Comprehension:
Principles of Inheritance
Read the following passage carefully and answer the given questions.
Gregor Mendel conducted hybridisation experiments on garden peas for seven contrasting characters. He crossed tall and dwarf pea plants to study the inheritance of characters. Mendel observed that all the F1 progeny plants were tall. Mendel then self-pollinated tall F1 plants and surprisingly observed the proportion of tall and dwarf plants in 3:1 ratio. Hence, the tall and dwarf traits were identical to their parental type and did not show any blending that is all the offspring were either tall or dwarf and none were of in between height. This study explains the Law of Segregation that the alleles do not show any blending and that both the characters are recovered as such in the F2 generation though one of these is not seen at the F1 generation. Further, Mendel observed the four different phenotypes from the cross of two contrasting characters (seed shape and seed color) which led to the conclusion that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters as stated by Law of Independent Assortment. Walter Sutton and Theodore Boveri noted that the behaviour of chromosomes was parallel to the behaviour of genes and used chromosome movement to explain Mendel's law. The important things to remember that chromosomes as well as genes occur in pairs. The two alleles of a gene pairs are located on homologous sites on homologous chromosomes.
In the dihybrid cross between the yellow color and round shaped seeds with green coloured and wrinkled shaped seeds, what will be the probability of getting RrYy genotype in F2 generation?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1/4
The two pairs of genes are inherited independently, so each pair can be worked out alone, and the two results are multiplied.
The parents are RRYY (round, yellow) and rryy (wrinkled, green). The F1 plants are all RrYy.
For seed shape, Rr × Rr gives 1 RR : 2 Rr : 1 rr. So the chance of Rr is 2/4 = 1/2.
For seed colour, Yy × Yy gives 1 YY : 2 Yy : 1 yy. So the chance of Yy is also 1/2.
\( \displaystyle P(RrYy) = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} \)
Of the 16 boxes of the Punnett square, 4 are RrYy, which is 4/16 = 1/4. The fraction 9/16 is the share of the round yellow phenotype, and 3/16 is the share of each of the two phenotypes with one recessive character.
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.