Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

9. A biochemical pathway for making pigments in a plant shows the following sequ

ID: 195274 • Letter: 9

Question

9. A biochemical pathway for making pigments in a plant shows the following sequential color conversions, each catalyzed by separate enzymes, A, B, E and H. Each enzyme is encoded by a separate, independent gene. Wild type plants are purple, plants with no colors at all are white yellow red Reading comprehension check: what color are completely wild type plants? colorless blue a. Give colors for each genotype. Assume any mutant allele is recessive and null (missing enzyme). AABBEEHH aaBBEEHH aabbeehh AABBeehh aaBBeeHH AAbbEEHH AABBeeHH AABBEEhh b. Give genotypes and phenotypes for the offspring of this cros: purpe (AABBEBExbueeehih) x aaee F2:

Explanation / Answer

In the given cross Enzyme A and B work independently. The A allele produce a certain phenotype regardless of the allele of the other locus in this case A is said to be epistasis to the B locus. The allele E produce colour when allele A is present and similarly allele H produce colour when allele B is present

AABBEEHH- Red because A is epistastic to allele B

aabbeehh- colourless because all the alleles are in recessive condition

AABBeehh-Yellow because A is epistastic to allele B and allele E and H is in the recessive condition

aaBBeeHH-Blue because allele A is in the recessive condition that’s why allele B shows its effect

aaBBEEHH- Blue because allele A is in the recessive condition that’s why allele B shows its effect

AAbbEEHH- Red because allele B is in the recessive condition that’s why allele A shows its effect

AABBeeHH- Yellow because A is epistastic to allele B and that’s why allele A expresss as yellow

AABBEEhh- Red because allele B is in the recessive condition that’s why allele A shows its effect

2). AABBEEHH x aabbeehh

F1=(Purple-AAEE) x (Red-aaee)=AaEe (all purple)

F2= AaEe x AaEe=12:3:1

AAEE(1), AAEe(2), AaEE(2), AaEe(4), AAee(1), Aaee(2)=12 (purple)

aaEE(2), aaEe(1)=3 (Red)

aaee(1)=1 (white)

Genotype=1:2:2:4:2:3:1

Phenotype=12:3:1