Determine whether the statements below are true or false. If a statement is false, provide the correct information or revise the statement to make it correct.
If a dihybrid plant is self-fertilized,
(1) 9/16 of the progeny will have the same phenotype as the F₁ parent.
(2) 1/16 of the progeny will be true-breeding.
(3) 1/2 of the progeny will be heterozygous at one or both loci.

Question

Determine whether the statements below are true or false. If a statement is false, provide the correct information or revise the statement to make it correct.

If a dihybrid plant is self-fertilized,

(1) 9/16 of the progeny will have the same phenotype as the F₁ parent.

(2) 1/16 of the progeny will be true-breeding.

(3) 1/2 of the progeny will be heterozygous at one or both loci.

Accepted Answer

Everyone. Let's take a look at this practice problem together. A cross was performed between individuals with the big A big A big B, big B and little, a little, a little B, little B Gina types. The F one progeny of this cross were then self fertilized. What is the probability that the F2 progeny will have the same phenotype as their parents? They answer options are a 1/ B 3/16 C 6/16 and D 9/16. Now recall that the parental generation produces offspring called the F one progeny. And the F one progeny is offspring is called the F two progeny. For this problem, we are working from the top and going down. We know that the parental Gina types are big, a big A big B, big B and little, a little, a little B, little B. Therefore, we can solve for the F- one genotype and phenotype. Then we will self fertilize that F1 progeny To determine the F two genotype. At that point, we can then count how many of the F2 progeny share the same phenotype as that F1 progeny. So let's get started. Let's determine the F one genotype and phenotype by completing that di hybrid cross between the parents. We have a big, a big, a big B, big B crossed with individual little, a little, a little B, little B. The results of this cross will produce offspring all with genotype, big a little A big B, little B. The rationale for that is the homos, I guess dominant parent can only pass on dominant alleles and the homos, I guess recessive parent can only pass on recessive alleles. Therefore, all of the offspring would inherit one dominant A will and one recessive A will for both traits. That means that all of the F1 progeny are hetero zegas for both traits and the phenotype displayed would be the dominant. So the dominant genotype now we can self fertilize the F one progeny and that cross will be big a little A big B, little B cross with another big, a little, a big B, little B individual. But first, we need to determine the possible gametes that each of these parents can produce on the screen is the result of the punnett square that solves for the possible gametes on the top are the Lil's for the first trait. Big a little A and on the side are the alleles for the second straight. Big B, little B. There are four possible gametes. Big A Big B big, a little B, little, a little B and little A little B. Now we can complete that planet square for the F one progeny. Self fertilization, the results of that cross are on the screen. The next step is to analyze this planet square and determine how many of the offspring will display that dominant phenotype. Because remember the F one progeny displayed the dominant genotype for both traits and those individuals are being highlighted in yellow. So they are the individuals that display the dominant phenotype for both treats. And if we count those, there are nine out of 16 offspring that display that dominant phenotype. Therefore, the correct answer is option D 9/16. Alright. Everyone, I hope you found this helpful and I'll see you soon for the next practice problem.

Verified video answer for a similar problem:

Key Concepts

Dihybrid Cross

True-Breeding

Heterozygosity