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- Sickle cell hemoglobin DNA CACGTAGACTGAGG ACAC.. Sickle cell hemoglobin MRNA Sickle cell hemoglobin AA sequence 4. What type of mutation is this? Please explain why.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 PBM4_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 Leadple
- he Sequence below comes from the alpha-2 globin of the human hemoglobin gene cluster found in chromosome 16. The globin region of the hemoglobin protein itself consists of 2 alpha chains and 2 beta chains. 1 actcttctgg tccccacaga ctcagagaga acccaccatg gtgctgtctc ctgccgacaa 61 gaccaacgtc aaggccgcct ggggtaaggt cggcgcgcac gctggcgagt atggtgcgga 121 ggccctggag aggatgttcc tgtccttccc caccaccaag acctacttcc cgcacttcga 181 cctgagccac ggctctgccc aggttaaggg ccacggcaag aaggtggccg acgcgctgac 241 caacgccgtg gcgcacgtgg acgacatgcc caacgcgctg tccgccctga gcgacctgca 301 cgcgcacaag cttcgggtgg acccggtcaa cttcaagctc ctaagccact gcctgctggt 361 gaccctggcc gcccacctcc ccgccgagtt cacccctgcg gtgcacgcct ccctggacaa 421 gttcctggct tctgtgagca ccgtgctgac ctccaaatac cgttaagctg gagcctcggt 481 agccgttcct cctgcccgct gggcctccca acgggccctc ctcccctcct tgcaccggcc 541 cttcctggtc…The coding sequences of Gene F and Gene G are shown by the double-stranded DNA below: Gene F 5' ATGGGAGCACCAGGACAAGATGGATATCATTAG 3' 3' AGTTACCCTC GT GG TCCTGTTCTACCTATAGTAS Gene G Questions: 1. Write down the messenger RNA sequence when Gene F is transcribed. 2. Write down the polypeptide chain when Gene F is completely expressed. 3. Write down the messenger RNA sequence when Gene G is transcribed. 4. Write down the polypeptide chain when Gene G is completely expressed.Cynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-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. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X G
- What would the amino acid sequence be for the following DNA Transcript? 5’AAGCCATTTAAAGGC 3’ 3’ TTCGGTAAATTTCCG 5’ Phe Gly Lys Phe Pro Phe Leu Lys Phe Val Lys Phe Phe Lys Pro Lys Pro Phe Lys Gly More information is needed5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. Write the resulting amino acid sequence using the 3 letter code. Write the answer in a all capital letters. Leave a space between the amino acids. Do not write 5' and 3'. 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a T, then the result will be A) A nonsense mutation B) A frameshift mutation C) A silent substitution D) A missense mutation 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a A, then the result will be A) A nonsenese mutation B) A frameshift mutation C) A silent substitution D) A missense mutationGiven Sequence: 3’-TACGGTCCGGATTCGGTAGCTAGCATC-5’ Provide: Complementary Strand: Transcript Amino Acid Sequence 2.Given Sequence: 5’-GGGCATATGCCGTTTACCGGTTTGACTAAATAACCA-3’ Provide: Complementary Strand: Transcript Amino Acid Sequence 3.Given Sequence: 3’-AAC CAA TAC GTG AGG ATA CCA AGT AAC ACT CCC-5’ Provide: Complementary Strand: Direct Transcript: Transcript for Translation: Amino Acid Sequence:
- EboV RNA from Guinea Pig EboV RNA from Guinea Pig GAU ACG UUC GUC AAU EboV DNA CTA TGC AAG CAG TTA Translated amino acids Asp Thr Phe Val Asn EboV Strain 1 EboV Strain 1 (Circle the mutated bases) CCA TGT AAG TGG TTA mRNA Protein Of the mutations you identified, how many are: _______substitutions __________ frameshifts The result of the caused by the mutation is (silent, missense, nonsense). Underline your answer.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)Fig 3.16 EcoRI SacI KpnI |ampR Aval Xmal Smal lacZ BamHI Xbal SalI Accl HincII PstI Sphl HindIII Bam H1 Bam H1 Bam H1 1 2 3 4 The ends of the double-stranded DNA fragment above are blunt ends. The directionality of genes contained within the fragment is from left to right (arrow). After digestion by BamH1, which fragment can be inserted into the vector cut with BamH1 and Sma 1 maintaining the same directionality as the lacz promoter (green segment with arrow on vector)? 3 O 4 O1 2
