You have isolated (1) a streptomycin-resistant mutant (s...

Question

You have isolated (1) a streptomycin-resistant mutant (strᴿ) of Chlamydomonas that maps to the chloroplast genome and (2) a hygromycin-resistant mutant (hygᴿ) of Chlamydomonas that maps to the mitochondrial genome. What types of progeny do you expect from the following reciprocal crosses?

mt⁺ strᴿ hygˢ× mt⁻ strˢ hygᴿ
mt⁺ strˢ hygᴿ× mt⁻ strᴿ hygSˢ

Accepted Answer

Hi, everyone. Let's look at our next problem. It says, Consider that there are five genes, PQXYZ determine the sequence of genes on the chromosomes. If the crossover percentage between linked genes is as follows. Number one P and Y 25%. Number two PNZ, 15%,, Number three P and Q 15%. Number four Y and X 10%. Number five Z and Q, 30%. Number six Z and Y 10%. Then my answer choices are choice A P Q X Y Z choice B Z Y X P Q choice C X Y Z P Q N choice D P Q Z Y X. So when I'm trying to map the sequence of genes, I always like to start with the largest distance between two genes. So when I look at all my different crossover percentages, recalling that recombination frequency, crossover percentage is proportional to the physical distance between the genes. So I can use that as a measure of distance. And when I look over these values, I see my largest number is number five, the distance between Z&Q is 30%. So I'm gonna start by putting a lime with Z on the left and Q on the right And mark that distance between those 30%. It's my greatest distance. So I'm gonna start by positing that the other genes are in between those two. Let's see if that works. So my next step will be to look for the two genes that are closest to Z and Q on the end. So I see that my next small number here is in number six that says Z&Y have a crossover percentage of 10%. So we'll start by putting Y in the middle here just a little bit to the right of Z. I'll label y here noting that that difference is 10%. Well, let's see if that works. So now we'll look for the closest gene to Q and when I look at that, I have that number three, P and Q are 15% recombinant frequency. So I'll put P on the inside to the left of Q Noting that it's halfway in between since the uh my line is represents 30%. So I'll put p there and then note that the difference between P&Q is 15%. Well, let's see if my arrangement makes sense. I can check that with number one, which has the distance between P and Y that I've both put in the middle here. Number one says that P and Y are 25%. However, that means that this arrangement won't work. P and Y are not far enough apart. I do have the information that number two p and Z are 15% recombination. So that confirms that P does belong in the middle. It's halfway in between Q and Z. So I can, I know that that's correct. So if that distance between PNZ is Then and the distance between PM Y is with that space between Y and Z is 10% then Y must belong on the other side of Z. So I'm gonna erase Y from where I put it there and put it on the other side of Z. So extend my line a little to the left and add Y before Z noting that distance as 10% at that point, my P Y distance distance works because I have P to Z 15% Then Z to Y of 10%, which correctly gives me the Distance from Y to P being 25%. So now I only have one other distance that I haven't looked at and one other gene that needs to be placed. And number four tells me that Y&X have a distance of 10%. Well, Y and Z also have a distance of 10%. So I'm gonna assume that X, the X gene is not right on top of the Z gene and must be on the other side of Y. So I've got to extend that line a little further to the left and put X first of all, With that difference distance between X&Y of 10%. So now I've placed all five of my jeans on my line. I have X and then Y and then Z and then P and then Q and that corresponds to choice C which is indeed in that word X Y Z P Q see you in the next video.

Key Concepts

Chloroplast and Mitochondrial Genetics

Reciprocal Crosses

Cytoplasmic Inheritance

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