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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.
What is the DNA template of the following DNA coding:
ATGGCTAACCTTGTA
Step by step
Solved in 3 steps
- Given below is the DNA template. What are the gene products? 3’ TACCGGCCTATCTAGGGCCATGGCTTAATTCCC 5’ 5’ ATGGCCGGATAGATCCCGGTACCGAATTAAGGG3’The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all 5 groups and translate. Group A 5’-GGCAATGGGTTTGTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTTTCAAAAATTAAG-5’ Group B 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’ Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGFor the following short sequence of double stranded DNA and the given primers, there will be one major duplex DNA product after many cycles (imagine 10 cycles) of PCR. Provide the sequence of this one major duplex product and label the 5’ and 3’ ends of each strand. Sequence to be amplified: 5’- GGTATTGGCTACTTACTGGCATCG- 3’ 3’- CCATAACCGATGAATGACCGTAGC- 5’ Primers: 5’-TGGC-3’ and 5’-TGCC-3’
- Provide the complementary strand and the RNA transcription product for the following DNA template segment:5'-AGGGGCCGTTATCGTT-3'A DNA strand was sequenced using the Sanger method (https://www.youtube.com/watch?v=KTstRrDTmWI). The reaction tube contained the DNA strand, fluorescently labelled dideoxynucleotide triphosphates (ddATP – yellow, ddGTP – green, ddCTP – blue, ddTTP - red), deoxynucleotide triphosphates, DNA polymerase, or its Klenow fragment. Synthesis of DNA is allowed to proceed, and the results are shown on the right: 15 14 13 12 11 10 (a) What is the sequence of the copy and the template strands? (b) If the template strand were in the 5'-3' direction, what will be the sequence of the DNA copy? Nucleotide LengthGiven the DNA template strand 3' GCATTCAAG 5', write the amino acid sequence in the N‑terminal to C‑terminal direction. Note: Enter the amino acids using their three-letter designations. Put a hyphen between each amino acid.
- The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGWhat restriction enzyme (or enzymes) would you use to cut the following... (Gene of Interest is Bolded) 1 tctagagtca tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg tatgccaatg…The following is the base sequence of DNA that codes for first eight amino acids of the β chain of hemoglobin. The β chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. DNA Template Strand: TACCACGTGGACTGAGGACTCCTC 1. What is the minimum number of DNA nucleotides in this whole gene? 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 3. What mRNA will be formed from the template strand of DNA?
- given the following DNA template, write out the cDNA, mRNA, tRNA anticodons, and give the amino acid sequence. what are the three possible outcomes if there ws a base substitution mutation were to occur to the template? template TAC CGC TCC GCC GTC GAC AAT ACC ACT#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:Given the template DNA strand 3’-TACCCTCAAGGGCAAACT-5’, provide the complimentary DNA strand, mRNA, tRNA, and protein using the figure that will post here