Question

Wild type blue-eyed Mary has blue flowers. Two genes control the pathway that makes the blue pigment: The product of gene W turns a white precursor into magenta pigment. The product of gene M turns the magenta pigment into blue pigment. Each gene has a recessive loss-of-function allele: w and m, respectively. A dihybrid (WwMm) is self-pollinated. What proportion of the offspring will be magenta?

Answer 1

Answer:

3/8

Explanation:

Two genes are involved in the determination of flower color, W/w and M/m.

The W allele uses the white pigment as a substrate to produce the magenta pigment, and the M allele uses magenta to produce blue pigment.

The recessive alleles w and m produce a non-functional protein, so the white and magenta pigments can't be transformed into their respective products if the genotype is homozygous recessive.

Please see the first attachment for a representation of the flower color pathway.

We have different genotype possibilities:      

• W_M_ = blue flowers, because the metabolic pathway can go all the way.
• W_mm = magenta flowers, because the W allele can turn white pigment into magenta, but the mm genotype can't produce a functional protein and the magenta pigment can't turn into the blue one.
• wwM_ = white flowers; the first enzyme of the pathway can't change the white pigment into magenta, so it doesn't matter that the M allele is present.
• wwmm = white flowers, for the same reason as wwM_.

To analyze the offspring of a dihybrid cross, we need to do the Punnett Square (see second attachment).

The genotypes that determine the magenta color are WWmm and Wwmm.

From the Punnett Square, we can see that 3/8 individuals of the offspring will be W_mm and therefore magenta.