In experiments published in 1918 that sought to verify and expand the genetic linkage and recombination theory proposed by Morgan, Thomas Bregger studied potential genetic linkage in corn (Zea mays) for genes controlling kernel color (colored is dominant to colorless) and starch content (starchy is dominant to waxy). Bregger performed two crosses. In Cross 1, pure-breeding colored, starchy-kernel plants (C1 Wx/C1 Wx) were crossed to plants pure-breeding for colorless, waxy kernels (c1 wx/c1 wx). The F₁ of this cross were test-crossed to colorless, waxy plants. The test-cross progeny were as follows:
In Cross 2, plants pure-breeding for colored, waxy kernels (C1 wx/C1 wx) and colorless, starchy kernels (c1 Wx/c1 Wx) were mated, and their F₁ were test-crossed to colorless, waxy plants. The test-cross progeny were as follows:
Calculate the recombination frequency for each of the progeny groups.

Question

In experiments published in 1918 that sought to verify and expand the genetic linkage and recombination theory proposed by Morgan, Thomas Bregger studied potential genetic linkage in corn (Zea mays) for genes controlling kernel color (colored is dominant to colorless) and starch content (starchy is dominant to waxy). Bregger performed two crosses. In Cross 1, pure-breeding colored, starchy-kernel plants (C1 Wx/C1 Wx) were crossed to plants pure-breeding for colorless, waxy kernels (c1 wx/c1 wx). The F₁ of this cross were test-crossed to colorless, waxy plants. The test-cross progeny were as follows:

In Cross 2, plants pure-breeding for colored, waxy kernels (C1 wx/C1 wx) and colorless, starchy kernels (c1 Wx/c1 Wx) were mated, and their F₁ were test-crossed to colorless, waxy plants. The test-cross progeny were as follows:

Calculate the recombination frequency for each of the progeny groups.

Table displaying phenotypes and their corresponding numbers from genetic mapping experiments in corn.

In Cross 2, plants pure-breeding for colored, waxy kernels (C1 wx/C1 wx) and colorless, starchy kernels (c1 Wx/c1 Wx) were mated, and their F₁ were test-crossed to colorless, waxy plants. The test-cross progeny were as follows:

Table displaying phenotypes and their corresponding numbers: Colored, waxy 340; Colored, starchy 115; Colorless, waxy 92; Colorless, starchy 298.

Calculate the recombination frequency for each of the progeny groups.

Accepted Answer

Hello, everyone and welcome to today's video. So, in a genetic study of peas, a researcher is interested in two traits, seed color and the shape. The researcher starts with two pure breeding pea plants, one with yellow round seats and the other with green wrinkled seeds. The researcher performs a di hybrid cross between the two plants and obtains an F one progeny that are all yellow and brown. The researcher then crosses the F one progeny to a pure breeding pre plan with green wrinkled seeds, seeds and obtains the following F two progeny. And so we are given the phenotypic numbers for the yellow and brown, yellow and wrinkled, green and brown and green and wrinkled phenotypes. Considering the given data, the recombination frequency of the seed color is blank and the recombination frequency for this seed shape is blank. So as answer choice A, we have 30% comma 41% B is 6.25% comma 5% C, 15% comma 20% D is 31% comma 10%. But in order to solve this problem, we first need to remember or recall from previous videos, what is going to be the formula forges the recombination frequency between these genes. And this is going to be that the recombination frequency is going to be equal to the number of recombinant divided by the to number of offsprings times 100%. So we obtain a percentage. So for example, if we need to calculate the recombination frequency of say seed color, this is going to be remember that the recombinant for seed colors are going to be these green and brown and these yellow and wrinkled, these are going to be our recombinant w green and wrinkled and yellow and brown are going to be what we call the parent phenotypes. And so five plus 20 which is the total number of recombinant divided by 400 individuals which we have in a population times 100% is going to be equal to 6.25%. And this is going to be the real combination of frequency of see color moving on. Let's calculate their combination of frequency of C shape. Now, the rec convenience for the shape are go is only going to be this yellow in wrinkled, which is a total of 20 individuals. And therefore we have the 20 or the number of recoin times or divided by the total number of offspring, which is 400 times 100%. And this is going to be equal to 5%. And that is exactly the recommendation frequency that we're going to be adding here. And therefore the correct answer for this question is going to be b 6.25% comma 5%. I really hope this helped you and I hope to see you on the next video.

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Key Concepts

Genetic Linkage

Recombination Frequency

Test Cross