Question

In the Mendelian dihybrid cross, out of 256 individuals obtained in the F2 generation, how many are pure homozygous for both traits?

• A. 32
• B. 16
• C. 8
• D. 48

Answer 1

The correct option is A

32

The correct option is A

Explanation of the correct option:

1. A simple cross between two pure species using two gene pairs is known as a dihybrid cross.
2. We may infer from the monohybrid cross that genotypically, one or two Y alleles will cause the pea to phenotypically turn yellow, whereas genotypically (yy) will cause the pea to phenotypically turn green.
3. The pea's form, however, is determined by two alleles. They may have wrinkles or be rounded. The form of the pea can be altered by the genotypically dominant gene R.
4. The pea will exhibit wrinkling due to the recessive gene r.
5. Therefore, the gametes produced by a cross between a homozygous dominant round yellow plant (RRYY) and a homozygous recessive wrinkled green plant (RRYY) will be RY and RY. Round and yellow phenotypes will be shared by all plants, but genotypically, they will all be heterozygous for both traits (RrYy).
6. Therefore, if we cross two F1 plants, the potential gametes will be genotypically RY, Ry, and Ry.
7. Round yellow to wrinkled yellow to round green to wrinkled green will be distributed in a 9:3:3:1 ratio in the F2 generation, correspondingly.
8. Observing this, we arrive at the ratio shown below:

1:2:1:2:4:2:1:2:1 (1 RRYY : 2 RRYy : 1 RRyy : 2RrYY : 4 RrYy : 2Rryy : 1rrYY : 2 rrYy : 1 rryy respectively)

Hence 2 out of 16 have the pure homozygous condition.

Therefore 1/8th of 256= 32.

Final answer: Pure homozygous offspring will be 32.