Question

Agapanthus is a dominant, homozygous lethal trait among Telorites from the planet Teleflora. Subjects with this condition love to send flowers. From a group of Telorites heterozygous at the Agapanthus locus (with two alleles: A = dominant (trait) allele, a = recessive (wild-type) allele), 88% of the Telorites have the Agapanthus phenotype, and 12% look wild-type.

Also, true-breeding wild-type Telorites (aa genotype) have 0% penetrance.

A Telorite with the Agapanthus phenotype mates with a true-breeding Telorite with the wild-type phenotype. One of the F₁ children is randomly selected and is mated back to the true-breeding wild-type parent. Out of twenty F₂ individuals, we observe the following counts:

1. Eight F₂ individuals have the Agapanthus phenotype;
2. Twelve F₂ individuals have the Wild-type phenotype.

The genotype of the randomly-selected F₁ subject must be:

A. Aa or aa.

B. We cannot say with certainty. We need to perform a Chi-square goodness of fit test.

C. AA.

D. aa.

E. Aa.

Answer 1

Answer: "E" (Aa)

The genotype of the randomly-selected F₁ subject must be: "Aa".

Explanation:

Agapanthus is a dominant, homozygous lethal trait. Agapanthus loci have the two alleles.

A: Dominant allele

a: recessive allele

Individual phenotype AA: Lethal

Aa: Agapanthus phenotype

aa: Wild type (True breeding phenotype)

Aa x aa

  ↓

F1 1Aa: 1aa

One of the F₁ children is randomly selected and is mated back to the true-breeding wild-type parent.

Out of twenty F₂ individuals, we observe the following counts:

1. Eight F₂ individuals have the Agapanthus phenotype;
2. Twelve F₂ individuals have the Wild-type phenotype.

Condition 1: If the child is "aa"

aa x aa

   ↓

F2 all are wild type.

So condition 1 is not true when compared with the given data. In this condition, Agapanthus phenotype was not observed. SO the selected F1 is not "aa".

Condition 2: If the child is "Aa"

Aa x aa

   ↓

F2 1Aa (50% Agapanthus phenotype): 1 aa (50% wild type). There is 0% penetrance in "a" allele.

Expected outcome: out of twenty, 10 Agapanthus phenotype: 10 wild type

but observed is: out of twenty, 8 Agapanthus phenotype: 12 wild type

By chi-square test: The given hypothesis is accepted (Chi-square analysis was performed)

Degree of freedom: n-1: 2-1=1

Critical value at 0.05 is 3.84. Tota X² value is 0.8 (0.4+0.4). 0.8 less than 3.84. So a given hypothesis was accepted.