You obtained the sequence of the frog gene X you amplified in Question #16 through a process called automated sequencing. In automated sequencing, you are given a printout of the sense strand of your DNA. The printout is shown below. 'The first thing you need to do is use the correct reading frame. Having done this, the next thing to do is to write out the mRNA sequence using this sense strand reading frame. The last thing to do is to translate the sequence. The reading frame DNA sequence is:The mRNA sequence is:The polypeptide sequence is:A disease in frogs which causes their tongue to fall out of their mouths is killing the frog population in LA County. You obtain a dead frog and isolate its gene Xf. When you sequence this mutated gene, you find that the last ‘G’ at the end of the first line of this sequence has been deleted (i.e. the G at position 86). In order to determine how this mutation changes the resulting polypeptide, write the mutated polypeptide sequence in the space below. What kind of mutation was produced?The mutated polypeptide sequence is now: This deletion mutation was a __________________________________ mutation File: gvf11b11.y1Machine: Cochise-1414-017MatricLane: 93Primer: DT3730POP7(BDV3).mobComment: 1778D3Spacing: 15.06Signal: C: 1728 A: 3254 G: 2538 T: 2619Bases: 903Gel name:Gel start date: 3 Jan 2004Gel start time: 4:1231020304050607080NNACTCA TCTGGTGGA T TC C TA TC CTG AC A AG TG ATGTG CAAAC TG G TAACTC TG AG GCAG A TA AC CA G GG CAAA AAG GTG TATAAG90TAAAACA AAGCCTAACT TG GA AGA AAGAGAGA AAAAGT TCAGTT TTACACA TCT TTA TA TGAAGCAGA AGTC CG GA AA ATCA TT TA AAGAT100110120130140150160170230180TAT TCTAT TA ATGTCCTGT TA TATT TGT CA TATTCATA CAGT TGTCA CAGTA TA TT TCAAAC CA A C TG T TTAAAA ACA AAC TGAAATAAA190200210220240250260270AAATTTAAA TACCCT TA TG TA AA ATAG GC T TC CCTG GTG GC TCAG GG GTAAAAA ACTCGC CCG C CA ACGCAGGAG ATGTAGATTTGATC CC T280290300310320330340350360GG GT TAG GA AGATCC CCTGGAGA AG GAAATGAAAAAC CACTCTAGTATTCT TGC CTG GG AA ATC CCA TG GACAG AG GAGCGTGGAGG GC370380390400410420430440450460470480490500510520530TACAGTCCATGG GAGT CGCAAA AGAGT TG GACATG ACTAAACAACAACATATAAAATA ACCT TACTC CATAATGTCAAACT TATGTCACACwww mhma m wmmm590630540AAA ATGCA AAGT TCT TACATCTATTA ACTTTTATGGT TAAATATAACCTAATGCACTGTTT TATACAGCA ACAACTACTTTTT TATTT TAAA550560570580600610620wwhwmmm mmm mmpiyng710720TG GTGAAAAAT TAGTGAACAGAG G GTCTCAGAGATGCTAAACTTATGACT TATAAT TAACAAGATGTTÄTAGTTCTAG ACAT TTTTGATAT640650660670600690700

A linear polymer of nucleotides is called as nucleic acid. DNA (Deoxyribonucleic acid) and RNA are…

Answered: You obtained the sequence of the frog…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Question

You obtained the sequence of the frog gene X you amplified in Question #16 through a process called automated sequencing. In automated sequencing, you are given a printout of the sense strand of your DNA. T

he printout is shown below. '

The first thing you need to do is use the correct reading frame.

Having done this, the next thing to do is to write out the mRNA sequence using this sense strand reading frame. The last thing to do is to translate the sequence.

The reading frame DNA sequence is:

The mRNA sequence is:

The polypeptide sequence is:

A disease in frogs which causes their tongue to fall out of their mouths is killing the frog population in LA County. You obtain a dead frog and isolate its gene X_f. When you sequence this mutated gene, you find that the last ‘G’ at the end of the first line of this sequence has been deleted (i.e. the G at position 86). In order to determine how this mutation changes the resulting polypeptide, write the mutated polypeptide sequence in the space below. What kind of mutation was produced?

The mutated polypeptide sequence is now:

This deletion mutation was a __________________________________ mutation

File: gvf11b11.y1
Machine: Cochise-1414-017
Matric
Lane: 93
Primer: DT3730POP7(BDV3).mob
Comment: 1778D3
Spacing: 15.06
Signal: C: 1728 A: 3254 G: 2538 T: 2619
Bases: 903
Gel name:
Gel start date: 3 Jan 2004
Gel start time: 4:123
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NNACTCA TCTGGTGGA T TC C TA TC CTG AC A AG TG ATGTG CAAAC TG G TAACTC TG AG GCAG A TA AC CA G GG CAAA AAG GTG TATAAG
90
TAAAACA AAGCCTAACT TG GA AGA AAGAGAGA AAAAGT TCAGTT TTACACA TCT TTA TA TGAAGCAGA AGTC CG GA AA ATCA TT TA AAGAT
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TAT TCTAT TA ATGTCCTGT TA TATT TGT CA TATTCATA CAGT TGTCA CAGTA TA TT TCAAAC CA A C TG T TTAAAA ACA AAC TGAAATAAA
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AAATTTAAA TACCCT TA TG TA AA ATAG GC T TC CCTG GTG GC TCAG GG GTAAAAA ACTCGC CCG C CA ACGCAGGAG ATGTAGATTTGATC CC T
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GG GT TAG GA AGATCC CCTGGAGA AG GAAATGAAAAAC CACTCTAGTATTCT TGC CTG GG AA ATC CCA TG GACAG AG GAGCGTGGAGG GC
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TACAGTCCATGG GAGT CGCAAA AGAGT TG GACATG ACTAAACAACAACATATAAAATA ACCT TACTC CATAATGTCAAACT TATGTCACAC
www mhma m wmmm
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AAA ATGCA AAGT TCT TACATCTATTA ACTTTTATGGT TAAATATAACCTAATGCACTGTTT TATACAGCA ACAACTACTTTTT TATTT TAAA
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wwhwmmm mmm mmpiyng
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TG GTGAAAAAT TAGTGAACAGAG G GTCTCAGAGATGCTAAACTTATGACT TATAAT TAACAAGATGTTÄTAGTTCTAG ACAT TTTTGATAT
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