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A: The process of synthesizing mRNA strand with the help of template stand of DNA is called…
WILD-TYPE MC1R GENE (LIGHT COAT-COLOR
|
DNA |
CGG |
GAC |
CGG |
TGG |
GCC |
CAC |
TGA |
CAC |
|
mRNA |
||||||||
|
Amino Acid |
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- BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation LeadpleEcoRI --- 5' G - AATTC 3' 5' AGAATTCCGACGTATTAGAATTCTTAT CCGCCGCCGGAATTCT CATCA 3' 3' TCTTAAGGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5' Number of pieces of DNA , and type of fragment .#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
- This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' (i) Draw the structure of hairpin loop that will be formed during the end of transcription. (ii) Describe the function of the hairpin loop during transcription.Figure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5' TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5' 46 5 AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5 77 90 110 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3' ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. Illustrate how termination of transcription occurs in the gene above. (Hint: position from 156 to 180)COMPLEMENTARY DNA SEQUENCE OF GACGGCTTAAGATGC
- Q3. Cystic fibrosis is an autosomal recessive inherited disorder that causes severe damage to the lungs, digestive system and other organs in the body. Cystic fibrosis affects the cells that produce mucus, sweat and digestive juices. The disease occurs when there is a mutation in the cystic fibrosis transmembrane conductance regulator (CFTR) gene, which is the gene responsible for the movements of negatively charged particles known as chloride ions into and out of cells. Using the base pairing rules and the codon table, determine the mRNA and protein sequences produced by the CFTR gene's segment (below). Transeription begins at and includes the bold and underlined G nucleotide. CFTR gene's segment. - GCGATGTACAACCGAGGGTAAAAAA - 3' coding sequence a) The DNA template sequence 3'.. ...5' b) Fill in the first 9 nucleotides of the primary/ nascent mRNA transcribed from Gene A. 5'-.... -3' c) Protein N-term... . . C- terminus d) Exposure to cigarette smoke (a known mutagen) deletes base #9…Table 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?#1 HindII --- 5’ GTC ↓ GAC 3’ 5’ ACGACGTAGTCGACTTATTAT GTCGACCCGCCGCGTGTCGACCATCA 3’ 3’ TGCTGCATCAGCTGAATAATACAGCTGGGCGGCGCACAGCTGGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
- 11.5 A A Aa-AE-E-¹5- U - abe X₂ X² A-ay-A- Font Ulla Unigriffin DNA: mRNA: amino acids: traits: DNA: traits: mRNA: amino acids: · DNA: mRNA: to search #N O E Et CE- Paragraph $ 15 Ser 1. CAT AGG GAG | CAA GGG TGA CTT TTT | AAT AAT GAC GGG | A E ALT | CAA TTG TTA CGG | AAA AGA CCC | GCC ATA ACA TTT | % STP | CAC CGT CGA | GTA GTA | AGA GGG CAT | TTG TAA GGA GGG GGG TGT | 16 AaBbCcDc AaBbCcl AaBbCcL Aa BbCcDc 1 Normal 1 Body Text 1 List Para... 1 No Spac... W] Tyr 17 Val & 7 Gly E CO OM no num lk T Aa Bb Cc 1 Table Pa (p)) Styles 12 PFigure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5 TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5 46 77 90 5' AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 110 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5' 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3 ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. (i) Identify both the hexameric sequences of the promoter region in the coding strand above.48 Second letter If any single nucleotide is deleted from the DNA sequence shown below, what type of mutation is this? UUU U UUC UUA UCU Phe UCC UAU UGU Cys ANTISENSE 5' GGACCCTAT3' UAC Tyr UGC UAA Stop UGA Stop UAG Stop UGG Trp Ser UCA Leu UUGL" UCG CUU CU CAU) His CAC) CGU CGC CGA Arg CUC C Leu Pro CAA1 Gin CAG) CUA CCA CUG CG CGG AAU AAC Asn AGC AAA AAG Lys AGG Arg AUU ACU AGU Ser AUC lle ACC Thr AUA ACA AGA AUG Met ACG GCU GCC GUU GAU] GGU GUC Val GAC Asp GGC Ala Gly GUA GCA GAA GGA Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a FRAMESHIFT SILENT NONSENSE MISSENSE Third letter First letter