Read the paragraph given below and answer the question.
Sutton and Boveri argued that the pairing and separation of a pair of chromosomes would lead to the segregation of a pair of factors they carried. Sutton united the knowledge of chromosomal segregation with Mendelian principles and called it the chromosomal theory of inheritance.
Following this synthesis of ideas, experimental verification of the chromosomal theory of inheritance by Thomas Hunt Morgan and his colleagues, led to discovering the basis for the variation that sexual reproduction produced. Morgan worked with the tiny fruit flies, which were found very suitable for such studies. They could be grown on simple synthetic medium in the laboratory. They complete their life cycle in a short span of time and a single mating could produce a large number of progeny flies. Also, there was a clear differentiation of the sexes - the male and female flies are easily distinguishable. Also, it has many types of hereditary variations that can be seen with low power microscopes.
Cross performed between yellow bodied, white eyed females and brown bodied red eyed males of Drosophila results in:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(F_2\) ratio deviating from 9 : 3 : 3 : 1
Morgan crossed yellow-bodied, white-eyed females with brown-bodied, red-eyed males of Drosophila and then crossed the F1 flies among themselves to get the F2. This is a dihybrid cross, like Mendel’s cross of round-yellow and wrinkled-green peas.
For two genes on different chromosomes, the F2 ratio is 9 : 3 : 3 : 1 because the genes assort independently. Here both genes (body colour and eye colour) lie on the same chromosome, the X chromosome. Genes on one chromosome tend to be inherited together; this is linkage.
So most F2 flies showed the parental combinations (yellow-white and brown-red), and only a few showed new combinations (yellow-red and brown-white), formed by recombination (crossing over). For yellow body and white eye the recombinants were only about 1.3 per cent. The F2 ratio therefore deviated strongly from 9 : 3 : 3 : 1.
Option (1) 9 : 3 : 3 : 1 \(F_2\) ratio — Incorrect. 9 : 3 : 3 : 1 is obtained only when the two genes assort independently; these two genes are linked on the X chromosome.
Option (2) 1 : 2 : 1 \(F_1\) ratio — Incorrect. 1 : 2 : 1 is the F2 ratio of a monohybrid cross with incomplete dominance; it is not an F1 ratio, and this is a dihybrid cross.
Option (3) \(F_2\) ratio deviating from 9 : 3 : 3 : 1 — Correct. Linkage makes parental combinations far more common than the 9 : 3 : 3 : 1 ratio predicts.
Option (4) \(F_1\) ratio deviating from 1 : 2 : 1 — Incorrect. The deviation is seen in the F2, where the two traits separate; the F1 is not where a 1 : 2 : 1 ratio is expected.
Note: The passage is taken from NCERT. NTA has put lines of the NCERT text together to make it. Most CUET (UG) questions come straight from NCERT, so study the book line by line, including figures and their titles.
NCERT Extract:
Linkage and Recombination – “Morgan carried out several dihybrid crosses in Drosophila to study genes that were sex-linked. The crosses were similar to the dihybrid crosses carried out by Mendel in peas. For example Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F1 progeny. He observed that the two genes did not segregate independently of each other and the F2 ratio deviated very significantly from the 9:3:3:1 ratio (expected when the two genes are independent). Morgan and his group knew that the genes were located on the X chromosome (Section 4.4) and saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type.”