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
B. Ovulation would not happen
DNA is the main type of genetic material found in a cell. In addition, it is found in the nucleus of the cell, so (D) is correct. DNA in the nucleus is used in replication (through mitosis and meiosis) via daughter cells to continue cell growth.
