Plant Species Chromosome Numbers

Let's analyze the chromosome numbers for the given plant species and the different ploidy and aneuploidy conditions described in the statements.

We are given two plant species:

• Species 1: Diploid ($\text{2n} = \text{6}$), chromosomes AABBCC. This means the haploid number ($\text{n}$) for species 1 is $\text{3}$. One haploid set is ABC.
• Species 2: Diploid ($\text{2n} = \text{4}$), chromosomes PPQQ. This means the haploid number ($\text{n}$) for species 2 is $\text{2}$. One haploid set is PQ.

Analyzing Statement A: Autotriploid of Species 1

An autotriploid arises from a single species where the chromosome number is tripled ($\text{3n}$).

For Species 1, the haploid number ($\text{n}$) is $\text{3}$.

An autotriploid of Species 1 would have $\text{3} \times \text{n} = \text{3} \times \text{3} = \text{9}$ chromosomes.

Statement A says an autotriploid of species 1 will have $\text{12}$ chromosomes. This is incorrect.

Analyzing Statement B: Allotetraploid Involving Species 1 and 2

An allotetraploid is formed by combining the complete diploid sets of two different species. It can be represented as $\text{2n}_1 + \text{2n}_2$ or $\text{2(n}_1 + \text{n}_2)$.

Diploid number for Species 1 ($\text{2n}_1$) is $\text{6}$.

Diploid number for Species 2 ($\text{2n}_2$) is $\text{4}$.

An allotetraploid involving Species 1 and 2 would have $\text{2n}_1 + \text{2n}_2 = \text{6} + \text{4} = \text{10}$ chromosomes.

Alternatively, the hybrid ($\text{n}_1 + \text{n}_2$) would have $\text{3} + \text{2} = \text{5}$ chromosomes. Doubling this gives $\text{2} \times \text{5} = \text{10}$ chromosomes.

Statement B says an allotetraploid involving species 1 and 2 will have $\text{16}$ chromosomes. This is incorrect.

Analyzing Statement C: Monosomy in Species 1

Monosomy is a type of aneuploidy where a single chromosome is missing from a diploid set. The chromosome number is $\text{2n} - \text{1}$.

Diploid number for Species 1 ($\text{2n}$) is $\text{6}$.

A monosomy in Species 1 would have $\text{2n} - \text{1} = \text{6} - \text{1} = \text{5}$ chromosomes.

Statement C says a monosomy in species 1 will generate $\text{5}$ chromosomes. This is correct.

Analyzing Statement D: Double Trisomy in Species 1

Trisomy is a type of aneuploidy where there is an extra copy of a single chromosome, resulting in $\text{2n} + \text{1}$.

Double trisomy means there are extra copies of two different chromosomes, resulting in $\text{2n} + \text{1} + \text{1}$.

Diploid number for Species 1 ($\text{2n}$) is $\text{6}$.

A double trisomy in Species 1 would have $\text{2n} + \text{1} + \text{1} = \text{6} + \text{2} = \text{8}$ chromosomes.

Statement D says a double trisomy in species 1 will generate $\text{8}$ chromosomes. This is correct.

Analyzing Statement E: Nullisomy in Species 2

Nullisomy is a type of aneuploidy where a pair of homologous chromosomes is missing from a diploid set. The chromosome number is $\text{2n} - \text{2}$.

Diploid number for Species 2 ($\text{2n}$) is $\text{4}$.

A nullisomy in Species 2 would have $\text{2n} - \text{2} = \text{4} - \text{2} = \text{2}$ chromosomes.

Statement E says a nullisomy in species 2 will generate $\text{2}$ chromosomes. This is correct.

Summary of Statements

• Statement A: Incorrect ($\text{9}$ chromosomes)
• Statement B: Incorrect ($\text{10}$ chromosomes)
• Statement C: Correct ($\text{5}$ chromosomes)
• Statement D: Correct ($\text{8}$ chromosomes)
• Statement E: Correct ($\text{2}$ chromosomes)

The statements with all correct answers are C, D, and E.