In mice, apricot eyes is recessive to black eyes. Tail length is governed by another gene, linked to the eye color gene. Long tails is dominant to short tails. To determine the distance between the two genes, a double heterozygote is mated in a testcross and the classes of progeny produced were as follows: Apricot eyes, Long tails 33
Apricot eyes, Short tails 20
Black eyes, Long tails 17
Black eyes, Short tails 30
Determine whether the heterozygous parent is in the cis or trans arrangement.
a. Cis
b. Trans

trans

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.