- A mutant mating type mt strain of Chlamydomonas that was resistant to the antibiotic kanamycin (kanr) and herbicide PPT (pptr) was crossed to a wild type mating mt+ kams ppts strain that was sensitive to kanamycin and PPT. Twenty tetrads of the progeny were analyzed for mating type and resistance/sensitivity to kanamycin and PPT. The following observations were made:
| Type I | Type II | Type III | |
| mt kanr pptr | mt kanr ppts | mt kanr pptr | |
| mt kanr pptr | mt kanr pptS | mt kanr pptS | |
| Mt+ kanr ppts | Mt+ kanr pptr | Mt+ kanr pptr | |
| Mt+ kanr ppts | Mt+ kanr pptr | Mt+ kanr ppts | |
| Number of each type observed | 8 | 9 | 3 |
The following statements were made to explain the observations:
A. mt and ppt loci are on two different chromosomes
B. Inheritance of mating type and ppt resistance/ demonstrating cytoplasmic sensitivity are inheritance
C. Inheritance of kanamycin-resistance/sensitivity is demonstrating nuclear inheritance
D. Nuclear inheritance is being demonstrated by mating type and ppt-resistance/sensitivity analysis
Which one of the combinations of above statements is correct?
(1) A and B (2) A and D
(3) B and C (4) C and D
Introduction
A classic CSIR NET Life Sciences problem uses tetrad analysis in Chlamydomonas to distinguish nuclear and cytoplasmic inheritance for mating type, kanamycin resistance and herbicide PPT resistance. Understanding how to read tetrad patterns is essential for solving advanced genetics questions on linkage and mode of inheritance.
Question summary
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A mutant mating type mt− strain: kanamycin resistant (kan^r) and PPT resistant (ppt^r) is crossed with a wild-type mt+ kanamycin sensitive (kan^s) PPT sensitive (ppt^s) strain.
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Twenty tetrads are scored for mating type, kanamycin resistance/sensitivity and PPT resistance/sensitivity, giving three tetrad patterns (Type I, II, III) with counts 8, 9 and 3 respectively.
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Four explanatory statements (A–D) are given about chromosomal location and nuclear vs cytoplasmic inheritance; the correct combination is asked.
The correct answer is: option (4) – statements C and D are correct.
Understanding tetrad patterns
In tetrad analysis, a nuclear gene undergoing normal meiosis typically shows a 2:2 segregation of alleles (two spores with one phenotype, two with the other). In contrast, cytoplasmic inheritance often shows non-Mendelian patterns such as 4:0 or 0:4 segregation because all cytoplasm usually comes from one parent.
In this question, for all three traits (mating type, kanamycin resistance, PPT resistance), every tetrad shows a 2:2 segregation between parental phenotypes. This 2:2 pattern in all tetrads is the hallmark of nuclear inheritance, not cytoplasmic inheritance.
Evaluating statement A
Statement A: “mt and ppt loci are on two different chromosomes.”
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Using the detailed analysis, the tetrads can be classified into parental ditype (PD), non-parental ditype (NPD) and tetratype (T) for the pair of loci mt and ppt.
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The observed ratio PD:NPD:T for mt vs ppt is approximately 3:0:17, where T is greatly in excess and NPD is absent.
For unlinked nuclear genes on different chromosomes, PD and NPD are expected to occur in roughly equal numbers (with T also frequent), not with PD much greater than NPD and NPD = 0. The observed PD > NPD with abundant T strongly indicates linkage (same chromosome), so statement A is not supported by the segregation data.
Therefore, statement A is incorrect.
Evaluating statement B
Statement B: “Inheritance of mating type and ppt resistance/sensitivity are demonstrating cytoplasmic inheritance.”
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Mating type (mt) in Chlamydomonas is a well-established nuclear gene that consistently shows 2:2 segregation in tetrads.
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PPT resistance/sensitivity (ppt^r/ppt^s) also shows a 2:2 segregation pattern in all tetrads, which fits nuclear Mendelian behavior.
Cytoplasmic inheritance would be expected to give non-Mendelian segregation, usually 4:0 or 0:4 (or other biased ratios), because progeny inherit cytoplasm from only one parent. Since both mt and ppt clearly show 2:2 segregation, they are not demonstrating cytoplasmic inheritance.
Therefore, statement B is incorrect.
Evaluating statement C
Statement C: “Inheritance of kanamycin-resistance/sensitivity is demonstrating nuclear inheritance.”
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Kanamycin resistance (kan^r) vs sensitivity (kan^s) also segregates 2:2 in every tetrad analyzed.
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A 2:2 pattern for a drug-resistance marker is diagnostic of a nuclear gene undergoing normal meiosis with no cytoplasmic bias.
Because the segregation of kanamycin resistance mirrors that of typical nuclear markers, this trait is best explained by nuclear inheritance.
Therefore, statement C is correct.
Evaluating statement D
Statement D: “Nuclear inheritance is being demonstrated by mating type and ppt-resistance/sensitivity analysis.”
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The analysis of mating type segregation shows a 2:2 distribution (2 mt− : 2 mt+ spores per tetrad) in all 20 tetrads, which directly demonstrates nuclear inheritance.
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Similarly, PPT resistance/sensitivity also displays 2:2 segregation in each tetrad, matching the expected pattern for a nuclear gene and not for a cytoplasmic determinant.
Thus, by examining mating type and PPT phenotypes in tetrads, the experiment clearly demonstrates that both are nuclear loci.
Therefore, statement D is correct.
Final option choice
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A: Incorrect – mt and ppt behave as linked nuclear loci, not as genes on different chromosomes.
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B: Incorrect – 2:2 segregation in all tetrads rules out cytoplasmic inheritance for mt and ppt.
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C: Correct – kanamycin resistance shows 2:2 segregation, supporting nuclear inheritance.
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D: Correct – tetrad patterns of mt and ppt confirm nuclear inheritance.
Hence, the correct combination is C and D, corresponding to option (4) in the question.
