Question

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12 - 14: Consider three unlinked genes: a, b, c are the recessive alleles, and A, B, C are the dominant alleles. Now if two triple heterozygote individuals (i.e. Aa/Bb/Cc) are crossed; *12. What is the probability of getting an offspring that was recessive for all three genes? a) 3/4 = 0.75 b) 1/64 = 0.0156 c) 1/8 = 0.125 d) 1/4 = 0.25 e) 27/64 = 0.422 *13. What is the probability of getting an offspring with the recessive phenotype for the A locus, but dominant phenotype for the B and C loci? a) 3/4 b) 1/4 c) (1/2)(1/4)(3/4) = 3/32 d) (1/4)(1/4)(1/2) = 1/32 e) (3/4)(3/4)(1/4) = 9/64 %3D * 14. If you were to take all of the dominant phenotype offspring and randomly cross them, what proportion of those offspring would have the dominant phenotype at the A locus? a) 1/9 b) 2/3 c) 4/9 d) 1/3 e) Not enough information provided

Answer 1

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12. - (b) 1/64 = 0.0156 lsbablity of getting a single recessive gene=1/4 for each. ie. XX XX X X xx Out of 4 only 1 is recessive. So sobaklity of recessive endividual in triple "heterozygote 13. Probablihy you dominant phenobape Probablity for reccesive pherelipe Requised phenotype = secessive A = YA - Dominant B = 3/4 = Dominant c = 3/4 Regused probablety = (3/4 x (3²) x /6 - 9 . (C) = 6 092)) = %64 14. (@) Not enough information provided