Question

In tigers, 24 units of enzyme activity are needed for the wild-type (striped) phenotype. Less than 24 units of activity results in a solid-colored tiger.

A1 is the WT allele and produces an enzyme with 16 activity units

A2 is the mutant allele and produces an enzyme with 10 activity units

Which allele is dominant, and what is the phenotype of a heterozygous tiger? Explain your answer and state whether this is an example of haplosufficiency or haploinsufficiency (referring to the wild-type allele).

Answer 1

Given, 24 units of enzyme activity are needed for the wild-type (striped) phenotype.

Less than 24 units of activity results in a solid-colored tiger.

A1 is the WT allele and produces an enzyme with 16 activity units

A2 is the mutant allele and produces an enzyme with 10 activity units.

The wild-type phenotype of a homozygous genotype A1A1 thus the result of the expression of two fully functional A1 alleles that produce 16 + 16 (= 32) activity units .

Then, a heterozygous genotype A1A2 with a functional allele and a non-functional allele produces 16 + 10 (= 26) activity units, resulting in wild-type phenotype.

The solid coloured tiger of a homozygous genotype A2A2 of two non-functional allele that produces 10 + 10 (= 20) activity units, results in a non-standard phenotype.

Here,

* allele A1 is dominant (since, it is producing a wild-type allele even in the presence of A2 allele)

* Phenotype of heterozygous: wild-type (striped).

* Wild-type allele can provide sufficient activity units to produce the wild-type phenotype. Therefore, it is HAPLOSUFFICIENT.