1 Below are 210 consecutive base pairs of DNA that includes

1. Below are 210 consecutive base pairs of DNA that includes only the beginning of the sequence of gene X. The underlined sequence (from position 20-54) represents the promoter for gene X and the underlined and italicized sequence (from position 71-90) encodes the gene X ribosome binding (RBS) site. Transcription begins at and includes the T/A base pair at position 60 (underlined).

a) What are the first 6 nucleotides of the mRNA from gene X?
b) What are the first 4 amino acids encoded by gene X? Use the one letter amino acid code to indicate your answer. Ex. TRCG
c) You have found a mutant of gene X. The mutation represents an SNP which changes the G/C base pair at position 110 (underlined) to T/A. Would the mRNA expressed from this version of gene X be longer, shorter, or the same as that produced from the normal gene X?
d) If the mutant mRNA can be translated, would you expect the protein to be longer, shorter, or the same as that produced from the normal gene X?
e) You have found another mutant of gene X. The mutation represents an SNP which changes the G/C base pair at position 110 (underlined) to C/G. Do you expect that the protein produced will have a similar level of activity as the normal protein X?
f) What is the length of the 5’ UTR?

PLEASE ANSWER ALL AND EXPLAIN I WANT TO UNDERSTAND!

10 20 30 40 50 60 70 5, ATCGGTCTCGGCTACTACATAAACGCGCGCATATATCGATATCTAGCTAGCTATCGGTCTAGGCTACTAC 3, TAGCCAGAGCCGATGATGTATTTGCGCGCGTATATAGCTATAGATCGATCGATAGCCAGATCCGATGATG Promo ter 80 90 100 110 120 130 140 5, CAGGTATCGGTCTGATCTAGCTAGATGCTCTTCTCTCTCGAACCCGCGGGGGCTGTACTATCATGCGTCG 3, GTCCATAGCCAGACTAGATCGATCTACGAGAAGAGAGAGCTTGGGCGCCCCCGACATGATAGTACGCAGC 150 160 170 180 190 200 210 TATCGATATCT 5, TCTCGGCTACTACGTAAACGCGCGCATAAGCTAGCTATCGGTCTCGGCTACTACGTAAA 31 AGAGCCGATGATGCATTTGCGCGCGTATATAGCTATAGATCGATCGATAGCCAGAGCCGATGATGCATTT 71CG 41G 5-GC 31 21TA 11 S5_CG 1-AT

Solution

3\'ATCCGA5\' (DNA at position 60)

1) 5\' UAGGCU 3\'

2) 5\' UAGGCUACUACC3\' so,

N met arg arg leu C, so, code - MAAL

3) As, It is a SNP, so, there is neither deletion nor addition of the base pairs, so, the number of base pairs remains the same therefore, the mRNA is of same length.

4)As, It is a SNP, so, there is neither deletion nor addition of the base pairs, so, the number of amino acid remains the same therefore, the mutated protein is of the same length.