47. A species of plant (species 1) is diploid (2n = 6) with chromosomes AABBCC and a related species (species 2) is also diploid (2n = 4) with chromosomes PPQQ. The following statements were given by students regarding the chromosome numbers involving these plant species:
A. Autotrip/oid of species 1 will have 12 chromosomes
B. Allotetraploid involving species 1 and 2 will have 16 chromosomes
C. A monosomy in species 1 will generate 5 chromosomes
D. A double trisomy in species I will generate 8 chromosomes
E. A nullisomy in species 2 Will generate 2 Chromosomes
The combination of statements with all correct
(1) A, B and C (2) A, C and C
(3) C, D and E (4) D, E and A
Given data
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Species 1: diploid, 2n = 6 → n = 3 (chromosomes A, B, C).
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Species 2: diploid, 2n = 4 → n = 2 (chromosomes P, Q).
Use standard aneuploidy and polyploidy formulas:
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Autotriploid: 3n
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Allotetraploid from species 1 and 2: 2n1+2n2
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Monosomy: 2n−1
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Double trisomy: 2n+1+1=2n+2
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Nullisomy: 2n−2
Checking each statement
A. Autotriploid of species 1 will have 12 chromosomes – False
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For species 1, n=3.
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Autotriploid: 3n=3×3=9 chromosomes, not 12.
B. Allotetraploid involving species 1 and 2 will have 16 chromosomes – False
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Combine complete diploid sets of each: 2n1+2n2 = 6+4=10.
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16 is incorrect.
C. A monosomy in species 1 will generate 5 chromosomes – True
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Monosomy: 2n−1.
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For species 1: 2n=6→ 6−1=5 chromosomes.
D. A double trisomy in species 1 will generate 8 chromosomes – True
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Double trisomy: extra chromosomes for two different pairs → 2n+1+1=2n+2.
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For species 1: 6+2=8 chromosomes.
E. A nullisomy in species 2 will generate 2 chromosomes – True
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Nullisomy: loss of both members of one homologous pair → 2n−2.
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For species 2: 2n=4→ 4−2=2 chromosomes.
Evaluating options
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(1) A, B and C – includes false A and B.
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(2) A, C and C – A is false; also malformed.
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(3) C, D and E – all three are correct.
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(4) D, E and A – includes false A.
So the combination of correct statements is C, D and E (option 3).
