Name and explain the three possible consequences of having a mutation somewhere in the region from 61-75
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Gene Interactions
When the expression of a single trait is influenced by two or more different non-allelic genes, it is termed as genetic interaction. According to Mendel's law of inheritance, each gene functions in its own way and does not depend on the function of another gene, i.e., a single gene controls each of seven characteristics considered, but the complex contribution of many different genes determine many traits of an organism.
Gene Expression
Gene expression is a process by which the instructions present in deoxyribonucleic acid (DNA) are converted into useful molecules such as proteins, and functional messenger ribonucleic (mRNA) molecules in the case of non-protein-coding genes.
Name and explain the three possible consequences of having a mutation somewhere in the region from 61-75.
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- Based on these sequences, how do you decide decided which fragments were the beginning, middle, and end. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTC AATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTC GCGCCGAAAAAGATATGGHpaI --- 5' GTT - AAC 3'5' GGATGTTAACAATCTCTACGGGTTAACACCCTTGGGTTAACATCCGCGG 3' 3' CCTACAATTGTTAGAGATGCCCAATTGTGGGAACCCAATTGTAGGCGCC 5' Number of pieces of DNA____Transcribe the DNA into mRNA. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGG
- Sequence 1 : TACGCTACGGTAATC Sequence 2: TACGCTACTATCGTAEcoRI --- 5' G - AATTC 3' 5' AGAATTCCGACGTATTAGAATTCTTAT CCGCCGCCGGAATTCT CATCA 3' 3' TCTTAAGGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5' Number of pieces of DNA , and type of fragment .Find 5’ UTR and 3’UTR of Mrna 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’
- If we have the following mutations, find the type of the mutation (silent or missense or nonsense?) 17C=U 36G=A 49G=U 115A=C 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’You are given the sequence below. Which one would be an appropriate SaCas9 (PAM = ‘NNGRRT’) protospacer? (Multiple possible) 5’-CATGATCTGGGTCATCTTCTCGCGGTTGGCCTTGGGATTGAGGGGGGCCTCGGTGAGCAGGGTGGGG-3’ 5’-AGGCCAACCGCGAGAAGATG-3’ 5’-CATCTTCTCGCGGTTGGCCT-3’ 5’-GAGGGGGGCCTCGGTGAGCA-3’ 5’-CATGATCTGGGTCATCTTCT-3’ 5’-TTGAGGGGGGCCTCGGTGAG-3’Transcribe the following DNA sequence into RNA, and then into amino acids 5’-GTATACTTGTGGGCCAGGGCATTAGCCACACCAGCCACCACTTTCGGATCGGCAGCC-3’ 3’-CATATGAACACCCGGTCCCGTAATCGGTGTGGTCGGTGGTGAAAGCCTAGCCGTCGG-5’
- Sequence A: TCT/ TCC/ CTC/ CTA/ AAC/ GTT/ CAA/ CCG/ GTT/ CTT/ AAT/ CCG/ CCG/ CCA/ GGG/ CCC/ CGC/ CCC/ TCA/ GAA/ GTT/ GGT AGA Sequence B: TCA/ GAC/ GTT/ TTT/ GCC/ CCG/ TAA/ CAA/ CTT/ GTT/ ACA/ ACA/ TGG/ TCA/ TAA/ ACG/ TCA/ GAG/ ATG/ GTC/ AAT/ CTC/ TTA/ ATG/ ACT Sequence C: TAT/ ATA/ CAT/ GTA/ AAC/ ACA/ TAC / TCA/ GTG/ GAC/ CAA/ CTC/ AAC/ ATA/ AAC/ CAA/ ACA/ CCG/ CTC/GCG/ CCG/ AAA/ AAG/ ATA/ TGG STEP 1: Transcribe each of the 3 DNA sections into nRNA. Write out the complementary CODONS directly below the DNA strands. STEP 2: Clearly identify the three fragments as the FIRST, SECOND and THIRD fragments by clearly labeling any obvious PROMOTERS, START and STOP codons. STEP 3: Number the the START codon #1 and number the codons in the correct order until the STOP codon is reached. STEP 4: Codons 14- 64 (including 14 & 64) represent an intron. Draw a single red line through these codons to indicate that they will NOT be translated. STEP 4: In the space provided, write out the…5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC Find start codon and stop codon UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’**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’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAG