Northern blot analysis is performed on cellular mRNA iso...

Question

Northern blot analysis is performed on cellular mRNA isolated from E. coli. The probe used in the northern blot analysis hybridizes to a portion of the lacY sequence. Below is an example of the gel from northern blot analysis for a wild-type lac⁺ bacterial strain. In this gel, lane 1 is from bacteria grown in a medium containing only glucose (minimal medium). Lane 2 is from bacteria in a medium containing only lactose. Following the style of this diagram, draw the gel appearance for northern blots of the bacteria listed below. In each case, lane 1 is for mRNA isolated after growth in a glucose-containing (minimal) medium, and lane 2 is for mRNA isolated after growth in a lactose-only medium.

lac⁻ bacteria with the genotype I⁺ P⁻ Oᶜ Z⁺ Y⁺

Accepted Answer

Hi, everyone. Here's our next problem. Suppose the researcher includes a sample from an E coli strain with a deletion mutation in the lac eye gene in a northern blood analysis along with a sample from a wild type strain. How would the expression pattern of LAC Z M R N A differ in this strain compared to the wild type strain choice. A, the mutant strain would show higher levels of LA C than the wild type choice. B, the mutant strain would show higher levels of lack Y M R N A than the wild type choice C. The mutant strain would show higher levels of black O M R N A than the wild type or choice D. All options are correct. So we're looking at the expression of lack Z M R and A. Well, with that, we can actually eliminate all but one of these answers right away because choices B and C have the wrong M R N A referred to choice B says the mutant strain would show higher levels of lack Y M R N A. But we're not looking at the expression of lack Y M R N A in our blood, we're looking for lack Z. So choice B is not correct twice. C says the mutant strains would show higher levels of lack O M R N A. But again, we're not looking for that. That's not what our probe is for. So choice C would not be correct. And then choice D, all options are correct. Well, we've got two incorrect answers right already. So choice D is not correct. We've only got one left choice A but to learn from this, let's look at choice. A to understand why it is our answer. So we have the wild type stream which we know a wild type strain. And we're looking at the expression of LAX C. That's one of the structural genes in the lac operon and that is an inducible operon. So the lack for pressor protein prevents the expression of that lay gene in the absence of lactose. When lactose is present, the repressor is released and the gene is expressed. So that is a condition in the wild type stream. Our mutant strain has a deletion in the lack igene. Well, the lack gene is the gene for the lac repressor protein. So with the deletion of that gene, the result is there is no repressor protein. And that means that lack Z is continuously expressed regardless of the presence of lactose. So when we look at the expression of black C, we'll see higher levels in the mutant sample where LAX C is continuously expressed versus the wild type where it's only expressed in the presence of lactose. So that's why choice A the mutant strain would show higher levels of LA Z M R N A expression than the wild type is our answer. See you in the next video.

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