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DerKrebs [107]
2 years ago
11

La polidactilia en la especie humana se debe a un alelo dominante. Entre Francisco y Martha, ambos abuelos de Santiago, uno de e

llos tenía polidactilia, sin embargo Juan su hijo (papá de Santiago) nació normal. A pesar de ello Santiago es polidactílico. La explicación a esta situación es que
Biology
1 answer:
lesya [120]2 years ago
6 0

Answer:

  • Abuelo/a con polidactilia heterocigoto, Pp
  • Abuelo/a normal, pp
  • Juan normal, pp
  • Esposa de Juan con polidactilia, PP o Pp
  • Santiago con polidactilia, Pp

Explanation:

<u>Datos disponibles</u>:

  • Polidactilia es causada por un alelo dominante, P.
  • El alelo recesivo p produce numero normal de dedos.
  • Francisco y Martha → Abuelos → uno de ellos con polidactilia
  • Juan → Hijo → Normal, pp
  • Santiago → Nieto → Polidactílico

Uno de los padres, ya sea Francisco o Martha, es polidactílico, lo que implica que ésta persona tiene, por lo menos, un allelo dominante en su genotipo, P- (El simbolo "-" esta representando un alelo dominante o uno recesivo).

Si Juan es normal, entonces su genotipo es pp, lo que significa que recibió un alelo recesivo de su padre y otro de su madre. Es decir que el parental policactílico es heterocigota, de forma que pudo heredarle un alelo recesivo a Juan. El otro parental probablemente sea normal, pp, pudiendo solo heredarle a Juan el alelo recesivo.

Entonces, uno de los padres tiene genotipo Pp (heterocigota) y el otro pp (homocigota recesivo).

Juan es norma, pp.

Santiago, hijo de Juan, es polidactílico, lo que implica que lleva en su genotipo por lo menos un alelo dominante, P. Pero Juan solo pudo heredarle un alelo recesivo, p. Por lo tanto, Santiago es heterocigota para la característica, Pp, y heredó el alelo dominante de su madre, que puede ser homcigota dominante PP o heterocigota Pp.

Supongamos que Francisco es polidactílico y Martha normal.

<u>1º Cruza</u>: Francisco x Martha

Parentales) Pp    x    pp

Gametas) P    p     p      p

F1) 1/2 = 50% Pp

    1/2 = 50% pp → Juan

<u>2º Cruza</u>: Juan  x Esposa (opción uno: esposa homocigota dominante)

Parentales) pp    x     PP

Gametas)   p    p      P     P

F2) 4/4 = 100% Pp → Santiago

<u>2º Cruza</u>: Juan  x  Esposa (opción dos: esposa heterocigota)

Parentales)  pp    x     Pp

Gametas)   p    p      P     p

F2) 1/2 = 50% Pp → Santiago

      1/2 = 50% pp

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