Function Posted April 10, 2014 Share Posted April 10, 2014 (edited) Hello everyone This is probably my last question concerning genetics. It's a question about recombination frequence, and thus also the centimorgan (cM). Given out of my book for preparation on the acceptance exam most of you probably already know the existance of: "With tomatos, a red colour is dominant over a yellow colour. The allele for a big plant is dominant over that for a small plant. A farmer wants to optimalize his harvest. In the future, he'd like to get only plants that are both big and produce red tomatos. He asks you to do a test-experiment on his land. He gives you a seed that is the result of the crossing of a small plant of pure race for red tomatos with a big plant with yellow tomatos. Of this last plant, he's sure that it's the result of a big plant and a small plant, both of pure race. You decide to perform a testcross as first experiment. If you know that the genes for the colour and size are on the same, autosome on a distance of 18 cM, then what is, after your testcross, the percentage of big plants with red tomatos?" (Sorry if there're lots of mistakes in this text, I translated it quickly) Now, I know the answer is 9%. On first sight, I'd just say 18%, which is wrong, so I started wondering how I could get to 9%. So I drew a square for dihybride crossing: RRgg x rrGg. In this square, I could see that 1/2 of all possibilities, working with mendelian genetics, are big plants with red tomatos (RrGg) (pretty logic: RR x rr always gives Rr and Gg x gg gives in 1/2 of all cases Gg). And then I started reasoning: Mendel assumed that 100%*1/2 of all F1 are RrGg, but he didn't bring the distance between genes into account. 1/2*18cM = 9cM = 9% = the correct answer. Is this way to solve it the correct way? In other words; is this correct: [math]P(X)=P(X_{\text{mendelian genetics}})\cdot (\text{distance between the two alleles resulting in X})[/math] Thanks. Ta-ta Function P.S. Could anyone of the moderator team please change the title to "Morgan's genetics: the cM"? Thanks. Edited April 11, 2014 by Function Link to comment Share on other sites More sharing options...
Function Posted April 12, 2014 Author Share Posted April 12, 2014 (edited) So, actually, this is my reasoning: Of all F1-products, 18% is recombinant (something else than RRgg and rrGg) Of all recombinants (since Mendel is in casu only having recombinants), 50% is what's being asked. The total amount of what's being asked is thus 18%*50% = 9%. But, let's imagine the crossing of AaBb x AaBb, with a distance between A/a and B/b of 20 cM. Mendelian genetics: AABB: 1/16 AABb: 2/16 AaBB: 2/16 AaBb: 4/16 AAbb: 1/16 Aabb: 2/16 aaBB: 1/16 aaBb: 2/16 aabb: 1/16 --- Now, you'd like to know how big the chance is to get, for example, Aabb, which is a recombinant. On first sight, I'd say 2/16 * 20% = 12,8%*20% = 2,56% But, in this Mendelian genetics, 4/16 is not recombinant! Do we have to take this into account? So there are 16 F1-products, of which 12 recombinant. Of this 12, 2 have the asked genotype (Aabb) So on second sight, I'd say 2/12 * 20% = 3,33% 1) Which is the correct one? --- 2) Is it, by the way, correct, that the total chance of getting a recombinant genotype, is 12/16 * 20% = 15%? --- 3) But then, what's the chance to get a non-recombinant genotype (i.e. AaBb)? Is it just 80%? For there are 12 recombinants, of which 12 recombinant (you wouldn't say this, huh? ), so, as we stated, 12/12 * 20% = 20% recombinant, so 80% must be non-recombinant? --- Thank you in advance for helping me understand genetics as I have to get it Edited April 12, 2014 by Function Link to comment Share on other sites More sharing options...
Function Posted April 12, 2014 Author Share Posted April 12, 2014 (edited) Hmm.. I start doubting: If the recombination frequence is, let's say 20 cM, and in the mendelian, ideal square, 4/16 is not recombinant (e.g. crossing of GgVv x GgVv), so 12/16 is recombinant, then is the total chance to get Ggvv 20%*1/12 or 20%*1/16? Edited April 12, 2014 by Function Link to comment Share on other sites More sharing options...
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