A cross between a blue blahblah bird (B) & a white blahblah (B*) bird produces offspring that are silver. The color of blahblah birds is determined by just two alleles. What would be the phenotypic ratios of offspring produced by crossing two silver blahblah birds?

Answer 1

To determine the phenotypic ratios of offspring produced by crossing two silver blahblah birds, we first need to understand the inheritance pattern involved in the coloration of the blahblah birds.

From the information provided, we have:

• The blue blahblah bird is represented by allele B.
• The white blahblah bird is represented by allele B*.
• The combination of these alleles (B and B*) produces silver offspring.

Assuming that the silver color is the result of a heterozygous genotype (Bb*), we can set up our examination of the possible offspring from two silver blahblah birds.

When we cross two silver birds (both Bb*), we can use a Punnett square to visualize the potential offspring. Here’s what the square looks like:

       B      b*
    ----------------
B  |  BB   |   Bb*  |
    ----------------
b* |  Bb*  |  b*b*  |
    ----------------

From this Punnett square, we get the following combinations:

• BB (homozygous blue)
• Bb* (heterozygous; silver)
• bb (homozygous white)

Now, let's tally the phenotype ratios from these combinations:

• 1 BB (blue)
• 2 Bb* (silver)
• 1 bb (white)

So the phenotypic ratio is:

• 1 Blue : 2 Silver : 1 White

Thus, the phenotypic ratios of offspring produced by crossing two silver blahblah birds would be 1 Blue : 2 Silver : 1 White.