In rats, the following genotypes of two independently assortin...

Question

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

Genotype and phenotype combinations for coat color in rats, including gray, yellow, black, and 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

Hi everyone. Let's take a look at this practice problem. Together we have true breeder round and yellow seeded pea plants were crossed with a true breeder wrinkled and green seeded pea plant to raise F one offspring. What is true for the F. One generation of this cross? Note the little big Y. Which produces the yellow color of the seed. In the pea plant is completely dominant over little Y. Which makes the seed green. The little for the round shape, big R. Is completely dominant over the wrinkle shape. Little R. So we need to determine the F. One generation offspring of a cross between a true breeder round in yellow pea plant with a true breeder wrinkled and green pea plant. But let's first write out our key more clearly we have big are for around little R. For wrinkled, then we have big Y. For yellow and little I for green. Now recall that true breeder means that the parent is homogeneous for each trait. So we needed to determine the Jena types for each parent and parent one is round and yellow represented by big R. Big R. Big Y. Big Y. Parents too will be wrinkled and green. So little r. Little R. And little lie little lie. So these are the gina types. They're homos I guess for each trait. And then now we need to determine the gametes. So we know that each gamete, the parent one producers will look exactly the same. There will be four of them and they will be big are big Y. And for a parent to there will be four gametes and they'll be little our little by. So when we do a cross between these two parents, our punnett square will look like the following. And we can see that all offspring are going to be big R big R big r little R big y. Little lie. So now that we know that remember that we are looking for the answer that is true for the F. One generation. So let's take a look at our answer choices. We have a all F. One offspring will be mono hybrid. Well mono means one and hybrid means hetero. So this answer is saying all the offspring, We'll have one treat and be hetero psychosis. Well we have two traits for each offspring. And so we can eliminate option A. If we look at option B. Half of the F. One offspring will be mono hybrid while the other half will be true breeders again like an answer A. We are looking at more than one trait. Mono hybrid is incorrect and therefore we can eliminate option B. As well right away. Then let's look at option C. All F. One offspring will show the wrinkled green phenotype. Okay, all our offspring are big R little r big y little Y. So they'll display the dominant males which will be round and yellow. So C. Is incorrect as well. Lastly let's look at auction. D All F one offspring will be die hybrid. We can see that each of our offspring have two traits, and each trait is hetero status. Therefore, this is called die hybrid, and D. Will be our correct answer. Alright, everyone. I hope you found this helpful, and I'll see you for the next practice problem.

Key Concepts

Independent Assortment of Genes

Epistasis and Gene Interaction

Genotype to Phenotype Mapping

Watch next