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The R and S loci are 35 m.u. apart. If a plant of genotype is selfed, what progeny phenotypes will be seen and in what proportions

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marianaegarciaperedo

Complete question:

The R and S loci are 35 m.u. apart. If a plant of genotype RS/rs is selfed, what progeny phenotypes will be seen and in what proportions

Answer:

  • R-/S- = 0.6052
  • R-/ss = 0.1443
  • rr/S- = 0.1443
  • rr/ss = 0.1056

Explanation:

Available data:

  • R-S are 35 mu apart
  • Cross: RS/rs   x   RS/rs

First, we need to recognize the parental gametes and the recombinant ones.

Cross:

Parentals)     RS/rs    x      RS/rs

Gametes) RS  Parental   ⇒  Equal to the parental genotype

                 rs   Parental   ⇒  Equal to the parental genotype

                 Rs  Recombinant   ⇒  Product of recombination

                 rS   Recombinant   ⇒  Product of recombination

We know that genes are 35 MU apart from each other. The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.  

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.  

According to this information, if 1 MU = 1% recombination frequency, then

35MU --------- 35% recombination frequency = 0.35

Now, let us calculate the frequency of each gamete type.

35 map units = 35 % of recombination in total  

                      = % Rs + % rS  

                      = 17.5% Rs + 17.5% rS

Now, if the recombination frequency equals 35% then the parental frequency is 100% - 35% = 65%

65 % parental frequency = % of RS + % rs  

                                          = 32.5% RS + 32.5% rs

The frequency of each gamete is

  • RS  ⇒ Parental  ⇒  32.5% (65% / 2) = 0.325
  • rs  ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325
  • Rs  ⇒ Recombinant  ⇒ 17.5% (35% / 2) = 0.175
  • rS  ⇒  Recombinant ⇒  17.5% (35% / 2) = 0.175

F1 Genotypes and Proportions

  • RS/RS = 0.325 x 0.325 = 0.1056    
  • RS/Rs = 2x 0.325 x 0.175 = 0.1137
  • RS/rS = 2x 0.325 x 0.175 =0.1137  
  • rs/rs = 0.325 x 0.325 = 0.10562
  • rs/RS = 2x 0.325 x 0.325 = 0.2112
  • rs/Rs = 2x 0.325 x 0.175 = 0.1137
  • rs/rS = 2x 0.325 x 0.175 = 0.1137
  • Rs/rS = 2x 0.175 x 0.175 = 0.061
  • Rs/Rs = 0.175 x 0.175 = 0.0306
  • rS/rS = 0.175 x 0.175 = 0.0306

F1 Phenotypes and Proportions

  • R-/S- = 0.1056 + 0.1137 + 0.1137 + 0.2112 + 0.061 = 0.6052
  • R-/ss = 0.1137 + 0.0306 = 0.1443
  • rr/S- = 0.1137 + 0.0306 = 0.1443
  • rr/ss = 0.1056

luisvaschetto

If R and S loci are 35 m.u. apart and a plant of genotype RS/rs is selfed, then the progeny phenotypes will be 0.606 RS; 0.106 rs; 0.144 Rs and 0.144 rS.

In this case, it is imperative to estimate gamete frequencies

  • RS = 0.65% / 2 = 0.325
  • rs  = 0.65% / 2 = 0.325
  • Rs  (recombinant) = 0.35 / 2 = 0.175
  • rS (recombinant) = 0.35 / 2 = 0.175

  • In consequence, the F1 genotypes will be RS/RS = 0.1056  (i.e., 0.325 x 0.325); RS/Rs = 0.1137; RS/rS = 0.1137; rs/rs = 0.10562; rs/RS = 0.2112; rs/Rs = 0.1137; rs/rS = 0.1137; Rs/rS = 0.06; Rs/Rs = 0.0306; and rS/rS = 0.0306.

  • Finally, it is possible to obtain phenotypic frequencies by summing genotypic frequencies for each case in particular.

In conclusion, if R and S loci are 35 m.u. apart and a plant of genotype RS/rs is selfed, then the progeny phenotypes will be 0.606 RS; 0.106 rs; 0.144 Rs and 0.144 rS.

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