TBR Genetics Question

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EnginrTheFuture

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I had basic genetics in intro to bio and delved in deep to genetics for TBR but this question is blowing my mind.

Ch. 9 Passage 6 Question 34 (page 340)

It is about genetic map distance:

Based on the information in experiment II, estimate the minimum genetic map distance between the two genetic loci. (Note: one genetic map unit (m.u.) gives a recombinant frequency (RF) of 1 percent.)

A. 17 mu
B. 34 mu
C. 52 mu
D. 68 mu


Sorry I don't have any way of getting the passage info on here right now. Does anything like this have an even remote chance of making its way onto the actual exam? The usual stuff like progeny probability questions, genotype/phenotype frequency of previous or future generations... that's all cake now, but this? Way beyond what I have ever been tested on before and it makes me nervous I have a weakness going in
 
I had basic genetics in intro to bio and delved in deep to genetics for TBR but this question is blowing my mind.

Ch. 9 Passage 6 Question 34 (page 340)

It is about genetic map distance:

Based on the information in experiment II, estimate the minimum genetic map distance between the two genetic loci. (Note: one genetic map unit (m.u.) gives a recombinant frequency (RF) of 1 percent.)

A. 17 mu
B. 34 mu
C. 52 mu
D. 68 mu


Sorry I don't have any way of getting the passage info on here right now. Does anything like this have an even remote chance of making its way onto the actual exam? The usual stuff like progeny probability questions, genotype/phenotype frequency of previous or future generations... that's all cake now, but this? Way beyond what I have ever been tested on before and it makes me nervous I have a weakness going in

Seems legit if they give you the conversion factors. It's based on the fact that close loci are linked, far loci are not.
 
its based on the passage and this question owned me today. There were a few passages in that chapter that really got me.
 
If anyone else has means of putting up a picture of the data/passage that would be super helpful. I read and reread the answer key and it just confused me more.
 
I remember doing some pretty challenging recombination problems in my intro genetics class. I might not be grasping some nuance of the problem, but generally you're going to see which alleles tend to be together over 50% and then you're going to divide the number of recombinants (the ones who have less common pairings) by the total number of individuals from a mating. That's frequency X 100 MU (or centimorgans) = the map distance.

So for the easier types of problems you know the parents' genotypes. Say you do a dihybrid cross AaBb x AaBb. When the genes are unlinked, AB, aB, Ba, and aa are all equally probable (1/4) and thus each genotype should be seen at a 1/16th frequency. So, for any genotype you can find the expected frequency through probability as in a homozygote is going to be 1/4*1/4=1/16. Something like AABb is 1/2*1/2*(2 * 1/2 * 1/2) = 1/8 (you multiply by two because you could get Bb from Bb or bB). If you aren't getting those ratios in the offspring, the genes must be linked.

Say out of 100 offspring you are getting

30 AABB
30 aabb
30 AaBb
2 AABb
2 aaBb
3 AaBB
3 Aabb

This is because the parents' chromosomes were either AB or ab. The last 4 genotypes are the result of recombination during prophase I. since you have 3+3+2+2=10 recombinants out of 100 progeny, you have 10% recombination so the genes are 10cm (aka 10mu) apart. Hopefully someone else can double check my crappy example here as I am rusty as hell on these types of problems.

EDIT and unless there is a subtlety I'm not considering here, 50 map units (centimorgans) means the genes are unlinked. A number over 50mu is meaningless, so you should be able to discard those answers immediately. I realize that saying that a 75% chance of genes being unlinked is kind of like saying they're 25% unlinked though, so it could be a trick. I kind of would need to see a problem.
 
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I remember doing some pretty challenging recombination problems in my intro genetics class. I might not be grasping some nuance of the problem, but generally you're going to see which alleles tend to be together over 50% and then you're going to divide the number of recombinants (the ones who have less common pairings) by the total number of individuals from a mating. That's frequency X 100 MU (or centimorgans) = the map distance.

So for the easier types of problems you know the parents' genotypes. Say you do a dihybrid cross AaBb x AaBb. When the genes are unlinked, AB, aB, Ba, and aa are all equally probable (1/4) and thus each genotype should be seen at a 1/16th frequency. So, for any genotype you can find the expected frequency through probability as in a homozygote is going to be 1/4*1/4=1/16. Something like AABb is 1/2*1/2*(2 * 1/2 * 1/2) = 1/8 (you multiply by two because you could get Bb from Bb or bB). If you aren't getting those ratios in the offspring, the genes must be linked.

Say out of 100 offspring you are getting

30 AABB
30 aabb
30 AaBb
2 AABb
2 aaBb
3 AaBB
3 Aabb

This is because the parents' chromosomes were either AB or ab. The last 4 genotypes are the result of recombination during prophase I. since you have 3+3+2+2=10 recombinants out of 100 progeny, you have 10% recombination so the genes are 10cm (aka 10mu) apart. Hopefully someone else can double check my crappy example here as I am rusty as hell on these types of problems.

EDIT and unless there is a subtlety I'm not considering here, 50 map units (centimorgans) means the genes are unlinked. A number over 50mu is meaningless, so you should be able to discard those answers immediately. I realize that saying that a 75% chance of genes being unlinked is kind of like saying they're 25% unlinked though, so it could be a trick. I kind of would need to see a problem.


Thanks so much for the reply, it's starting to sink in but I'm still a little confused in how you knew which one's were "recombinants" and how you knew the parents were the genotypes that you said they were (AB or ab)


If you have a dihybrid cross AaBb x AaBb I get the theoretical spread of:



AABB AABb AaBB AaBb
AABb AAbb AaBb Aabb
AaBB AaBb aaBB aaBb
AaBb Aabb aaBb aabb


But what is happening because of the inseparability of the alleles is?

AaBb x AaBb

.....AB ab
AB
ab

AABB AaBb
AaBb aabb


I think a major assumption in this problem lies in the fact that they are (I believe) assuming that the F1 generation AaBb that is being crossed with an aabb is A/B a/b but there's no reason to believe that it shouldn't start off as A/b a/B before we cross it with a/a b/b.

I'm really really confused about this. I get that normally everything is independent AaBb means you can give the combo of Aa Ab aB or ab from your set and so can the other person but now there seems to be all these crazy assumptions as to what's linked to what to start? Gah I can't even put this confusion into words
 
Sorry these are really hard to explain especially because the problems can be set up in a bunch of different ways, and it's a rare case where a trihybrid cross actually makes things easier to understand. Your first red flag that they're recombinants in the admittedly crappy practice problem I wrote up is because you see them at a low frequency.

Look at how the genotypes of the low % offspring and how they differ from the high % ones. In each one you have an individual that's heterozygous for one trait and homozygous for the other. I actually think I screwed up and there should be something more like below because each parent could give AB or ab without recombination (in this corrected example, the recombination frequency is 20% instead of 10).

20 AABB
20 aabb
40 AaBb
4 AABb
4 aaBb
6 AaBB
6 Aabb

So in each one of the low frequency offspring, there is an Ab or aB whereas you can infer from the higher % nonrecombinants that the 1:2:1 pattern means that the parents are contributing AB or ab. It's kind of something you just know from common cross ratios. You basically ignore the low frequency offspring first and just ask yourself "what gene linkage from the parents produces this genotype frequency in offspring?" Then you looks to the recombinant ones and say "okay, these are low frequency, I know there is recombination, what does each individual show that the nonrecombinants don't?"

In this case, it's the fact that they must have gotten one dominant allele and one recessive from one of their parents. This could be the case with the AaBb ones as well, but you infer that it's not or can be ignored by knowing that they are being produced at a % that's consistent with crossing 2 linked gene parents.
 
Aha, that gave me the insight I needed!

So essentially I can backtrack the linked alleles from the high % guys by asking "what can give me this combination of high yield progeny". Makes a ton of sense! In the TBR problem they don't give genotypes obtained in the final offspring but rather observed phenotypes and the crossings we did to get there. This makes things a bit crazier BUT... if I start from the P generation, cross homoDom (A/B A/B) with homoRec (a/b a/b) assuming complete conservation (no recombination) that gives me an F1 gen with A/B a/b only. Then when I cross those with aabb, if I assume no recombination again, I get two equally likely possibilities A/B a/b OR a/b a/b these are my "high yielders" if no recombination occurs. Of course some is going to occur and the closer I get to independent assortment the more recombinants I will get

Tricky part is that I have to extract out the number of actual high yielder genotypes are in the observed phenotypes since the low yield recombinant genotypes can look like high yield genotypes... this is where things turn into assumptions for "lowest number" answer and BS that TBR pulls. I don't care about that intricate of a detail, I wanted to understand a general approach to tackling/ thinking about this and you definitely got me there I think. Thanks so much!
 
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