Question

The number of ${{F}_{2}}$ plants similar to ${{F}_{1}}$phenotype and genotypes respectively in a total of 128 plants in ${{F}_{2}}$ progeny of a mendelian dihybrid cross are
A. 48, 32
B. 72, 32
C. 48, 24
D. 72, 48

Answer 1

Hint: Dihybrid cross is the cross between two individuals having two characters with two different alleles. Both the characters have one dominant and one recessive allele in the characters under study.

Complete Answer:
If for example the primary cross is carried out between parents that are homozygous for the two traits under study i.e. one parent carries all dominant alleles (RRYY) while the other carries all recessive traits (rryy) then the ${{F}_{1}}$ progeny will all be genotypically heterozygous (RrYy) for the phenotypically dominant characters.
If these ${{F}_{1}}$ progeny are crossed with each other, then progeny of the following genotypes as expressed in the punnett square can be obtained.

PARENTS	♂️
	RY	Ry	rY	ry
♀️	RY	RRYY	RRYy	RrYY	RrYy
	Ry	RRYy	RRyy	RryY	Rryy
	rY	rRYY	rRYy	rrYY	rrYy
	ry	rRyY	rRyy	rryY	rryy

Amongst the 16 plants produced as progeny in the ${{F}_{2}}$ generation, 9 plants have phenotype similar to that of the parent plants i.e. both characters express dominant traits. Within these 9 plants, 4 plants have genotypes similar to the parent plants.
Applying the values obtained in the example to the total number of plants in the question; we obtain the following values:-
$\Rightarrow {{F}_{2}}$ plants similar to ${{F}_{1}}$ phenotype
$\Rightarrow \left( \dfrac{9}{16} \right)\times 128=9\times 8=72$
$\Rightarrow {{F}_{2}}$ plants similar to ${{F}_{1}}$ genotype
$\Rightarrow \left( \dfrac{4}{16} \right)\times 128=4\times 8=32$

Therefore, the correct answer to this question is (B) 72, 32.

Note: Before making the dihybrid cross, Mendel already knew about the dominant and recessive traits. The main purpose as to why the dihybrid cross was made was to study whether the inheritance of the genes were interdependent or not.