Function Posted April 10, 2014 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
Function Posted April 12, 2014 Author 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
Function Posted April 12, 2014 Author 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
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now