In pea plants, the appearance of flowers along the main ...

Question

In pea plants, the appearance of flowers along the main stem is a dominant phenotype called 'axial' and is controlled by an allele T. The recessive phenotype, produced by an allele t, has flowers only at the end of the stem and is called 'terminal.' Pod form displays a dominant phenotype, 'inflated,' controlled by an allele C, and a recessive 'constricted' form, produced by the c allele. A cross is made between a pure-breeding axial, constricted plant and a plant that is pure-breeding terminal, inflated.

Suppose that all of the F₂ progeny with terminal flowers, i.e., plants with terminal flowers and inflated pods and plants with terminal flowers and constricted pods, are saved and allowed to self-fertilize to produce a partial F₃ generation. What is the expected phenotypic distribution among these F₃ plants?

Accepted Answer

Hey, everyone. Let's take a look at this question together. A male is heterozygous for Pheno Keena. If he marries a female, that is also heterozygous for P K U and has four Children. What is the probability of having at least one child affected by P K U? Is it answer choice? A 76% answer choice. B 68% answer choice. C 50% or answer choice. D 80%. Let's work this problem out together to try to figure out which of the following answer choices is the correct answer. So we know that we have a male who is heterozygous for Pheno Ketonuria or P K U. And we have a female that is also heterozygous, which means that both parents r heterozygous and will be represented as big a little A. And we know that pheno Ketonuria or P K U is an autosomal recessive disorder, which that autosomal recessive means that the genotypes, homozygous dominant as well as that heterozygous genotype do not express the condition since it is autosomal recessive, we need that homozygous recessive genotype in order to express pheno ketonuria. And so now that we know that both parents are heterozygous. And that P K U is inherited in that Azal recessive inheritance pattern. Let's do a puit square to determine if the parents had four Children, what the probability of having at least one affected child would be. So, here we have the punnet square of the two heterozygous parents in which we can see that we have one out of four homozygous dominant, two out of four heterozygous and one out of four homozygous recessive, which as you can recall, and you can see on the screen that the homozygous dominant as well as the heterozygous do not express the condition, which means that we have those unaffected individuals making up three out of four in our puit square. And we have those affected individuals making up one out of four in the punt square. So in order to determine what the probability of having at least one child affected by P K U, if the parents had four Children, we take this unaffected number of three out of four And we take it to the fourth power because they want four Children. And we end up with 81 divided by 256 unaffected individuals. And in order to determine that probability, we take this 81 divided by 256 and subtract it from one. So we have 1 -81 divided by 256, which equals 0.68 or 68 as the probability of having at least one child affected by P K U, which means that answer choice B 68% is the correct answer. Because when you have two parents that are both heterozygous for P K U, which has an autosomal recessive inheritance pattern, we know that the unaffected individuals represent three out of four. And so we take that three out of four and set it to the power of four since parents have four Children and we end up with 81 divided by 256 which we then subtract that from one and we are given 0.68 or 68% which is the probability of having at least one child affected by pheno or P K U. So answer choice B 68% is the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Dominant and Recessive Alleles

Punnett Square and Genetic Crosses

Self-Fertilization and Generational Analysis

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