Answer:
trans
Step-by-step explanation:
From the given information:
The study observes the genes present in mice for eye color and tail length. Since both genes are linked, it implies that they exist in the same chromosomes.
Black eyes is dominant over apricot eyes
Let Black eye be B and apricot eyes be b
Long tail is dominant over short tail
Let long tail be L and short tail be l
If double heterozygote(homoozygous-recessive) engage in the testcross
Then:
From the result given:
The parental combinations are:
Apricot eyes, Longtail (bL / bl) = 33
Black eyes, Short tails (Bl / bl) = 30
The recombinant genes are:
Black eyes, Long tails (BL / bl) = 17
Apricot eyes, Short tails (bl / bl) = 20
The recombination frequency relates to the distance between the two genes which can be computed as:
= (20+17)/100
= 37%
Thus; the heterozygous parent is in trans arrangement.