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.
If the plants with terminal flowers produced by the cross in part (c) are saved and allowed to self-fertilize, what is the expected phenotypic distribution among the progeny?

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.
If the plants with terminal flowers produced by the cross in part (c) are saved and allowed to self-fertilize, what is the expected phenotypic distribution among the progeny?

If the plants with terminal flowers produced by the cross in part (c) are saved and allowed to self-fertilize, what is the expected phenotypic distribution among the progeny?

Accepted Answer

Welcome back, everyone. Let's look at our next problem. An individual with genotype heterozygous for genes A and B is test crossed with an individual homozygous recessive for genes A and B. If the two traits are completely linked, what is the expected phenotypic ratio of the offspring? And our answer choices are A 9 to 3 to 3 to choice B 1 to 1 choice C, 3 to 1 or choice D, 1 to 2 to one. Well, in this problem, we are talking about a dy hybrid test cross since we have a parent who's a die hybrid heterozygous for two traits. However, we have traits that are completely linked. So if we imagine our cross here, here's our heterozygous parent capital, a lowercase A capital B, lowercase B with our homozygous recessive parent. Well, our homozygous recessive parent only has one type of gamete, little, a little B, that's the only gammy they can pass on to their offspring. If the traits were not linked, our heterozygous parent would have four possible gametes, dominant for both dominant. A recessive B, dominant B, recessive A and then recessive for both. That would be if we had an independent assortment. But we don't, we're told our traits are completely linked. They're right next to each other on the chromosome, we can imagine and always get passed together. Now, there's one little wrinkle here, we have to assume what the parental genotypes were of our header as I get. But again, since we know that our traits are completely linked and inherited together, then we can make that assumption that the parents here were homozygous dominant, crossed with homozygous recessive to give us our heterozygote. So again, since our traits are completely linked, the only gas possible from our heterozygous parent are then dominant for both recessive for both. So I'm going to cross out those gametes that would show homozygous for one and recessive for the other. So when we think about the possibilities for phenotypes for our offspring, 100% of the time they inherit the recessive all from the homozygous recessive parent, 50% of the time they inherit the two dominant alleys and 50% of the time they inherit the two recessive alleles. So that means one half of the offspring will be heterozygous by inheriting the dominant allies from the heterozygous parent and the recessive allele from the homozygous recessive parents and half of the offspring will be homozygous recessive. And what does that mean for the phenotype? Well, half the offspring will show the dominant phenotype since they're heterozygous and half of the offspring will show the recessive phenotype. So that gives us a phenotypic ratio of choice B 1 to 1 equal amounts showing the dominant and recessive phenotypes. Let's look at why our other answer choices are incorrect. Well, choices C and D would refer to ratios of a mono hybrid cross choice C 3 to 1 is going to refer to the phenotypic ratio and choice D 1 to 2 to one, the genotypic ratio of a mono hybrid cross. And that would be heterozygote cross with heterozygote. So like say heterozygous for a cross with heterozygous for a and our choice A 9 to 3 to 3 to 1 is our phenotypic ratio for a die hybrid cross with independent assortment. And importantly, when both parents are heterozygous for both traits. So a different kind of cross than we're looking for, which is a test cross, a di hybrid test cross. So that's why AC and D are not correct. And again, if we have a heterozygous individual test crossed with a homozygous recessive individual and the two traits are completely linked. Our expected phenotypic ratio of the offspring is choice B 1 to 1. See you in the next video.

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Key Concepts

Mendelian Genetics

Punnett Square

Phenotypic Ratios