CGATGAGCGA GGCGCGTTCG TAGGGGCCAG GCGGACAACG ATAGGGAGAA GGTGGTGCGG

361 TGATCAGAAC CAGGCCGCCG TCGGGCATCG CGCGGATCTG CCGGGCGAGC AAAGCGGTCT

421 GGGGGCCGGC CTTCCAGGCA TGGGGAACGA ACCGGCTCGC CGCCTCGTCA TAGCCCATTA

481 TCGCCTCCCA GCGAAAGTCG ATGCCGGGGG AGAGGACTAG CCTGTCGTAG GTGACCTCGG

541 CACCATCGTT CAGTATCACA CGCCGTCGCT GCCGATCCAG GCGGGCGACC CGGGCGGTTA

601 CCTTTCCTAT ATCCAGCTCC CGCCGCAACC AGTCATAAGA CCGCGCGAGA GTCCCCATGT

661 CCCCGAGGCC GGCGACTGCT TCGTTCGATC CGGGGCAGGA AAGATAAGTC TCCTGCGGTT

721 CGATCAATGT GATCGTCAGA CCGGGATTCA TCTGCTTCAG ATAGCGGGCG GCCGTGGCAC

781 CGCCATAACC GCCACCGACG ACCACGACCG GCCGCCTGAG CGAAGACCTT GCCGGAAGCG

841 CAGCCACCCA GTCCCAGCCC ATGCCGCAGA CTGCCAGCAG GCGCAGGAAC CGCCGCCGGC

901 GGATCATGGC CTCTTTCCCA GAAAGCCGGC GATCGCTTCA ATGTCCTGGT TCGTGAGTCC

961 GGCAGCGATC CGGTTCATGA CCGTGCCCGA TCTTTTTCCT TCACGGTACT CCCGCAACAG

1021 AGATGCCATC TCCTTCGCAT CGAAGCGGCG TAACGATGCC GGTTCGGGAA TCTGCTCCTC

1081 CTCGTCGGCA TGGCAGCCGA GGCAACCGAG CGCAGCCAAA ACCATGTCCG GTTTTTCGGC

1141 CCGGGCCGGG AAACAGAGAG CGACAGCGAT CGTACCGATC TGGACGCGTC TCACAAACGA

1201 CAAAACGTCA CGATGGGTGT TCGGTAGCTG AGTCACGGGG ATTTGTAGAA GTATAGGACC

1261 GACGGATTTT ATGCAAGCAT GTCGCTTTGA CCAAGCCGGG ATTCCATGGA AGGGATGTCA

1321 TCGGGAGAGT TATTTATGTC GTTGATTTAT AAGAAACTAC CCCTGCGTCA AAATGTCGCA

1381 GATTTTTCTT GACAGTTTGG GGGAGGGTGA TAGATCCTCC ACCGATGGAC CGGTACCGCC

1441 TCTGTTGCGG GGTCCATGAA ATGCCCGTTA GAGGCAGAAC CGATAGGGAA TTAGAGAAGC

1501 GGGCGTCGGC GCCGAATGCC GGCCCCTGTC AACCATCACT TTAGGAGGAA CAAACAATGG

1561 CAGCAACAAC CATTGGTGGT GCAGCTGCGG CGTAAGCGCC GCTGCTGGAC AAGAAGTGGC

1621 TCACGTTCGC ACTGGCGATT TACACCGTGT TCTACCTGTG GGTGCGGTGG TACGAAGGTG

1681 TCTATGGCTG GTCCGCCGGA CTGGACTCGT TCGCGCCGGA GTTCGAGACC TACTGGATGA

1741 ATTTCCTGTA CACCGAGATC GTCCTGGAGA TCGTGACGGC TTCGATCCTG TGGGGCTATC

1801 TCTGGAAGAC CCGCGACCGC AACCTGGCCG CGCTGACCCC GCGTGAAGAG CTGCGCCGCA

1861 ACTTCACCCA CCTGGTGTGG CTGGTGGCCT ACGCCTGGGC CATCTACTGG GGCGCATCCT

1921 ACTTCACCGA GCAGGACGGC ACCTGGCATC AGACGATCGT GCGCGACACC GACTTCACGC

1981 CGTCGCACAT CATCGAGTTC TATCTGAGCT ACCCGATCTA CATCATCACC GGTTTTGCGG

2041 CGTTCATCTA CGCCAAGACG CGTCTGCCGT TCTTCGCGAA GGGCATCTCG CTGCCGTACC

2101 TGGTGCTGGT GGTGGGTCCG TTCATGATTC TGCCGAACGT GGGTCTGAAC GAATGGGGCC

2161 ACACCTTCTG GTTCATGGAA GAGCTGTTCG TGGCGCCGCT GCACTACGGC TTCGTGATCT

2221 TCGGCTGGCT GGCACTGGCC GTCATGGGCA CCCTGACCCA GACCTTCTAC AGATTCGCTC

2281 AGGGAGGGCT GGGGCAGTCG CTCTGTGAAG CCGTGGACGA AGGCTTGATC GCGAAATAAG

2341 GAGCTTGTCA GGCTGGCGGG CCGGTTTTTT CGGGGTTCTG GCGACAGGCC CCGGGGAACC

You decide to give the Heiress and her guests cards printed with all of the codons and their corresponding amino acids. As you go through the sequence, each person will bring their codon card to the front of the room and to show how the DNA and mRNA sequences encode a protein with a specific sequence of amino acids. This will be great ..you think!! You begin reading the sequence at site 1557 which is the position that GenBank indicated as the first nucleotide of the start codon of pmoC1. As you are underlining the sets of three nucleotides and discussing how each corresponds to a codon in the mRNA and an amino acid in the protein, you read ahead and notice!! Your student helper must have made a mistake!! The mistake in the DNA sequence! DARN IT, is this going to change the protein product?? Will my audience notice?? What a mess....

You need to determine if the helper made a mistake. One quick way to do this is to use a BLAST Search to compare the student’s sequence to the sequence that is actually in the database. Go to http://blast.ncbi.nlm.nih.gov/Blast.cgi. Use the Specialized BLAST option: Align two (or more) sequences using BLAST (bl2seq). Copy the DNA sequence above(student’s sequence) and paste it into the box designated as the Query Sequence (don’t worry about the numbers or about ‘tidying up’ the sequence by removing spaces...BLAST will do this for you). Find the correct sequence for the pmoCAB operon from http://www.ncbi.nlm.nih.gov/nuccore/14456718, Copy this sequence and paste it into the Subject Sequence box on the bl2seq form. Run the BLAST comparison (click the BLAST button) and note any differences/non-alignment. Now you are ready to go to the PAK2.2 questions.

1. If the nucleotides in the gene were changed as indicated in the mistake sequence, what type of mutation would this be? Explain.

2. The DNA sequence shown above is the coding strand of the double stranded DNA that makes up the pmoCgene, expected to encode PmoC. What would the mRNA sequence be when the template strand DNA is transcribed? (give first 60 bases)

3. What would the amino acid sequence be when the mRNA is translated?

4. How is the protein product of this DNA sequence different from the PmoC that would be made if the gene did not have a mutation?
Make a prediction as to the functionality of this altered protein compared to PmoC.

ANSWERS

1) If the student has made a mistake, it can be of two types.

- Point mutation is a type of mutation that affects only one or very few nucleotides in a gene sequence. Thus it may change a few amino acids in the protein i.e. change an histide residue into alanine residue.

-frame shift mutation is caused byinsertions or deletions of a number of nucleotides in a DNA sequence that is not divisible by three. This results in a complete change in the amino acid sequence of the protein.

2) the following is the mRNA sequence AUGGCAGCAACAACCAUUGGUGGUGCAGCUGCGGCGUAAGCGCCGCUGCUGGACAAGAA

3) the dollowing is the protein sequence

MAATTIGGAAAA.APLLDK?

4)Due to the mutation, a stop codon is present on the mRNA. Therefore when the mRNA is translated, a protein of only 12 amino acids will be produced instead of the full length protein.

As a truncated version of hte protein will be produced, the chances are that this protein will not be functional.

Explanation / Answer

Ans: 2. The DNA sequence shown above is the coding strand of the double stranded DNA that makes up the pmoCgene, expected to encode PmoC. What would the mRNA sequence be when the template strand DNA is transcribed? (give first 60 bases)

The DNA sequence shown above: What he mean is the coding strand (sense strand) is given. A sense strand, or coding strand, is the segment within double-stranded DNA that runs from 5' to 3', and which is complementary to the antisense strand of DNA, or template strand, which runs from 3' to 5'. The the coding strand OR sense strand is the strand of DNA that has the same sequence as the mRNA, which takes the antisense strand as its template during transcription, and eventually undergoes (typically, not always) translation into a protein. The antisense strand is thus responsible for the RNA that is later translated to protein, while the sense strand possesses a nearly identical makeup to that of the mRNA. Note that for each segment of dsDNA, there will possibly be two sets of sense and antisense, depending on which direction one reads (since sense and antisense is relative to perspective).

In simple terms the sequence given above is the actual mRNA strand in which you replace T with U and from triplet codon you write down amino acids up to 60bp (60/3 = 20 amino acids, but one stop codon hence 19 amino acids)

1557 ATG G CA GCA ACA AC C ATT GGT GGT GCA GCT GCG G CG TAA GCG CC G CTG CTG GAC AAG AAG 1617

mRNA AUG GCA GCA ACA ACC AUU GGU GGU GCA GCU GCG GCG UAA GCG CCG CUG CUG GAC AAG AAG

Protien : MAATTIGGAAAA (STOP) APLLDKK

check once your second answer, I think you copied 59 bases only instead of 60. One more amino acid K should come in the last, if I am wrong pardon.

Hope you understood, good luck

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