46. Two near inbred parental lines P1 and P2 of an angiosperm species are crossed to produce F1 seeds in which. theplojdv of the endosperm is 6N. If plants generated from these F1 seeds are backcrossed with P1. what be the ploidy of the somatic cells in the next generation ?
(1) 2N          (2) 4N
(3) 5N          (4) 6N

Step 1: Use endosperm ploidy to infer gamete ploidy

In typical angiosperms:

• Endosperm ploidy = (polar nuclei) + (one sperm nucleus).

• Genetically: $\text{Endosperm}=2\times (\text{female gamete})+1\times (\text{male gamete})$.

• For a normal diploid species: egg = n, polar nuclei together = 2n, sperm = n → endosperm = 3n.

Here, in the F1 seeds, endosperm = 6N.

Let each reduced gamete (egg or sperm) have ploidy x.
Then:

$$\text{Endosperm}=2x+x=3x=6N\Rightarrow x=2N$$

So both male and female gametes that formed the F1 have ploidy 2N.

That means each parental sporophyte (P1 and P2) was 4N (because meiosis halves ploidy: 4N → 2N gametes).​

Step 2: Ploidy of the F1 sporophyte

The F1 zygote is formed by 2N (from P1) + 2N (from P2):

$$\text{F1 somatic ploidy}=2N+2N=4N$$

So plants grown from these F1 seeds are tetraploid (4N).

Step 3: Backcross F1 (4N) with P1

P1 is one of the original parents that produced 2N gametes, so its sporophyte ploidy is 4N as well.

• Gamete from F1: 2N

• Gamete from P1: 2N

Zygote ploidy in the backcross generation:

$$\text{Next-generation somatic ploidy}=2N+2N=4N$$

Thus, somatic cells in the next generation are 4N.

Option-wise explanation

1. 2N – would require one gamete to be N and the other N, which contradicts the 6N endosperm information.

2. 4N – correct – consistent with 2N × 2N gametic fusion deduced from the 6N endosperm.

3. 5N – impossible from combining whole-number gametes of 2N each.

4. 6N – would require 3N × 3N gametes or similar, incompatible with the 3x = 6N relationship.

Therefore, the ploidy of somatic cells in the backcross generation is 4N (option 2).