In rats, the following genotypes of two independently assorting a...

Question

In rats, the following genotypes of two independently assorting autosomal genes determine coat color:

A— B— (gray)
A — bb. (yellow)
aaB— (black)
aabb (cream)

A third gene pair on a separate autosome determines whether or not any color will be produced. The CC and Cc genotypes allow color according to the expression of the A and B alleles. However, the cc genotype results in albino rats regardless of the A and B alleles present. Determine the F₁ phenotypic ratio of the following crosses:

AaBBCc×AaBBCc

Accepted Answer

everyone. Let's take a look at this question together. Blue eyes with black coats are not often seen on wolves, assume that normal coat color dominates over black and that brown eyes dominate over blue and alpha male is mated with an alpha female? What will be the probability of a wolf's offspring having blue eyes and a black coat? If both parents are hetero ziggy's for eye and coat color, note that the alleles for coat color and eye color are not linked. So since the alleles for coat color and eye color are not linked, we have to do each punnett square separately. Let's go ahead and write that out. So that way we can figure out the combinations the father, the mother. So let's start off with coat color. So we see that in normal coat color dominates over black. So we have hetero ziggy's for both. We have big and little end and big and little. And then for coat color we see that brown eyes dominates over blue. So we also have Big B. Little B. And Big B. Little B. So now we can go ahead and cross and get our different genotype parents. We have big and big and little again little end little and little end. And then over here we have Big B. Big B. Big B. Little B. Big B, little B. Little B, little B. Which we know this is pretty standard for a heterocyclic crossing. And so since we want to find out the probability of the offspring having blue eyes and a black coat, which we know that both blue eyes and a black coat are the recessive trait. We have 1/4 recessive or both because we can see that here is that recessive trait. And so since these are both in independent of each other, we go ahead and multiply both ratios and that gives us 1/16, which is answer choice C. The correct answer. Because when we cross that heterocyclic spare for both coat color and eye color, we find out that 1/4 of offspring will have that recessive genotype. So we multiply them together and we get our answer of 1/16. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Independent Assortment

Genotype and Phenotype

Epistasis

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