In experiments published in 1918 that sought to verify a...

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:

Phenotype Number
Colored, waxy 310
Colored, starchy 858
Colorless, waxy 781
Colorless, starchy 311
2260

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:

Phenotype Number
Colored, waxy 340
Colored, starchy 115
Colorless, waxy 92
Colorless, starchy 298
845

Taken together, are the results of these two experiments compatible with the hypothesis of genetic linkage? Explain why or why not.

Accepted Answer

Hello, everyone and welcome to today's video. So in a study of the inheritance of two genes in pea plants, a researcher crossed a pure breeding plant with yellow brown seats to a pure breeding plant with green wrinkled seats. All of the F one Austrian had yellow round sets. The researcher then crossed the F one to each other to produce an F two generation. The results of the F two generations are as follows. We have the observed phenotype number for yellow and round, yellow and wrinkled, green and brown and green and wrinkled for a total of 1600 asp. So using the K square test, which of the following statements can be true as answer choice. A we have that the genes are linked b the genes are close to each other. C the genes are in the same chromosome D. All options can be true. So in order to calculate this chi square value, we need the sum of the observed phenotypes value, which what we are provided in the problem minus D expected, which we need to calculate somehow divided by the expected and then this is going to be squared and then this continues until we get the sum of all four of these phenotypes. So how do we obtain these expected values? Well, remember that in this cross of the F one generation, we're co we're performing a die hybrid cross, we are performing a cross between two individuals which are heterozygous for two genes. And this cross is going to have a 9 to 3 to three to one phenotypic ratio out of 16. So nine out of 16 individuals are supposed to be yellow and round or are expected to be yellow and round three out of 16 are supposed to be yellow and wrinkled. Three out of 16 are supposed to be green and round and one over 16 is supposed to be rain and wrinkled. We multiply that by the total number of individuals and we're going to obtain the expected number for each phenotype and for the yellow and brown, it is going to be 900 for the yellow and wrinkled. It is going to be 300 the same is going to be for the green and brown and for the green and wrinkled, we're going to have it be 100. Now we have the expected as well as the observed value. So we can calculate these guys core value. Now, in order to do this, we as we said, need to calculate the sum of the observer minus the expected square divided by the expected as we see here in this formula. So let's do that for each of the phenotypes. So for the yellow and round, the observed is going to be 602 minus the expected which is 900 score over 900. Now, let's keep going. We have now for the yellow arm wrinkled which is minus 300 square over 300. Let's keep going. Now we have for the green and round which is minus 300 square over 300. And finally, we have the ring and wrinkled, which is going to be minus 100 square divided by 100. And this is going to give us a K square value of 132 co two with three degrees of freedom. And we find that the P value for these is going to be or the critical value for P equals 0.5 is going to be 7.8 15. Since our calculated K square value is much larger than the critical value, we can reject the hypothesis, we can reject the hypothesis of independent assortment and conclude that the two genes are going to be linked. And therefore, because all the two genes are going to be linked, we can say that first of all, as Spencer choice, a the genes are linked because they are linked, they're going to be inherited together. So they're going to be close to each other. And because they're close to each other, they're going to be in the same chromosome. And therefore all options can be true. Or answer choice. D is the right answer for this question. I really hope this video helped you and I hope to see you on the next one.

Key Concepts

Genetic Linkage

Recombination Frequency

Test Cross

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