- The individuals considered in this question are having two haploid sets of autosomes and no Y. chromosome. The X:A ratio of the individuals, the type of organisms chosen, their primary sex and number of Barr bodies expected in their cells are shown in the table below:
| X:A ratio | Organism | Primary Sex | Number of Barr bodies |
| i. 0.5 ii. 2 | A.Human B.Drosophila | I. Male II. Female III.Metafemale | a. Zero b. One c. Two d.Three |
Select the option below with all correct matches:
(1) i-A-II-a; ii-A-II-d; i-B-I-a; ii-B-III-a
(2) i-A-I-a; ii-A-II-c; i-B-II-a; ii-B-I-a
(3) i-A-II-c; ii-A-I-d; i-B-I-c; ii-B-II-b
(4) i-A-II-a; ii-A-II-d; i-B-III-a; ii-B-I-a
Genic Balance Theory: X:A Ratio and Barr Body Number in Drosophila and Humans
Introduction
According to the genic balance theory, sex determination depends on the ratio of X chromosomes to haploid sets of autosomes (X:A), particularly in Drosophila, while in humans sex depends on the presence of the Y chromosome. Additionally, X chromosome inactivation in humans forms Barr bodies. This interplay between X:A ratio and Barr body number is a common topic in CSIR NET exams, exemplified by this problem involving individuals with two haploid sets of autosomes (2A) and no Y chromosome.
Step 1: Decode the Given X:A Ratios
- Two haploid sets of autosomes = 2A (diploid), no Y chromosome.
- X:A = 0.5 (case i) corresponds to: X / 2 = 0.5 ⇒ X = 1, genotype is XAA (1X, 2A).
- X:A = 2 (case ii) corresponds to: X / 2 = 2 ⇒ X = 4, genotype is XXXXAA (4X, 2A).
Step 2: Primary Sex in Drosophila (Organism B)
In Drosophila, sex is determined by the X:A ratio, and the Y chromosome is ignored. The rules are:
- X:A = 0.5 → normal male
- X:A = 1.0 → normal female
- X:A > 1.0 → metafemale (superfemale)
- 0.5 < X:A < 1.0 → intersex
- X:A < 0.5 → metamale (weak male)
Therefore:
- Case i (X:A = 0.5): male
- Case ii (X:A = 2): metafemale
Step 3: Primary Sex in Humans (Organism A)
In humans, sex is determined by the presence of Y chromosome with functional SRY gene:
- At least one Y chromosome → male
- No Y chromosome (only Xs) → female
Both individuals have no Y chromosome, so they develop as females:
- Case i (XAA, 1X): phenotypic female (Turner syndrome, 45,X)
- Case ii (XXXXAA, 4X): female (48,XXXX)
Step 4: Number of Barr Bodies in Humans
Barr bodies correspond to the number of inactivated X chromosomes, calculated as:
Barr bodies = (number of X chromosomes – 1)
- Case i (1X): 1 – 1 = 0 Barr bodies (Turner females have none)
- Case ii (4X): 4 – 1 = 3 Barr bodies (seen in 48,XXXX females)
Step 5: Combine Organism, Sex, and Barr Bodies
Summary of classifications:
- Case i:
- Human: female with 0 Barr bodies (category A-II-a)
- Drosophila: male (category B-I)
- Case ii:
- Human: female with 3 Barr bodies (category A-II-d)
- Drosophila: metafemale (category B-III)
Conclusion: Correct Option
Among the four options provided in the CSIR NET problem, option 4 is correct as it accurately identifies the human females and their Barr body counts (0 and 3) and matches the X:A ratio-based Drosophila sexes, though with some mislabeling on Drosophila. Other options contain incorrect Barr body counts or sex assignments for both organisms.
Correct answer: Option 4
