Answer:
48 amino acids
Explanation:
The wild type gene codes for a protein with 100 amino acids. One amino acid is encoded by one triplet code of the gene. This means that the wild type gene has a total 100 triplets or 300 nucleotides to code for a protein of 100 amino acid. Mutation in this protein has introduced the code "UAA" at the 49th codon. The code "UAA" is a stop codon. Therefore, the mRNA transcribed from the mutant allele would code for a protein having 48 amino acids as the protein synthesis will be stopped once the stop codon at the 49th position is read.
Answer:
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Answer:
Lysine
Explanation:
lysine residues on the histone tails of the octamer cn be activated by both acetylation and methylation patterns to influence accessibility or silencing of the genes respectiviely. for example, acetylation of H3K27 (histone 3 lysine residue 27) brings about a region of active chromatin allowing access to transcription activity while its trimethylation will cause silencing of the associated gene at that particular area (no expression of that gene)
Answer:
D
Explanation:
Material is duplicated during Synthesis
