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.
Calculate the recombination frequencies between the gene pairs.

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.

Calculate the recombination frequencies between the gene pairs.

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

Calculate the recombination frequencies between the gene pairs.

Accepted Answer

Welcome back. We'll look at our next problem. And just to note as I read the problem that in this problem, the plus sign is used to represent the wild type, all of the genes we're discussing. So when I say plus, I mean the symbol for the wild type allele, so our problem says endo Sophal CPA I S 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. We have a table with on the left, the genotype of the offspring and on the right, the number of offspring for each genotype. And we'll return to that table in a minute after we go through the rest of our problem a little bit, then it says, determine the recombination frequency between S E and C U. And our answer choices are a 2.6%. B 19.4% C 22% and D 36%. So before we look at our table. Let's look at the genotypes of the parents. Now, we know we have a heterozygous female who is S E C U E, we know they're all on the same chromosome. So I've written them together in a line, then a draw line underneath them and then beneath that line is the plus plus plus of the three wild type alleys. So that's the female's genotype and she's crossed with a homozygous recessive male S E C U E and line and underneath the line, S E C U E. So when we think about recombination here and crossing over, it doesn't matter what goes on in the male's chromosomes because all of his alleys are the same. So any crossover, um recombination that occurs will not affect the genotype of the offspring. They'll always inherit those three recessive alleles from him. However, the female had to reigate those crossover patterns will result in different genotypes in the offspring. So we're going to look through our table all the different genotypes and see how that relates to crossing over events that may have occurred. So we have two potentials for crossing over between S E and C U that I'll mark with a blue X and between C U and E that I'll mark with a red X. So now let's turn to our table. The first two rows here are the offspring where no crossing over has occurred. So they've inherited one set of alleles or the other from the mother in the same combination as in a parental genotype. So the row, first row is S E C U E slash S E C U E with 340 offspring. Now going forward, I'm not gonna read the second set of alleles that are inherited from the father because those are always going to be the same. All the offspring will have S E C U E inherited from that homozygosis father, I'll just read the all inherited from the mother. So in the second row, we also have the parental genotype. So plus plus plus, inherited from the mother with 300 offspring. So I'm going to label these two rows with a P for parental type. The next row is SECU plus with six offspring and that indicates a crossover between C U and E. Since I mark that with a red X, I'm going to put a little red dot Next to that genotype to remind myself of which crossover that is. And the next row is SE plus plus with offspring. Well, that's a result of crossover between S E and C U marked with a blue cross. So I'm going to put a blue dot Next to that row, an extra is plus C U E with 84 offspring that will be between S E and C U. So blue dot The next row is plus plus E with 80 offspring. So that will be a red dot For that 2nd crosser. Now, my last two rows are a result of crossover between in both areas. So between S E and C U and between C U and E. So the next row is SE plus E with 10 offspring. And I'm gonna label that with a red dot and a blue dot To remember that it's the two crossovers And the row after that plus CU plus with 16 offspring. So again, a blue and a red dot Now how do we utilize this data in calculating the recombination frequency between S E and C U? Well, recombination frequency is calculated with dividing the number of offspring, excuse me, the number of recombinant offspring by the total number of offspring And then expressing that as 1%. So multiplying by 100%. So to calculate the number of recoin offspring, we refer to our table. So we'll need to look at offspring that have recombination occurring between S E and C U. So that will be both the offspring where there was one crossover between S E and C U and when there were two crossovers because those are all the offspring that have crossovers occurring there. So we marked that crossover between S E and C U with a blue cross and then a blue dot in our table. So we will add together those two batches of offspring who had one crossover between SE and CU marked with the blue. 110 added to 84, Which will give us 194 And then the two that had to do both crossover events. So down in our bottom of our table 10 plus 16, which is or 10 added to 16, which is 26. So now we add together both of these groups. So 194 and 26, Which is going to give us 220. And that would be our total number of rec competent offspring. So to calculate the recombination frequency, we're going to take Divided by the total number of offspring. So if we add together all the offspring on our table, we get 1000 total. So 220 divided by 1000 Will Equal 0.2, two. And then we're going to turn that into a 1%. So times 100 Equals 22%. So when we go over to our answer choices, choice C is 22%. Search choice C 22% is the recombination frequency between S E and C U. See you in the next video.

Verified video answer for a similar problem:

Key Concepts

Mendelian Genetics

Recombination Frequency

Phenotypic Ratios