Question

In a flowering plant, gene "A" encodes an enzyme responsible for the presence of dots on the flowers’ petals. A1, A2, and A3are the three known alleles of this gene; A1 is the wild type, A2 is a null allele, while A3 has a mutation in the promoter region of the gene, which results in the synthesis of very little gene product.If the "A" gene is haploinsufficient, what is the predicted phenotypic ratio in the F1 of a cross between a wild type and a A2/A3 heterozygous plant ? (please include your explanation)1. 100% of the F1 plants will have flowers with dots.2. 25% of the F1 plants will have flowers with dots, and 75% will have flowers with no dots3. 50% of the F1 plants will have flowers with dots, and 50% will have flowers with no dots.4. 50% of the F1 plants will have flowers with a few dots, 25% will have flowers with normal amounts of dots, and 25% will have flowers with no dots.5. 100% of the F1 plants will have flowers with no dots.

Answer 1

Answer:

5. 100% of the F1 plants will have flowers with no dots.

Explanation:

When a gene is haploinsufficient, the wild type allele needs to be present in two copies to produce the phenotype. In heterozygous condition, the phenotype wont be produced even though one wild type allele is present. The product produced by single wild type allele is insufficient to produce the phenotype.

Here, gene A is haploinsufficient. Wild type allele A1 will produce flowers with dots when it is in two copies. Null allele A2 and mutant allele A3 wont produce flowers with dots. If a wild type (A1A1) is crossed with A2A3 :

        A1        A1

A2  A1A2   A1A2

A3  A1A3   A1A3

Since all the plants in F1 progeny have only one wild type A1 allele, its product will be insufficient to produce dots pattern. 100% of the plants will have flowers with no dots.