Question

Question 6:

(1 mark)

An mRNA has the following codon:

5? GCA 3?.

What tRNA anticodon will bind to this codon?

The tRNA anticodon 5'  3' will bind to the mRNA codon 5' GCA 3'.

**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**

1 points   

QUESTION 7

Question 7:

(1 mark)

A tRNA anticodon has the following sequence:

3' CUA 5'.

What amino acid does it carry?

The amino acid carried on a tRNA with the anticodon 3' CUA 5' is .

**Hint: You will need to consult the genetic code to answer this question. And watch your 5' to 3' direction.**

1 points   

QUESTION 8

Questions 8 to 13 are based on the following information.

Question 8:

The top strand of the following segment of DNA serves as the template strand:

3’ TACACCTTGGCGACGACT 5’

5’ ATGTGGAACCGCTGCTGA 3’

We will refer to this segment of DNA as the original (or unmutated) sequence.

Please answer the following questions:

(a) What is the mRNA sequence?

The mRNA sequence is  5'  3'.

**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**

(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.

The amino acid sequence for this protein is N-terminus  C-terminus.

**Please note**

The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.

i.e. input the amino acids in the order that they would be translated.

If a codon encodes for a stop codon, type STOP.

When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).

You will need to consult the genetic code to answer this question.

4 points   

QUESTION 9

Questions 9 to 13 are in reference to the DNA sequence shown in Question 8.

Question 9:

The original (unmutated) DNA sequence (shown above in Question 8) has been mutated to the following (this represents the template strand):

3’ TACATCTTGGCGACGACT 5’.

We will refer to this sequence as mutation #1.

Please note that for simplicity only the template strand for this mutated segment of DNA is shown.

Answer the following questions:

(a) What is the complete mRNA sequence for the mutated segment mutation #1?

The mutated mRNA sequence is  5'  3'.

**Please enter your sequence in the 5' to 3' direction. Deductions will be made if a sequence is inputted in the wrong direction.**

(b) Using the mRNA sequence you determined in part (a) of this question, give the sequence of the protein that would be translated.

The amino acid sequence for this protein is N-terminus  C-terminus.

**Please note**

The N-terminus refers to the beginning of the primary sequence for a protein, and the C-terminus refers to the end of the primary sequence for a protein.

i.e. input the amino acids in the order that they would be translated.

If a codon encodes for a stop codon, type STOP.

When inputting your sequence, separate each amino acid with a hyphen (e.g. Ser-Tyr-STOP).

You will need to consult the genetic code to answer this question.

Answer 1

Answer for Question 6: Using the principle of base complementarity, Guanine (G) will be paired with Cytosine (C) and Adenine (A) will be paired with Uracil (U). Accordingly, for an mRNA codon sequence 5’-GCA-3’, the corresponding tRNA anticodon sequence will be 5’-UGC-3’. This will code for the amino acid alanine.

Answer to Question 7: Since a mRNA is read in the 5’-3’ direction, the orientation of a tRNA-based anticodon will be complimentary or 3’-5’ direction. Accordingly, for the anticodon 3’-CUA-5’, the corresponding mRNA will be 5’-GAU-3’, and the corresponding amino acid will be Aspartate.

Answer to Question 8:

1. As per the standard principle of transcription, the DNA strand is read in the 3’-5’ direction and accordingly a mRNA sequence in the complimentary direction (5’-3’) is generated. Accordingly, for the following DNA sequence:

3’ TACACCTTGGCGACGACT 5’

5’ ATGTGGAACCGCTGCTGA 3’

The corresponding mRNA sequence will be:

          3’ TACACCTTGGCGACGACT 5’ (DNA sequence)

          5’ AUGUGGAACCGCUGCUGA 3’ (mRNA sequence)

Please note that as Uracil is incorporated in the RNA sequence, instead of thymine.

2. As per the standard principles of translation, mRNA sequence is read in a 5’-3’ direction by a tRNA. A codon is a sequence of 3 mRNA nucleotides that is recognized by an anticodon on the tRNA and the corresponding amino acid is encoded.

          3’ TACACCTTGGCGACGACT 5’ (DNA sequence)

          5’ AUG UGG AAC CGC UGC UGA 3’ (mRNA sequence)

          N-Met–Trp–Asn–Arg–Cys–STOP-C (Protein sequence)

AUG - Met – Methionine

UGG – Trp - Tryptophan

AAC – Asn – Asparagine

CGC – Arg – Arginine

UGC – Cys – Cysteine

UGA – STOP

Answer to Question 9:

1. As per the standard principle of transcription, the DNA strand is read in the 3’-5’ direction and accordingly a mRNA sequence in the complimentary direction (5’-3’) is generated. Accordingly, for the following mutated DNA sequence:

3’ TACATCTTGGCGACGACT 5’       

The corresponding mRNA sequence will be:

          3’ TACATCTTGGCGACGACT 5’ (DNA sequence)

          5’ AUGUAGAACCGCUGCUGA 3’ (mRNA sequence)

Please note that as Uracil is incorporated in the RNA sequence, instead of thymine.

2. As per the standard principles of translation, mRNA sequence is read in a 5’-3’ direction by a tRNA. A codon is a sequence of 3 mRNA nucleotides that is recognized by an anticodon on the tRNA and the corresponding amino acid is encoded.

          3’ TACATCTTGGCGACGACT 5’ (DNA sequence)

          5’ AUG UAG AAC CGC UGC UGA 3’ (mRNA sequence)

          N-Met–STOP -C (Protein sequence)

AUG - Met – Methionine

UAG – STOP

Since a stop codon (UAG) is generated owing to a mutation, translation will be terminated prematurely.