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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 4 steps
- A certain mRNA strand has the following nucleotide sequence: 5AUGACGUAUAACUUU3 What is the anticodon for each codon? What is the amino acid sequence of the polypeptide? (Use Figure 13-5 to help answer this question.) Figure 13-5 The genetic code The genetic code specifies all possible combinations of the three bases that compose codons in mRNA. Of the 64 possible codons, 61 specify amino acids (see Figure 3-17 for an explanation of abbreviations). The codon AUG specifies the amino acid methionine and also signals the ribosome to initiate translation (start). Three codonsUAA, UGA, and UAGdo not specify amino acids; they terminate polypeptide synthesis (stop).Given the following mRNA, write the double-stranded DNA segment that served as the template. Indicate both the 5 and the 3 ends of both DNA strands. Also write out the tRNA anticodons and the amino acid sequence of the protein encoded by the mRNA message. DNA: mRNA: 5-CCGCAUGUUCAGUGGGCGUAAACACUGA-3 protein: tRNA:Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also, write out the amino acid sequence of the protein encoded by this message. tRNA: UAC UCU CGA GGC mRNA: protein: How many hydrogen bonds would be present in the DNA segment?
- Consider the mRNA sequence below. Assume that the following mRNA segment has been translated. 5'-GCAAGUCUUAAU-3' Note for numbers 1 and 2: Use the three-letter abbreviation of amino acid; separate amino acids with a hyphen: do not include the stop codon. Example: ala-cys-glu 1. Using the table of the genetic code, determine the sequence of amino acids. ala-ser-leu-asn 2. If mutation occurs by substitution of the 6th nucleotide with adenosine-5'- monophosphate, what is the resulting amino acid sequence? 3. What type of mutation occurred? Choose from same sense, missense and non-sense.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 codonConsider the mRNA sequence below. Assume that the following mRNA segment has been translated. 5’-UACCGAAUGUCU-3’ Note for letters a and b: Use the three-letter abbreviation of amino acid; separate amino acids with a hyphen: do not include the stop codon. Example: ala-cys-glu a. Using the table of the genetic code, determine the sequence of amino acids. b. If mutation occurs by substitution of the 12th nucleotide with cytidine-5’-monophosphate, what is the resulting amino acid sequence? c. What type of mutation occurred? Choose from same sense, missense and non-sense.
- Translation is the process by which the sets of 3 bases (codons) of the mRNA are read to specify the sequence of amino acids for the protein to be produced. Using the genetic code data provided, find the sequence of amino acids that would correspond to the MRNA codons shown. Codons 1 3 MRNA A UGUGGAUC CGAG UCACG Amino acid SECOND LETTER A U UUU Phenylalanine UCU UCC Serine (S) UAU Tyrosine (Y) UAC TAA stop codon UAG stop codon UGU Cysteine (C) UGC TỮA Leucine (L) TGA stop codon UGG Tryptophan (W) F UCA UUG UCG I H CUU CCU CAU Histidine (H) R CGU CỨC Leucine (L) CỦA CCC Proline (P) ССА CCG CGC Arginine (R) CGA CAC "CAA Glutamine CAG (Q CUG CGG G D A AUU L AUC Isoleucine (1) AAU Asparagine AAC (N) ÄÄÄ Lysine (K) AGU Senine (S) ACU ACC Threonine ACA (T) AGC E AUA AGA Arginine (R) E ACG T AUG stat codon (M) AAG AGG TG GƯỮ GAU Apartic acid GAC (D) "GAÄ Glutamic acid GCU GGU GUC Valine (V) GUA GGC Glycine (0) GCC Alanine (A) CE GCA GGA R GUG GCG GGG GAG (E) The start codonencodes the amino…On the mRNA codon table below, the first nucleotide in mRNA is to the left, the second is above, and third is to the right. On the sequence, the 5’cap is indicated by (5’). The poly (A) tail is not shown. Use the codon table to translate this short mRNA. Mark the codons and write the amino acid sequence beneath them. (5’)CGUUACAAUGUAUCGCGCGGUACUCGGCAAAGUGCCCUGAAUAGAGUUGGUA(3’) In a previous round of replication, DNA polymerase made a mistake and added a C on what is now the DNA template strand. In the space on the mRNA sequence below, write the added base. Mark the codons again and write the amino acid sequence beneath them. What do you observe? (5’)CGUUACAAUGUAU CGCGCGGUACUCGGCAAAGUGCCCUGAAUAGAGUUGGUA 3’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’
- Indicate 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’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…Using the provided coding strand below: 5-ATCAGATGGCCGGGCCAATAGAATAGCTGT-3 Provide the mRNAstrand for the coding strand above. Type your answer in the box below. Use two dash lines in between the 5 and your first base, and two dash line between your last base and before 3' (follow formatting above). What is your Start codon? What is your Stop codon? 5-AUGGCCGGGCCAAUAGAAUAG-3 AUGGCCGGGCCAAUAGAAUAG