Start Learning Free

Before You Begin

Question #45824

Difficulty: 8/10 | 🟢 Free

The Question

A geneticist conducts a test cross to determine the recombination frequency between two linked genes in a Citrus variety. Upon analyzing the germinated seeds, the phenotypic ratio completely collapses, yielding an overwhelming excess of progeny that are genetically identical to the maternal parent. What is the most accurate biological explanation for this failure of linkage mapping?

Options

  • A.

    The genes exhibit complete linkage due to their extreme proximity, completely suppressing recombinase activity during the pachytene stage.

  • B.

    Diploid nucellar cells divided mitotically and protruded into the embryo sac, forming apomictic embryos that bypassed meiotic crossing over.

  • C.

    The zygote underwent repeated mitotic cleavages to form multiple embryos, which dominantly masked the expression of recombinant phenotypes.

  • D.

    The primary endosperm nucleus (PEN) differentiated into multiple diploid embryos, skewing the genetic ratio towards the maternal genotype.

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Comparative_Analysis

Sharpen your ability to evaluate multiple scenarios and distinguish between competing potential answers.

Concepts Behind This Question

Recombination Frequency|Polyembryony Mechanisms|Embryo Sac Nuclei Count

Why This Question Is Tricky

Common Mistakes

Conceptual Pivot: Students often compartmentalize Mendelian Genetics (Linkage) and Plant Reproduction (Polyembryony).

The trap is failing to synthesize that nucellar polyembryony is a form of apomixis (mitotic cloning of maternal tissue).

Because linkage mapping strictly requires meiotic recombination (chiasmata formation), the presence of apomictic embryos fundamentally invalidates Recombination Frequency (RF) calculations.

Quick Facts About This Question

Question Information

SubjectBiology
ChapterPrinciples of Inheritance and Variation - Class 12
ConceptRecombination Frequency (RF)
Difficulty8
Cognitive Load5
Target Solve Time120 sec
AccessFree