Question

In: Biology

1) Given your understanding of transcription and translation, fill in the items below with the proper...

1) Given your understanding of transcription and translation, fill in the items below with the proper sequences. Please align your answers under the nontemplate DNA strand. Ignore RNA stop codons if they are present.

Nontemplate strand of DNA:                 5′- A T G T A T G C C A A T G C A -3′

     Template strand of DNA:       

                                 RNA:       

             Anticodons on tRNA:      

            Amino acid sequence:

2) Original template strand of DNA:  3′- T A C G C A A G C A A T A C C G A C G A A -5′

a. Transcribe this sequence:

                                      RNA:

b. Translate the RNA sequence. Ignore start/stop codons if present.

                 Amino acid sequence:

3) The table below lists five single-base point mutations that may occur in DNA. What happens to the amino acid sequence as a result of each mutation? (Position 1 refers to the first base at the 3′ end of the DNA strand. Position 21 would refer to the last base at the 5’ end.). Note that amino acids are numbered from L à R as 1-7.

Original template strand:  3’ TACGCAAGCAATACCGACGAA 5’

                   RNA strand:

     Amino acid sequence:                                                            (number aa’s 1-7 L-R)

Mutation

Effect on amino acid sequence. Write ~3 amino acids around the mutation site to show a tripeptide sequence with the change. Indicate the aa numbers of the new tripeptide.

i. Substitution of T for G at position 8.

ii. Addition of T between positions 8 and 9.

iii. Deletion of C at position 15.

iv. Substitution of T for C at position 18.

v. Deletion of C at position 18.

vi.   Which of the mutations above produces the greatest change in the amino acid sequence of the polypeptide coded for by this 21-base-pair gene? That is, which will have the largest effect on structure? Why is this?

Solutions

Expert Solution

1) Non-Template strand of DNA : 5' - ATGTATGCCAATGCA - 3'

Template strand of DNA: 3' - TACATACGGTTACGT - 5'

RNA: 5' - AUGUAUGCCAAUGCA - 3'

Anti codons on tRNA:

3' - UAC - 5'

3' - AUA - 5'

3' - CGG - 5'

3' - UUA - 5'

3' - CGU - 5'

Amino acid sequence: MYANA (Met-Tyr-Ala-Asn-Ala).

2) Original template strand 3′- TACGCAAGCAATACCGACGAA -5′

a) Transcribed sequence

RNA: 3' - AUGCGUUCGUUAUGGCUGCUU - 5'

b) Tramslation of RNA.

Amino acid sequence: MRSLWLL (Met-Arg-Ser-Leu-Trp-Leu-Leu).

3) Original template strand 3’ TACGCAAGCAATACCGACGAA 5’

RNA strans 5' - AUGCGUUCGUUAUGGCUGCUU - 3'

Amino acid sequence: 1=> M1 R2 S3 L4 W5 L6 L7 <=7(Met-Arg-Ser-Leu-Trp-Leu-Leu).

(i) Substitution of T for G at position 8 in the Original template strand leads to sequence

3’ TACGCAATCAATACCGACGAA 5’

causing non sense mutation at position 3. S in position 3 is replaced by Stop codon, The amin acid sequence being only:

M1 R2 Stop.

(ii) Addition of T between positions 8 and 9 of original template strand would lead to sequence:

3’ TACGCAAGTCAATACCGACGAA 5’

That will lead to a frameshift mutation (L4V) leading to amino acid sequence of:

M1 R2 S3 V4 M5 A6 A7

(iii) Deletion of C at position 15 would lead to the original template sequence:

3’ TACGCAAGCAATACGACGAA 5’

leading to a frameshift mutation (W5C) and reduction in the amino acid sequence length by 1. The amino acid sequence obtained is:

M1 R2 S3 L4 C5 C6

(iv) Substitution of T for C at position 18 will lead to the original template sequence:

3’ TACGCAAGCAATACCGATGAA 5’

leading point silent mutation (L6L) where the amino acid sequence obtained is same:

M1 R2 S3 L4 W5 L6 L7

(v) Deletion of C at positionn 18 would lead to the original template sequence:

3’ TACGCAAGCAATACCGAGAA 5’

leading to the frameshift mutation (L6L) where, L7 is removed and no change in rest of the sequence.

M1 R2 S3 L4 W5 L6

(vi)The mutation that produces the greates chenge in the amino acid sequence of poplypeptide coded by 31-pair gene is the deletion of C at position 15 (iii), which is causing frameshift mutation and appearance of Cysteine in place of Tryptophan at position 5 of amino acid.

This changes the structure in great way since Tryptophan (W) is an aromatic (non-polar) amino acid, and Cysteine has polar R group in addition to which it consists of S-H linkage. These tend to form disulfide bridges that hold the secondary structure and improve rigidity of protein.


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