Question

For both genes A and B, the dominant, normal alleles are haplosufficient. Enzymes made by gene A change red pigment to purple; enzymes made by gene B change purple pigment to blue, a and b represent loss of function alleles. An individual pure- breeding for normal alleles at both genes is crossed with an individual pure-breeding for mutant alleles at both genes. The F1 are then crossed with each other. What proportion of offspring will be red? O 1 O 1/2 09/16 04/16 01/16 03/16 Oo

Answer 1

ANSWER : The haplas ufficient conditions produce to produce the as seen in the genes, in heterozygous sufficient gene product same, standard phenotype homerygous condition . Hence, the dominant alleles A and B haple sufficient, the inheritance pattern illustreated as follows : being can be Purple Bis Blue Red pigment pigment. pigment Paxant : AABB qa Rabb Gametes: fi : AaBb selfing

- | AB AB AABB (Blue) Ab Ab I aB I ab AABb | AaBB AaB b. (Blue) Blue) (Blue) Anbb Aa Bb Hablo (Purple) (Blue) I (Purple) Aa Bb aa BB aaBb (Blue) Creed) I Creed) Aabb aa Mb aahh (Purple) Creed Creed). AABB (Blue) AaBB (blue) Aa Bb (blue) aB ab. fa phenotypic reatic = Blee : Purple : Red = q : Š : 4 • The genotypes containing both dominant alleles A and B will presduce blue pigments. The allele À will turn reed pipment into purple, which will subsequently be converted into blue pegment by alléle B'. The genotypes containing only the deminant allele À will preeduce pwepte pigments since the people pigments produced by A can't be converted into blue due to absence of allele B.

The genetypes lacking will preoduce reed can't be modified of allele A. the dominant allele A pigments, since they Rwether due to absence # Hence, the this much correct answere is 4/16 ie proportion of offeprings will be reed.