Having correctly established the F₂ ratio in Problem 18, predict ...

Question

Having correctly established the F₂ ratio in Problem 18, predict the F₂ ratio of a 'dihybrid' cross involving two independently assorting characteristics (e.g., P₁ = WWWWAAAA x wwwwaaaa).

Accepted Answer

Hey everyone, let's take a look at this question together. The genes for wing color and size and beetles assort independently the red color, a little of the wing is dominant over the white color alil. Whereas the normal wing allele is dominant over the short wing will determine the fanatic ratio of red to white winged offspring produced when a beetle with normal red wings is mated with a beetle having short white. We so let's set up upon a square to determine the fanatic ratio of red to white winged offspring produced. When a beetle with normal red wings is made with a beetle having short white wings are beetle that has normal red wings. Is hetero ziggy's for both traits which would look something like this. We have big r. Little R and big L. Little L. And we made that with a Beatle who has short white wings, which we know that the white color as well as the short wings are both excessive. So that would look like little our little our little l, little L. And so when we distribute that and set up our punnett square, it looks something like this. So here we have that beetle with the normal red wings. And then here we have the beetle with the short white wings. And so let's go ahead and fill out the rest of this punnett square. See what types we end up with. Now that we can see all the different Jenna types. Let's figure out what each phenotype is To figure out the phenotype IQ ratio of red to white winged offspring. So now we can see that these have the red wings, whereas these down here have the white wings. And so the ratio of red two white is 8/16 to 8/ because eight out of the 16 possible phenotype are all red and then the remaining eight are all white. And so this equals a perfect 1 to 1 ratio or answer choice A. The correct answer because when we made that normal redwing beetle with the short white wing beetle, the phenotype ratio is 8/16 to 8/16, which equals 1 to 1 when reduced. So answer choice A is the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

Having correctly established the F₂ ratio in Problem 18, predict the F₂ ratio of a 'dihybrid' cross involving two independently assorting characteristics (e.g., P₁ = WWWWAAAA x wwwwaaaa).

Key Concepts

Mendelian Genetics

Dihybrid Cross

Independent Assortment

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