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hieunguyen288 hieunguyen288
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10 years ago
Two corn homozygous, white-seeded corn lines were crossed and all progeny were red-seeded. These red-seeded F1, progeny were selfed, and the population segregated 9 red-seeded : 7 white-seeded. Explain these results by determining the number of genes controlling seed coat colors, give the genotypes of the parents and perform a dihybrid cross of parents

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10 years ago
Two different, independently segregating genes are involved. Let's call them "C" (for allows color), "c" (for does not allow color), "R" (for red), and "r" (for not red). "C" is dominant over "c". "R" is dominant over "r". The genes that allow color in general are dominant over the genes for a specific color.

On natural way to think of this is that the C gene makes a colorless protein. Just having the gene is enough to get it made, so CC and Cc both make protein, and cc doesn't. The R gene turns the C protein into something red. Again, just having the gene is enough to turn C into something red, so RR and Rr both make C red, but rr leaves it white. However, R can't do anything if C isn't made first!

Genotypes:

One parent is CCrr. They allow color, but have no color genes.

The other parent is ccRR. They don't allow color, but have red genes.

F1 cross:

Any cross between these do, in the F1, will have the genotype CcRr. This means that they will allow color, and they have a red color gene. They will all be red-seeded.

F2 cross:

A cross between CcRr and CcRr produces a big mix. It's hard to draw a Punnett Square here.

That first Punnet Square is a dihybrid heterozygous cross with genes "A" and "a", and "C" and "c". Replace the "A" with "R" and the "a" with "r", and you get exactly this cross.

It works out to 9 red-seeded (have the C gene at least once and the R gene at least once) and 7 white-seeded (either have "cc", or "rr") for the 16 possible mixes. This means that a perfect population would have a 9:7 ratio, just like the F2 progeny that you describe.
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