A geneticist from an alien planet that prohibits genetic research brought with him to Earth two pure-breeding lines of frogs. One line croaks by uttering 'rib-it rib-it' and has purple eyes. The other line croaks more softly by muttering 'knee-deep knee-deep' and has green eyes. With a newfound freedom of inquiry, the geneticist mated the two types of frogs, producing F₁ frogs that were all utterers and had blue eyes. A large F₂ generation then yielded the following ratios:
27/64 blue-eyed, 'rib-it' utterer
12/64 green-eyed, 'rib-it' utterer
9/64 blue-eyed, 'knee-deep' mutterer
9/64 purple-eyed, 'rib-it' utterer
4/64 green-eyed, 'knee-deep' mutterer
3/64 purple-eyed, 'knee-deep' mutterer
Of these, how many are controlling eye color? How can you tell? How many are controlling croaking?

Question

A geneticist from an alien planet that prohibits genetic research brought with him to Earth two pure-breeding lines of frogs. One line croaks by uttering 'rib-it rib-it' and has purple eyes. The other line croaks more softly by muttering 'knee-deep knee-deep' and has green eyes. With a newfound freedom of inquiry, the geneticist mated the two types of frogs, producing F₁ frogs that were all utterers and had blue eyes. A large F₂ generation then yielded the following ratios:

27/64 blue-eyed, 'rib-it' utterer

12/64 green-eyed, 'rib-it' utterer

9/64 blue-eyed, 'knee-deep' mutterer

9/64 purple-eyed, 'rib-it' utterer

4/64 green-eyed, 'knee-deep' mutterer

3/64 purple-eyed, 'knee-deep' mutterer

Indicate the genotypes of the six F₂ phenotypes.

27/64 blue-eyed, 'rib-it' utterer

12/64 green-eyed, 'rib-it' utterer

9/64 blue-eyed, 'knee-deep' mutterer

9/64 purple-eyed, 'rib-it' utterer

4/64 green-eyed, 'knee-deep' mutterer

3/64 purple-eyed, 'knee-deep' mutterer

Of these, how many are controlling eye color? How can you tell? How many are controlling croaking?

Accepted Answer

Hi everyone. Let's take a look at this practice problem together. A geneticist cross fertilized to pea plants and obtain the following F. One progeny round seeded equal 77. Wrinkle seeded equal 77 The lil for around us. Big art is completely dominant over the little. For wrinkle nous little are. What is the probability of having a homogeneous round offspring from a cross between F one round and wrinkled. So we need to determine the probability of homos, I guess round offspring. And those would look like big R. Big R. By crossing two P plants from the F. One generation. So the f. one round and wrinkled. Now let's talk about how we're going to approach this problem before we even do the cross to figure out the F. One cross between the round and wrinkled plants. We need to know what the jena types are. So what are their Jenna types? We're not given that information. So how are we going to figure that out? Well, note that the phenotype is occurring in a 1-1 ratio. This is going to be important information and we're going to refer back to that in a bit. So to determine the F1 Generation Gina types, we're going to work backwards. So let me write out the steps of how we're going to do this. So step one, we are going to determine the parent gina types and we can do that using The F one generation phenotype information that is given to us. Step two will be to cross the parents and figure out the F one generation Jenna types. Then using that will determine the F two Generation Jenna types. And lastly we're going to figure out the probability of having a homogeneous round offspring. So step one determine the parent Jenna types. So let's run through some possible scenarios for the parents. So the parents, well, if both parents were homos, I guess dominant, all offspring would be homos. I guess dominant and they would all appear round at the same time. If both parents were homos, I guess recessive, all offspring would be homos I guess recessive and appear wrinkled. So we know that that's not a correct scenario. And so parent one at least one parent has to be hetero zegas. So they will appear big R. Little R. Now the second parents, genotype could be big R. Little R. It could be big R. Big R. Or little R. Little R. If you did the process for each type, you would see that the only combination that gives us offspring in a phenotype. IQ ratio of 1 to 1 would be a hetero ziggy's dominant parent with a homogeneous recessive parent. So that is going to look like the following will have a hetero Seaga's parent with the homos I guess recessive parent. And let me draw off this punnett square. And so this is our F1 generation offspring and we can see that there would be two that appear round and two that would appear wrinkled. And so that is that 1 to 1 phenotype ratio that we were given in The problem. If you did any of the other crosses, you would see that you would not get that phenotype IQ ratio appearing and so are round plant in our F one generation are going to be big r little R genotype And our wrinkle plant will be little r little r genotype. So now we can do step three determine the F. Two generation Jenna types. We'll take a note. The Jenna types for the F one generation are the same exact Jenna types that we did The cross to figure out the F. one generation. So our offspring for our F two generation are going to look exactly like our F. One generation. So F two generation, you would have two big R. Little R. Into little R. Little R. And so there are zero that appear to be home aside this round. And so that tells us our answer. Our answer is a 0%. Alright everyone, thanks for hanging in with me. I know this explanation was lengthy. Feel free to watch this video again. Um It may take you a second time for it to really sink in. But I hope this was helpful and I'll see you for the next practice problem

Verified video answer for a similar problem:

Key Concepts

Dihybrid Cross and Mendelian Ratios

Gene Interaction and Phenotypic Ratios

Determining Number of Genes from Phenotypic Classes