The pair-rule gene fushi tarazu is expressed in the seven even-numbered parasegments during Drosophila embryogenesis. In contrast, the segment polarity gene engrailed is expressed in the anterior part of each of the 14 parasegments. Since both genes are active at similar times and places during development, it is possible that the expression of one gene is required for the expression of the other. This can be tested by examining the expression of the genes in a mutant background—for example, looking at fushi tarazu expression in an engrailed mutant background, and vice versa. Based on your prediction, can you predict the phenotype of the fushi tarazu and engrailed double mutant?

Question

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next question. It says the expression of certain genes in an organism can be interdependent. And this can be tested by test examining their expression in a mutant background. Which of the following is true. Regarding the phenotype of a Shiraz and engrail double mutant choice. A, this phenotype can be accurately predicted without additional information. Choice B loss of both genes will not result in any developmental defects. Choice C compensatory mechanisms may arise leading to a viable but abnormal phenotype and choice D loss of both genes will lead to a normal developmental phenotype. Well, before we can think about what we predict for the phenotype of a double mutant. For these two genes, we need to think about what these genes do. And Shi Taraz, which is abbreviated as F T Z is a pair rule jean while grilled, abbreviated E N is a segment polarity gene. And we can recall that these are active in the very early developmental stages of the des Sophal embryo. The para genes of course, break the embryo into those stripes or segments. Where is the segment polarity genes then interact with the expression of those para rule genes to establish polarity within each of those stripes that will then lead to determining the fate of each of the cells in that segment in terms of developing into anatomical structures. So these are pretty fundamental in early development and their expression is interdependent. So they interact with each other. So the F T Z gene is necessary for the expression of the engrail gene. But then the engrail gene interacts with that gene product where it's expressed to repress proteins in the certain parts of the segment and allow them to be expressed in others. In the case of engrail, this confers a posterior identity. But due to its interdependent expression, the two genes affect each other and their patterns of expression. Uh Another thing to note is the F T Z gene is present in the even numbered segments of the embryo. And the alternative parable gene is the even skipped genes that skips the even numbered segments. So when we look at our answer choices to think about which is correct, we'll actually start by ruling out two that are extremely unlikely. So choice B says that the loss of both genes will not result in any development of defects. Well, anytime we talk about genes that are involved in the very earliest stages of development, setting up the very basic body pattern, the likelihood that losing two genes from this early stage won't cause any effects is just so unlikely. We can go ahead and cross out choice B and that's the same with choice D, which says that loss of both genes will lead to a normal developmental phenotype. So those two, we can rule out right away. Finally, they mean they have a choice that we can accurately predict what the phenotype will be. That's choice A or that choice C, we might see some compensatory mechanisms that arise leading to a viable but abnormal phenotype. Well, this is a little bit tricky because again, when we talk about this early stage of development, a lot of defects in these early genes are not viable. So that might lead me to feel a little questionable about choice C. But since my alternative is choice A, which says that I can accurately predict the phenotype without any more information that I'm given in this problem that helps me zero in on choice C. Because two such um early acting genes that interact with each other and with other proteins in the cell in the emb early embryo in such a complex way means we wouldn't have a simple answer to what the phenotype would be in a double mutant. So I'm going to eliminate choice A and select choice C because even though we have this double mutant, we definitely will see an abnormal phenotype. We aren't going to see zero effect of this. But you could end up. And if we think about all the different mutant phenotypes we have in Drosophila, you could end up with compensatory mechanisms allowing the embryo to survive just in a very abnormal way. So choice C is going to be our correct answer here. See you in the next video.

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Key Concepts

Gene Expression in Development

Mutant Analysis

Phenotypic Prediction