Consider a true-breeding plant, AABBCC, crossed with ano...

Question

Consider a true-breeding plant, AABBCC, crossed with another true-breeding plant, aabbcc, whose resulting offspring are AaBbCc. If you cross the F₁ generation, and independent assortment is operational, the expected fraction of offspring in each phenotypic class is given by the expression N!/M!(N−M)! where N is the total number of alleles (six in this example) and M is the number of uppercase alleles. In a cross of AaBbCc×AaBbCc, what proportion of the offspring would be expected to contain two uppercase alleles?

Accepted Answer

Hi, everybody. Let's take a look at this practice problem together. A palla genic tree is controlled by two gene pairs in a cross between big, a little, a big B, little B and big, a little A big b, little B. What proportion of the offspring would you expect to have to additive? Little's apologetic tree is what recall that it's a trait controlled by two or more genes. Now, for our question, we're told that the palla genic tree is controlled by two gene pairs. And in palla genic traits, you can have examples of incomplete dominance recall that incomplete dominance is the phenotype ipic blending of two or more traits. Now, an example of this, the classic example of this is taking flower color. So you can have a red flower crossed with a white flower and your offspring produced are pink. So that's the classic example of incomplete dominance. Now, in the case of additive alleles, the dominant alleles represent the incomplete dominance. So the additive lil's are the dominant capital letter a little. Therefore, what this means is that the greater the number of dominant alleles, the more of the dominant phenotype is going to be expressed. So for our question, we're asked how many will have to additive alleles? So how many will have to capital letter L ils since capital letters represent the additive dominant alleles. So if we did the punnett square for our parents that are given, this is the result we would get now. So we are looking for offspring that have two capital letter alleles. And if we go through and count them, you can see that there will be six of them. So six total, we'll have to additive, the Lil's and there are 16 offspring. Therefore, what this means is that our answer is D six out of 16 offspring will have to additive. Lil's. Alright. Everybody, I hope you found this helpful and I'll see you soon for the next practice problem.

Key Concepts

Mendelian Genetics

Independent Assortment

Binomial Probability

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