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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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- 5. When can a mutation on the DNA cause shortening of the translation product? Give a specific example to support your answer.1. A portion of the template strand of a DNA molecule that codes for the 5'-end of an mRNA has the following nucleotide sequence. Give the primary structure of the polypeptide (use 3-letter amino acid codes), beginning with the most common translation initiation codon, that will be specified by this portion of the gene. Be sure to label the ends of the polypeptide. 3'-TTTTACGGGAATTAGAGTCGCAGGATG-5'1. When can a mutation on the DNA cause shortening of the translation product? Give a specific example to support your answer. 2. When can a mutation on the DNA cause production of a longer translation product? Give a specific example to support your answer.
- 8. The following diagram illustrates a step in the process of translation. fMet Pro mRNA UAC GGG AUGCCCACG UAG a) Identify the following elements on the diagram. Aminoacyl site (A) Peptidyl site (P) Exit site (E) Amino end of the newly synthesized polypeptide chain Carboxyl end of the newly synthesized polypeptide chain Approximate location of the next peptide bond that will be formed19. Predict the effect of a mutation that would alter the Shine-Dalgarno (SD) sequence T / F Initiation of translation would probably be affected T / F Elongation of translation would be faster since ribosome will no longer stall at the SD sequence T / F mRNA levels maybe lower T / F replication would not be affected T / F RNA polymerase would not bind as efficiently2. Do you think mutation has any direct relationship with codon? Why/Why not? Please explain in your own words. [Max 200 words]
- 7. The large subunit of the ribosome contains the E, P, and A sites. Discuss what occurs at each site in a single round of translation elongation.6. Similar to the class notes (Intro to Genetics), a segment of DNA (shown below) contains a promoter segment (the first 9 base pairs), a ribosome binding segment (the next 6 base pairs), and a segment that codes for protein synthesis which is started by the rest of the base pairs. ACTCCATTGAACCATTTCTATGATCCGCTAACG-... TGAGGTAACTTGGTAAAGATACTAGGCGATTGC-... A. When the DNA is induced to be copied to mRNA, the top strand is coding, meaning that the mRNA makes an identical copy of the lower strand (replacing T with U) The mRNA copy starts with the ribosome binding sequence. What is the sequence of the mRNA that will go to the ribosomes? B. What are the first 6 amino acids of the protein that are coded for by the mRNA? C. What would the amino acid sequence be if... i. a transition mutation occurred on the final G in the mRNA? ii. all of the G & C bases in the protein synthesis portion had transition mutations? iii. a point deletion mutation occurred in the ATA sequence (in the lower strand…6. Below are several DNA sequences that are mutated compared with the wild-type sequence: 3’-T A C T G A C T GA C G A T C-5’. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5’ and 3’ ends) for each mutated DNA sequence, then transcribe (indicating 5’ and 3’ ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the N and C termini). What type of mutation is each? 6.d. Mutated DNA Template Strand #4: 3’-T A C G A C T G A C T A T C-5’Complementary DNA sequence:mRNA sequence transcribed from template:Amino acid sequence of peptide:Type of mutation: 6.a. Mutated DNA Template Strand #1: 3’-T A C T G T C T G A C G A T C-5’Complementary DNA sequence:mRNA sequence transcribed from template:Amino acid sequence of peptide:Type of mutation: 6.b. Mutated DNA…
- 6. Below are several DNA sequences that are mutated compared with the wild-type sequence: 3'-TACT GACTG ACGAT C-5'. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5' and 3' ends) for each mutated DNA sequence, then transcribe (indicating 5' and 3' ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the N and C termini). What type of mutation is each?6. Below are several DNA sequences that are mutated compared with the wild-type sequence: 3'-TACT GACTG ACGAT C-5'. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5' and 3' ends) for each mutated DNA sequence, then transcribe (indicating 5' and 3' ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the N and C termini). What type of mutation is each? 6.a. Mutated DNA Template Strand #1: 3'-TACTGTCT GACGATC-5' Complementary DNA sequence: mRNA sequence transcribed from template: Amino acid sequence of peptide: Type of mutation: 6.b. Mutated DNA Template Strand #2: 3'-TACG GACT GAC GATC-5' Complementary DNA sequence: mRNA sequence transcribed from template: Amino acid sequence of peptide: Type of mutation:2. Transcribe the following gene sequence to the appropriate MRNA codons: TCA TTT CCA CAG ACA * AGU AAA GGU GUc UGU 3. Translate the mRNA codons (i.e. your answer to question 2) into the appropriate amino acids using the following Universal Genetic Code Chart. * Universal Genetic Code Chart Messenger RNA Codons and the Amino Acids for Which They Code SECOND BASE A UCU UAU UGU UUU UUC UUA PHE TYR CYS UCC UCA UAC UGC SER UAA UGA STOP LEU STOP UUG UGG} TRP UCG UAG CUU CCU CAU CGU HIS CUC CCC CAC CGC LEU PRO ARG CCA CCG CUA CAA CGA GLN CUG CAG CG AUU AUC ACU AAU AGU AGC AGA ASN SER ILE ACC AAC THR AAA AAGS ACA AUA AUG MET or START LYS ARG ACG AGG S GCU GCC GUU GAU GGU ASP GUC GAC GGC VAL ALA GLY GUA GCA GCG GAA GGA GGG GLU GUG GAG トエーCロ ASE A, L-RST BASE