In tomato plants, the production of red fruit color is under the control of an allele R. Yellow tomatoes are rr. The dominant phenotype for fruit shape is under the control of an allele T, which produces two lobes. Multilobed fruit, the recessive phenotype, has the genotype tt. Two different crosses are made between parental plants of unknown genotype and phenotype. Use the progeny phenotype ratios to determine the genotypes and phenotypes of each parent.
Cross 1 progeny:
3/8 two-lobed, Red
3/8 two-lobed, yellow
1/8 multilobed, Red
1/8 multilobed, Yellow
Cross 2 progeny:
1/4 two-lobed, Red
1/4 two-lobed, yellow
1/4 multilobed, Red
1/4 multilobed, yellow

Question

In tomato plants, the production of red fruit color is under the control of an allele R. Yellow tomatoes are rr. The dominant phenotype for fruit shape is under the control of an allele T, which produces two lobes. Multilobed fruit, the recessive phenotype, has the genotype tt. Two different crosses are made between parental plants of unknown genotype and phenotype. Use the progeny phenotype ratios to determine the genotypes and phenotypes of each parent.
Cross 1 progeny:
3/8 two-lobed, Red
                    3/8 two-lobed, yellow
                    1/8 multilobed, Red
                    1/8 multilobed, Yellow
Cross 2 progeny:
1/4 two-lobed, Red
                     1/4 two-lobed, yellow
                     1/4 multilobed, Red
                     1/4 multilobed, yellow

Cross 1 progeny:

3/8 two-lobed, Red

3/8 two-lobed, yellow

1/8 multilobed, Red

1/8 multilobed, Yellow

Cross 2 progeny:

1/4 two-lobed, Red

1/4 two-lobed, yellow

1/4 multilobed, Red

1/4 multilobed, yellow

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together. Any fish, gold skin color. Capital G is dominant to black skin color, lowercase G and split tail fin. Capital S is dominant to single tail fin. Lowercase s breeding between two fish of unknown genotypes results in the following phenotypic ratios in the progeny. 1 4th gold split tail fin, 1 4th gold single tail fin, 1 4th black, split tail fin and 1 4th black, single tail fin. What are the genotypes of the two fish used for breeding? Our answer options are a capital G capital G, capital S capital S for both parents, B capital G, lowercase G, capital S and lowercase S and capital G capital G, capital S capital S C capital G, lowercase G capital S lowercase S and lowercase G, lowercase G, lowercase S, lowercase S and option D lowercase G, lowercase G, lowercase S lowercase S for both parents. So from the question, we know that the phenotypic ratio is 1-1-1-1 in the offspring. We know that the recessive phenotype which is the black skin color and the single tail fin. We know those are expressed in the progeny. We also know that the only way a recessive phenotype is expressed is when both ails are recessive. Therefore, each parent must have at least one recessive A lil for each trait. So a recessive A lil for skin color, lowercase G, one lowercase G and the recessive A lil for fin type one case. S So again, each parent must have at least one recessive ali for each trait Knowing this, we can eliminate option A. Since both parents in option A are homozygous dominant. For both traits, there are no recessive ails in the genotypes. Now, furthermore, we know that the dominant phenotype is also expressed the gold skin color and the split tail fin are expressed in the progeny. Now remember the dominant phenotype can be expressed with just one dominant a little in the trait. Therefore, this means that one parent must have at least one capital G or one dominant A lil for skin color and one dominant A lil for the fin or one capital S. Now we know that we can eliminate option D, since both genotypes in this option only contain recessive ally. So we're left with either option B or option C. Now note that in option B, the second parent capital G, capital G, capital S, capital S does not have any recessive allys. Remember from the question that the recessive phenotype is expressed for both traits. This means that option B is incorrect there are no recessive ails in that second parent or the second parent's genotype to pass on. Therefore, all offspring of that cross would display only the dominant trait. That means the correct answer is option C Now, I love these types of questions because if you wanted to be 100% sure, without a doubt, you could take the genotypes of both parents in option C and fill out a punt square and confirm that the offspring will have a phenotypic ratio of 1 to 1 to to 1. So that's just no way for you to know on how to confirm that you have the correct answer. But for us, since we've deduced it together, we know that the answer must be option C capital G, lowercase G, capital S, lowercase S and lowercase G, lowercase G, lowercase S, lowercase S All right. Everyone. I hope you found this helpful and I'll see you soon for the next practice problem.

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Key Concepts

Mendelian Genetics

Phenotypic Ratios

Genotype Determination