Concept explainers
The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate.
FIND THE POSSIBLE MUTATIONS
Group B - MUTATION
5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’
3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’
Group C-
5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’
3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’
The genetic information of all living organisms (except some viruses) is stored in the cell in the form of DNA. This stored information is converted into the protein, via two main steps, 1. Transcription and 2. Translation.
In the transcription, the information present in the DNA is copied in the form of mRNA. Transcription takes place in the nucleus of the cell (In eukaryotes). In translation, the copied information in the form of mRNA is used to make proteins.
The triplet codon (genetic code) is the group of three consecutive nucleotides that code for a specific amino acid. Such amino acids come together to form the polypeptide chain that folds into the unique confirmation and act as protein.
There are 61 different types of codons that code for 20 amino acids and three stop codons that terminate the peptide chain formation. The one amino acid can be coded by more than one codon, this phenomenon is called redundancy of genetic codon.
The mutation is the change in the nucleotide sequence of DNA. The mutation can take place due to some replication error or due to environmental factors such as UV exposure or chemical exposure. The main three types of mutations that can be seen in the DNA sequence are 1. Base substitution mutation (Point mutation, silence mutation, missense mutation, and nonsense mutation) 2. Deletion, 3. Insertion.
Step by stepSolved in 2 steps
- 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.arrow_forward**ALL GROUPS HAVE A MUTATION MAKE SURE TO IDENTIFY THEM** 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’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGarrow_forwardBelow is a set of sequence reads from an individual aligned to a reference sequence. AGACACTCCACC CGATCAGAATCG GATCATCATCGA CGACCTGAAAAG CGAATATGCTGG AGAG ACACTCCACCTG ATCAGAATCGAA TCATCATCGACT ACCTGAAAAGAT AATATGATGGTA AG AG ACTCCACCTGCCAT GAATCGAAGА АТСАТСGAСТСG СТGAААAGATCG ТАТСАТGСТАGAG CTGAAAAGATCG TATGATGGTAAGAG AGAC TCCAАССТСССАТСА АТCGAAGAТС САТСGAСТСGAC GAААAGATCGAA САТCGACTCGAC GAAAAGATCGAA GTAAGAG AGACAC CACCTGCGATCAGA CGAAGATCAT TCGACTCGACCT AAAGATCGA СТССТААGAG AGACACTC ССTGCGATCAGAAT AAGАTСАТСА GACTCGACCTGA AGATCGAАТАТG GGTAAGAG AGACACTCCА TGCCATCAGAATCG GATCATCAТС СТCGACСTGААА АТСGAАТАTGAT TАAGAG AGACACTCGACCTGCCATCAGAATCGAAGATCATCATCGACTCGACCTGAAAAGATCGAATATGATGGTAAGAG Looking at the alignment what are the number of the following? Homozygous SNPS Heterozygous SNP5 Homozygous Insertions Heterozygous Insertions Homozygous Deletions Heterozygous Deletionsarrow_forward
- TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTTarrow_forwardWhat would be the percentage of G, C, A, and T in each column? Have to calculate the nucleotide frequency from my sequence.arrow_forwardYou have the following DNA sequence: 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G that is underlined changes to to a C the result will be - A) A nonsenese mutation B) A frameshift mutation C) A silent substitution D) A missense mutation You have the following DNA sequence: 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G that is underlined is deleted, then the result will be A) A nonsense mutation B) A frameshift mutation C) A silent substitution D) A missense mutatio If there are 3000 bases in the coding region of a gene, the gene will have A) 3000 amino acids B) 6000 amino acids C) 1000 amino acids D) 3000 codonsarrow_forward
- Give the RNA molecule sequence transcribed from the following DNA sequence of a eukaryotic gene and with the correct 5' and 3' ends. DNA: 5'-ATAGGGCATGT-3' 3'-TATCCCGTACA-5' <--- template strand Group of answer choices 5'-ATAGGGCATGT-3' 3'-UAUCCCGUACA-5' 5'-AUAGGGCAUGU-3' 3'-TATCCCGTACA-5'arrow_forwardTable I CACGT A GA CTGAGG ACTC CACGTAGACTGAG G ACAC Wild-type beta-globin gene fragment Sickle-cell beta-globin gene fragment > Circle the mutation in DNA of the sickle-cell beta-globin gene fragment Compare fragments of DNA the wild-type and mutant beta-globin genes in the Table I above, what are the similarities and differences you observe?arrow_forwardWhich reading frame gives you the longest amino acid sequence? 5'3 Frame 1 5'3 Frame 2 5'3 Frame 3 3'5 Frame 1 3'5 Frame 2 3'5 Frame 3arrow_forward
- The DNA sequence of a short gene from a sea slug, and the mature RNA synthesized from this gene, are shown below. DNA sequence: 5’ - AGCATCTCATGTGCGAGTCCTGACGCTGACTAGC – 3’ 3’ - TCGTAGAGTACACGCTCAGGACTGCGACTGATCG – 5’ mature mRNA: 5’ – cap-AUCUCAUGUGCGAACGCUGACUAGAAAAAAAAAA- 3’ How many amino acids are in the peptide encoded by the gene? _____________arrow_forwardThe 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.arrow_forwardYou are screen sharing Here is part of a gene: 5' TTTAATGGTAACCGTATTGCAGCTATTAGCATAAATG 3' AAATTACCATTGGCATAACGTCGATAATCGTATTTAC 5' 3' If the bottom strand of the DNA is the template strand, what will be the mRNA sequence and the amino acid sequence that are present in the protein? Copyright © 2010 Pearson Education, Inc.arrow_forward
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