Question

1. Please note that Questions 15 to 17 are connected questions.

   Question 15:

   The following shows a partial DNA sequence from the wild-type (normal) allele for the human leukemia-linked apoptotic gene.  

   5' ATGCGATTAATCGGTAAA 3' (non-template strand)

   3' TACGCTAATTAGCCATTT 5' (template strand)

   Please answer the following questions:

   (a) If the bottom strand serves as the DNA template for transcription, what is the resulting mRNA sequence?

   The mRNA sequence is  5'  3'. (2 marks)

   5' AUG CGA UUA AUC GGU AAA 3' ?

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

   Carefully check that the sequence you input contains no typographical mistakes.

     

   (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. (2 marks)

   Met-Arg-Leu-Ile-Gly-Lys ?

   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. (Do not translate any codons after a STOP codon, as a STOP codon signals to the translation machinery to disassemble.)

   • 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 (one is shown above in Question 10, or you can consult Figure 11.5 in your textbook).

4 points   

QUESTION 16

1. Please note that Questions 15 to 17 are connected questions.

   Question 16:

   The following shows the same partial DNA sequence that was shown above in Question 15, however, this DNA sequence is for themutant allele for the human leukemia-linked apoptotic gene. This mutated allele is suspected to be linked to some forms of cancer, most notably leukemia.

   5' ATGCGATTGATCCGGTAAA 3' (non-template strand)

   3' TACGCTAACTAGGCCATTT 5' (template strand)

   Answer the following questions:

   (a) If the bottom strand serves as the DNA template for transcription, what is the resulting mRNA sequence for this mutant allele?

   The mutated mRNA sequence is  5'  3'. (2 marks)

   5' AUG CGA UUG AUC CGG UAA A 3' ?

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

   Carefully check that the sequence you input contains no typographical mistakes.

   (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. (2 marks)

   Met-Arg - Leu- Ile - Arg ?

   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. (Do not translate any codons after a STOP codon (if one exists), as a STOP codon signals to the translation machinery to disassemble.)

   • 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 (one is shown above in Question 10, or you can consult Figure 11.5 in your textbook).

4 points   

QUESTION 17

1. Please note that Questions 15 to 17 are connected questions.

   Question 17:

   Compare both the nucleic acid and protein sequences you determined above in Questions 15 and 16 for the human leukemia-linked apoptotic gene.

   You should have noticed that there are two separate mutations present in the mutant allele compared to the wild-type (normal) allele (if you cannot identify both of these mutations, please go back and check your work).

   Refer to the first mutation you encounter in the nucleic acid sequence as mutation #1 and the second mutation you encounter in the nucleic acid sequence as mutation #2 (remember we read DNA/RNA in the 5' to 3' direction).

   Answer the following questions:

   (a) Clearly identify where each mutation is located in the mutant allele compared to the wild-type (normal) allele.  (2.5 marks)

   WT AAT mutated to AAC and CCA in WT to GCC. ?

   (b) Classify each mutation according to the classification systems presented to you above in Questions 13 and 14.  (2.5 marks)

   1st mutation is silent mutation - Leu to Leu. 2nd mutation Frameshift mutation "G" is added in the mutated sequence.

   (c) Describe the effect, if any, each of these mutations would likely have on the function of the protein within the cell. (2.5 marks)

   1st mutation have no effect in the function of protein but the 2nd mutation shows greater variation in the function of protein due to frame shift mutation?

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Answer 1

Answer:

$\rightarrow$WILDTYPE SEQUENCE:-

• DNA - 3' TAC GCT AAT TAG CCA TTT 5'

$\rightarrow$Mutated

• DNA - 3' TAC GCT AAC TAG GCC ATT T 5'
• mRNA - 5' AUG CGA UUG AUC CGG UAA A 3'
• Protein N' Met-Arg - Leu- Ile - Arg C'

A):

• WT AAT mutated to AAC and CCA in WT to GCC.

B):

• 1st mutation is silent mutation - Leu to Leu
• 2nd mutation Frameshift mutation "G" is added in the mutated sequence.

C):

• 1st mutation have no effect in the function of protein but the 2nd mutation shows greater variation in the function of protein due to frame shift mutaion.

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