Mendelian Genetics Question?

Question

In some plants a red pigment, cyanidin, is synthesized from a colorless precursor. The addition of a hydroxyl group (OH-) to the cyanidin molecule causes it to become purple. In a cross between two randomly selected purple plants, the following results were obtained:
94 purple 31 red 43 white
a. How many genes are involved in the determination of these flower colors? b. Which genotypic combinations produce which phenotypes? Diagram the purple x purple cross.

Answer 1

This is an example of epistasis. Two genes are involved.

One gene encodes an enzyme that adds hydroxyl to cyanidin. If the plant has a dominant version, the plant is purple. If the plant is homozygous recessive, the plant is red. Lets call this the P gene.

The second gene for the enzyme that makes cyanidin determines whether or not there is any color. In homozygous recessive individuals there is no color regardless of what the genotype of the hydroxl adding enzyme gene is. Lets call this the C gene

So the original cross must have been PpCc x PpCc, becuase if it were anything different it would not have been possible to see the recessive phenotypes (red plants and white plants)

Look at the C gene first, you expect: 1/4 CC, 1/2 Cc, 1/4 cc.
This means that 1/4 of the plants (anything with cc) will be white, no matter what the genotype for the P gene is.

Look at the P gene now. You expect 1/4 PP, 1/2 Pp, 1/4 pp.
Phenotypically that means 3/4 purple and 1/4 red.

Now put the two together:

3/4 colored plants:
3/4 of these are purple (3/4 x 3/4 = 9/16). The genotypes are PPCC, PpCC, PPCc and PpCc
1/4 of these are red (3/4 x 1/4 = 3/16). The genotypes are ppCC and ppCc

1/4 white plants, with the genotypes of PPcc, Ppcc and ppcc.