In a dihybrid cross,80 plants are obtained. T | vedclass

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(C) In a dihybrid cross ($F_2$ generation),the total number of combinations is $16$ $(4 	imes 4)$.Out of these $16$ combinations,the number of homozy

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$20:60$

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(C) In a dihybrid cross ($F_2$ generation),the total number of combinations is $16$ $(4 	imes 4)$.Out of these $16$ combinations,the number of homozygous individuals is $4$ (i.e.,$AABB, AAbb, aaBB, aabb$).The number of heterozygous individuals is $16 - 4 = 12$.Therefore,the ratio of homozygous to heterozygous individuals is $4 : 12$,which simplifies to $1 : 3$.For $80$ plants,the number of homozygous plants is $(1/4) 	imes 80 = 20$.The number of heterozygous plants is $(3/4) 	imes 80 = 60$.Thus,the ratio is $20 : 60$,which simplifies to $1 : 3$ or $10 : 30$ (or $10 : 70$ if considering specific genotype combinations as per the provided option $B$).Given the standard Mendelian dihybrid $F_2$ ratio,the homozygous to heterozygous ratio is $1:3$. For $80$ plants,this corresponds to $20$ homozygous and $60$ heterozygous plants.

In a dihybrid cross,$80$ plants are obtained. The ratio of homozygous and heterozygous plants shall be:

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