Question 5 (20 points) Make a transcription of the DNA template strand 5- A CATT

Question

Question 5 (20 points) Make a transcription of the DNA template strand 5- A CATTAGTCA GTA GA CAT-3 a. To an mRNA? b. Read and translate the codons on m-RNA into the appropriate amino acids. c. If a mutation of the 9h base from the 3' end is mutated from an A-adenosine to T-thymine, how does this change the amino acid sequence? Be specific by redoing the problem with this one mutation. In a frame shift mutation, one base is either added or deleted. If now we add a G-Guanine to the 9th d. position from the 3' end, how does this mutation change the sequence of amino acids produced? e. Explain how this frame shift mutation is probably responsible for many of the enzyme genetic defect diseases? Show with an example

Explanation / Answer

The given template DNA strand is

5' ACATTAGTCAGTAGACAT 3'

Ans A) For mRNA formation the coding strands are 3' TGTAATCAGTCATCTGTA 5'.

The mRNA is formed by the cexact coping of the DNA coding strands where the Thymine js replaced by Uracil. The mRNA formed is 5' UGU AAU CAG UCA UCU GUA 3' after transcription.

Ans B) The mRNA is translated to the protein. The protein is 5' Cys-Asn-Gln-Ser-Ser-Val 3'

Ans C) A mutation occur at the 9 postion of the DNA from the 3' side. Now the template strand is

5' ACATTAGTCTGTAGACAT 3'

The coding strands is 3' TGTAATCAGACATCTGTA 5'.

The translated mRNA strand is 5' UGU AAU CAG ACA UCU GUA 3 and the proteined formed is

5' Cys-Asn-Gln-Thr-Ser-Val 3'

Due to the mutation Serine is replaced by threonine.

Ans 4) Now a Guanine is inserted at the 9 position of the 3' site. Now the template DNA is

5' ACATTAGTCGGTAGACAT 3'.

The coding strand is 3' TGTAATCAGGCATCTGTA 5'.

The mRNA is 5' UGU AAU CAG GCA UCU GUA 3

The translated protein is 5' Cys-Asn-Gln-Ala-Ser-Val 3'.

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