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Exercício de Mendelismo
Exercício:
Assinale a alternativa que preenche corretamente as lacunas do texto abaixo, na ordem em que aparecem.
De acordo com a hipótese de segregação de fatores proposta por Mendel, a produção de uma geração F3 a partir da autofertilização da geração F2 resultante do cruzamento de ervilhas parentais homozigotas amarelas (AA) e verdes (aa) produziria …….. de ervilhas amarelas e …….. de ervilhas verdes.
(A) 3/8 – 5/8
(B) 4/8 – 4/8
(C) 5/8 – 3/8
(D) 6/8 – 2/8
(E) 7/8 – 1/8
Alternativa D.
Vamos a explicação:
A geração F2 surgiu do cruzamento de AA*aa, que resulta em: Aa,Aa,Aa,Aa.
Já a geração F3 surgiu da autofecundação da geração acima, ou seja, do cruzamento Aa*Aa.
Assim, têm-se: AA, Aa, Aa, aa.
Analisando acima, vê-se que teriam-se 3/4 de ervilhas amarelas e 1/4 de ervilhas verdes. Como o problema não oferece essa opção, infere-se que a alternativa seja aquela que possui os múltiplos do´s valores encontrados, que seriam, respectivamente: 6/8 e 2/8. Letra D
eh verdade…..
fiz os calculos eh…..
deu na msm!!!
♥♥♥
P. AA x aa = 100% Aa F1. Aa x Aa = AA Aa Aa aa F2. por auto fecundação seria (AA x AA = 100% AA) / (Aa x Aa =3/4 amar 1/4 verd) /(Aa x Aa =3/4 amar 1/4 verd) ( aa x aa = 100% verd) pegando os tirando os resultados dos homozigotos e usando os dos heteros teremos 3/4 + 3/4 = 6/8amarela e 2/8 verd ..
calculo louko …
Segundo o gabarito da UFRGS a alternativa correta é a letra C.
A DETERMINAÇAO DA COR AVERMALHADA OU AMARELADA DA PARTE SUCULENTA DO CAJU(PSEUDOCARPO) É DEVIDA A UM PAR DE GENES ALELOS,SENDO QUE O GENE RECESSIVO DETERMINA A COR AMARELADA. SE UM CAJUEIRO RESULTAR DE UM RETROCRUZAMENTO,DEVERÁ PRODUZIR:??????
RESOLUÇAO CORRETA E DE ACORDO COM O GABARITO:
1-Geração parental: AA x aa
2- F1: Aa Aa Aa Aa
3- F2: Aa x Aa
4- F3: AA Aa Aa aa
A PARTIR DA AUTO FERTILIZAÇAO DA GERAÇAO F3 OBTEM-SE:
AAxAA Aa x Aa Aa x Aa aaxaa
4 AA 1AA 2Aa 1aa 1AA 2Aa 1aa 4aa
Total: 16 Proporçao:
amarelas: 10 Amarelas: 10/16 Letra”c”
verdes: 6 Verdes: 6/16
A resposta correta é a letra D. O resultado da F1 foi de 100% amarela, da F2 – 75% amarela e 25% verde e a F3 – 48/64 amarelas e 16/64 verdes, o que corresponde a 6/8 e 2/8, respectivamente.
RESOLUÇAO CORRETA E DE ACORDO COM O GABARITO:1-Geração parental: AA x aa2- F1: Aa Aa Aa Aa3- F2: Aa x Aa4- F3: AA Aa Aa aaA PARTIR DA AUTO FERTILIZAÇAO DA GERAÇAO F3 OBTEM-SE:AAxAA Aa x Aa Aa x Aa aaxaa4 AA 1AA 2Aa 1aa 1AA 2Aa 1aa 4aaTotal: 16 Proporçao:amarelas: 10 Amarelas: 10/16 Letra”c”verdes: 6 Verdes: 6/16
essa resposta está correta, no inicio achi q era a D mas nao levei em conta a autofertilizaçao q o exercicio pediu, por isso mt gente errou tbm! bjos
mas gente pera ai. sie que esse topico ta meio velho mas sente só.
a questao nao fala nada de autofertilizaçao de F3, e sim na produçao de uma F3 a partir da autofertilizaçao de F2. e perai, f2 num eh a mesma coisa q a F1 da resolucao desse gabarito ai?
eu tbm tava achando que eh alternativa D
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