In parakeets, two autosomal genes that are located on different chromosomes control the production of feather pigment. Gene B codes for an enzyme that is required for the synthesis of a blue pigment, and gene Y codes for an enzyme required for the synthesis of a yellow pigment. Green results from a mixture of yellow and blue pigments, and recessive mutations that prevent production of either pigment are known for both genes. Suppose that a breeder has two green parakeets and mates them. The offspring are green, blue, yellow, and albino (unpigmented). Based on this observation, what are the genotypes of the green parents? What genotypes produce each color in the offspring? What fraction of the progeny should exhibit each type of color?

Verified step by step guidance

1. Identify the dominant and recessive alleles: Let B represent the dominant allele for blue pigment production and b the recessive allele that prevents blue pigment production. Similarly, let Y represent the dominant allele for yellow pigment production and y the recessive allele that prevents yellow pigment production.
2. Determine the genotype of green parakeets: Since green is a mixture of blue and yellow, and both parents are green, they must be able to produce both pigments. Therefore, the genotype of each green parent must include at least one dominant allele of each gene, making them BbYy.
3. Analyze the possible gametes and offspring: The green parents can produce gametes BY, By, bY, and by. Using a Punnett square, cross the gametes of both parents to find the genotypes of the offspring: BBYY (green), BBYy (green), BbYY (green), BbYy (green), BByy (blue), Bbyy (blue), bbYY (yellow), bbYy (yellow), and bbyy (albino).
4. Calculate the phenotypic ratios of the offspring: From the Punnett square, 9/16 of the offspring are green (BBYY, BBYy, BbYY, BbYy), 3/16 are blue (BByy, Bbyy), 3/16 are yellow (bbYY, bbYy), and 1/16 are albino (bbyy).
5. Summarize the findings: The green parents have the genotype BbYy. The genotypes that produce green are BBYY, BBYy, BbYY, BbYy; blue are BByy, Bbyy; yellow are bbYY, bbYy; and albino are bbyy. The expected fraction of progeny is 9/16 green, 3/16 blue, 3/16 yellow, and 1/16 albino.