Researchers cross a corn plant that is pure-breeding for the dominant traits colored aleurone (C1), full kernel (Sh), and waxy endosperm (Wx) to a pure-breeding plant with the recessive traits colorless aleurone (c1), shrunken kernel (sh), and starchy (wx). The resulting F₁ plants were crossed to pure-breeding colorless, shrunken, starchy plants. Counting the kernels from about 30 ears of corn yields the following data.
What is the interference value for this data set?

Question

Researchers cross a corn plant that is pure-breeding for the dominant traits colored aleurone (C1), full kernel (Sh), and waxy endosperm (Wx) to a pure-breeding plant with the recessive traits colorless aleurone (c1), shrunken kernel (sh), and starchy (wx). The resulting F₁ plants were crossed to pure-breeding colorless, shrunken, starchy plants. Counting the kernels from about 30 ears of corn yields the following data.

What is the interference value for this data set?

Table displaying kernel phenotypes and their corresponding counts from a trihybrid cross in corn plants.

What is the interference value for this data set?

Accepted Answer

Hi, everyone. Let's look at our next problem. And before I read it, just to note that in this problem, the notation of the plus sign is used to represent the wild type alleles for a gene. So when I read plus, it refers to that symbol for the wild type allele. So here's our problem. Andro Sophal eyes, lowercase S E, curled wings, lowercase C U and ebony body, lowercase E are encoded by recessive genes Found on chromosome three. A researcher crosses S E C U E slash plus plus plus females with S E C U E slash S E C U E males and obtains the following progeny data. And then we have a table with genotype of offspring on the left side, all the different possible genotypes from this cross and number of offspring for each type on the right side. And we'll come back to this table after looking through the rest of the question here, determine the recombination frequency between C U and E. And our answer choices are a 2.6%. B 14%, 16.6% and D 19.4%. So let's look out, look at the crops that we're making here and the potential for recombination. And then we'll look through the table since that lists all the genotypes that result from all the different recombination possibilities. So we have a heterozygous female who has these recessive alls one copy of each S E C U and E our question tells us they're all on the same chromosome. So we write them all in a row above a line And the other copy of the gene she has are three pluses, three wild type, all plus plus plus. And in a typical task cross, she's crossed with a male whose homozygous recessive for all of the, all, all S E C U E and then over S C C U E. So we only need to look at recombination of the female because the male being homozygous recessive, any recombination that occurs will always result in offspring who have the same uh parental genotype and the alleys they inherit from him. So we have two possibilities for locations of crossovers and our heterozygous female. The first being between S E and C U. So I'm gonna mark a blue X in that spot to represent crossing over there. And the second being between the C U and the E all and I'm using a red X to mark crossing over there. And then of course, we could also have a situation where there were crossovers in both places. So with all that in mind, let's look over at our table and read through it seeing what the different genotypes are a result of. So, in the first two rows, we have offspring whose genotype are the same as the parental genotype. So in the first row, the offspring have inherited all the recessive alleles from the mother with no recombination. So their genotype is S E C U E slash S E C U E and that's 340 offspring. And in the rest of the rows, I won't read the second pair of alleles because we know that that homozygous recessive male will always give the three recessive alls. So all of them have S E C U E as that second, a set of values there. So I won't read them. We'll take that as a given. I'm just going to read the set of alls they inherited from the mother, which can vary In the second row. We have plus plus plus all the wild type alls and 300 offspring with that genotype. So these two rows, I'm going to notate with a letter P because they have the parental genotype with no accommodation. So now let's move on to the 3rd row which has S E C U plus with offspring. Well, that would result from a crossover between the C U and the E gene which we've marked with a red X. So I'm gonna put a red dot Next to that genotype to remember that, that takes place from that second crossover In the next row, I have SE plus plus with 110 offspring. Well, that would result from a crossover between S E N C U which I marked with blue. So I'm gonna put a blue dot next to that genotype. The next row after that plus C U E with offspring, that genotype would result from also a crossover between S E and C U. So I've marked it with a blue dot The next row is plus plus E with 80 offspring. That of course will result from a crossover between C U and E. So I mark that with a red dot Then our bottom two rows are going to be the result of both crossovers taking place, crossovers between S C and C U and C U and E. So I have S E plus E with 10 offspring and I'm gonna go ahead and mark that with two dots, red and blue. Do you remember that I have both crossovers there? And in the final row plus CU plus with offspring and again, mark that with the red and the blue dot So that's my entire table there. And I'm looking for the recombination frequency between C U and E. So recombination frequency we can recall can be calculated by taking the number of recombinant offspring divided by the total number of offspring. And then we express that as a 1%. So after we do that division, we will multiply by 100%. So now we're looking for the number of Recombinate recombinant offspring which represent a recombination having occurred between C U and E. So we will look at that we marked with a red X. So we will look for the number of offspring with red dots that's going to include both the offspring with just a single crossover between those two and the offspring where double crossovers have occurred because we're looking for anywhere that we see that crossing over. So as we saw in the, where the double crossovers are, you have plus the plus all and the E all together and in the other combination, the C U all and the plus all. So there has been recombining Between those two alleys. So first, let's add together the Offspring with just one crossover. That will be 60 for the S E C U plus and 80 from the plus plus E Genotypes. So 60 added to 80 gives us 140. And then we also need to add in the offspring that had two crossovers And in our last two rows, that is 10 and 16 when added together and make 26. So now our total number of recombinant offspring, we're gonna add those two numbers together 140 and 26, which will give us 166 total recombinant offspring. And the last thing we need is just the total number of offspring which we get by adding together all of the numbers of offspring in there the entire table. And when I do that, I get 1000 total offspring. So going down to my formula, I have the number of recombinant offspring, which is 166, Divided by the total number of offspring, 1000. And that will equal 0.16, six Times Equals 16.6% as my recombination frequency. So I'm going to go back to my answer. Choices and choice C was 16.6%. So choice C will be my answer for the recombination frequency between C U and E see you in the next video.

Verified video answer for a similar problem:

Key Concepts

Trihybrid Cross

Phenotypic Ratios

Genetic Interference