0400sw 40043 3. You are now to the lab and your job is to test a new antibody ma

Question

0400sw 40043 3. You are now to the lab and your job is to test a new antibody made against the lab' s guerkin. You run a series of Western blots using collular extracts from a varit tissues to see if the antibody recognizes the 40 KD guerkin protein. In almost al observe a strong band at 40 kD. However, in extracts from skelotal muscle, you always band. You're pretty sure this band is specific to the guerkin protein because you skeletal muscle samples from mice lacking the guerkin gene. Your advisor and wants you to look at the genomic structure of the guerkin gene to see for this observation the lab' favorite protein of different of the extracts, you mouse see a 70 kD if you can find an e anation Genomic organization of guerkin open boxes indicate exons and the filled boxes are introns. Nucleotide number is shown. 34 200 302 A) Provide an explanation for the generation of 40 kD and 70 kD proteins that react with the anti-guerkin antibodies. Be certain to illustrate the molecular differences between the two proteins. [Hint: the average molecular weight of an amino acid is 110 Daltons] (6 points) B) The antibodies recognize both the 40 and 70 kD proteins. In what region of the two proteins would you predict the antibodies to bind in order to recognize both? [Please provide a protein domain or regior in your answer, not a DNA domain] (2 points) Total Points

Explanation / Answer

A single gene can produce multiple transcript/proteins via alternate splicing.
It involves,
1. Use of alternative start codon
2. Retention of introns
3. Skipping of exons.

Guerkin = 40 kDa
Average MW of an amino acid = 110 Da
Number of amino acids = 40000/110
= 363

Number of nucleotides = 363 X 3
= 1089

Alternative form of Guerkin = 70 kDa
Average MW of an amino acid = 110 Da
Number of amino acids = 70000/110
= 636

Number of nucleotides = 636 X 3
= 1908

1st exon size = 200-134 = 66
2nd exon size = 1052-302 = 750
3rd exon size = 1985-1736 = 249
4th exon size = 2536-2236 = 300
5th exon size = 3546-3000 = 546
Total CDS length = 66+750+249+300+546 = 1911
1 stop codon = 3 ntds
So, effective length = 1911 - 3 = 1908 ntd
= 1908/3 = 636 amino acids
= 636 X 110 = 70 kDa

So, a 70 kDa protein is produced if all the exons are translated.

Difference in protein size = 70 - 40 = 30 kDa
= 30000/110 = 272 aa
= 272 X 3 = 816 ntd

Size of first two exons = 66 + 750 = 816 ntd
So, a 40 kDa protein is produced when an alternate translation start codon is utilized.

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