II. Use the eukaryotic gene DNA sequence below to answer the following questions: 1 11 21 31 CGACTTACTG 51 TGGACGCGCC 101 GTCATTCAGC GGCGTATAAA GCGACGACTG TAGACTGATG AGCCTATCCA 61 71 111 81 121 41 ATGGCCCTGT AAGCGGTGCG ATGCAATAAA ACGCGTATCA 131 91 141 GTAGTCTGAT GCCAGTCGAC TGCATTGGAC ACCGGTTACA
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- summarize these results using concise language in a neat table; Control : 5’ ATGTACGCGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ This is the coding strand of DNA and hence this DNA sequence is similar to mRNA sequence. So the mRNA sequence is : 5’ AUGUACGCGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ Mutant 1: 5’ ATGTACGAGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ mRNA sequence 5’ AUGUACGAGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ The bold Adenine is the mutated base which is substituted in place of Cytosine. So the codon change from GCG to GAG. GCG codes for Alanine but GAG codes for Glutamic acid. So the amino acid sequence changes. Hence this mutation is missense mutation where a base substitution results in change in amino acid sequence. Mutant 2: 5’ ATGTATGCGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ mRNA sequence: 5’ AUGUAUGCGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ In this mutation, Cytosine is replace by Thymine and hence the codon…iiBelow is the DNA sequence of a patient with overlapping genes (a single mRNA has multiple initiation points for translation) for two different proteins (DADαs and AMA): 5’- GTCCCAACCATGCCCACCGATCTTCCGCCTGCTTCTGAAGATGCGGGCCCAGGGAAATCTCTAACG-3’ 1. Indicate the DNA sequence coding for RNA. 2. Indicate the amino acid sequence of each of them.
- D. The sequence of a eukaryotic gene is given below, where in boed an inset containing: 5'GA TTATGGAATTCACCTAT GATCGCAT GGCCATTGAACCT 3 3CTAATACETTAAGTGGATA CTAGCGTA CCGGTAACIT GGA 5 A. Write the sequence of the WRNA produced by its process of transcription and of wRNA that is transferred to ribosomes in order to produce the peptide B. A mutation has occured in the above gene. The GIC pair highlighted replaced by T/A. Cells that are homozygous for that mutation become Cancerous. Do think this gene is a you proto-oncogene? or a tumor suppressor gene and why? describes D₂. The following family tree is given which the way inheritance of a dease. It is noted that only one of its two persons generation I is heterozygous and that one of the individuals in subsequent generations exhibits unexpected Phenotype. Individuals II4 and III2 show its phenotype disease. The remaining individuals have normal phenotype. αAnswer the following whether it is TRUE or FALSE: 1. For each DNA segment 3'-ACCTGCCTACCCG-5' the sequence of the mRNA molecule synthesized is 5'-TGGACGGATGGGC-3' 2. In the template strand TACCGAGGTATGTAC, the coding strand is 5'-ATGGCTCCATACATG-3'. 3. In the template strand TACCGAGGTATGTAC, the coding strand is 5'-AUGGCUCCAUACAUG-3'. 4. The template strand is the strand of DNA used for RNA synthesis. 5. Transcription forms a messenger RNA molecule with a sequence that is identical to the DNA template from which it is prepared.What is the lagging strand sequence if the leading strand of DNA is TTA CCG ATC GAA? How many amino acids can be identified if the coded mRNA sequence is GCAUUACAUGGCGGA
- 17) Synthesis of the mRNA starts at the boxed A/T base pair indicated by the box and proceeds left to right on the sequence below. Transcribe and translate this bacterial gene. 5'-GGACCGCGGGGCAGGATTGCTCCGGGCTGTTTCATGACTIGICAGGTGGGATGACTTGGATGGAAAAGTAGAAGGTCATG-3 3'-CCTGGCGCCCCGTCCTAACGAGGCCCGACAAAGTACTGAACAGTCCACCCTACTGAACCTACCTTTTCATCTTCCAGTAC-5′ 1 -+--at - --+-- 80What is the sequence of the mRNA transcript that will be produced from the following sequence of DNA? The top strand is the template strand, the bottom strand is the coding strand. 5’ – TCGGGATTAGACGCACGTTGGCATACCTCG – 3’ 3’ – AGCCCTAATCTGCGTGCAACCGTATGGAGC – 5’ Enter the mRNA sequence here (pay close attention to the direction of the molecule!): 5'-_____-3'4a) Write out the protein sequence (the amino acids, in order) encoded by the mRNA sequence: 5'AUGCGACCUAGCUAUGGA3'
- iii and ivUsing the genetic code below, decipher the following mRNA sequence. 5'CCCGGGAUGCGGUGUUGGUUUUAACCCGGG3' (show all work)Consider the following mature mRNA from a human cell: 5' UAAUGUCGCAAUAACC 3¹ What is the sequence of amino acids in the translated protein? Second letter A First letter С U d A G บบบ UUC P U UUA UUG Leu CUU CUC CUA CUG GUU GUC GUA GUG -Phe Met-Ser-Gln Met-Arg-Lys-Ser Leu Stop-Cys-Arg-Asn-Asn C AUU AUC lle AUA ACA AUG Met ACG Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC GCU GCC GCA GCGJ Ser Pro Thr Ala UAU UAC Tyr UGU UGC. UAA Stop UGA Stop UAG Stop UGG Trp CAC CAGGin CAU His Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. AAU Asn AAA Lys AAG. There is no start codon, resulting in no translated protein. GAU ASP GAC Glu GAA GAG G CGU CGC CGA CGG Cys AGU AGC AGA AGG Arg GGU GGC GGA GGGJ Arg ser Gly MCAG U UCAG с А SCAG U SCAG U Third letter