A wild-type trihybrid soybean plant is crossed to a pure...

Question

A wild-type trihybrid soybean plant is crossed to a pure-breeding soybean plant with the recessive phenotypes pale leaf (l), oval seed (r), and short height (t). The results of the three-point test cross are shown below. Traits not listed are wild type.

Phenotype Number
Pale 648
Pale, oval 64
Pale, short 10
Pale, oval, short 102
Oval 6
Oval, short 618
Short 84
Wild type 98
1630

What are the alleles on each homologous chromosome of the parental wild-type trihybrid soybean plant? Place the alleles in their correct gene order. Use L, R, and T to represent dominant alleles and l, r, and t for recessive alleles.

Accepted Answer

Hey, everyone. Let's take a look at this question together. What is the likelihood of having offspring with the genotype homozygous dominant A B and C in a tri hybrid cross involving three independently assorting genes. Is it answer choice? A one out of four answer choice B, one out of eight, answer choice C one out of 16 or answer choice D one out of 64. Let's work this problem out together to try to figure out what the likelihood of having offspring with the genotype homozygous dominant for all three genes would be. So first, we will assume that each parent in this tri hybrid cross is heterozygous for all three genes. And after that assumption, we know that the probability of the homozygous dominant for A is one out of four, since each parent is heterozygous for A and since they are also heterozygous four B and C, that means that the probability of that homozygous dominant four B is also one out of four. And the probability of the homozygous dominant or C is also one out of four. And since we want to know what the probability is of the offspring having the genotype of that homozygous dominant. For all three genes, we are going to use the product which states that the probability of each individual gene can be multiplied to get that total probability. So we have one out of four, multiplied by one out of four, multiplied by one out of four. Since this is the probability of homozygous dominant. A homozygous dominant B and homozygous dominant C which when we multiply one out of four by one out of four, by one out of four, we end up with a total of one out of 64 which is that probability of that homozygous dominant genotype or A B and C making answer choice. D one out of 64 the correct answer since we want to know the probability of the offspring having that Homozygous dominant. A homozygous dominant B and Homozygous dominant C. And we can assume that each parent is heterozygous, which the probability of that homozygous dominant for each gene is one out of four, for a one out of four, for B and one out of four for C. So we multiply one out of four by one out of four by one out of four, which gives us a probability of one out of 64 having that genotype of Homozygous dominant for A B and C. So answer choice, D one out of 64 is the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Trihybrid Cross

Phenotypic Ratios

Linkage and Gene Order

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