Question

Below is the initial part of the gene sequence for  Glut1 as reported in Flybase. It includes the 3' untranslated sequence and the first part of the translated sequence. The promoter for this gene is TATAT. Transcription begins with the third base after the promoter. For each answer portion, in your response, write each one starting on a new line and please write A, B, C, etc. as you go.

A)   Transcribe and B). translate this sequence into its final peptide product (but just include first four amino acids).   Make sure you show (write sequence of) the the mRNA product of transcription (A) as well as the final protein product (B) . A codon table is provided . Do not try to show any processing of the mRNA. Do labels the ends of the mRNA product as well as the ends of the peptide for parts A and B.

5'- GGCTATATACGATATGGAGCGCCAAGTTCGCTACGATAGGCCTCGG

3'- CCGATATATGCTATACCTCGCGGTTCAAGCGATGCTATCCGGAGCC

C. The sequence below is identical to that of the the gene you just translated above, except that here you must assume that a chemical mutagen caused a C to convert into a G (the bolded T in the lower strand), and the base on the upper strand consequently changed to C. What is the consequence of this substitution mutation in terms of the final protein product? What do you think the functional consequences will be?

D. What is this type of a mutation called (most specific name you can use)


5'- GGCTATATACGATATGCAGCGCCAAGTTCGCTACGATAGGCCTCGG

3'- CCGATATATGCTATACGTCGCGGTTCAAGCGATGCTATCCGGAGCC

E.   There are six different mutant alleles of  Glut1 listed in Flybase. One of them exhibits no base changes in the region of the gene that is eventually translated. In this mutant glucose transport into cells is decreased 85% in all cells. List two possible sites on the DNA template strand where a mutation could cause the mutant phenotype.

F. Given the function of this protein, are the cells containing it likely to need the help of the Na+ K+ ATPase pump

Answer 1

A. Transcript:-

5' GAUAUGGAGCGCCAAGUUCGCUACGAUAGGCCUCGG 3'

B. After translation it will give polypeptide of 12 amino acid. The first four amino acids will be

Aspartic acid, Methionine, Glutamic acid,Argnine

C.The mutation causes change in nucleotide cytosine to guanine i.e , transversion and it also causes change in the amino acid after translation i.e, from Glutamic acid to glycine

D. It is a type of missense mutation.

E. The mutation that result into such phenotype:-

1. Entry of a nucleotide base after promotor region leading to frame shift mutation in the translation region. This do not involve any type of change in nucleotide of translation region

2. Change of nucleotide base of termination region changing it from stop codon to functional codon so that translation leads to formation of protein with few more amino acids.