NEET Class neet Answered

Given:

DNA sequence-

3′ TACATGGGTCCG 5′

The sequence of mRNA is a direct copy of the sequence of deoxyribonucleotides in the sense strand of DNA with U in place of T.

So, the mRNA transcribed would have the sequence-

3′ UACAUGGGUCCG 5′

During translation, the mRNA moves through the ribosome in one direction only, i.e., from the 5′ end first. This means that the ribosome moves 5′→3′ on the mRNA.

Therefore, in our anticodon strand, 3′ UAC¦AUG¦GGU¦CCG 5′, the ribosome would move from right to left.

In the given sequence, AUG is the start codon. Therefore, translatin would begin at AUG in the given sequence and the first coded amino acid would be ‘b’ in our case.

As the ribosome shift from right to left, the next codon would be UAC which would code for amino acid ‘a’.

Then the next codon would be CCG which would code for ‘d’ followed by GGU which would code for amino acid ‘c’.

Thus, the correct order of binding of amino-acyl tRNA complexes would be b, a, d, c.