A plasmid is transfected into a cell line with the aim of blocking the expression of a wild-type chromosomal gene called X1. In order for this to be accomplished, the plasmid would have to contain:
A. a mutant X1 gene sequence with a strong promoter.
B. a mutant X1 gene sequence with a weak promoter.
C. a wild-type X1 gene sequence that is inverted with respect to its promoter.
D. a wild-type X1 gene sequence that is separated from its promoter.
Answer: C
No idea what this question is asking. anyone wanna give it a try?😕
Let us look at each answer choice.
A. If a cell line was transfected (btws...why does transfected get auto corrected on iPhone lol?) with a mutant x1 gene with a strong promoter, then each cell will take up the plasmid and express the mutant x1 gene. The goal is to block expression of wild type expression of x1, so cells containing the plasmid will express both wild type and mutant forms of x1. This is only inhibitory to expression if x1 gene product is some form of activator or deactivator and the mutation constitutes constitutive activation of the deactivating nature or constitutive repression of the activating nature. If this information is not given in the passage, then it's safe to assume that this is not the case. So let's look at B.
B. answer choice b states that the cells will contain a plasmid with a weak promoter of mutant gene x1. If this is the case, then little gene product will be made from the mutant gene, so this answer really can have no inhibitory affect on gene x1 expression, unless mutant gene x1 is in some way a super repressor of a functional product formation, similar to super repressor lacI in the lac operon. But if this info is not given in the passage, it's safe to assume this is not the case and this choice must not be correct.
C. If the cells contain a wild type x1 gene sequence that is inverted with respect to its promoter, then the actual coding sequence 5 prime to three prime is still the same, however the promoter is on the opposite side of the gene. There are multiple ways to think of this, but this way seems easiest to me. So, if the promoter is on the opposite side of the gene and switched in direction (hence inverse, NOT REVERSE) then the promoter can still initiate transcription of the wild type gene. HOWEVER, because the wild type gene is still in the same direction but the direction of transcription is now reversed, and mRNA is always transcribed 5' to 3', the primary mRNA transcript of the transcribed plasmid x1 gene will actually be the complementary sequence to the genomic wild type mRNA transcript. This means that in the cytoplasm, these two sequences will bind together, making translation impossible, thus knocking down expression.
D. Transfecting a cell line with a wild type gene separated from its promoter is pointless. The plasmid gene will not be transcribed and nothing will inhibit transcription of the genomic copy of the gene.
Thus, C is the only plausible answer choice with the given information.
