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.
Why are these data consistent with genetic linkage among the three genes?

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.

Why are these data consistent with genetic linkage among the three genes?

Table displaying kernel phenotypes and their corresponding counts from corn plant crosses.

Why are these data consistent with genetic linkage among the three genes?

Accepted Answer

Hi, everyone. Let's take a look at this question together. Assuming we have three link genes or a specific plant. The allele for purple flowers are dominant over the white flowers. The allele for tall plants is dominant over short plants and the allele for smooth stems is dominant over the hairy stems. A test cross is performed by crossing a plant with heterozygous P, heterozygous T and heterozygous S genotype with the true breeding plant with homozygous recessive P. Homozygous recessive T and homozygous recessive S genotype. They yield the following progeny. Purple, tall and smooth, 32 purple, tall and hairy. 29 purple, short and smooth, 50 purple, short and hairy, two white, tall and smooth, 10 white, tall and hairy, 45 white, short and smooth, 20 white, short and Harry 22 with a total progeny of 210 which of the following genes are more closely linked to each other. Is it answer choice. A gene P and gene T, answer choice B, gene P and gene S, answer choice C gene S and gene T or answer choice D, the three genes are of the same distance from each other. Let's work this problem out together to try to figure out which of the following answer choices contains the genes that are more closely linked to each other. So in order to figure out which of the following genes are more closely linked to each other, we have to figure out the recombination frequency of the different genes. And before doing so, we have to identify which option spring contain the parental alleles and which offspring are the double crossover allele combinations. So we can go ahead and identify our parental and double crossover alleles and we can identify each phenotype by a shorthand abbreviation. So our offspring with the parental alleles are purple, short and smooth that have 50 white, tall and hairy that have 45. Since the parental allele offspring are the ones that are most numerous. And our double crossovers are purple, short and hairy with two and white tall and smooth with 10. Since the double crossover allele combination offspring are those that are least numerous. And we also know that our recombination frequency formula, which we can abbreviate as RF is equal to the number of recombinant divided by the total number of offspring. And this is multiplied by 100%. And so we cross the two parental offspring, which we will use the shorthand abbreviation and properly arrange it. So we have parent one which is dominant, uppercase P, dominant, uppercase S and lower case recessive T. And we have our parent two, which is lowercase recessive P, lower case recessive S and dominant, uppercase T. And we want to use the frequencies from the combinations of P and S. So we have dominant P, recessive S as well as recessive P and dominant S which looking at our chart, we can see that the individuals that have either combination of P and S are the purple, tall and hairy, purple, short and hairy, white, tall and smooth and lastly white, short and smooth. And we add the value for those individuals. So we have 29 plus two plus 10 plus 20 which equals 61 divided by the total, which is 210 multiplied by 100% which means that the recombination frequency of P and S is equal to 0.29 or 29%. And next, we go ahead and do the same thing but for the alleles S and T, so we have our parent one and our parent two and we want to know the frequency for all combinations of S and T. So we have that uppercase dominant, S and uppercase dominant T as well as lower case recessive S and lower case recessive T, which are those individuals that are purple, tall and smooth, purple, short and hairy, white, tall and smooth and white, short and hairy. So we add 32 plus two plus 10 plus 22 which equals 66 divided by that total of 210 multiplied by 100% which means that the recombination frequency of S and T is equal to 0.31 or 31%. And so the recombination frequency of PNS is 29% whereas SNT is equal to 31%. And so the genes P and S are more closely linked than genes S and T, which means that answer choice B, gene P and gene S is the correct answer because we know that the recombination frequency of genes P and S is equal to 29% which is lower than the recombination frequency of gene S and T, which is equal to 31%. And so that means that gene P and gene S are more closely linked to each other than genes S and gene T as well as gene P and GT and the three genes are not the same distance from each other because gene P and gene S are more closely linked. So answer choice B 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

Phenotypic Ratios

Recombination Frequency