In a diploid plant species, an F₁ with the genotype Gg Ll Tt is t...

Question

In a diploid plant species, an F₁ with the genotype Gg Ll Tt is test-crossed to a pure-breeding recessive plant with the genotype gg ll tt. The offspring genotypes are as follows: Genotype Number Gg Ll Tt 621 Gg Ll tt 3 Gg ll Tt 64 Gg ll tt 109 gg Ll Tt 103 gg Ll tt 67 gg ll Tt 7 gg ll tt 626 1600 Calculate the recombination frequency between each pair of genes.

Table showing offspring genotypes and their counts for a trihybrid cross in genetics.

Accepted Answer

Hi, everyone. Welcome back. Here's our next problem. Just a quick note before I read it when I read the word plus, uh assume that that refers to the wild type genotype in the problem. I'm reading in roopa C P I S gene, S E curled wings, gene C U and ebony body gene 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 now we have a table on the left side is listed. The genotype of the offspring on the right side is the number of offspring. I am not going to read the entire genotype of each offspring genotype. Since we know that the male being homozygous recessive can only pass on the recessive alleles. So that set of all from the male will be the same in all offspring. The only thing that will vary is which genes they inherit from the heterozygous female parent. So I'm just going to read the all they inherited from the heterozygous parent. So here's our table, S E C U E 340 plus plus plus 300 S E C U plus 60 S E plus plus plus C U E 84 plus plus E 80 S E plus E 10 and plus C U plus 16. Then our problem says, determine the number of double crossovers with answer choices. A 26 B 140 C 194 and D 360. So there's actually a really quick shortcut to this, that we can go over when you have a list of genotypes like this. And you have three genes. I had her, I get in the three traits and you have a table with all the genotypes and number, we know that the least frequent occurrence will be a double crossover because you have to have a crossover event occurring between all three genes or twice between the two, first two genes and the second two genes. So that will be less common than a single crossover or no crossovers. And therefore the two genotypes that represent the result of a double crossover will give us the lowest number of offspring. So we can look at our table and pick the two lowest numbers and assume they represent double crossovers. So in our table, we have the bottom two rows, S E plus E with 10 and plus C U plus with 16. So we can assume those are double crossovers And if we add them together, 10 added to 16, we get a total of 26 offspring with double crossovers. That's choice. A here 26. If we just want to be very thorough or we're a little unsure, we can look at our initial genotype again, we can discount the homozygous recessive male and just look at the heterozygous parent. So we'll write S E C U E representing one set of values and a line and then underneath plus plus plus for the three wild type alls. So let's represent the two crossovers occurring between S E and C U drawn X to represent the first crossover between C U and E A second X to represent the second crossover. And then let's list the resulting genotypes of both crossovers occurring. And that would be indeed S E plus E N plus C U plus. And we note that a double crossover leaves the two flanking alleles being the same as in the parental genotype and just swaps the middle alley. And we confirm that those are the genotypes listed as the two lowest numbers on our table. So again, the number of double crossovers would be 10 added to 16 resulting in choice A 26 being our answer. See you in the next video.

Key Concepts

Test Cross

Recombination Frequency

Linkage and Genetic Mapping

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