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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.
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- Consider the following DNA sequence: CATGTGTAGTCTAAA. Address the followin questions: AWrite the sequence of the DNA strand that would be replicated from this one. GTACACATCAGATT 2. White the sequence of the messenger RNA (MRNA) molecule that would be transcribed from the DNA strand. 30 GUACACAUCAGAU00 37 38 39 3. State how many codons the sequence specifies. 40 41 42 43 44 4. State how many amino acids the sequence specifies. 45For each of the following items, fill in either the DNA strand, the MRNA codons, the tRNA anticodons, or the amino acid sequence that have been left blank. If several sequences might work, choose only one. Furthermore, circle the start and the stop codons of each mRNA sequence. 1. DNA (3'-5') ACG TAC GGC CGG TTA AAG CAT ACT TTC TTG MRNA TRNA Amino Acid 2. DNA (3'-5') MRNA AUG ACU AGC UGG GGG UAU UAC UUU UAG AAA TRNA Amino Acid 3. DNA (3'-5') MRNA TRNA GCU CCU UAC CAC ССС CGU AUG GCU GGG AUC Activate Go to Sett Amino AcidGiven the following protein, which of the following sequences of TEMPLATE strand DNA would code for it? Pay attention to the polarity of the polypeptide and the strands of DNA that you choose. Use the codon chart to the right. AUG = met AAA = lys GCU = ala | CUU = leu ACU = thr -lys - thr - ala - leu - met (amino end) 5' TAC GAA CGA TGA TTT TAC ATT 3' 5' ATG CTT GCT ACT AAA ATG TAA 3' (carboxyl end) met 5' TAC TTT TGA CGA GAA TAC ATT 3¹' 3' TAC TTT TGA CGA GAA TAC ATT 5¹ 5' ATG AAA ACT GCT CTT ATG TAA 3¹ 3 TAC GAA CGA TGA TTT TAC ATT 5'
- Below is a double stranded DNA that contains an ORF. Identify the ORF and in the spaces below, first provide the mRNA sequence that would result (in the 5' to 3' direction). Then give the protein sequence this mRNA would encode (in the N to C direction using the 3-letter abreviations and a single space in between). 5'-CTGCGC TAGGCGCAGTAGGGCATCCCACGACA-3' 3'-GACGCGATCCGCGTCATCCCG TAGGG TGC TGT-5' Codon Table: บบบ UUC JUA UUG CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val mRNA sequence: UCU UCC UCA UCG protein sequence: CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala UAU UAC UAA UAG CAU CAC CAA CAG Tyr Stop GAU GAC His Gin AAU Asn AAC AAAG} LUS Asp GAA GAG} Glu UGU Cys UGC UGA Stop UGG Trp CGU CGC CGA CGG AGU AGC Arg Ser AGG} Arg GGU GGC GGA GGG (DO NOT include the 5' and 3' numbers or the N and C. Write only the nucleotides and the abreviations for the amino acids. Include no spaces in the mRNA and a single space in between the amino acid…Give only typing answer with explanation and conclusion which of these choices represents one possible corresponding mRNA sequence that can be transcribed by RNA polymerase, and later translated by ribosomes from the following DNA template? 5'- CTGTATCCTAGCACCCAAATCGCAT - 3'; A. 5'- CTA GCA CCC AAA TCG CAT TAG - 3', B. 5' - AUG CGA UUU GGG UGC UAG - 3', C. 5' - AUG CGA UUU GGG UGC - 3', D. 5- ATG CGA TTT GGG TCG TAG - 3'2) Create an MRNA strand based on the given DNA template strand: TACTTCCTATTITCTTGTCA CCGCACT 3) Using the mRNA codon chart, determine the amino acid sequence for the MRNA sequence determined in question 3. 4) Consider the following double-stranded DNA molecule: Complementary Strand: ATGTGTAGTGCGAGTTGA Template Strand: TACACATCACGCTCAACT a) What would be the amino acid sequence coded for by the template strand of the DNA molecule above?
- If DNA segments changes from GCATAG to GCATGG, this is a: MRNA Codon/Amino Acid Chart First Base Second Base Third Base A G UUU1 Phenylalanine UCUT UAUT FTyrosine (Tyr) UAC UGU, U FCysteine (Cys) UGCJ UUCJ (Phe) UCC Serine (Ser) UCA U UGA - Stop A UUA1 FLeucine (Leu) UUG- UAA1 FStop UAGJ UCG- UGG - Tryptophan (Trp) G CUU CAU1 CCU1 CGUT Histidine (His) CAC- CUC CCC Proline (Pro) CCA CGC FLeucine (Leu) CUA FArginine (Arg) CGA CAA1 Glutamine A CAGJ (Glu) CGG- CUG- CCG- ACU AAUJ Asparagine AUUT AGUT FSerine (Ser) AGC- U AUC FIsoleucine (lle) ACC AACJ(Asn) Threonine A AUA- (Thr) AAA1 FLysine (Lys) AAG- АСА AGA, FArginine (Arg) AGG- A Start Methionine AUG- (Met) ACG- G GUU GCU, GAU1 Aspartic Acid GGU U GAÇJ(Asp) GUC Fvaline (Val) GUA GCC GGC G Alanine (Ala) GCA Glycine (Gly) A GAAJ Glutamic Acid GGA GAG- (Glu) GUG- GCG- GGG- G1) Complete the following tables by filling in the DNA sequence, MRNA codons, and the amino acids expected in the final protein (using the MRNA codonstable): DNA TAC CGC TCC GCC GTC GAC AAT ACC ACT MRNA Amino Acid DNA TGA C АTG ATC MRNA AUG ACU AGC UGG GGG UAU UAC U00 UAG Amino Met Ser Trp Tyr Phe Acid DNA TAC CAC CGT ATG GCT GGG AAT ATC MRNA Amino AcidAssume the first nucleotide in the sequence is at the +1 position. Transcribe the DNA sequence into mRNA, and then translate it into the polypeptide. Give the polypeptide sequence in the following form: Met-Thr-Trp-Tyr-Val etc. 3' TCACAATACAAAGGTGTACTGATCTCATCTCCATAA 5'
- Use a codon chart determine the amino acid sequence. Remember to read through the strand and ONLY start after the promoter and STOP when it tells you to stop. Follow example below: Example: DNA AGA TATA TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC MRNA O protein AUG GAG GCC ACC CAC GAA CAG ACA UAG GAA GAG UCA UAG start-glu-ala-thre-hist - asp-glu-threo-stop met DNA CCT ATA TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCC ATA ATC mRNA DGGA UAU) AUG uGul Gcc nccl cAul GCol protein ly Tur MeT cys AlA ser HIJ Ala 2 3 4 DNA AGA ACT ATA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA mRNA protein DNA TAT ATAC CTT GGG GAA TAT ACA CGC TGG CTT CGA TGA ATC CGT ACG GTA CTC GCC ATC mRNA protein D DNA TAA ACT ATA TAC CTA GCT TAG ATC TAA TTA CCC ATC mRNA protein Auu UGA UAU AGU GAUCGA AUC MAG Auu AAU leu Stop. TRY-Met-Asp- ARG-Isle-Stop-Ile. Asn DNA CTA TTT ATA TAC TAG AGC GAA TAG AAA CTT ATC ATC mRNA protein D DNA CAT ATA TAC CTT AGT TAT CCA TTG ACT CGA ATT GTG CGC TTG…Answer 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.A Section of a GeneAAG ATA CAG GCT CGG TAA For the DNA sequence shown above, identify the following: mRNA codons tRNA anticodons amino acids The section of a gene shown above (AAG ATA CAG GCT CGG TAA) is called the Answer (template or non-template) strand. It is also known as the Answer (sense or antisense) strand. The amino acids are determined from the Answer (DNA or mRNA or tRNA) strand