Use the genetic code below to determine the third amino acid in the protein translated from the following mRNA: 5'-CCAAAUGCCAUCGUAA-3' HINT: remember where a ribosome starts translation!!!
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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.
Step by step
Solved in 3 steps
- The following segment of DNA codes a protein. The uppercase letters represent Exons, the lowercase letters introns. Draw the pre- mRNA, the mature mRNA and translate the codons using the genetic code to form the protein. Identify the 5’UTR and 3UTR 5’- AGGAAATGAAATGCCAgaattgccggatgacGGTCAGCaatcgaGCACATTTGTGATTTACCGT-3’Refer to the information on the genetic code. Use this information to determine how many amino acids are coded for by the mRNA sequence AUGCGCAGUCGGUAG. The genetic code Second letter of codon UAU UAC JUU Phenylalanine uCU UUC Phe) UUA Leucine (Leu) UUG Tyrosine (Tyr) GCysteine (Cys) UGC 1oStop codon |UGG Tryptophan (Trp) CGU CGC UcC Serine (Ser) UCA ucc CCU cC Proline (Pro) Stop codon UAG Stop codon CAU Histidine His) CU CUC CUA CUG Arginine (Arg) Leucine (Leu) cca CAA CCA CGA Glutamine (Gin) CAG AUU AUC AUA ACU Isoleucine (le) AAU AAC AGU AGC Asparagine (Asn) Serine (Ser) ACC Threonine (Thr) ACA Methicnine ACC start codon GCU Lysine (Lys) AGA Arginine (Arg) ARC AGS GAU Aspartic acid (Asp)G0 GAC GUU GUC Valine (Val) GCC Alanine (Ab) GG Glycine (Gly GUA GUG GCA GCG GA Glutamic acid (Glu) GA GGG GAG 4 15 First letter of codon Third letter of codonIndicate the amino acid sequence of the protein encoded by the following mRNA molecule. Use the genetic code table and assume that the very first “AUG” the ribosome encounters will serve as the start codon and specify methionine. 5’-AAUUCAUGCCCAAAUUUGGGGCACGAAGCUUCUUAGGCUAGUCCUAAAAAA-3’
- Find 5’ UTR and 3’UTR of Mrna 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’Write the sequence of the mRNA molecule synthesized from a DNA template strand having the sequence 5'-ATTACAGGCGGT-3' 5'- UAAUGUCCGCCA Incorrect Write the amino acid sequence encoded by the mRNA base sequence 5'-GAGUUAGUUUGUAAGUGC-3' Assume the reading frame starts at the 5' end. Refer to the codon table . Amino acid sequence: Glu-Leu-Val-Cys-Lys-Cys What is the sequence of the polypeptide formed on addition of poly(UUAC) to a cell-free protein-synthesizing system that does not require a start codon? Enter an amino acid sequence of four amino acids using the three-letter abbreviations. Polypeptide sequence: Poly( -3' Glu-Leu-Val-Cys-Lys-CysGiven the following DNA sequence: 3'-TACTTNGTNCTNTCN-5' where N stands for any nucleotide, give the complementary mRNA sequence. Indicate direction of strand as 3'--> 5' or 5'--> 3' as in the given sequence above. Give the amino acid sequence of your mRNA sequencelin No. 1. Indicate direction of strand as above. Use all lowercase letters, 3-letter name of amino acid separated by a hyphen (-), no spaces in-between.
- An mRNA transcript is listed below and contains both start and termination codons. Assume that the initial methionine will stay on the polypeptide in this case. What amino acid sequence will be specified during translation? List the amino acids. The start codon is highlighted. 5’ – CAGCCAAGCAUGCUCGCAAAUGGACGUUGAUAUUUUGUC – 3’For the messenger RNA sequence below, find the beginning of the amino acid coding sequence and translate the sequence using the genetic code provided below. 5' - AAUUAUGGGCAAUAUGCCGGGCcGGUUAAGCG - 3' Second Letter A UGU cys u UGC Phe UCU UU U UUC UUA UAU Tyr Ser UAC UAA UAG Leu UCA Stop UGA Stop UUG UCG Stop UGG Trp CUU CU CAU His CGU c cuc Leu ccc ССА CCG Pro CÁC CAA CAG CGC CGA CGG Arg CUA CUG Gin 1st 3rd letter Ser u letter AUU ACU AAU AAC AAA AAG Asn A AUC AUA AUG lle ACC ACA AGU AGC AGA AGG Thr Lys Arg Met ACG GUU G GUC GUA GUG GCU GAU GAC GAA Asp GGU Val GCC Ala GGC Gly GCA GGA Glu GCG GAG GGG G5’ AUG UUA CGU AAU GCU GUC GAA UCU AUU UGC UUU ACA UAA 3' Write the sequence of the DNA template (antisense) strand from which the mRNA was synthesized. Write the sequence of the DNA coding (sense or informational) strand complementary to the template strand. Write the sequence of tRNA anticodon that corresponds to the given mRNA molecule. Write the amino acid sequence of the peptide synthesized from the given mRNA nucleotide sequence.
- Based on sequences A,B,C. Provide an anticodon sequence that would build this protein. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGGUsing the genetic code table provided below, identify the open reading frame in this mRNA sequence, and write out the encoded 9 amino acid long peptide sequence: 5'- CGACAUGCCUAAAAUCAUGCCAUGGAGGGGGUAACCUUUU C A G U UUU Phe UCU Ser UUC Phe UCC Ser UAC UCA Ser UAA UCG Ser UAG UUA Leu Leu G C CUU Leu CUC Leu CCC CUA Leu CUG Leu AUU lle AUC lle AUA lle AUG Met ACG ACU Thr ACC Thr ACA Thr Thr A UAU Tyr UGU Cys Tyr UGC Cys CCU Pro CAU His CGU Arg Pro CAC His Pro CAA Gln CGC Arg CGA Arg CCA CCG Pro CAG Gln CGG Arg GUU Val GCU Ala GAU GUC Val GCC Ala GAC GUA Val GCA Ala GAA GUG Val GCG Ala GAG Stop UGA Stop UGG AAU Asn AAC AAA AAG AGU Asn AGC G Lys Lys Asp Asp Glu Glu Stop A Trp Ser Ser AGA Arg AGG Arg GGU Gly GGC Gly UCAG GGA Gly GGG Gly с U C A G U C A G U C A GA segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: GGCTAGCTGCTTCCTTGGGGA CCGATCGACGAAGGAACCCCT Note out the mRNA sequence generated by the template strant to produce that polypeptide chain Label each stran with its correct polarity (5' and 3' ends on each strand)