1. The Liliputian gene is known to be repressed 5 fold when cells are grown in g

Question

1. The Liliputian gene is known to be repressed 5 fold when cells are grown in glucose as compared to growth in glycerol. To study the expression of this gene, a northern analysis was performed for a wild type strain as well as various mutants. Blots were simultaneously probed for the gene shown above, as well as a house keeping gene; GAPDH. A densitometric analysis was then performed. The densitometric values obtained for a wild type strain of the organism grown in glucose were 500 for both the Liliputian mRNA and the GAPDH mRNA. Indicate the expected densitometric value for the Liliputian mRNA in each of the following scenarios:

a) Wild type grown in glycerol (GAPDH value = 100)
b) Mutant strain with a mutation which enhances the promoter's activity 2 fold grown in glycerol (GAPDH value = 500)
c) Mutant strain with a mutation which changes the ATG codon to CTG grown in glycerol (GAPDH value = 250)
d) Mutant strain with a mutation which represses the promoter's activity 5 fold grown in glucose (GAPDH value = 500)

PLEASE EXPLAIN I WANT TO UNDERSTAND

Explanation / Answer

Answer:

a. The densitometric values obtained for a wild type strain of the organism grown in glucose were 500 for both the Liliputian mRNA and the GAPDH mRNA.

So, Liliputian mRNA : GAPDH mRNA = 1 : 1

Liliputian mRNA : GAPDH mRNA (in glycerol) = 5 :1 (Since, Liliputian expression is repressed 5 fold in glucose)

GAPDH mRNA = 100. Thus,  Liliputian mRNA = 100 * 5 = 500

b. Promoter activity gets doubled. Here, GAPDH = 500 (Considering that the mutation does not change the expression of GADPH)

Liliputian mRNA in glucose = 500 * 2 = 1000

Liliputian mRNA in glycerol = 5 * 1000 = 5000

c. CTG codes for leucine (L). If the ATG codon mentioned here is the start codon, then a truncated product (if an ATG later in the sequence can initiate transcription) would be transcribed. So, we should not expect the transcribed product of the mutant gene.

d. GAPDH = 500

So, for  Liliputian mRNA in glucose (repressed 5 times) = 500/5 = 100

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