Answer:
The question is incomplete, it lacks the mRNA sequence. The sequence is as follows:
5′−AUGGCAAGAAAA−3′
The answer is Met-Ala-Arg-Lys
Explanation:
Gene expression in living organisms involves the process of transcription and translation. Transcription is the synthesis of a complementary strand of mRNA from a DNA template while translation involves using the transcibed mRNA as a template to synthesize amino acid sequence (proteins).
In the RIBOSOME, where the synthesis of protein occurs, the mRNA nuceleotide sequence is read in a group of three nucleotides called CODON. Each codon specifies a particular amino acid. The collection of all codons is the genetic code. Hence, for a specific mRNA sequence that reads 5′−AUGGCAAGAAAA−3′. The nucleotides will be read three at a time starting with AUG which is a codon that encodes METHIONINE.
Next, GCA is a codon that encodes ALANINE
Next, AGA is a codon that encodes ARGININE
Finally, AAA is a codon that encodes LYSINE.
Hence, the amino acid sequence using the above mRNA sequence, will read: Met-Ala-Arg-Lys
The base pairs are put into different orders and these orders code for different amino acids.
It would be 10
Atomic number = Number of protons
so, 10 = 10
Option D is your answer.
Hope this helps!
Answer:
Main protein in ending high fidelity in E. Coli is the Tus protein that binds to Ter sequences in order to prevent replication forks from passing through the end region. In the Ter sequences, the Tus protein blocks replication by establishing a close association with a particular G-C base pair.
The main protein in human cells is telomerase, which contains an RNA primer and is required to extend the synthesis of lagging strands in linear chromosomal telomeres.
