There is a population of alien kitties, species Cattus really-defty, that live o
ID: 161565 • Letter: T
Question
There is a population of alien kitties, species Cattus really-defty, that live on the planet Scratch. Their genomes and biology are amazingly identical to the genomes and biology of earth cats. Consider a di-hybrid cross, where the first locus has alleles A and a, and second locus has alleles B and b. Each additional copy of either an A or B allele in the two-locus genotype increases the adroitness of the cat. Adroitness is measured by the Liverpool-LI Diagnostic Juggling Measurement Instrument (LLDJMI). There are five categories:
0 = Minimally adroit (has difficulty walking across the back lawn);
1 = Somewhat adroit (can walk, but has difficulty catching mice);
2 = Adroit (can catch mice, birds, small animals easily);
3 = Very adroit (can join the circus and get a high-paying job).
4 = Extremely adroit (usually works as a Formula I race-car driver).
As an example, an alien cat with a two-locus genotype of AAbb is categorized as the Adroit phenotype (Category = 2), since there is two copies of A and zero copies of B, adding to a score of 2 for the LLDJMI.
Similarly, an alien cat with a two-locus genotype of AaBB has a Very adroit phenotype (Category = 3) (one copy of A + two copies of B = three copies total of A or B).
In the following mating: AaBb x AaBb, what is the expected proportion of offspring that will be Adroit?
Explanation / Answer
As the score of 2 gives Adroit phenotype, by crossing AaBb x AaBb we will focus on genotypes having score of 2 to give the desired phenotype.
AaBb x AaBb = 1 AABB, 2AABb, 1AAbb, 2AaBB, 4AaBb, 2Aabb, 1aaBB, 2aaBb, 1aabb = 16 genotypes.
Of these 1AAbb, 4AaBb, 1aaBB will give Adroit phenotype. So in total 6/16 ~0.375 offspring will be adroit.
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