Question

In: Biology

You cross a true-breeding dark orange flowered poppy with crinkly petals to a true-breeding white-flowered poppy...

You cross a true-breeding dark orange flowered poppy with crinkly petals to a true-breeding white-flowered poppy with smooth petals. The F1 poppies are light orange with smooth petals. In the F2 generation you get 201 orange crinkly poppies; 400 light orange smooth poppies; and 199 white smooth poppies. Explain this observation using all appropriate terminology. Also provide a breeding diagram that clearly links genotypes to phenotypes.

This is all that is given.

Expert Solution

The crossing is between true breeding dark orange flowered poppy with crinkly petals to a true breeding white-flowered poppy with smooth petals.

Genotype of dark orange flowered poppy with crinkly petals can be represented by Ar/Ar and genotype of white flowered poppy with smooth petals can be represented by aR/aR as these genes are linked.

Here orange colour is a dominant character and crinkly petals is a recessive character.Also white flowered poppy is a ressive character and smooth petal is a dominant character.

By crossing we get

Parents Ar/Ar (Dark orange flower crinkly petals) x aR/aR(white flower smooth petals)

Gametes Ar aR

F1 Ar/aR (Light orange flower ,smooth petals)

Dark orange colour is incompletely dominant over white colour and smooth petal is dominant trait

F2 by selfing F1.

F2 Ar/aR (Light orange flower ,smooth petal) x Ar/aR (Light orange flower,smooth petal)

The progenies are

Ar/Ar (Orange crinkly poppies) , Ar/aR (light orange smooth poppies)

aR/Ar (light orange smooth poppies), aR/aR (white smooth poppies)

They are in the ratio 1:2:1

Ar/Ar 201 orange crinkly poppies

Ar/aR 400 light orange smooth poppies

aR/aR 199 white smooth poppies

The result obtained is aslo in 1:2:1 ratio

gladiator answered 2 years ago

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