D. Fill in the table below. Determine the correct template and coding strands to properly answer this. _C_ __A T_C AT_ _GT DNA double helix T__ _T_ _CA mRNA __A AC_ GC_ __G_C___ tRNA anticodon GCA Polypeptide (Stop) Trp
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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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- Analysis of a mRNA sample showed that 18% of the nitrogen-base molecules present were uracil molecules. The DNA molecule that was transcribed to form the mRNA sample would most likely contain Select one: a. 18% adenine b. 18% cytosine c. 32% thymine d. 32% adenine O O OUse 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…C. Convert the DNA template to mRNA. Then, convert the mRNA to tRNA. Based from the resulting sequence in the anticodons of tRNA, determine the appropriate Amino acid sequence that will be synthesized. Refer to the genetic code. 1. DNA Template: TAC - GGC - TAC - CAT - ATG - GAG mrNA: tRNA: Amino acid sequence: 2. DNA Template: TTA - CAT - CAT - ATC - GAT - GAC mrNA: tRNA: Amino acid sequence: 3. DNA Template: CTA - GCG - ATA - AAA - TTT - ATT mrNA: tRNA: Amino acid sequence:
- For each mutant, state what change has occurred in the DNA, whether it was a substitution by transition or transversion, sense mutation, nonsense or reading frame change. It must present the codon sequence. Normal nucleotide sequence starting from the third codon: CCC-ACG-GUG-ACG-ACA-CGG-UGG Please show the codon and nucleotide sequence of the mutation.Using a codon table, complete the DNA triplets, mRNA codons, tRNA anticodons, and amino acids in the table below. DNA triplet mRNA codon tRNA anticodon Amino Acid AAG GGC CAG UUA AAA GTA CUC ACA TAT AGC AUU CCA GGChe sequence is read from left to right. The table below shows which mRNA codons code for each type of amino acid. UUA - Leu | UCA - Ser UAA - Stop | UGA-Stop UUU - Phe | UCU - Ser UAU- Tyr UGU- Cys CỦA - Leu CCA - Pro CAA - Gln | CGA - ArgA UUG - Leu UCG - Ser| UAG-Stop UGG- TrpG A DNA sequence before and after replication IS SHO Second mRNA base G DNA sequence before replication: UUC - Phe UCC -Ser UAC U TACCTAGCT Туг UGC Cys DNA sequence after replication: A TACCTCGCT Leu CCU - Pro CAU - His CGU. CUU Arg U - Pro CAC - His CGC- ArgC CUC - Leu ССС Pro CAG - Gln | CGG - Arg G CUG - Leu CCG Thr AAU - Asn AGU - Ser Ile ACC - Thr AAC- Asn AGC Ile ACU AUU AUC Ser Lys AGA Arg Thr AAG - Lys AGG - AUA Ile ACA - Thr AAA- A mutation occurred in the DNA sequence during replication. Which of the following, A-D, Arg Asp GGU-Gly AUG - Met ACG GUU - Val GCU - Ala GAU - GỤC - Val GCC - Ala GAC - Asp GGC - Gly Val GCA -Ala GAA - Glu GGA - Gly Val GCG - Ala GAG-Glu GGG-Glv describes the result of the…
- CTA GCC CTC CGT TAC TAG TTA CCT ACT TAT TCA ATT TTG TAA ACG CTC ATC CGA ACC CGC TTT TAA TTG CCC ACT TAG TCG ATT ACC CGT TTA TGT TAA TTA CCT ATC 1. Build the mRNA molecule matching the RNA nucleotides to the DNA nucleotides properly letter by letter. (assume that the mRNA is bacterial there are not intros to cut out) 2. Figure out the tRNA triplet (codons) that would fit the mRNA triplets. (letter by letter) 3. Look up for each tRNA codon and find the corresponding symbol and amino acid abbreviations. The symbols should spell out a meaningful English message.CTA GCC CTC CGT TAC TAG TTA CCT ACT TAT TCA ATT TTG TAA ACG CTC ATC CGA ACC CGC TTT TAA TTG CCC ACT TAG TCG ATT ACC CGT TTA TGT TAA TTA CCT ATC 1. Build the mRNA molecule matching the RNA nucleotides to the DNA nucleotides properly letter by letter. (assume that the mRNA is bacterial there are not intros to cut out)CTA GCC CTC CGT TAC TAG TTA CCT ACT TAT TCA ATT TTG TAA ACG CTC ATC CGA ACC CGC TTT TAA TTG CCC ACT TAG TCG ATT ACC CGT TTA TGT TAA TTA CCT ATC 2. Figure out the tRNA triplet (codons) that would fit the mRNA triplets. (letter by letter)
- B. One strand of a section of DNA isolated from E. coli reads: (Assume no start codon is required as is true under certain test tube conditions). 5' GTAGCCTACCCATAGG 3' What is the complementary DNA strand? 2. Suppose mRNA is transcribed from this DNA using the complementary strand as a template. What will be the sequence of the mRNA? 3. What would be the corresponding anticodons? 4. What peptide would be made if translation started exactly at the 5' end of this mRNA?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. αBelow 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.