Final answer:
When two white horses that are heterozygous for the coat color, pigment production, and spotting genes are crossed, the proportion of live progeny that will be black in color and have spots is 50%.
Step-by-step explanation:
To determine the proportion of black-colored horses with spots in the live progeny, we need to consider the genotypes of the parents and use Punnett squares. The parents in this case are two white horses that are heterozygous for the coat color genes (Ww), the pigment production gene (Bb), and the spotting gene (Oo).
Step 1: Determine the possible genotypes of the parents. The possible genotypes for each gene are:
- Coat color gene: WW, Ww, and ww
- Pigment production gene: BB, Bb, and bb
- Spotting gene: OO and Oo
Step 2: Construct Punnett squares for each gene combination and determine the ratios of the different genotypes in the offspring.
- Coat color Punnett square: WW x ww
- Possible genotypes: WW, Ww, and ww
- Ratios: 1:1:0 (50% WW, 50% Ww, and 0% ww)
- Pigment production Punnett square: BB x bb
- Possible genotypes: BB and Bb
- Ratios: 0.5:0.5 (50% BB and 50% Bb)
- Spotting Punnett square: OO x OO
- Possible genotypes: OO
- Ratios: 1 (100% OO)
Step 3: Multiply the ratios from each Punnett square to determine the overall proportion of black-colored horses with spots in the live progeny. Multiplying the ratios of the coat color, pigment production, and spotting genes, we get:
- Black coat color and spots: (1:1:0) x (0.5:0.5) x (1)
- Ratio: 0.5:0.5:0 (50% black coat color and spots, 50% black coat color without spots, 0% other coat colors with or without spots)
Therefore, 50% of the live progeny will be black in color and have spots when two white horses heterozygous for these three genes are crossed.