Homework problem concerning map units

[virtuoso735]

Hi everyone, I hope you guys can help me with this simple genetics problem. I have the solutions manual for the problems, but this particular problem is not explained very well, and I’m not understanding how they come up with certain numbers.

Here is the problem:

The R and S loci are 35 m.m apart. If a plant of genotype

R……………S
——————
——————
r…………….s

is selfed, what progeny genotypes and phenotypes will be seen and in what proportions?

I figured that since R and S are 35 mu apart, there will be a 35% chance of recombination. Therefore, likelihood of getting a RS or rs is .5(1-.35) = .325. The likelihood of getting a recombination (Rs or rS) is .5(.35) = .175.

The genotypes possible (with prob. given by book next to them) are

RS/RS .1056
rs/rs .1056
RS/rs .2113
RS/rS .1138
RS/Rs .1138
rs/rS .1138
rs/Rs .1138
Rs/Rs .0306
rS/rS .0306
Rs/rS .0613

How did they come up with those numbers? I get the ones with .1056. For RS/RS, the likelihood of RS is .325^2 because they don’t need to be recombined. But RS/rs doesn’t need to be recombined either, so why isn’t it also .325^2? rs is also a nonrecombinant gene like RS, so why isn’t its probability the same? And for rs/rS, why isn’t it .325 x .175? How are these probabilities obtained?

[Cat]

"RS/rs doesn’t need to be recombined either, so why isn’t it also .325^2"

but it is. The only difference is that there are 2 possibilities: RS/rs or rs/RS that are the same no matter which way to write it. So, RS/rs is 2x(0.325^2).

The same applies to the recombinant. If you do the cross you will realize that there are also two possibilities of the rs/rS and, therefore, 0.325 x 0.175 x 2 = 0.1138 etc.

[Author]

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