Assuming a single diallelic gene coding for each trait, expressing complete dominance, and with no interaction between genes, the most probable genotypes of the parents are BbRr. Both parents must be dihybrids.
In this example, two features are considered,
After the cross, the results of the progeny were as follows,
Total number of pigs among the progeny = 19 + 6 + 7 + 1 = 34 pigs.
We will assume that of these traits are coded by a single diallelic gene each, that express complete dominance, and do not interact with each other.
If we perform a cross considering this two genes, we will expect to get 16 possible combinations of the parental gametes.
So, the 34 pigs from the progeny are included in these 16 possible genotypes.
34 pigs ---------------------- 16 combinations
19 black and rough ----- X = (19 x 16) / 34 = 8.94 ≅ 9
6 black and smooth ----- X = (6 x 16) / 34 = 2.82 ≅ 3
7 white and rough ------- X = (7 x 16) / 34 = 3.2 ≅ 3
2 white and smooth ----- X = (2 x 16) / 34 = 0.94 ≅ 1
So, by looking at the phenotypic proportions of the progeny, we can suggest the possible parental genotypes.
Phenotypic proportion
These proportions are the result of a cross between two dihybrid individuals, meaning that both parents are heter0zyg0us for both traits.
Also, according to these numbers, we can tell that the dominant traits are black and rough over white and smooth, respectively.
So let us make this cross assuming two dihybrid parents,
Parentals) BbRr x BbRr
Gametes) BR, Br, bR, br
BR, Br, bR, br
Punnett square) BR Br bR br
BR BBRR BBRr BbRR BbRr
Br BBRr BBrr BbRr Bbrr
bR BbRR BbRr bbRR bbRr
br BbRr Bbrr bbRr bbrr
F1) Genotypes
Genotypic ratio ⇒ 1:2:1:2:4:2:1:2:1
Phenotypes
Phenotypic ratio ⇒ 9:3:3:1
In conclusion, The most probable genotypes of the parents are BbRr, being dihybrid both of them.
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