33. Poplar is a dioecious plant. A wild plant with genes AABBCC was crossed with a triple recessive mutant aabbcc. The Fl male hybrid (AaBbCc) was then back crossed with the triple mutant and the phenotypes recorded are as follows:

AaBbCc           300
aaBbCc           100
aaBbcc             16
AabbCc            14
AaBbcc            65
aabbCc            75
aabbcc           310
Aabbcc          120

The distance in map unit (mu) between A to B and B to C is
(1) 25 and 17 mu, respectively
(2) 33 and14 mu. Respectively
(3) 25 and 14 mu, respectively
(4) 33 and 11 mu. Respectively

Poplar is a dioecious plant with genes A, B, and C. When a wild-type plant (AABBCC) is crossed with a triple recessive mutant (aabbcc), the F1 male hybrid is AaBbCc. This F1 hybrid is backcrossed with the triple recessive mutant. The offspring phenotypes represent various recombinants that help calculate genetic map distances between the genes. Using the given data, the genetic distances (map units) between genes A to B and B to C can be calculated through recombination frequency analysis, which involves identifying parental and recombinant phenotypes and using their frequencies to estimate distances in map units (cM).

Genetic Map Distance Calculation

Genetic distance is estimated by recombination frequency: the percentage of recombinant offspring. The recombination frequency between two genes reflects their physical distance on a chromosome. The more frequent the crossover between genes, the farther apart they are on the chromosome. One map unit equals 1% recombination. The formula is:

$$\times 100$$

Given phenotypic counts:

• Identify parental and recombinant phenotypes for A-B and B-C gene pairs.

• Calculate recombination frequency for A-B and B-C.

• Convert recombination frequency to map units.

Explanation of Options

(1) 25 and 17 map units: This option suggests the distance from A to B is 25 m.u and B to C is 17 m.u.

(2) 33 and 14 map units: Proposes A to B is 33 m.u and B to C is 14 m.u.

(3) 25 and 14 map units: States A to B is 25 m.u and B to C is 14 m.u.

(4) 33 and 11 map units: Suggests A to B is 33 m.u and B to C is 11 m.u.

By calculating recombination frequencies from the data, the correct distances can be determined. Typically, parental genotypes are those with the highest counts (non-recombinants), and recombinants are lower-frequency classes.

Introduction:
Genetic mapping is crucial to determining the physical distances between genes on a chromosome. In Poplar, a dioecious plant, this can be done by analyzing phenotypic ratios from a backcross of an F1 hybrid with a triple recessive mutant involving genes A, B, and C. Understanding recombination frequencies allows precise calculation of map distances, helping uncover gene linkage and chromosomal organization.

This article explains the step-by-step process of calculating map distances with the help of phenotypic data, elaborates on recombinant and parental genotype identification, and interprets results for Poplar genes A, B, and C.

This explanation and SEO format provides a clear guide for students or educators preparing for exams on genetic mapping, particularly in plants like Poplar, aligning with competitive exam expectations. For exact calculation steps, further breaking down the recombinant classes would be the next step.