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:
Merge the two sets of progeny data and determine the combined recombination frequency.

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:

Merge the two sets of progeny data and determine the combined recombination frequency.

Table displaying phenotypes and their corresponding numbers from genetic experiments on 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.

Merge the two sets of progeny data and determine the combined recombination frequency.

Accepted Answer

Hey, everyone. Let's take a look at this question together in an experiment. Researchers are studying two genes in mice that control fur color and eye color genes are located on the same chromosome and the alleles for the genes are as follows. In one black fur is dominant to brown fur and gene two red eyes are dominant to black eyes. In the F one generation of a cross between two pure breeding parents, all the offspring had black fur and red eyes. The F one mice were then crossed with homozygous recessive B and homozygous recessive R mice to produce the F two generation out of 100 F two mice. The researchers observed the following phenotypes, black fur with red eyes, 48 black fur with black eyes, 16, brown fur with red eyes, 20 and brown fur with black eyes. 16. What is the recombination frequency between the two genes? Is it answer choice? A 16% answer choice. B 20% answer choice C, 36% or answer choice D 64%. Let's work this problem out together to try to figure out which of the following answer choices represents the recombination frequency between the two genes. Since we're trying to find out the recombination frequency between the two genes, we must first recall what the recombination frequency formula is. So we have the recombination frequency which we can abbreviate as R F is equal to the number of recombinant divided by the total number of offspring multiplied by 100%. And in order to figure out our number of recombinant, we must first identify our offspring that have the parental alleles, which we know are the individuals that have black fur and red eyes as well as brown fur and black eyes. So that means that our individuals with black fur and black eyes and brown fur and red eyes are the recombinant. And our other individuals are the ones that have the parental els. And so that means that our number of recombinant is 16 plus 20. So our recombination frequency or R F is equal to 16 plus 20 Divided by the total number of offspring which we know it is out of 100. So we have 100 as the total offspring Multiplied by 100%. So our recombination frequency or RF is equal to 36 divided by 100 multiplied by 100%, which we note that 36 divided by 100 is equal to 0.36 multiplied by 100%. So that means that our recombination frequency is equal to 36%. So that means that answer choice C 36% is the correct answer because we know that the recombination frequency or R F is calculated as the number of recombinant divided by the total number of offspring multiplied by 100%. And the number of recombinant is 16 plus 20 and our total number of offspring is 100. So our recombination frequency is calculated as 36 divided by 100 multiplied by 100%, which gives us our recombination frequency of 36%, which is our final answer. So answer choice C36% is the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

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Key Concepts

Genetic Linkage

Recombination Frequency

Test Cross